Structures: Pearson New International Edition [7 ed.] 1292040823, 9781292040820

Citation preview

Structures Schodek

Structures Daniel Schodek

Seventh Edition

9 781292 040820

Copyright © 2013. Pearson Education Canada. All rights reserved.

Bechthold

ISBN 978-1-29204-082-0

Martin Bechthold Seventh Edition

P E A

R

S

O

N

C U

S T O

M

L

I

B

R

A

R Y

Table of Contents

1. Structures: An Overview Daniel L. Schodek/Martin Bechthold

1

Introductory Concept Daniel L. Schodek/Martin Bechthold

27

2. Principles of Mechanics Daniel L. Schodek/Martin Bechthold

29

3. Introduction to Structural Analysis and Design Daniel L. Schodek/Martin Bechthold

87

4. Trusses Daniel L. Schodek/Martin Bechthold

121

Analysis and Design of Structural Elements Daniel L. Schodek/Martin Bechthold

169

5. Funicular Structures: Cables and Arches Daniel L. Schodek/Martin Bechthold

171

Copyright © 2013. Pearson Education Canada. All rights reserved.

6. Beams Daniel L. Schodek/Martin Bechthold

211

7. Members in Compression: Columns Daniel L. Schodek/Martin Bechthold

275

8. Continuous Structures: Beams Daniel L. Schodek/Martin Bechthold

299

9. Continuous Structures: Rigid Frames Daniel L. Schodek/Martin Bechthold

323

10. Plate and Grid Structures Daniel L. Schodek/Martin Bechthold

351

11. Membrane and Net Structures Daniel L. Schodek/Martin Bechthold

383

I

12. Shell Structures Daniel L. Schodek/Martin Bechthold

399

13. Structural Elements and Grids: General Design Strategies Daniel L. Schodek/Martin Bechthold

419

Principles of Structural Design Daniel L. Schodek/Martin Bechthold

453

14. Structural Systems: Design for Lateral Loadings Daniel L. Schodek/Martin Bechthold

455

15. Structural Systems: Constructional Approaches Daniel L. Schodek/Martin Bechthold

483

Copyright © 2013. Pearson Education Canada. All rights reserved.

16. Structural Connections Daniel L. Schodek/Martin Bechthold

505

Index

513

Pearson New International Edition Structures

Copyright © 2013. Pearson Education Canada. All rights reserved.

Daniel Schodek Martin Bechthold Seventh Edition

Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS.

Copyright © 2013. Pearson Education Canada. All rights reserved.

All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners.

ISBN 10: 1-292-04082-3 ISBN 10: 1-269-37450-8 ISBN 13: 978-1-292-04082-0 ISBN 13: 978-1-269-37450-7

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

Structures: An Overview

Copyright © 2013. Pearson Education Canada. All rights reserved.

1

IntroductIon

Definitions are a time-honored way to start any book. A simple definition of a structure, in a building context, is “a device for channeling loads that result from the use or presence of the building in relation to the ground.” The study of structures involves important and varying concerns, one of which is gaining an understanding of the basic principles that define and characterize the behavior of physical objects subjected to forces. More fundamentally, the study involves defining what a force itself is because this familiar term represents an abstract concept. The study of structures also involves dealing with much broader issues of space and dimensionality: size, scale, form, proportion, and morphology are all terms commonly found in a structural designer’s vocabulary. To begin the study of structures, consider again the definition of a structure in the previous paragraph. Although valuable because it defines a structure’s purpose, that definition provides no insight into the makeup or characteristics of a structure: What is this device that channels loads to the ground? Using the complex and exacting style of a dictionary editor, a structure can be defined as a physical entity having a unitary character that can be conceived of as an organization of positioned constituent elements in space in which the character of the whole dominates the interrelationship of the parts. Its purpose was defined earlier. It might be hard to believe, but a contorted, relatively abstract definition of this type, which is almost laughable in its academic tone, does have some merit. First, it states that a structure is a real physical object, not an abstract idea or interesting issue. A structure is not a matter of debate; it is something that is built and it is implied that a structure must be dealt with accordingly. Merely postulating that a structure can carry a certain type of load or function in a certain way, for example, is inadequate. A physical device that conforms to basic principles governing the behavior of physical objects must be provided to accomplish the desired behavior. Devising such a structure is the role of the designer. The expanded definition also makes the point that a structure functions as a whole. This point has fundamental importance, but it can be easily forgotten when one is confronted with a typical building composed of a seemingly endless array of individual beams and columns. In such cases, there is an immediate tendency to think of the structure only as an assembly of individual, small elements in which From Chapter 1 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Structures: An Overview each element performs a separate function. In actuality, all structures are, and must be, designed primarily to function as an overall system and only secondarily as an array of discrete elements. In line with the latter part of the expanded definition, these elements are positioned and interrelated to enable the overall structure to function as a whole in carrying vertically or horizontally acting loads to the ground. No matter how some individual elements are located and attached to one another, if the resultant configuration and interrelation of all elements does not function as a system and channel all anticipated types of loads to the ground, the configuration cannot be called a structure. The reference to anticipated types of loads is important because structures are normally devised in response to a specific set of loading conditions and function as structures only with respect to those conditions; structures are often relatively fragile with respect to unanticipated loads. A typical building structure capable of carrying normally encountered occupancy and environmental loads cannot, for example, be simply picked up by a corner and transported through space. It would fall apart because its structure was not designed to carry the unique loadings involved. So much for Superman carrying buildings around! To highlight yet another formal definition, the act of designing a structure also can be defined in complex language, but the result also has value. Designing a structure is the act of positioning constituent elements and formulating interrelations, with the objective of imparting a desired character to the resultant structural entity. The notions that elements are positioned and that relationships exist among these elements are basic to the concept of designing a structure. Elements can be positioned in various ways to carry loads, and many types of relationships may exist. For example, a block arch is made of carefully positioned elements. A beam may be related to a column simply by resting on top of it, or it may be rigidly attached to the column, with radically different structural actions ensuing.

2

Copyright © 2013. Pearson Education Canada. All rights reserved.

2.1

General types of structures primary classifications

Introduction. Fundamental to understanding any field is gaining knowledge of how groups within that field are systematically distinguished, ordered, and named. It also is important to know the criteria or presumed relationships that form the basis for such classifications. This section introduces one method for classifying structural elements and systems: according to their shape and basic physical properties of construction. (See Figure 1.) This classification scheme implies that complex structures are the result of only additive aggregations of elements; the scheme is inherently simplistic. In aggregations, only the additive nature of the elements is significant. In structures, it is significant that the elements are also positioned and connected to give the structure certain load-carrying attributes. The simpler classification approach illustrated in Figure 1 is useful as an introduction. Geometry. In terms of their basic geometries, the structural forms at the left in Figure 1 can be classified either as line-forming elements (or composed of line-forming elements) or as surface-forming elements. Line-forming elements can be further distinguished as straight or curved. Surface-forming elements are either planar or curved. Curved-surface elements can be of either single or double curvature. Strictly speaking, there is no such thing as a line or surface element because all structural elements have thickness. Still, it is useful to classify any long, slender element (such as a column whose cross-sectional dimensions are small with respect to its length) as a line element. Similarly, surface elements also have thickness, but this thickness is small with respect to length dimensions. Closely coupled with whether an element is linear or surface forming is the material or method of construction used. Many materials are naturally line forming.

Structures: An Overview

Copyright © 2013. Pearson Education Canada. All rights reserved.

fIGure 1 Classification of basic structural elements according to geometry and primary physical characteristics. Typical primary structural units and other aggregations also are illustrated. The schema is limited and suggests nothing about the importance of properly positioning constituent elements to make feasible structural assemblies.

Timber, for example, is inherently line forming because of how it grows. It is possible, however, to make minor surface-forming elements directly from timber (as evidenced by plywood) or larger surface-forming structures by aggregating more elemental pieces. Other materials, such as concrete, can be line forming or surface forming with equal ease. Steel is primarily line forming, but it can also be used to make directly minor surface-forming elements (e.g., steel decking). stiffness. Figure 1 illustrates a second fundamental classification: the stiffness characteristics of the structural element. The primary distinction is whether the element is rigid or flexible. Rigid elements, such as typical beams, do not undergo appreciable changes in shape under the action of a load or under changing

Structures: An Overview

fIGure 2

Nonrigid and rigid structures.

Copyright © 2013. Pearson Education Canada. All rights reserved.

loads. [See Figure 2(a).] However, they are usually bent or bowed to a small degree by the load’s action. Flexible elements, such as cables, assume one shape under one loading condition and change shape drastically when the loading nature changes. [See Figure 2(b).] Flexible structures maintain their physical integrity, however, no matter what shape they assume.1 Whether an element is rigid or flexible is often related to the construction material used. Many materials, such as timber, are inherently rigid; others, such as steel, can be used to make either rigid or flexible members. A good example of a rigid steel member is the typical beam (an element that does not undergo any appreciable change in shape under changing loads). A steel cable or chain, however, is clearly flexible because the shape that it and similar elements assume under loading is a function of the exact pattern and magnitudes of the load carried. A steel cable thus changes shape with changing loads. Whether a structure is rigid or flexible, therefore, depends either on the inherent characteristics of the material used or on the amount and microorganization of the element’s material. Many specific structures that are usually classified as rigid are so only under given loading conditions or under minor variations of a given loading condition. When loading changes dramatically, structures of this type become unstable and tend to collapse. Structures such as arches made by aggregating smaller rigid elements (e.g., blocks) into larger shapes are often in this category. one-Way and two-Way systems. A basic way to distinguish among structures is according to the spatial organization of the system of support used and the relation of the structure to the points of support available. Two primary cases of importance are one- and two-way systems. In a one-way system, the structure’s basic load-transfer mechanism for channeling external loads to the ground acts in one direction only. In a two-way system, the load-transfer mechanism’s direction is more complex but involves at least two directions. A linear beam spanning two support points is an example of a one-way system. (See Figure 3.) A system of two crossed elements resting on two sets of support points not lying on the same line and in which both elements share the external load is an example of a two-way system. A square, flat, rigid plate resting on four continuous edge supports also is a two-way system: An external load cannot be simplistically assumed to travel to a pair of the supports in one direction only. The distinction between one- and two-way structural actions is of primary importance in a design context. As is discussed in more detail later, there are 1

Common English-language connotations of the terms rigid and flexible are evoked here. In some more advanced structural theory applications, these terms are not used in their literal sense. Rather, distinctions are made among stiffness, strength, and stability.

Structures: An Overview

One-way structures

One-way beam

Two-way structures

One-way plate

Two-way beams

Typical rigid structures

Post and beam

Load-bearing walls

Three-hinged arch

Two-hinged arch

Copyright © 2013. Pearson Education Canada. All rights reserved.

Arch with tie-rod

Typical flexible structures

Cable-stayed

fIGure 3 Types of structural elements.

Two-way plate

Structures: An Overview situations typically involving certain patterns in the support system used that often lead to specific advantages (in terms of the efficient use of materials) in using a twoway system compared to a one-way system. Other patterns in the support system, however, often lead to the converse result. For this reason, it is useful early on to begin distinguishing between one- and two-way systems. Materials. A common classification approach to structures is by the type of material used (e.g., wood, steel, and reinforced concrete). A strict classification by materials, however, is somewhat misleading and is not adopted here because the principles governing the behavior of similar elements composed of different materials (e.g., a timber and steel beam) are invariant and the differences are superficial. General descriptions have a more intrinsic value at this stage. As one takes a closer look at structures, however, the importance of materials increases. One reason stems from the close relationship between the nature of the deformations induced in a structure by the action of the external loading and the material and method of construction that is most appropriate for use in that structure. Steel can be used under all conditions. Plain concrete can be used only where the structure is compressed or shortened under the action of the load. Concrete cracks and fails when subjected to tensile forces that elongate the material. Concrete reinforced with steel, however, can be used where elongating forces are present because the steel can be designed to carry those forces.

2.2

primary structural elements

Copyright © 2013. Pearson Education Canada. All rights reserved.

elements. Common rigid elements include beams, columns or struts, arches, flat plates, singly curved plates, and shells having a variety of different curvatures. Flexible elements include cables (straight and draped) and membranes (planar, singly curved, and doubly curved). In addition, several other types of structures (frames, trusses, geodesic domes, nets, etc.) are derived from these elements. Assigning a specific name to an element having certain geometrical and rigidity characteristics is done for convenience only and has its basis in tradition. Naming elements in this way can, however, be misleading because it is easy to assume that if two elements have different names, the way they carry loads also must be different. This is not necessarily so. Indeed, a basic principle is that all structures have the same fundamental load-carrying mechanism. At this point, however, it is still useful to retain and use traditional names to gain familiarity with the subject. Beams and columns. Structures formed by resting rigid horizontal elements on top of rigid vertical elements are commonplace. Often called post-and-beam structures, the horizontal elements (beams) pick up loads that are applied transversely to their lengths and transfer the loads to the supporting vertical columns or posts. The columns, loaded axially by the beams, transfer the loads to the ground. The beams are bowed or bent as a consequence of the transverse loads they carry (see Figure 3), so they are often said to carry loads by bending. The columns in a beam-and-column assembly are not bent or bowed because they are subjected to axial compressive forces only. In a building, the possible absolute length of individual beams and columns is rather limited compared with some other structural elements (e.g., cables). Beams and columns are therefore typically used in a repetitive pattern. Continuous beams often exhibit more advantageous structural properties than simpler single-span beams supported only at two points. frames. The frame, illustrated in Figure 3, is similar in appearance to the postand-beam type of structure but has different structural action because of the rigid

Structures: An Overview joints between vertical and horizontal members. This rigidity imparts stability against lateral forces that is lacking in the post-and-beam system. In a framed system, both beams and columns are bent or bowed as a result of the load’s action on the structure. As with the post-and-beam structure, the possible lengths of individual elements in a frame structure are limited. Consequently, members are typically formed into a repetitive pattern when they are used in a building.

Copyright © 2013. Pearson Education Canada. All rights reserved.

trusses. Trusses are structural members made by assembling short, straight members into triangulated patterns. The resultant structure is rigid as a result of how the individual line elements are positioned relative to one another. Some patterns (e.g., a pattern of squares rather than triangles) do not necessarily yield a structure that is rigid (unless joints are treated the same way as they are in framed structures). A truss composed of discrete elements is bent or bowed as a whole under the action of an applied transverse loading in much the same way that a beam is bent or bowed. Individual truss members, however, are not subject to bending, but are only either compressed or pulled on. arches. An arch is a curved, line-forming structural member that spans two points. The common image of an arch is a structure composed of separate, wedgeshaped pieces that retain their position by mutual pressure induced by the load. The shape of the curve and the nature of the loading are critical determinants as to whether the resultant assembly is stable. When shapes are formed by stacking rigid block elements, the resultant structure is functional and stable only when the load’s action induces in-plane forces that make the structure compress uniformly. Structures of this type cannot carry loads that induce elongations or any pronounced bowing in the member. (The blocks pull apart and the structure fails.) Block structures can be strong when used properly, as their extensive historical usage attests. The strength of a block structure is due exclusively to the positioning of individual elements because blocks are typically either rested one on another or mortared together. (Mortar does not appreciably increase the structure’s strength.) The positioning, in turn, depends on the type of loading involved. The resultant structure is thus rigid under only particular circumstances. The rigid arch is frequently used in modern buildings. It is curved similarly to  block arches but is made of one continuous piece of deformed rigid material (Figure 3). If properly shaped, rigid arches can carry a load to supports while being subject only to axial compression, and no bowing or bending occurs. The rigid arch can better carry variations in the design loading than its block counterpart made of individual pieces. Many types of rigid arches exist, and they are often characterized by their support conditions (e.g., fixed, two hinged, and three hinged). Walls and plates. Walls and flat plates are rigid, surface-forming structures. A load-bearing wall can typically carry vertically acting loads and laterally acting loads (e.g., wind and earthquake) along its length. Resistance to out-of-plane forces in block walls is marginal. A flat plate is typically used horizontally and carries loads by bending to its supports. Plate structures are normally made of reinforced concrete or steel. Horizontal plates can also be made by assembling patterns of short, rigid line elements. Three-dimensional triangulation schemes are used to impart stiffness to the resultant assembly. Long, narrow, rigid plates can also be joined along their long edges and used to span horizontally in beamlike fashion. These structures, called folded plates, have the potential for spanning fairly large distances. cylindrical shells and Vaults. Cylindrical barrel shells and vaults are examples of singly curved-plate structures. A barrel shell spans longitudinally such that the

Structures: An Overview curve is perpendicular to the span’s direction. When fairly long, a barrel shell behaves much like a beam with a curved cross-section. Barrell shells are made of rigid materials (e.g., reinforced concrete and steel). A vault, by contrast, is a singly curved structure that spans transversely. A vault is basically a continuous arch. spherical shells and domes. A wide variety of doubly curved surface structures are in use, including structures that are portions of spheres and those that form warped surfaces (e.g., the hyperbolic paraboloid). The number of shapes possible is boundless. The most common doubly curved structure is the spherical shell. It is convenient to think of it as a rotated arch. The analogy, however, is misleading regarding how the structure carries loads because loadings induce circumferential forces in spherical shells, and such forces do not exist in arches. Domed structures can be made of stacked blocks or a continuous rigid material (reinforced concrete). Shells and domes are highly efficient structures capable of spanning large distances with a minimum of material. Dome-shaped structures can also be made by forming short, rigid line elements into repetitive patterns. The geodesic dome is such a structure.

Copyright © 2013. Pearson Education Canada. All rights reserved.

cables. Cables are flexible structural elements. The shape they assume under a loading depends on the nature and magnitude of the load. When a cable is pulled at either end, it assumes a straight shape. This type of cable is often called a tie-rod. When a cable is used to span two points and carry an external point load or series of point loads, it deforms into a shape made up of a series of straight-line segments. When a continuous load is carried, the cable deforms into a continuously curving shape called a catenary. The self-weight of the cable produces such a catenary. Other continuous loads produce curves that are similar in appearance to, but not exactly the same as, the catenary. Suspension cables can be used to span extremely large distances. They are often used in bridges, where they support a road deck, which in turn carries the traffic loading. Moving traffic loads ordinarily cause the primary support cable to undergo changes in shape as load positions change. The changing cable shape would lead to undesirable changes in the shape of the road surface. In response the horizontal bridge deck is made into a continuous rigid structure so the road surface remains flat and the load transferred to the primary support cables remains constant. Cable-stayed structures are used to support roof surfaces in buildings, particularly in long-span situations. Membranes, tents, and nets. A membrane is a thin, flexible sheet. Common tents are made of membrane surfaces. Simple and complex forms can be created using membranes. For surfaces of double curvature, however, such as a spherical surface, the surface must be an assembly of much smaller segments because most membranes are available only in flat sheets. (A spherical surface is not developable.) A further implication of using a flexible membrane to create a surface is that it has to be either suspended with the convex side pointing downward or, if used with the convex side pointed upward, supplemented by some mechanism to maintain its shape. Pneumatic, or air-inflated, structures are the latter type. The internal air pressure inside the structure maintains the shape of the membrane. Another mechanism is to apply external jacking forces that stretch the membrane into the desired shape. Various stressed-skin structures fall into this category. The need to pretension the skin, however, imposes several limitations on what shape can be formed. Spherical surfaces, for example, are difficult to pretension by external jacking forces, while others, such as the hyperbolic paraboloid, are handled with comparative ease. Nets are three-dimensional surfaces made up of a series of crossed, curved cables. Nets are analogous to membrane skins. By allowing the mesh opening to vary as needed, a wide variety of surface shapes can be formed. An advantage of using crossed cables is that the positioning of the cables mitigates fluttering due to

Structures: An Overview wind suctions and pressures. In addition, tension forces are typically induced into the cables by jacking devices so the whole surface is turned into a type of stretched skin. This also gives the roof stability and resistance to flutter.

2.3

primary structural units and aggregations

Copyright © 2013. Pearson Education Canada. All rights reserved.

Introduction. While many of the basic elements discussed in the preceding section can function in isolation as load-carrying structures, some must be combined with others to create a structure that encloses or forms a volume. In this respect, structures used in buildings are often distinct from those used for other purposes. Building structures are typically volume forming in nature; others are not necessarily so. Bridge structures, for example, are used to form or support linear surfaces. In this context, it is useful to introduce the notion of a primary structural unit, which is a discrete, volume-forming structural element or assembly of structural elements used in building design. Four columns supporting a rigid planar surface at its corners, for example, form a primary unit. Such units can be stacked or placed side by side to form a connected series of volumetric units. When placed side by side, columns are typically shared between units. Primary units are often an intermediate step between a series of discrete elements (e.g., beams and columns) and an entire building complex. The way discrete elements can be conceptually assembled into units and then aggregated often, but certainly not always, reflects the way building complexes are constructed. The importance of considering structures of this type of unit is most apparent in preliminary design stages. The idea’s usefulness stems from the fact that a unit’s dimensions must invariably be related to the programmatic requirements of the building considered. Many buildings, for example, are considered to consist of a cellular aggregation of volumetric units of sizes related to the intended occupancy. Housing is such a building type. In this case, the dimensions of the primary structural unit are directly related to the functional dimensions of the housing unit. The primary structural unit, however, could be larger and encompass several functional units. It could not be smaller than the minimum functional subdivision of a unit. The point is that the primary structural unit dimensions are either the same as or a multiple of the critical functional dimensions associated with the building occupancy. In some cases, the building can be defined as consisting of one large functional unit (e.g., a skating rink) and not an aggregation of cellular volumes. These simple, immensely valuable concepts are useful in early design stages. structural units. Primary structural units may be made using different combinations of the elements discussed previously. (See Figure 4.) With common cellular units, it is useful to distinguish among the horizontal spanning, vertical support, and lateral support systems. With planar surfaces, horizontal spanning systems may have one- or two-way spanning elements. A hierarchy is often present in systems made of one-way spanning elements. For example, short-span, surface-forming plank or decking elements are periodically supported by closely spaced secondary beams, which in turn may be supported by other beams. Loads acting on the surface, such as snow, are first picked up by the decking and then transferred to the secondary beams. The secondary beams then transfer the loads to the vertical support system. Forces are transferred from one member to another via the development of reactive forces at member supports. Consequently, loads and related internal forces build up in members in lower layers of the hierarchy, which must be made larger and stiffer than others. Hierarchies of any number of layers may be used, but one, two, and three layers are most common. In short-span situations, beam-and-decking systems are common. As the length of the spans increases, trusses or cables might be used for

Structures: An Overview

Copyright © 2013. Pearson Education Canada. All rights reserved.

fIGure 4 Typical structural units.

secondary and primary spanning elements. Other spanning systems, such as arches and barrel shells, could also be used. Plate-and-grid systems may also be used for horizontal spanning. For short spans, often only one layer is present when plates are used. As the length of the spans increases, a hierarchical system of plates and beams may be used. In common cellular assemblies, vertical support systems are composed of loadbearing walls or columns. Load-bearing walls may be used to receive loads along their length (e.g., from a horizontal plate). Columns receive concentrated forces, typically from the ends of beams. Therefore, a close relationship exists between the pattern of the vertical support system and the nature of the horizontal spanning system.

Structures: An Overview Horizontally acting forces (e.g., from wind and earthquake) can cause structures to collapse laterally. Wall structures are inherently resistant to these forces. Beam-and-column systems, however, need cross bracing. Rigid-frame systems that are also resistant to lateral forces provide an alternative to beam-and-column assemblies. Stability responses of this type are explored later in the chapter.

3 3.1

analysIs and desIGn of structures: BasIc Issues fundamental structural phenomena

The preceding section discussed the nature of structural forms in broad terms. Specific forms mentioned are acted on by applied forces that can cause the form to slide or overturn as a whole or to collapse internally. Components also could break apart or deform. Forces causing overturning or collapse come from the specific environmental or use context (e.g., effects of wind, earthquakes, and occupancies) or from the selfweight of the form. These same applied loadings produce internal forces in a structure that stress the material and may cause it to fail or deform. Failure can occur in several ways. (See Figure 5.) A first set of concerns is the overall stability of a work. As a whole unit, a structure might overturn, slide, or twist about its base—particularly from horizontally acting wind or earthquake forces. Structures that are relatively tall or have small bases are prone to overturning effects. Forces from earthquakes cause overturning or sliding actions, but their magnitude depends on the weight of the structure because of the inertial character of earthquake forces. Overturning or twisting is not caused only by horizontally acting forces: A work could be out of

Copyright © 2013. Pearson Education Canada. All rights reserved.

Wind or earthquake forces

Types of failures

Structural responses for preventing failures

Gravity loads

fIGure 5 Structural phenomena and general design responses.

Structures: An Overview balance under its self-weight and overturn. The use of wide, rigid foundations helps prevent overturning, as does using special foundation elements such as piles to carry tension forces. A second set of concerns deals with internal, or relational, stability. If a structure’s parts are not properly arranged or interconnected, an entire assembly can collapse internally. Such collapses involve large relative movements within the structure. Assemblies may be stable under one loading condition and unstable under another. Horizontally acting wind or earthquake forces, in particular, cause collapses of this kind. Several basic mechanisms—walls, frame action, cross bracing—can be used to make an assembly internally stable. In the next section, we explore the issue of stability in detail. A third set of concerns deals with the strength and stiffness of constituent elements. Many structural issues revolve around the strength of a structure’s parts. The failure of parts, which might lead to total collapse, might be caused by excessive tension, compression, bending, shear, torsion, bearing forces, or by deformations that develop internally in the structure because of the applied loadings. Associated with each force state are internal stresses that exist within the fabric of the material. By carefully designing components in response to the force state present, the stresses developed in the components can be limited to safe levels.

Copyright © 2013. Pearson Education Canada. All rights reserved.

3.2

structural stability

A fundamental consideration in designing a structure is assuring its stability under any loading condition. All structures undergo some changes in shape under load. In a stable structure, the deformations induced by the load are typically small, and internal forces generated by the action of the load tend to restore the structure to its original shape after the load is removed. In an unstable structure, the load-induced deformations are typically massive and tend to increase while the load is applied. Such unstable structures do not generate internal forces that restore the structure to its original configuration. Unstable structures often collapse completely and instantaneously when a load is applied to them. It is the structural designer’s core responsibility to ensure that a proposed structure forms a stable configuration. Relational stability is a crucial issue in the design of structures assembled of discrete elements. For example, the post-and-beam structure illustrated in Figure 6(a) appears stable. Any horizontal force, however, causes deformations of the type indicated in Figure 6(b). The structure cannot resist horizontal loads, and it has no mechanism to restore it to its initial shape after a horizontal load is removed. The large changes in angle that occur between members characterize an internally unstable structure that is beginning to collapse. This structure will collapse instantaneously under load; this particular pattern of members is called a collapse mechanism. Only a few fundamental ways can be used to convert a self-standing structure like that shown in Figure 6(b) from an unstable to a stable configuration. These are illustrated in Figure 6(d). The first is to add a diagonal member to the structure so it cannot undergo the “parallelogramming” indicated in Figure 6(b) without a dramatic release in the length of the diagonal member. (This would not occur if the diagonal were adequately sized to take the forces involved.) Another method used to assure stability is through shear walls—rigid planar surface elements that resist shape changes of the type illustrated. A reinforced concrete or masonry wall, either full or partial, can be used. (The required extent of a partial wall depends on the magnitudes of the forces.) A final method to achieve stability is by stopping the large angular changes between members that are associated with collapse. Such stability is achieved when connections between members are such that their angular relationship remains constant under any loading. This is done by making a rigid joint between members. A typical table, for example, is stable because the rigid joint between each leg and the top maintains a constant angular relationship between the elements. Structures that use rigid joints to assure stability are called frames.

Structures: An Overview

Copyright © 2013. Pearson Education Canada. All rights reserved.

fIGure 6 Stability of structures.

There are, of course, variants on the basic methods of assuring stability. Still, most structures composed of discrete elements rely on these basic approaches. More than one approach can be used (e.g., a structure having both rigid joints and a diagonal), but some redundancy may be involved. With the assemblies diagrammed in Figure 7, one of the lateral-stability devices must be used to prevent lateral collapse. A single volumetric element may be designed with one stability device in one direction and a different one in the other direction. A larger aggregation of volumetric units might have stability devices only along the external periphery (instead of around each unit) or at a few locations internally.

3.3

forces, Moments, and stresses in Members

An external force on a structure due to its environment or use produces internal forces within that structure. Common internal force states are tension, compression, bending, shear, torsion, and bearing (Figure 8). Internal stresses and strains are associated with each of these force states. Stress is a measure of internal force intensity per unit area [typically stated in lb/in.2 or N/mm2 (MPa)], and strain is a measure of deformation (in./in. or mm/mm). Tension forces pull an element apart. The strength of a tension member depends on the cross-sectional area of a member and its material. Members in tension can be strong, as affirmed by the many cables in long-span structures. A tension member’s strength is independent of its length. Tension stresses are uniformly distributed across the cross section of the member 1stress = force>area or f = P>A2. If the internal stress developed is greater than an experimentally determined failure stress, the

Structures: An Overview

Copyright © 2013. Pearson Education Canada. All rights reserved.

fIGure 7 Common structural options for a typical volumetric unit.

member will fail. Safety factors are often used to limit stresses to a comfortable, allowable level. Compression forces can crush or buckle an element. Short members crush and have high strengths comparable to members in tension. A long compression member’s load-carrying capacity, however, decreases with increasing lengths. Long compression members may become unstable and suddenly snap out from beneath a load at critical load levels. This inability to carry additional load occurs without evident material distress and is called buckling. Because of it, long compression members are incapable of carrying very high loads. The stresses induced by compression forces are uniformly distributed across the cross section of the member, whether it is long or short. Short compression members can carry large stresses before crushing. Long compression members can buckle and fail at low stress levels. Tension or compression forces or stresses may develop in structural surfaces as well as linear members. When they act in an in-plane direction within the surface (rather than transversely), as might occur within a simple balloon, they are called membrane forces, and stresses are biaxial in nature. Bending is a complex force state associated with the bowing of a member such as a beam. Bending results from an applied external loading or force that acts transversely to the member’s long axis. These external forces produce internal bending moments that have a rotational sense and in turn cause the bending or bowing. Depending on the structure’s geometry and the applied external loadings or forces, these effects may vary along the length of the beam (these actions are described more in the following section). The bending action makes fibers on one face of the member elongate, and hence be in tension, and fibers on the opposing face compress. Thus, both tension and compression stresses are developed at the same cross section. (This complex force state is not given by the formula stress = force>area.) These stresses act perpendicularly to the section’s face. A member subject to bending can carry only a small load relative to its size and in comparison to a member carrying purely tensile forces. The strength of a

Structures: An Overview

Tension or compression

Tension members fail by pulling apart.

C

Short compression members fail by crushing.

T Membrane forces

Long compression members can fail by buckling at low force levels. Membrane stresses Tension or compression stresses are uniformly distributed.

Axial forces: External forces that act along the length of a member cause the member to pull apart or compress.

Bending

In-plane tension or compression forces and stresses in surface structures

Bending stresses act perpendicular to the beam cross section and vary from compression on the top to tension on the bottom Bending failure mode

Bending moments: Transversely acting loads and forces cause internal bending moments and related bowing in the structure. Vertical shear

Copyright © 2013. Pearson Education Canada. All rights reserved.

Shear forces: Transversely acting loads and forces can tend to cause parts of a structure to slide with respect to one another.

Other important considerations in analyzing and designing members

Horizontal shear

Shear stresses act parallel to faces and vary from compression on the top to tension on the bottom.

Torsion

Torsion is a twisting action.

Bearing

Bearing stresses cause localized crushing.

Deflections

Deflections can be excessive.

fIGure 8 External forces cause internal shears, bending moments, and other force states to develop. Compression, tension, bending, shear, bearing, or torsional stresses consequently develop.

Structures: An Overview

Copyright © 2013. Pearson Education Canada. All rights reserved.

fIGure 9 Dominant stress states: bending, shear, and axial forces and stresses in common structural forms under primary loadings.

member in bending is highly dependent on the amount and distribution of material in a cross section and on the type of material. Special responses to carrying bending forces include the ubiquitous wide-flange shape in steel or the common reinforced concrete beam (where the steel carries the tension forces). Bending may also develop in rigid plate surfaces. In this case, bending moments may be biaxial and occur in orthogonal directions. Shear is a force state associated with the action of opposing forces that cause one part of a structure to slide with respect to an adjacent part. Stresses that act tangentially to the sliding surface are developed. It is interesting that shear stresses develop along vertical and horizontal planes in a typical beam member, are highest at the central part of the cross section, and get smaller toward top and bottom member faces. Forces that cause shear stresses may vary along the length of the beam. Torsion is twisting. Tension and compression stresses normally develop in a member subjected to torsion. Bearing stresses exist at the interface between two members when forces are transferred from one member to another. Bearing stresses develop, for example, at the ends of beams where they rest on walls or on columns. The stresses act perpendicularly to the members’ faces. Deflections caused by loads on members must be limited to allowable values. Other, more complex stresses and stress interactions also develop in members. The forces and stresses noted are common in structural members. Not all of these actions, however, are present in all types of structures. A common beam, for example, carrying transversely applied loadings, is primarily subjected to bending and shear actions that produce bending and shear stresses and may or may not be subject to axial forces except in special conditions. By contrast, members of a truss structure are primarily in a state of either axial tension or compression. Bending or shear forces are typically not present. Figure 9 illustrates some common elemental structural forms and notes the primary internal force states developed within them because of the external forces associated with normal primary loading conditions. As noted, beams are subjected primarily to bending and shear. In related frame structures, bending and shear forces dominate the design, but some levels of axial tension and compression forces also can exist. Trusses are remarkable because only tension or compressive forces develop. Arches carry design loads through the development of compressive forces only (albeit off-balanced loadings can cause bending). Cables are subject only to tension forces. Biaxial bending dominates the design of rigid-plate or crossed-beam structures, although shear forces also are present. In so-called space

%HDPVDQGIUDPHV SULPDULO\LQEHQGLQJ ZLWKVKHDUDQGD[LDO IRUFHV

7UXVVHV D[LDOWHQVLRQ7 DQG FRPSUHVVLRQ& LQ PHPEHUV

$UFKHV FRPSUHVVLRQ&  XQGHUSULPDU\ ORDGLQJV

&DEOHV WHQVLRQ7

)ROGHGSODWHV EHDPOLNHDFWLRQLQ EHQGLQJZLWKVKHDU

3QHXPDWLFVWUXFWXUHV LQSODQHELD[LDOWHQVLRQ PHPEUDQHVWUHVVHVLQ VXUIDFH

0HPEUDQHVWUXFWXUHV LQSODQHELD[LDOWHQVLRQPHPEUDQH VWUHVVHVLQVXUIDFHPDVWVLQ FRPSUHVVLRQDQGWLHEDFNVLQWHQVLRQ

3ODWHV SULPDULO\ELD[LDO EHQGLQJZLWKVKHDU

6SDFHIUDPHV D[LDOWHQVLRQ7 DQG FRPSUHVVLRQ& LQ PHPEHUV

6KHOOV LQSODQHPHPEUDQH VWUHVVHVLQVXUIDFH 7RU& PLQRUEHQGLQJ DWERXQGDULHV

)UHHIRUPULJLG VKDSHVSULPDULO\LQ EHQGLQJZLWKVRPH LQSODQHWHQVLRQRU FRPSUHVVLRQ

Structures: An Overview frames, individual members are subject only to tension or compression—they behave similar to trusses. A folded plate structure carries loads primarily by bending, although shear forces also are present. In pneumatic structures, only biaxial membrane stresses in tension develop. In other kinds of stretched-skin structures, only biaxial membrane forces in tension are developed as well; however, special elements such as masts may be in a state of compression or even bending. Specially shaped rigid surfaces of double curvature, such as domes, have membrane stresses only in either tension or compression developed within them, although high-axial tension or compression forces can develop in related elements (such as tension rings at the base of domes). In free-form geometrical shapes, loads are primarily carried via the bending and shear forces, although some minor in-plane tension and compression forces may be present. Structures in which bending develops are less efficient than those in which only tension or compression forces exist. A good design principle is to explore shapes that minimize bending and shear forces. The funicular structural shapes (including arches and cables) discussed later in this chapter seek to minimize bending.

Copyright © 2013. Pearson Education Canada. All rights reserved.

3.4

Basic structural analysis and design process

All of the phenomena noted in the previous sections (sliding, overturning, racking, twisting, bending, shear, torsion, and buckling) occur when a force acts on a whole structure or some specific component within it. It is necessary to understand quantitatively or numerically the type and magnitude of these forces to determine whether a structure could fail in any of the modes noted previously, or, alternatively, to determine the size of a member that is expected to carry forces safely. This understanding is accomplished through structural analysis and design, which is briefly described here. The first step is to determine the types and magnitudes of the forces acting on the whole structure. The forces may result from external loading or may be caused by the weight of the structure. External forces are called live loads and result from the occupancy of the building (by people, furniture, etc.), from environmental forces (e.g., wind, snow) that impinge on the surfaces of the building, and from earthquake-induced forces that are associated with rapid ground movements. The weights of the structural elements and other fixed elements (such as roofing or insulation) are referred to as dead loads. Once the external forces are known, the next step is to determine how these applied forces might cause the structure to overturn, slide, or rack. At the same time, an analysis is made of how these same external forces cause internal forces to develop within the elements of the structure. These steps demand a study of the equilibrium of both the overall structure and each of its parts. This study is carried out by applying a branch of mechanics known as statics. The laws of statics derive from Newtonian physics and state that any object at rest must be in a state of equilibrium with respect to all forces and torques acting on it. In the simple column shown in Figure 10, it is obvious that it holds up the end of the beam, so forces are developed within it. The live and dead loads associated with the beam generate a downward force acting on top of the column, and, in accordance with Newton’s basic laws, the column in turn generates an equal and opposite force acting upward on the end of the beam. This same force is transmitted internally through the column to the foundation and, together with the additional dead weight of the column, exerts a downward force on the foundation. Again, in accordance with Newton’s laws, an equal and opposite force is exerted upward. The process is repeated with the foundation and ground. The equal and opposite forces generated at the connection points are called reactions. With respect to the column, as a formal statement of static equilibrium, we say that the sum of all forces (including reactions) acting on the column must not cause it to translate in any direction; thus, the sum of all forces acting in the

Structures: An Overview

fIGure 10 Basic statics (force and moment equilibrium). 5RRIORDGVDQG EHDPZHLJKWWB :H[HUWVDIRUFHRQ % FROXPQDQGFROXPQ H[HUWVHTXDODQG RSSRVLWHUHDFWLYHIRUFH FDOOHGDUHDFWLRQ RQWKHHQGRIWKHEHDP

F W B 

&ROXPQZHLJKWWC

W C

(TXDODQGRSSRVLWH UHDFWLYHIRUFHV WW B C )RXQGDWLRQZHLJKWW

F

 W W B C

WF

F

(TXDODQGRSSRVLWH UHDFWLYHIRUFHV WWW B F C D %HDPDQGFROXPQ VWUXFWXUH



F  W W W  B C F

E )RUFHWUDQVIHU,QWHUIDFHIRUFHVDFWRYHU FRQWDFWDUHDV

F 6LPSOLILHGPRGHORIIRUFHV

0RPHQWRI)RUFHF DERXWSRLQW M F[d 

MF

R

FR

)RUFHF d

&ROXPQZHLJKW WC 0RPHQWRIWC DERXW

M W[d  wR C

Copyright © 2013. Pearson Education Canada. All rights reserved.

d

d

M

wR

d

3RLQW 0RPHQWFHQWHU

G $IRUFHFDQDOVRSURGXFHDURWDWLRQDOHIIHFWFDOOHGD PRPHQWIRUFH[GLVWDQFH DERXWDQ\SRLQW,QWKLVFDVH WKHPRPHQWDVVRFLDWHGZLWKWKHKRUL]RQWDOIRUFHRYHUWXUQV WKHFROXPQ7KHPRPHQWDVVRFLDWHGZLWKWKHFROXPQ ZHLJKWUHVLVWVWKHRYHUWXUQLQJ

H ,QWKLVFDVHWKHRYHUWXUQLQJ PRPHQWFDXVHVWKHFROXPQ WRWRSSOH

vertical direction must sum to zero (ΣF = 0). The isolated diagram with all forces acting on it is called a free-body diagram or an equilibrium diagram. An externally acting horizontal force on the column, associated with either wind or earthquake, causes the same column to overturn. (See Figure 10.) The force’s tendency to produce a rotation about a point is called a moment. The magnitude of the moment is the product of the force times the perpendicular distance from its line of action to the point of rotation (M = F * d). For the column not to overturn, a counterbalancing rotational moment must be present; in this case, it is provided by the column’s dead weight. (See Figure 10.) In formal terms, we say that the rotational moments acting on the column must sum to zero (ΣM = 0).

Structures: An Overview

%DVLFVWUXFWXUH

'HWHUPLQDWLRQRIYHUWLFDOORDGLQJV OLYHDQGGHDGORDGV&KDSWHU 'HWHUPLQDWLRQRIZLQGRU HDUWKTXDNHIRUFHV&KDSWHU

6KHDUGLDJUDPV&KDSWHUVDQG

3URSHUWLHVRID FURVVVHFWLRQ &KDSWHU

0RPHQWGLDJUDPV&KDSWHUVDQG 5HDFWLRQV

&ROXPQ FUXVKLQJEXFNOLQJ DQDO\VHV &KDSWHU

6KHDUVWUHVVHV EHQGLQJVWUHVVHV DQGRWKHUVWUHVVHV &KDSWHU

$[LDOVWUHVVHV&KDSWHUVDQG

6WDWLFDQDO\VHVXVHRIEDVLFSULQFLSOHVRIVWDWLFVΣF ΣM  WRDQDO\]H WKHVWDELOLW\RIWKHVWUXFWXUHDQGWRGHWHUPLQHUHDFWLYHIRUFHV&KDSWHU

0HPEHUDQDO\VHVGHWHUPLQDWLRQRIPDJQLWXGH DQGGLVWULEXWLRQRIIRUFHVVWUHVVHVDQGGHIRUPDWLRQV LQFRQVWLWXHQWPHPEHUV

Figure 11 is a diagrammatic analysis of a simple beam-and-column system. In the overall analysis, live and dead loads are first determined. A series of equilibrium diagrams showing applied and reactive forces is drawn, and the magnitudes and directions of all reactive forces are determined by applying the basic principles of statics. Once the reactive forces are known, various other analyses are made to determine the distribution of internal forces within each member. This is often done through developing shear and moment diagrams. This same information is then used in connection with a detailed analysis of certain properties of members (centroids, moments of inertia) to determine stress and deformation levels in each member. Stress (expressed as force per unit area) measures the intensity of a force at a point. Knowing from experimentation what stress levels a particular material can withstand, the engineer can determine the relative safety of a member. Alternatively, appropriate sizes can be determined.

4

Copyright © 2013. Pearson Education Canada. All rights reserved.

4.1

funIcular structures Basic characteristics

Many whole structures can be characterized as being in a state of pure tension or compression. These interesting structures deserve special treatment. Consider a simple flexible cable spanning two points and carrying a load. This structure must be exclusively in a state of tension because a flexible cable cannot withstand compression or bending. A cable carrying a concentrated load at midspan would deform, as indicated in Figure 12. The whole structure is in tension. If this shape were inverted and loaded in the same way, it is evident by analogy that the resulting structure would be in a state of pure compression. If the loading condition were changed to continuous, a flexible cable carrying this load would deform into the parabolic shape indicated in Figure 12. Again, the whole structure is in tension. If this exact shape were inverted and loaded with the same continuous load, the resulting structure would be in a state of compression. The common arch is such a structure. Structures with shapes derived in this way, wherein only a state of tension or compression is induced by the loading, are called funicular structures. The easiest way to determine the funicular response for a particular loading condition is by identifying the exact shape to which a flexible string would deform under a load. Such a shape is called the tension funicular. Inverting this shape yields a compression funicular. A given loading condition has only one funicular shape. Bending develops in any structure whose shape deviates from the funicular one for

fIGure 11 Typical structural analysis process.

Structures: An Overview

fIGure 12 Typical funicular structures.

Concentrated loads

Distributed loads

Forces at support

Families of funicular shapes

the given loading. Figure 13 illustrates an early, but nonetheless latter-day, analysis of a well-known structure based on the idea that an arch can be conceived of as an inverted catenary. Funiculars need not be only two-dimensional structures; they also can be three dimensional. (See Figure 12.) The shape a flexible membrane assumes under a uniformly distributed load is of special interest. The funicular response is not spherical, but rather parabolic, so it follows that the spherical shell often thought ideal as a structural form for this type of loading is not a funicular response for the vertical loads that commonly act on buildings. (It is, however, for a radial loading.) This shortcoming does not limit its usefulness because other mechanisms at work (circumferential forces) in

fIGure 13 St. Peter’s Dome, Rome. Construction of the thrust line as an inverted catenary.

Copyright © 2013. Pearson Education Canada. All rights reserved.

(Source: Poleni, Memorie istoriche della Gran Cupola del Tempio Vaticano, 1748.)

Structures: An Overview

Copyright © 2013. Pearson Education Canada. All rights reserved.

fIGure 14 Use of threedimensional funicular models. The suspended weights are carefully modeled to represent the actual weights involved.

(a) Inverted photo of the funicular model for the Colonia Guel chapel by Antonio Gaudi

(b) Sketch of the exterior of the chapel drawn from an inverted photo of the model

(c) Inverted photo of the interior of the model

(d) Companion sketch for the interior of the chapel

a shell structure of this type cause it to be highly efficient. Figure 14 illustrates a wellknown example of applying three-dimensional funicular shapes as the basis for determining the form of a structure.

4.2

structural Behavior

Many variants are possible in the way funicular structures are used in practice. They can be used in their pure forms as arches or cables. However, the basic behavior of numerous different structural forms can be described in terms of funicular action. These structures then can be viewed as special forms of funicular structures. Some basic transformations of simple funicular shapes into other structural types are illustrated in Figure 15.

Structures: An Overview

Concentrated loading

Uniformly distributed loading

Foun-

comhori-

Uniformly distributed loading

Copyright © 2013. Pearson Education Canada. All rights reserved.

Concentrated loading

volume-

comp-

fIGure 15 Funicular structures: transformations derived from basic shapes.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Structures: An Overview Figure 15(a) illustrates basic funicular responses for a concentrated load and a uniformly distributed load. Figure 15(b) illustrates the movement that can be expected at the support points if the structure were not constrained at these points. A tension funicular pulls inward and downward. The support or foundation system must apply an outward and upward force on the structure at each end to maintain its shape and keep it from collapsing. These forces are reactive in nature and are referred to as reactions. [See Figure 15(c).] Note that the reactions exert an equal and opposite pull on the structure. The applied forces are thus balanced by equal and opposite reactive forces. A compression funicular produces a converse effect. The structure moves outward and downward. Therefore, an outward and downward force is exerted on the foundation, which must then exert an equal and opposite set of forces on the structure to maintain its shape. The combinations of applied forces acting on the foundation are called thrusts. The foundation must contain these thrusts. The final or resultant direction of the thrusts associated with a funicular structure is along the tangent to the slope of the structure where it meets the support. This follows from the fact that because bending is not present in the structure, all internal forces are directed axially along the length of the member. As Figure 15(d) indicates, this can have important bearings on the design and shaping of the foundation structure. Using built-in massive foundations is not the only way to handle the thrusts developed in a funicular structure. Part of the thrusts can be absorbed into the structure by adding a member. Figure 15(e) illustrates how this is done for several structures. In tension funiculars, the tendency of the two ends to move inward can be restrained by introducing a linear member capable of carrying a compressive force (a strut) between the two points. The net result would be to relieve the foundation of the necessity of providing restraint to the structure’s inward-pulling forces. Hence, the foundation could be designed to carry only vertical forces. The same is true for compressive funicular structures (e.g., arches). By tying the two ends of an arch together, the outward-spreading tendency of the structure is eliminated, and the foundation can be designed to carry vertical forces only. A tension force would develop in the tie-rod connecting the ends of the arch. Several points of interest arise in connection with the preceding discussion. One is that the force the compressive strut or tension tie-rod must carry is exactly equal to the horizontal component of the total force developed by the structure at the foundation. Using struts or ties does not eliminate this force; it handles it in a different way. In some circumstances, it might be preferable to use a strut or tie rather than absorbing the horizontal thrust with a massive foundation. Designing and building foundations capable of handling horizontal forces is not easy. It is easy, however, to design and build a foundation capable of carrying vertical forces only. For this reason, ties or struts are frequently used. Ties, in particular, are quite efficient for taking up the horizontal component of the thrust in a compression funicular because they can be long tension members. Struts for cable structures are less desirable because the element is a long member in compression and potentially susceptible to buckling. The remaining illustrations in Figure 15 indicate further evolution of the funicular shapes into other forms. In some triangulated configurations of linear members (e.g., trusses), the structure’s primary action can also be discussed in terms of funicular shapes. In the truss shown to the left in Figure 15(f), a close inspection reveals that the center diagonals function exactly like a cable carrying a concentrated load, and the member across the top acts as a compression element serving the function previously described. The two end verticals translate the whole assembly upward and are thus in compression. Other truss configurations can be discussed in similar ways. The analogy, however, does not extend to all conceivable shapes of triangulated bar networks Figures 15(h), (i), and (j) illustrate other transformations of the basic funicular structure into forms that could be aggregated to enclose volumes. The method

Structures: An Overview of taking up the horizontal thrusts generated by the structure is a crucial determinant of the exact characteristics of the structures developed. The arch-shaped assembly shown in Figure 15(i), which carries a uniformly distributed load, is a timehonored way to take care of horizontal thrusts by adding buttressing elements. More preferable would be lining up the buttressing elements with the slope of the primary structure at the point of connection. Thus, a wide variety of structures are apparently different but are related in terms of their internal structural behavior.

5

other classIfIcatIons

To summarize some of the concepts discussed, a classification of structures that reflects their internal structural behavior is illustrated in Figure 16. The natures of the loading and boundary conditions are important, although they were not reflected in the

fIGure 16 Classification of typical structures according to their basic load-carrying actions.

Typical structures in bending

Continuous edge supports Continuous edge supports

Uniformly distributed loads

Copyright © 2013. Pearson Education Limited. All rights reserved.

Concentrated loads

One-way structures

Two-way structures

Typical funicular structures One-way structures

Two-way structures

Structures: An Overview classification discussed and illustrated in Figure 1. A close comparison of the two schemes indicates other differences, but there are also overlaps. The detail associated with the classification scheme in Figure 16 is limited. Many of the entries could be expanded into a categorization of their possible transformations discussed and illustrated in Figure 16. Still, the scheme is useful as a conceptual device. The two methods of classification discussed so far are not the only ways structures could be classified. Other classification schemes could be developed on the basis of span or load-carrying capabilities. An interesting way to classify structures is according to energy considerations. Although not yet discussed, energy is stored in a structure as work is done on it (in the form of applied loads). Considerations like this are used to characterize structures into categories such as low-energy or high-energy structures. These and other classification schemes are worth reviewing or delving into to develop an appreciation for the wide variety of structures in use and the ways they are interrelated.

QuestIons 1. Determine how the load-carrying capacity of a long, slender column varies with length; for example, as a column’s length is doubled, what happens to its load-carrying capacity? Do this by measuring the load required to buckle a series of slender compression elements of similar cross sections but varying lengths. A simple bathroom scale can be used to measure loads. Square basswood pieces 1 18 in. * 18 in. 2 are recommended. Support the ends without restraining rotations. Concentrate on long pieces only. (As pieces get shorter, the failure load is more difficult to measure.) Plot your results graphically (buckling load vs. column length). 2. Take a thin sheet of rubber and inscribe a grid on it. Subject the sheet to different loading conditions and different types of support. Do the following: a. Take a square sheet, continuously support its edges, and pile sand on it. b. Repeat, except support the sheet only at its corners. c. Repeat (a) and (b), using a concentrated load at midspan.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Note how the surface deforms in each case by studying how elemental parts of the grid deform. Identify the most highly deformed areas in each case (regions of highest stress). Sketch the deflected shapes obtained. 3. Find photographs of a series of buildings that represent each of the primary structural types described in this chapter. For example, identify one that uses trusses as the primary supporting structures. Make a copy and clearly mark and label the truss structures. Repeat for structures that primarily use arches, cables, plates and space frames, and other structures. 4. Consider Figure 14, which illustrates the hanging-weight model used by Gaudi for the Colonia Gell chapel. Were similar model analyses used for other Gaudi works? Was the structure built? Is the chapel illustrated a singular example of Gaudi’s concern for structure, or are there other examples? Consult your library. 5. Draw a diagram along the lines shown in Figure 4 of a typical wood building of your choice. 6. Obtain and study a photograph of the well-known architectural icon Crown Hall on the IIT campus in Chicago by Mies van der Rohe. What kind of primary structure or structures do you see? Identify them by their full name. Draw a framing diagram of the whole roof structure showing primary and secondary elements. Consult your library. 7. Obtain and study photographs of the Experience Music Theater in Washington by Frank Gehry and Associates (be sure to obtain photographs of the inside of the building during construction). Can the roof structure that is visible from the outside be described as some type of funicular or shell structure, or is it better described some other way?

Copyright © 2013. Pearson Education Limited. All rights reserved.

InTroducTory concepT

Copyright © 2013. Pearson Education Limited. All rights reserved.

This chapter is a self-contained overview of the field and discusses different ways to classify structural elements and systems.

From Part I of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Principles of Mechanics

Copyright © 2013. Pearson Education Limited. All rights reserved.

1 Introduction Mechanics is the branch of applied science dealing with forces and motions. Fundamental to the field is equilibrium, the condition existing when a system of forces acting on a body is in a state of balance. The term statics is used to describe the part of mechanics concerned with relations between forces acting on rigid bodies that are in equilibrium and at rest. The term dynamics refers to the part of mechanics dealing with rigid bodies in motion. If correctly placed inertial forces are taken into account, bodies in motion can also be considered to be in equilibrium. The field of study referred to as strength of materials is an extension of mechanics that addresses the relationship between applied or external forces acting on a body and the internal effects produced by those forces in the body. The study of the deformations produced in a body by a set of external forces is an integral part of the field of strength of materials. These distinctions reflect how the study of the subject has evolved in the engineering disciplines. In most engineering curricula, statics, dynamics, and ­ strength of materials are treated as separate topics presented sequentially under the umbrella of mechanics. During the process of analyzing and designing structures in buildings, however, professionals freely use ideas and elements from each of these fields (as well as others) as tools in a nonsequential manner. At this point, however, it is pedagogically useful to retain traditional engineering distinctions. In the first part of this chapter, we introduce fundamental ideas in statics. In the second part, we focus on basic elements of strength of materials. The following presentation provides only an overview of the basic issues involved in statics and the strength of materials, so topics are presented succinctly. The reader is referred to any of several basic texts that treat the subject matter in more detail.1

Note to readers and instructors: Although presented in a logical continuum of ­topics, the reader may not want to cover all of the topics in the order they are ­presented because they are often abstract.

1

Students who want more extensive coverage of the application of structural principles to sculpture should see Schodek, Structure in Sculpture. Cambridge, Massachusetts: MIT Press, 1993.

From Chapter 2 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Principles of Mechanics It is not necessary to know everything about shear and moment diagrams or material properties that are presented later in the chapter to understand topics on load modeling, trusses, cables, and arches, but some understanding is essential for dealing with beams and columns. Depending on the curricula, for example, topics such as shear moment diagrams are best covered after truss analysis and before beam analysis. The topics in this chapter are presented to be drawn from as needed.

2  Forces and Moments 2.1 Analysis Objectives and Processes Section 2 explores forces, moments, and equilibrium concepts. Before developing these concepts, it is useful to look at a simple example of where the discussion will lead. Concepts of force, moment, and equilibrium are given visual expression in the mobile by Alexander Calder illustrated in Figure 1.2 The mobile is in careful balance, as is each of its parts. Any typical arm of the mobile experiences a set of forces acting on it. These forces consist of external, or applied, forces—for example, one of the weights—and internal forces, or reactions, that develop within the structure at connection points. The arm must be in equilibrium with respect to the various forces (assuming that it is not sliding in space or spinning madly). The net translational effect of all forces or their components acting on an object must have a sum of zero along any axis or in any direction. As developed in Section 3, this fundamental requirement is expressed as follows: g F = 0 (read as “the sum of forces equals zero”). Likewise, there must be no net rotational effects of the forces about

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 1  Basic equilibrium diagrams for each arm of the mobile are shown to the right. The diagram for arm D is shown in more detail.

W

2

Ferdinand P. Beer and E. R. Johnston, Vector Mechanics for Engineers, 3rd ed. New York: McGraw-Hill, 1977.

Principles of Mechanics the point of suspension. Note that these rotational effects can be quantified as a product of the magnitude of the force times its distance from the point of suspension 1F * d2. Such a rotational effect is called a moment 1M = F * d2. Moments can act in clockwise or counterclockwise directions. For the arm to be in rotational equilibrium, the net total of the rotational moments acting on it must be zero, or g M = 0. In the mobile, a clockwise moment acting on an arm is exactly balanced by a moment that acts in a counterclockwise direction. Note that a small force acting a long distance from the suspension point can have the same rotational effect as a large force acting over a short distance. Analytical drawings that illustrate force systems that act on an object are called equilibrium diagrams (also called free-body diagrams; see Section 3.3). Constructing these kinds of diagrams is a first step in making a static analysis of a structure. Using equilibrium concepts, it is then possible to determine numerical values for the reactions (force interactions generated by the action of one object on another; see Section 3.3) that occur at supports or connections. Subsequently, internal forces (shears and bending moments; see Section 4.3), stresses (internal forces per unit area), and strains (deformations caused by stresses) can be determined (see Section 6). The following sections explore these concepts in detail. What happens, for example, if the forces applied to an object are inclined? What if the forces are distributed over a surface rather than acting at a point?

2.2 Forces

Copyright © 2013. Pearson Education Limited. All rights reserved.

Fundamental to the field of mechanics is the concept of force and the composition and resolution of forces. A force is a directed interaction between bodies. Force interactions cause changes in the shape or motion (or both) of the bodies involved. The basic concept of force is likely familiar to the reader and intuitively clear. Viewed from a historical perspective, however, the idea of force and its characterization in terms of magnitude, sense, and direction was hardly obvious. The precise formulation of these concepts is a remarkable accomplishment in view of the degree of abstraction involved. Indeed, the distinction between force and weight, as well as the notion of a nonvertical force, was only just beginning to be appreciated by scholars in the Middle Ages. The name Jordanus de Nemore is repeatedly connected to the emergence of these concepts. Once force was conceived in vectorial (directional) terms, the problem of the components of a force and the general composition and resolution of forces were addressed by several individuals, including Leonardo da Vinci, Stevin, Roberval, and Galileo. This problem, often termed the basic problem in statics, was finally solved by Varginon and Newton.

2.3 Scalar and Vector Quantities A distinction is made in the study of mechanics between scalar and vector quantities. Scalar quantities can be characterized by magnitude alone. Vector quantities must be characterized in terms of magnitude and direction. Forces are vector quantities. Any vector quantity can be represented by a line. The direction of the line with respect to a fixed axis denotes the direction of the quantity. The length of the line, if drawn to scale, represents the magnitude of the quantity. [See Figure 2(a).] The line of action of a force has an indefinite length, of which the force vector is a segment. Most structures are rigid bodies that deform only slightly when a force is applied, so it can be assumed that the point of application of a given force may be transferred to any other point on the line of action without altering the translatory or rotational effects of the force on the body. Thus, a force applied to a rigid body may be regarded as acting anywhere along the line of action of the force.

Figure 2  Free vectors, force interactions, resultant forces, and the parallelogram of forces.

Principles of Mechanics

Figure 3  Parallelogram methods of finding the resultant force R of two ­concurrent forces. R can be found algebraically or graphically.

2.4  Parallelogram of Forces Essential to a study of structural behavior is knowing the net result of the interaction of several vector forces acting on a body. This interaction can be studied in terms of the laws of vector addition. These laws and fundamental postulates are based on experimental observation. Historically, the first method of adding vector quantities was based on the parallelogram law. In terms of force vectors, the law states that when the lines of action of two forces intersect, a single force, or resultant, that is equivalent to the two forces can be represented by the diagonal of the parallelogram formed by using the force vectors as its sides. [See Figure 2(b)–(d).] In general, a resultant force is the simplest force system to which a more complex set of forces may be reduced and still produce the same effect on the body on which those forces acted. Figure 3 illustrates a numerical example. A technique for finding the resultant force of several force vectors whose lines of action intersect is illustrated in Figure 4. The individual vectors, drawn to scale, are joined in tip-to-tail fashion. The order of combination is not important. Unless the resultant force is zero, the force polygon thus formed does not form a closed figure. The closure line is identical to the resultant force of the several individual vectors (i.e., the resultant is the vector that extends from the tail of the first vector to the tip of the last vector in the group). The resultant closes the force polygon. This general technique follows from the parallelogram law. An algebraic method for finding the resultant of several forces acting through a point is discussed in Section 6. Although conceptually simple, graphical approaches to finding the resultants of force systems are powerful structural analysis aids. Historically, they found wide usage because of the ease of their application. Early investigators used graphical techniques extensively in their attempts to understand the behavior of complex structures. Figure 5 is a latter-day analysis of a gothic structure by graphic techniques. The sphere model of an arch in Figure 6 also utilizes graphic techniques. The model, based on the parallelogram law, is still an elegant way to look at arches. Note that the techniques are simple and based on the concepts discussed in this chapter. Graphical techniques are no longer used extensively but remain an elegant way to look at structures and are useful in developing an intuitive feeling for the flow of forces in a structure.

2.5 Resolution and Composition of Forces

Copyright © 2013. Pearson Education Limited. All rights reserved.

A process that follows directly from the fundamental law concerning the parallelogram of forces is that of breaking up a single force into two or more separate forces that form a system of forces equivalent to the initial force. This process is usually called

Figure 4  Graphical methods of finding the final resultant force RB of a concurrent force system.

Principles of Mechanics

Figure 5  Application of graphic methods to the analysis of a gothic structure (Amiens). The dead weights of the vertical buttresses and pinnacle help turn the thrusts of the flying buttresses downward through the middle portion of the buttresses. The vertical buttresses are stable and not prone to overturning or cracking when the force resultants pass through the middle portions of these buttresses.

resolving a force into its components. The number of components into which a single force can be resolved is limitless. (See Figure 7.) In structural analysis, it can be convenient to resolve a force into rectangular, or Cartesian, components. By utilizing right angles, components can be found by means of simple trigonometric functions. When a force F is resolved into components on the x- and y-axes, the components become Fx = F1cos u2 and Fy = F1sin u2. The process is reversed if Fx and Fy are given and the resultant force wants to be known: F :F = 2F 2x + F 2y and u = tan-1 1Fy >Fx 2.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 6  Early sphere model of an arch: If a series of spheres is stacked as illustrated, the assembly is stable. Note that the shape is the inversion of a freely hanging chain made of similar spheres. The assembly will collapse if either the loading or the positioning of the spheres is changed.

Principles of Mechanics

Figure 7  Resolution of a force into components.



Using Cartesian components is a matter of convenience. A right angle is only a special form of a parallelogram. Figure 8 illustrates several manipulations with components. For example, a force of F = 500 lb that acts 30° to the horizontal has components Fx = 500 cos 30° = 433 lb and Fy = 500 sin 30° = 250 lb. Similarly, the resultant of two forces of 100 and 200 lb acting orthogonally to one another is given by R = 2F 2x + F 2y = 211002 2 + 12002 2 = 223.6 lb, with u = tan-1 1100 > 2002 = 26.6°. Components could be found in three dimensions as well as two. Example Determine the components on the x- and y-axes of a force F of 1000 lb that acts at an angle of f = 60° to the x-axis. Solution: Fx = F cos f = 1000 cos 60° = 500.0 lb Fy = F sin f = 1000 sin 60° = 866.6 lb

 



Alternatively, if the components of a force F were known to be orthogonal and have values of 866.6 lb and 500 lb on the x- and y-axes, respectively, you would find the magnitude and direction of F with the following formulas:

and

F = 4F 2x + F 2y = 31500.02 2 + 1866.62 2 = 1000 lb tan f x = 500/866.6 = 0.92 6 f x = 60°

or f x = tan-1 866.6>500 = 60°   

and F = Fx >cos f = 500>cos 60° = 1000 lb

Copyright © 2013. Pearson Education Limited. All rights reserved.

2.6 Statically Equivalent Systems Implicit in the previous section’s discussion is the notion of static equivalency. When a system of forces applied to a body can be replaced by another system of forces applied to the same body without causing any net change in translational or rotational effects on the body, the two force systems are statically equivalent. A resultant force, for example, is statically equivalent to the force system from which it was derived. The diagrams in Figure 8 illustrate a process for determining the statically equivalent single resultant force of a series of coplanar concurrent forces. If an equilibrating force F that is equal and opposite to the resultant force R is applied to the same point, then the point is not subjected to any net force. (The point is said to be in equilibrium; see the next section.) The algebraic process illustrated depends on resolving each force into components (Fx and Fy forces) and summing components acting in the same direction. The resultant force is then given by R = 21 g Fx 2 2 + 1 g Fy 2 2 and its orientation by ux = tan-1 1 g Fy > g Fx 2. Concurrent forces act through the same point and do not produce rotational effects about that point. (The moment arms are zero.) A single statically equivalent force for a nonconcurrent force system (in which forces do not intersect at a common point) can also be found.

Principles of Mechanics

Figure 8  Resultants and equilibrating forces. The equilbrating force F is The resultant force R closes the force polygon. equal and opposite to R.

Force polygon 100 P

R = 116.1 P @ 54°

60 P

60 P

F = 116.1 P @ 54° R = 116.1 P

100 P

A

R = 116.1 P

54°

70 P

70 P

Three concurrent forces

54°

Alternative graphical solutions yield same value of R.

Graphical (tip to tail) determination of the resultant force R or the equilibrating force F of the three forces

(a) Resultants and equilibrating forces of concurrent forces systems: graphical methods 100 P R = 116.1 P @ 54°

60 P

Resultant force R = 116.1 P @ 54°

= 54°

A

A 70 P

The resultant force produces the same effect on point A as the three forces

54°

A

Equilibrating force F = 116.1 P

The equal and opposite equilibrating force makes point A not subject to any net force (and is in a state of equilibrium)

(b) Resultants and equlibrating forces

 R

Copyright © 2013. Pearson Education Limited. All rights reserved.

(c) Resultants of concurrent force systems: algebraic method

2.7 Moments Moment of a Force.  A force applied to a body causes the body to translate in the direction of the force. Depending on the force’s point of application on the body, however, the force may also cause the body to rotate. This tendency to produce rotation is called the moment of the force. (See Figure 9.) With r­espect to a point or line, the magnitude of this turning or rotational tendency is equal to the product of the magnitude of the force and the perpendicular distance from the line of action of the force to the point or line under consideration. The ­moment M of a force F about a point O is MO = F * r, where r is the perpendicular distance from the line of action of F to O. The distance r is often called the ­moment arm of the force. A moment has the units of force times distance (e.g., ft-lb or N # m). A force of 1000 lb acting 5 ft away from a point produces a moment of M = F * r = 1000 lb * 5 ft = 5000 ft@lb.

Principles of Mechanics

Line of action of F

Figure 9  Moment of a force

F2

Line of action of F

Moment center

r

F1

F

d2

=

Moment arm

Force F

MA = F x d1

d1

Moment arm

Mo = F x r

d1 MA = ( F1 x d 1 ) +(F xd ) 2 2 Point A

Point A

(a) Rotational moment of a force about a point (moment = force × distance)

(b) Moments cause members to rotate

(c) Moment of several forces

L

w lb/ft or w kN/m

Uniformly distributed load acting along the length of a member Equivalent concentrated = w L force

M A = ( w L )(L /2 )

= Ld /2

Moment arm Point A

(d) Member with a uniformly distributed load acting on it

(e) Moment of a uniformly distributed as modeled by the action of an equivalent concentrated load

Example

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 10  Rotational balance

The moment concept can be used to understand the behavior of many familiar objects. In the Calder mobile (Figure 1), the moments on each side of the point of suspension must be identical for balance to occur. Assume that W = 100 lb, FE 1the weight of the lower arm2 = 150 lb, a1 = 4.5 ft, and a2 = 3 ft. Demonstrate that the arm is in ­balance about the point of suspension. (See Figure 10.) Solution: The rotational moment of W about the point of suspension is given by M1 = W * a1 = 100 lb * 4.5 ft = 450 ft@lb This rotational effect must be balanced by the moment produced by the other force: M2 = 150 lb * 3 ft@lb = 450 ft@lb

Because W * a1 = FE * a2, the arm is in rotational balance 1 gM = 02.

Moment of a Distributed Load.  Loads are frequently uniformly distributed along the length of a member (e.g., w lb>ft). The total force on the member equivalent to this uniformly distributed load is wL. The moment of the distributed force is found by imagining the uniformly distributed load concentrated into an equivalent point load of magnitude wL located at the point of symmetry of the loading and multiplying it by the distance to this point (L>2 for a uniform load). This is illustrated in Figure 9. For example, for the member in Figure 9, a uniform

Principles of Mechanics loading of 100 lb>ft on a 20-ft-long member produces a moment about point O of MO = 1wL21L>22 = 1100 lb>ft2120 ft2110 ft2 = 20,000 ft@lb. If a triangular load distribution (which varies from zero to a maximum value of w) is present, the equivalent concentrated load is wL>2, and it acts at the two-thirds point of the loading. Modeling loads in this way is valid only for calculating reactions, not for constructing shear and moment diagrams or for other purposes. The foregoing method is too simple and is difficult to apply when loads are more complex and when they vary along the length of a member. Moments Due to Multiple Forces.  The total rotational effect produced by ­several forces about the same point or line is the algebraic sum of their individual moments about that point or line. Thus, MO = 1F1 * r1 2 + 1F2 * r2 2 + c + 1Fn * rn 2.

Moments about a Line.  The rotational effect on a rigid body caused by multiple forces acting about a line, but not in the same plane, is the same as what would result if all the forces were acting in the same plane. Moment of a Couple. A couple is a force system made up of two forces equal in magnitude, but opposite in sense, and with parallel lines of action that are not on the same straight line 1M2. A couple causes only rotational effects on bodies and does not cause translation. The moment of a couple is the product of one of the forces and the perpendicular distance between the two forces. It can be shown, however, that the moment of a couple is independent of the reference point selected as a moment center. The magnitude of the rotational effect produced by a couple on a body also is independent of the point of application of the couple on the body.

3 Equilibrium

Copyright © 2013. Pearson Education Limited. All rights reserved.

3.1 Equilibrium of a Particle A body is in equilibrium when the force system acting on it produces no net translation or rotation of the body and the body is in balance. Equilibrium exists in concurrent force systems (systems in which all forces act through a single point) when the resultant of the force system equals zero. A concurrent force system having a nonzero resultant force can be put in equilibrium by applying another force (called an equilibrant) that is equal in magnitude and on the same line of action but of opposite sense. If a force system is in equilibrium, its resultant must be zero, and it follows that g Fx = 0 and g Fy = 0. Thus, the algebraic sum of the components of the forces applied to a particle in the x direction must be zero and likewise for the y direction. Note that x and y need not be horizontal and vertical, respectively: The previous statement is true for any orthogonal set of axes, no matter what their orientation. More generally, the conditions g Fx = 0, g Fy = 0, and g Fz = 0 are necessary and sufficient to ensure equilibrium in a concurrent force system. A force system satisfying these conditions will not cause the particle to translate. (Rotation is not a problem because all forces act through the same point in concurrent force systems.)

3.2 Equilibrium of a Rigid Member General Equilibrium Conditions.  When a nonconcurrent force system acts on a rigid body, the potential for both translation and rotation is present. For the rigid body to be in equilibrium, neither must occur. With respect to translation, this implies that, as in the case of concurrent force systems, the resultant of the

Principles of Mechanics force system must be zero. With respect to rotation, the net rotational moment of all forces must be zero. The conditions for equilibrium of a rigid body are, therefore, g Fx = 0, g Fy = 0, g Fz = 0 and g Mx = 0, g My = 0, g Mz = 0. Sign Conventions.  In working with general force systems, sign conventions are always problematic. One classic approach to length measurements is to use x, y, and z values measured from a reference point on the extreme left of the structure. Thus, distances measured from this point to the right are always positive 1 +x2. For forces, a similar convention can be used. A force acting vertically upward would be positive 1 +F2 and one acting downward would be negative 1 -F ). Components of forces would be similarly designated. For rotational moments, note that with respect to a reference point on the extreme left of the structure, an upward force 1 +F c 2 located to the right of a positive distance 1 +d2 would cause a positive moment about the point that acts in a counterclockwise direction, or M = 1 +F21 +d2 = +Fd. A downward force would be negative 1 -F2 and cause a negative moment that acts in the clockwise direction, or M = 1 -F21 +d2 = -Fd. While useful, this convention can get cumbersome and confusing (particularly regarding rotational moments) when nonvertical forces are present. Most individuals use a convention directly relating to the rotational sense of the moment effect. Thus, moments that produce counter+ M2, and those that produce clockwise clockwise rotations are considered positive 1⤺ + M2. These conventions are for equilibrium calrotations are considered negative 1 culations involving external forces only. Other conventions will be developed later for describing internal forces and moments that act within the structure. Figure 11  Equilibrium of a beam

Example Determine the unknown forces FA and FB in the structure shown in Figure 11 so that it is in a state of static equilibrium. Solution: Force equilibrium in the vertical direction, gFy = 0: FA + FB -4P = 0. This expression cannot be solved yet because we have only one equation and two ­unknown force values. Force equilibrium in the horizontal direction, gFx = 0: No forces act in the horizontal (or x) direction.

Moment equilibrium about point A using force and distance algebraic conventions:

Copyright © 2013. Pearson Education Limited. All rights reserved.

gMA = 0

1 +FA 2102 + 1 -4P21 + 152 + 1 +FB 21 + 202 = 0

FA 102 - 14P21152 + FB 1202 = 0 6 FB = 3P c

Moment equilibrium about point A using counterclockwise rotations as positive: gMA = 0

-4P1152 + FB 1202 = 0

From gFy = 0: FA + FB - 4P = 0:

6 FB = 3P c

FA + 3P = 0 6 FA = 1P c

In the next section, we see that the forces FA and FB in situations similar to that shown typically occur at structural supports and are reactive. By contrast, there might be situations where the forces are applied. The point is that no matter what their origin, they must have the numerical values noted for the structure to be in equilibrium.

Principles of Mechanics

Figure 12  Two- and three-force members.

Two-Force Members.  Several special equilibrium cases exist. When a rigid member is subjected to only two forces, the forces cannot have arbitrary magnitudes and lines of action if the member is to be in equilibrium. Consider the member shown in Figure 12. By summing moments about point A, it can be seen that the structure cannot be in rotational equilibrium unless the line of action of the force at B passes through point A. In a similar way, it is necessary that the line of action of the force at A pass through point B if the structure is to be in rotational equilibrium about point B. Thus, the two forces must be collinear. They must also be equal in magnitude but opposite in sense. This result is particularly important in the analysis of trusses. Three-Force Members.  As with two-force members, three forces acting on a member cannot have random orientations and magnitudes if the member is to be in equilibrium. This is seen in Figure 12. For the member to be in rotational equilibrium, the lines of action of all three forces must pass through a common point. Figure 13 shows an analysis of the flying buttress of a gothic structure. The lines of action of an assumed point load and the two reactions must pass through the same point.

3.3 Applied and Reactive Forces

Copyright © 2013. Pearson Education Limited. All rights reserved.

Forces and moments that act on a rigid body can be divided into two types: applied and reactive. In common engineering usage, applied forces act directly on a structure (e.g., the force produced by snow). Reactive forces are generated by the action of Figure 13  Application of graphical methods to the analysis of a gothic structure. The weight FB of the buttress is equilibrated by its two reactions R1 and R2, so the lines of ­actions of all three forces meet in one point.

R1 R2 FB

Principles of Mechanics one body on another and typically occur at connections or supports. The existence of reactive forces follows from Newton’s third law, which, broadly speaking, states that to every action, there is an equal and opposite reaction. More precisely, it states that whenever one body exerts a force on another, the second always exerts a force on the first that is equal in magnitude, opposite in direction, and has the same line of action. In Figure 14(b), the force on the beam causes downward forces on the foundation, and upward reactive forces are developed. A pair of action and reaction forces thus exists at each interface between the beam and its foundations. In some cases, moments also form part of the reaction system. (See Figure 14(c).) The diagrams in Figure 14 that show the complete system of applied and reactive forces acting on a body are called free-body diagrams. If a body (such as those illustrated) is in a state of equilibrium, the general conditions of equilibrium for a rigid body that were stated in the previous section must be satisfied. The magnitude and direction of any reactive forces that are developed must be such that equilibrium is maintained and is dependent on the characteristics of the applied force system. The whole system of applied and reactive forces acting on a body (as represented by the free-body diagram) must be in a state of equilibrium. Free-body diagrams are, consequently, often called equilibrium diagrams. Drawing equilibrium diagrams and finding reactions for loaded structural members is a common first step in a complete structural analysis. Support Conditions.  The nature of the reactive forces developed on a loaded body depends on how the body is either supported by or connected to other bodies. Figure 15 illustrates relations between the type of support condition present and the type of reactive forces developed. Basic types of support conditions are indicated; others are possible. Of primary importance are pinned connections, roller connections, and fixed connections. In pinned connections, the joint allows

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 14  Reactions and free-body (or equilibrium) diagrams.

Principles of Mechanics

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 15  Types of support conditions: idealized models.

attached members to rotate freely but does not allow translations in any direction. The behavior is similar to that of a simple hinge. The joint cannot provide moment resistance but can provide force resistance in any direction. A roller connection also allows rotations to occur freely. It resists translations, however, only in the direction perpendicular to the face of the support (either into or away from the surface). It does not provide any force resistance parallel to the surface of the support. A fixed joint completely restrains rotations and translations in any direction. Therefore, it can provide moment resistance and force resistance in any direction. Other types of supports include a cable support and a simple support. These are similar to the roller connection, except that they can provide force resistance in one direction only. The connections shown are, of course, idealized. For a structure to be stable, the supports must provide a specific minimum number of force restraints. A simple beam loaded with both downward and horizontal forces must have three such restraints, corresponding to the three conditions of equilibrium for that type of structure: g Fx = 0, g Fy = 0, and g M = 0. One way to meet this requirement is to use a fixed connection. Another is to use a pinned connection on one end and a roller connection on the other. Connections that offer more degrees of restraint than the minimum required may be used. Structures having connections or supports that provide more than the minimum needed for stability are referred to as statically indeterminate externally. Because there are more unknown constraining forces than equations of equilibrium, it is not possible to solve for the magnitudes of these constraining forces by statics alone.

Principles of Mechanics Reactions for Typical Load and Support Conditions.  The examples that follow illustrate how reactions are calculated for common situations. In the first example, the reactions must act only in the vertical direction because the beam sits on just two supports. (Note that the beam would be unstable and slide if a horizontal force were applied to it.) In the second example, the applied loads act vertically downward, and the structure has pin and roller support conditions. The preceding discussion argued that a reactive force can develop in any direction at a pinned connection but only perpendicular to the plane of support in a roller joint. Common sense correctly suggests, however, that when all applied loads act vertically, all reactive forces, including the one at the pin, also act vertically. From a calculational perspective, it is still necessary to assume initially that the reaction at the pin is inclined. The reaction analysis should then predict that the force is indeed vertical. (The horizontal component should turn out to be zero, meaning that the force acts vertically.) In the third example, the applied force is inclined. An inclined reactive force then develops at the pin. This result is common because when a structure carries any applied force with a horizontal component, the reactive force at the pin is also inclined to obtain necessary equilibrium balances. The examples include cantilever structures, which illustrate that roller condition can develop either compressive or tension (tie-down) reactions. Example Determine the reactions RA and RB for the beam shown in Figure 16, which rests on two supports. Figure 16  Reactions of a beam resting on two supports. P = 4000 lb

A 12 ft L = 30 ft

P = 8000 lb

=

B 9 ft

Copyright © 2013. Pearson Education Limited. All rights reserved.

P = 4000 lb

P = 8000 lb

RA

9 ft

9 ft

Assume that the reactions act only in the vertical direction.

12 ft

9 ft

RB

L = 30 ft

Solution: The reactions must act only in the vertical direction for these support conditions. The loading is not symmetrical, so it is necessary to determine one of the reactions by moment equilibrium first. Point A is chosen for convenience. Counterclockwise moments are assumed to be positive. The other reaction is found by vertical force equilibrium. A static check is shown to verify results. Moment equilibrium about point A: gMA = 0:

-9 ft 14000 lb2 - 21ft 18000 lb2 + 102RA - 30RB = 0

6 RB = 6800 lb Force equilibrium in the vertical direction: gFy = 0:

-4000 lb - 8000 lb + RA + RB = 0 -12000 + RA - 6800 = 0 6 RA = 5200 lb

Check: Moment equilibrium about point A: gMB = 0:

+21 ft 14000 lb2 + 9 ft 18000 lb2 + 102RB - 30RA = 0

6 RB = 5200 lb

Principles of Mechanics

Example Determine the unknown reaction forces RA and RB in the structure in Figure 17. Demonstrate that the horizontal component of RB is zero for this loading. Use x and y components of RB in the equilibrium analysis. Solution: The reactive force roller RA at the left must, by definition, act perpendicularly to the support plane. The pinned condition on the right can provide a reactive force resistance RB in any direction, so it is shown acting at an angle and broken into x and y components. By inspection, however, and noting that the external forces act only in the vertical direction, it can be surmised that the direction of the reactive force at the pin is vertical only. If a horizontal force component were present, the structure could not be in equilibrium in the vertical direction. Use force components for RB in the following analysis:

Force equilibrium of all the forces acting in the vertical direction, gFy = 0 c + : RA + RBy - 1P - 4P = 0 or RA + RBy = 5P

While valid, this equation cannot yet be solved, due to the presence of two unknowns. + : Force equilibrium in the horizontal direction, gFx = 0 S

0 + RBx = 0 Because no forces act in the horizontal direction, RBx must be zero, which in turn implies that the resultant RB acts vertically and is equal to RBy. Moment equilibrium about point A, gMA = 0: The convention that moments acting in the counterclockwise direction are positive is used. RA 102 - 11P2152 - 4P1152 + 20RBy + RBx 102 = 0 or - 5P - 60P + 20RBy = 0 6 RBy = 3.25P c

Note that, because RBx = 0, RB = RBy. RA + 3.25P = 5P 6 RA = 1.75P c

Copyright © 2013. Pearson Education Limited. All rights reserved.

Example Determine the reactions to the structure in Figure 18. The structure has an inclined applied force.

Figure 18  Reactions for a beam with inclined load.

Solution: For convenience, the applied load is broken into horizontal and vertical components. The components are then considered applied loads. The vertical reactions are found first. Because gFy = 0 can rarely be directly solved first, use gMA = 0, which will immediately yield the vertical component of the reactive force at B.

Figure 17  Reactions for a simple beam with vertical loads.

Principles of Mechanics Moment equilibrium about point A, gMA = 0 ⤺ + : RA 102 - 4P sin 60°152 + RBy 1102 + RBy 102 = 0 or - 3.46P152 + 10RBy = 0

RBy = 1.73P

6

RBy is the component of the reactive force at B in the vertical direction. To find the reactive force at A that acts in the vertical direction—which must act vertically because of the roller condition—sum forces in the vertical direction. Force equilibrium of all the forces acting in the vertical direction, gFy = 0

c +:

-4P sin 60° + RAy + RBy = 0 -3.46P + RAy + 1.73P = 0

Force equilibrium in the horizontal direction, gFy = 0

RA = 1.73P

6 S

:

4P cos 60° - RBx = 0 2.0P - RBx = 0

RBx = 2.0P

6

Resultant reactive force RB at B: To find the resultant force at B, use the known components, RBx and RBy: tan-1uB = 1.73>2.0 RB = 1.73P>sin u = 2.64P

uB = 40.85°

6 or

= 2.0P>cos u = 2.6

Thus, RB = 2.64P at 40.85° to the horizontal. All reactions are now known. Note that the reaction at B must be inclined so that it provides a resistive force to balance the horizontal component of the applied load. If the reactive force acted vertically, as in the previous example, the structure would not be in equilibrium. Note also that if two rollers were present, instead of a pin and a roller, it would be impossible for the reactions to provide equilibrium restraints, and the structure would slide to the right. In the moment calculations, several forces went through the moment center. Moment arms were thus of zero length, indicating that the forces produced no rotational effects about the point considered.

Example

Copyright © 2013. Pearson Education Limited. All rights reserved.

Determine the unknown reaction forces RA and RB in the structure in Figure 19.

Figure 19  Reactions for a cantilevering beam loaded with a point load.

Solution: The pinned connection on the right can provide a force RB that acts in any direction. No external forces act horizontally, however, so it is evident that the force must act only vertically, and RBx = 0. (If RB had a component in the x direction, gFx = 0 could not be satisfied.) The roller on the left can transmit a force RA in the vertical direction only, but this force may be directed upward or downward. Assume that it is upward leads us to the following calculations: Moment equilibrium about A, gM = 0 ⤺ + : A

1RB * 102 - 14P * 152 = 0

Find RA from gFy = 0

6 RB = 6P

c+:

RA + sRB - 4P = 0 RA + 6P - 4P = 0 RA = -2P c

or

+2P c

The negative sign in the solution indicates that the direction assumed for RA was incorrect and that it acts in the direction shown (as a tie-down force). Rollers can be designed to act as tie-downs.

Principles of Mechanics

Example

Figure 20  Reactions for a cantilevering beam loaded with two point loads.

Determine the unknown reaction forces RA and RB in the structure in Figure 20. Solution: This problem is similar to the last but shows that the direction of the force at the roller is ­dependent on the magnitudes and locations of the loadings present. +: Moment equilibrium about point A, gM = 0 ⤺ A

- 6P152 - 4P1152 + 10RB = 0

6

RB = 9P c

Force equilibrium of all the forces acting in the vertical direction, gFy = 0: RA + RBy - 6P - 4P = 0

6

+c

RA = 1.0P c

Example Determine the unknown reaction forces RA and RB in the structure in Figure 21. Solution: The pinned connection on the right can provide a force resistance in any direction, as reflected in Figure 21 by an unknown force RB acting at the connection at an arbitrary angle. The roller on the left can transmit forces in the vertical direction only. Components of RB are initially used. Force equilibrium of all the forces acting in the vertical direction, gFy = 0

c +:

Figure 21  Reactions for an L-shaped beam with horizontal and vertical point loads.

RA + RBy - 4P = 0 or RA + RBy = 4P

+ Force equilibrium in the horizontal direction, gFx = 0 S :

2P - RBx = 0

6

Moment equilibrium about point A, gMA = 0 ⤺ + :

RBx = 2P

RA 102 - 14P21102 + RBy 1202 + RBx 102 - 12P2152 = 0 6 RBy = 2.5P

Find RA from gFy = 0: RA + RBy = 4P:

RA + 2.5P = 4P 6 RA = 1.5P c Find the resultant force RB from RBx and RBy: RB = 2R2Bx + R2By = 3.2P

Copyright © 2013. Pearson Education Limited. All rights reserved.

tan-1u = 2.5>2.0

6 u = 51.3°

Example Determine the unknown forces RA and RB in Figure 22. The uniformly distributed load is first converted into an equivalent concentrated load for equilibrium calculations. Figure 22  Uniformly loaded beam.

Equivalent total load: wL = (50 lb/ft)(20 ft) = 1000 lb.

wL = 50 lb/ft = RA

L = 20 ft

RB

L /2 = 10 ft R A = wL /2 = (50 lb/ft)(20 ft)/2 = 500 lb.

L /2 = 10 ft R B = wL /2 = (50 lb/ft)(20 ft)/2 = 500 lb.

Principles of Mechanics + : Moment equilibrium about A, gMA = 0 ⤺ Solution:

gM = 0: RA L - 1wL21L>22 = 0 RA = wL>2 = 150 lb>ft2120 ft2 >2 = 500 lb

Force equilibrium in the vertical direction, gFy = 0

gF = 0: RA + RB - wL = 0

+ c: 6

RB = 500 lb

Note that by symmetry, RA = RB. Thus, it is possible to solve for the unknown forces by considering equilibrium in the vertical direction only: gF = 0; RA + RB = total downward load = wL. Thus, RA = RB = wL>2 = 500 lb. Figure 23  Reactions for a beam loaded partially with uniform load.

Example Determine the reactions for the structure shown in Figure 23. Note that 1 kip = 1000 lb. Solution: To determine the reactions, the uniformly distributed load that acts over part of the structure is modeled as a statically equivalent concentrated load. Moment equilibrium about point A, gMA = 0:

[12 kips>ft2110 ft2]115 ft2 - 20RB

6

RB = 15 kips

f

x

moment arm equivalent concentrated load 1to loading center2

Force equilibrium in the vertical direction, gFy = 0: 5 ft

+RA + RB - [12 kips>ft2110 ft2] = 0

5 ft

6 RA = 5 kips

Example Determine the reactions for the structure shown in Figure 24. Use components.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 24  Beam with roller resting on a sloped support.

Solution: Equilibrium in the vertical direction, gFy = 0: RAy + RBy - wL = 0. The total load acting downward due to the distributed loading is the load per unit length multiplied by the length over which the load acts. Note that the direction of the unknown

Principles of Mechanics reaction RB is initially known because roller joints can transmit loads only perpendicular to the surfaces on which they roll. Hence, RBy must be RB sin 60°. The direction of RA is not known a priori. Moment equilibrium about point A, gMA = 0:

L + RBy 1L2 - 1wL2 a b = 0 2

or

RBy =

wL 2

Hence, RAy = wL>2 from gFy = 0. All other unknown components of the two reactive forces pass through the moment center and consequently have zero moment arms and drop out of the equation. The moment produced by the uniformly distributed load was found by imagining it to be concentrated at its center of mass and finding the moment produced by this concentrated load about point A. (See Section 7.) Final reactions: Because RBy is known, RB can be calculated next. Thus, RBy = RB sin 60° or RB = RBy >sin 60° = 1wL>22 >sin 60° = 0.58 wL. RBx is RB cos 60°, or 0.29 wL. From gFx = 0, it is seen that RBx = RAx. Hence, RAx = 0.29 wL. Because RAy = wL>2, it follows that RA = 0.58 wL.

Principle of Superposition.  For rigid structures that carry multiple loadings, it is possible to determine the reactions for each individual load and then add all the results obtained. This approach is often useful for conceptualizing the behavior of structures under different loading conditions.

Example Determine the reactions to the structure in Figure 25, using a superposition technique.

Figure 25  Principle of superposition. 4000 lb

Copyright © 2013. Pearson Education Limited. All rights reserved.

4000 lb

1500 lb

3500 lb

1000 lb

Solution: The reactions for each loading are determined first (as shown to the right in Figure 25) and are then added.

Forces and Reactions Associated with Overturning.  In many situations, when applied forces act horizontally or have significant horizontal components, they cause a structure to overturn. Determining reactions for these situations is straightforward and no different in principle from determining reactions in any previous example. Care must be taken to use the correct moment arms associated with the overturning and reactive forces. The next example illustrates how a rigid structure resists overturning via the use of tie-down supports. The example after that illustrates how a structure with no tie-down supports can resist overturning via its own dead weight.

3000 lb

Principles of Mechanics

Example Determine the reactions for the structure shown in Figure 26. Figure 26  Reaction for a truss-like structure.

Solution: Moment equilibrium about point A, gMA = 0 ⤺ + : - P1L2 - 2P1L2 + RBy 1L2 = 0, or RBy = 3P c

Equilibrium in the vertical direction, gFy = 0

c +:

- RAy + RBy - 2P = 0; hence, RAy = P T

+ Equilibrium in the horizontal direction, gFx = 0 S :

-RAx + P = 0, or RAx = P Resultant force at A:

Copyright © 2013. Pearson Education Limited. All rights reserved.

RA = 1.4P @ 45°

Figure 27 illustrates the rotational stability analysis of a block with a large dead weight that is subjected to an overturning force. In this case, there are no active mechanisms such as pins to prevent overturning. As will be seen, when the dead weight is high, the applied overturning moment is less than the moment available to resist overturning (which is associated with the dead weight of the structure), and the structure is stable. If the applied moment is greater than the resisting moment, the structure overturns. If the applied and resisting moments are exactly equal, the structure is in neutral equilibrium. The example is a simplification of the buttress analysis shown in Figure 5(b). Many high-rise buildings, however, that have no tiedown piles and rely on their own dead weights and proportions to resist overturning due to wind or earthquake forces, can be analyzed similarly. General equilibrium conditions based on a moment analysis are shown at the top of the figure. If the overturning moment of the horizontal load exceeds the resisting moment associated with the weight and size of the block, then the block will overturn. If not, then the block is either in neutral equilibrium or stable. The lower part of the figure uses an algebraic procedure and a graphically oriented approach similar to that shown in Figure 5. As long as the inclined force (which is statically equivalent to the horizontal and vertical forces combined) passes through the point of tipping or to its right, the block will not overturn. (Note that the block could be stable, but cracking might still occur in a masonry structure.) The example demonstrates that increasing resistance to overturning can be accomplished by either increasing the dead weight of the structure, increasing the width of the base footing, decreasing the height at which the applied horizontal force acts, or applying some combination of these techniques.

Principles of Mechanics

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 27  Block stability analysis with respect to point A.

Special Types of Reactions: Fixed-End Moments.  Many structures, especially cantilevered beams, have their ends rigidly attached to walls or other supports. This connection restrains the member from either rotating or translating, and it can thus project from a wall or column face. Reactive forces that are developed at the support include the usual vertical and horizontal forces that prevent the member from translating as well as restraining moments that prevent the end of the member from rotating. These restraining moments, which balance the external applied moments at the same point, are typically called fixed-end moments and are a special form of reaction. One kind of restraining moment can be felt simply by extending one’s arm and noting what happens at the shoulder joint. Restraining moments in structures are provided by physical mechanisms that depend on the type of structure used. In a wide-flange beam cantilevered from a column, for example, the top and bottom flanges of the member are welded to the column face. Force couples are developed in the welds that provide the restraining fixedend moment. If a structure has multiple fixed ends (e.g., it is fixed on both ends) or many other reaction points, it becomes statically indeterminate, and values cannot be found by the techniques presented in this chapter. Example Determine the fixed-end moments for the two beams in Figure 28.

Figure 28  Fixed-end moments.

Principles of Mechanics Solution: For the beam to the left with the concentrated load P, the rotational moment at support A that is associated with the applied load is given by Mapplied = PL. The balancing fixed-end moment is equal in value but of opposite sense; thus, M FA = PL. For the beam to the right with the uniformly distributed loading, the rotational moment at support A that is associated with the applied load is given by Mapplied = 1wL21L>22 = wL2 >2. The balancing fixed-end moment is equal in value but of opposite sense; thus, M FA = wL2 >2.

Cables.  Many structures are supported on one end or the other by cables. An example of a cable-supported structure is shown in Figure 29. An analytical objective would be to identify the forces in the cables and, ultimately, determine their required diameters. A related objective would be to identify the reactive forces exerted by the cables and the floor and roof systems on the adjacent main structure so the cable can be properly sized. In a real structure, this process involves first estimating dead and live loads and determining how those loads are carried by the framing structure and related cables. Briefly, in this example, roof loads are picked up by the facing crossbeam, which in turn carries its loads to the corner columns. Forces are carried through the columns to the end cable-supported beams. Floor loads are also picked up by the lower crossbeam and carried directly to the beam end. A simplified load model is shown in Figure 29. Once this model is developed, the principles of statics—the focus of this chapter—can be used to determine forces in the cables and connection points. It is interesting that the use of cables as shown normally means that horizontal as well as vertical forces are developed on the adjacent structure. Several examples of finding forces in cables are given subsequently. Reactive forces are developed within the cables that provide equilibrium for the supported structure. As noted in Figure 15, the forces developed in a cable are in tension (cables cannot provide compressive force resistance) and directed along the length of the cable. Thus, the direction of a cable defines the direction of the reaction it provides. The reactive force developed is equivalent to the internal tension force in the cable.

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 29  Determination of the loading model for a cable-supported structure. The analysis is simplified and based on several assumptions.

.

.

.

Principles of Mechanics

Example Determine the reactions to the cable-supported structure shown in Figure 30. Figure 30  Cable-supported structure. C

 = 30o

P

A

B

Cantilever beam supported by a cable. An internal force T is developed within the cable. This force T can be found directly by summing moments about point A.

L/2

L/2

TBCy

Lsin30 = 0.5L

RA

TCB =

P



P

TBC

RAx

P

=

RA cos 

= 0.866 P

TBC

x

ΣMA = 0 TBC (0.5L) – P(0.5L) = 0 TBC = P (Tension)

TCB

= TBC sin 30 = P sin 30 = 0.5P

RA

Note that the lines of action of the three forces meet at a point.

=

=

TBC cos 30 = 0.866 P

RA sin  = 0.5P

Σ Fy = 0 TBC sin 30 + RA – P = 0 RA = 0.5 P y Σ Fx = 0 – TBC cos 30 + RA = 0 RA = 0.866 P x

P

TBC = P

RA = P @  = 30 Final equilibrium diagram showing final forces

Copyright © 2013. Pearson Education UK. All rights reserved.

RA = P

Solution: The steps in the determination of forces are shown in Figure 30. First, an equilibrium diagram is drawn in which the direction of the reaction provided by the cable is shown as coincident with the location and direction of the cable. The internal force in the cable, TBC, is treated as an external reaction for calculating the equilibrium of the beam AB. Moments are summed about point A to obtain TBC. Note that the moment arm of TBC is perpendicular to the line of action of TBC. Once TBC is determined, forces are summed in the horizontal direction to obtain the horizontal component of the left reactive force RA. (The horizontal component of the cable force is equal to the horizontal component of the left reaction.) Summing forces in the vertical direction yields the vertical component of RA. RA can then be found from its components. A final equilibrium diagram is drawn. The three forces acting on the structure meet at a point. (See Section 3.2 on three-force members.) Note that using a cable induces a compressive force in beam AB equivalent to the horizontal component of the cable force. Note also that the reactive force at C is equal and opposite to the force in the cable.

Example A stayed-cable mast is shown under two different loading conditions in Figure 31. Determine the forces present in the cable and at the support points. Solution: The steps of the solution are shown in the figure. In the first loading condition, where the force is applied at the top of the mast, the load P produces a clockwise overturning moment that is balanced by a resisting moment associated with force TBC developed in cable BC. The cable force is assumed to be in tension, as it must be, and acts in the counterclockwise direction. The moment arm for the cable force is the perpendicular distance from the base pin A to the

Principles of Mechanics

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 31  Cable-stayed column with two loading conditions.

cable BC. The cable force TBC is found by summing moments about A. Thus, 5TBC = 10160002, or TBC = 12,000 lb. Note that the reaction at C is numerically equal to the force TBC developed in the cable. Equilibrium in the vertical and horizontal directions is considered next to find the reactive force at A. Although initially assumed to be inclined, this force acts only vertically. Because the reaction is vertical (in this case), the compressive force developed in the mast has an equivalent numerical value. In the bottom example, the location of the applied horizontal force has been lowered and the same analysis repeated. In this case, the reaction at A was found to have a horizontal component. It cannot be assumed that the reactive force at A is oriented in the same way as the mast. In this case, the force in the mast is numerically equivalent to the vertical component of the inclined reactive force at A. The mast is also subject to bending by force P.

3.4 Complete Static Analyses The next example illustrates a complete static analysis for a cable-stayed structure. Reactions for the overall structure are determined first. The structure is then ­decomposed into its fundamental components, each of which is shown with the complete set of external and internal or reactive forces acting on it. Reactions for one member become applied forces on the adjacent member (e.g., the reactions to the beam become equal and opposite forces applied to the mast). The equilibrium of different components can be considered in turn until all the unknown forces at connections are found. Note how forces at connections are shown as equal and ­opposite, a necessary condition that follows because these forces are reactive in nature and internal to the overall structure.

Principles of Mechanics Example For the structure shown in Figure 32(a), determine the reactions at the base of the structure, the internal force in the stabilizing cable CA, and the internal force in cable CE that supports the projecting member. Figure 32  (a) Structure; (b) final analysis results.

 



Figure 33  Cable-stayed structure: the lines of actions of reactions and load meet in one point. For the Whole Structure: Find TCA and RA (Figure 33): gMB = 0:

TCA 1h sin u2 - Pa = 0



TCA 1h sin u2 = Pa

TCA 120 sin 25°2 = 5110002 TCA 18.452 = 5110002

TCA = 592 lb @ 75° to horizontal

Copyright © 2013. Pearson Education UK. All rights reserved.

gFx = 0:

Note that RA = TCA

- RA sin 25° + RBx = 0 RBx = 250 lb

gFy = 0:

- RA cos 25° + RBy - 1000 = 0 RBy = 1536 lb

U-

u = tan-1 11536>2502 = 80.7° to horizontal



RB = 1536>sin u = 1556 lb

For the Beam: Find cable force TEC and RD (Figure 34):

gMD = 0: TEC[7.07] - 5110002 = 0 TEC = 707 lb

Find RDx and RDy :

gFx = 0: -707.2 cos 45° + RDx = 0, or RDx = 500 lb gFy = 0: - 707.2 sin 45° + RDy = 0, or RDy = 500 lb

Thus, RD = 500>sin 45° = 707 lb at 45°

Final equilibrium diagrams are shown in Figure 32(b). All members should be in equilibrium. (Check to make sure that gFx = 0 and gFy = 0).

Figure 34  Detailed study of the horizontal element of the cable structure shown in figure 32.

Principles of Mechanics

4  Internal Forces And Moments Forces and moments can be either external or internal. Forces or moments that are applied to a structure (e.g., a weight attached to the end of a rope) are called external. Forces and moments that are developed within a structure in response to the external force system present in the structure (e.g., the tension in a rope resulting from the pull of an attached weight) are called internal. Internal forces and moments are developed within a structure due to the action of the external force system acting on the structure. Such forces serve to hold together, or maintain the equilibrium of, the constituent particles or elements of the structure. These internal forces or moments that develop within a member are in direct response to externally applied forces and are typically in tension, compression, shear, or bending. They are rarely constant throughout a specific structure but vary from point to point. Only in a simple structure, such as a rope with a weight on the end, are the forces constant within the structure. In a typical beam, however, the external forces acting on the structure generate internal forces that vary from point to point. This section begins a general study of these internal forces and their distributions by considering simple members in a state of pure tension or compression, in which the external forces are applied along the length of the structure (so-called axial forces). Next, is a series of sections on shear forces and bending moments that are developed within a structure. An ability to determine their magnitudes and distributions is fundamental to analyzing and designing structures. These axial forces, shear forces, and bending moments in turn produce stresses and corresponding deformations within the material fabric of the structure. Section 5 introduces the concept of stress. For now, we focus on how to determine internal force and moment states in structures.

4.1 Axial Forces (Tension and Compression) Consider the system shown in Figure 35. A tension force has developed in the cable supporting the block. This force has a magnitude equal to the weight of the block. As illustrated, the equilibrium of the block is maintained by the development of an internal force, Ft, in the cable. In this case, Ft is equal to the weight of the block.

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 35  Internal tension and compression forces in members.

Principles of Mechanics Internal compressive forces are similar in character but opposite in sense. Figure 35 illustrates a member in which the internal compressive force varies due to the character of the external force system present. The magnitude and direction of the internal forces developed are such that all parts of the structure are in a state of equilibrium, as they must be no matter what part of the structure is considered.

4.2 Shear and Moment Basic Phenomena.  This section begins a study of the internal forces and moments generated in a member carrying an external force system that acts transversely to the axis of that member. The concepts of shear and moment in structures introduced here are fundamental to the importance of understanding of the behavior of structures under load. They also provide the basis for developing tools for designing structures. These concepts also are later developed in the context of looking at a particular structural type (trusses). The material is presented at this time because the concepts of shear and moment apply to all specific structural elements (e.g., beams, trusses, cables). Consider the loaded cantilever member illustrated in Figure 36. The member might fail in two primary ways because of the applied load. One potential failure is for the load to cause two contiguous parts of the member to slide relative to each other in a direction parallel to their plane of contact. This is called a shear failure. The internal force developed in the member and that is associated with this phenomenon is called an internal shear force. This force is developed in response to the components of the external force system that act transversely to the long axis of the member and that cause the transverse sliding indicated in Figure 36(c). Internal shear forces resist or balance the net external shearing force that causes the sliding. Failure of this type occurs when the member no longer can provide an equilibrating internal shear force. The second possible mode of failure is illustrated in Figure 36(b). This failure is associated with the tendency of the transverse external forces to cause part of the structure to rotate, or bend. Because free rotation cannot occur in a rigid body

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 36  Shear and moment in structures.

Principles of Mechanics (unless a pin connection is present), an internal resisting or balancing moment equal and opposite to the applied moment (the moment associated with the rotational tendencies of the external forces) must be developed within the structure. Failure of this type occurs when the structure no longer can provide a resisting moment equal to the applied moment. At any section of the loaded member, internal shear forces and moments are developed simultaneously. If the member is decomposed at this point into two parts, the forces and moments developed internally serve to maintain the translational and rotational equilibrium of each part. They also represent the internal actions and reactions of one part of the member on the other part. As will be seen, one objective of the structural design process is to create a configuration capable of providing these internal shears and moments in an efficient way and with factors of safety sufficient to prevent shear and moment failures. Internal shear forces and moments of the type just described are developed in any structure carrying transverse loads. Determining the magnitude of these quantities is a straightforward process based on the proposition that any structure, or any part of any structure, must be in a state of equilibrium under the action of the complete force system (including internal as well as external forces and moments) acting on it. The external force system is typically known. Parts of the force system not initially known, such as reactions at supports, can be readily calculated by methods previously discussed. If the equilibrium of an isolated portion of a structure is considered, the unknown internal shears and moments that must be present at the point of decomposition can be determined through equilibrium considerations involving known parts of the external force system.

Copyright © 2013. Pearson Education UK. All rights reserved.

4.3 Distribution of Shears and Moments Shear and Moment Diagrams.  A variation in the magnitude (and often the sense) of the shears and moments is commonly present at different sections in a structure. The distribution of these shears and moments is found by considering, in turn, the equilibrium of different elemental portions of the structure and calculating the shear and moment sets for each elemental portion. To help visualize the distribution of these shears and moments, the values thus found can be plotted graphically to produce what are called shear and moment diagrams. These diagrams are invaluable aids in the analysis and design of structures; a working knowledge of what they mean and how to draw them quickly is of permanent importance. Note that their use is not restricted to beams; they represent a graphic way to look at the effects of an external force system acting on any type of structure. The sections that follow discuss in detail how to construct shear and moment diagrams for any loading condition. It is useful, however, to first talk through the general nature and uses of these diagrams. Figure 37 includes shear and moment diagrams for a simple beam structure carrying a concentrated loading. At any section in the structure, the external force system causes shearing and bending actions to develop throughout the structure. For vertically acting forces, at any section of the structure along its length, a net external shear force is present that is the algebraic sum of all upward- and downward-acting forces (applied forces, loads, and reactions) on the section. This external shear force is balanced by an internal resisting shear force that maintains the translational equilibrium of the section. In the upper part of Figure 37, note that the directional sense of the internal shear forces varies, depending on the location considered. The shaded diagram shown below is a graphical depiction of the magnitude of the external shear force present at each section of the structure along its length. This diagrammatic convention is discussed in detail subsequently. (See Figure 39.) The resulting diagram suggests that different parts of the structure have different shear magnitudes and different relative movement tendencies. Figure 37 also indicates the tendency of the

Principles of Mechanics

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 37  General nature of externally applied and internally resisting shear forces and bending moments in structures and illustrations of how different types of structural configurations provide the resisting or balancing shears and moments via forces or stresses developed within members.

s­ tructure to bend underneath the load. In this case, it bows into a concave curve. At any section of the structure, a net external bending moment causes some bowing at the section. The external bending moment is the algebraic sum of the rotational effects produced by all external forces (applied forces, loads, and reactions) to the left (or right) of the section considered. This external bending moment is balanced by an internal resisting bending moment that maintains the rotational equilibrium

Copyright © 2013. Pearson Education UK. All rights reserved.

Principles of Mechanics of each part of the structure. The shaded diagram beneath the diagram showing the deformed shape of the beam depicts the magnitude of the external bending moment present at each section of the structure along its length. The bending moment diagram varies from zero to some maximum value and then decreases, which indicates that the external bending moment (and balancing i­nternal resisting moment) varies in the same way. The maximum tendency of the structure to fail in bending typically occurs where the diagram peaks. The specific sign convention used is shown in Figure 39, where it is noted that a value is considered positive when compression exists on the top face of the structure and tension exists on the bottom—a condition associated with a concave shape—and is considered negative when reverse bending occurs. Figure 37 also suggests that the specific mechanism by which a structure provides the necessary internal resisting shear and moment values to balance the externally applied shear forces and bending moments at a point varies with the type of structure used. In a typical truss structure, internal resisting shear forces are provided by components of internal forces developed in specific members (often the diagonals—particularly when upper and lower chords are parallel and horizontal). The resistance to external bending moments is provided by a force couple (oppositely acting force pairs that are separated by a moment arm corresponding to the depth of the structure at the point considered) generated by internal forces developed in truss members (particularly the upper and lower chords but normally involving horizontal components in the diagonals). In a beam structure, internal resisting shear forces are provided by tangential shear stresses acting over the face of the section. (Stresses acting over an area produce a force.) Bending-moment resistance is provided via tensile and compressive bending stresses that act perpendicularly to the face of the structure, over the upper and lower faces of the beam, thereby producing a resisting moment couple. (Note, however, that these stresses are not uniformly distributed and vary in intensity across the face.) In a reinforced concrete beam, the tensile force part of the force couple is carried by the reinforcing steel. Figure 37 also suggests how a cable structure provides internal resisting shear and moment values. The variations in sense and intensity of shear and moment values along the length of a structure have profound design implications. The magnitudes of shears and moments represented in the diagrams are used to determine the required size of a beam or the appropriate depth and member sizes in a truss. Many members with a constant-cross-section or constant depth are loaded such that shear and moment values vary in intensity along the structure. These members are sized for peak values at specific points and are oversized almost elsewhere else, thus using more material than structurally necessary. Alternatively, the shapes of the diagrams can be used to fashion a structure whose cross-sectional size or depth varies along its length so that material is used to a high level of efficiency. Many designers shape trusses or beams along their lengths to ­reflect the external force distributions shown in the diagrams. These are but a few of the uses of shear and moment diagrams. Here, the next objective is to learn how to draw the diagrams. Determination of Shears and Moments at a Cross Section. The beam shown in Figure 38 is in equilibrium, and reactions have been determined. Assume that we want to know the nature of the internal forces at an arbitrary ­location, such as section x–x located at midspan. To find these internal forces, the structure is decomposed into two parts at this location. Examine the left portion of the beam [Figure 38(b)]. This part of the structure does not appear to be in ­equilibrium if only the effects of the external force system acting on the part are considered. The net effect of the portion of the external force system acting on the left part of the structure causes it to translate vertically downward and rotate in a

Principles of Mechanics

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 38  Shear forces and bending moments in a beam. Values are calculated at section x–x in (a) to (d). Figures (f) and (g) represent values throughout the beam as calculated according to the methods presented in Section 4.3.

Principles of Mechanics clockwise direction. The net downward force is called the external shear force 1VE 2 at the section. In this case, at x = L>2, it is given by VE = +3P>4 - P = P>4 T . For this part of the beam to be in equilibrium, the structure must somehow provide an internal resisting shear force 1VR 2 equal in magnitude but opposite in sense. Thus, VE = VR and it follows that VR = P>4 c . The net rotational effect associated with the external force system acting on the part of the beam considered (the portion to the left of midspan) is called the external bending moment 1ME 2. If sign conventions are temporarily ignored (a new convention, described in the next section, is used), the external bending moment at x = L>2 is given by ME = 13P>421L>22 - 1P21L>42 = PL>8. For equilibrium to be maintained, the structure must provide an internal resisting moment 1MR 2 equal in magnitude, but opposite in sense, at this location. Because MR = ME, MR = PL>8. A more direct approach is to write immediately the equations of equilibrium for the complete free-body diagram of the part of the structure considered. [See Figure 38(d).] Thus, at x = L>2, g Fy = 0:

g Mx = 0:

+3P>4 - P + VR = 0

6

VR = P>4 c

+ 13P>421L>22 - 1P21L>42 - MR = 0

6

M = PL>8

Copyright © 2013. Pearson Education UK. All rights reserved.

This is a conceptually different approach. It does not, however, stress that the external force system creates a net translatory force and applied moment at the section, which must be balanced by numerically similar internal forces and moments acting in the opposite senses. These internal shear and moment values also could have been found by considering the equilibrium of the right rather than left portion of the structure. The external shear force for this part is given by VE = P>4 c . Because VE = VR, VR = P>4T . Numerically, this is the same result found previously. The sense of the force, however, is opposite. In a like vein, the external moment on the right part is given by ME = 1P>421L>22 = PL>8; then, because ME = MR, MR = PL>8. This is equal to the moment found previously but again of opposite sense. Note that these shear and moment values are valid only for the specific section considered. Values for other points are shown graphically in Figure 38(f) and (g). Sign Conventions.  It is reasonable that the shears and bending moments found by considering alternate parts of the structure should be numerically equal, but ­opposite in sense. After all the forces found are internal to the structure and represent the action and reaction of one part of the structure on the other! This brings up a difficulty, however, regarding sign conventions. Designating the internal shear or moment as either positive or negative by its direction only, as was done for calculating reactions, is misleading because a value would be positive with reference to one part of the structure and negative with reference to the other part. The same shear forces and moments, however, are involved. To resolve this problem the conventions usually employed are based on the physical effects of forces and moments on structures. The bending moments present in the beam shown at the top of figure 39, for example, cause the beam to develop concave curvature as viewed from the bottom (or from the right if the member was vertical) at the section shown. The bending moment at this section is said to be positive. If the external force system produces a convex ­curvature (concave downward) at a section, the internal resisting moment at that section is said to be negative. Another way to visualize positive moments is to say that positive moment diagrams are always plotted on the compression face of the structure. Conventions of this type give the same sign to the moment developed at a section, regardless of the orientation of the structure.

Principles of Mechanics

Figure 39  Sign convention used for shear and moment diagrams. Top surface in compression

V

E

Bottom surface in tension

Net of transverse forces pushes section to left upward.

V

E

Net external shear force V E acts upward on left section of member.

Section

Positive bending moment (+) Top surface in tension

Up-down slip tendency relative to section

Positive shear force (+)

V

E

Net of transverse forces pushes section to left downward.

V

Down-up slip tendency relative to section

E

Net external shear force V E acts upward on left section of member.

Bottom surface in compression

Negative bending moment ()

Negative shear force ()

For shear, the typical convention is to call shearing forces positive if the e­ xternal forces on the left part of a structure have a net resultant that acts v­ ertically upward. Positive shear is associated with the tendency of the external forces to produce relative movements of a section of the type illustrated to the left in Figure 39(b). An “up–down” relative movement is positive. The conventions just described are used in the United States. The opposite convention is used in Europe and elsewhere, where positive moments are plotted on the ­tension face of the structure. Diagrams resulting from the two conventions are similar in shape and mean the same thing but look like inversions of one another. (See Figure 40.) Distribution of Shears and Moments.  This section discusses how to construct diagrams that visualize the magnitudes and distributions of shears and moments along the length of a structure. Algebraic expressions for how external shear forces and bending moments vary along the length of a structure. Expressions are written for the variation of the external shear force VE and the external bending moment ME as a function of a variable distance x along the length of the member, with the left end of the member chosen as a reference point. The sign conventions discussed earlier are used. For legibility, the areas beneath the plotted graphs are often shaded. Once familiar with the formal methods of constructing

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 40  Sign conventions used in different countries. Compression

Tension

Bending moment diagram for simply supported beam

+ Point of inflection (point of reverse curvature)

Bending moment diagram for beam with cantilever

Compression

Tension

Tension

Compression

+ Positive moment

 Negative moment

Sign convention used in United States and other countries

Sign convention used extensively in Europe and other countries

Principles of Mechanics shear and moment diagrams, the reader will rapidly find a series of “shortcuts”— notably, finding values that are critical points—that will facilitate construction of these diagrams. Example Draw shear and moment diagrams for the beam shown in Figure 41(a).

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 41  Shear and moment diagrams for a simply supported beam carrying a concentrated load.

Solution: Vertical reactions are calculated first. gM = 0: P12L>32 - L1RB 2 = 0;  6 RB = 2P>3. gF = 0: RA + RB = P; RA = P>3. Next, consider the free-body diagram of an elemental portion of the beam (having a length x) to the left of the load. As shown in the figure, the rotational effect of the set of external forces acting on the left portion of the beam about the cross section defined by the distance x is given by ME = 1P>321x2. The net translatory effect is given by VE = P>3. Thus, if x = 0, ME = 0, and VE = P>3. If x = 2L>3, ME = 2PL>9, and VE = P>3 (if the section is infinitesimally to the left of the load). Once x moves to the right of the load, new equations for shears and moments are needed because a new force now acts on

Principles of Mechanics the elemental portion of the beam considered. Thus, ME = 1P>321x2 -P[x - 2>31L2] and VE = + 1P>32 - P = 1- 2>3 2 P. These shears and moments can be plotted as illustrated in Figure 40(c).

Example Draw shear and moment diagrams for the structure in Figure 42. Figure 42  Shear and moment diagrams for a beam loaded with two different point loads.

Solution: Reactions are calculated first. Next, determine equations for shear and moment at a distance of x from the left reaction by considering the equilibrium of the portion of the structure to the left of x: For 0 6 x 6 L>3,  For L>3 6 x 6 2L>3, 

Copyright © 2013. Pearson Education UK. All rights reserved.

VE =

4P 3

ME = a

VE = 4P bx 3

ME = a

4P P - P = 3 3

For 2L>3 6 x 6 L, VE =

L 4P bx - P ax - b 3 3

4P 5P - P - 2P = 3 3 ME = a

L 4P bx - P ax - b 3 3

-2P ax -

2L b 3

These expressions can be plotted to obtain shear and moment diagrams, as illustrated in Figure 41. The shear forces are constant between loads, whereas the moments vary linearly. Note that critical values of the moment occur at x = L>3, where M = 4PL>9, and at x = 2L>3, where M = 5PL>9. Major changes in the shear also occur at these points. It is interesting to note that the maximum value of the moment occurs, not at midspan, but where the shear diagram passes through zero.

In structures carrying concentrated loads, critical moment and shear values occur directly beneath the concentrated load application points. Utilizing this observation, one can draw the moment diagram more easily by considering sections beneath the load points. Thus, the moment at x = L>3 (the first load point) is given by ME = 14P>32 1L>32 = 4PL>9, and that at x = 2L>3 (the second load

Principles of Mechanics point) is given by ME = 14P>3212L>32 - 1P21L>32 = 5PL>9. The latter is even easier to find if the equilibrium of a portion of the structure to the right, instead of left, of the section studied is considered. Thus, ME = 15P>321L>32 = 5PL>9. The values of 4PL>9 and 5PL>9 define critical points on the moment diagram. Straight lines can be drawn between them because the moments vary linearly ­between critical points. Two sign conventions for the moment diagram are shown in Figure 42. The upper one is used in the United States. The lower one is widely used elsewhere. Example Draw shear and moment diagrams for the two cantilevered beams shown in Figure 43. Do not write formal equations, but determine values as directly as possible. Observe that the structures are mirror images.

Figure 43  Shears and moments in similar cantilever structures oriented differently. Section 1 Section 2

10 ft

Section 3 Section 4

5 ft 4P

Deflected shape

5 ft

4P

C B RB = 6P

A RA = 2P

10 ft Deflected shape

RB = 6P

SHEAR FORCE DIAGRAM

RA = 2P

SHEAR FORCE DIAGRAM VC = +4P

6P VA = −2P Negative shear

Positive shear

Negative shear

BENDING MOMENT DIAGRAM

MB = −2P (10) = −20P (calculated from left)

Copyright © 2013. Pearson Education UK. All rights reserved.

6P

VA = +4P

MB = −4P (5) = −20P (calculated from right)

Bending moments are negative for both beams because each structure has a concave shape (tension on top, compression on bottom).

VC = +4P Positive shear

BENDING MOMENT DIAGRAM

MB = −4P (5) = −20P (calculated from left) Tension

MB = −2P (10) = −20P (calculated from right)

-M

Compression

Solution: Once the reactions are determined, the shear diagrams can be quickly drawn by noting that applied forces “push” the shear diagram up and down. (See Figure 43.) Because there is no variation in load between vertical forces, the shear diagram is constant and horizontal between load points. The moment diagram for the structure to the left can be quickly drawn by noting that bending moments must be zero at member ends and the peak value must be at the interior support. Values can be found by multiplying a single force and a distance on either the left or right side of the support. The moment diagrams must vary linearly between zero and the maximum values calculated because there are no intervening forces. In the two cases shown, observe that the moment diagrams are similar in algebraic sign because the deformed shape is the same in both cases. The shear diagrams of the mirror-image structures have similar numerical values but with opposite algebraic signs because the sense of shear deformation differs in each case. This would have no effect on sizing the beam!

Principles of Mechanics

Example Draw shear and moment diagrams for the cantilever structure shown in Figure 44. Figure 44  Uniformly loaded cantilevering beam.

Solution: Shears and moments at any section defined by a distance x from the left end of the member are determined by considering the equilibrium of the beam to the left of x. The moment of the uniform load to the left of the section being examined is found by replacing the uniformly distributed load by the statically equivalent concentrated load acting at the center of mass of the portion of the load considered. (See Section 5.) For 0 6 x 6 L, VE = - wx x wx2 ME = - wx a b = 2 2

Copyright © 2013. Pearson Education UK. All rights reserved.

Critical values occur at x = L, where VE = - wL and ME = - wL2 >2. Note that the shear forces vary linearly according to the first power of x and that the moments vary according to the second power of x.

Example Draw shear and moment diagrams for the uniformly loaded, simply supported structure shown in Figure 45. This type of beam is extremely common in buildings. Solution: As before, the net effect of the external force system acting on an elemental portion of the structure is to produce a translational shear force VE and a rotational moment ME at the section considered. These external shears and moments must be balanced by an internal resisting shear force VR and internal resisting moment MR provided by (or developed in) the structure to maintain equilibrium. As shown in Figure 45, the shear reaches a maximum value wL>2 at x = 0 and a minimum at x = L>2. The moment is zero at x = 0 and reaches a maximum value of wL2/8 at x = L>2. Note that the shear diagram varies linearly and has maximum values at x = 0 and x = L, where VE = wL/2 and VE = - wL/2, respectively. The moment diagram varies according to the second power of x and reaches a maximum at midspan (where the shear is zero).

Principles of Mechanics

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 45  Uniformly loaded beam.

The maximum moment present is M = wL2 >8. This is one of the most frequently encountered expressions in structural theory. It is used in the analysis and design of simple beams, but also crops up in other, less expected places.

Example Draw shear and moment diagrams for the partially loaded beam shown in Figure 46. Solution: Reactions are calculated first. Next, equations for shear and moment are determined and plotted. For 0 … x … 12,     For 12 … x … 16, VE = 15 - 2x     VE = 15 - 21122

Principles of Mechanics

Figure 46  Beam loaded partially with a uniform load.

Copyright © 2013. Pearson Education UK. All rights reserved.

x ME = 15x - 2x a b 2

ME = 15x - 211221x - 62

Note that where uniform loads are acting on the member, the shear diagram varies linearly and the moment diagram parabolically. Determining the maximum moment present in the structure could be tedious because it is not intuitively obvious where this maximum moment occurs. A trial-and-error process of assuming different values of x would work, but it is easier to use the fact that the moment has a critical value where the shear is zero. An alternative, practical approach to constructing the shear diagram is to note that the left reaction pushes the diagram initially up to a level corresponding to the left reaction (indicating that a shear force of a value equal to that of the reaction exists in the beam immediately to the right of the reaction—this result could have been found by passing a section through that point). The effect of the uniform load is to cause the shear diagram to decrease at a rate per unit length equal to the load per unit length. (This also is evident by looking at the preceding shear equation. The shear diagram thus has a slope equal to the uniform load. This information can then be used to find where the shear diagram passes through zero. Hence, 15 = 2x, or x = 7.5. This is equivalent to setting VE = 0 in the original shear equation (i.e., 0 = 15 - 2x). By noting that the moment is a critical value at points of zero shear, one can calculate the moment at x = 7.5 to yield the maximum ­moment present. Thus, ME = 1517.52 - 217.5217.5>22 = 56.25.

Principle of Superposition.  Complex loading conditions can be separated into simpler loading cases. (See Figure 47.) Individual shear and moment diagrams

Figure 47  Principle of superposition.

Principles of Mechanics can then be constructed and algebraically combined to create a composite diagram of the original complex loading. Superposition principles are valid for common rigid structures where the deformation of the structure does not significantly affect the location of the load.

Copyright © 2013. Pearson Education UK. All rights reserved.

Shapes of Diagrams.  Some summary observations can be made about the shapes of shear and moment diagrams. (See Figure 48.) Concentrated loads generally produce shears that are constant in magnitude along sections of a structure between those loads, so the shear diagram consists of a series of horizontal lines. Moments in such structures vary linearly (according to the first power of x) between concentrated loads. Moment diagrams are therefore composed of sloped lines. Uniformly distributed loads produce linearly varying shear forces. Shear diagrams correspondingly consist of a sloped line or a series of sloped lines. Uniformly distributed loads produce parabolically varying moments, and moment diagrams are correspondingly curved. Combined loads produce combined shapes. While not rigorously provable at this point, it is also generally true that a point of zero shear corresponds to a location of a maximum or minimum moment. This latter observation is particularly valuable for locating critical design moments in structures having unusual loading conditions. Diagramming the magnitudes of forces and moments along the member provides a useful visual and quantitative guideline when deciding on the shape of members. This is particularly true where the expression of the internal workings of the structural elements is a desired aspect of the architectural language. Of particular importance for shaping beams and other horizontal members are ­moment diagrams, as demonstrated in the Maillard’s bridge example in Figure 49. The example also shows the importance of superposition in drawing internal moment diagrams. Relation of Moment Diagram to Deflected Shape of Structure. As previously discussed, transverse loads produce a bowing in the structure. Depending on how the loads act, the bowing can be concave upward or the reverse. The overall deflected shape of the structure is intimately related to the nature of the bowing and, consequently, to the moments that are present. Consider the member shown in Figure 50. The loads on the ends of the projecting cantilevers produce a bowing that is concave downward in the left and right regions of the member. The moment diagram is negative in these zones, which are called negative moment regions in U.S. practice. Associated with the concavedownward bowing is a stretching of the upper fibers of the structure, which puts them in a state of tension, and a shortening of the lower fibers, which are consequently put in a state of compression. Converse phenomena occur in the middle portion of the member, in the positive moment region. Between the positive and negative moment regions, transition points must occur. Because of the reversal of curvature, no bowing or bending is present at the transition points so they are points of zero moment. These are called points of inflection. A numerical example of a cantilevered beam is illustrated in Figure 51. The point of inflection indicating a point of reverse curvature corresponds to a point of zero moment but not necessarily to a point of zero shear. The point of zero moment could be found by writing an equation for the moment values that are present in terms of some variable x and then equating the expression to zero.

Principles of Mechanics

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 48  Shear and moment diagrams for typical structures.

Principles of Mechanics

Figure 49  Shaping a bridge structure along its length according to the distribution of bending moments. Salginatobel Bridge Schiers, Switzerland Design and engineering: Robert Maillard Completed 1930

A 3-hinged arch supports the concrete deck. The clear span is 295 ft (90 m). The shape of the bridge is derived from the bending moment diagram shown below. The hinge support at the arch base (the image on the left shows a detail of a similar bridge by Maillard) is created by cork pieces on the outside that weaken the cross section. Reinforcing bars on the interior cross over. Point loads

Uniform load

+M

Uniform load

Copyright © 2013. Pearson Education UK. All rights reserved.

–M

+M

System diagram with typical loads. Swiss codes required a truck load on the third points. The shaping of the bridge members closely follows the moments diagram.

Point loads

+M

Moment diagram for an asymmetrical point load on the right. The arch portion on the left now shows negative bending moments.

Figure 50  Relationship between the moment diagram and the deflected shape of a structure.

Principles of Mechanics

Figure 51  Deflected shape, shear, and moment diagrams for a cantilevered beam.

Copyright © 2013. Pearson Education UK. All rights reserved.

4.4 Relations among Load, Shear, and Moment in Structures Relationships of the following type exist among the load, shear, and moment in a B structure: w = dV>dx, V = dM>dx, VB - VA = 1AB w dx, and MB - MA = 1A V dx . Briefly, these relationships can be found by looking at the equilibrium of an infinitesimal length of a structure and considering translatory and rotational equilibrium. (See Figure 52.) Shear forces V and M, and their incremental changes dV and dM, are considered. Thus, from g Fy = 0, we obtain +V - 1V + dV2 + wdx = 0, or w = dV>dx; and from g Mx = 0, we obtain +M + V dx + 1w dx2 1dx>22 - 1M + dM2 = 0, or V = dM>dx (when terms of negligible magnitude are dropped during the solution). These principles have many applications. The value of the distributed load at a point is equal to the slope of the shear diagram at that point. The value of the shear at a point is equal to the slope of the moment diagram at that point. The change in moment between points on a structure is represented by the area under the shear diagram between the same two points. (Points of reference and boundary conditions must be carefully considered.) Techniques like this are useful to check the accuracy of shear and moment diagrams (Figure 52). Such techniques point out that the maximum or minimum value of the moment in a structure occurs when the shear diagram passes through zero (because dM>dx = 0 at that point). (Note that not all of these relations apply when European sign conventions are used.)

5  Introduction To Stresses Associated with each of the states of tension, compression, shear, or bending moments within a structure are corresponding internal stresses. Stress, or force intensity per unit area, measures how a force is distributed over an area. A force acting over a small area produces higher unit stresses than does the same force acting over a

Principles of Mechanics

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 52  Relations among loads, shears, and moments.

larger area. These stresses cause a material to fail in tension, compression, or via a shearing action. It has been experimentally determined that different materials have different capacities for carrying different stresses. Thus, knowing the present stress level, one can determine that a member is safe or adequately or inadequately sized. Corresponding to these stresses are related strains, or deformations per unit length of material. Strains contribute to the overall deformed shape of a structure, including its total deflections. Tension and Compression Stress.  This section considers the internal forces and corresponding stresses in simple tension and compression members. These stresses are easy to determine because they are uniform across the crosssectional surface of a member. More complex stress states associated with shear and bending are not uniformly distributed across a cross section and their determination is correspondingly more complex. Following the discussion of simple tension and compression stresses is an introduction to the properties of materials. Consider a simple tension member illustrated in Figure 53. The internal tension that is present is not concentrated at a specific spot (as the arrows symbolizing internal force in Figure 53(b) seem to indicate) within the cross section of the

Principles of Mechanics

Copyright © 2013. Pearson Education UK. All rights reserved.

Figure 53  Tension members.

member, but rather, it is distributed over the entire cross section. The total internal force necessary to equilibrate the external force acting on the member is the resultant of all distributed forces, or stresses, acting at the cross section. With respect to a simple element carrying a tension force, it is reasonable to assume that if the external force acts along the axis of the member and at the centroid, or point of symmetry of the cross section, the stresses developed at the cross section are of uniform intensity. Their resultant would have the same line of action as the external force that is present. When stresses are uniformly distributed, their magnitude is given by stress =

force area

or

f =

P A

where f is the stress (force intensity per unit area), P is the axial force, and A is the area of the cross section considered. Stresses of this type are often called axial, or normal, stresses. The stresses developed in a member loaded in direct compression can be similarly described. The assumption that stresses associated with axial loads are uniformly distributed across a cross section is reasonable when the load is applied in an axial way and the member involved is straight and of uniform cross section. At a unique point, such as the point at which the external load is applied or a discontinuity in the member, a more complex stress pattern may occur. The assumption, however, is good and is acceptable as a working hypothesis for preliminary designs. This approach is not valid for beams and other members in bending, which have more complex stress distributions.

Principles of Mechanics

Figure 54  Prestressing a pile of stones. The tension rod shown in (b) and (c) artificially increases the internal compressive stress.

Example A member having a square cross section that measures 2 in. * 2 in.150.8 * 50.8 mm2 carries a tension load of 24,000 lb (106.75 kN). What is the stress level present at a typical cross section? Solution: Assuming that the internal stress is uniformly distributed, we have

Free-body diagram

(a) Stone pile without prestressing; Internal stresses equilibrate only the weight of the stone.

Free-body diagram

(b) The prestressing force leads to an increased compressive stress in the stone.

Copyright © 2013. Pearson Education UK. All rights reserved.

Free-body diagram

(c) The prestressed stone pile can resist a certain amount of lateral force. Compressive stresses in the stone vary in magnitude as they equilibrate both dead loads and the lateral loads.

force P 24,000 lb or f = = = 6000 lb>in.2 area A 2 in. * 2 in. 106.75 kN f = = 0.0414 kN>mm2 = 41.4 N>mm2 = 41.4 MPa 50.8 * 50.8 mm stress =

Note that the axial or normal stress in a member depends only on the force applied and cross-sectional area involved. The actual stress developed does not depend on the type of material used to make the member. Bending, shear, and other types of stresses are nonuniformly distributed and not described by f = P>A.

Stress Combinations: Prestressing.  In some circumstances, it can be advantageous to induce compressive stresses in a member artificially by pulling the ends of the member toward each other. In the element that performs the pulling action, tensile stresses are generated. The approach is called prestressing, implying that internal stresses are present even without any external loads. Instead, these stresses are generated by artificially tightening one or more tension members such as steel cable or rods, and using this tensile stress to push compressive material together and induce compressive stress. Applications of prestressing in beams are widespread. This section explores the fundamental principle by looking at the example in Figure 54. Here a pile of stone or other material is compressed artificially by tightening a tensile rod that is embedded in the middle. Note that the concentrated force of the rod is distributed uniformly onto the stone at the top of the pile. The pile of stone would work fine—at least up to a certain height—without this tensile rod. However, when subject to a lateral force—for example, a person leaning against the pile—the stones could easily slide or tip over. This instability can be overcome by compressing the stones together through the internal tension rod. Note that the rod has to be firmly anchored at the base, preventing it from pulling out. The prestressed pile of stones in Figure 54(c) can resist a certain amount of lateral force because the friction between the rocks is now higher, thus preventing the stones from sliding. At the base, a force couple develops with two forces in opposite directions. One force is the downward pull of the tension rod; the other is the upward reaction of the base pushing against the stone. The resulting moments resist the overturning moment generated by the lateral force. Strength of Tension Members.  The strength of a tension member depends on several factors. The probable mode of failure is a pulling apart of the member at the weakest location along its length. The load-carrying capacity of a member subjected to pure tension is independent of the length of the member if the member has a uniform cross section throughout (in terms of both area and material). The presence of a single weak spot, such as a point of reduced cross-sectional area as, for example, at the location of bolt holes, determines the capacity of the whole member. At such locations, the actual stress level, as given by f = P>A, is higher than elsewhere because of the smaller value of A. The locally reduced area is commonly referred to as the net area of the member.

Principles of Mechanics When a member is axially loaded, stress levels defined by f = P>A develop. If the material used can sustain this stress intensity, the member will carry the load. As stress levels increase—due to increasing loads, for example—there comes a point where the present stress intensity exceeds the capability of the material used to withstand the pulling apart. This is the failure stress level of the material and is a property of the material used. Failure stress levels are experimentally determined for various materials. When the actual stress level, given by f = P>A in a member exceeds the failure stress level for the material that is used, the member will pull apart. For certain types of steel, experiments have shown that the apparent tension stress level associated with the beginning of the material’s pulling apart, or yielding, is approximately Fyield = 36,000 lb>in.2 (248 N>mm2 or 248 MPa). In Section 6, we explore material properties in greater detail. Design Methods.  To determine whether a member is of adequate size to support a given tension load, it is necessary first to determine the actual stress level in the member 1f = P>A2. If this actual stress level is less than the experimentally determined failure stress level for the material, the member can support the loading involved. From a design point of view, however, it is desirable that safety ­factors be included when determining the adequate size of a member. Loads and failure stresses can never be predicted with absolute certainty, and a conservative note should be introduced in designing members. Two alternative design approaches are commonly used for steel and timber members. The first, referred to as Allowable Strength Design (ASD), divides the failure stress by a safety factor, and uses the resulting allowable stress fallowable to determine if members are adequately sized for tension. Safety factors are thus exclusively assigned to the material properties. Ftallowable =

Fyield Safety Factor

The second approach assigns most safety factors to the loads, thus artificially ­increasing them beyond magnitudes that can realistically be expected over the lifespan of the structure. The increased loads are usually referred to as ultimate loads Pultimate.

Copyright © 2013. Pearson Education UK. All rights reserved.

Pultimate = P * load factor Stresses close to the failure stress levels can then be safely used to determine the size and adequacy of a tension member. This approach is called Load and Resistance Factor Design (LRFD). Allowable Strength Design.  U.S. steel design practice typically uses a ­factor of safety of 1.67 for tension members. Thus, the allowable tensile stress for a steel member is given by Ft = 136,000 lb>in.2 2 >1.67 = 21,600 lb>in.2 or Ft = 1248 N>mm2 2 >1.67 = 149 N>mm2. This value can be used in evaluating the ­acceptability of a given member or as a tool for determining the required size of a new member, given the load it must carry. In sizing a new member, the required cross-sectional area for a member in tension carrying a load P is given by Arequired =

force P = allowable stress Ft

Load and Resistance Factor Design.  The uncertainties of structural loads depend on the types of loads present. Live

Principles of Mechanics loads from occupants, for example, are typically factored with 1.6 in U.S. design practice, whereas dead loads are only increased by a factor of 1.2. A working live load of 10,000 lb (44.5 kN) thus presents an ultimate load of 16,000 lb (7.2 kN). Recognizing the fact that construction practice and material production processes present additional risks of variability stresses slightly below the experimentally determined failure stresses are used for design. Yield stresses for steel in U.S. practice, for example, are factored with 0.9 for design purposes. The ultimate load carrying capacity Pultimate of tension members can be found with Pultimate = 0.9 Pnominal = 0.9 (fyield Area). In sizing a new member the required cross-sectional area in ­tension for a load P is given by Arequired =

Pultimate P * load factor = 0.9 * yield stress 0.9 * fyield

Example What diameter of steel rod is required to support a tension load of 10,000 lb (44.5 kN)? Assume that the yield strength for the steel is ft = 36,000 lb>in.2 1248 MPa or 248 N>mm2 2.

Solution Allowable Strength Design: Area required Arequired = Arequired = Diameter

P 10,000 lb = = 0.46 in.2 or Ft, all 136,000 lb>in.2 2 >1.67

44,500 N = 299 mm2 1248 N>mm2 2 >1.67

pd2 = 0.46 in.2 and d = 0.77 in. or 299 mm2 and d = 19.5 mm 4

Solution Load and Resistance Factor Design: Area required: A=

1.6 * P 1.6 * 10,000 lb 1.6 * 44,500 N = = 0.49 in.2 or A = = 318 mm2 0.9 * fyield 0.9 * 136,000 lb>in.2 2 0.9 * (248 N>mm2)

Copyright © 2013. Pearson Education UK. All rights reserved.

The required diameter is 0.79 in. or 20.1 mm.

6  Mechanical Properties of Materials 6.1 Introduction In this section, we briefly cover the various properties of materials that are interesting from a structural design viewpoint. Of primary interest are the strength and load-deformation properties of a material. In the final analysis, these properties are really explicable only in terms of the internal forces that act between the constituent parts of the material (i.e., the molecules or, in some cases, the atoms). This extent of such an investigation, however, is beyond the scope of this book. Instead, the coverage will be primarily descriptive in tone and will not dwell on the reasons underlying a particular material’s behavior. Readers new to the subject may find it better to omit this section temporarily and return to it later.

226

Theory of Machines

Base circle of rotation.

The base circle is the smallest circle that can be drawn to the cam profile from the centre

Prime circle The prime circle is the smallest circle drawn to the pitch curve from the centre of rotation of the cam. Pitch point The pitch point is a point on the pitch curve having the maximum pressure angle. Pitch circle The pitch circle is the circle drawn through the centre and pitch point. Trace point The trace point is a reference point on the follower and is used to generate the pitch curve. In the case of a knife edge follower, it is the knife edge and in the case of a roller follower, it is the centre of the roller. Pitch curve The pitch curve is the curve generated by the trace point as the follower moves relative to the cam. Cam angle

The cam angle is the angle turned through by the cam from the initial position.

Pressure angle The pressure angle is the angle between the direction of the follower motion and a normal to the pitch curve. Lift Lift is the maximum travel of the follower from the lowest position to the topmost position. It is also called throw or stroke.

7.4

FOLLOWER MOTION

The follower can have following type of motions: 1. Uniform velocity 2. Simple harmonic motion (SHM) 3. Uniform acceleration and deceleration 4. Cycloidal motion We shall discuss these motions now.

7.4.1

Simple Harmonic Motion

Copyright © 2013. Pearson India. All rights reserved.

For the SHM shown in Fig.7.4.

Fig.7.4

Let

SHM of follower

lift of the follower angle turned through by the crank from given datum displacement of follower

Cams

Then

OA

max

cos 2

d d

d d

Velocity

1

OB

2

d d

227

sin

2

where is the angular speed of the cam. Uniform velocity of a particle moving on the circumference of a circle 2

Maximum velocity of follower during ascent Maximum velocity of follower during descent

2

1

2

1

2

3

(7.1b)

2

Acceleration, Maximum acceleration,

(7.1a)

cos

2 2 max

2

Centripetal acceleration during ascent

2

2

2

2

(7.2a) 2

2

1 2

Centripetal acceleration during descent

2

3

Copyright © 2013. Pearson India. All rights reserved.

The motion of the follower is shown in Fig.7.5.

7.4.2

Motion with Uniform Acceleration and Deceleration

Let

d d

Then

uniform acceleration or deceleration

Integrating, we have 1

where

1

is a constant of integration.

(7.2b) (7.2c)

228

Theory of Machines

Fig.7.5

Copyright © 2013. Pearson India. All rights reserved.

If at

0

0, then

1

Motion of follower moving with SHM

0. Hence d d

Now Integrating again, we have

2 2

2 where As

2

is another constant of integration. 0 at 0, therefore 2 0. Hence 2

2 Average velocity,

avg

Cams

Maximum velocity,

2

2

max

2

during descent

3

where

1

is the angle of ascent, and

3

Maximum acceleration,

during ascent

1

229

(7.3a) (7.3b)

that of descent. 2

max

4

max

2

4

2

during ascent

2 1

during descent

2 3

(7.4a) (7.4b)

The motion of the follower is shown in Fig.7.6.

7.4.3

Motion with Uniform Velocity

Let If

follower rise angle through which the cam is to rotate to rise by

Then so that and

(7.5)

Velocity,

d d (7.6)

Acceleration,

d d

0

The displacement, velocity and acceleration are shown in Fig.7.7.

7.4.4

Parabolic Motion

Copyright © 2013. Pearson India. All rights reserved.

Let for the first half of the motion,

be equal to

For

2.

2

2 2

Hence

or

4

2

4

2

2

2 4

2 (7.7) (7.8)

2 2

constant

(7.9)

Copyright © 2013. Pearson India. All rights reserved.

230

Theory of Machines

Fig.7.6

Motion of follower moving with uniform acceleration and deceleration

Cams

231

Copyright © 2013. Pearson Education Limited. All rights reserved.

Introduction to Structural Analysis and Design replaces the effect of the entire loaded horizontal beam system acting on the column. The modeling in this case is simple. Note, however, that it is assumed that the beams are resting one on top of the other and on the column. The joint is assumed to transmit vertical forces only. This is true only under certain conditions, notably when the connection between members is not rigid. Thus, the column does not restrain the free rotation of the end of the beam. If it did, the modeling shown would be incorrect, and a more complex model of the type illustrated in Figure 9 would be necessary. This type of structure is called a frame. As is evident from the preceding discussion, effective modeling depends on identifying the exact structural nature of the joints between members. For analytical convenience, connections are modeled as one of the basic types (e.g., pins, rollers, and rigid joints). Determining the most appropriate model for an actual connection in practice is no easy matter and involves judgment. Several connections and their equivalent models are illustrated in Figure 10. The first step in analyzing a joint is to determine whether the nature of the joint is such that the rotations induced by a load acting on one member are transmitted to the other member through the joint. If the joint does not transmit rotations, it is usually modeled as a pin or roller. The choice between the two depends on whether the joint can transmit forces in only one direction or in any direction. If forces can be transmitted in any direction, the joint is considered pinned. Pinned joints allow relative rotations to occur between members, but not translations, and can transmit forces in any direction. If forces can be transmitted in one direction only, the joint is considered a roller. The latter allows not only rotations to occur between members but also translations in the direction perpendicular to the transmitted force. In many situations, pinned joints are made with a pin connecting two members, and a roller connection is made with rollers. Older bridges and buildings were often done this way—hence, the names of the joints (Figure 11). Such literal connections are occasionally still made in very large structures. If the joint does transmit rotations, and hence moments, between members, it is considered a rigid joint. A rigid joint always maintains a fixed angle between two members. When a rigid joint of the type shown in Figure 10(f) is part of a frame, however, it can translate and rotate as a whole unit. When a member joins a firm foundation with a connection that does not allow any rotation or translation to occur between the member end and the foundation, the resulting joint is called a fixed-ended connection. [See Figure 10(o).] Forces acting in any direction can be transmitted by this type of joint. Quite often, the difference between a pinned joint and a rigid one is difficult to determine immediately. Usually, if one member is connected to another at only one point, the joint is pinned. If the member is connected at two widely separated points, the joint is typically rigid. It is virtually impossible for rotations in one member to be transmitted to another through a connection that can be idealized as a point, as long as the member is large with respect to the size of the point. Figures 10(e) and (f) illustrate two steel wide-flange members connected in these two ways. Figure 10(e) represents a pin connection because the members are joined essentially at only one point. In Figure 10(f), the welding that joins the top and bottom flanges of one member to the other makes this connection rigid. In real structures, roller joints may or may not resist uplift forces. As Figure 10(g) illustrates, however, they can be designed to do so. When loadings are known to be only downward, actual joints are sometimes not designed to resist uplifting. Here, it is assumed that a roller joint can resist either downward or uplifting forces because if they must resist upward forces, they can be designed to do so. The number and type of joints connecting a structure to the ground have minimum requirements. The joints must provide sufficient constraints to satisfy the basic equations of equilibrium—ΣFx = 0, ΣFy = 0, and ΣM0 = 0. As illustrated in Figure 12(c), for example, a simple beam cannot rest on two rollers because

Introduction to Structural Analysis and Design

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 10 Types of connections and idealized models.

Introduction to Structural Analysis and Design

Figure 11 Many early structures used connections that were literally pinned. The term is still used to describe conceptually similar connections in modern structures.

(a) Early lenticular bridge located in Lowell, Massachusetts, 1882

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 12

(b) Typical pinned connection

Use of roller connections.

any horizontal force would translate it horizontally. An alternative way of looking at this constraint is to say that ΣFx = 0 could never be satisfied. Thus, at least one of the two joints must be pinned. Both joints also could be pinned. Typically, one of the joints is not restrained horizontally, however, because making it into a roller serves the useful function of allowing unrestrained expansions and contractions due to thermal effects. Pinning both ends would cause a large buildup of forces as the structure tried to change length with a change in temperature. The forces involved can be enormous and lead to serious failures. Thus, the support condition illustrated in Figure 12(b) is typical.

5

lOAd MOdeling And reACtiOns

Loadings in a building are typically either concentrated or uniformly distributed over an area. The former needs no modeling other than that necessary to characterize them as a force vector. In the latter, however, some modeling is needed when the area considered is made up of an assembly of one-way line and surface elements. These elements would pick up different portions of the total load acting over the surface, depending on their arrangement.

Introduction to Structural Analysis and Design Consider the simple structural assembly shown in Figure 13(a). Because all connections in this illustration are simply supported, the structure can be decomposed as indicated. For the wide plank elements, the reactions are better characterized as line reactions [as illustrated in Figure 13(b)] than as point reactions. The reactions from all the planks supported by a beam then become loads acting on the beam. Note that these loads form a continuous line load. Loads of this type are expressed in terms of a load or force per unit length (e.g., lb>ft or kN>m) and are

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 13 Two approaches to modeling loading conditions. The model shown on the left most correctly reflects how a beam system picks up surface loads. The model on the right, which is based on the concept of contributory areas, is often used for convenience. Both approaches yield the same loading on the beams.

Introduction to Structural Analysis and Design commonly encountered in the structural analysis process. These loads can be calculated by first determining the reactions of the planks and then considering them as loads on the supporting beam. [See Figure 13(c).] Another way to look at this same loading is to think in terms of contributory areas. By a symmetry argument, each of the beams can be considered as supporting an area of the extent indicated in Figure 13(d). The width of each area is often called the load strip. When the member is viewed in elevation, the load acting over the width of the load strip is considered transferred to the support beam. If the uniformly distributed load is constant and the load strip is a constant width, the amount of load carried per unit length by the support beam is the load per unit area multiplied by the width of the load strip. This process is illustrated in Figure 13(e). The result is again a continuous line load describable in terms of a load per unit length. The process is valid for symmetrical loads only. Both methods just described should yield the same load on the supporting beams. [See Figure 13(f).] The former is a more correct model of how the structure works and how loads are carried to the ground. The latter, however, is often more convenient, so it is used as a shorthand way of determining loads. The model can be as simple or as sophisticated as the analyst deems necessary. In early design stages, the approximation previously discussed is adequate for most purposes. The loading considered should include both live- and dead-load components. The value of the latter is found only through a detailed consideration of the size and unit weights of the elements of the assembly. Determining these values can be tedious. An alternative is to use an approximate equivalent weight, expressed as a force per unit area, to represent the weight of an entire assembly of some complexity. Thus, in Figure 14, the average dead load of the complete deck, joist, and beam system can be expressed as 1w2 2 lb>ft2 or 1w1 2 kN>m2, where the value of w is estimated or found from empirical data. Live loads are also expressed in terms of a force per unit area, so the calculation process is facilitated because both loads

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 14

Load modeling.

Introduction to Structural Analysis and Design can be considered simultaneously. Procedures of this type are fine for preliminary purposes, but more detailed takeoffs should be made at later stages in the analysis process. Especially when using design methods based on factored loads, it is important to maintain the distinction between different types of loads throughout the process. Dead and live loads, for example, are factored in different ways, and for the sake of transparency, it is often helpful to keep the distinction visible throughout the loads’ modeling process. exAMPle Assume that the combined average live and dead load of the floor system shown in Figure 14 is 50 lb>ft2. Determine the load strip widths and associated loading models for beams 1, 2, and 3. solution: Reactions: beam 1 RA = RB = wL>2 = 1125 lb>ft2118 ft2 >2

= 1125 lb 1by symmetry2

Or, directly, we have ΣMA = 0

+ RB 118 ft2 - 1125 lb>ft2118 ft219 ft2 = 0

(++)++* ()* equivalent total load

ΣFy = 0 RA + RB = 0

6 RB = 1125 lb

moment arm to equivalent total load

6 RB = 1125 lb

Reactions: beam 2 RA = RB = wL>2 = 1250 lb>ft2118 ft2 >2

= 2250 lb 1by symmetry2

Reactions: beam 3 Use equivalent concentrated loads and argue that, by symmetry, RA = RB: RA = RB = [1250 lb>ft216 ft2 + 1125 lb>ft216 ft2 + 1250 lb>ft216 ft2]>2 Copyright © 2013. Pearson Education Limited. All rights reserved.

(++)++* equivalent concentrated load

(++)++* equivalent concentrated load

(++)++* equivalent concentrated load

RA = RB = 1875 lb Note that the load per unit length carried by an end beam is typically one-half that of an interior beam and that the presence of a hole in the deck has a significant effect on the loading model.

exAMPle Assume that the average dead plus live load on the structure shown in Figure 15 is 60 lbs>ft2. Determine the reactions for Beam D. This is the same structure shown in Figure 1. solution: Note carefully the directions of the decking span. Beam D carries floor loads from the decking to the left (see the contributory area and load strip), but not to the right, because the center decking runs parallel to Beam D and is not carried by it. Beam D also picks up the end of Beam G and thus also carries the reactive force from Beam G. It is therefore necessary

Introduction to Structural Analysis and Design

Figure 15 Load modeling and reaction determination. Live and dead load Decking

Beam G carries distributed loads only. Beam F

Decking

Beam D

Beam G

Assume wDL+LL = 60 lbs/ft 2 . Beam E

Opening

Decking Beam C

12 ft

8 ft

Beam A

12 ft

w = 6 ft (60 lbs/ft2) = 360 lb/ft Beam G RG

Beam B 12 ft

Find reactions for Beam G.

12 ft

RG

1

2

RG = wL/2 = (360 lb/ft)(12 ft)/2 = 2160 lbs 1

Load strip for Beam D 2 = 6 ft (60 lbs/ft ) = 360 lb/ft Contributory load area for Beam D

RG = wL/2 = (360 lb/ft)(12 ft)/2 = 2160 lbs 2

Beam D carries both distributed loads and the reaction RG1 from Beam G. RG = 2160 lbs 1

Beam D Opening

RD

1

12 ft

RD

2

20 ft

Beam D

Beam G

Σ MD = 0 1

– (12 ft)(2160 lb) – (360 lb/ft)(20 ft)(20 ft/2) + 20 RD = 0

Contributory load area for Beam G

Load strip for Beam G 2 = 6 ft (60 lbs/ft ) = 360 lb/ft

RD = 4896 lb

2

2

Σ Fy = 0 RD + RD = (360 lb/ft)(20 ft) + 2160 lb 1

2

RD = 4464 lb 1

to analyze Beam G first to determine the magnitude of this force. The analysis appears in Figure 15. The reactive force from Beam G of 2160 lbs is then treated as a downward force acting on Beam D. The load model for Beam D thus consists of distributed forces from the decking plus the 2160-lb force. It is then analyzed using the equations of statics to obtain reactive forces of 4896 lbs and 4464 lbs at its ends.

Copyright © 2013. Pearson Education Limited. All rights reserved.

exAMPle Determine the reactions for a typical interior joist in the floor framing system shown in Figure 16(a). Assume that the spans and loads are as shown. Use the load factors 1.2 for dead loads and 1.6 for live loads. Note that this load calculation is to be used with a Load and Resistance Factor Design Approach. solution: Typical joist span = L = 16.0 ft = 4.88 m Typical center@to@center spacing = a = 16 in.11.33 ft2 = 0.406 m Live load = wL = 40 lb>ft2 = 1.915 kN>m2 Dead load = wD: Finished flooring Rough flooring Sheet rock ceiling Joists (estimated)

= 2.5 lb>ft2 = 2.5 lb>ft2 = 10 lb>ft2 = 9 lb>ft2

= 0.1197 kN>m2 = 0.1197 kN>m2 = 0.4789 kN>m2 = 0.4309 kN>m2

Total dead load = 24.0 lb>ft2 = 1.1491 kN>m2 Factored total loads = wD + wL = (24.0 lb>ft2)(1.2) + (40.0 lb>ft2)(1.6) = 92.8 lb>ft2 = (1.1491 kN>m2)(1.2) + (1.915 kN>m2)(1.6) = 4.443 kN>m2

Introduction to Structural Analysis and Design

Figure 16 Example of floor framing system.

Reactions:

gF y = 0: RA + RB - 1w = 0 D + wL 21a21L2 (+1)1+* ()* force per unit length

joist

length (+1)1+* total downward force

gMA = 0:

01RA 2 + L1RB 2 - 1wD + wL 21a21L21L>22 = 0 RB =

(+1+)1++* (1)1*

From gF y = 0:

Copyright © 2013. Pearson Education Limited. All rights reserved.

moment arm

total downward force

RA = RB = =

RA = 1wD + wL 21a21L2 >2

192.8 lb>ft2 211.33 ft2116.0 ft2 2

2

= 987.4 lb

14.4433 kN>m2 210.406 m214.88 m2 2

1wD + wL 21a21L2

= 4.402 kN

A typical joist reaction is therefore 987.4 lb (4.402 kN).

exAMPle Determine the reactions for a typical interior joist in the floor framing system that supports a partition wall, as shown in Figure 16(b). Assume that all the loads and spans are the same as those used in the preceding example and that the only difference is the addition of the wall. Do not factor the loads. solution: Partition: Assume that the partition is 8.0 ft (2.44 m) high and weighs 20 lb>ft2 10.958 kN>m2 2. The concentrated force on a joist exerted by the partition is computed as follows: The joists are

Introduction to Structural Analysis and Design

Figure 17 An equivalent uniformly distributed load can be used to approximate a series of closely spaced concentrated loads.

Load = w lb/ft2 or w kN/m2

spaced 1.33 ft (0.406 m) on centers, and, as the partition is 8.0 ft (2.44 m) high, the area of the partition that bears on one joist is 8.0 * 1.33 = 10.67 ft2 1or 0.406 * 2.44 = 0.99 m2 2. The concentrated force on a single joist is therefore 10.67 ft2 * 20 lb>ft2 = 213.4 lb 1or 0.99 m2 * 0.958 kN>m2 = 0.95 kN2. This force is located as shown in Figure 15(b). All other loads (i.e., wD + wL) are as in the preceding example. Reactions:

gMA = 0: 01RA 2 + L1RB 2 - [1WD + WL 21a21L2]

L L - 1P2 = 0 2 3

16.01RB 2 - [164.0 lb>ft2 211.33 ft2116.0 ft2]18.0 ft2 - 116.0>3 ft21213.4 lb2 = 0 RB = 753 lb

2

4.881RB 2 - [13.064 kN>m 210.406 m214.88 m2]12.44 m2 - 14.88 m>3210.95 kN2 = 0 RB = 3.352 kN

Copyright © 2013. Pearson Education Limited. All rights reserved.

gFy = 0: RA + RB - 1wD + wL 21a21L2 - P = 0 RA = 824 lb = 3.668 kN

When a beam is supporting a series of closely spaced joists that in turn support a plank deck, it is evident that each joist carries a uniform load per unit length. Figure 17 illustrates the process. The load can be found by multiplying the width of the load strip carried (i.e., the joist spacing) by the load per unit area. This method is used whether the planks over the joists are continuous over several joists or span from joist to joist. Theoretically, the two cases have some differences but none sufficient to change the model for a simple joist system. Joist reactions¡ can be found next. The collector, in turn, carries the reactions of the joists. The collector thus carries a series of closely spaced concentrated loads. For calculational convenience, however, these loads on the collector beam are often replaced with a uniformly distributed load per unit length, found by considering the load strip width of the contributory area of the surface supported by the member—a reasonable model if the concentrated loads are closely spaced. exAMPle Determine the loads on columns M and N in the structure shown in Figure 18. Assume that the loading is given by wT lb>ft2 and that this value reflects both live and dead loads. Assume further that L = 20 ft, a = 6 ft, and wT = 50 lb>ft 2.

Introduction to Structural Analysis and Design

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 18 Example structure.

Introduction to Structural Analysis and Design solution: The first step is to determine how the surface load is channeled to the columns. This is best done by drawing free-body diagrams for each element in the structure. (See Figure 18.) To determine the magnitudes of the column loads, it is necessary first to calculate the load carried by each joist, then to calculate the reactions for each joist, and finally to calculate the reactions of the beams that carry the joists. The latter reactions are the column loads. The process of starting with an analysis of the smallest members picking up the loading and tracing the analysis through by considering each collecting member is common in most analytical problems. Basic behavior The decking transfers loads to the joists. The joist reactions become forces applied on the two transverse collector beams. The reactions of the two transverse beams become the forces exerted on the columns, which in turn become forces on the foundations. Joist reactions: members A and D The load per unit length carried by each joist is found by considering the width of the load strip carried by each joist. The end joists, A and D, carry load strips a>2 in width. If the total load per unit area is the wT load per unit length carried by joists A and D is simply wT 1a>22. By symmetry, RA1 = RA2 = wT 1a>221L2 , 2

6 RA1 = RA2 = 0.25wT aL

RA1 = RA2 = 0.2515021621202 = 1500 lb

Or, by summing moments, gMRA1 = 0:

-[wT 1a>221L2]1L>22 + 1.0 L 1RA2 2 = 0

(+11)1111* total load

(1)1*

moment arm

RA2 = 0.25wT aL

(+)11* (1)1*

moment unknown arm reaction

-[150216>221202]120>22 + 1.01202RA2 = 0 RA2 = 1500 lb

gFy = 0:

+RA1 + RA2 - wT 1a>221L2 = 0 (1)1*

(1)1* (+1)11+*

unknown calculated reaction above

total downward load

RA1 + 1500 - 5016>221202 = 0

RA1 = 1500 lb

RD1 = 0.25wT aL = 1500 lb and RD2 = 0.25wTaL = 1500 lb

Joist reactions: members B and C The load per unit length carried by joists B and C is not constant, due to the presence of the opening. In sections where the decking is continuous, the width of the load strip is a. The load per unit length is consequently wT 1a2. In sections where the hole is present, each joist carries a load strip of one-half that in continuous sections, or The load per unit length on these sections is wT 1a>22. gMRB1 = 0:

- [wT 1a210.7 L2]10.7 L>22 - [wT 1a>2210.3 L2]10.7 L + 0.3L>22 + RB2 11.0L2

(+11)1111* (11)11* moment arm

load

(+111)11111* (+111)11111* partial load

moment arm

}

Copyright © 2013. Pearson Education Limited. All rights reserved.

Similarly,

RA1 = 0.25 wT aL

(1)1*

reaction moment arm

- [5016210.721202]10.72120>22 - [5016>2210.321202][0.71202 + 0.3120>22 + RB2 1202 = 0 RB2 = 0.3725wTaL = 2235 lb

gFy = 0 : R

+ R

- [w 1a210.7L2] - [wT 1a>2210.3L2] = 0 (+1111)111111*

B1 B2 T (+111)11111* ()* ()*

reaction

reaction

load

RB1 = 0.4775wTaL

partial load

RB1 + 2235 - [5016210.721202] - [150216>2210.321202] = 0 RB1 = 2865 lb

Introduction to Structural Analysis and Design Similarly, RC1 = 0.4775wTaL = 2865 lb and RC2 = 0.3725wTaL = 2235 lb Column forces: The forces on columns M and N are the reactions of the transverse beam-carrying joist loads (reactions): RA1, RB1, RC1, and RD1. By symmetry, RM = RN = 1RA1 + RB1 + RC1 + RD1 2 , 2 = 11500 + 2865 + 2865 + 15002 >2 = 4365 lb

or, by summing moments. Beam MN ΣMRM = 0: - RA1 102 - RB1 1a2 - RC1 12a2 - RD1 13a2 + RM 102 + RN 13a2 = 0

- 1500102 - 128652162 - 2865122162 - 1500132162 + RM 102 + RN 132162 = 0 RN = 0.7275wTaL = 4365 lb and

ΣFy = 0:

- RA1 - RB1 - RC1 - RD1 + RM + RN = 0

RM = 0.7275wTaL = 4365 lb The forces on columns O and P are the reactions of the second transverse beam. By symmetry, RO = RP = 1RA2 + RB2 + RC2 + RD2 2 , 2 = 11500 + 2235 + 2235 + 15002 , 2 = 3735 lb

Equilibrium checks: The total load that acts downward on the structure is wT * A, where A is the area of the surface. The forces developed on the four foundation points must sum to this same value (which provides a good overall check on your results): RM + RN + RO + RP = wT * A 4365 + 4365 + 3735 + 3735 = 50 lb>ft2 1324 ft2 2

Copyright © 2013. Pearson Education Limited. All rights reserved.

16,200 lb = 16,200 lb

Alternatively, using metric units, assume that L = 6.1 m, a = 1.83 m, and wT = 2.394 kN>m2. The force on column M is then given by 0.727512.39 kN>m2 211.83 m216.1 m2 = 19.41 kN.

exAMPle Determine the forces on a typical interior truss in the structure illustrated in Figure 19. Also determine the reactions for the truss analyzed. Assume that L1 = 60 ft = 18.3 m and L2 = 25 ft = 7.6 m and that the loads are as shown in the following solution. solution: Loads: Assume that the live load is 35 lb>ft2 11.676 kN>m2 2. Assume the following dead loads: Roofing Sheathing Joists and beams Truss Total

7.0 lb>ft2 2.5 8.0 4.0 21.5 lb>ft2

= = = = =

335 N>m2 120 383 191 1.029 kN>m2

Introduction to Structural Analysis and Design

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 19 Loading models.

The dead weights of the joists, beams, and trusses were estimated in terms of an approximate equivalent distributed load. This is often done in preliminary design stages. More detailed takeoffs are made in successive stages. Using the equivalent distributed load figures and the assumed live loading, we find that the total distributed load (live plus dead) is given by 35 + 21.5 = 56.5 lb>ft2 or

1.676 + 1.029 = 2.705 kN>m2.

Introduction to Structural Analysis and Design Forces on truss: A unit area of distributed live load is eventually carried to the supports as illustrated in the free-body diagrams in Figure 19. The load is first picked up by the roof decking, which carries it to adjacent joists. The reactions of the decking become forces on the joists. The reactions of the joists in turn become forces on the transverse beams. The reactions of the transverse beams become forces on the panel points of the trusses. The magnitudes of these forces could be determined by calculating each reaction involved in turn. This precise method accurately reflects how the structure carries loads, but it is cumbersome. Forces on the panel points of the truss can be determined more quickly by an alternative approach, which involves the models in Figure 19(c). These models are based on the concept of contributory areas. As the illustration suggests, it is possible to determine the force present at a truss panel point by finding the relative portion of the roof that serves as the contributory area for the panel point and then multiplying this area by the magnitude of the distributed uniform loading. We have the following calculations: contributory area for a typical interior panel point: = a

L1L2 L1 b 1L2 2 = 4 4

interior-panel-point force = Rp = =

156.5 lb>ft22160 ft2125 ft2 wTL1L2 = = 21,187.5 lb 4 4

12.705 kN>m2 2118.3 m217.6 m2 4

= 94.05 kN

contributory area for a typical exterior panel point: = a exterior-panel-point force =

L1L2 L1 b 1L2 2 = 8 8

Rp wTL1L2 = 8 2

= 10,593.75 lb = 47.03 kN

Copyright © 2013. Pearson Education Limited. All rights reserved.

The final free-body diagram for the truss is shown in Figure 19(c). Reactions: The reactions of the truss can be obtained by using the free-body diagram in Figure 19(c) with the forces just found. Because of symmetry, a typical reaction is given by 1Rp >2 + Rp + Rp + Rp + Rp >222 = 42,375 lb = 188,106 N. Alternatively, the reactions of the truss can be found directly (without finding the panel-point forces) by considering the loading model in Figure 19(d). The figure shows the relative portion of the total roof surface carried by a typical interior truss. This is the contributory load area for the truss and can be used to find the total load carried by the truss and then the reactions. The figure indicates that each reaction picks up one-half of the total contributory load area: contributory area for a typical truss = L1L2 total downward force on truss = wTL1L2 = 156.5 lb>ft2 2160 ft2125 ft2 = 84,750 lb

= 12.705 kN>m2 2118.3 m217.6 m2 = 376.2 kN

Because each truss has two reactions that are symmetrically placed with respect to the load, each reaction is one-half the total load, or 84,750>2 = 42,375 lb or 376.2>2 = 188.12 kN. These are the same values noted earlier. For finding truss reactions, this procedure saves time. Note that the use of loading models based on the notion of contributory areas is valid only when there is no extreme asymmetry in either the loading condition or the structure. For partial loading conditions, it is necessary to calculate the reactions for each element and consider them as forces acting on the elements supporting it. This process must be repeated until the reactions of the final collecting element on the foundation are found. Some structural systems carry loads over longer spans and to fewer supports. An example is shown in Figure 20. Here, floor and roof loads as well as other loads are collected by four steel arches that transfer them onto eight vertical supports at the base.

Introduction to Structural Analysis and Design

Figure 20 Complex load paths in a multistory building. Broadgate Exchange House, London Architect and structural engineer: Skidmore, Owings & Meril Completed 1990

Detail connection at arch base: the centroids of arch, tension rod, and the column meet in one point.

Floor Dead Load 150 Ib/ft2 Live Load 50 Ib/ft2 ______________________ 210 Ib/ft2

Roof Dead Load 120 Ib/ft2 Live Load 30 Ib/ft2 ______________________ 150 Ib/ft2

Contributory area: 19.7 x 25 = 492.5 ft2

Copyright © 2013. Pearson Education Limited. All rights reserved.

Load for each office floor: 492.5 ft2 x 210 Ib/ft2 = 130,425 Ib = 130 k Roof load: 492.5 ft2 x 150 Ib/ft2 = 73,875 Ib = 74 k

The 10-story office building spans 256 ft (78 m) across the underground railway tracks of Liverpool Street Station. Four large steel arches carry the loads to eight vertical supports. The arches follow a parabolic, funicular shape for equally spaced point loads. They were built as a series of welded polygonal structural steel shapes. Vertical steel members transfer the floor loads to the arch. Above the arch, these members are in compression; below the arch, they are in tension. The total gravity load, the sum of compression and tension load, remains constant across the arch. The horizontal thrust of the arches is resisted by four tension rods.

Contributory Area

36.9 k 4  51.7 k in compression

The load of contributory areas is split between two adjacent arches.

Total point load: 553.9 k

Point load onto a facade arch: Floor load: 130,425 Ib / 2 10 floors: 10 x 51.7 k Roof load: 73,875 Ib / 2 Total point load:

= 51.7k = 517 k = 36.9 k = 553.9 k

This is the typical point load onto a side arch. The arches in the middle are loaded from both sides, so their loads double.

Horizontal thrust is contained by tension rods. 6  51.7 k in tension

Vertical reaction at edge support (14  553.9 k) / 2  3876.9 k

Introduction to Structural Analysis and Design

exAMPle Determine the forces on the structural frames shown in Figure 21 due to the action of wind impinging on the face of the building. Assume that w = 20 lb>ft2, h = 15 ft, and L = 25 ft. Figure 21 Modeling wind loads on rigid frames.

Copyright © 2013. Pearson Education Limited. All rights reserved.

.

Introduction to Structural Analysis and Design solution: The forces can be determined either by tracing how specific elements, such as girts, are loaded and how their reactions become loads on the frames or by using a contributory-area concept. Both methods are illustrated in Figure 13. Numerical values can be found by substitution (e.g., whL>4: 1875 lb).

QuestiOns 1. Find two bridges in your area and identify the support conditions present in them (e.g., hinged, roller). Sketch the supports and include a diagram of the symbol that represents them. 2. Find two different trusses used in buildings in your area and identify the support conditions present in them. Sketch the end condition and include a diagram of the symbol that represents the support. 3. Consider a typical corridor in the building where you work or in one nearby. What do you estimate the actual live load to be during normal traffic conditions? During fire conditions? 4. What snow loads and wind loads are specifically recommended for buildings in your area? Consult the local building code. 5. For the floor system shown in Figure 13, assume that the combined live and dead load that is present is equal to 80 lb>ft2. Assume a beam span of 16 ft and a beam spacing of 3 ft. Determine the reactions for beams A, B, and C in the floor system. (First, determine the load strip widths, then determine the appropriate loading model for each beam, and finally determine reactions for each beam by a statics analysis.) Answer: Beam A: 960 lb each end; beam B: 1920 lb each end; beam C: 960 lb each end 6. Determine the reactions to Beam D in Figure 22. Assume that the average dead and live load is 60 lbs>ft 3.

Figure 22 Example.

12 ft

Beam B 12 ft

Decking Beam F

Beam G

Column 2

Beam E

Opening

Decking

Beam D

Column 3

Copyright © 2013. Pearson Education Limited. All rights reserved.

Beam A

Decking Beam C

12 ft

8 ft

Column 1

Column 4 12 ft

Answers: 4896 lb, 4464 lb 7. Find the forces in Columns 1, 2, 3, and 4 of the structure shown in Figure 22. This is the same structure shown in Figures 1 and 15. Assume that the average dead load and live load is 60 lbs>ft2. Answers: Column 1, 8496 lb; Column 2, 8496 lb; Column 3, 12,384 lb; Column 4, 12,384 lb 8. For the beam shown in Figure 21(a), determine the reactions if the live load is changed to 30 lb>ft. Repeat the example analyzed in Figure 21, assuming that w = 25 lb>ft2, h = 12 ft, and L = 28 ft. Show how loadings are traced through. 9. Model the loads on an interior truss of the Institute of Contemporary Art building in Boston, Massachusetts, which is shown in Figure 23. Assume a combined dead and live load of 190 lb> ft2 for the roof and 240 lb>ft2 for the floor, and ignore the dead weight of the trusses.

Figure 23

System diagram for problem no. 9. 7 x 23' 41'

23' 60'

Trusses

Copyright © 2013. Pearson Education Limited. All rights reserved.

1

IntroductIon

Although structures made of jointed members have been constructed throughout history, the conscious exploitation of structural advantages that arise when individual linear members are formed into triangulated patterns is relatively recent. Structures of this type, commonly called trusses, were built quite early in the history of construction. Simple trusses using relatively few members often appeared in common pitched roofs. More complex trusses were used in isolated instances. A bridge using a form of timber truss, for example, was built across the Danube River by the Romans as early as 500 BC. Such examples, however, had little impact on the building methods of the time. The Italian architect Andrea Palladio (1518–1580) illustrated a correctly triangulated truss and indicated that he had some knowledge of its potential and the way it carried forces. Trusses were occasionally used afterward in large public buildings such as Independence Hall, Philadelphia, but again without having much impact as a structural innovation. It was the bridge builders of the early nineteenth century who first systematically explored and experimented with the potential of the truss. This was in response to the demands of rapidly expanding transportation systems of the time. Emiland Gauthey’s Traitè de la Construction des Ponts (posthumously published between 1809 and 1813 by his nephew, the famous mathematician Louis Navier of the École Polytechnic in Paris) provided a foundation for many subsequent theoretical works in the area. Gauthey’s treatise included a discussion of what he termed the principles of equilibrium of position and equilibrium of resistance. The former was an attempt to resolve bridge loads into components in individual members. The latter dealt with material properties and the sizing of truss members. Later important contributions include Squire Whipple’s classic of structural engineering, A Work on Bridge Building, published in 1847. The development of the truss was thus fostered by a tentative, but rapidly expanding, body of theoretical knowledge. This contrasts with other structural forms that typically developed slowly over time in a strictly empirical way. The truss soon became a common structural form used in civil engineering structures spanning long distances. The use of trusses in buildings also increased, although more slowly, due to different traditions and needs, until they became common in modern architecture (Figure 1). The emergence of the truss as a major structural form has been rapid and its impact significant. The remainder of this chapter explores what a truss is, how it works, and why it is important. The presentation inquires into a specific structural element, but From Chapter 4 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Trusses

FIGure 1 Trusses are versatile structural elements that can even adjust to complex geometry and loading.

the analysis and design methods for trusses are used to demonstrate principles that are broadly applicable to the analysis and design of a wide range of other structural forms.

2

Copyright © 2013. Pearson Education Limited. All rights reserved.

2.1

General PrIncIPles triangulation

A truss is an assembly of individual linear elements arranged in a triangle or a combination of triangles to form a rigid framework that cannot be deformed by the application of external forces without deformation of one or more of its members. The individual elements are typically assumed to be joined at their intersections with pinned connections. Members are customarily arranged so that all loads and reactions occur only at these intersections. The primary principle of using the truss as a load-carrying structure is that arranging elements into a triangular configuration results in a stable shape. Consider the two pin-connected structures shown in Figures 2(b) and (c). Applying a load to the structure shown in Figure 2(b) causes the massive deformation indicated. This unstable structure forms a collapse mechanism under external loading. Such a structure may be deformed without a change in length of any of its individual members. The triangulated configuration of members in Figure 2(c) could not deform or collapse in a similar manner. This configuration is thus stable. Any deformations that occur in it are relatively minor and are associated with small changes in the lengths of members caused by forces generated by the external load. Similarly, the angle formed between any two members remains unchanged under load in a stable configuration of this type. This is in contrast to the large changes in angle that occur between members in an unstable configuration. [See Figure 2(b).] The external force causes forces to be developed in members of the stable triangulated structure. As is explained in detail in Section 3.2, these forces are either purely tensile or purely compressive. Bending is not present, nor can it be developed, as long as external loads are applied at nodal points.

Trusses

FIGure 2 Typical trusses and basic triangulation principles.

Copyright © 2013. Pearson Education Limited. All rights reserved.

O

O

A basic triangle of members is a stable form, so it follows that any structure made of an assembly of triangulated members also is a rigid, stable structure. This idea is the principle underlying the viability and usefulness of the truss in building because larger rigid forms of any geometry can be created by the aggregation of smaller triangular units. Again, the effect of external loads produces a state of either pure tension or pure compression in the individual members of the assembly. For common trusses with vertically acting loads, compressive forces are usually developed in upper chord members and tensile forces in lower chord members. Either type of force may develop in an interstitial member, although an alternating pattern of tensile and compressive forces is often present. It is important that trusses be loaded only with concentrated loads that act at joints in order for truss members to develop only tensile or compressive members. If loads are applied directly onto the truss members, bending stresses develop in the loaded members in addition to the basic tensile or compressive stresses already present, with the consequence that member design is greatly complicated and the overall efficiency of the truss is reduced.

2.2

Member Forces: Qualitative analyses

As discussed in sections that follow, the forces in the members of any truss can be determined by applying the basic equations of equilibrium. For some simple truss configurations, however, the basic sense (tension, compression, zero) of the forces in many members can be determined by less involved techniques that might help visualize how certain trusses carry external loads.

Trusses One way to determine the sense of the force in a truss member is to imagine that the member is removed and then visualize the structure’s probable deformed shape. The force in the member can be predicted because its role is to prevent the deformation visualized. Consider the diagonals shown in truss A in Figure 3(a). If the diagonals were removed, the assembly would dramatically deform [see Figure 3(b)] because it is a nontriangulated configuration. For the diagonals to keep this type of deformation from occurring, the left and right diagonals must prevent points B–F and points B–D, respectively, from drawing apart. Consequently, diagonals between these points would be pulled upon, and tension forces would develop in the diagonal members. The diagonals shown in truss B in Figure 3 must be in a state of compression because their function is to keep points A–E and C–E from drawing closer together. With respect to member BE in both trusses, it is easy to imagine what would happen to points B and E if member BE were removed. In truss A, points B and E would tend to draw together; hence, compressive forces would develop in any member placed between the two points. In truss B, however, removing member BE leads to no change in the gross shape of the structure (it remains a stable triangulated configuration). Therefore, the member serves no role in this loading; it is a zero-force member. Note that members AF, FE, ED, and DC in truss B could also be removed without altering the basic stability of the remainder of the configuration; thus, these members are also zero-force members. This is not true for the same members in truss A. The final forces in both truss A and truss B are illustrated in Figure 3(c). A different way to visualize the forces developed in a truss is to use an arch-and-cable analogy. Truss A, for example, can be conceived of as a cable with

FIGure 3 Forces in truss members: The senses of the forces in some simple truss configurations can be determined through intuitive approaches. More complex trusses require quantitative approaches.

Copyright © 2013. Pearson Education Limited. All rights reserved.

they were removed and

or arch analogy can also

is

cable configuration.

Trusses supplementary members. [See Figure 3(d).] Truss B can be conceived of as a simple linear arch with supplementary members. Diagonals in truss A are consequently in tension, while diagonals in truss B must be in compression. The forces in other members can be determined by analyzing their respective roles in relation to containing arch or cable thrusts or providing load transfer or reactive functions. As Figures 4 and 5 indicate, special forms of more complex trusses can also be visualized this same way.

FIGure 4 Cable analogy in truss analysis: Many complex truss forms can be imagined to be composed of a series of simpler basic cable units. If the directions of the diagonals were all reversed, an arch analogy could be used to analyze the structure. This analogy approach, however, is useful for only a few limited truss forms, typically parallel cord trusses supported at their ends.

cable unit: The diagonal in tension.

because it resists the

Copyright © 2013. Pearson Education Limited. All rights reserved.

imagining that the is carried

because increased portions of

accordingly.

individual top chords are combined into one member.

Trusses

FIGure 5 Use of the cable analogy in determining member characteristics in parallel cord trusses. Equivalent cable systems can be used to determine force characteristics of diagonal members. Hanging cable system

Both the methods discussed for visualizing forces in trusses become difficult or even impossible to apply when complex triangulation patterns are present. A method to visualize forces in such trusses on the basis of joint equilibrium considerations is considered next. In general, however, complex truss forms must be mathematically analyzed to obtain correct results. The applicable methods are reviewed in the following section.

3 Downward slope in tension

P' P

FIGure 6 Internal stability. Nontriangulated forms may collapse.

Copyright © 2013. Pearson Education Limited. All rights reserved.

The nontriangulated

The bar pattern

is still stable.

has more than the

3.1

analysIs oF trusses stability

The first step in the analysis of a truss is to determine whether the truss is a stable configuration of members. It is usually possible to tell by inspection whether a truss is stable under external loads by considering each joint in turn to determine whether the joint will maintain a fixed relation to other joints under any loading condition applied to the truss. In general, any truss composed of an aggregation of basic triangular shapes will be a stable structure. Nontriangular shapes in a bar pattern are an obvious sign that the truss should be carefully inspected. The truss in Figure 6(a) is unstable and would collapse under load in the manner illustrated. This truss does not have a sufficient number of bars to maintain a fixed geometrical relationship between joints. Assuming that the remaining truss members are adequately designed for the loads they carry, adding a member from B to E, as indicated in Figure 6(b), would make this configuration stable. It is possible that a truss can include one or more figures that are not triangles and still be stable. Study Figure 6(c), which illustrates this type of truss. It consists of groups of rigid triangulated bar patterns connected to form nontriangular, but still stable, figures. The triangles between A and C form a rigid shape, as do those between B and C. Joint C is thus fixed in relation to joints A and B in the same way as in a simpler triangular figure. The group of triangles between A and C can be considered a member, as can those between B and C. In a single truss, it is possible to have more than the minimum number of bars necessary for a stable structure. One of the two middle diagonal members of the truss in Figure 6(d) could be considered redundant. Either member BE or member CF could be removed, and the resultant configuration would remain stable. Removing both would result in the structure becoming a collapse mechanism. It is of utmost importance to determine whether a configuration of bars is stable or unstable because few errors are more dangerous. Total collapse occurs immediately when an unstable configuration is loaded. To help ascertain the stability of configurations, expressions have been developed that relate the number of joints present in a truss to the number of bars necessary for stability. A simple basic planar triangle has three bars 1n = 32 and three joints 1j = 32 Adding a single new connection beyond the original three requires that two new bars or members be added. Hence, the total number of members 1n2 in a truss having a number of connections 1 j2 is given by n = 3 + 21j - 32. (Three members are originally present, and for each new joint j - 3 two new members are added.) This expression can be rewritten more simply as n = 2j - 3. A similar expression can be found for three-dimensional bar networks. A simple tetrahedron is known to be stable. Three bars are required to form each new node added to this basic form. There are six bars to begin with, and new bars are added at the rate of three for each node beyond the original four. Thus, n = 6 + 31j - 42 = 3j - 6. Consider the planar truss in Figure 6(b). In this case, j = 6 so n must be 2162 - 3, or 9. This is the minimum number of bars required for stability. The truss has that number of bars, so it is stable. Generally speaking, fewer bars than

Trusses are given by this expression result in an unstable structure; more may indicate a structure has redundant members. The expression, however, is not foolproof and should not be used to replace a careful visual inspection of the truss. The expression is more an indicator of whether the internal forces in a structure can be calculated by the equations of statics alone, rather than being a predictor of stability. However, it is a useful tool for stability assessments because the forces in an unstable structure cannot be calculated by the equations of statics. Another aspect of stability is illustrated in Figure 7. Truss configurations may be used to stabilize structures with respect to laterally acting loads. Doing so using rigid members is straightforward. In certain cases, however, cables may be used in lieu of rigid members when the cables are subjected to tension forces only. Stability considerations thus far have assumed that all truss members can carry tension and compression forces equally well. Cable members do not meet this assumption; they buckle out of the way when subjected to compressive forces. As illustrated, whether tension or compression is developed in a diagonal depends on its orientation. When the loading comes from either direction, tension or compression may develop. A structure with a single diagonal cable could be unstable, which is not predicted by the previous expressions relating the number of joints and members. If cables were to be used, a cross-cable system would be required wherein one cable takes up the horizontal force while the other buckles harmlessly out of the way.

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 7 Diagonal bracing. The white arrows indicate the deformation tendencies of the system.

Trusses

3.2

Member Forces: General

Copyright © 2013. Pearson Education Limited. All rights reserved.

This section considers classical analytical methods to determine the nature and distribution of forces in the members of a truss that has known geometrical characteristics and carries known loads. The principle underlying the analytical techniques to be developed is that any structure or any elemental portion of any structure must be in a state of equilibrium. This principle is the key to common truss analysis techniques. The first step in analyzing a truss of many members is to isolate an elemental portion of the structure and consider the forces acting on that element. If some of the forces are known, it is usually possible to calculate the others by using the basic equations of statics because it is known that the element must be in equilibrium. These equations are formal statements that any set of forces, including both those externally applied and those internally developed, must form a system whose net force is zero. The portion of the structure selected for study is not restricted. A whole segment consisting of several members and joints could be considered, or attention could be limited to a single joint or member. To find the internal forces present at a location in a structure, it is necessary that the structure be decomposed at least at that point. Figures 8(b)–(e) illustrate free-body diagrams for typical elemental pieces of the truss shown in Figure 8(a). Each of the pieces must be in a state of equilibrium under the action of the force system present on the piece. The force system considered consists of not only any external loads applied to the piece but also those that are internal to the entire structure. The latter are forces developed in members in response to the external loading on the whole truss. Such forces are necessary to maintain the equilibrium of elemental portions of the truss. Regarding an element’s equilibrium, it is useful to think of these internal forces as applied forces. At points where the truss is decomposed, internal forces are equal in magnitude, but opposite in sense, in adjacent elements. Consider Newton’s third law: The forces of action and reaction between elements in contact with each other have the same magnitude and line of action but are of opposite sense. Members of a truss are pin connected and their ends are free to rotate, so only forces, and not moments, can be transmitted from one member to another at the point of connection. Thus, only vector forces are indicated in the free-body diagrams shown. Because only forces, not moments, can be transmitted at the connections, the forces at either end of a member must act collinearly (assuming that the member is not subjected to external loads) if the member is to be in a state of equilibrium. If the member itself aligns with these forces, then the member is only axially loaded and not subject to bending moments.

3.3

equilibrium of Joints

equilibrium of Joints within trusses. The fact that any portion of any structure must be in a state of equilibrium forms the basis for all analysis techniques directed at finding forces in truss members. In analyzing a truss by the classical method of joints, the truss is assumed to be composed of a series of members and joints. Member forces are found by considering the equilibria of the various joints, which are idealized as points. Each of these joints or points must be in a state of equilibrium. Figure 8(e) illustrates a truss that has been decomposed into a set of individual linear elements and a set of idealized joints. Free-body diagrams for all the members and joints are shown. The joints show that the system of forces acting on a joint is defined by the bars attached to it and by any external loads that might occur at the joint’s location. As shown in Figure 8(e), the forces on a joint are equal and opposite to those on the connecting members. Each joint must be in a state of equilibrium. The forces

Trusses

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 8 Typical free-body diagrams for elemental truss pieces. These diagrams are based on the fundamental principle that any structure or any portion of any structure must be in a state of equilibrium. The freebody diagrams in (e), (f), and (g) are used to solve bar forces by the method of joints.

applied to the joint all act through the same point. From an analytical perspective, we are interested in the equilibrium of a point. This requires considering translational equilibrium only. Rotational equilibrium is not a concern because all forces act through a common point and produce no rotational effects. This is the key to analyzing trusses by the method of joints. For planar structures, two independent equations of statics exist for a concurrent force system ( g Fx = 0 and g Fy = 0). Thus, two unknown forces can be found by applying these equations to the complete system of forces represented in the free-body diagram of a joint. If a joint with

Copyright © 2013. Pearson Education Limited. All rights reserved.

Trusses a maximum of two unknown forces is considered first, it is possible to calculate these forces. The starting point to analyze the forces in a truss is often at a support where the reaction was determined by considering the rigid-body equilibrium of the whole structure. Once all the forces acting on the initial joint (and thus also the forces in the bars attached to the joint) have been found, it is possible to proceed to another joint. Because the bar forces previously found can now be treated as known forces, it is convenient to consider an adjacent joint next. A subsequent example clarifies the process. A few words should be said about the arrow convention used. Arrows represent the nature and direction of the forces developed on an element. Thus, the arrows on member DE in Figure 8(e) indicate that forces causing the element to be in a state of compression are developed because of the loads on the larger structure. Note that the arrows seem to subject the member to a compressive force. Conversely, the arrows on member BC in Figure 8(e) indicate that a state of tension exists in the element. With respect to the joints, these arrows are shown to be equal and opposite. Thus, the action of member DE (in a state of compression) on joint E apparently is a pushing against the joint. In actuality, a reaction is developed. In an analogous way, the action of member BC (in a state of tension) seemingly pulls on joint C. It is useful to visualize a joint as being in a state of equilibrium when the pushes and pulls of the members framing into the joint balance each other. Analyzing bar forces in a truss by the method of joints with hand-calculation techniques is generally straightforward for trusses with few members. For the truss shown in Figure 8, the first step is to draw a set of free-body diagrams like those in Figure 8(e). Alternatively, draw simplified diagrams of only the forces on the joints. [See Figure 8(f).] The equations of translational equilibrium ( g Fx = 0 and g Fy = 0) are then applied in turn to each joint. In drawing the free-body diagrams and writing equilibrium equations, it is necessary to assume that an unknown bar force is in a state of either tension or compression. The state of stress assumed can be arbitrary. Whether the force is in the state of stress assumed will be evident from the algebraic sign of the force found after making equilibrium calculations. A positive sign means the initial assumption was correct, while a negative sign means the opposite assumption holds. To develop a more intuitive feeling for the force distribution in a structure, it is useful to try to determine whether a member is in tension or compression by a careful qualitative inspection of each joint’s equilibrium. Consider joint A in Figures 8(e)–(g), where it can be seen qualitatively that the directions assumed are reasonable. The reaction is known to act upward. For equilibrium in the vertical direction to be obtained, there must be a force acting downward. Only force FAE of the two unknown member forces has a component in the vertical direction and would be capable of providing the downward force necessary to balance the upward reaction. Member AB is horizontal and thus has no component in the vertical direction. Force FAE must therefore act in the downward direction shown. Thus, member AE must be in a state of compression. If force FAE acts in the direction shown, it must have a horizontal component acting toward the left. For the joint to be in horizontal equilibrium, there must be some other force with a horizontal component acting to the right. The reaction acts only vertically, so it does not enter into consideration. Force FAE must therefore act to the right to obtain equilibrium in the horizontal direction. Member AB is hence in a state of tension. Looking next at joint E and noting that member AE is in compression, it is evident that member EB must be in tension to provide a downward component necessary to balance the upward one of the force in member AE. Member ED is horizontal and thus can contribute nothing in the vertical direction. If forces FAE and FAE act in the directions shown, both have a component acting to the right in the horizontal direction. The force in member ED must therefore act to the left to balance the sum of the horizontal components of the forces in the two diagonals. Member ED is thus in compression. Because the truss is symmetrical, the state of

Trusses stress in the remaining members can be found by inspection. (Member DC must be in compression, BC in tension, and DB in tension.) Note that joint B is also in a state of equilibrium. Thus, the state of stress in all the members can be qualitatively determined without resorting to calculation. This process is not possible with all trusses, but is successful enough to make the attempt worthwhile. Such a qualitative approach does not yield numerical magnitudes of bar forces. These can be found only by formally writing the equations of equilibrium and solving for the unknown forces. The mathematical process, however, is similar to the qualitative one. Both are based on the principle that any element of a structure must be in equilibrium. To solve for numerical magnitudes, each joint is considered in turn. In the following example, the reference axes used are vertical and horizontal. FIGure 9

exaMPle Determine all the member forces in the truss shown in Figure 8. solution: Draw joint equilibrium diagrams for the whole structure, and then consider the equilibrium of each joint in turn. (See Figure 9.) Joint A Equilibrium in the vertical direction: gFy = 0 c + :

FIGure 10(a)

+ 0.5P - FAE sin 45° = 0

()*

reaction RAv in vertical direction

(+1)111*

vertical component of force in member AE; the member is assumed to be in compression

FAE = + 0.707P 1compression2

Member AE is in compression, as assumed, because the sign is positive. [See Figure 8(a).] Equilibrium in the horizontal direction: gFx = 0 S + :

- FAE cos 45° (+1)+1*

Copyright © 2013. Pearson Education Canada. All rights reserved.

horizontal component of force in member AE

+ FAB "

6 FAB = + 0.5P 1tension2

force in member AB assumed to be in tension

Member AB is in tension, as assumed, because the sign is positive. Next, proceed to an adjacent joint. Joint B involves three unknown forces for which only two independent equations are available for solution. Joint E, however, involves only two unknown forces. So go to joint E next. Joint E Equilibrium in the vertical direction: gFy = 0 c + :

+ FAE sin 45° - FEB sin 45° = 0 Because FAE is known, FBE can be solved for: FEB = + 0.707 P 1tension2

Because these are the only two forces at this joint having components in the vertical direction, the components must be equal. Because the member angles are the same, the member forces must also be the same. Thus, the force in member EB could have been found by inspection. [See Figure 10(b).]

FIGure 10(b)

Trusses Equilibrium in the horizontal direction: gFx = 0 S + :

+FAE cos 45° + FEB cos 45° - FED = 0 FED = 1.0 P 1compression2

FED is the sum of the horizontal components in the two diagonals. Next, proceed to an adjacent joint (either B or D). Joint B Equilibrium in the vertical direction: gFy = 0 c +:

FEB sin 45° + FBD sin 45° - P = 0 Because FEB is known, FBD can be solved for:

FIGure 10(c)

FBD = 0.707P 1tension2

A symmetry argument would yield the same result—that is, FBD must be similar to FEB, because the truss is geometrically symmetrical and symmetrically loaded. Otherwise, there is no a priori reason to believe the forces are the same. [See Figure 10(c).] Equilibrium in the horizontal direction: gFx = 0 S + :

Joint D

-FAB + FBC - FEB cos 45° + FBD cos 45° = 0 or FBC = 0.5 P 1tension2

Equilibrium in the vertical direction: gFy = 0 c +:

-FBD sin 45° + FDC sin 45° = 0

FIGure 10(d)

FDC = 0.707 P 1compression2

Equilibrium in the horizontal direction: gFx = 0 S + :

+FDE - FDB cos 45° - FDC cos 45° = 0 Copyright © 2013. Pearson Education Canada. All rights reserved.

1.0P - 0.707P cos 45° - 0.707P cos 45° = 0 All forces are known. The equation sums to zero. This is a good check on the accuracy of calculations. [See Figure 10(d).] FIGure 10(e)

Joint C Equilibrium in the vertical direction: gFy = 0 c +:

-FDC sin 45° + 0.5P = 0 or -0.707P sin 45° + 0.5P = 0 Again, this is a check because all forces are known. [See Figure 10(e).] Equilibrium in the horizontal direction: gFx = 0 S + :

-FBC + FDC cos 45° = 0 or 0.5P + 0.707P cos 45° = 0 Once more, this is a check. All member forces are known.

Trusses analysis steps. The next example emphasizes the general sequence of steps involved in analyzing a truss by the joint equilibrium approach, rather than the numerical analyses. The analyst should “think through” sequences like this before starting an analysis of any truss. exaMPle Determine the member forces in the cantilevered truss shown in Figure 11, using a joint equilibrium approach. solution: As illustrated in Figure 11, reactions must first be determined by using appropriate techniques. Next, the analyst identifies a node where no more than two unknown forces are present— typically, a support connection at which there is a known reaction and two unknown member forces. (Recall that member forces cannot be found directly if more than three unknowns are present.) In the case shown, the starting node connection is A. One then proceeds to an adjacent node with no more than two unknown forces and repeats the process. Determine Reactions:

gMA = 0 gFy = 0

gFx = 0

- 11000212L2 + RDy 1L26 RDy = 2000 lb

- 110002 + RAy + RDy = 0 6 RAy = 1000 lb

RDx = 0

Find Member Forces: Use gFy = 0 & gFx = 0 at Each Node: Step 1—Node A:

The sense of the forces (tension, compression) in members FAB and FAD is assumed. The force FAB in the diagonal is assumed to be in tension, so that it has an upward component to balance the downward reaction. This implies that the force FAD in the lower chord should be in compression, so that it has a leftward component to balance the rightward horizontal component of the diagonal force in FAB. The analyst then sums forces in the vertical direction to find FAB. Note that the vertical component of FAB equals the vertical reaction, or SFy = 0: - 1000 + FAB sin 30 = 0, or FAB = 2000 in tension. FAD can be found next because it is equal to the horizontal component of FAB, or SFx = 0: -2000 cos 30 + FAD = 0, or FAD = 1732 in compression. Step 2—Node B:

Copyright © 2013. Pearson Education Canada. All rights reserved.

Because FAB is known, forces can be summed in the vertical direction to obtain FBD and in the horizontal direction to obtain FBC. The approach is similar to that in Step 1. Step 3—Node C: Because FBC is known, forces can be summed in the vertical direction to obtain FCD. All member forces are now known. Step 4—Node D: Because all member forces are known, summing forces in the vertical and horizontal directions provides a good calculational check. If balances are not obtained, an error was made in earlier steps.

complex trusses. Many trusses have geometries that make their analysis difficult. The next example solves simultaneous equations because individual equilibrium expressions cannot be solved directly. A way around this complexity is then presented. exaMPle Determine the forces in members FFC and FFG in the truss shown in Figure 12. Start at the right support, where the reaction is known. Assume that vertical members are evenly spaced.

FIGure 11 Cantilevered truss.

Trusses

FIGure 12 Use of rotated reference axes.

solution: Joint E

gFy = 0

gFy = 0

Joint D

- FEF sin 30° + 3P/4 = 0 or FEF = 1.5P FEF cos 30° - FDE = 0 or FDE = 1.3P

gFx = 0

gFy = 0k

(a) Truss System

Joint F

gFy = 0

gFx = 0

1compression2 1tension2

FDE - FCD = 0 or FCD = 1.3P 1tension2 - P + FFD = 0 or FFD = 1.0P 1tension2

-FFD + FEF sin 30° - FFG sin 30° + FFC sin 30° = 0 - FEF cos 30° + FFG cos 30° + FFC cos 30° = 0

FFD and FEF are known. Each equation involves two unknown forces (FFG and FFC). Solving the equations simultaneously yields FFG = 0.5P 1compression2 FFC = 1.0P 1tension2

(b) Joint E

The need to solve two equations simultaneously can be prevented by using a rotated reference axis system 1m=n= 2 rather than the traditional vertical and horizontal (xy) system. Thus, with respect to m=n=, gFn= = 0

gFm= = 0

- FFD cos 30° + FFC cos 30° = 0 FFC = 1.0P

FFD sin 30° + FFC sin 30° - FEF + FFG = 0 FFG = 0.5P

An easier method of analyzing this kind of problem is discussed in a later example, using the equilibrium-of-sections approach. (See Section 3.4.)

Copyright © 2013. Pearson Education Canada. All rights reserved.

(c) Joint D

(d) Joint F

simplifying conditions. Occasionally, truss analysis by the method of joints can be facilitated by paying attention to some special conditions that frequently arise. One of these is the appearance of zero-force members. Consider joint B in the truss shown in Figure 13. By inspecting the equilibrium of this joint in the vertical direction, it is possible to determine that no internal force is developed in member BI. There is no applied external load and members BC and BA are horizontal, having no component in the vertical direction. If member BI had any force, the joint would not be in equilibrium; hence, member BI can have no force. It is a zero-force member. The same is true for member HC. By isolating joint J and considering equilibrium the horizontal and vertical directions, it can be seen that both members framing into J (JA and JI) also are zero-force members. The same is true of the members framing into joint F. This approach of picking out selected members and joints for a quick determination of forces also can be applied to some joints with external loads applied. Isolating joint D (see Figure 13) and inspecting equilibrium in the vertical direction indicates that the force developed in member DG must be equal and opposite to the applied load. Thus, member DG carries a force of 2P and is in tension.

(e) Joint H (f) Rotated reference axis system for joints H (left) and F (right)

Trusses As another example of simplifying conditions, consider the truss in Figure 12, which was analyzed previously. Inspection of joint B indicates that member BH must be a zero-force member. This implies that member HC also is a zero-force member. (Think in terms of the rotated axis system shown in Figure 12, and consider equilibrium in the m direction.) Member HC cannot have any force in it if joint H is to be in equilibrium in this direction (as it must be in any direction). Inspection of joint D indicates that DF must carry a tension force of P. This implies that member FC carries a compressive force of P. Again, this is easily understood in terms of a rotated axis system, as indicated in Figure 12. Because FDF is shown to be equal to P in tension, the component of this force in the m= direction must be balanced by a similar component in member FC. The members make a similar angle with the m= axis, so it follows that the forces must be numerically the same if the components are the same; thus, FFC = P. Simplifying conditions of the type described facilitate calculations enormously, but the technique is most valuable in developing a more intuitive understanding of the force distributions present in a truss. It is recommended to identify zero-force members prior to other analysis steps. Graphic statics. The truss in Figure 8 also can be analyzed by using graphic statics techniques based on joint equilibrium. These approaches are based on the tip-to-tail graphic techniques, in which reactions and member forces are graphically depicted to scale. Such approaches were the basic method to analyze trusses before the advent of computers. They are still favored by many because of their basic elegance and visualization advantages. Nonetheless, remember that these graphic approaches are often difficult to apply to complex trusses and have limited usefulness regarding indeterminate trusses. (See Section 3.6.) Graphical solution techniques are illustrated in Figure 14. Also remember that a closed tip-to-tail polygon drawn for any node means that it is in translatory equilibrium in the x and y directions.

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 14 Graphical method of analysis.

FIGure 13 Simplifying conditions in truss analysis.

Trusses Using this approach, it is best to first redefine members and loads in terms of adjacent spaces, as shown in Figure 14. Only spaces separated by forces are designated. The load P becomes ba the reaction RAV becomes ac, the force FAE becomes c1, and so forth. Using this nomenclature, the upper-right side of Figure 14 illustrates typical underlying polygons for the tip-to-tail approach. For joint A, the known reaction ac is laid off to scale. Proceeding clockwise around the joint, the directions of the unknown forces in c1 and a1 are drawn through c and a, respectively. The location of point 1 is the intersection of the lines of action of these two forces so that a closed polygon (necessary for equilibrium) is formed. Unknown forces in c1 and 1a can be measured directly. With c1 known, a similar process can be repeated at joint E and other joints. The steps described are a graphic solution of the equilibrium of joints approach. As shown in the lower half of Figure 14, this process can be streamlined in a composite diagram by first drawing a force polygon for the reactions and loading (simply a vertical line) and drawing the diagram for A. Because c1 is common in the solution of both joints A and E, a composite diagram for A and E may be drawn next. Proceeding to joint B on the composite diagram, we locate point 3 by extending lines of action of 23 and b3 from known points. Proceeding to joint D, we repeat the process to form the final diagram. With practice, diagrams of this type may be drawn without the intermediate steps shown at the top in Figure 14 to obtain the last diagram in the series, which is often called a Maxwell diagram after its developer, the English engineer James Clerk Maxwell. Forces may be measured directly from this scaled drawing. The sense of each force may be determined by looking at the forces in sequence around any particular node, observing the same sequence of letters on the Maxwell diagram, and keeping in mind the tip-to-tail graphic convention that underlies the diagrams. On the last diagram, force ac is the left upward reaction (at node A). The top of line ac is thus the head of the force arrow representing the reaction. Line c1 on the last diagram represents the force in the left diagonal, and its tail is connected to the head of ac. Thus, it acts downward and to the left. Looking at node A on the truss diagram, we see that that force is pushing on the joint and is thus in compression. On the Maxwell diagram, line 1a represents the force on the lower-left chord member. Its tail is connected to the head of the force in member AE (c1) Looking at node A on the truss diagram, we see that it is pulling on the joint and is therefore in tension. Note that ca, and 1a thus form a closed tip-to-tail polygon, which means that the joint is in equilibrium. The sense of the member forces at other joints can be similarly determined.

Copyright © 2013. Pearson Education Canada. All rights reserved.

3.4

equilibrium of sections

In our discussion of the equilibrium of joints, the elemental portions of the truss defined for equilibrium were the joints themselves, but any portion of a structure must be in a state of equilibrium. It follows that the limits of the portion considered can be extended to a complete subassembly consisting of several joints and members. A consideration of the equilibrium of a larger subassembly can then be used to find unknown bar forces. Considering the equilibrium of a portion of a structure larger than a point is a powerful concept and forms the basis for the analysis and design of many structures other than trusses. Solution of bar forces by this approach is best illustrated by example. Again, consider the truss in Figures 8 and 15. Assume you want to know the forces in members ED, BD, and BC. The truss is first considered to consist of two subassemblies, free-body diagrams that are shown in Figure 15. The internal set of forces shown is developed because of the external loading of the structure. These internal forces maintain the equilibrium of the elemental subassemblies of the truss. The complete force system acting on a subassembly, consisting of internal forces and any external forces present, must form a system whose net force is zero. Translational equilibrium and rotational equilibrium must be considered because the forces acting

Trusses

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 15 Free-body diagrams for solution of forces in members FED, FBD, and FBC by the method of sections.

on the body form a coplanar, but nonconcurrent and nonparallel, force system. Before solving mathematically for the bar forces in the example by applying the equations of statics, it is useful to determine the senses of the member forces by qualitative inspection. Study the equilibrium requirements in the subassembly shown to the left in Figure 15, for example. The force in member BD must act upward as shown to supply the vertical component necessary to balance the difference between the upward reaction of 0.5P and the downward force of P acting on the subassembly (a net of 0.5P acting downward). Thus, the member must be in a state of tension. Because the subassembly to the left must also be in rotational equilibrium, the moment produced by the external forces must be balanced by the moment developed by the internal forces. (See Figure 15.) By summing moments about point B, it can be seen that the sense of the force in member DE must be in the direction shown if moment equilibrium is to occur about point B. Remember that the sum of the moments produced by all forces must be zero about any point. Hence, member DE must be in a state of compression. The forces shown on the left subassembly are equal and opposite on the right subassembly. Because the right subassembly must also be in translatory and rotational equilibrium, the sense of the force in member BC can be found by summing moments about point D. For moment equilibrium to obtain about this point, force FBC must act in the direction shown and so be in a state of tension. Thus, the states of stress in the unknown bar forces can be qualitatively determined. The mathematical process to determine the numerical magnitudes of the forces is similar to the process just described.

Trusses

exaMPle Determine the forces in members DE, BD, and BC of the truss shown in Figure 15. solution: Left Subassembly Translatory equilibrium in the vertical direction: gFy = 0 + c : +0.5P ()*

- " P

reaction RAV in vertical direction

+

external load

F(+)11* BD sin 45° = 0 6 FBD = 0.707P 1tension2

vertical component of force in member BD; the member is assumed to be tension

Member BD is in tension, as assumed, because the sign of FBD is positive. We could next try gFx = 0, but this would involve two unknown forces (FED and FBC), and the equation could not be solved directly. Try to find an equation involving only a single unknown force by considering moment equilibrium about a point. By selecting point B, we make one unknown force, FBC, act through the moment center and consequently fall out of the moment equation (because its moment arm is zero), leaving an equation involving only the remaining unknown force and known external forces. Moment equilibrium about point B: ΣM = 0: ⤺ + B

- 10.5P ()*

reaction DAV

*

0.5L2 ()* moment arm

+

1F" ED

member force

* " h2

= 0 or

moment arm

-0.25PL + FED 10.25L2 = 0

or 0.25PL ()*

=

Copyright © 2013. Pearson Education Canada. All rights reserved.

“applied” moment produced by external forces

F(11)11* ED 10.25L2

6

“resisting” moment developed by internal forces

FED = +P 1compression2

The member is in compression, as assumed. The step where the moment developed by the internal forces is shown equal to that produced by the external forces is not actually necessary. It does, however, represent a fundamentally important way of looking at structural behavior and is important for design purposes. This way of looking at structures will be discussed further in Section 3.8. Now that two of the forces acting on the subassembly are known, the remaining unknown force can readily be found by applying the last unused equation of statics: gFx = 0. Translatory equilibrium in the horizontal direction: S gFx = 0 + :

-FED + FBD cos 45° + FBC = 0 or 6 FBC = +0.5P

-P + 10.707P2 cos 45° + FBC = 0 1tension2

Member BC is in tension, as assumed. All unknown forces acting on the left subassembly have now been found. These forces act in an equal and opposite way on the right subassembly, which, obviously, should also be in equilibrium. It is a good idea to check whether it in fact is. Right Subassembly Moment equilibrium about point D: gMD = 0 ⤺ + :

- 1FBC * 0.25L2 + 10.5P * 0.25L2 = 0

- 10.5P * 0.25L2 + 10.5P * 0.25L2 = 0

Check!

Trusses Translatory equilibrium in the vertical direction: gFy = 0 + c :

- FBD sin 45° + 0.5P = 0 - 0.707P sin 45° + 0.5P = 0

Check!

Translatory equilibrium in the horizontal direction: S gFx = 0 + : - FBC - FBD cos 45° + FED = 0 - 0.5P - 0.707P sin 45° + 1.0P = 0

Check!

The right-hand subassembly is in a state of translational and rotational equilibrium.

The approach just illustrated can be used to find member forces in planar trusses when no more than three unknowns are involved, because there are only three independent equations of statics available for analyzing the equilibrium of planar rigid bodies. Care must be taken in isolating subassemblies so that only three unknowns are present. The decomposition shown in Figure 8(d), for example, is perfectly valid, and each subassembly is indeed in a state of translatory and rotational equilibrium. However, the decomposition is not useful from an analytical viewpoint, because there are more unknown forces acting on each segment than could be solved for by using the basic equations of statics. The descriptive name associated with this approach—the method of sections—stems from one particular way of looking at how subassemblies are separated from the remainder of the truss. An imaginary section line is passed through the truss, dividing it into two segments. In actuality, it does not follow that the section line must be straight. The method of joints could quite easily be thought of in terms of passing a section line around each joint. [See Figure 8(b).] Both can be used interchangeably in analyzing trusses. The method of joints is often considered preferable for use when all the bar forces in a truss must be determined. The method of sections is particularly useful when only a limited number of forces must be found.

exaMPle

Copyright © 2013. Pearson Education Canada. All rights reserved.

Determine the forces in members FG and FC in the truss shown in Figure 16 (the same truss previously analyzed by the method of joints). solution: To determine the force in member FC, the truss is decomposed as shown in Figure 16(b). Point E is selected as the moment center as a matter of convenience, because the unknown

FIGure 16 Truss analysis by the method of sections.

Trusses forces FFG and FCD pass through that point and hence would not enter into the moment equation because their moment arms are zero. gME = 0:

-FFC 1a2 + P1a2 = 0 or FFC = 1.0P

1compression2

To find the force in member FG, it is more convenient to decompose the structure as shown in Figure 16(c) and use point C as a moment center. gMc = 0:

P +FFG 1a2 - a b 12a2 = 0 or FFG = 0.5P 4

1compression2

The results are obviously the same as those obtained by the method of joints, but are found in a considerably more direct way.

exaMPle Determine the forces in members MN, ML, and KL of the truss shown in Figure 17(a).

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 17 Solution of bar forced by the method of sections.

solution: A section line is first passed through these three members and the truss is decomposed into subassemblies. The left subassembly is shown in Figure 17(a). Left Subassembly

gFy = 0 + c : +3.5P - P - P - P - FML sin 45° = 0 6 FML = 0.707 p

gMm = 0:

1compression2

- 13.5P * 3a2 + P12a2 + 1Pa2 + FKL 1a2 = 0 6 FKL = 7.57P 1compression2 S gFx = 0 + : - FKL + FMN - FML cos 45° = 0 - 7.5P + FMN - 0.707P cos 45° = 0 6 FMN = 8.0P

1tension2

Trusses Finding the same forces in the last example by the method of joints would be a long and tedious process, because each joint would have to be considered in turn from one of the ends. The example also illustrates some very important points about forces in a truss. The three-member forces found are clearly dependent on the overall geometry and dimensions of the truss and the loading condition, as well as on the local geometry of the three members themselves. Thus, the force in the top chord (member KL) is directly related to the height of the truss. If the height were doubled, the member force would be halved, and vice versa. The force in the diagonal is also related to the height. Increasing the truss height would decrease the force in the diagonal, because the sine of the angle involved is increased. The force in the lower chord would also decrease as the truss height is increased. It is very important to note that these forces, which are sensitive to the local geometry at the section considered, are independent of many other physical aspects of the truss. Insofar as these three forces are concerned, it makes no difference whether the remainder of the diagonals have the original orientation or an alternative orientation. Indeed, the geometry of the truss at other sections could be drastically altered, and the three forces would not be affected, as long as their local geometry remained constant. The preceding discussion illustrates the power of the method of sections as a conceptual tool. The observations presented could have been made through analysis by the method of joints, but doing so would have been cumbersome.

Copyright © 2013. Pearson Education Canada. All rights reserved.

3.5

shears and Moments in trusses

At this point, it is useful to introduce a particular way of looking at how trusses carry loads, which is of fundamental importance and will prove useful in designing them. This involves studying a truss in terms of sets of externally applied forces and moments and responding sets of internal resisting forces and moments. The following section focuses on applications of these concepts to truss analysis. Consider the truss shown in Figure 18 and the diagram of the portions of the structure to the left of section A–A. Note that, as yet, the diagram in Figure 18(b) is not an equilibrium diagram, because only the set of external forces, consisting of applied loads and the reaction at the support, are shown. Study of this diagram indicates that, insofar as translation in the vertical direction is concerned, the net effect of this set of forces is to produce an upward acting force of 0.5P. [See Figure 18(c).] This net force is often referred to as the external shear force (VE) A study of the rotational effects about A–A produced by the set of external forces reveals that the net effect is a rotational moment of 4Pa. [See Figure 18(c).] This net moment is often referred to as the external bending moment (ME) present at the section. The descriptive term bending comes from the tendency of the net external moment to produce bowing in the whole truss. The function of the set of forces developed internally in members of the truss can now be discussed in terms of the external shear force and bending moment present at the section. For vertical equilibrium to be obtained in the portion of the structure shown, a set of internal forces must be developed in the structure whose net effect is a resisting force that is equal in magnitude, but opposite in sense, to the applied external shear force. This net force is referred to as the internal resisting shear force (VR) In a like manner, the net rotational effect of the set of internal forces must be to provide an internal resisting moment (MR) equal in magnitude, but opposite in sense, to the external bending moment (or applied moment) present on the portion of the structure considered, so that the total rotational moment is zero, as it must be for equilibrium. [See Figure 18(d).] Thus, ME = MR, or ME - MR = 0. In the structure shown (with respect to equilibrium in the vertical direction), the top and bottom chords cannot provide any of the internal resisting shear force necessary to balance the external shear force, because they are horizontal and can have no force components in the vertical direction. The entire resisting shear force

Trusses

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 18 Shears and moments in trusses. These diagrams illustrate the development of shearing forces and moments at a section in a truss through the action of the external force system acting on the truss. A resisting set of shears and moments is developed at the same section through the action of the internal forces developed in the bars.

must be provided by the vertical component of the force in the diagonal member. This way of looking at the structure could also be used as either a way of calculating the force in the member or checking the force found by some other method. With respect to moment equilibrium, there must be a couple 1M2 developed by the set of internal forces to provide an equilibrating moment to the applied external bending moment. As indicated in Figure 18(e), the horizontal forces in the chord members act with the horizontal component of the force in the diagonal to provide the couple that balances the external bending moment.

Trusses Looking at the behavior of structures this way, it is easier to see that the basic structural function of any configuration of bars used in a truss is to provide a resistance to the external shears and moments present. This, then, is the thread of similarity that conceptually ties structures together. The exact way different structures respond to the same loading may vary, but each provides (and must provide) the same internal resisting shears and moments, no matter what truss configuration is used. This way of thinking about the behavior of trusses is also important in a design situation: One must determine the structural configuration that most efficiently provides the necessary set of internal resisting shears and moments for given external shear forces and bending moments.

3.6

statically Indeterminate trusses

In all the trusses previously discussed, it was possible to calculate member forces by applying the equations of statics. Trusses of this type are often referred to as being statically determinate. Other truss structures exist where this is not possible, because of the number of external supports or number of truss bars. Consider the planar truss shown in Figure 19. It is not possible to calculate the forces in the middle members by the equations of statics alone, because there are four unknown bar forces and only three equations of statics available to use in solving for these unknowns. The calculational difficulties are evident when one considers equilibrium in the vertical direction for the left subassembly. Thus, g Fy = 0 + c :

Copyright © 2013. Pearson Education Canada. All rights reserved.

+ 1P>32 - FBE sin u - FFC sin u = 0 or FBE sin u + FFC sin u = P>3

Clearly, this equation cannot be solved. Nor will applying the other equations of equilibrium for a planar rigid body, namely, g Fx = 0 and g M0 = 0, yield equations that could be solved, either alone or simultaneously. Structures of this type are referred to as being statically indeterminate internally. Note that all the principles of statics discussed are still valid and applicable to such structures. For example, it is still true that any elemental portion of the truss is in translational and rotational equilibrium (i.e., g Fx = 0, g Fy = 0, and g M0 = 0 still apply), but some other method must be used to calculate member forces. One way to interpret the physical meaning of a structure being statically indeterminate, however, is that the senses and magnitudes of the internal forces that are present in members of the structure are dependent on the actual physical properties of the members themselves. Stiffer members generally carry greater internal forces than do less stiff members. The matrix analysis procedures discussed in Section 3.10 are frequently used to analyze trusses of this type. The truss can still be simply analyzed, however, if the crossed diagonals are made with cables.

FIGure 19 Truss that is statically indeterminate internally. The magnitudes of the forces present depend on the physical properties of the members.

Because there are four

applying the equations

Trusses

Copyright © 2013. Pearson Education Canada. All rights reserved.

3.7

use of special tensile Members: cables

In all the trusses discussed previously, it was assumed that an individual truss member was capable of carrying the tensile or compressive force generated in the member. Other types of members, however, can be useful. One such member is commonly referred to as a cable, which is capable of resisting tensile forces only. Physically, members of this type are typically thin steel rods or actual wound cables. Such members cannot resist compressive forces, but are often used when a truss member is known from analysis to always be in a state of tension and need never carry compressive forces. Members carrying only tension forces can have much smaller cross sections than those carrying compressive forces and, for that reason, are often considered desirable. A cable could be used wherever analysis has demonstrated that a tension force is present in a member. One of the crucial issues pertaining to using cables as elements in a truss is stability. With cables, the repetitive element in the truss is no longer a basic triangular shape composed of rigid members and inherently stable under any loading condition, but is a special shape, stable only under particular loading conditions. Care must be taken to ensure that the cable member, intended to carry tensile forces only, is never expected to carry compressive forces. Stress reversals of this type would usually cause the whole truss to become unstable. Stress reversals in members are often caused by a change in the location of the external loads applied to the whole structure. With respect to the truss shown in Figure 20(a–1), note that if the load were applied at joint E rather than joint F, the force distribution in the truss would be as shown in Figure 20(b–1). If an attempt were made to use the cable arrangement shown in Figure 20(a–2) to carry a force at E as indicated, the truss would become unstable and collapse under the second loading condition. This is because member BE, previously in tension and thus designed as a cable, would have to be capable of resisting a compressive force when the truss is subjected to its new loading. Clearly, it cannot, and the member would “buckle” out of the way. The truss would consequently become unstable and begin collapsing, as indicated in Figure 20(b–2). Potential instability due to stress reversals is therefore a critical issue in using cables in trusses. If cables are used, it is mandatory to review every possible way that the truss could be loaded, to determine whether a cable member might be subjected to compressive forces. If it might, a cable cannot be used at that location, and a member capable of carrying compressive loads must be used instead. In some specific situations, however, an alternative to designing a member that is capable of carrying either tension or compressive loads is to use additional cables. In the set of diagrams shown in Figure 20, it is evident that if a crossed-cable system is used in the middle [see Figure 20(e)], the truss illustrated is stable under varying load conditions. When the external load is at joint F, the cable between B and E acts in tension and makes the truss stable, while the cable between F and C merely buckles (harmlessly) out of the way and carries no load. The converse is true when the external load is at E. Thus, crossed cables are useful in assuring the stability of trusses under varying load conditions. The same truss appears to be statically indeterminate (as described earlier in this section), but in this case it can still simply be analyzed via basic statics. Any cable is assumed to buckle harmlessly out of the way when it is subjected to compressive tendencies and is thus known a priori to be a zero-force member. That is, the truss reverts to a statically determinate form.

3.8

space trusses

The stability inherent in triangulated patterns of bars is also present when the structure is extended into the third dimension. Whereas the simple triangle is the basic repetitive element in planar trusses, the tetrahedron is the basic repetitive element in three-dimensional trusses. The principles developed for analyzing planar trusses are generally applicable to space trusses. Again, stability is a primary consideration. The presence of

Trusses

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 20 Cables in trusses: effect of changing load conditions and the use of crossed cables. Parts (a) and (c) illustrate trusses that are stable under a particular loading condition. Parts (b) and (d) illustrate the instability that could occur if loading conditions are changed. Part (e) illustrates the use of crossed cables as a device for making a truss using cables that are stable under several loading conditions.

nontriangulated patterns is a sign that each joint in turn should be looked at closely to see whether it will always maintain the same fixed geometric relationship to other joints when a load is applied to any point on the structure. The forces developed in the members of a space truss can be found by considering the equilibrium in space of elemental portions of the space truss. Obviously, the equations of statics that are used must be a full set of equations that are developed for the equilibrium of a rigid body in three dimensions, rather than the simplified set for the equilibrium of a rigid body in two dimensions employed in the previous section on planar truss analysis. The relevant equations are g Fx = 0,

g Fy = 0,

g Fz = 0 and

g Mx = 0,

g My = 0,

g Mz = 0

While straightforward, applying these equations by hand is usually tedious, due to the large number of joints and members that are typically present in a large

Trusses

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 21 Three-dimensional bar configurations.

three-dimensional truss. Accordingly, such an analysis will not be undertaken in this text. Figure 21, however, illustrates its results for a simple configuration. Computer formulations are usually used to analyze more complex configurations. (See Section 3.10.)

3.9

Joint rigidity

Up to now, it has been assumed that the trusses studied were modeled according to the assumptions stated in the opening sections of this chapter. All joints between bars have been taken to be idealized pin connections. In many cases, however, actually making a connection may be neither possible nor desirable. If the actual joint conditions are such that the ends of the bars are not free to rotate, local bending moments may develop in the bars, in addition to any axial loads that are present. If these bending moments were too large, the member would have to be designed so that it would still be safe under the action of the combined axial and bending stresses. The magnitudes of bending stresses typically induced by rigid joints are rarely more than 10 to 20% of the axial stresses that are normally developed in the structure. For preliminary design purposes, these secondary bending stresses are often simply ignored, but at some later point in the design process they would be

Trusses taken into account. One positive effect of increasing joint rigidity is to increase the overall resistance of the truss to deflections. Making joints rigid is rarely an influential factor affecting the general shaping of the final truss.

Copyright © 2013. Pearson Education Canada. All rights reserved.

3.10

computer-aided Methods of analysis

The numerical analysis of multimember trusses by hand calculation can be tedious. Fortunately, a number of analytical methods have been developed for truss analysis that are particularly suitable for use with a computer. Currently, many formulations exist wherein a user can input the geometry of a truss via a graphical interface, specify loads on nodes, and quickly get a range of analytical results, including the magnitudes and directions of forces in all members, as well as nodal displacements. The latter can be used to create overall deflection patterns. Both statically determinate and indeterminate trusses can be analyzed. Typically, the user must specify node locations, the geometry of how members connect to nodes, and the types and directions of loading conditions. Normally, an initial estimate must be made of member sizes and materials. (This information is obviously necessary for indeterminate structures and for calculating truss deflections.) Briefly, two primary analytical approaches that underlie these computerbased formulations are used. One uses what are typically called force methods, wherein member forces are treated as primary unknown values. A brief inspection of the method of joints, for example, reveals that a set of equations could be written for the vertical and horizontal equilibrium of each node of a truss under the action of impinging member forces. This complete set of equations could then be solved to find forces in individual members. Force methods, however, have been largely replaced by what are called displacement or stiffness methods, according to which the actual displacements of the nodes are treated as the primary unknowns. Displacement approaches tend to be more powerful and robust than force approaches. They can be used indistinguishably for determinate and indeterminate trusses. In particular, matrix displacement methods are in wide use and underlie many commonly used analysis programs. The reader should keep in mind that most of these formulations are analytical only; they tell a user not how to design a truss, but rather how to analyze a given configuration. In addition to the particular method of analysis noted, many other analytical approaches are available, including finite-element techniques. There are many implementations as well, each with differing ways of specifying input and output parameters. Figure 22 shows the same truss previously analyzed in Figure 16. A widely available analysis program has been used to create the analytical results shown.1 Input geometry for nodes can be specified by coordinate data or by an interactive graphic interface. Support conditions were defined via various joint restraint options. Member end conditions were specified as pinned. As with most programs, it is then necessary to specify “member sizes” before an analysis can take place. While forces in this particular structure do not depend upon initial size estimates, forces in indeterminate trusses (or any other type of indeterminate structure) do depend upon such estimates. In this simple case, virtually any default size can be specified. In more complex cases, it would be necessary to guess at sizes, analyze the structure, improve size estimates, and repeat the process until satisfactory results are obtained. In any event, starting from any default estimate will point the user in the right direction. Analytical results typically include information on force distributions and deflected shapes (via numerical tabulations and often graphical plots). Different loading cases can be quickly analyzed. Influences of variations in shape can also be quickly analyzed by changing the input geometry.

1

Multiframe, by Formation Design Systems.

Trusses

FIGure 22 Results obtained from a computer-based structural analysis program.

4

Copyright © 2013. Pearson Education Canada. All rights reserved.

4.1

desIGn oF trusses objectives

There are many aspects to the design of trusses. Broad issues include the overall external configuration of a truss, the pattern of its internal triangulation, and attitudes toward the choice of materials and the design of members. Important dimensional variables include the spans and depths of the trusses, the lengths of specific truss members (particularly compression members), the spacing of trusses, and transverse beam spacing (which in turn dictates the way loads are brought onto trusses and, frequently, the placement of nodes within a truss). Broad contextual issues or other design objectives may well force many of these parameters, while others may be directly manipulated. Criteria used to develop designs vary. Structural efficiency objectives are common and are reflected in design procedures that seek to minimize the total amount of material used in a truss in order to support a given loading over a given span while providing adequate safety and stiffness. Overall external configurations and related internal triangulation patterns are manipulated with this objective in mind. Internal triangulation patterns are similarly affected. Truss depths are surely an important variable in minimizing volume requirements. Member design attitudes are also affected. Each individual truss member might be sized precisely to carry its internal force. Because forces vary among members, so, then, will member sizes. Rods or cables might well be used. Member design attitudes might also affect overall configurations and internal patterns (e.g., efficient patterns may be developed

Trusses that minimize the length of long compression members and maximize the length of tension elements). Alternatively, design criteria might be based on construction efficiency considerations related to the fabrication and making of a truss. Responding to these objectives frequently leads to trusses with simple external configurations (often, parallel upper and lower chords or simple gabled forms). Equally simple internal triangulation patterns are used as well, with the objective of making all members the same length. Member design attitudes are also affected: Instead of making different truss members different sizes in response to the varying forces that are present, the maximum force present anywhere might be identified and used as a basis for sizing all members throughout the structure (even when any forces that are present are less than the size the structure can carry), simply because it might be easier to fabricate a truss in this way. Having all members be identical surely makes connecting joints easier than when adjacent members are differently sized and shaped. A common variant of this attitude is to use three different sizes: one for the upper chord, another for the lower chord, and a third for interstitial members. Although other design objectives could be noted as well, and although the characterization just presented is somewhat simplistic, structural efficiency versus constructional efficiency objectives provide a useful point of comparison in subsequent discussions. Often, these objectives are conflicting in the sense that they lead to quite different design responses.

4.2

configurations

Copyright © 2013. Pearson Education Canada. All rights reserved.

Probably the single question most asked in a design context is, “What shape should it be?” Unfortunately, there is no easy answer to this question. Several common configurations are shown in Figure 23. External configurations vary, as do internal patterns. Configurations are influenced by both broad external factors and direct structural or construction considerations. Different configurations are useful for different purposes and may be suitable for different span lengths or loading conditions. external Factors. External factors are not immediately a concern of this book, but should still be noted, because they often are primary influences on truss configurations. For example, the shapes shown in Figures 23(c) and (d) are largely responses to external factors. The common pitched roof is ingrained in our building traditions for a number of very good reasons, and roof trusses are shaped accordingly. The northlight truss shown in Figure 23(y) is commonly configured so that light may be admitted through upper zones into spaces below. The scissors truss shown in Figure 23(w) is often used to open up space beneath. Trusses of many different shapes may be made to work, but this does not mean that they are particularly efficient or attractive from a structural or constructional point of view. The scissors truss noted, for example, is quite susceptible to large deflections (its actual structural depth is quite shallow), and member sizes must be quite large if the truss is to work for anything other than minor spans. Basic Forms. From both structural and constructional perspectives, the parabolically shaped truss and the parallel chord truss shown in Figure 23 provide useful discussion referents. Assume that they are subjected to a series of identical point loads across their tops. A useful way of discussing the two configurations is through a consideration of the external shears and moments that are developed in them in response to external loadings. (See Section 3.5.) Internal forces in truss members are developed in response to these external shears and moments as shown in Figure 24. Maximum bending typically occurs at the center of a simply supported truss with uniform loads and decreases toward the ends, whereas maximum shear occurs at either end and decreases toward the center. Upper and lower chord members often carry primarily (but not exclusively) moments, and forces are

Copyright © 2013. Pearson Education Canada. All rights reserved.

Trusses

FIGure 23

Common types of trusses.

developed within them in direct proportion to the moment that is present and the height of the structure. (During a given moment, a shallow structure will undergo high forces, and vice versa.) Typically, Mexternal = Cd = Td, where C and T are the gross horizontal internal forces that are present and d is the depth of the truss. External shears are resisted by vertical components in inclined members. Parallel chord trusses. In a parallel chord truss, external moments are largely  resisted by couples directly formed by the forces in the upper and lower

Trusses

Shear forces are

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 24 Shaping of trusses based on internal shears and moments.

chords: Mexternal = Cd = Td. Because the truss depth d is constant, C and T must vary with different values of Mexternal, as shown in Figure 24. Thus, the forces in chord members generally vary as shown. The external shear forces are carried exclusively by the diagonals, because the chords are horizontal and can contribute no vertically acting resisting forces. The forces in the diagonals then must generally vary as does the external shear force. The ensuring wide variation in member forces influences member design attitudes, because members must either be designed in response to specific force levels that are present or be designed on the basis of maximum values. (See Section 4.4.) Different diagonal organizations are possible, with different sloping patterns yielding different force states within them. In order to make the most efficient use of tension or compression elements, a typical objective might be to try to cause as many as possible of the longest members of a truss to be in tension rather than compression. This can often be done by paying careful attention to how certain members are oriented in the truss. Figure 26 shows an example of a truss with tension diagonals. Parallel-chord trusses having bar patterns of the type illustrated in Figure 25(a) invariably have all diagonals in tension (for the loading shown), whereas trusses having patterns of the type illustrated in Figure 25(f) have all interior vertical members in tension while diagonals are in compression. The reversal of the direction of diagonals at midspan in such trusses is characteristic of designs for symmetrical loadings. Pratt and Howe were nineteenth-century bridge designers who developed and popularized the forms shown.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Trusses

FIGure 25 Parallel chord trusses: force distributions, use of cables, and effects of changing load conditions.

Whether cables can be used for tension elements in these trusses depends largely on whether the loading condition on the truss is invariant. In the diagrams shown, it is obviously suggested that cables could be suitable for use as tension elements. If members of this type are used, extreme care should be taken to ensure that these tensile members are never expected to carry compressive forces. As previously noted, stress reversals of this type could cause the truss to become unstable. Stress reversals are usually associated with a change in the pattern of external loads carried

Trusses

FIGure 26 Parallel chord truss with cables for the diagonal members. Typical force characteristics are shown.

C C

C T

C

C

T

C

T C

T

T

Copyright © 2013. Pearson Education Canada. All rights reserved.

T

by the truss. Consider what would happen to the same two trusses under discussion if they were expected to carry the single load shown, a case that might occur if the other loads were simply removed. Analysis of the member forces due to these loads results in the distributions shown. Clearly, some of the members previously in tension are now in compression. If a cable were used in members previously in tension, but now in compression, the trusses would become unstable and collapse as indicated. This illustrates a fundamental design consideration: Each element in a structure must be designed to carry the forces that might develop in it under any possible loading condition. Satisfying this requirement could involve a large number of different load combinations. The trusses shown can be made stable for the new loading condition by using members that are capable of carrying the compressive forces indicated, rather than using cables. Alternatively, the addition of crossed cables would cause the truss shown in Figure 25(b) to be stable. The reader should carefully inspect the truss shown in Figure 25(g) to see if the addition of a crossed cable leads to stability. Clearly, it does not. For this reason, trusses having this bar pattern rarely use cables, or they are used only when the nature of the external loads are so predictable that stress reversals need not be feared. This is sometimes the case when the dead load carried by a truss significantly exceeds the live load or when there is no live load at all, because the former is usually fixed and determinable while live loads are not. It is useful to highlight briefly two other general types of bar configuration often encountered in a constant-depth truss. The first is an internally statically indeterminate configuration, shown in Figure 27(a). An approximate force distribution for the loading indicated, based on the assumption that the diagonals are of similar stiffnesses and share equally in carrying the shear forces that are present, is also shown. The reader should study this truss closely and determine whether it is indeed stable under loading conditions other than the one illustrated. (It is, but being able to explain why is of crucial importance in gaining a thorough understanding of the structural behavior of trusses.) A second typical type of parallel chord truss has all the diagonals oriented in one direction [Figure 27(b)]. From a study of the previous diagonal organizations, it can be seen that in such cases diagonals on one-half of the structure are in compression and those in the other half are in tension. It makes little sense to use cables for diagonals in such a situation.

Trusses

FIGure 27 Force distributions in parallel chord trusses using rigid diagonals that are capable of carrying either tension or compression forces.

Copyright © 2013. Pearson Education Canada. All rights reserved.

N

Funicularly shaped trusses. Shaped trusses of the type shown at the right in Figure 24 are often quite efficient in a structural sense. These trusses are funicular shaped for the loading, reflecting the shape assumed by a flexible cable under the same loading (albeit inversely). The configuration responds to the magnitudes of the shears and moments that are present. As the bending moment varies from a maximum at the center to a minimum at the ends, the depth of the structure varies as well. (See Figure 28.) Nodes are consequently at varying heights and connecting members are sloped. This shaping leads to forces of more or less similar magnitude being developed in upper and lower chord members. (Forces in upper members are slightly higher, and not quite constant, due to their sloping.) Interestingly, if the structure is shaped precisely, the sloped chord members may also carry the shear forces involved via their vertical components. Note that the slope of the chord members increases toward the ends of the structure, where the shear forces are highest. Vertical resisting components in sloped members are thus highest as well at these points. Interstitial members carry only small forces. In fact, if shaped precisely, all interior members turn out to be zero-force members, no matter what specific internal triangulation pattern is used. Figure 28 illustrates variations on the truss discussed. Conceptually, the zeroforce members could simply be removed to form a nontriangulated configuration without changing the ability of the structure to carry design loads to the ground. Pinned structures without diagonals are, however, of dubious practicality, because they are stable only under the exact loading patterns shown. If the loading pattern were changed in any way (e.g., if even one of the loads were removed or its magnitude altered), the configurations would no longer be stable and would collapse. This is because the required shear and moment resistance at the different locations is now no longer the same as that for which the structure was originally designed and the structure cannot intrinsically satisfy these new requirements as it did for the design loading. The same overall configuration with diagonals, however, is stable under not only the design loading, but any other loading condition. A function of the interstitial members is to carry forces generated by other nonuniform loads the structure must occasionally bear. If the loads were suspended from the lower chord panel points instead of being applied to the upper chord, the vertical interior members would serve as suspension rods transferring the applied loads to the upper chord members. One final transformation of the truss shown in Figure 28(a) is of interest. The lens-shaped structure shown in Figure 28(c) has the same structural depth at all sections as the original structure. Its outer configuration is clearly part of the

Trusses

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 28 Funicular trusses: transformations of a funicular truss into related forms.

same family of shapes as that present in the original truss. An analysis of this truss would reveal that the interstitial diagonals are zero-force members and thus serve only the function of stabilizing the assembly under variant loading conditions. The verticals transfer loads such that the upper and lower chord members are similarly loaded (a condition that must be met for the similarity of shape to be correct). Instead of conceiving of this lens-shaped structure as a special form of truss, it could equally well be conceived of as a combination arch-cable structure. In this conception, the outward thrusts of the upper arch are balanced by the inward

Trusses

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 29 Lenticular truss: the Smithfield Stress Bridge across the Monongahela River in Pittsburgh, Pennsylvania (circa 1883).

pulls of the cable, with the consequence that there is no net lateral force present at the reaction, a result long acknowledged to be advantageous in foundation design. The lens-shaped form was commonly applied to many lenticular trusses built in the last century for use as bridges. (See Figure 29.) Additional diagonal bracing was typically used to allow the structures to carry nonuniform loading patterns. Funicular-shaped trusses are interesting. Obviously, what is happening in these trusses is that the specific members organized along the lines of the funicular shape for the loading involved have become the dominant (actually, the only) mechanism for transferring the external loads to the supports. Other members are zero-force members serving a bracing function at most. A remarkably simple load-carrying action is thus displayed by these trusses. In order for the rather remarkable load-carrying actions described to occur, it is necessary that the basic shape of the structure correspond exactly to the funicular shape. Relative heights of the interior points on the trusses shown in Figures 28, 29, and 30, for example, must be determined exactly. The relative heights are clearly a function of the loads and their locations. In general, the external moments present at different sections are calculated first. A maximum depth dmax is selected, and corresponding maximum forces in chords are calculated: T max = C max = M max >d max. The depth dx of the structure at other sections is then varied directly in response to the moment Mx, on the assumption that the horizontal components of the chord forces are maintained constant: dx = Mx >T max = Mx >C max . By knowing the different structural depths, member slopes can be found and final resultant forces in members calculated on the basis of known slopes and horizontal components. The vertical components of these forces will be found to balance the external shear forces present at the sections. Because different loading conditions lead to different overall configurations (as can be appreciated from the deflected string method of arriving at funicular

Trusses

FIGure 30 Funicular trusses: Trusses based on the funicular shape for the applied loading demonstrate a uniquely simple load-carrying action.

Copyright © 2013. Pearson Education Canada. All rights reserved.

.

shapes), a shape that is funicular for one loading will not be funicular for any other. This implies that forces are developed in interior members of trusses designed to be funicular for one loading when nondesign loads are present. The load-carrying action would again be more complex in such a case. Comparing a truss not having an overall funicular shape for the loading involved with a funicular-shaped one (see Figure 30) reveals that deviating from the funicular shape not only serves to make the total member length longer, but subjects more members to primary forces (so that they would have to be designed as loadcarrying, rather than zero-force, members). The consequence is that a strategy of using a funicular shape for the overall envelope of a truss may be more likely to lead to a lighter weight or minimum-volume solution than using a nonfunicular shape would. other special shapes. Other truss configurations that have been specially shaped to carry loads in basically a funicular way are illustrated in Figure 23. Under primary design loadings, the diagonals in these forms are zero-force members, but they are obviously needed for stability under varying loading conditions if joints are pin connected. Some actual structures reflecting such forms have been made without the aid of diagonals by using rigid, rather than pinned, connections and can carry off-balanced loads via bending in members. (Strictly speaking, these then become rigid-frame structures rather than trusses.)

Trusses Figure 23(r) is funicular shaped in response to the moment diagram, but the forces naturally vary slightly along the lower chord. Figure 23(s) seeks constant forces along the lower chord, which can be accomplished by inclining the uprights in a precise way. For sloped roofs, the shape shown in Figure 23(e) seeks to maintain similar forces in upper and lower chords. (Robert Maillart used a shape such as this in his building at Chiasso in 1924.) The cantilevered form shown in Figure 23(x) seeks to maintain similar and nearly constant forces in opposing chords. (This form is obviously reminiscent of the Eiffel Tower, shaped primarily by wind loads.)

Copyright © 2013. Pearson Education UK. All rights reserved.

4.3

depths of trusses

One of the most frequently encountered questions during the design of a truss is how deep it should be—that is, whether there is an optimal depth for a truss. It would seem that this should be a relatively easy design question to answer, but it is, on the contrary, one of the more difficult. As already noted in Section 3.4, the member forces, and hence volumes, generated in a truss by an external load are directly dependent on the dimensions, including the height (d), of the truss. As can be appreciated, determining an optimum height that minimizes the total volume of a truss is no easy matter, but the process is conceptually straightforward. In practice, truss heights will depend on many nonstructural factors as well— such as clear heights that need to be maintained, as well as visual and spatial relationships between the truss and other elements of the structure, to just name a few. For each truss member, an expression for the volume, based on the member forces written as a function of the variable height d, is first determined, and an expression for the total volume of the entire set of members is obtained. This expression is then minimized with respect to the variable height. In general, the optimization process will reveal that trusses which are relatively deep in relation to their span are most efficient and that shallow trusses are less so. Angles formed by the diagonals with respect to the horizontal are typically from 30 to 60°, with 45° often a good choice for determining triangulization geometries. Over the years, rules of thumb have evolved that help estimate what the depth of a truss should be. Trusses that carry relatively light loads and are closely spaced often have depths approximately 1>20 of their span (e.g., roof load-transfer members). Secondary collector trusses that carry reactions produced by load-transfer members are usually deeper, because the loads carried are larger. The depths of these trusses can be on the order of 1>10 of their span. Primary collector trusses, which support huge loads (e.g., a truss carrying the column loads from a multistory building over a clear span on the ground floor), are usually very deep and are often made equal to the depth of a story, say, 1>4 or 1>5 of the span of the truss. Rules of thumb of this nature should not be taken as final truths. Some of the best and most efficient uses of trusses in building contexts have occurred when heights varied drastically from the preceding ratios. Still, when one is in doubt, they are a place to start. Figure 31 shows an example of a primary collector truss with a span-to-depth ratio of 1>3 for a cantilevering part.

4.4

Member design Issues

critical loadings. The selection of physical members for use in a given truss configuration is one of the more straightforward aspects of truss design. The crucial issue is predicting what force a member should be designed to carry. In earlier sections, it was shown that the nature and magnitude of the force present in a particular member are dependent on the specific loading condition on the whole truss. It is thus necessary to consider the whole spectrum of possible loading conditions and analyze the forces present in each of the members under each of the conditions. A particular member would then be designed in response to the maximum

Trusses

Institute for Contemporary Art, Boston Architect: Diller, Scofidio & Renfro Structural Engineer: Ove Arup & Partner 41'

60'

Image: Courtesy of James Moorhead

Loads (facade loads not included): Roof: Dead Load 160 Ib/ft2 Live Load 30 Ib/ft2 _________________________

The cantilevering truss marked in bold is one of four trusses that carry the upper level of the building. Secondary trusses span between the primary ones. For the simplified analysis shown, the total load is modeled as concentrated at the nodes. In the real building, the secondary trusses load the primary truss members also in between the nodes, thus generating bending moments and shearing forces in addition to the axial forces shown below. For the critically important cantilever, the span to depth ratio is approximately L /3.

190 Ib/ft2 Floor: Dead Load 160 Ib/ft2 Live Load 80 Ib/ft2 ______________________ 240 Ib/ft2 Truss Reactions: Moment equilibrium around support A or B RA = 1,292 k RB = 236 k For full dead and live load, there is no uplift at support B!

Truss geometry and loads

11.5 22 K 69 K

98 K

104 K

6 x 23' 110 K 116 K

122 K

11.5 96 K 33 K

23'

57 K

121 K

A

128 K

143 K

69'

B

151 K 158 K

92'

Analysis steps: Identify tension, compression, and zero-force members in the truss. The arch and cable analogy cannot be used because of the cantilever condition. In a first study, it would be interesting to identify the members with the largest force and size them provisionally. This initial study indicates the feasibility of the truss for the given loads. Overall deflections and members need to be checked in later steps. The tension member with the largest force is the upper chord member over support A. Use the method of section to find the member force. Pass a vertical section through support A and sum up the moments around A:

((57k  22k)69ft)  (69k57.5ft)  (98k34.5ft) 23ft

F=

(104k11.5ft)(121k46ft)(128k23ft)



23ft

Copyright © 2013. Pearson Education UK. All rights reserved.

= 935k

Deflections with undeformed shape superimposed

Member with the largest tension force F = 935 k Force characteristics and reactions

0

T

C T C C

T T C T C

T C C

T C

T C

Relative magnitudes of member forces

C

0

C

C T

C T

T

T

T CC T RB = 236 K

RA = 1,292 K

(critical) force possible. Consequently, individual members quite often are designed in response to different loading conditions (e.g., the diagonals of a truss may be designed for one external loading condition and the chord members for another). Only in instances where the loading condition is invariant are all truss members designed in response to the forces generated by the same load.

FIGure 31 Example of a cantilevering truss.

Trusses

Copyright © 2013. Pearson Education UK. All rights reserved.

Member design. Once the critical force in a member has been found, the problem becomes one of choosing an appropriate material and a set of cross-sectional dimensions for a member of known length having simple pin-ended connections and subjected to a known tension or compressive force. There is usually no great difficulty in designing the member. There are two alternative design methods for steel and timber tension members, Allowable Strength Design (ASD) and Load and Resistance Factor Design (LRFD). Of primary importance is that members carrying only tension forces can usually be designed to have much smaller cross sections than those carrying compressive forces of a similar magnitude. The cross-sectional area of a tension member is only dependent on the member force present and the strength of the material used. It is independent of the member length. For tension members the fundamental formula area required = tension force>stress is adapted to the design procedure chosen (ASD or LRFD). For members in compression, it is necessary to take into account the possibility of a buckling failure, which can occur in long members subjected to compressive forces. In long members, the load-carrying capacity of the compressive member is inversely proportional to the square of the member’s length. If compressive members are relatively short and below a certain maximum length, buckling is not a problem, and the cross-sectional area is then directly dependent only on the magnitude of the force involved and the crushing stress of the material. Stresses in compression are independent of the specific length of the element. The general implication is that long compression members that may be subject to buckling require a greater crosssectional area to support a given compressive load than would a short member that is not subject to buckling. In truss design, either members in tension or short compressive members are considered more desirable than long compressive members, because less material is required to carry a given force. Thus, a truss design objective is often used to create a situation in which as many members as possible carry only tension forces or are short compressive members, because in such a situation less total material is required to support the given external loads. Techniques for doing this will be discussed shortly. constant- Versus Variable-size Members. The relative sizes of members, designed in response to the forces that are actually present in two typical trusses are shown diagrammatically in Figure 24. Obviously, the illustration is not meant to represent real members, but is intended to convey a graphic sense of how a material is distributed over a structure. Instead of varying the size of each individual member in response to specific forces, it may be more convenient or less costly to make several pieces (e.g., the entire top or bottom chord) out of constant-cross-sectional continuous members. Whether to do so is a design decision that must be evaluated in light of other factors. When the top chord is designed to be a continuous member of a constant cross section, for example, the cross section used must be designed to carry the maximum force present in the top chord. Because this would usually occur only in one segmental portion of the entire top chord, this same cross section would be excessively large for the remaining segments and so potentially inefficient from a material’s use viewpoint. Whether to size each truss member in exact response to the forces present or to base all member sizes on certain critical elements is not an easy choice to make. The latter is often done when a large number of trusses are used repetitively, spaced closely together, and designed to carry relatively light loads (e.g., mass-produced bar joists). The members of major trusses carrying large dead loads are often designed more in response to the actual forces carried.

Trusses

Copyright © 2013. Pearson Education UK. All rights reserved.

Buckling considerations: effects on Patterns. The dependency of the load-carrying capacity of a compressive member on its length and the associated design objective of trying to make such members relatively short often influence the pattern of triangulation used. Consider the truss shown in Figure 32(a) and the force distribution in its members, shown in Figure 32(b). Assuming that the members in compression are relatively long members, Figure 32(a) also indicates how the compression members might buckle. (The directions of buckle shown are arbitrary, because there is no way of predicting in which direction the member would actually buckle.) The tension members would exhibit no such tendency to buckle. Note that the top chord members are the longest in the truss and carry relatively high compressive forces. Members could undoubtedly be sized to carry these forces, but with a few relatively minor alterations to the bar pattern, it is possible to reduce the total amount of material required in the top chord below what would be required for the configuration shown in Figure 32(a). This can be done by locating those members which will serve as bracing for these top chords. There are four zero-force members in the original configuration. A close look at the truss reveals that these members are not necessary to the stability of the truss (as some zero-force members occasionally are) and could be removed. Or they could be relocated as shown in Figure 32(c). They would now serve as midpoint bracings, thereby effectively halving the effective lengths of top chord members. This type of bracing has an enormous influence on the load-carrying capacity of the top chord members. Recall that halving the length of a compressive member increases the load required to cause buckling by a factor of four. A much smaller member could be used to support the axial load shown than would be the case in the original configuration. This is satisfactory, because the cross-sectional areas of tension members required to carry given loads are not dependent on member length. Short compression members and longer tension members often characterize trusses that have been designed with care. This principle is often the design determinant underlying the configurations of many commonly used trusses. It might be thought that, with the preceding approach, the bracing members would have to be very large to serve their function. Usually this is not so, because there is relatively little force exerted on a brace member, so they can be quite small. Sizing these members exactly is a difficult problem beyond the scope of this book. Often, they are simply given a certain minimum size that is empirically known to be adequate, or, more likely, those same members carry some relatively large load when the external loading pattern on the truss changes. Sizes for other loads are usually more than adequate when the member serves only a bracing function.

FIGure 32 Member buckling: relation to triangularization pattern.

Trusses effect of lateral Buckling on Member design. The preceding discussion treated only the behavior of a truss in its own plane. Even in a planar structure, however, the out-of-plane direction is extremely important. Consider the planar truss shown in Figure 33, and assume that the top chords are made of members symmetrical about both the y- and z-axes. A square or round member (including a pipe) would exhibit a symmetry of this type. More precisely, the members of interest are those having equal moments of inertia about both axes, and they need not be geometrically symmetrical. If a top chord of this type carries an axial compressive load, it is equally apt to buckle in the horizontal xz-plane. This implies that a crucial design issue is to ensure that this mode of buckling does not adversely affect the load-carrying capacity of the truss. Assume that transverse beams are used to carry the loads from the decking to the truss, as shown in Figure 33(b). These beams inherently brace the top chord in the horizontal plane at the points of attachment. Because the chords are braced only at those points, it is still possible for the top chords to buckle in the horizontal plane, as shown in Figure 33(b). The effective lengths of the top chord members with respect to buckling is then 2a, not just a. Without transverse bracings in the horizontal plane at other points, minimizing top chord length in the vertical plane to gain design advantages is an exercise in futility, as the members will merely buckle in the horizontal plane. Members in the vertical plane do nothing to prevent this type of buckling. From a design point of view, it is necessary to brace the truss as shown in Figure 33(a). This design makes the effective length of the top chords the same in both directions. An alternative would be to change the cross-sectional geometry of the top chord members, making them more inherently resistive to buckling in the horizontal plane. This approach is diagrammatically illustrated in Figure 33(b), in which the member is made stiffer in the horizontal (xz) plane. Using various techniques, one can design member characteristics so that the potential for buckling in the vertical plane (about the weak axis of the member shown, but with an effective member length of a) is exactly identical to that for buckling in the horizontal plane (about the stronger axis of the member, but with an effective member length of 2a). Such a member is diagrammatically

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 33 Lateral buckling of truss members: use of transverse members for bracing.

Trusses

FIGure 34 Lateral buckling of planar truss without any bracing.

Copyright © 2013. Pearson Education UK. All rights reserved.

illustrated. Many member configurations could provide the necessary balance in stiffnesses. Which of the two approaches just discussed is preferable for controlling the potential for member buckling in the horizontal plane depends on many other design considerations. Of primary concern would be the effect of the transverse beam spacing in the design of the roof decking. One spacing or the other might prove to be more desirable in this respect. A detailed look at the design of the decking is necessary to answer this question. effects of lateral Buckling of Bar assemblies. While there is usually the possibility for bracing the top chord in the horizontal plane, it is possible to have a truss that is essentially freestanding (Figure 34). In structures of this type, there is, in addition to the possibility that individual members of the type discussed previously will buckle, the possibility that very large portions of the structure may buckle laterally. This type of buckling, which often makes the use of freestanding trusses undesirable, is illustrated in Figure 34. In the truss shown, all of the top chord members are in a state of compression, so the entire top of the truss can buckle as indicated. Preventing this phenomenon is no small design task. The whole length of the top of the truss can be thought of as the effective length of a long, composite compression member. The type of bracing provided when transverse beams are attached to the top chord, however, generally prevents this type of buckling. When trusses must be freestanding and hence cannot be braced by transverse members, it is necessary to make the top chord of the structure sufficiently stiff in the transverse direction to resist lateral buckling. This can be done by increasing the lateral dimension of the top chord or by adopting a three-dimensional triangular pattern (Figure 35).

4.5

FIGure 35 Preventing lateral buckling of bar assemblies in freestanding trusses.

Width

Planar Versus three-dimensional trusses

A number of issues should be considered in making a decision to use either a planar or a three-dimensional structure. Which uses a greater amount of material to support a given load in space, or are planar and three-dimensional structures similar in this respect? Quantitative resolution of this very complex question is unfortunately beyond the scope of this book. Note, however, that nothing a priori suggests that three-dimensional trusses necessarily use less material for equivalent loads and spans than do planar trusses. In fact, it can be demonstrated that when trusses are

Sloped

Trusses

FIGure 36 Deflections and axial forces in a three-dimensional warren truss. Compression forces are shown dark gray; tension forces are shown light gray.

Copyright © 2013. Pearson Education UK. All rights reserved.

(a) Loads onto the joints of the upper chord split into components aligned with each inclined truss side. The interstitial members in the plane of the lower chord remain zero force members. Horizontal as well as vertical reactions develop.

(b) Loads onto the joints of the lower chord shorten the distance between the outer lower chord members. The short interstitial members in the plane of the lower chord prevent this action through their internal compression force.

(c) Assymmetrical loads onto one side result in torsion. One truss side shows the typical force pattern for gravity loads, and the lower horizontal truss plane resists the tendency of the truss to twist.

used as one-way spanning elements, a planar truss usually requires less volume of truss material than does a comparable three-dimensional truss serving the same function. Thus, when trusses are used as one-way spanning elements, a series of planar trusses is often preferable to a series of three-dimensional trusses, particularly when they are used on the interior of a building and lateral bracing is intrinsically provided by the roof framing system or some other element. Three-dimensional configurations often prove more efficient when trusses are used to form two-way systems. Three-dimensional trusses also prove efficient when the trusses are used in a freestanding way (without transverse beams framing into their top chords). In these cases, the forms are inherently resistant to simple lateral overturning. Their compression zones are also naturally resistant to lateral buckling of whole bar assemblies. The top plane of the three-dimensional truss shown in Figure 35(b) inherently provides a strong resistance to lateral buckling and is one of the primary reasons such configurations are often used. A truss of the type shown in Figure 35(c) also inherently provides a measure of bracing against this phenomenon, owing to the orientation of the diagonals. In these two cases, the resistance to lateral buckling provided by three-dimensional structures is largely dependent on the spacing of members in the third dimension, with a larger spacing more desirable than a very small one. The determination of an optimum separation is a difficult problem that is beyond the scope of the book. The spacings illustrated are, however, quite reasonable. On occasion a three-dimensional truss may be chosen because of its inherent capability to resist torsional effects as may result from asymmetrical loadings. Figure 36 shows a three-dimensional Warren truss with a variety of loading conditions. Doing these types of analyses for three-dimensional systems can be lengthy. In practice engineers are typically using computer programs to analyze those systems.

Trusses

QuestIons See Figure 37. 1. For the truss shown in Figure 37(Q1), quantitatively determine the magnitudes of the forces that are present in all of the truss members. A method-of-joints approach is suggested. Answer: FAH = FFE = 70.7 compression, FAB = FBC = FCD = FDE = FFD = FHB = 50 tension, FHC = FGC = FFC = 0, and FFG = FGH = 50 compression. 2. For the truss shown in Figure 37(Q2), qualitatively determine the nature of the force that is present in each of the members. Numerical values are not required. 3. For the truss shown in Figure 37(Q2), quantitatively determine the magnitudes of the forces that are present in all the truss members. A method-of-joints approach is suggested. 4. For the truss shown in Figure 37(Q4), quantitatively determine the magnitudes of the forces that are present in all the truss members. Use a method-of-joints approach. 5. For the truss shown in Figure 37(Q5), quantitatively determine the magnitudes of the forces that are present in all the truss members. A method-of-joints approach is suggested. 6. Determine the force in member GH in the truss shown in Figure 37(Q4) by using a method-of-sections approach. Answer: FGH = 4P. 7. For the truss shown in Figure 37(Q7), quantitatively determine the magnitudes of the forces that are present in all the truss members. A method-of-joints approach is suggested. 8. Quantitatively determine the forces in members JI, JE, and ED in the truss shown in Figure 37(Q5) by a method-of-sections approach. How would the force in member JI be affected if the overall depth of the structure were doubled? How would the forces in members JI and JE be affected if the depth of the truss were continually decreased until it approached zero? 9. Consider the parallel chord truss shown in Figure 37(Q9). Obviously, the truss needs diagonal elements for stability. Add the diagonal elements in an arrangement such that all the diagonals are in a state of tension under the loading condition indicated. 10. Draw a diagram of the relative sizes of the individual members of the truss shown in Figure 37(Q5). Assume that cables are used for tension elements.

Copyright © 2013. Pearson Education UK. All rights reserved.

11. Qualitatively determine the force characteristics for the truss shown in Figure 37(Q11). Additionally, identify the members in the upper and lower cords with the largest tension and compression forces. Use the method-of-sections to determine the magnitude of forces in those members. 12. A parallel chord truss with a depth of 7 ft is loaded with five equal point loads. Each point load P weights 2 k. The vertical members of the truss are evenly spaced at 8 ft. Please find the force characteristics (qualitatively) in the truss. Additionally, determine the force magnitudes in all truss members. 13. Find the qualitative force characteristics (compression, tension, or zero force) for the truss in Figure 37(Q13). Also find the magnitudes of forces in members AB, CE, and DE. 14. Determine the force characteristics and force magnitudes of each member of the truss in Figure 37(Q14). 15. Assume that the lenticular truss shown in Figure 37(Q15) is funicularly shaped. Please find the correct depth of the truss at the vertical members BH and CI. All loads P are equal. 16. The triangulated supports of the Exhibition Hall in Hannover act as a trussed support for the suspended roof. Analyze a typical support structure and qualitatively determine all force characteristics. Also determine the force magnitudes in member AE using the method of joints.

Trusses

FIGure 37

Problems.

15' Q11

15'

15'

15'

8'

C

15'

2k 8'

D

2k 8'

E

2k 8'

F

2k 8'

G

2k 8'

H

I

7' N

A

L

M

2k

2k

2k

2k

2k

40k

D

B

J

K

B

30k

20k

F

Q13

6' A

E

C 5'

5'

10' Copyright © 2013. Pearson Education UK. All rights reserved.

15'

12'

2k

Q12

15'

5'

10'

5'

10'

B

A

5'

C 10'

Q14 E

D

F

50k a

Q15

a =10' =P

a

a

a

D

C

B

A

a

a

E

F G

H

I

K

L

M

8'

Trusses

FIGure 37 Continued A  Q16 

B

E

D

C

B

A

G Q17

6' F

C A

E

F P P = 20 k

8'

8'

D E

8'

D

C

8' A

10k

30k

B

17. Please determine the magnitude of force in member AG of the truss shown in Figure 37(Q17). 18. Use a computer analysis program to determine the force characteristics in all members of the truss shown in Figure 37(Q18). Step 1: Assign the Same cross section to all members, and use pinned connections throughout. Step 2: Change the cross section of the diagonals in the bay on the left, significantly increasing their stiffness. Comment on your observations and include a discussion of force distribution and deflections.

Copyright © 2013. Pearson Education UK. All rights reserved.

H F P

P

D

8'

A

C

B

6'

Q18

G

H

5' E

Copyright © 2013. Pearson Education UK. All rights reserved.

Copyright © 2013. Pearson Education UK. All rights reserved.

AnAlysis And design of structurAl elements design endeavors may be classified into first-, second-, and third-order activities, generally characterizing the sequence in which design decisions are encountered and must be made. first- and second-order activities deal with establishing relations between broader design intents, specific morphological characteristics of the building, and the type and attributes of different structural systems (materials, systems, structural hierarchies, bay geometries). third-order activities focus on the design of members and connections within the constraints previously established by earlier decisions. A thorough knowledge of the analysis and design of such members as arches, cables, beams, columns, trusses and frames, plates, shells, and membranes is a necessary precursor to understanding first- and second-order design activities; for this reason, the analysis and design of elements are discussed first, even though the design process does not proceed in that sequence.

From Part II of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Copyright © 2013. Pearson Education UK. All rights reserved.

Funicular Structures: Cables and Arches

Copyright © 2013. Pearson Education UK. All rights reserved.

1

IntroductIon to FunIcular StructureS

Few structures have consistently appealed to the imagination of builders as have the hanging cable and the arch. These two apparently different structures share some fundamental characteristics that cause them to be more closely related than might initially appear, particularly in terms of their basic structural behavior. A cable subjected to external loads will deform in a way that is dependent on the magnitude and location of the external forces. The form acquired is often called the funicular shape of the cable. (The term funicular is derived from the Latin word for rope.) Only tension forces will be developed in the cable. Inverting the structural form obtained yields a new structure that is analogous to the cable structure, except that compression rather than tension forces are developed. Theoretically, the shape found could be constructed of nonrigidly connected stacked elements (a compression chain), and the resulting structure would be stable. Any slight variation in the applied loading, however, would cause the structure to stop being funicular in shape. Bending would develop under the new loading, and complete collapse could occur because the nonrigidly connected blocks cannot resist such bending. Because the forms of the tensile and compressive structures that are derived in the manner just described are related to the notion of a loaded hanging rope, they are collectively called funicular structures.

2

General PrIncIPleS oF FunIcular ShaPeS

In the study of arches and cables, it is important to know what exact curve or series of straight-line segments defines the funicular shape for a given loading. The funicular shape is naturally assumed by a freely deforming cable subjected to loading. A cable of constant cross section carrying only its own dead weight naturally deforms into a catenary shape (Figure 1). A cable carrying a load that is uniformly distributed along the horizontal projection of the cable, like the primary loading in a suspension bridge supporting a horizontal bridge deck, will deform into a parabola. Cables carrying concentrated point loads will (ignoring the dead weight of the cable itself) deform into a series of straight-line segments. While only one general shape of structure is funicular for a given loading, invariably a family of structures has the same general shape for any given condition. From Chapter 5 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Funicular Structures: Cables and Arches

FIGure 1 Funicular structures.

Copyright © 2013. Pearson Education UK. All rights reserved.

:

:

The structures in Figure 1—(e) and (f), for example—are funicular for the loadings indicated. All structures in a group have the same shape, but the physical dimensions are different. Within a family, the relative proportions of all the shapes are identical. It is obvious that such a family could be obtained by using a series of flexible cables of different lengths. All would deform in a similar way under the action of the load, but the amount that the structure would sag would be different. The magnitude of the forces developed in the arch or cable, however, are dependent on the relative height or depth of the funicular shape in relation to both its length and the magnitude and location of the applied loads. The greater the rise of an arch or the sag of a cable, the smaller are the internal forces developed in the

Funicular Structures: Cables and Arches structure, and vice versa. Reactive forces developed at the arch or cable ends also depend on these parameters. End reactions have vertical and horizontal thrusts that must be resisted by the foundations or some other element, such as a compressive strut or tie-rod. Figure 2 shows an arch bridge with pinned connections designed to resist the inclined load from the arch. Note that in a funicular structure, its shape always changes beneath an external load. Where the structure is not loaded, it remains straight. The funicular shape appropriate for a continuous load must therefore change continuously. Similarly, if the shape of the structure changes when there is no load change, bending will be present. The presence of bending may or may not be problematic, depending on how the structure is built. A masonry structure subject to bending might develop cracks and collapse. (See Section 5.1.) A steel structure, however, could be designed to accommodate both the axial forces and any bending that is present. It would have larger dimensions, however, than a pure funicularly shaped form with only axial forces present. It is interesting to note that if the shape of a funicular structure is imagined and superimposed on the actual structure considered, the amount of deviation of the actual structure from the funicular shape generally reflects the severity of the bending present in the structure. [See Figures 3(g) and (h).] Later sections revisit and elaborate on these points.

3

Copyright © 2013. Pearson Education UK. All rights reserved.

3.1

analySIS and deSIGn oF cable StructureS Introduction

The flexible suspension bridge, which was initially developed in China, India, and South America, is of great antiquity. While most early bridges used rope, often made from bamboo, there is recorded evidence of bridges made with the use of chains in China as early as the first century ad. In addition to the varieties of tents that used ropes, cable structures also were used in some major buildings. A rope cable structure, for example, was used in about ad 70 to roof a Roman amphitheater (Figure 4). Still, it was in connection with suspension bridges that the use of cables as theoretically understood structural elements was developed. This theoretical understanding is rather recent because the suspension bridge remained relatively unknown in Europe (although a type of chain-suspended structure was built in the Swiss Alps in 1218), where most developments in structural theory were occurring. Fausto Veranzio was one of the first to publish drawings of a suspension bridge in 1695, and it was not until 1741 that a permanent iron chain footbridge, located in Durham County, England, was finally built. This bridge was probably the first significant suspension bridge in Europe. It failed, however, to establish a precedent. A major turning point in the evolution of the suspension bridge occurred in the early part of the nineteenth century in America, when James Findley developed a suspension bridge capable of carrying vehicular traffic. His initial bridge, built in 1801 over Jacobs Creek in Uniontown, Pennsylvania, used a flexible chain of wrought-iron links. Findley’s real innovation was not the cable itself, however, but the introduction of a stiffened bridge deck in which the stiffening was achieved by longitudinal trusses made of wood. The use of a stiffened deck kept the supporting cable from changing shape and, consequently, from changing the shape of the road surface it supported, by distributing concentrated vehicular loads over a larger portion of the cable. With this innovation, the modern suspension bridge was born. The work of Findley became known to others, possibly even to the great English builder Thomas Telford, who designed the bridge over the Menai Strait in Wales (1818–1826). Louis Navier, the great French mathematician, discussed Findley’s work in his classic book on suspension bridges, Rapport et Mémoire sur les Ponts Suspendus, published in 1823, in which he gave Findley credit for the introduction

FIGure 2 Arch bridge. Point loads from the deck are closely spaced. The reaction forces are transferred via pinned connections to the abutments.

Funicular Structures: Cables and Arches

FIGure 3 Funicular and nonfunicular structural shapes.

I I

Copyright © 2013. Pearson Education UK. All rights reserved.

A

A

of the stiffened bridge deck and provided a way for other bridge builders to become aware of his work. Other great suspension bridges were built in rapid succession, including the beautiful Clifton Bridge in England by Isombard Brunel, John Roebling’s Brooklyn Bridge in 1883, and several other major bridges. Two significant modern bridges are the Verrazano-Narrows Bridge in New York, with a middle span of 4260 ft (1300 m),

Funicular Structures: Cables and Arches

FIGure 4 Cable roof structure over Roman Colosseum, circa ad 70. Rope cables anchored to masts spanned in a radial fashion across the open structure supported a movable sunshade that could be drawn to cover the arena. The span of the structure was 620 ft (188 m) along the major axis and 513 ft (156 m) along the minor axis. (From Dürm.)

and the new Kobe Bridge in Japan. Basic structural issues in designing a suspension bridge are illustrated in Figure 5. Applying cables to buildings other than tents developed more slowly because of the lesser need to span large distances and the intrinsic problems of using cables. James Bogardus submitted a proposal for the Crystal Palace of the New York Exhibition of 1853 in which the roof of a circular cast-iron building, 700 ft (213 m) in diameter, was to be suspended from radiating chains anchored to a central tower; however, the pavilion structures of the Nijny Novgorod exhibition that were designed by V. Shookhov in 1896 marked the beginning of modern applications. Subsequent structures include the Locomotive Roundhouse Pavilion at the Chicago World’s Fair in 1933 and the Livestock Judging Pavilion built at Raleigh, North Carolina, in the early 1950s. Since then, several significant cable-supported buildings have been constructed. One of the longest suspension bridges built to date, the Akashi Kaikyo Bridge in Japan, has a span of 1991 m. At these dimensions, the curvature of the earth must be considered during the design and construction process.

Copyright © 2013. Pearson Education UK. All rights reserved.

3.2

Suspended cable Structures: concentrated loads

Reactions are developed at cable supports such that the overall cable is in a state of equilibrium. The cable itself normally exerts an inward and downward force on the supports. The foundation’s reactive forces are equal in magnitude and opposite in sense. It is usually not possible to calculate these reactions directly by considering only the equilibrium of the whole cable. Reactive forces are normally considered in terms of their components in the horizontal and vertical directions. Because each reactive force has two components, a total of four unknown components is present, but only three independent equations of statics are available to aid in their determination. As will be shown subsequently, the equilibrium of an isolated portion of the structure must be considered in order to find reactive forces. Normally, an accompanying task is to determine the precise geometry of the cable. The exact shape of a loaded cable depends on the loading conditions that are present, and the designer may not arbitrarily define it a priori. A maximum sag or some other value may be specified beforehand but not the exact shape of the cable. Finding the exact shape of the cable is a fundamental objective of analysis procedures. Internal forces in cable members are in a state of pure tension. A cable can be conceived of as a continuous series of discrete elements connected to one another by hinged connections—a chain is an obvious image. Hence, each discrete element is free to rotate as it will under the action of a load. The connections are such that local internal bending moments cannot be transmitted from one element to another. It then follows that the sum of all rotational effects produced about any such location by the external and internal forces must be zero. (This is the principle

FIGure 5 Importance of rigid bridge decks for carrying vertical live loads.

Funicular Structures: Cables and Arches underlying approaches to calculating cable geometries and reactive forces.) All internal forces that exist in a cable, therefore, must be in a state of pure tension. If the exact geometry of the cable were known or somehow established, then the magnitudes of internal forces could be calculated via an approach similar to the joint equilibrium approach described for trusses. (Any point on a cable must be in a state of vertical and horizontal translatory equilibrium under the action of all of the internal and external forces acting on the point.) Any structure’s function can be defined as that of carrying the external shears and moments generated by the effects of applied loads. In a cable or funicular-shaped arch, the external shear at a section is balanced by an internal resisting shear force that is provided by the vertical component of the internal, axial cable force. The overall external bending moment at the same section is balanced by the force couple consisting of the horizontal component of the internal cable force and the horizontal component of the reaction at support of the structure. This latter force may be the horizontal thrust developed at the foundation or the internal force present in a tie-rod or compression strut used between end points of an arch or cable structure, respectively. The next example illustrates how cable geometries, reactions, and internal forces can be determined by applying the principles described.

examPle Figure 6 illustrates a cable structure carrying two loads. Assume that the deformed structure has the maximum depth shown under the first load. The deformed structure has another critical depth under the second load. The latter depth is not precisely known beforehand, but it must be a value that directly depends on the maximum depth of the other point, which is a given. Solution: The first step in the analysis is to determine the reactions at the ends of the member. Vertical components can be found by considering the moment equilibrium of the whole structure about one or the other reaction, much like the procedure used for straightforward beams. Horizontal components can be found by passing a section through the structure at a location of known structural depth and considering the equilibrium of either the right or left portion of the structure. Thus, we have the following equations: Vertical reactions:

gMA = 0 ⤺ +:

Copyright © 2013. Pearson Education UK. All rights reserved.

gFy = 0 + c :

- 151102 - 30152 + 45RBv = 0 - 10 - 5 + RAv + RBv = 0

6 RBv = 6.7

6 RAv = 8.3

Horizontal reactions: Pass a vertical section through the structure immediately to the right of the maximum depth, and consider the equilibrium of the left portion of the structure: gMC = 0 ⤺ +:

For the whole structure,

gFX = 0 S : +

-1518.32 + 101RAH 2 = 0 -RAH + RBH = 0

6RAH = 12.5

6 RBH = 12.5

Depth of structure at point D: Pass a vertical section immediately to the left of point D, and consider the equilibrium of the right portion of the structure. Let hD equal the height of the structure at this point. Thus: gMD = 0 ⤺ +:

-hD 112.52 + 1516.72 = 0

6 hD = 8.3

Note that no other structural depth would satisfy rotational equilibrium considerations at point D.

Funicular Structures: Cables and Arches

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 6 Analysis of a cable supporting concentrated loads.

Cable forces: Forces in individual segments in the cable can now be found by using the method of joints. The results of this process are illustrated in Figure 6(c). Note that the absolute magnitude of the forces in different segments varies: Cables are not constant-force structures!

Funicular Structures: Cables and Arches

examPle An alternative method to analyze the same structure (Figure 6) is to first draw shear and moment diagrams for the external loading condition and then use a method-of-sections approach. Vertical reactions are found first, as before. Solution: The horizontal reactions can be found by observing that the structure must provide an internal moment resistance equal and opposite in sense to the external moment at every point. The internal moment is provided by a couple formed between the end horizontal reaction and the horizontal component of the cable force. Considering the structure at point C, we have 10R

AM (+)1*

internal resisting moment

125 "

=

or RAM = 12.5

external applied moment (see moment diagram)

This is the same process as described before but is a conceptually different way to think about it. Depth of structure at point D: As before, hD 112.52 (+)1* internal resisting moment

=

100 " or hD = 8.3

external moment

Cable forces:

Copyright © 2013. Pearson Education UK. All rights reserved.

Sections are passed through points immediately to the right of each critical point, and equilibrium of the left portion is considered. [See Figure 6(e).] Note that in each case, the internal shear resistance that balances the external shear force is provided by the vertical component of the force in the cable. The vertical and horizontal forces are components of the actual forces in the cables and can be used to calculate these forces. The second method presented is useful for conceptualizing how cable or arch structures provide a mechanism for carrying the shear and moment associated with the external load. It also shows particularly well how the reactions and internal forces depend on the height of the structure. Doubling the initial maximum depth would decrease horizontal thrusts, for example, by a factor of 2. Cable forces would thus be reduced. Still, the new combination of cable height and force would yield the same internal balancing shear and moments as just discussed.

examPle Alternatively, a graphical technique may also be used to analyze the same cable (Figure 7). While seemingly simple, such techniques are subtle and must be used with care. The process is only briefly described herein. FIGure 7 Use of graphic statics to determine cable forces and stresses.

Funicular Structures: Cables and Arches Solution: First, redefine members and loads in terms of adjacent spaces, and construct a force polygon, as shown. The total downward force is equal to the line connecting the starting and ending points. If the vertical reactions are first calculated via statics, they also are marked on the line. To find the shape of the structure, the individual forces P1 and P2 are replaced by component pairs because graphic statics work best with inclined forces. First, the known inclination of the cable segment 0-3 (right cable segment connecting the support B with the known sag at D) is transferred into the force diagram on the left. Its intersection with the horizontal that aligns with the reaction point is the pole point 0. Next, point 0 is connected to points 1 as well as points 2 in the force diagram. These newly drawn force polygons are equivalent to the original loads (e.g., forces 0-1 and 0-2 are components of P1). Forces in cable members 0-1 and 0-2 balance P1 (from joint equilibrium considerations), while 0-2 and 0-3 balance P2. Forces 0-4 and 4-1 must equal the horizontal and vertical components of the final force in cable member 1-0 (so the left node point is in translatory equilibrium). Forces 0-5 and 5-3 must be components of the final force in cable member 0-3. For the whole structure to be in horizontal equilibrium, 0-5 and 0-4 must be identical in magnitude (as shown on the diagram) but opposite in sense. Forces are then drawn through known points on their respective lines of action to determine the cable shape. Changing the position of the pole point 0 would change the height of the structure but not its shape. Initially establishing the location of the pole point 0 is equivalent to a design decision establishing cable slopes. With this technique, it is possible to stipulate the desired cable sag only at one point. That point may or may not represent the maximum sag. Instead of finding vertical reactions first through statics, they may be found by arbitrarily choosing any pole location, drawing the components of P1 and P2 as before, and then drawing a closing line to determine the directions of the components of the reactions (see Figure 8). Next, this closing line is drawn on the polygon to determine the reactions. A new polygon with a horizontal closing line is then drawn, and the process proceeds as before. Techniques of this type may be used with any number of loadings acting in any direction and with supports on different levels.

3.3

FIGure 8 Graphic statics analysis of a cable with the supports on different levels. Once the funicular shape has been found, reactions and cable forces can be determined.

P2

P1

B

A (a) Support locations and loads are given. P1

e 0

f P2

g

(b) Force polygon: A pole point 0 is chosen, and temporary lines e, f, and g are drawn to connect the tips and tails of the load vectors to 0.

B

Suspended cables: uniformly distributed loads

Cables or arches carrying uniformly distributed loads can be analyzed in much the same way as for concentrated loads. Because the funicular shape is constantly curving, however, a variant of the method of sections is the exclusive analytical technique used.

A Temporary baseline

e

h1

h2

g

f

examPle Consider the cable shown in Figure 9. Assume the maximum cable sag to be hmax. First, calculate the vertical reactions that are formed. Do this by considering the equilibrium of the overall structure. The steps that follow are to calculate horizontal reactions, find the shape of funicular curve, and then determine the internal forces in the cable.

(c) Keeping their inclinations, e, f, and g are transferred into the system diagram. The sags h1 and h2 can be measured off the temporary baseline. Line of action B for P1  P2

Copyright © 2013. Pearson Education UK. All rights reserved.

Solution: A

Vertical reactions: RAv = RBv =

wL 2

Pass a section through the structure at midspan (the location of the known sag), and consider the equilibrium of the left portion of the structure to find RAH: - RAHhmax - a

wL L wL L ba b + a ba b = 0 2 2 2 4

Alternatively, using shear and moment diagrams directly yields RAHhmax = (11)11* internal resisting moment

wL2 wL2 6 RAH = 8 8hmax (11)11* external applied moment

h2

g'

e'

by inspection

Horizontal reactions:

gML>2 = 0:

h1

6 RAH =

wL2 8hmax

(d) The sags h1 and h2 are transferred into the system diagram, and the funicular shape can be drawn over the baseline AB. Forces can now be found by drawing a force polygon. Note: The inclinations of e' and g' can also be found directly in Step b. The extension of e' and f' intersect with the line of action of the equivalent point load.

Funicular Structures: Cables and Arches

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 9 Analysis of a cable supporting a horizontally distributed load.

Shape of cable: The shape of the cable can be found by considering the equilibrium of different sections of the structure. Consider a section of the structure defined by a distance x from the left connections. Let y be the depth of the cable at that point. Then gMx = 0:

gives

a

wL2 wL x b 1x2 + wx a b = 0 b(y2 - a 8hmax 2 2

6 y =

4hmax 1Lx2 - x2 L2

Alternatively, equating the internal resisting moment to the external applied moment

a

wL2 wL x bx - 1wx2 a b by = a 8hmax 2 2

6 y =

4hmax 1Lx - x2 2 L2

Funicular Structures: Cables and Arches The equation found is a parabola. The absolute value of the slope at either end is given by 1dy>dx2 x = 0, x = L = ux, L = 4hmax >L. This slope also could have been found by considering the ratio of the horizontal and vertical reactions: u = 1wL>221wL2 >8hmax 2 = 4hmax >L. Cable forces:

Because the slope of the cable is zero at midspan, where both the external and internal shear forces also are zero, the cable force can be seen to be identical to the horizontal reaction by considering the horizontal equilibrium of the section to the left or right of midspan. Thus, the cable force at midspan becomes TL>2 = wL2 >8hmax. The force in the cable at either end connection is found by joint equilibrium considerations: gFx = 0: T0, L cos u = wL2 >8hmax, where u is given by 4hmax >L and cos u by 1> 21 + 16h2max >L2. Thus, T0, L = 1wL2 >8hmax 2 21 + 16h2max >L2. Alternatively, T0, L = 21RAH 2 2 + 1RAv 2 2 = RAH 21 + 1RAv >RAH 2 2 = 1wL2 >8hmax 2

21 + 16h2max >L2. The force in the cable at the ends exceeds that at midspan.

examPle A series of cables spaced at 15 ft (4.6 m) on center span 100 ft (30.5 m) between abutments carries a live load of 50 lb>ft2 (2.39 kN>m2). Each cable has a maximum sag of 20 ft (6.1 m). Assume that the dead load of the whole structure is 50 lb>ft2 (2.39 kN>m2). Assume also that the live load and dead load are horizontally projected. Find the minimum required cable diameter, assuming the allowable cable stress is 75,000 lb>in.2 1517 MPa2. Solution: Reactions: vertical = RAv = =

312394 + 23942 Pa]14.6 m2130.5 m2 2

horizontal = RAH = =

3150 + 50) lb>ft2 4115 ft21100 ft2 wL = = 75,000 lb 2 2 = 334 kN

3150 + 502 lb>ft2 4115 ft21100 ft2 2 wL2 = = 93,750 lb 8h 8120 ft2

312394 + 23942 Pa414.6 m2130.5 m2 2 816.1 m2

= 417 kN

Copyright © 2013. Pearson Education UK. All rights reserved.

Maximum cable force (at ends): T = 2R2AV + R2AH = 175,0002 + 93,7502 2 1>2 = 120,000 lb = 13342 + 4172 2 1>2 = 534 kN

Alternatively,

T = RAH 21 + 16h2 >L2 = 93,750 lb c 1 + 16 a Cable diameter: A =

= 417 kNc 1 + 16 a

1>2 202 bd = 120,000 lb 2 100

1>2 6.12 bd = 534 kN 2 30.5

T 120,000 534,000 N = = 1.60 in.2 = = 1033 mm2 fallowable 75,000 517 MPa

Twisted-strand cable would probably be used, in which the total diameter is made up of a series of individual wire strands of smaller diameters. The gross area of the total diameter therefore does not equal the net area of the cross sections of individual strands. The resisting area of the whole cable is about two-thirds of its gross area. Thus, the gross area of the cable needed for this example is 1 3>2 211.602 = 2.4 in.2, or 1 3>2 2110332 = 1550 mm2. This leads to a cable diameter of 1 3>4 in., or 45 mm.

Funicular Structures: Cables and Arches

3.4

cables with Varying Support levels

Many cable structures have end supports at different elevations. The essential analytical problem here is to determine the exact shape of the cable and the location of the curve’s low point. In the previous formulation for level supports, it was known that the low point was at midspan. A section could thus be passed through this point of known sag and the forces subsequently determined. Finding the exact shape of the curve and forces within cables with varying support levels, however, is difficult.1 For initial design purposes, however, the low point can be estimated or determined experimentally, or an approximate curve can be fitted in. Once the low point is known, forces can be determined as before. Figure 10 illustrates this process for a cable-supported exhibition hall in Hanover, Germany, with points of attachment 88 and 47 ft off the ground. Flat

FIGure 10 Cable structure with supports at different levels. ([KLELWLRQ+DOO +DQQRYHU*HUPDQ\ $UFKLWHFW7KRPDV+HU]RJ 6WUXFWXUDOHQJLQHHU 6FKODLFK%HUJHUPDQQ 3DUWQHU 3KRWRJUDSK1LFR.LHQ]O 5RRIGHWDLO 3O\ZRRGVWUHVVHGVNLQER[ FRQWDLQLQJLQVXODWLRQSOXVVDQG OD\HU IW

5$

6ROLGVWHHOVHFWLRQWKHFDEOH LQ[LQ

[

5$

&DEOHVSDFLQJ IWRQFHQWHU IW

\

5%

/RZHVW3RLQW

5%

[

$VVXPHDGHDGORDGRIOEVIW   DQGDOLYHORDGRIOEVIW /RDGSHUIRRWRQDW\SLFDOFDEOH Z OEVIWOEIW [IW OEIW

\

IW

(TXLOLEULXPRIOHIWVHFWLRQ

IW

5$

[

Copyright © 2013. Pearson Education UK. All rights reserved.

5$

/RDGGRZQZDUGRQVHFWLRQWRWKHOHIW RIWKHORZSRLQW OEIW[IW OE Σ) \ 

N

\

/RZHVW3RLQW

5$

N

\

5 $ IW Σ0 5$ R  ±IW

IW

\

[

IW N  7KXV 5 $ N

R

[

(TXLOLEULXPRIZKROHDVVHPEO\ Σ) \ 

N N

N $ %

N

N

Σ) \ 

5$

\

 5% \

OEIW[IW

N 7KXV 5 % \ 5 % N [

5HVXOWDQWIRUFHDW$PD[LPXPLQFDEOH  5HVXOWDQWRIN N N $FWXDOVWUHVVLQFDEOH

$

%

1

N

I )$ NLQ[LQ  NVL /HVVWKDQDOORZDEOHRI) D NVL

A technique for determining shapes and forces in structures of this type is contained in D. Schodek, Structures, 4th ed., Prentice Hall, Inc., 2001, pp. 199–200.

Funicular Structures: Cables and Arches steel sections were used as cable elements. These elements do not have the tensile strength of bundled wire strands often used in bridges and buildings, but it is easy to provide a large cross-sectional area with them that works well with the detailing of the remainder of the roof structure.

3.5

cable lengths

Cable length can be evaluated by considering the basic expression for the deformed shape of a cable. For a uniformly loaded cable with a horizontal chord (both supports on the same level), let Ltotal be the total length of the cable, Lh the span, and h the maximum sag. The total cable length can be shown to be approximately Ltotal = Lh 11 + 8>3h2 >L2h - 32>5h4 >L4h 2. For the derivation for this and other important expressions defining the elongation of a cable, the reader is referred to other references.2

Copyright © 2013. Pearson Education UK. All rights reserved.

3.6

Wind effects

A critical problem in the design of any cable roof structure is the dynamic effect of wind, something that does not significantly affect an arch structure. Consider the simple roof structure supported by cables in Figure 11. As the wind blows over the top of the roof, suction is created. If the magnitude of the suction force due to the wind exceeds the dead weight of the roof structure, the roof surface will rise. As it begins rising and dramatically changing shape, the forces on the roof begin to change because the magnitude and distribution of wind forces on a body depend on the exact shape of the body. Because the wind forces change due to the roof ’s changing shape, the flexible structure itself again changes shape in response to the new loading. The process is cyclical. The roof will not remain in any one shape but will move, or flutter, as long as the wind is present. While the preceding description of wind effects is useful, a more rigorous way to understand wind effects is to study vibratory phenomena in cables. A vibration in a structure that is induced by wind (or earthquake) effects is a reciprocating or oscillating motion that repeats after an interval of time, which is called the period of vibration. The frequency of vibration is equal to the reciprocal of the period. All suspension structures (and other structures) have a natural frequency of vibration when subjected to an externally applied force. When an external dynamic force acting on the structure comes within the natural frequency range, a state of vibration may be reached wherein the driving-force frequency and the body’s natural frequency are in tune, a condition referred to as resonance. At resonance, the structure undergoes violent vibration, resulting in structural damage. The natural frequency of a suspended cable is given by fn = 1Np>L2 2T> 1w>g2, where L is the cable length, N is any integer, w is the applied load per unit length, T is the cable tension, and g is the acceleration due to gravity. The first three modes of vibration are shown in Figure 11. The cable could vibrate in these or other modes, depending on the frequency of the excitatory force. Resonance in a cable occurs when the external excitatory force has a frequency exactly equal to the first natural frequency of the cable or any of its higher frequencies. Unfortunately, for many cable structures, the frequency of wind forces is often close to the natural frequency of the cable structure. Major cable structures have been destroyed because of this phenomenon; a case in point is the well-known Tacoma Narrows failure. (See Figure 12.)

2

See, for example, C. H. Norris and J. B. Wilbur, Elementary Structural Analysis, 2nd ed., New York: McGrawHill Book Company, 1960; or J. Scalzi, W. Poddorny, and W. Teng, Design Fundamentals of Cable Roof Structures, U.S. Steel Corporation, Pittsburgh, 1969.

FIGure 11 Dynamic effects of wind on flexible roof structure.

Funicular Structures: Cables and Arches

FIGure 12 Single-cable suspension systems can be susceptible to uplift and oscillations from wind loads. Left: Twisting oscillations of the Tacoma Narrows Bridge before its collapse in 1940. The deck initially began heaving in vertical wave motions in winds of 25 to 40 mph. Rhythmic twisting of increasing amplitudes followed until a 600 ft-foot section collapsed into the Narrows. The failure was due to vortex shedding associated with wind action. The shape and extreme flexibility of the deck contributed to the collapse. The depth of the deck plate girder was only 8 feet or 1/350 of the span. The bridge was then reconstructed with a deep and rigid truss in place of the girder (right). [(Left) Courtesy of Special Collections Divisions, University of Washington Libraries; (Right) Courtesy of the Department of Transportation, State of Washington.]

There are only a few fundamental ways to combat flutter due to wind forces. One is to increase the dead load on the roof, thereby also increasing cable tensions and changing natural frequencies. Another is to provide the cable anchoring guy cables at periodic points to tie the structure to the ground. A related method is to use a crossed-cable or double-cable system. The latter method is interesting because it can be used to create an internally self-damping system. This approach is discussed in more detail in the next section.

Copyright © 2013. Pearson Education UK. All rights reserved.

4

deSIGn oF cable StructureS

Although most of the discussions thus far have focused on the cable as a simple, singly curved or draped suspension element, cables are used in a variety of ways. Cable structures are more correctly categorized into suspension structures or cablestayed structures. Suspension structures can be subclassified into the following: (1) single-curvature structures, in which roofs are made by placing cables parallel to one another and using a surface formed by beams or plates to span between cables, (2) double-cable structures, in which interconnected double cables of different curvatures are used in the same vertical plane, and (3) double-curvature structures, in which a field of crossed cables of different and often reverse curvatures makes up the primary roof surface. Cable-stayed structures typically use vertical or sloping compression masts, from which straight cables run to critical points or horizontally spanning members. Stay cables are also frequently used to tie back the supporting mast of suspension structures. Cable structures are commonly used for relatively long spans and in situations that allow the structure to be shaped with a significant depth. Efforts to design the cable system itself are the focus of this chapter’s discussion, but it is important to remember that the structure’s overall behavior, appearance, and economy may depend as much on the supporting elements as they do on the cable system. Of particular concern are the reaction forces at the connections of the cables to the supporting structure and especially the horizontal components of reactions. These can be dealt with either internally in the structural system or by appropriately designing tiebacks and large foundations. Resolving these horizontal forces within the structure includes using compression struts or rings. As regarding all systems subject to compression, these elements must be designed to prevent buckling.

Section 5-6  Exponents and Scientific Notation    277 9  × ​10​​−3​ __ 9 (b) ​​ _______  ​   = ​   ​  × ​10​​−3−5​ = 4.5 × ​10​​−8​​ 2  × ​10​​5​ 2 1.2  × ​10​​−5​ ___ 1.2 (c) ​​ _________  ​   = ​   ​  × ​10​​−5−(−4)​ = 0.25 × ​10​​−5+4​ = 0.25 × ​10​​−1​​ 4.8  × ​10​​−4​ 4.8

This time, the answer needs to be rewritten to put it in proper scientific notation. The decimal point will be moved one place to the right, so we need to subtract one from the exponent (we made the decimal part larger). ​0.25  × ​10​​−1​ = 2.5 × ​10​​−2​​

Try This One 4. Perform operations with numbers in scientific notation.

7

Find each quotient. Round the decimal part of the answer to two decimal places. 3.2  × ​10​​7​ 9.6  × ​10​​−5​ 2.2  × ​10​​−6​ _________ _________ (a) ​​ ________  ​​      (b) ​​       ​​     (c) ​​   ​​  8  × ​10​​4​ 1.3  × ​10​​−8​ 4.8  × ​10​​5​

Applications of Scientific Notation As we pointed out earlier, many areas in science and economics require the use of very large or very small numbers. This is when scientific notation really comes in handy. Many calculators display only 8 to 12 digits, so numbers with more digits can only be entered in scientific notation.

Example 8

Applying Scientific Notation to Astronomy Earth’s orbit around the sun is not circular, but it’s reasonably close to a circle with radius 9.3 × 107 miles. How far does Earth travel in one year? (The formula for circumference of a circle is C = 2πr.)

SOLUTION The distance Earth travels in one year is one full turn around a circle with radius 9.3 × 107 miles. Using the formula for circumference, and approximating π with 3.14, ©Getty Images/Digital Vision RF

C ≈ 2(3.14  )  (9.3  × ​10​​7​  ) ​​ ​        ​​​ ​ ≈​  58.4  × ​10​​7​​ ​ = 5.84  × ​10​​8​  miles

In decimal form, this is 584,000,000, or 584 million miles!

Try This One

8

The speed of light in glass is about 6.56 × 108 feet per second, and the circumference of Earth is about 1.31 × 108 feet. Using the formula ©Lawrence Lawry/Getty Images RF

5. Use scientific notation in applied problems.

distance ​Time  = ​ _______  ​​    speed find how long it takes light to circle the globe in a glass fiber-optic cable.

278   Chapter 5  The Real Number System

Answers to Try This One 1

(a) 729 1 (b) ​​ ___  ​​  64 (c) 1

3

(a) 66 = 46,656 1 (b) ​​10​​−3​  = ​ _____    ​​  1,000 (c) 46 = 4,096 (a) 5.16 × 108 (b) 1.62 × 10−4

4

(a) 96,100,000,000 (b) 0.00000277

2

5

(a) 6 × 109 (b) 1.55 × 1010

6

(a) 8.4 × 10−6 (b) 3.72 × 10−2 (c) 5.39 × 103

7

(a) 4 × 102 (b) 7.38 × 103 (c) 4.58 × 10−12

8

About 0.2 seconds

Exercise Set 5-6

Writing Exercises 1. Describe how exponents are defined using multiplication. 2. What is scientific notation? How do you convert a number from decimal to scientific notation? 3. Explain how to convert a number from scientific to decimal notation. 4. What are some of the advantages of using scientific notation? 5. Describe in your own words how to multiply two numbers in scientific notation.

6. Describe in your own words how to divide two numbers in scientific notation. 7. When a number written in scientific notation has a positive exponent on 10, what can we say for sure about the number? Why? 8. If a number between 0 and 1 is written in scientific notation, what can we say about the exponent on 10? Why?

Computational Exercises For Exercises 9–18, evaluate each without a calculator. 9. 35 14. (−4)0 4 10. 6 15. 3−5 0 11. 8 16. 6−4 0 12. 9 17. 2−6 0 13. (−5) 18. 7−2 For Exercises 19–38, simplify the expression using rules for exponents, then evaluate the resulting expression. ​25​​3​ ​34​​ ​ 19. 34 · 32 26. ​​ ___ ​​  34. ​​ __6 ​​  25 ​3​​ ​ 20. 53 · 53 2 3 4 3 ​100​​4​ 27. (5 ) ____ 21. 4  · 4 35. ​​   ​​  4 2 7 28. (4 ) 1 ​ 00​​ ​ 22. 26 · 24 2 −4 2 4 29. 3  · 3 1 ​ 3​​ ​ ​3​​ ​ 36. ​​ ___5 ​​  23. ​​ __2 ​​  30. 4−3 · 45 ​13​​ ​ ​3​​ ​ 31. 35−3 · 35−2 ​92​​ ​ ​__ 65​​ ​ __ 37. ​​  6 ​​  24. ​​  3 ​​  32. 163 · 16−3 ​3​​ ​ ​6​​ ​ 5 5 2 ​ ​​ ​ ​59​​ ​ __ ​12​​ ​ 33. ​​  7 ​​  38. ​​ ___5  ​​  25. ​​ ___4 ​​  ​2​​ ​ ​25​​ ​ ​12​​ ​

In Exercises 39–46, simplify the expression using rules for exponents. ​a10 ​​ ​ ⋅ ​a12 ​​ ​ ​x−5 ​​ ​ ⋅ ​x3​​ ​ 39. (a4)5 43. ​​ _______  ​​    45. ​​ ______  ​​    8 ​a​​ ​ ​x−9 ​​ ​ 40. (k−3)2 ​​ ​ ​y4​​ ​ ⋅ ​y−8 ​b7​​ ​ ⋅ ​b5​​ ​ _____ 41. (m−4)2 46. ​​ ______  ​​    44. ​​   ​​     −5 6 5 6 ​y​​ ​ ​b​​ ​ 42. (b ) For Exercises 47–56, write each number in scientific notation. 47. 48. 49. 50. 51.

625,000,000 9,910,000 0.0073 0.261 528,000,000,000

52. 53. 54. 55. 56.

2,220,000 0.00000618 0.0000000077 43,200 56,000

For Exercises 57–66, write each number in decimal notation. 57. 5.9 × 104 58. 6.28 × 106 59. 3.75 × 10−5

60. 9 × 10−10 61. 2.4 × 103 62. 7.72 × 105

Section 5-6  Exponents and Scientific Notation    279 63. 3 × 10−6 64. 4 × 10−9

65. −1.46 × 10−8 66. −2.26 × 10−7

For Exercises 67–80, perform the indicated operations. Write the answers in scientific notation, rounding to two decimal places. Then write the rounded answer in decimal notation as well. 9  × ​10​​6​ 67. (3 × 104)(2 × 106) 76. ​​ _______2 ​​  3  × ​10​​ ​ 68. (5 × 103)(8 × 105) 4.2  × ​10​​−2​ 69. (6.2 × 10−2)(4.3 × 10−6) 77. ​​ _________  ​​  7  × ​10​​−3​ 70. (1.7 × 10−5)(3.8 × 10−6) 6.4  × ​10​​8​ 71. (4 × 104)(2.2 × 10−7) 78. ​​ ________  ​​  5 −4 72. (2.2 × 10 )(3.6 × 10 ) 8  × ​10​​−2​ 6.6  × ​10​​3​ 73. (5 × 10−2)(3 × 10−8) 79. ​​ ________5 ​​  5 −6 74. (4.3 × 10 )(2.2 × 10 ) 1.1  × ​10​​ ​ 4 5  × ​10​​ ​ 3  × ​10​​7​ 75. ​​ ________   ​​  80. ​​ _________    ​​  2 2.5  × ​10​​ ​ 1.5  × ​10​​−5​

For Exercises 81–86, write each number in scientific notation and perform the indicated operations. Leave answers in scientific notation. Round to two decimal places. 81. (63,000,000)(41,000,000) 82. (52,000)(3,000,000) 600,000,000 83. ​​ ___________ ​​  25,000,000 32,000,000 84. ​​ __________ ​​  64,000,000 85. (0.00000025)(0.000004) 0.0000036 86. ​​ _________  ​​    0.0009

Applications in Our World 87. When a new rumor about a team’s star running back starts to spread across a large college campus, the number of students that have heard the rumor d days after it starts can be described by the formula N = 50d4. The number of students that have heard the rumor but don’t believe it can be described by the formula D  =  40d2. Find a formula that describes the percentage of students that have heard the rumor that don’t believe it in terms of days after it starts to spread. What happens to that percentage for days 1, 2, 3, and 4 after the rumor starts? 88. The total box office revenue for a new movie can be described by the formula M  =  200w4, where M is in dollars, and w is the number of weeks since the movie debuts. The total number of people that have seen the movie in theaters can be described by the formula T  =  300w3. Find a formula that describes the average revenue per person in terms of weeks since the movie debuted. Then find the average revenue for weeks 4, 6, 8, and 10. What trend do you notice? 89. Light travels through air at 1.86 × 105 miles per second. How many miles does light travel in 10 minutes? Write your answer in decimal notation. 90. There are about 1 × 1014 cells in the human body. About how many body cells are there in the United States? (Use 324 million as the population of the United States.) 91. The mass of a proton is about 1.67 × 10−24 grams. There are 1.27  ×  1026 molecules in a gallon of water, and 10 protons in one molecule of water. Also, 1 gallon of water weighs 3,778 grams. Based on this information, what percentage of the weight of water comes from protons? 92. It has been estimated that there are 1  ×  1020 grains of sand on the beach at Coney Island, New York. By one estimate, the beach has an area of 3.3 million square feet. How many grains of sand are there on average per square foot of beach?

For Exercises 93–96, use the following information. The number of miles that light travels in one year is called a light-year. One light-year is equal to 5.88 × 1012 miles. 93. The nearest star (other than the sun) to Earth is Proxima Centauri, at 4.2 light-years away. How far is this in miles? 94. The star Gruis is 280 light-years from Earth. How far is this in miles? 95. The (former) planet Pluto is 4,681 million miles from Earth. Write this number in scientific notation. How far is that in light-years? 96. The average distance from Earth to the sun is 93 million miles. Write this number in scientific notation. Use the number of miles in a light-year to find how many minutes it takes light to reach Earth from the sun. 97. One atom is 1 × 10−8 centimeters in length. How many atoms could be laid end-to-end on a meter stick (which is 100 cm in length)? 98. One grain of spruce pollen weighs about 7 × 10−5 grams. How many grains are there in a pound of spruce pollen? (One pound is about 454 grams.) 99. The planet Venus is about 67 million miles from the sun. Assuming that its orbit is approximately circular, how far does it travel in one year? (One Venus year, not one Earth year.) 100. Each red blood cell contains 250 million molecules of hemoglobin. There are about 25 trillion red blood cells in the average human. How many molecules of hemoglobin are there in an average human? 101. The largest bank in the United States, JP Morgan Chase, announced a profit of 2.44 × 1010 dollars in 2015. How much on average did the company earn every day in that year? 102. Refer to Exercise 101. If the CEO of JP Morgan Chase had gone stark raving mad and decided to equally distribute the company’s 2015 profit to the 324 million

280   Chapter 5  The Real Number System people in the United States, how much would each person have received? 103. Between the 2016 and 2017 seasons, the Los Angeles Dodgers committed a total of $1.92 × 108 to new player salaries. The team was projected to have an average ticket price of $33 in 2017. How many tickets would the team have to sell to cover the cost of those salaries?

104. At the end of the college football season, there are three bowl games that are part of the college football playoff. ESPN signed a contract to televise these three games from 2014 through 2025 at a cost of 4.7  ×  108 dollars per year. In 2016, an estimated 45.7 million viewers watched the games. If viewership continues at that rate, how much is ESPN spending per viewer?

Critical Thinking 105. Read the second column of the table below from top to bottom. Using inductive reasoning, complete the missing entries in the table. Then explain how that justifies our definitions of zero and negative exponents. Exponent (n)

 2 n

5

32

4

16

3

8

2

4

1

2

0

?

−1

?

−2

?

106. The quotient rule for exponents is am/an = am − n, where a is any nonzero number, and m and n are integers. Use the definition of exponents from the beginning of the section and reducing fractions to justify this rule. Does your justification require any restrictions on m or n? Explain. 107. The power rule for exponents is (am)n  =  am·n, where a is any nonzero number, and m and n are integers. Use

Section 5-7 LEARNING OBJECTIVES 1. Define arithmetic sequence. 2. Find a particular term of an arithmetic sequence. 3. Find the sum of n terms of an arithmetic sequence. 4. Define geometric sequence. 5. Find a particular term of a geometric sequence. 6. Find the sum of n terms of a geometric sequence.

the definition of exponents from the beginning of the section and multiplication to justify this rule. Does your  justification require any restrictions on m or n? Explain. 108. We have already developed rules for multiplying and dividing numbers in scientific notation. In this exercise, we’ll develop a power rule for scientific notation. (a) Using the definition of exponent from the beginning of this section and our rule for multiplying numbers in scientific notation, compute (2 × 104)3. (b) Repeat for (a × 10n)3. (c) Use the results of (a) and (b) to verbally describe a rule for raising a number in scientific notation to a power. Does your rule work for zero and negative powers? 109. Refer to the result of Example 8 on page 277. How fast is Earth traveling around the sun in miles per hour? If you could fly that quickly, how long would it take to get from New York to Los Angeles (2,462 miles)? 110. Is there a largest number that you can write in decimal  form using less symbols than it would take to write it in scientific notation? If there is, find it. If not, explain why.

Arithmetic and Geometric Sequences For most of the first decade of the 21st century, if you had a job and a pulse, you could get a home loan. Then the banks started finding out that giving loans to people that couldn’t afford them wasn’t such a smart idea. The housing market collapsed, millions of people lost their homes, and major financial institutions went belly-up. By the beginning of 2009, it was very difficult to get a home loan unless you had a significant down payment. This forced a lot of people to rent housing and try to save ©Stockbyte/Getty Images RF money for a home. Suppose that this is the predicament a young couple find themselves in. They determine that they can afford to sock away $400 each month. They don’t trust banks, so they plan to put $400 in cash in their cookie jar each month, which, by the way, is not the brightest idea I’ve ever heard. The amount they save would grow like this as the months pass: $400, $800, $1,200, $1,600, $2,000. . . . This is a special type of sequence that we will study in this section.

Section 5-7  Arithmetic and Geometric Sequences    281

A sequence is a list of related numbers in a definite order. Each number in the sequence is called a term of the sequence. Note: In many cases, there is a rule, or formula, that determines the terms of a sequence. But not always.

Arithmetic Sequences In the sequence defined by saving money for a down payment above, every term after the first can be obtained by adding 400 to the previous term. This is an example of an arithmetic sequence. A sequence is arithmetic when every term after the first is obtained by adding the same fixed number to the previous term. This amount is called the common difference. Examples of Arithmetic Sequences

Common Difference

1, 3, 5, 7, 9, 11, . . .

2

2, 7, 12, 17, 22, 27, . . .

5

100, 97, 94, 91, 88, 85, . . .

−3 _​​  1 ​​ 

_​​  1 ​ , ​ _2 ​ , ​ _3 ​ , ​ _4 ​ , ​ _5 ​ , ​ _6 ​ ,  …​

2

2 2 2 2 2 2

Example 1

Finding the Terms of an Arithmetic Sequence Write the first five terms of an arithmetic sequence as described. (a) The value of an account that starts with $700 and has $80 deposited each month (b) First term 9 and common difference 7 (c) First term __ ​​  161  ​​ and common difference ​− ​ _18 ​​ 

SOLUTION (a) We begin at $700, and add $80 to get each of the next four terms: ​$700, $780, $860, $940, $1,020​ (b) We begin at 9, and add 7 to get the next four terms: ​9, 16, 23, 30, 37​ (c) This time we will start at __ ​​ 161  ​​ and subtract _​​  18 ​​  to get the next four terms: it will be helpful to write _​​  18 ​​  as __ ​​  162  ​​ so we can perform the subtractions easily. 1 1 3 5 7 ​​ ___  ​,  − ​ ___  ​,  − ​ ___  ​,  − ​ ___  ​,  − ​ ___  ​​  16

16

16

16

16

Try This One 1. Define arithmetic sequence.

1

Write the first five terms of an arithmetic sequence as described. (a) The weight of an infant that was 7 lb at birth and has been gaining 1.3 lb per week (b) First term 6 and common difference 10 (c) First term _​​  38 ​​  and common difference ​− ​ _12 ​​  We often use the letter a with a subscript to represent the terms of a generic sequence. The symbol a1 represents the first term, a2 the second term, a3 the third, and so on. So an arbitrary sequence looks like a1, a2, a3, a4, a5, . . . . We call an the nth term of the sequence.

282   Chapter 5  The Real Number System

Math Note A finite sequence has a finite number of terms; an infinite sequence has infinitely many terms, like 2, 4, 6, 8, 10, . . .

When a sequence is arithmetic with first term a1 and common difference d, we can find a formula to quickly calculate any term of the sequence without having to find all of the terms that come before it. In this case, the sequence looks like ​​a​  1​​,   ​a​  1​​  +  d,   ​a​  1​​  +  d + d,   ​a​  1​​  +  d + d + d,  . . .​ Doing some simplifying, we find that ​​a​  1​​  = ​a​  1​​   ​a​  2​​  = ​a​  1​​  +  d  ​a​  3​​  = ​a​  1​​  +  2d  ​a​  4​​  = ​a​  1​​  +  3d​ Every term an looks like a1 plus some coefficient times d, and the coefficient is one less than the subscript of that term in the sequence. This gives us a formula:

Finding the nth Term of an Arithmetic Sequence The nth term of an arithmetic sequence can be found using the formula ​​a​  n​​  = ​a​  1​​  + (n − 1 ) d​ where n is the number of the term you want to find, a1 is the first term, and d is the common difference.

Example 2 Math Note You could check your answer by writing out the first 14 terms of the sequence, but that kind of defeats the purpose of developing a formula for the nth term in the first place.

Finding a Particular Term of an Arithmetic Sequence Find the value of the account in Example 1 after 14 months.

SOLUTION The account values were an arithmetic sequence with first term $700 and common difference $80. So we substitute into the nth term formula using a1 = $700, d = $80, and n = 14. ​ ​  n​​ = $700 + (14 − 1 ) ($80 ) a ​a​  14​​ = $700 + 13($80 ) ​​     ​       ​   ​​ ​​ ​ ​ = $700 + $1,040 ​ = $1,740 The account is worth $1,740 after 14 months.

Try This One

2

Use the nth term formula to find the weight of the baby in Try This One 1 after 20 weeks.

Example 3

Finding a Particular Term of an Arithmetic Sequence Find the 10th term of an arithmetic sequence with first term ​​ _17 ​​  and common difference _​​  25 ​​ .

Section 5-7  Arithmetic and Geometric Sequences    283

SOLUTION Substitute in the formula using ​​a​  1​​  = ​ _17 ​ ,  n = 10,​and ​d = ​ _25 ​​ . ​a​  n​​ = ​a​  1​​  +  (n − 1 ) d 1 2 ​a​  10​​ = ​ __ ​  + (10 − 1 ) ​​(​​​ _ ​​ )​​​ 7 5 1 2 ​ = ​ __ ​  +  (9  ) ​​(_ ​​​   ​​ )​​​ 5​  ​​        ​    ​  7  ​​ ​​ ​ 5 126 ​ = ___ ​    ​  + ​ ____ ​  35 35 131 ​ = ____ ​   ​  35 ©Thinkstock Images/Jupiterimages RF

The trombone plays a different note when the musician changes the length of the instrument while setting up a vibrating column of air through the mouthpiece. For a given length of trombone L, it will resonate for notes with wavelengths 2L 2L ___ 2L ___ 2L ___ 2L ​​  ___ ​,   ​  ___ ​,    ​   ​,    ​   ​,    ​   ​, . . . .​    1 2 3 4 5 (This is true of other brass and woodwind instruments as well.) What is the nth term of this sequence?

Try This One

3

Find the requested term of the arithmetic sequence with the given first term and common difference. (a) First term _​​  19 ​ ,​ common difference _​​  13 ​  :​find the seventh term. 1 (b) First term ​− ​ _56  ​,​ common difference ​− ​ __ 10  ​  :​find the sixth term. Now let’s change our savings account scenario a bit. Suppose that you put aside $10 this month, then $15 next, $20 the following, and so on. This time it’s the amount you’re saving that forms an arithmetic sequence (because it’s going up by $5 each month) rather than the value of the account. In this case, it would be of interest to find how much you’d put aside in, say, the 14th month, but it would be a whole lot more useful to find the SUM of the contributions for the first 14 months. The point is that sometimes it’s more useful to find the sum of terms in an arithmetic sequence rather than the terms themselves. Our next formula accomplishes just that. (If you’re interested, we’ll work out a proof of this formula in Project 3 at the end of this chapter.) Finding the Sum of an Arithmetic Sequence

2. Find a particular term of an arithmetic sequence.

The sum of the first n terms of an arithmetic sequence is given by the formula n(​a​  1​​  + ​a​  n​​  ) ​​S​  n​​  = ​ ________  ​​    2 where a1 is the first term of the sequence and an is the nth term.

Example 4

Finding the Sum of an Arithmetic Sequence Find the sum of the first 14 terms in the sequence of deposits described in the text above the sum formula.

SOLUTION First, we need to find the 14th term. Using the formula shown previously, ​​a​  n​​  = ​a​  1​​  +  (n − 1) d​ where a1 = $10, n = 14, and d = $5.

284   Chapter 5  The Real Number System ​a​  14​​ = $10 + (14 − 1 ) $5 ​​ ​ ​       ​​ ​ ​ =​  $10 + 13($5 ) ​ = $75 Next, substitute into the formula for finding the sum. n(​a​  1​​  + ​a​  n​​  ) ​S​  n​​ = ​ ________  ​     2 ​​    ​    ​  14($10 + $75 )  ​​ ​ ​ ​ = ___________ ​     ​   2 ​ = $595 The total of all deposits over the first 14 months is $595.

Try This One 3. Find the sum of n terms of an arithmetic sequence.

4

Find the sum of the first 12 terms of each sequence. (a) 5, 12, 19, 26, 33, . . .   (b)  −1, −3, −5, −7, −9, . . .   (c) ​​ _15 ​ , ​ _25 ​ , ​ _35 ​ , ​ _54 ​ ,   . . .​

Geometric Sequences Remember the geniuses from the beginning of the lesson that decided to open a savings account in their sock drawer? Suppose that a month after starting to save $400 a month, the wife gets discovered by a major modeling agency, and the cash starts rolling in. Instead of adding $400 to their savings every month, now they’re able to double the total each month. The amount would now look like $400, $800, $1,600, $3,200, $6,400, .  .  .  . This is an example of a geometric sequence. A sequence is geometric when every term after the first is obtained by multiplying the previous term by the same fixed nonzero number. This multiplier is called the common ratio. Examples of Geometric Sequences 1, 3, 9, 27, 81, 243, . . . 2, 10, 50, 250, 1,250, . . . 5, −10, 20, −40, 80, . . . 1 __ 1 ___ 1 ​1, ​ _14 ​ , ​ __    ​ ,  …​ 16  ​ , ​  64  ​ , ​  256

Sidelight

Common Ratio r = 3 r = 5 r = −2 ​r = ​ _14 ​​ 

The Fibonacci Sequence

Sometime around the year 1200, an Italian mathematician named Leonardo Pisano, who went by the name Fibonacci, discovered the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . . Do you see the pattern? After the first two terms, each successive term comes from adding the two previous terms. Not exactly earthshattering, right? The truly amazing thing about this innocent little list of numbers is that numbers from this sequence are found all over the place in nature. The number of petals that many flowers have correspond to Fibonacci numbers. Lilies and irises have 3 petals, buttercups have 5, delphiniums have 8, and marigolds have 13. Different varieties of daisies have 21, 34, 55, or 89 petals. The seeds on the head of a sunflower spiral out from the center in clockwise and counterclockwise directions: some sunflowers have 21 spirals in one direction and 34 in the other. A giant sunflower has 89 in one direction and 144 in the other, and these are Fibonacci numbers as well. Speaking of spirals, the beautiful shell of the chambered nautilus, pictured here, is a smooth spiral that can be built using a series of squares with areas matching—you

guessed it—the Fibonacci sequence. Interestingly, Fibonacci numbers have a very strong connection to the golden ratio we studied in the Sidelight on page 256. This connection will be explored in Project 2 at the end of this chapter. Even more surprising, ratios of Fibonacci ©Kaz Chiba/Getty Images RF numbers are used by some stock analysts to find ideal times to buy or sell a certain stock. The theory is based on the idea that the golden ratio is not just naturally attractive to the human eye, but to the human subconscious as well. (Now that’s deep.) A Google search for “Fibonacci” and “stock market” yields a veritable cornucopia of pages that explain the technique (as well as a bunch that claim it’s bunk . . . you be the judge).

Section 5-7  Arithmetic and Geometric Sequences    285

Example 5

Finding the Terms of a Geometric Sequence Write the first five terms of a geometric sequence with first term 4 and common ratio −3.

SOLUTION Start with first term 4, then multiply by −3 to obtain each successive term: ​4, − 12, 36, − 108, 324​

4. Define geometric sequence.

Try This One

5

Write the first five terms of a geometric sequence with first term _​​ 21 ​​  and common ratio 4. We can also find a formula to quickly calculate any term of a geometric sequence without having to find all of the terms that come before it. A geometric sequence with first term a1 and common ratio r looks like ​​a​  1​​,   ​a​  1​​  ⋅  r,   ​a​  1​​  ⋅  r ⋅ r,   ​a​  1​​  ⋅  r ⋅ r ⋅ r,  . . .​ Doing some simplifying, we find that ​​a​  1​​  = ​a​  1​​   ​a​  2​​  = ​a​  1​​  ⋅  r  ​a​  3​​  = ​a​  1​​  ⋅ ​r​​2​   ​a​  4​​  = ​a​  1​​  ⋅ ​r​​3​​ Every term an looks like a1 times some power of r, and the power is one less than the subscript of that term in the sequence. This gives us a formula: Finding the nth Term of a Geometric Sequence The nth term of a geometric sequence can be found using the formula ​​a​  n​​  = ​a​  1​​  ⋅ ​r n−1 ​​ ​​ where n is the number of the term you want to find, a1 is the first term, and r is the common ratio.

Example 6

Finding a Particular Term of a Geometric Sequence Find the 12th term of a geometric sequence with first term ​​ _12 ​​  and common ratio 6.

SOLUTION Substitute in the formula using ​​a​  1​​  = ​ _12 ​ ,  r = 6​and n = 12. ​​ ​ ​a​  n​​ = ​a​  1​​ ​r n−1 1 ​a​  12​​ = __ ​   ​ ​ (6)​​12−1​ 2 ​​  ​  ​  ​    ​  ​​ 1 11 __ ​ = ​   ​ (​6​​ ​  ) Don't 2 ​ = 181,398,528

forget order of operations!

The twelfth term is 181,398,528. Example 6 illustrates the distinguishing feature of geometric sequences: they can grow VERY quickly.

286   Chapter 5  The Real Number System

Sidelight

Sequences and the Planets

In the late 1700s, the world of scientists became very excited when two German astronomers discovered a mathematical sequence that actually predicted the average distance the then-known planets were from the sun. This distance is measured in what are called astronomical units. One astronomical unit (AU) is equal to the average distance Earth is located from the sun (about 93 million miles). The sequence, called the Titius-Bode law (named for its discoverers in 1777), is 0, 3, 6, 12, 24, 48, 96, 192, . . . . When 4 is added to each number and the sum is divided by 10, the result gives the approximate distance in AUs each planet is from the sun as shown.

Planet

Sequence

AU

Mercury

(0 + 4) ÷ 10

0.4

Venus

(3 + 4) ÷ 10

0.7

Earth

(6 + 4) ÷ 10

1.0

Mars

(12 + 4) ÷ 10

1.6

_____

(24 + 4) ÷ 10

2.8

Jupiter

(48 + 4) ÷ 10

5.2

Saturn

(96 + 4) ÷ 10

10.0

_____

(192 + 4) ÷ 10

19.6

Between Mars and Jupiter, no planet exists, and it looks like the sequence breaks down. However, an asteroid belt is located between Mars and Jupiter, and some astronomers thought that this was once a planet. More amazing was the fact that in 1781 William Herchel discovered the planet Uranus, which is located at 19.2 AU from the sun! Unfortunately, the next two planets discovered didn’t fit the pattern at all. The predicted locations are 38 and 76.4 AU: Neptune is just 30.1 AU from the sun, and Pluto is 39.5. And now poor Pluto isn’t even considered a planet anymore. So maybe the whole thing was just a big coincidence to begin with. This is a cautionary tale about thinking you’ve found a definite pattern from looking at a small handful of numbers. Sometimes the pattern continues, and sometimes it does not.

Try This One 5. Find a particular term of a geometric sequence.

6

Find the ninth term of a geometric sequence with first term 3 and common ratio (−2).

There is also a formula for adding terms of a geometric sequence, which we’ll prove in Project 3. Finding the Sum of a Geometric Sequence ​a​  1​​(1  − ​r n​​ ​  ) The sum of the first n terms of a geometric sequence is S​  ​​ n​​  = ​ _______     ​,  ​ where a1 is the first 1 − r term and r is the common ratio.

Example 7

Finding the Sum of a Geometric Sequence Find the sum of the first 15 terms of a geometric sequence with first term 8 and common ratio  ​− ​ _12 ​​ .

Section 5-7  Arithmetic and Geometric Sequences    287

SOLUTION Using a ​​ ​  1​​  =  8, r = − ​ _12 ​ ,​ and n = 15, substitute into the formula ​a​  ​​(1  − ​r​​n​  ) ​S​  n​​ = _______ ​  1  ​     1 − r ​​    ​    ​  __________  ​​ 8[1 − ​​ ​​​​ ​  ] ​ (​​− ​ _12 ​​ )​15 ​S​  15​​ =    ​   ​   1 _ [1 − ​​(​​− ​  2 ​​ )​​​  ] ​ ≈ 5.33

A calculator is a big help here.

The sum is about 5.33.

Try This One 6. Find the sum of n terms of a geometric sequence.

7

Find a decimal approximation for the sum of the first 20 terms of a geometric sequence with first term 6 and common ratio _​​ 14 ​​ .

Applications of Sequences EXAMPLE 8

Applying Arithmetic Sequences to Saving for a Home We began this section with an example in which a couple is saving money for a down payment on a house. Suppose that your plan is similar, but you decide to put away $150 the first month, $175 the second, $200 the third, and so on. Would you save more or less in the first 2 years than if you put away $400 each month?

SOLUTION If we list the amount saved each month, we get an arithmetic sequence with first term 150 and common difference 25. To find the amount after 2 years, we need to find the sum of the first 24 terms, or S24, for this sequence. But first we need to know a24 since it’s part of the formula for S24. ​a​  ​​ = ​a​  ​​  +  (n − 1 ) d    ​​  n​  ​  1 ​ ​​ ​a​  24​​ = 150 + (24 − 1 ) 25 = 150 + 23 ⋅ 25 = 725 Now we use the formula for the sum, Sn. n(​a​  1​​  + ​a​  n​​  ) S​ ​  n​​ = ​ ________  ​     2    ​​  ​  ​  ​ ​​ 24(150 + 725 ) ​S​  24​​ = ___________ ​     ​ = 12(875 ) = $10,500   2 If you had put away $400 each month, the total after 2 years would simply be 24 × 400 = $9,600. The first plan saves more money in 2 years.

Try This One

8

If you’re offered a 1-year job where you get paid just $20 the first week, but $20 is added to your pay each week, how much money would you earn in that year?

EXAMPLE 9

Applying Geometric Sequences to Salary When a worker earns a 4% raise each year, to find his or her salary in any given year, you multiply the salary from the year before by 1.04. If you take a job with a starting annual salary

288   Chapter 5  The Real Number System

Math Note In calculus, we study infinite series, which is what results when you add ALL of the terms of an infinite sequence. Believe it or not, often the result is a finite number! The sum of the infinitely many terms of the sequence described in Example 7 is 16/3, for example. See Problems 81 and 82 for more on this topic.

of $40,000 and earn a 4% raise each year, how much money would you earn in your first 10 years?

SOLUTION Since the salary gets multiplied by 1.04 each year, the annual salaries form a geometric sequence with first term 40,000 and common ratio 1.04. We can use the formula for the sum of a geometric sequence with n = 10: ​a​  ​​(1  − ​r​​n​  ) ​S​  n​​ = _______ ​  1  ​     1 − r 10 ​ ​​    ​    ​  _______________ ​  ​​ 40,000(1  − ​1.04​​ ​  ) ​S​  10​​ =    ​      ​ 1 − 1.04 ​ = $480,244.28

Try This One

9

For the job in Example 9, if you’re offered the option of starting at $30,000 and taking an 8% raise each year, would that earn you more or less in the first 10 years? By how much?

Answers to Try This One 1

78 __ (a) 7, 8.3, 9.6, 10.9, 12.2     4 (a) 522  (b) −144  (c) ​​    8 $27,560 5 ​​          1 _ (b) 6, 16, 26, 36, 46      5 ​​  2 ​  ,  2,  8,  32,  128​              9  $45,647.41 less 13 (c) ​​ _38 ​ , − ​ _18 ​ , − ​ _58 ​ , − ​ _98 ​ , − ​ __  ​​         6 768 8

2

31.7 lb         7  About 8

3

19 (a) ​​ __   9 ​​  

(b) ​− ​ _43 ​​ 

Exercise Set

5-7

Writing Exercises 1. If you’re given a list of numbers, explain how to tell if the list is an arithmetic sequence. 2. If you’re given a list of numbers, explain how to tell if the list is a geometric sequence. 3. For an arbitrary sequence, explain what the symbol an represents. 4. For an arbitrary sequence, explain what the symbol Sn represents. 5. Give an example of a quantity in our world that might be described by an arithmetic sequence, and justify your answer.

6. Give an example of a quantity in our world that might be described by a geometric sequence, and justify your answer. 7. What’s the difference between a sequence and a set of numbers? 8. A sequence is called alternating if its signs alternate from positive to negative. Is it possible for an arithmetic sequence to be alternating? What about a geometric sequence?

Computational Exercises For the arithmetic sequence in Exercises 9–24, find each:

(a) the first five terms (b) the common difference



(c) the 12th term (d) the sum of the first 12 terms

Section 5-7  Arithmetic and Geometric Sequences    289 9. a1 = 1, d = 6 10. a1 = 10, d = 5 11. a1 = −9, d = −3 12. a1 = −15, d = −2 13. ​​a​  1​​  = ​ _14 ​ ,  d = ​ _38 ​​  14. ​​a​  1​​  = ​ _37 ​ ,  d = ​ _17 ​​  15. ​​a​  1​​  =  4,  d = − ​ _13 ​​  16. ​​a​  1​​  = ​ _52 ​ ,  d = − ​ _14 ​​ 

17. 5, 13, 21, 29, 37, . . . 18. 2, 12, 22, 32, 42, . . . 19. 50, 48, 46, 44, 42, . . . 20. 12, 7, 2, −3, −8, . . . 35 __ 67 17 __ __ 21. ​​ _18 ​ , __ ​  19 24 ​ , ​  24 ​ , ​  8 ​,  ​  24 ​ , …​ 17 __ 21 __ 22. ​​ _21 ​ , __ ​  109  ​,  __ ​  13 10 ​ , ​  10 ​ , ​  10 ​ , …​ 23. 0.6, 1.6, 2.6, 3.6, 4.6, . . . 24. 0.3, 0.7, 1.1, 1.5, 1.9, . . .

For the geometric sequence in Exercises 25–40, find each: (a) the first five terms (b) the common ratio (c) the 12th term (d) the sum of the first 12 terms 25. a1 = 12, r = 2 31. ​​a​  1​​  =  100, r = − ​ _14 ​​  1 26. a1 = 8, r = 3 32. ​​a​  1​​  =  10, r = − ​ __ 10  ​​  27. ​​a​  1​​  =  −  5, r = ​ _14 ​​  33. 4, 12, 36, 108, 324, . . . 28. ​​a​  1​​  =  −  9, r = ​ _23 ​​  34. 6, 12, 24, 48, 96, . . . 1 __ 1 29. ​​a​  1​​  = ​ _16 ​ , r = − 6​ 35. ​​ _12 ​ , ​ _14 ​ , ​ _18 ​ , ​ __ 16  ​,  ​  32  ​,  . . .​ 2 __ 2 ___ 2 30. ​​a​  1​​  = ​ _37 ​ , r = − 3​ 36. ​​ _32 ​ , ​ _29 ​ , ​ __    ​   ,  ​     ​   ,  ​  27 81 243   ​,  . . .​ 37. −3, 15, −75, 375, −1,875, . . . 38. −3, 12, −48, 192, −768, . . . 39. 1, 3, 9, 27, 81, . . . 1 40. 8, 2, _​​  12 ​ , ​ _18 ​ , ​ __ 32  ​,  . . .​ For Exercises 41–48, determine whether each sequence is an arithmetic sequence, a geometric sequence, or neither. 41. 5, −15, 45, −135, 405, . . . 42. 42, 35, 28, 21, 14, . . . 43. 2, 4, 8, 14, 22, . . . 44. 2, 4, 12, 48, 240, . . . 45. 6, 2, −2, −6, −10, . . .

9 ___ 3 ___ 81 27 ____ 1 __ 46. ​​ __    ​,  . . .​ 10  ​,  ​  40  ​,  ​  160   ​,  ​  640  ​,  ​  2,560 5 _ 3 7 1 11 _ _ _ __ 47. ​​  8 ​ , ​  8 ​ , − ​  8 ​ , − ​  8 ​ , − ​  8 ​, . . .​   9 81 27 ___ __ __ 48. 4, ​−  3, ​  16  ​,  − ​  64 ​ , ​  256  ​,  . . .​

In Exercises 49–60, find and simplify a formula for an, the nth term of the given sequence. (Hint: Decide what kind of sequence it is, then use information from within the section.) 49. 3, 12, 48, 192, 768, . . . 50. −1, 2, −4, 8, −16, . . . 51. 11, 19, 27, 35, 43, . . . 52. −8, −5, −2, 1, 4, 7, . . . 53. 40, 30, 20, 10, 0, −10, . . . 54. 8, 2, −4, −10, −16, −22, . . . 55. 0.5, 0.1, 0.02, 0.004, 0.0008, . . . 56. 243, 81, 27, 9, 3, 1, . . . 15 ___ 45 135 ____ 405 57. 5, ​​ ___ ​, ​     ​ , ​ ____ ​, ​     ​ , . . .​ 4 16 64 256 1 1 1 1 58. ​−  2, − ​ __ ​ , − ​ ___  ​,  − ​ ____    ​,  − ​ ____    ​,  . . .​ 3 18 108 648 59. 4, 2.4, 1.44, 0.864, 0.5184, . . . 60. −2, 0.8, −0.32, 0.128, −0.0512, . . . 61. If the fifth term of a geometric sequence is 81, and the common ratio is ​​ _34 ​ ,​find the sum of the first 20 terms to two decimal places. 62. If the fifth term of a geometric sequence is __ ​​ 64 27 ​​  and the 256 ___ sixth term is ​​ 81 ​ ,​find the sum of the first 50 terms to the nearest whole number. 63. If the tenth term of an arithmetic sequence is 10 and the common difference is 3, find the sum of the first 30 terms. 64. If the eighth term of an arithmetic sequence is _​​ 52 ​​  and the common difference is ​− ​ _32 ​ ,​find the sum of the first 100 terms.

Applications in Our World 65. A new car that cost $27,000 originally depreciates in value by $3,500 in the first year, $3,000 in the second year, $2,500 in the third year, and so on. (a) How much value does it lose in the seventh year? (b) How much is the car worth at the end of the seventh year? 66. A large piece of machinery at a factory originally cost $50,000. It depreciates by $1,800 the first year, $1,750 the second year, $1,700 the third year, and so on. (a) What is the amount of depreciation in the fifth year? (b) What is the value of the machinery after the fifth year? 67. At one university, the first parking violation in the union lot results in a $5 fine. The second violation is an $8 fine, the third $11, and so on. One particular student has the fines sent home to her parents, who inform her that if she racks up over $100 in fines, they’re selling her car on eBay.

If she has been given eight tickets, will she be able to keep her car? 68. A company decided to fine its workers for parking violations on its property. The first offense carries a fine of $25, the second offense is $30, the third offense is $35, and so on. What is the fine for the eighth offense? 69. A contractor is hired to build a cell phone tower, and its price depends on the height of the tower. The contractor charges $1,000 for the first 10 feet, and after that the price for each successive 10-foot section is $250 more than the previous section. What would be the cost of a 90-foot tower? 70. At one plant that produces auto parts, management had to lay off 25 workers in the first quarter of 2016, 30 workers in the second quarter, 35 in the third, and continued this pattern through 2017. If the plant originally had 700 workers, how many remained at the end of 2017?

290   Chapter 5  The Real Number System 71. A bungee jumper reaches the bottom of her jump and rebounds 80 feet upward on the first bounce. Each successive bounce is ​​ _12 ​​  as much as the previous. If she bounces a total of 10 times before coming to a stop, what is the total height of all bounces? 72. A ball that rebounds _​​ 78 ​​   as high as it bounced on the previous bounce is dropped from a height of 8 feet. How high does it bounce on the fourth bounce and how far has it traveled after the fourth bounce? The fourth bounce is over when the fifth bounce begins. 73. A desperate gambler named Phil borrows $1,200 from Vito the loan shark. Vito informs him that he will be charging 10% interest each month, and Phil must pay off his debt in 1 year or his thumbs will be broken. How much cash does Phil have to cough up at the end of 1 year to save his thumbs? 74. A student deposited $500 in a savings account that pays 5% annual interest. At the end of 10 years, how much money will be in the savings account? 75. A contestant on a quiz show gets $1,000 for answering the first question correctly, $2,000 more for answering the second question correctly, $4,000 more for the third question, $8,000 for the fourth, and so on. The contestants can stop at any time, but if they get a question wrong, they lose it all. Monica’s goal is to make enough money to pay cash for a $225,000 house. How many questions does she have to answer correctly? 76. An eccentric business owner gives a new employee a choice: he can work for $4,000 per month, or he can get 1 cent the first day, 2 cents the second, 4 cents the third,

8 cents the fourth, and so on. The employee, completely offended at the thought of working all day for a measly penny, chooses the $4,000 without a second thought. If there are 22 workdays in an average month, how much money per month did his haste cost him? 77. Suppose that you’re offered two jobs: both have a starting salary of $30,000 per year. The first includes a raise of $2,000 each year, and the second a raise of 4% each year. Which job will pay more in the 10th year, and by how much? 78. Your financial advisor calls, offering you two can’tmiss propositions if you can scrape together an initial investment of $5,000. The first investment will increase in value by $1,000 per year, every year. The second will increase at the rate of 10% every year. Which is the better investment if you plan to keep the money invested for 12 years? 79. Rhonda and her coworkers receive a 4% raise every year. Rhonda’s starting salary was $38,000 per year. One day, she sneaks a peek at the pay stub of Keiko (one of her coworkers) and finds that Keiko is in her sixth year at the company, and her current pay is $4,100 per month. Did the coworker start at a higher salary than Rhonda? If so, by how much? 80. A utility company claims that it has been cutting its emissions by 9% each year for the last 4 years. This year, the EPA takes measurements and finds that its carbon dioxide emissions are at 84% of the acceptable maximum. Assuming that the company’s claim is accurate, how far  above or below the acceptable maximum was it 4 years ago?

Critical Thinking When a geometric sequence has infinitely many terms, if the and they’re both of reproductive age. Do you think that a common ratio is between −1 and 1, you can still find the sum sequence describing the population is more likely to be ​a​  1​​ of all the terms. The formula for the sum in this case is ​S = ​ ___ arithmetic or geometric? Explain your answer. 1 − r  ​ .​  Use this formula for Exercises 81 and 82. 84. In most cases, reducing fractions is a good idea, but not always. When trying to find a pattern in a sequence 81. A repeating decimal between −1 and 1 can be written of numbers to determine what kind of sequence it is, as the sum of an infinite geometric sequence as shown reducing fractions is notorious for obscuring the pattern. below. For each sequence below, if you “unreduce” some of the 3 3 3 fractions, you can recognize the type of sequence. Do so, ​0.333 . . . = ___ ​    ​ + ​ ____    ​ + _____ ​     ​  + . . .​ then find a formula for an, the nth term of the sequence. 10 100 1,000 1 7 5 13 Find a1 and r, then find the sum of all of the terms using (a) ​​ __ ​ , 1, ​ __ ​ , ​ __ ​ , ​ ___ ​, 4, . . .​   4 4 2 4 the sum formula above. 82. Refer to Problem 81. Write the repeating decimal 1 4 14 (b) ​2, ​ __ ​ , − ​ __ ​ , −  3, − ​ ___ ​, . . .​   0.151515 .  .  . as the sum of an infinite geometric 3 3 3 sequence and find the sum using the given sum formula. 83. Consider the following situation: in a mythical post25 ___ 93 ___ 86 ___ 79 ___ 72 (c) ​​ ___ ​, ​     ​, ​     ​, ​     ​, ​     ​, . . .​   apocalyptic world, there are just two people left to 2 8 8 8 8 repopulate. Fortunately, one is a man and one is a woman,

CHAPTER

Section

5 Summary Important Terms

Important Ideas

Natural number Factor Divisor Divisible Multiples Prime number Prime factorization Composite number Fundamental theorem of arithmetic Greatest common factor Relatively prime Least common multiple

The set of natural numbers is {1, 2, 3, 4, 5, . . .}. If we can write a natural number as the product of two numbers, we call those numbers factors of the original number. A natural number is prime if its only factors are one and itself and composite if it has other factors. Every composite number can be factored uniquely as the product of prime numbers—we call this the prime factorization for a number. This fact is known as the fundamental theorem of arithmetic. For a set of two or more numbers, we can find the greatest number that is a factor of each (the greatest common factor, or GCF). We can also find the smallest number that is a multiple of each (the least common multiple, or LCM).

Whole number Integer Opposite Absolute value Order of operations

The set of whole numbers is {0, 1, 2, 3, . . .}. The set of integers is {. . . − 3, −2, −1, 0, 1, 2, 3, . . .}. Every integer has an opposite, or negative. The absolute value of a negative integer is the opposite of that integer (making it positive). The absolute value of zero or any positive integer is itself. When we perform calculations that involve more than one operation, the order in which we perform them affects the outcome, so there is an order of operations that must be followed.

5-3

Rational number Proper fraction Improper fraction Mixed number Lowest terms Reciprocal Least common denominator Place value Terminating decimal Repeating decimal

Rational numbers can be written either as fractions or decimals. The decimals are either terminating or repeating; the fractions have integers in the numerator and denominator. A fraction is in lowest terms when the numerator and denominator have no common factors. The rules for performing operations on rational numbers are given in this section.

5-4

Irrational number Radical Perfect square Like radicals Rationalizing the denominator

Irrational numbers are nonterminating nonrepeating decimals. Square roots of numbers that are not perfect squares are irrational numbers. Rules for performing operations on square roots are given in this section.

Real number Property Identity

A real number is either rational or irrational. We studied 11 properties of the real numbers: the closure properties for addition and multiplication, the commutative properties for addition and multiplication, the associative properties for addition and multiplication, the identity properties for addition and multiplication, the inverse properties for addition and multiplication, and the distributive property.

5-1

5-2

5-5

291

292   Chapter 5  The Real Number System

5-6

5-7

Exponential notation Base Exponent Scientific notation

In order to write very large or very small real numbers without a string of zeros, mathematicians and scientists use scientific notation. Using powers of 10, scientific notation simplifies operations such as multiplication and division of large or small numbers. When the power of 10 is positive, the number written in scientific notation is greater than 1. When the power is negative, the number is less than 1.

Sequence Arithmetic sequence Common difference Geometric sequence Common ratio

A sequence of numbers is a list of related numbers in a definite order. We studied two basic types of sequences: arithmetic sequences and geometric sequences. Many problems in our world can be solved using sequences. A sequence is arithmetic if every member is obtained from adding the same number to the previous member. A sequence is geometric if every member is obtained from multiplying the previous member by the same number.

Math in  Government Spending REVISITED ©Photodisc/Punchstock RF 12

Question 1: Divide the total spending, 3.54 × 10 , by the 8

4

population, 3.24 × 10 , to get 1.092593 × 10 . This means the government spent $10,925.93 per person. Next divide the

$61,635 dollars from every man, woman, and child in the United States to pay off the national debt. Question 4: Divide the national debt by 535 to get

deficit, 1.298 × 1012, by the population to get $1.811 × 103. The

3.733 × 1010. This means that every representative in

government spent $1,811 per person more than it took in.

Congress would have to contribute about $37.3 billion to

Question 2: Divide the amount spent on interest, 11

12

2.15 × 10 , by the total spending, 3.54 × 10 , to get 0.061.

pay off the national debt. Nobody is that well off! Question 5: Subtract the actual deficit, 5.87 × 1011, from

This means that 6.1% of every dollar spent went just to pay

the rounded deficit, 5.9 × 1011. The result is 0.03 × 1011, or

interest on the national debt.

3 × 109. You just cost the American people 3 BILLION dollars.

Question 3: Divide the national debt, 1.544 × 1013, by the

So thanks for that.

4

population to get 6.1635 × 10 . This tells us that it would take

Review Exercises Section 5-1 For Exercises 1–4, find all factors of each. 1. 60 2. 45

3. 380 4. 650

17. An investor takes out two certificates of deposit. One matures every 18 months, the other every 22 months. He decides to continue rolling them over until they both mature at the same time. How long will it be until that happens?

For Exercises 5–8, find the first five multiples of each.

Section 5-2

5. 4 6. 32

For Exercises 18–27, perform the indicated operations.

7. 9 8. 60

For Exercises 9–12, find the prime factorization for each. 9. 96 10. 44

11. 250 12. 720

For Exercises 13–16, find the GCF and LCM. 13. 6, 10 14. 35, 40

15. 60, 80, 100 16. 27, 54, 72

18. −6 + 24 19. 18 − 32 20. 5(−9) 21. 32 ÷ (−8) 22. 6 + (−2) − (−3) 23. 6 · 8 − (−2)2 24. 4 · 3 ÷ (−3) + (−2) 25. 100 − {[6 + (2 · 3) − 5] + 4}

Chapter 5  Review Exercises   293 26. {8 · 73 − 55[(3 + 4) − 6]} + 20 27. (−5)3 + (−7)2 − 34 28. Ty has a separate checking account set aside for housing, utilities, and car payment. For fall semester, the account starts with $2,400. During the 5-month semester, Ty writes checks each month for rent ($340), utilities ($45), and car payment ($170). His parents make three deposits of $350 during the semester. How much money is left at the end of the semester?

Section 5-3

For Exercises 44–45, change each fraction to a decimal. 5 6 44. ​​ ___  ​​  45. ​​ __ ​​  16 7 46. Change __ ​​  236 ​​  into a mixed number, then into a decimal. For Exercises 47–50, change each decimal to a reduced fraction. ¯ 47. 0.6875 49. ​0.2​ ​​ 5   ¯ 48. 0.22 50. ​0.​ 45​​  

51. In the National Football League, _​​ 38 ​​  of the teams make the playoffs; in Major League Baseball, __ ​​ 154  ​​ make the playoffs, and in the National Basketball Association and the National Hockey League, __ ​​  158  ​​ do. There are 32 teams in the NFL, and 30 teams in MLB, NBA, and NHL. How many teams make the playoffs in each sport?

Section 5-4 ___

____ √___ 52. ​​√ 48 ​​       53. ​​    112 ​​ 

√ 

5 5 ___ ___ 54. ​​ _____    ​​       55. ​​   ​    ​ ​​   √ 12 ​  20 ​  For Exercises 56–61, perform the indicated operations. ___

__

___

​√ 20 ​   56. ​​√___  20 ​  + 2​√__  75 ​  − 3​√    5 ​​ ___ __ ​​  59. ​​ _____ √ 57. ​​√___  18 ​  − 5​√ ​√   5 ​ ___ 2 ​  + 4​  72 ​​  58. ​​√__  27 ​  ⋅ ​√ __ __ __  63 ​​  __ ___ ___ √ 14 ​  + ​√ 6 ​  ) (​√ 14 ​  − ​√ 6 60. ​​√   6 ​( ​√ 2 ​  + ​√ 5 ​    )​    61. ​ (​  ​  ​)

Section 5-5 For Exercises 62–67, classify each number as natural, whole, integer, rational, irrational, and/or real. 5 62. ​− ​ __ 16  ​​  63. 0.86 ¯ 64. ​0.3​ ​​ 7  

Section 5-6 72. 45 73. (−3)0 74. 3−4 75. 72 · 74

​5​​ ​ 76. ​​ __  ​​  ​5​​2​ 77. (34)2 78. 23 · 2−5 79. 622 · 6−3 6

For Exercises 80–83, write each number in scientific notation. Round to two decimal places. 80. 3,826 81. 25,946,000,000

82. 0.00000327 83. 0.00048

For Exercises 84–87, write each number in decimal notation. 84. 5.8 × 1011 85. 2.33 × 109

86. 6.27 × 10−4 87. 8.8 × 10−6

For Exercises 88–89, perform the indicated operations and write the answers in scientific notation. 88. (3.2 × 10−5)(8.9 × 10−7) 1.8  × ​10​​−5​ 89. ​​ _________  ​​    3  × ​10​​2​ 90. According to numerous websites, the average American drinks the equivalent of 5.97 × 102 cans of soda per year. According to the U.S. Census Bureau, the population of the United States at the end of 2016 was about 3.24 × 108. How many cans of soda were consumed in the United States in 2016? Write your answer in both scientific and decimal notation, then write how the decimal answer would be read aloud. 91. The speed of sound in air is about 1.126 × 103 feet per second. When the volcano Krakatoa erupted on August 26, 1883, the blast could be heard 3,000 miles away. How long would it have taken for the sound to reach that far?

Section 5-7

For Exercises 52–55, simplify each.

___

68. ​8  ⋅ ​ _18 ​  = 1​ 69. 3 + 5 = 5 + 3 70. 6 + 5 is a real number 71. 2(3 + 8) = 2 · 3 + 2 · 8

For Exercises 72–79, evaluate each.

For Exercises 29–31, reduce each fraction to lowest terms. 75 56 265 29. ​​ ___ ​​  30. ​​ ___ ​​  31. ​​ ____ ​​  95 64 30 For Exercises 32–43, perform the indicated operations. Write your answer in lowest terms. 1 5 2 3 __ 1 1 32. ​​ __ ​  + ​ __ ​​  38. ​​ __ ​​​  __ ​   ​  + ​   ​  − ​ __ ​  ​​​ 8 6 3(4 2 6) 2 3 2 1 7 3 33. ​​ ___  ​  − ​ __ ​  + ​ __ ​​  39. ​1 ​ __ ​  − ​ __ ​​   ​​​  ​ ​​ 10 5 4 8 (4 ) 5 3 6 1 1 34. ​​ __ ​  × ​ __ ​​  40. ​− ​ __  ​​ __ ​​   ​  +  2 ​ __ ​​  ​​ 9 7 7(2 3) 15 21 9 2 1 35. ​​ ___ ​  ÷ ​​ − ​ ___ ​  ​​​ 41. ​​ ___  ​  + ​ ​− ​ __ ​​​ − ​ __ ​​  ​​ 16 ( 40 ) 10 ( 5 4 ) 1 2 3 5 2 2 36. ​​ __ ​  ÷ ​​ __ ​   ​  + ​ __ ​  ​​​ 42. ​​ __ ​  − ​ __ ​​  ​−  1  + ​ __ ​​  ​​ 2 (3 4) 8 3( 5) 1 3 7 1 9 5 1 43. ​​ __ ​  − ​ __ ​  − ​ __ ​  ⋅ ​ __ ​​  37. ​​ ___  ​  × ​​ __ ​   ​  − ​ __ ​  ​​​ 2 4 8 6 10 ( 6 8 )



For Exercises 68–71, state which property of the real numbers is being illustrated.

___

65. ​​√ 15 ​​  66. 0 67. 16

For Exercises 92–93, write the first six terms of the arithmetic sequence. Find the ninth term and the sum of the first nine terms. 92. a1 = 8, d = 10

93. ​​a​  1​​  =  − ​ _15 ​ ,  d = ​ _12  ​​

For Exercises 94–95, write the first six terms of the geometric sequence. Find the ninth term and the sum of the first nine terms. 94. a1 = −3, r = 3 95. ​​a​  1​​  =  − ​ _25 ​ ,  r = − ​ _12 ​​  96. The population of the United States is increasing by about 2.2 million people per year. Given that the population was 324 million in 2016, find the expected population in 2025. 97. The net profit of a small company is increasing by 5% each year. If the net profit for this year is $20,000, find the projected profit for the sixth year of operation and the total amount of money the company can be expected to make in the next 6 years.

294   Chapter 5  The Real Number System

Chapter Test For Exercises 1–6, classify each number as natural, whole, integer, rational, irrational, and/or real. 1. −27 2. 8.6 3. ​​ _59 ​​  ¯ 4. ​0.6​ ___ ​​   2 5. ​​√ 50 ​​___   6. ​−​√ 25 ​​  For Exercises 7–8, find the GCF and LCM of each group of numbers. 7. 42, 56    8.  150, 175, 200 For Exercises 9–11, reduce each fraction to lowest terms. 15 9. ​​ ___ ​​  35 81 ____ 10. ​​    ​​  108 112 11. ​​ ____ ​​  175 For Exercises 12–13, simplify the radical. ___

____

√ 243 ​​  12. ​​√ 48 ​​       13. ​​   For Exercises 14–22, perform the indicated operations.

14. −5 · (−6) + 3 · 2 15. 18 − 32 − 42 + 6 ÷ 3 5 3 2 16. ​​​(__ ​​​   ​  ⋅ ​ __ ​​ )​​  ÷ ​ __ ​​​  6 4 3 1 __ 1 2 3 __ 17. ​​​(​​​   ​  + ​   ​​ )​​  − ​ __ ​  ⋅ ​ __ ​​​  7 9 3 4 1 __ 2 √___ __ 18. ​−  6  + ​   ​  ÷ ​   ​  + ​  81 ​​  4 3 19. [4 + (2 × 3) − 62] + 18 ___

__

__

20. ​​√ 27 ​  + ​√   3 ​(2 ​ √ 2 ​  − 1 )​ 16 _____ 21. ​​  ___   ​​  √ ​  32 ​  ___ ___ 22. ​2 ​√ 50 ​  − 3 ​√ 32 ​​ 

For Exercises 23–24, change each decimal into a reduced fraction. ¯ 23. 0.875   24. ​ 0.​  ​​ 2   For Exercises 25–30, state the property illustrated. 25. 0 + 15 = 15 + 0 26. 6 × 7 is a real number. 27. 0 + (−2) = −2 28. ​​ _15 ​  ⋅ 5 = 1​ 29. (4 × 6) × 10 = 4 × (6 × 10) 30. 6(5 + 7) = 6 · 5 + 6 · 7 For Exercises 31–35, evaluate each. 31. 84 32. 7−3 33. 60 34. 5−3 · 5−2 2−3 35. ​​ ___  ​​  ​22​​ ​ 36. Write 52,000,000 in scientific notation. 37. Write 0.00236 in scientific notation. 38. Write 9.77 × 103 in decimal notation. 39. Write −6 × 10−5 in decimal notation. 40. (5.2 × 108)(3 × 10−5) = ______ in scientific notation. 2.1  × ​10​​9​ 41. Divide ________ ​​   ​​  7  × ​10​​5​ 42. Write the first seven terms, the 20th term, and the sum of the first 20 terms for the arithmetic sequence where a1 = 1 and d = 2.5. 43. Write the first seven terms, the 15th term, and the sum of the first 15 terms for the geometric sequence where ​​a​  1​​  = ​ _34 ​​  and r​  = − ​ _16 ​​ . 44. A runner decides to train for a marathon by increasing the distance she runs by ​​ _12 ​​  mile each week. If she can run 15 miles now, how long will it take her to run 26 miles? 45. A gambler decides to double his bet each time he wins. If his first bet is $20 and he wins five times in a row, how much did he bet on the fifth game? Find the total amount he bet.

Projects 1. To add together the first hundred natural numbers, you can think of them as an arithmetic sequence with first term and common difference both equal to one. In this project, we’ll develop a different method. (a) Find the sum using the formula for the sum of an arithmetic sequence. (b) Write out the sum of the first 100 natural numbers. You don’t have to write out all 100 numbers, but write at least the first five and the last five with an ellipsis (. . .) between. (c) Write the same sum underneath the first one, but in the opposite order. What property of real numbers can you use to conclude that the two sums are equal? (d) Add the two sums together one term at a time. You should now have a sum that’s easy to compute using the fact that multiplication is repeated addition by the same number.





(e) The sum of those two lists is twice the number we’re looking for, so divide by two. Did you get the same answer as in part (a)? (f) Repeat steps (b) through (e), but this time find the sum of the first n natural numbers, 1 + 2 + 3 + . . . + n. The result should be a formula with variable n. (g) Use your answer from part (f) with n = 100. Do you again get the same answer as in part (a)? (h) Now compare your answer from part (f) to the formula for the sum of an arithmetic sequence with first term and common difference both equal to one. Do you get the same formula? 2. In the Sidelights on pages 256 and 284, we learned about the golden ratio and the Fibonacci sequence. Based on the way they’re defined, there appears to be absolutely no connection between them. But appearances can be deceiving, can’t they?

Chapter 5  Projects   295 Recall that the Fibonacci sequence starts out 1, 1, then each term after that is obtained by adding the previous two. (a) Write the first 12 terms of the Fibonacci sequence. (b) Now we’re going to define a new sequence Rn, with R standing for ratio. R1 is the ratio of the second term in the Fibonacci sequence to the first: that is ​​ _11 ​  . R ​ ​  2​​​ is the ratio of the third term to the second: _​​ 21 ​ .​ In general, Rn is the ratio of the n + 1st term in the Fibonacci sequence to the nth term. Using your answer to (a), find the first 11 terms of the sequence Rn. (c) The terms of the sequence Rn should be closing in on a certain number. What is the significance of that number? (d) Now let’s define a variation on the Fibonacci sequence. Rather than starting out 1, 1, start with 2, 2. Then find the other terms as usual, by adding the two previous. Write the first 12 terms of this sequence. (e) Repeat part (b), using the sequence you wrote in part (d). Do you get the same result? (f) Repeat parts (d) and (e), this time starting out your new Fibonacci sequence with any natural number you like (other than 1 or 2, obviously). Conclusions? (g) What if we define a Fibonacci sequence starting with a noninteger rational number? Try it with ​​ _12 ​ , ​ _12 ​ ,​ then _​​  53 ​ , ​ _53 ​ .​ What do you find? 3. (a) In this part, we’ll prove the formula for finding the sum of the first n terms of an arithmetic sequence: ​​ n(​a​  1​​  + ​a​  n​​  ) S​  n​​  = ​ _____   2 ​​  (i) The first term is a1; the second is a1 + d (where d is the common difference); the third is a1 + 2d; and in general, an = a1 + (n − 1)d. Using these expressions, write the sum of a1 through an. (ii) Use the commutative property to regroup, putting all of the a1’s next to each other. How many are there? How can you simplify that sum? (iii) All of the remaining terms have a d in them. Factor out the d; what remains is the sum of a







group of consecutive natural numbers. Using the Internet as a resource, find a formula for the sum of the first (n − 1) natural numbers and use it to replace that sum. (This formula was developed in Project 1 if you did it, so you won’t need the Internet.) (iv) You should now have a two-term expression, with a fraction in the second term. Get a common denominator and perform the addition. (v) Factor n out of the two terms in the numerator, then write 2a1 in the first term as a1 + a1; the sum of the second and third term can now be simplified using the formula listed above for the nth term of an arithmetic sequence, and if all went well, you’ve proved the formula for the sum. (b) Now we’ll prove the formula for the sum nof the first n a(1 − ​r​​ ​  ) terms in a geometric sequence: S​ ​​ n​​  = ​ _____   1 − r ​​  (i) The first term is a1; the second is a1r (where r is the common ratio); the third is a1r2; and in general, an = a1rn − 1. Using these expressions write the sum of a1 through an, and call this Sn. (ii) Find rSn by multiplying both sides of your equation from (i) by r. Make sure to use the distributive property on the right side! (iii) Subtract the equation from part (ii) from the equation from part (i). (Hint: Most of the stuff on the right side will subtract away to zero!) (iv) Solve the resulting equation for Sn using factoring and dividing, and you have proved the formula. Good job. (v) Extra credit: if the common ratio is a positive number less than one, what happens to the expression rn in the long run (as n approaches positive infinity)? What does this say about the sum of infinitely many terms in a geometric series with common ratio between zero and one?

Design elements: Front Matter, Chapter Opener, Summary and End Matter header design (random numbers background ­illustration): ©pixeldreams.eu/Shutterstock RF

CHAPTER

6 Topics in Algebra

Photo: ©Gennadiy Poznyakov/123RF

Outline 6-1 6-2 6-3

296

Applications of Linear Equations Ratio, Proportion, and Variation The Rectangular Coordinate System and Linear Equations in Two Variables

6-4 Functions 6-5 Quadratic, Exponential, and Logarithmic Functions Summary

Math in  Drug Administration Have you ever looked at the dosage information on a bottle

skills you learn in the chapter, you should be able to answer

of aspirin and thought, “It just doesn’t seem reasonable to rec-

the following questions.

ommend the same dosage for all adults”? People come in all

Suppose that a new pain-killing drug is being tested for

shapes and sizes, and the effect of a certain dosage is in large

safety and effectiveness in a variety of people. The recom-

part dependent on the size of the individual. If a 105-pound

mended dosage for an average 5′10″, 170-pound man is 400

woman and her 230-pound husband both take two aspirin the

mg. The manufacturer claims that the minimum effective dose

morning after their wedding reception, she is in effect getting

for a person of that size is 250 mg, and that anything over

more than twice as much medicine as he is.

1,500 mg could be lethal. Two of the patients in the test are

Because there’s a lot of variation in the world, the study

a 6′5″, 275-pound college football player recovering from a

of algebra was created based on a brilliantly simple idea:

major injury, and a 4′2″, 65-pound girl being treated for sickle-

using a symbol, rather than a number, to represent a quan-

cell disease. According to the manufacturer, dosages for this

tity that can change. Since the word “vary” is a synonym for

drug are proportional to body weight.

change, we call such a symbol a variable. The use of vari-



ables is the thing that distinguishes algebra from arithmetic and makes it extremely useful to describe phenomena in a

patient?

world in which very few things stay the same for very long.

2. If the medications are mixed up and the recommended dosage for the football player is given to the

An understanding of the basics of algebra is, in a very real sense, the gateway to higher mathematics and its applica-

1. What would the recommended dosage be for each

child, is she in danger of dying?

tions. Simply put, you can only go so far with just arithmetic.

3. Would the child’s recommended dosage be at all effective for the football player?

To model real situations, expressions involving variables are

For many medications, the effective dosage depends not just

almost always required.

on body weight but also on body surface area (BSA), which

In this unit, we’ll review some key ideas from algebra, but one thing we won’t do is find x just for the heck of it. A

accounts for the overall size of the patient. One model for body surface area is ____________

lot of students, maybe even most, believe that algebra was

√ ​  1.15 wh ​  ​BSA =  ​ ____________  ​​    60

conceived as an abstract study, with early nerds sitting around simplifying expressions and solving equations on scrolls. Then

where w is weight in pounds and h is height in inches.

a few centuries later, some genius said “Holy cow, we can use



this stuff to find when two trains left San Francisco!”

player compared to the girl. How does it compare to

Nothing could be farther from the truth. Algebra, and really all of classical math, was developed 100% as a tool for

the ratio of their weights?

solving very real problems. So our study of algebra will focus on modeling, which is using algebra to represent situations in

4. Find the ratio of body surface area for the football

5. Find the body surface area for an average 5′10″, 170-pound man.



6. If the dosages for the medication in the drug trial

our world, like dosage calculations. You’re accustomed to los-

are based on body surface area rather than weight,

ing points when you make a mistake in a calculation, but out-

rework Questions 2 and 3 to see if the child would

side the classroom, the stakes can be a LOT higher. The wrong

be in danger, or if the medication would be at all

dose of aspirin might upset your stomach, but more serious

effective for the football player.

drugs carry with them more serious consequences. In many cases, an incorrect dosage could lead to death. Using the



7. Write any conclusions about using body surface area rather than weight when doing dosage calculations.

For answers, see Math in Drug Administration Revisited on page 357 297

298   Chapter 6  Topics in Algebra

Section 6-1 LEARNING OBJECTIVES 1. Translate verbal expressions into mathematical symbols. 2. Solve problems using linear equations.

Applications of Linear Equations Two of the most common questions that students ask when learning about solving equations are “When would I ever use this?” and “Who cares what x is?” (If you want to drive your professor up the wall, ask these questions at least once a day.) Nobody will claim that you will solve multiple equations every day once you leave the hallowed halls of college. But the fact is that solving equations is a topic that is widely applied to almost every area of study. And even if it weren’t, the problem-solving skills that you learn and hone while solving applied problems are among the best “brain exercise” you can get. And what could possibly be more useful to your education than training your brain to work better? So in this section, we will solve problems that relate to issues in our world. In most cases, the plan is to write an equation that describes a situation, then solve that equation to find some quantity of interest. And even if you don’t find the situations applicable to your life, keep the bigger picture in mind: you’ll be practicing useful problem-solving skills with every question. Before we get into problem solving, let’s take a quick look at what makes algebra algebra. Think about this question for a second: what is a variable? If you answered something along the lines of “It’s a letter, like x,” then rest assured that (1) you’re not alone, and (2) your answer is (to put it politely) not very good. (To put it impolitely, it kind of stinks.) A lot of people think of algebra in those terms, but if you do, you’ll probably never get it. Look at the word itself: variable. Able to vary. THAT’S what a variable is: it’s a quantity that is able to vary. We typically use letters to represent variables, but the letter itself isn’t a variable: it’s the quantity that it represents. For example, if you work a part-time job, the number of hours you work each week probably changes. So the number of hours is a variable, and if you wanted, you could represent that variable with h, W, a smiley face, a picture of your mom, or any symbol that isn’t a numeral. I’ll stick to using letters because I don’t know what your mom looks like. In the box below, we’ve outlined a general strategy for attacking word problems using algebra. If you look close enough, you’ll notice that our strategy is based on Polya’s problemsolving strategy that we studied in Chapter 1, with a few elaborations matching this specific type of problem. A General Procedure for Solving Word Problems Using Equations Step 1 Read the problem carefully, but don’t read it all at once without doing anything! As you’re reading, write down any information provided by the problem that seems relevant. This will at least get you started. Make sure you carefully note what it is you’re being asked to find. Draw a diagram if the situation calls for one. Step 2 Assign a variable to an unknown quantity in the problem. Most of the time, the variable should represent the quantity you’re being asked to find. Step 3 Write an equation based on the information given in the problem. Remember, an equation is a statement that two quantities are equal, so keep an eye out for statements in the problem indicating two different ways to express the same quantity. Step 4 Solve the equation. Step 5 Make sure that you answer the question! The best approach is to reread the question, then write your answer in sentence form. Step 6 Check to see if your solution makes sense based on the original wording of the problem.

The step that almost everyone finds most challenging is Step 3. In order to write an equation that describes a situation, you have to translate verbal statements into mathematical symbols. For example, the verbal statement “six more than three times some number” can be written in symbols as “6 + 3x” or “3x + 6.” A careful read of Table 6-1 will help get you started on these types of translations. The table is also very useful to refer back to as you work on problems.

Section 6-1  Applications of Linear Equations    299

Table 6-1

Common Phrases That Represent Operations Phrases that represent addition 6 more than a number A number increased by 8 5 added to a number The sum of a number and 17

6 + x x + 8 5 + x x + 17

Phrases that represent subtraction 18 decreased by a number 6.5 less than a number 3 subtracted from a number The difference between a number and 5

18 − x x − 6.5 x − 3 x − 5

Phrases that represent multiplication 8 times a number Twice a number A number multiplied by 4 The product of a number and 19

8x 2x 4x 19x

3

3

_​​  2 ​​  of a number

_​​  2 ​  x​

Phrases that represent division A number divided by 5 35 divided by a number The quotient of a number and 6

Example 1

x ÷ 5 35 ÷ x x ÷ 6

Translating Verbal Statements into Symbols Translate each verbal statement into symbols. (a) (b) (c) (d) (e)

Math Note Remember, the actual letter you choose for a variable is unimportant. You can choose any letter (or other symbol) you like.

14 times a number A number divided by 7 10 more than the product of 8 and a number 3 less than 4 times a number 6 times the sum of a number and 18

SOLUTION (a) Using variable x to represent the unspecified number, we can write this as 14x. x (b) ​​ __  ​​  7 (c) 10 + 8x (d) It might help to reword this as 3 subtracted from 4 times a number: 4x − 3 (e) Parentheses are required here because the multiplication is 6 times the sum: 6(x + 18)

Try This One 1. Translate verbal expressions into mathematical symbols.

Translate each verbal statement into symbols. (a) (b) (c) (d)

100 divided by a number 5 more than the product of a number and 7 The difference between 25 and a number The product of 8 and the difference of a number and 4

1

300   Chapter 6  Topics in Algebra

Solving Word Problems We’ll begin our study of solving word problems with a basic translation problem. It’s not ­terribly realistic (to say the least), but gives you a start on the basic steps for solving.

Example 2

Solving a Basic Translation Problem If 8 times a number plus 3 is 27, find the number.

SOLUTION Math Note When translating a statement into an equation, the word “is” usually indicates where the equal sign should go.

Step 1 Write the relevant information: ​8 times a number plus 3 is 27​ Identify what we’re asked to find: that unknown number. Step 2 Use variable x to represent the unknown number. Step 3 Translate the relevant information into an equation: ​​8 times    a number plus 3 is 27​​​  8x     +  3 = 27

Step 4 Solve the equation:

Subtract 3 from both sides. 8x + 3 − 3 = 27 − 3 Simplify. 8x = 24    ​​  ​  ​   ​​ ​  ​ Divide both sides by 8. 8x ___ 24 ___ Simplify. ​   ​ = ​   ​  8x + 3 = 27

8

Math Note When checking your answer to a word problem, don’t just substitute it into the equation you wrote—if you wrote the wrong equation, you won’t know. Instead, check that it matches the verbal description of the problem.

8

x=3 Step 5 Answer the question: the requested number is 3. Step 6 Check: 8 times 3 is 24, and when you add 3, you get 27. This matches the description of the problem.

Try This One

2

Ten less than twice a number is 42. Find the number. In the next example, see if you can recognize the similarity to the abstract problem in Example 2.

Example 3

A Problem Involving Contract Negotiations Two basketball teams are interested in signing a free-agent player. An inside source informs the general manager of one team that the other has made an offer, and the player’s agent said “Double that and add an extra million per year, and you’re in our league.” According to a published report, the player is seeking a contract of $18 million per year. What was the rival team’s offer?

SOLUTION Step 1 Relevant information: twice the offer plus 1 million is 18 million. We’re asked to find the offer. Step 2 Use the variable x to represent the offer. Since the numbers are in millions, we’ll let x stand for the offer in million dollar units—that will keep the arithmetic simpler.

Section 6-1  Applications of Linear Equations    301 Step 3 Translate the relevant information into an equation: Twice the offer plus one is eighteen. ​​     ​  ​  ​  ​  ​  ​  ​   ​​​ 2x + 1 = 18 Step 4 Solve the equation:

Subtract 1 from both sides. 2x + 1 = 18 Simplify. 2x + 1 − 1 = 18 − 1 Divide both sides by 2. 2x = 17 ​​    ​  ​  ​  ​  ​​ 17 x = ___ ​   ​ or 8.5 2 Step 5 Answer the question: the team’s offer was $8.5 million. Only $8.5 million? That’s an insult! Step 6 Check: doubling $8.5 million gives $17 million, and adding 1 million more makes it $18 million as required. ©Purestock/SuperStock RF

Math Note Remember, in Example 3 we’re using million dollar units, so 1 million and 18 million are represented by 1 and 18.

Example 4

Try This One

3

A teacher with a fondness for joking around with students tells one that if he doubled his score on the last test and subtracted 12%, he would have just barely gotten an A. The syllabus says that the minimum cutoff for A is 92%. What was the student’s score? Sometimes when there are two unknowns in a word problem, one unknown can be r­epresented in terms of the other. For example, if I know that one number is 5 more than another number, then the first number can be represented by x and the second number can be represented by x + 5. Example 4 uses this idea.

An Application to Home Improvement Pat and Ron are planning to build a deck off the back of their house, and they buy some plans from the Internet. The plans can be customized to the required deck height, which in this case will be 92 inches. They call for support posts of two different heights. The taller ones are 8 inches longer than the shorter ones, and the plans say that the sum of the lengths should be the height of the deck. How long should the support posts be cut?

SOLUTION Step 1 Relevant information: the posts are 8 inches different in length, and the lengths should add to 92 inches. Step 2 We’ll call the length of the shorter posts x. The other posts are 8 inches longer, so they must be x + 8. ©Ryan McVay/Getty Images RF

Step 3 Translate the relevant information into an equation: ​​Length of     shorter post + length of longer post is 92.​​​

x      +         x + 8    = 92 Step 4 Solve the equation: x + x + 8 = 92 ​​ 2x + 8​  =​  92​​​      2x = 84 x = 42

Math Note We could also call the length of the longer posts x; in that case, the shorter ones would have length x − 8.

Step 5 Answer the question: this is where it becomes really important to reread the original question. We were asked to find two lengths, so x = 42 isn’t a valid answer. We used x to represent the length of the shorter posts, and found that it’s 42 in. The longer posts are supposed to be 8 in. longer, so the two lengths are 42 in. and 50 in. Step 6 Check: the two lengths are definitely separated by 8 inches and 42 inches + 50 inches = 92 inches, as required.

302   Chapter 6  Topics in Algebra

Try This One

4

The railings for the deck in Example 4 have three different lengths of board. The shortest is 10 inches less than the next shortest, which is 14 inches shorter than the longest. Their combined length is supposed to match the overall length of the deck, which in this case is 20 feet. How long should each piece be?

Example 5

An Application Involving Money After a busy Friday evening, the tip jar at an off-campus bar is stuffed full of dollar bills and quarters. The tradition is that the bartenders split the dollars, while the barbacks split the quarters. There’s $245 in the jar, with three times as many quarters as dollar bills (college students aren’t known to be the best tippers in the universe). How much money goes to the bartenders, and how much to the barbacks?

SOLUTION Step 1 Relevant information: three times as many quarters as dollar bills, and the total value is $245.

Math Note In this case, we could have used a variable like q to represent the number of quarters, but then the number of q dollar bills would be _​​ 3 ​ ,​ and we’d be introducing fractions. So it’s simpler to use d = the number of dollar bills.

Step 2 Use variable d to represent the number of dollar bills. ©Tom Grill/Corbis RF Then 3d is the number of quarters (because there are three times as many). Step 3 Translate the relevant information into an equation: the value in dollars of the quarters is the number of quarters (3d) times $0.25. The value of the dollar bills is the number of them (d). Total value is $245. ​​   ​​​ 0.25(3d )  +  d = 245 Step 4 Solve the equation: 0.25(3d )  +  d = 245 Multiply. 0.75d + d = 245 Combine like terms. 1.75d​  =​  245​  ​​      ​​ Divide both sides by 1.75. 245 d = ____ ​    ​ = 140 1.75 Step 5 Answer the question: there are 140 dollar bills, so the bartenders split $140. Three times as many quarters is 420 quarters; multiply by $0.25 to get $105 to be split by the barbacks. Step 6 Check: 420 quarters is three times as many as 140 dollar bills, and $140 + $105 = $245. Sounds like a winner to me.

Try This One 2. Solve problems using linear equations.

5

Corrine collected $137 in tips during a Friday evening shift waiting tables, split among one- and five-dollar bills. She got 53 more singles than fives. How many of each did she get? As you work the exercises, keep in mind that one of the main goals is to practice organized thinking and problem-solving skills. You shouldn’t worry too much about how realistic or interesting you think the situations are—focus on the process of setting up and solving the equations. But if you find the situations interesting and relevant, feel free to send me some cash. I prefer hundreds.

Section 6-1  Applications of Linear Equations    303

Answers to Try This One 2

100 (a) ​​ ____  ​  ​     (b) 5 + 7x  (c) 25 − x  (d) 8(x − 4) x 26

3

52% (ouch!)

1

Exercise Set

4

2 2 2 ​68 ​ __ ​  in., 78 ​ __ ​  in., 92 ​ __ ​  in.​ 3 3 3

5

67 singles and 14 fives

6-1

Writing Exercises 1. Describe in your own words what a variable in algebra is. 2. Write a real situation where you could represent a variable quantity with a symbol. 3. Write some reasons why it’s a bad idea to read an entire word problem without writing down any information. 4. Write some reasons why it’s a great idea to write your answer to a word problem in the form of a sentence.

5. What’s the difference between checking your answer when solving an equation, and checking your answer when solving a word problem? 6. How do you choose what the variable in a word problem should represent?

Computational Exercises For Exercises 7–26, write each phrase in symbols. 7. 3 less than a number 8. A number decreased by 17 9. A number increased by 9 10. 6 increased by a number 11. 11 decreased by a number 12. 8 more than a number 13. 6 subtracted from a number 14. 7 times a number 15. One-half a number added to that number 16. 5 more than 3 times a number 17. The quotient of 3 times a number and 6 18. 4 less than 6 times a number 19. The quotient of a number and 14 20. The product of 7 and a number, all subtracted from 10. 21. Triple the sum of a number and pi. 22. One-fourth the difference of 12 and a number. 23. Three times a number subtracted from the quotient of twice that number and the sum of that number and 8. 24. The sum of 146 and the product of a number raised to the third power and 18. 25. The square of the sum formed from adding five times a number to three times a different number.

26. The square root of the difference between half a number and two-thirds of a different number. For Exercises 27–36, solve each. 27. Six times a certain number plus the number is equal to 56. Find the number. 28. The sum of a number and the number plus 2 is equal to 20. Find the number. 29. Twice a number is 32 less than 4 times the number. Find the number. 30. The larger of two numbers is 10 more than the smaller number. The sum of the numbers is 42. Find the numbers. 31. The difference of two numbers is 6. The sum of the numbers is 28. Find the numbers. 32. Five times a number is equal to the number increased by 12. Find the number. 33. Twice a number is 24 less than 4 times the number. Find the number. 34. The difference between one-half a number and the number is 8. Find the number. 35. Twelve more than a number is divided by 2. The result is 20. Find the number. 36. Eighteen less than a number is tripled, and the result is 10 more than the number. Find the number.

Applications in Our World 37. A math class containing 57 students was divided into two sections. One section has three more students than the other. How many students were in each section?

38. In 2015, the Coca-Cola company and PepsiCo had combined revenues of $107.4 billion, with PepsiCo revenues $18.8 billion higher. Find the revenue for each company.

304   Chapter 6  Topics in Algebra 39. The cost, including sales tax, of a Ford Focus SE is $19,392.70. If the sales tax is 6%, find the cost of the car before the tax was added. 40. During the first day of a heat wave, an emergency room had three times as many patients as the day before. There were 48 patients seen total over the 2-day period. How many patients visited the E.R. on each day? 41. Three students that share a townhouse find that their electric bill for October is $2.32 less than the September bill. The total of both bills is $119.48, and each bill is split evenly among the roommates. How much did each owe in September? 42. If Marita invested half of her money at 8% and half at 6% and received $210 simple interest, find the total amount of money invested. 43. A basketball team played 32 games and won 4 more games than it lost. Find the number of games the team won. 44. The enrollment of students in evening classes at a local university decreased by 6% between the years of 2016 and 2017. If the total number of students attending evening classes in both years was 16,983, find how many students enrolled in evening classes in each of those years. 45. If a television set is marked _​​ 13 ​​ off and sells for $180, what was the original price? 46. A carpenter wanted to cut a 6-foot board into three pieces so that each piece is 6 inches longer than the preceding one. Find the length of each piece. 47. The Halloween Association reported that last year, Americans spent $0.68 billion more on candy than costumes for Halloween. If the total spent by Americans for both items was $3.18 billion, how much did Americans spend on each item? 48. In a charity triathlon, Mark ran half the distance and swam a quarter of the distance. When he took a quick break to get a drink of Gatorade, he was just starting to bike the remaining 15 miles. What was the total distance of the race? 49. A nurse is told to give a patient recovering from surgery a total of 21 units of a potent antibiotic over 3 days. The dosage should be cut in half the second day, then in half again for the third day. How many units should she administer on the first day? 50. An investor flips a house, selling it for $82,000. If her profit was 20%, how much did she pay for the house originally? 51. A father left ​​ _12 ​​  of his estate to his son, _​​ 13 ​​  of his estate to his granddaughter, and the remaining $6,000 to charity. What was his total estate? 52. In 2016, there were 104 female officials in Congress, and there were 64 more female members of the House of Representatives than female senators. Find the number of females in each house of Congress.

53. While shopping on BlueFly.com, Juanita notices a special where if she buys two items, the third will be half off. She buys one item, then another item that is half of that amount, and then a third item that is a quarter of the original item amount. The discount she is given is half off the cheapest item. She ends up spending $65 on the order (neglect taxes). What is the price of the first item she bought? 54. There were five winning lottery tickets for a total jackpot of $24 million. Three of the winners won twice as much as the other two. How much did each of the two that won the least get? 55. In Mary’s purse, there are $3.15 worth of nickels and dimes. There are 5 times as many nickels as dimes. The  vending machine is only taking dimes, and Mary needs 10 dimes for her purchase. Does Mary have enough dimes? 56. If the perimeter of a triangular flower bed is 15 feet with two sides the same length and the third side 3 feet longer, what are the measures of the three sides of the flower bed? 57. Last semester, Marcus’s tuition bill was 12% cheaper than this semester’s tuition bill of $640. How much did Marcus pay for tuition last semester? 58. Jane and her two friends will rent an apartment for $875 a month, but Jane will pay double what each friend does because she will have her own bedroom. How much will Jane pay a month? 59. In 2017 NBA all-star voting, 5,513,247 votes were cast for the top three vote-getters. LeBron James got 45,630 more votes than Stephen Curry, who got 76,746 votes more than James Harden. What percentage of the ballots cast for the top three vote-getters did each receive? 60. Three sisters inherited $100,000 from a rich uncle. The uncle’s favorite niece got twice as much as his second favorite niece and the second favorite niece got twice as much as the least favorite niece. How much did the favorite niece get? 61. A bounty hunter makes a base monthly salary of $1,700, plus $900 for every bail-jumper he brings in. Write an algebraic expression that describes his monthly earnings, using a variable that stands for the number of fugitives he captures. Then use your equation to find how many fugitives he needs to average per month in order to make $60,000 per year. 62. A telemarketer is paid $3.70 per hour, plus $0.30 for every caller she keeps on the line for at least a minute. Write an algebraic expression describing her hourly earnings, using a variable that stands for the number of callers she keeps on the line for at least a minute. Then use your equation to find how many callers she has to keep on the line for at least a minute over an 8-hour shift to make $104.

Section 6-2  Ratio, Proportion, and Variation    305

Critical Thinking 63. The temperature and the wind combine to cause body surfaces to lose heat. Meteorologists call this effect “the windchill factor.” For example, if the actual temperature outside is 10°F and the wind speed is 20 miles per hour, it will feel like it is −20° F outside, so −20° F is called the windchill. When it is 25°F outside and the wind speed is 40 miles per hour, it will feel like it is −35° F outside. From the information given, write a linear equation for determining the windchill using the actual temperature and the wind speed. Then use your equation to find the windchill factor on a day with temperature 30°F and 18 mile per hour winds. 64. Suppose your roommate brags that he made $250 in singles and fives one night waiting tables, and that he collected four times as many one-dollar bills as five-dollar bills. How can you tell that he’s not telling the exact truth? 65. The prevailing winds for air travel in the United States typically blow from west to east, slowing down travel to the west considerably. Last year, I flew from Cincinnati to Salt Lake City, a distance of 1,447 miles from east to west. The return flight took three-fourths as long as the flight out west. If the average speed of the plane in still air is 422 mph, what was the wind speed? (Assume it was the same for both flights.) (Hint: Use a variable to represent what you’re asked to find, and fill in the following chart. The information will help you to write an equation.) Distance

Speed

Time

66. In an attempt to conserve energy (and, let’s be honest, save some cash), Bob decides to ride his bike to work every day. He starts out by riding a half mile uphill, which slows him down by 4 miles per hour. The rest of the ride is a mile downhill, which speeds him up by 5 miles per hour. After several days, he notices that when he reaches the top of the hill, he’s exactly halfway there if you measure in terms of time. How fast would Bob be riding if the trip were on level ground? (Hint: Make a table similar to the one in Problem 65.) 67. Winona has two jobs: one pays $11.25 per hour, and the other pays $9.50 per hour plus an average of 60% of the hourly pay in tips and bonuses. Each pay period, 22% of her total pay goes to taxes. (a) Write and simplify an equation that describes Winona’s total take-home pay in terms of the number of hours worked at the first job and the number of hours worked at the second. (b) If Winona is committed to work 30 hours per week at the first job, how many hours per week would she need to work at the second job if she needs her biweekly take-home pay to be $800? 68. Refer to Problem 67. If Winona decides that she’ll work exactly 45 hours per week total and is not committed to a certain number of hours at either job, how many hours should she work at each job to earn $1,000 in take-home pay every two weeks? What about $1,200?

Cincinnati → Salt Lake Salt Lake → Cincinnati

Section 6-2 LEARNING OBJECTIVES 1. Write ratios in fraction form. 2. Solve proportions. 3. Solve problems using proportions. 4. Solve problems using direct variation. 5. Solve problems using inverse variation.

Ratio, Proportion, and Variation The most obvious way to compare the sizes of two numbers is to subtract them. But is that the best way? Suppose you’re comparing the cost of an item at two different stores, and you find that the item is a dollar more at Target than at Wal-Mart. If that item is a bottle of Coke, and the prices are $1 and $2, that dollar difference is significant. But if the item is a 55-inch curved-screen TV and the prices are $1,201 and $1,200, would you really care? It’s essentially the same price. If you divide the prices rather than subtract them, however, something interesting happens: $2 Coke: ​ ___ ​ = 2 $1 ​​  ​ ​​ $1,201     TV: ​ ______ ​  = 1.0008 $1,200 Do you see the point? For the Coke, a dollar more is twice as much—kind of a big deal. For the TV, when you divide the prices, you essentially get 1, meaning the two prices are pretty much the same. The most meaningful way to compare the sizes of two numbers is to divide them, forming what we called a ratio in Section 5-3.

306   Chapter 6  Topics in Algebra

Ratios A ratio is a comparison of two quantities using division. For example, at the end of 2016, about 44% of all smartphones in the United States were iPhones, and about 56% were some other model, so we would say that the ratio of iPhones to other smartphones was 44 to 56. For two nonzero numbers, a and b, the ratio of a to b is written as a:b (read a to b) or _​​  ab ​  ​. Ratios can be written using either a colon or a fraction as shown in the definition, but in math we’ll typically use the fraction so that we can do arithmetic with ratios.

Example 1 Math Note To set up a correct ratio, whatever number comes first in the ratio statement should be placed in the numerator of the fraction and whatever number comes second in the ratio statement should be placed in the denominator of the fraction.

Writing Ratios According to the Sporting Goods Manufacturers’ Association, 95.1 million Americans participate in recreational swimming, 56.2 million Americans participate in recreational biking, 52.6 million Americans participate in bowling, and 44.5 million Americans participate in freshwater fishing. Find each: (a) The ratio of recreational swimmers to recreational bikers (b) The ratio of people who fish to people who bowl

SOLUTION Number of swimmers ____ 95.1 (a) ​​ __________________        ​ = ​   ​​  Number of bikers 56.2 Number of people who fish 44.5 (b) ​​ ________________________         ​ = ____ ​   ​​  Number of people who bowl 52.6

Try This One

We omitted millions from each number because they would divide out.

1

From 1969 through 1977, there were 24 teams in Major League Baseball, and only 4 made the playoffs each year. Now, there are 30 teams, and 10 make the playoffs. Find the ratio of teams making the playoffs to those not making the playoffs in 1969 and today. Since ratios can be expressed as fractions, they can be simplified by reducing the frac10 __ tion. For example, the ratio of 10 to 15 is written 10:15 or __ ​​ 10 15  ​​ and the fraction ​​  15 ​​ can be reduced 2 _ to ​​  3 ​ .​So the ratio 10:15 is the same as 2:3. Some of the ratios we’ll be interested in compare different measurements, like lengths, weights, and others. In that case, we have to make sure that the units match, or we’ll get some really deceiving comparisons. A ratio is supposed to compare the sizes of two quantities, but you can’t always get the story from just numbers.

Example 2

Writing a Ratio Involving Units Suppose that we want to compare the lengths of a 2-foot board, and one that is 18 inches long. First, explain why the ratio below is deceiving. Then write a ratio that provides an accurate comparison. 2 ___ ​​    ​​  18

Section 6-2  Ratio, Proportion, and Variation    307

Math Note We could have written 18 inches as _​​  23  ​​feet in Example 2, and the ratio would have been _____ ​​  3 2/ 2feet  ​  ,​ which also feet  ­simplifies to _​​  43 ​​ .

SOLUTION This is deceiving because 18 is 9 times as big as 2, so this looks like the 18-inch board is a lot longer. Of course, it isn’t: 18 inches is less than 2 feet. To make the comparison accurate, we need to rewrite so that both measurements have the same units. We could convert both to feet, or both to inches: either is fine. But it’s easier to convert 2 feet to 24 inches, giving us the ratio 24 4 ___ ​​   ​ = __ ​   ​​  18 3 4 So the ratio of 2 feet to 18 inches is __ ​​   ​​ . 3

Try This One 1. Write ratios in fraction form.

2

Find the ratio of 40 ounces to 2 pounds. (There are 16 ounces in 1 pound.)

Proportions When two ratios are equal, they can be written as a proportion. A proportion is an equation in which two ratios are stated to be equal. For example, the ratios 4:7 and 8:14 are equal; this fact can be expressed as a proportion: 4 8 __ ​​   ​  = ___ ​    ​​  7 14 Two fractions, ​​ _ab ​​  and ​​ _dc  ​,​are equal if ad = bc. (This will be shown in Exercise 59.) The product of the numerator of one fraction and the denominator of the other fraction is called a cross product. For example, _​​ 34 ​  = _​  68  ​​since 3 · 8 = 4 · 6, or 24 = 24. This tells us that two ratios form a proportion if the cross products of their numerators and denominators are equal. For example, the two ratios _​​ 56 ​​  and __ ​​  15 18 ​​  can be written as a proportion since 5 15 __ ​   ​     ___ ​   ​  This is called cross multiplying. 6 ​​       ​  ​  ​  18 ​​ ​ ​ 5 ⋅ 18 = 6 ⋅ 15 90 = 90 So we can write 5:6 = 15:18, or _​​ 56 ​  = __ ​  15 18  ​​.

Example 3 Math Note Cross multiplying is a really convenient procedure for working with proportions, but there’s a catch: it ONLY works when the equation in question looks like single ­fraction = single fraction. Don’t try to apply cross multiplying to any equation that’s not of that form.

Deciding if a Proportion Is True Decide if each proportion is true or false. 3 9 (a) ​​ __ ​  = ___ ​    ​​   5 15

5 7 (b)  ​​ __ ​  = __ ​   ​​   3 2

14 7 (c)  ​​ ___ ​ = __ ​   ​​  16 8

SOLUTION In each case, we will cross multiply and see if the two products are equal. (a) 3 · 15 = 45; 5 · 9 = 45     The proportion is true. (b) 5 · 2 = 10; 3 · 7 = 21     The proportion is false. (c) 14 · 8 = 112; 16 · 7 = 112    The proportion is true.

308   Chapter 6  Topics in Algebra

Try This One

3

Decide if each proportion is true or false. 2 6 5 25 (a) ​​ __ ​  = ___ ​    ​​   (b)  ​​ __ ​  = ___ ​   ​​   9 25 2 4

11 55 (c)  ​​ ___ ​ = ___ ​   ​​  2 10

If there is an unknown value in a proportion, we can solve for the unknown value by cross multiplying as shown in Examples 4 and 5.

Example 4

Solving a Proportion Solve the proportion for x. 12 3 ___ ​​   ​ = __ ​   ​​  48

x

SOLUTION We begin by cross multiplying. 12 3 ___ ​   ​    __ ​   ​ 

©Arterra Picture Library/Alamy

The height of the people and the statue are in proportion. If we know the ratio of the heights and the height of the people, we can find the height of the statue.

Cross multiply.

48 x 12x = 3 ⋅ 48 ​​  12x    ​  =​  ​  144 ​   ​​ ​    ​ Divide both 12x 144 ____ ____ ​   ​  = ​   ​  Simplify. 12 12 x = 12

sides by 12.

12 3 ___ ​   ​ ​=? ​​​  ___ ​    ​ 

Check: ​​

48 12 ​  ​  ​  ​​ 1 1 __ ​   ​  ​=? ​​​  __ ​   ​  ✓ 4 4

Try This One

4

x 22 Solve the proportion: __ ​​    ​ = ___ ​   ​​  7 25

Example 5

Solving a Proportion x − 5 _____ x+2 Solve the proportion. _____ ​​   ​   = ​   ​​    10 20

SOLUTION

x−5 x+2 ____ ​   ​      ____ ​   ​     10

20

20(x − 5) = 10(x + 2)    ​​ 20x       ​  ​  − 100​  =​  ​  10x ​​ +​ 20​​ 10x − 100 = 20 10x = 120 x = 12

Cross multiply. Multiply out parentheses. Subtract 10x from both sides. Add 100 to both sides. Divide both sides by 10.

Section 6-2  Ratio, Proportion, and Variation    309

Check:

x − 5 ____ x+2 ____ ​   ​    = ​   ​    

10 20 12 − 5 12 +2 _____ ​   ​    ​=? ​​​  _____ ​   ​     10 20    ​​  ​  ​  ​  ​​ ​ ​  7 14 ___ ​    ​ ​=? ​​​  ___ ​   ​  10 20 7 7 ___ ​    ​ = ___ ​    ​  ✓ 10 10

Try This One 2. Solve proportions.

5

x + 6 _____ x−2 Solve the proportion:  ​​ _____  ​   = ​   ​​    15 5

Applications of Proportions Proportions have been around for a really long time in one form or another—a written record of their use goes back at least to 400 BCE or so, but most math historians feel that the idea is almost as old as formal numeric thought. This is because they’re very useful in solving problems in our world. Example 6 illustrates a procedure that works well for problems where ratios are provided in some way.

Example 6

Applying Proportions to Fuel Consumption While on a spring break trip, a group of friends burns 12 gallons of gas in the first 228 miles, then stops to refuel. If they have 380 miles yet to drive, and the SUV has a 21-gallon tank, can they make it without refueling again?

SOLUTION

©JG Photography/Alamy RF

Math Note When setting up a proportion, be sure to put like quantities in the numerators and like quantities in the denominators. In Example 6, gallons were placed in the numerators and miles in the denominators.

Step 1 Identify the ratio statement. The ratio the problem gives us is 12 gallons of gas to drive 228 miles. 12 gallons Step 2 Write the ratio as a fraction. The ratio is _________ ​​   ​​  . 228 miles Step 3 Set up the proportion. We need to find the number of gallons of gas needed to drive 380 miles, so we’ll call that x. The ratio we already have is gallons compared to miles, so the second ratio in our proportion should be as well. We have x gallons, and 380 miles, so the proportion is 12 gallons _____________ x gallons _____________ ​​   ​  = ​     ​​  228 miles 380 miles Step 4 Solve the proportion.

12 x ____ ​    ​  = ____ ​     ​ 

228 380 12 x ____ ​    ​       ​ ____    ​  228 380 ​​          ​  =​  ​  12 ​​ ​ ​  ​ 228x ⋅ ​  380 228x = 4,560 228x 4,560 _____ ​   ​ = _____ ​   ​  228 228 x = 20

Cross multiply. Simplify. Divide both sides by 228.

Step 5 Answer the question. The SUV will burn 20 gallons of gas to cover the last 380 miles, so they can make it without stopping.

310   Chapter 6  Topics in Algebra

Try This One

6

In 2016, roughly 13 of every 100 people in the United States were African-American. A marketing company wants to select a group of 250 people that accurately reflects the racial makeup of the country. How many African-Americans should be included?

In order to decide that a certain species is endangered, biologists have to know how many individuals are in a population. But how do they do that? It’s actually an interesting application of proportions illustrated in Example 7.

Example 7

Applying Proportions to Wildlife Population As part of a research project, a biology class plans to estimate the number of fish living in a lake thought to be polluted. They catch a sample of 35 fish, tag them, and release them back into the lake. A week later, they catch 80 fish and find that 5 of them are tagged. About how many fish live in the lake?

SOLUTION Step 1 Identify the ratio statement. Five of 80 fish caught were tagged. 5 tagged Step 2 Write the ratio as a fraction. ​​ ________ ​​  80 total Step 3 Set up the proportion. We want to know the number of fish in the lake, so call that x. 35 tagged The comparison in the lake overall is _________ ​​   ​  ,​  so the proportion is x total 5 tagged _____________ 35 tagged ​​ ____________ ​  = ​   ​​    80 total

Source: USDA

x total

5 35 ​ ___  ​ = ___ ​   ​ 

Step 4 Solve the proportion. ​​

80

x

5x = 35 ⋅ 80 ​  ​    ​ ​  ​  ​​ 5x = 2,800 2,800 x = _____ ​   ​   = 560 5

Step 5 Answer the question. There are about 560 fish in the lake.

Try This One 3. Solve problems using proportions

7

The staff biologists at a wildlife preserve tagged 74 protected woodpeckers shortly after hatching season. In midsummer, 30 of the birds were caught and 16 were tagged. Estimate the total population of these woodpeckers in the preserve.

Variation Two quantities are often related in such a way that if one goes up, the other does too, and if one goes down, the other goes down as well. For example, if you have a job that pays $95 a day, the amount you make goes up or down depending on how many days you work. This is an example of what is called direct variation. In this case, we can write a ratio statement based on the pay: ______ ​​ 1$95  .​We could then use this to write an equation that describes your total day  ​ pay depending on how many days you work: y​ = ______ ​ 1$95 day  ​ ⋅ x days,​or just y = 95x.

Section 6-2  Ratio, Proportion, and Variation    311

Sidelight

Proportions in My World

Proportions are a topic that it’s easy to write application problems about because they’re useful in a wide variety of everyday settings. In fact, I just used a proportion yesterday in my car. I have one of those cool touch screen systems

with GPS, entertainment options and so forth, and you can upload a picture to be the background for the home screen. The catch is that the screen resolution is 800 × 384 pixels, and to make a picture display right, it should be cropped to make it fit. But it doesn’t have to be exactly those dimensions—it just needs to have the same ratio of width to length. So the picture I wanted to use had a width of 1,164 pixels, and I had to figure out the height it should be cropped to. How did I do it? Using a proportion. The ratio of width to height needed to be the same as it is for the 800 × 384 screen, so I got 800 _____ 1,164 ____ ​​   ​ = ​   ​   ​  384

Photo of touchscreen by Dave Sobecki. Background photo of Jacobs Field source: Michael Frain, Jr./Wikipedia

x

The solution, 559 (rounded), told me the number of pixels to crop to, and I think the result looks pretty darn good. Yay proportions.

A quantity y is said to vary directly with x if there is some nonzero constant k so that y = kx. The constant k is called the constant of proportionality.

Example 8

Using Direct Variation to Find Wages Suppose you earn $95 per day. Write a variation equation that describes total pay in terms of days worked, and use it to find your total pay if you work 6 days and if you work 15 days.

SOLUTION Let  y = the total amount earned     x = the number of days you work     k = $95 per day (as we saw above) Then y = 95x is the variation equation. For    x = 6 days: y = 95 · 6 = $570 For    x = 15 days: y = 95 · 15 = $1,425

Try This One ©REB Images/Blend Images LLC RF

8

A 6-month-old Labrador puppy gets 4​  ​ _12 ​​  cups of food per day. Write a variation equation that describes how many cups of food she eats in terms of days, then use it to find how much she eats in 6 days and in 2 weeks. When two quantities vary directly, if we know the size of each for some specific case, we can use that information to find the constant of proportionality. That gives us the equation of variation, which we can then use to solve problems.

Example 9

Using Direct Variation to Find a Weight When utility cables are strung above ground, the weight is an important consideration. The weight of a certain type of cable varies directly with its length. If 20 feet of cable weighs 4 pounds, find k and determine the weight of 75 feet of cable.

312   Chapter 6  Topics in Algebra

SOLUTION Step 1 Write the equation of variation.

Step 2 Find k.



   y = kx where y = the weight x = length of cable in feet ​​     ​  ​   ​​       ​  ​  ​​ ​ k = the constant   ​​

4 lb  =  k ⋅​ ​20 ft Substitute y = 4 and x 4 lb = k ⋅ 20 ft Divide both sides by 20 4 lb k ⋅ ​20​ ​ft​ _________   ​​  =​  ___________ ​   ​   ​​ ​    ​    ​​ 20 ft ​20​ ​ft​ k = 0.2 lb / ft

= 20 ft.

Now we know that the equation of variation can be written as y = 0.2x.

Step 3 Solve the problem for the new values of x and y using k = 0.2. y = 0.2x 75​ ​​ ​​y​ ​  =​ ​  0.2 ⋅​  y = 15 pounds

Substitute x = 75

So 75 feet of cable will weigh 15 pounds.

Try This One 4. Solve problems using direct variation.

9

The weight (in pounds) of a hollow statue varies directly with the square of its height (in feet); i.e., y = kx2, where y = the weight and x = the height. If a statue that’s ​4 ​ _12 ​​  feet tall weighs 12 pounds, find the weight of a statue that’s 10 feet tall. Sometimes quantities that are connected vary so that if one goes up, the other goes down. Think about driving a certain distance. If your speed goes up, the amount of time it takes goes down, right? If you average 60 miles per hour, you’ll get where you’re going in half the time it would take if you average 30 miles per hour. This is an example of inverse variation. A quantity y is said to vary inversely with x if there is some nonzero constant k such that ​ y = _​ kx ​​ .

Example 10

Using Inverse Variation to Find Driving Time The time it takes to drive a certain distance varies inversely with the speed, and the constant of proportionality is the distance. A family has a vacation cabin that is 378 miles from their residence. Write a variation equation describing driving time in terms of speed. Then use it to find the time it takes to drive that distance if they take the freeway and average 60 miles per hour, and if they take the scenic route and average 35 miles per hour.

SOLUTION ©Photodisc/PunchStock RF

Let y = the time it takes to drive the distance      x = the average speed      k = 378 miles (the distance) k 378 Then the variation equation is ​y = __ ​   ​ ​ or y​ = ____ ​   ​  ​.  x x If they average 60 miles per hour: 378 ​y = ____ ​   ​ = 6.3 hours​ 60

Section 6-2  Ratio, Proportion, and Variation    313 If they average 35 miles per hour:

I vote for the freeway.

378 ​y = ____ ​   ​ = 10.8 hours​ 35

Try This One

10

A student lives 120 miles from the college she attends. Write a variation equation describing the time it takes to drive home for a weekend in terms of speed. Then use it to find driving time if she averages 72 miles per hour and manages to not get pulled over by the state patrol.

Example 11

Applying Inverse Variation to Construction In construction, the strength of a support beam varies inversely with the cube of its length. If a 12-foot beam can support 1,800 pounds, how many pounds can a 15-foot beam support?

SOLUTION Step 1 Write the variation equation. Let y = strength of the beam in pounds it can support x​ ​​ ​  =​ ​  length of the beam​          ​​​ k = the constant of proportionality The variation equation is y​ = __ ​ ​xk3​​   ​​, ​ since y varies inversely with the cube of x. Step 2 Find k. k y = ​ __3  ​  Substitute y =  1,800 ​x​​ ​ k 1,800 = ​ ___ 3 ​  Solve for k. ​​       ​  ​  ​12​​​ ​ ​  ​​

and x = 12

k 1,800 = ​ _____    ​  1,728 k = 1,800 ⋅ 1,728 = 3,110,400 Step 3 Substitute in the given value for x. In this case, the given length is 15. 3,110,400 y = _________ ​   ​     ​x3​​ ​ ​​   ​  ​  ​ ​​ 3,110,400 y = _________ ​   ​     = 921.6 ​15​​3​ A 15-foot beam can support 921.6 pounds.

Try This One 5. Solve problems using inverse variation.

11

If the temperature of a gas is held constant, the pressure the gas exerts on a container varies inversely with its volume. If a gas has a volume of 38 cubic inches and exerts a pressure of 8 pounds per square inch, find the volume when the pressure is 64 pounds per square inch.

314   Chapter 6  Topics in Algebra

Answers to Try This One

5

4 10 In 1969: ___ ​​    ​; ​ Today: ___ ​​   ​​  20 20 5 __ ​​   ​​  4 (a) False  (c) False  (b) True 154 ​x = ____ ​   ​​  25 x = 6

6

Either 32 or 33

1 2 3 4

Exercise Set

   7 139 woodpeckers 9    8 ​y = __ ​   ​  x;​In 6 days: 27 cups; In 2 weeks: 63 cups 2 7    9 ​59 ​ ___  ​  lb​ 27 120 10 ​y = ____ ​   ​  ; ​about 1.7 hours x 11 4.75 cubic inches

6-2

Writing Exercises

1. Write an example of a ratio in an applied situation. 2. Why are units important when writing ratios? 3. What’s the difference between a ratio and a proportion? 4. How are proportions used in solving problems? 5. Write an example of two quantities that vary directly in an applied situation. No using the ones in this section!

6. Write an example of two quantities that vary inversely in an applied situation. No using the ones in this section! 7. Describe the procedure for solving a proportion. 8. What is the constant of proportionality in a variation problem?

Computational Exercises For Exercises 9–18, write each ratio statement as a fraction and reduce to lowest terms if possible. 9. 18 to 28 10. 5 to 12 11. 14:32 12. 40:75 13. 12 cents to 15 cents 14. 18 inches to 42 inches 15. 3 weeks to 8 days 16. 2 pounds to 12 ounces 17. 5 feet to 30 inches 18. 12 years to 2 decades

For Exercises 19–28, solve each proportion. 3 14 x + 3 ___ 35 19. ​​ __ ​  = ___ ​   ​​  24. ​​ _____  ​   = ​   ​​  x 45 5 25 x 18 2 5 20. ​​ __  ​ = ___ ​   ​​  25. ​​ _____    ​ = _____ ​     ​​  2 6 x−3 x+8 5 x 4 16 21. ​​ __ ​  = ___ ​    ​​  26. ​​ _____    ​ = _____ ​    ​​  6 42 x−3 x−2 9 45 x − 3 _____ x+6 22. ​​ __ ​  = ___ ​   ​ ​ 27. ​​ _____  ​   = ​   ​​    8 x 4 20 x − 6 __ 1 x x−2 23. ​​ _____  ​   = ​   ​​  28. ​​ ___  ​ = _____ ​   ​​    12 3 10 20

Applications in Our World 29. The Information Resources Institute reports that one out of every five people who buy ice cream buys vanilla ice cream. If a store sells 75 ice cream cones in one day, about how many will be vanilla? 30. The U.S. Department of Agriculture reported that 57 out of every 100 milk drinkers drink skim milk. If a storeowner orders 25 gallons of milk, how many should be skim? 31. Under normal conditions, 1.5 feet of snow will melt into 2 inches of water. After a monster snowstorm, there were

3.5 feet of snow. How many inches of water will there be when the snow melts? 32. The Travel Industry Association of America reports that 4 out of every 35 people who travel do so by air. If there are 180 students who are traveling for spring break, how many of them will fly? 33. A gallon of paint will cover 640 square feet of wall space. If I plan to paint a room whose walls measure 2,560 square feet, how many gallons of paint will I need?

Section 6-2  Ratio, Proportion, and Variation    315 34. The American Dietetic Association reported that 31 out of every 100 people want to lose weight. If 384 students were surveyed at random in the student union, how many would want to lose weight? 35. The U.S. Census Bureau reported that 9 out of every 20 joggers are female. On a trail, there were 230 joggers on July 4. Approximately how many were female? 36. Angel took a 2-year lease on a new car and after 8 months of driving, he’d put 6,600 miles on it. The lease allows 10,000 miles per year. If his driving habits stay consistent, will he stay under the allotted mileage for 2 years? By how much? 37. Out of every 80 phone chargers sold by a discount website, 3 were returned as defective. If the website sold 1,000 phone chargers this holiday season, how many should they expect will be returned as defective? 38. The American Dietetic Association states that 11 out of every 25 people do not eat breakfast. If there are 175 students in a large lecture hall, about how many of them did not eat breakfast? 39. Mark is interested in measuring the height of a tree in his yard. He measures the length of the tree’s shadow at 18 feet at the same time his own shadow is 3 feet 6 inches. If Mark is 5′10″, how tall is the tree? 40. An online photo printing service will put any photo you like on a commemorative baseball. The preferred dimensions for photo uploads are 640  ×  480  pixels. Quan wants to have her son’s Little League photo put on a ball. The file she has is 1,100 pixels wide. To what height should she crop it so that the proportions fit the requirements? 41. A small college has 1,200 students and 80 professors. The college is planning to increase enrollment to 1,500 students next year. How many new professors should be hired, assuming they want to maintain the same ratio? 42. The taxes on a house assessed at $128,000 are $3,200 a year. If the assessment is raised to $160,000 and the tax rate did not change, how much would the taxes be now? 43. According to a poll conducted by the Pew Research Center in early 2011, for every 69 people who could correctly identify the religion practiced by the president of the United States, there were 131 who could not. At that time there were about 235 million adult Americans. How many could not identify the president’s religion? 44. Based on a combination of worldwide surveys conducted from 2008 to 2010 and reported at gallup.com, for every 29 people who would be happy to stay in the country they live in, 71 would prefer to move to another country. Among those who would prefer to move, 24 of 71 say they would like to move to the United States. The two next-highest proportions were for Canada and Great Britain, each at 7 of 71.



(a) Based on these numbers, if a random sample of 1 million people was chosen from around the globe, how many would like to move to the United States? (b) How many more would like to move to the United States than Canada and the UK combined? 45. According to a poll conducted by the Gallup Corporation in 2014, the metropolitan areas with the highest and lowest proportions of obese adults were Baton Rouge, LA, and Colorado Springs, CO, respectively. In Baton Rouge, for every 321 people who were not obese, there were 179 who were. In Colorado Springs, for every 49 obese people, there were 201 who were not obese. At that time, the estimated populations were 825,478 for Baton Rouge and 686,908 for Colorado Springs. How many more obese people were there in the Baton Rouge area than in the Colorado Springs area? 46. A survey showed that in 2016, 11 of every 25 employers in the United States were adding more workers to their workforce, 3 of 50 were reducing their workforce, and 1 of 2 were keeping it the same. (a) If there were 1,250 employers in one metropolitan area, and those who were hiring planned to hire an average of seven new employees, how many new jobs were about to become available? (b) If the employers who were cutting their workforce in that area were planning on letting go an average of nine workers, what would the net change in jobs for the area be? 47. The amount of simple interest on a specific amount of money varies directly with the time the money is kept in a savings account when the interest rate is constant. Find the amount of interest on a $5,000 savings account, if the interest rate is 6%, and the money has been invested for 4 years. 48. The number of tickets purchased for a prize varies directly with the amount of the prize. For a prize of $1,000, 250 tickets are purchased. Find the approximate number of tickets that will be purchased on a prize worth $5,000. Use the following information for Exercises 49–52. If everyone had the same body proportions, your weight in pounds would vary directly with the cube of your height in feet. According to the Centers for Disease Control, in 2016 the average height and weight for an adult male in the United States is 5 feet 9.4 inches and 196 lb. Use this information to write a variation equation, then use it to find the weight that each of the following famous athletes would be if they had the same body type as the average male. 49. Basketball player Lebron James: 6′8″ (Actual weight is 250 lb.) 50. Basketball player Roy Hibbert: 7′2″ (Actual weight is 270 lb.) 51. Jockey Pat Day: 4′11″ (Actual weight is 105 lb.)

316   Chapter 6  Topics in Algebra 52. Football player Tom Brady: 6′4″ (Actual weight is 225 lb.) 53. Under certain conditions, the pressure of a gas varies inversely with its volume. If a gas with a volume of 20 cubic inches has 36 pounds of pressure, find the amount of pressure 30 cubic inches of gas is under. 54. The strength of a particular beam varies inversely with the square of its length. If a 10-foot beam can support 500 pounds, how many pounds can a 12-foot beam support? 55. In karate, the force needed to break a board varies inversely with the length of the board. If it takes 5 lb of force to break a board that is 3 feet long, how many pounds of force will it take to break a board that is 5 feet long?

56. The weight of a body varies inversely with the square of the distance from the center of the earth. If the radius of the earth is 4,000 miles, how much would a 150-lb woman weigh 500 miles above the surface of the earth? 57. The time to complete a project is inversely proportional to the number of people who are working on the project. A class project can be completed by 3 students in 20 days. In order to finish the project in 5 days, how many more students should the group add? 58. The intensity of sound varies inversely as the square of the distance from the source. A sound with an intensity of 300  watts/m2 is heard from 20 feet away from a speaker. What is the intensity of the sound 50 feet away from the same speaker?

Critical Thinking

59. Starting with the equation _​​ ab ​  = ​ _dc  ​,​find the LCD and multiply both sides by it. Make sure you simplify fractions. This proves something that’s very important in working with proportions. What is it? 60. Write the generic variation equations for direct and inverse proportionality. Now fill in the blanks in the next two sentences, and explain how you can deduce your answer from the equations. When a quantity goes up in proportion to another going up, this is ________________ variation. When a quantity goes down in proportion to another going up, this is ________________ variation. Stores are required by law to display unit prices. A unit price is the ratio of the total price to the number of units.For Exercises 61–66, decide which is a better buy. These prices were obtained from actual foods. 61. Flour: 10 pounds for $3.39 or 25 pounds for $7.49 62. Candy: 20 ounces for $1.50 or 24 ounces for $1.75 63. Potato sticks: 7 ounces for $1.99 or 1.5 ounces for $0.50 64. Cookies: 7 ounces for $0.99 or 14 ounces for $1.50 65. Coffee: 11.5 ounces for $2.75 or 34.5 ounces for $7.49 66. Beggin’ Strips dog treats: $11.20 for 1.4 pounds, $2.75 for 8 ounces. 67. If a varies directly with the square of b, and b in turn varies inversely with d: (a) Give a verbal description of the variation of a with respect to d. (b) If a is 90 when d is 5, find a when d is 9.



(c) Given the information in (b), can you find b? Why or why not? 68. If x varies inversely with the cube of y, and y in turn varies inversely with the square root of z: (a) Give a verbal description of the variation of x with respect to z. (b) If x is 14 when z is 4, find x when z is 16. (c) If x is 20 when y is 2 and y is 16 when z is 4, find x when z is 100. In Exercises 69–72, decide if the two quantities are likely to vary directly or inversely, and explain your answer. 69. (a) The amount of time you spend on Facebook and your GPA (b) The square footage of an apartment and the monthly rent 70. (a)  The outdoor temperature and the total weight of clothing you wear (b) The number of miles you run per week and your weight 71. (a)  The age of your car and the total cost of maintaining it (b) The crime rate in a given area and the rate of unemployment in that area 72. (a)  The distance you drive in an hour and a half, and your average speed (b) The magnitude of an earthquake and the cost of property damage that it causes

Section 6-3  The Rectangular Coordinate System and Linear Equations in Two Variable     317

LEARNING OBJECTIVES 1. Plot points in a rectangular coordinate system. 2. Graph linear equations. 3. Find the intercepts of a linear equation. 4. Find the slope of a line. 5. Graph linear equations in slopeintercept form. 6. Graph horizontal and vertical lines. 7. Find linear equations that describe situations in our world.

The Rectangular Coordinate System and Linear Equations in Two Variable When the economic crisis hit our country in 2008, the unemployment rate became a big story. But what did unemployment look like before the crisis? Here’s a look at unemployment rates from 1992 to 2015, according to the U.S. Bureau of Labor Statistics. Year

′92

′93

′94

′95

′96

′97

′98

′99

′00

′01

′02

′03

Rate (%)

7.5

6.9

6.1

5.6

5.4

4.9

4.5

4.2

4.0

4.7

5.8

6.0

Year

′04 ′05 ′06 ′07 ′08 ′09 ′10 ′11 ′12 ′13 ′14 ′15

Rate (%)

5.5 5.1 4.6 4.6 5.8 9.3 9.6 8.9 8.1 7.4 6.2 5.3

There’s quite a lot of data here, and in order to make sense of the data, we’ll have to follow along the table one number at a time. The rate went down pretty steadily from 1992 to 2000, then started to rise. In 2004 it started to fall again before changing direction in 2008 (uh oh) and really taking off in 2009. Then it fell pretty steadily from 2010 to 2015. Now let’s look at the same information in graphic form: Unemployment Rate in the United States

12 10 Percent Unemployed

Section 6-3

8 6 4 2 0

′92

′94

′96

′98

′00

′02

′04

′06

′08

′10

′12

′14

′16

Year

Two things should jump out at you immediately. One, it’s much, much easier to see the trends in unemployment compared to when the information was in table form. Two, this graph looks kind of like a sea monster with a big nose, which is totally irrelevant but still pretty cool. The first observation is the important one because it illustrates the big advantage of being able to draw graphs: graphing data almost always makes it easier to understand their significance than simply looking at raw numbers. In this section, we’ll discuss the basics of graphing and study a particular kind of graph that we can use to explore many situations in our world. y

The Rectangular Coordinate System The foundation of graphing in math is a system for locating data points using a pair of perpendicular number lines. We call each one an axis. The horizontal line is called the x axis, and the vertical line is called the y axis. The point where the two intersect is called the origin. Collectively, they form what is known as a rectangular coordinate system, sometimes called the Cartesian plane. The two axes divide the plane into four regions called quadrants, which we number using Roman numerals I, II, III, and IV as shown in Figure 6-1.

Quadrant II

6 5 4 3 2 1

y axis Quadrant I Origin

x axis

–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –1 –2 –3 Quadrant III Quadrant IV –4 –5 –6

Figure 6-1

x

318   Chapter 6  Topics in Algebra

Sidelight

The Controversial Descartes

The rectangular coordinate system we use in graphing is called the Cartesian plane because the whole study of graphing was developed by the 17th-century philosopher René Descartes (pronounced “day-cart”). His work revolutionized math because he brought together the ancient fields of arithmetic, geometry, and algebra into a single subject that we now call analytic geometry. This alone makes Descartes one of the most important figures in the history of human thought. In 1637, he published a book with the lengthy and somewhat grandiose title Discourse on the Method of Rightly Conducting the Reason and Seeking for Truth in the Sciences. (Descartes was not the most humble guy in town.) In the book, he advocated the position that all knowledge should be devised using mathematical reasoning. Terrific for math, not so good for compassion. Because of his extreme feelings on the importance of reasoning in every area, Descartes felt that any being

y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 (–5, –4) –5 –6

(2, 5)

1 2 3 4 5 6

Figure 6-2

x

incapable of logical reasoning was merely a biological machine, also incapable of having feelings, emotions, or even feeling physical pain. In particular, his mistreatment of animals, some of which was far too graphic for a familyfriendly textbook, is used to this day as an argument in opposition to animal cruelty. Anyone that’s ever owned a pet will find it hard to believe Descartes was smart enough to develop an entire branch of mathematics, but dense enough to not recognize that animals can feel pain. In popular culture, Descartes is best known for his famous statement “I think, therefore I am,” which roughly means that thinking about existence is proof that one exists. And how did Descartes meet his end? One day, he walked into a bar and ordered an ale, and the bartender asked him “do you want some peanuts with that?” When Descartes replied “I think not,” he ceased to exist. (Actually, he died of pneumonia at age 53 in 1650, but my story is much more fun.)

When visualizing data, it’s often natural to have two quantities paired together, like the year 2015 and the number 5.3 in the unemployment table. We can associate pairs of numbers in a rectangular coordinate system with points by locating each number on one of the two number lines that make up the coordinate system. We call the two numbers coordinates of a point, and write them as (x, y), where the first number describes a number on the x axis and the second describes a number on the y axis. The coordinates of the origin are (0, 0). A point P whose x coordinate is 2 and whose y coordinate is 5 is written as P = (2, 5). It is plotted by starting at the origin and moving two units right and five units up, as shown in Figure 6-2. Negative coordinates correspond to negative numbers on the axes, so a point like (−5, −4) is plotted by starting at the origin, moving five units left and four units down. Example 1 illustrates the process of plotting points.

Example 1

Plotting Points

Math Note

Plot the points (5, −3), (0, 4), (−3, −2), (−2, 0), and (2, 6).

One common way to go wrong when plotting points is to mix up the order of the coordinates, so think alphabetical. If you’re thinking in terms of x and y axis, x comes first alphabetically, and the x coordinate is first. If you think horizontal and vertical axis, it still works.

y

SOLUTION To plot each point, start at the origin and move left or right according to the x value, and then up or down according to the y value.

Try This One

(–2, 0)

6 5 4 3 2 1

–6 –5 –4 –3 –2 –1 –1 –2 (–3, –2) –3 –4 –5 –6

(2, 6) (0, 4)

1 2 3 4 5 6

(5, –3)

1

Plot the points whose coordinates are (5, 3), (−1, 5), (0, −5), (−3, 0), and (−4, −2).

x

Section 6-3  The Rectangular Coordinate System and Linear Equations in Two Variable     319 Given a point on the plane, its coordinates can be found by drawing a vertical line back to the x axis and a horizontal line back to the y axis. For example, the coordinates of point C shown in Figure 6-3 are (−3, 4). This skill comes in VERY handy in interpreting information from graphs displaying real data, like the one that begins this section.

y

C

–6 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 –6

©Aaron Roeth Photography RF

Archaeological digs use a rectangular coordinate system to track where objects are found.

Example 2

6 5 4 3 2 1 1 2 3 4 5 6

x

Figure 6-3

Finding the Coordinates of Points Find the coordinates of each point shown on the plane below. y B

12 10 A 8 6 D 4 2

C –6 –5 –4 –3 –2 –1–2 –4 –6 E –8 –10 –12

1 2 3 4 5 6

x

SOLUTION The easiest way to make a mistake in graphing is to not be conscious of the scale marked on each axis. In this case, notice that every tick mark on the y axis represents TWO units, not one. 1. Plot points in a rectangular coordinate system.

A = (1, 8)   B = (−2, 12)   C = (−5, 0)   D = (0, 6)   E = (−4, −8)

Try This One

2

Find the coordinates of the points shown. y 6 5 4 3 2 1

G

C

©Blend Images/Getty Images RF

The interface between a mouse and a computer uses a coordinate system to match the cursor’s motion to the mouse’s motion.

–18 –15–12–9 –6 –3 –1

E

A B 3 6 9 12 15 18

–2 –3 D –4 –5 –6

F

x

320   Chapter 6  Topics in Algebra

Example 3

Identifying the Significance of the Signs of Coordinates Earlier we saw that the two axes divide a coordinate plane into four quadrants. Fill in each blank. ∙ If both coordinates of a point are positive, the point is in quadrant _____________. ∙ If both coordinates are negative, the point is in quadrant _____________. ∙ If the x coordinate is positive and the y coordinate is negative, the point is in quadrant _____________. ∙ If the y coordinate is positive and the x coordinate is negative, the point is in quadrant _____________.

SOLUTION Two positive coordinates puts a point in the upper right portion of the graph, which is quadrant I. Two negative coordinates is the bottom left portion, which is quadrant III. If the x coordinate is positive, a point is to the right of the origin; a y coordinate that’s negative means the point is below the y axis. So positive x and negative y is quadrant IV (lower right portion). A negative x coordinate (left of the origin) and a positive y coordinate (above the y axis) is the upper left portion, which is quadrant II.

Try This One

3

Fill in each blank. ∙ A point with first coordinate zero and positive y coordinate is on the _____ axis, between quadrants _____ and _____. ∙ A point with second coordinate zero and negative x coordinate is on the _____ axis, between quadrants _____ and _____.

Linear Equations in Two Variables y 16 14 12 10 8 6 4 2

–5 –4 –3 –2 –1

(4, 14) (3, 12) (2, 10) (1, 8) (0, 6)

1 2 3 4 5

Figure 6-4

Math Note Notice that in Figures 6-4 and 6-5, each box along the x axis represents one unit, but each box along the y axis represents two units. It’s very common to label the two axes with a different scale, especially in applied situations where the quantities represented by x and y could be totally different in size.

x

Suppose you get offered six bucks to stand on a street corner and hand out fliers for a restaurant, plus an extra $2 for every customer that brings your flier into the restaurant and orders a meal. If we use variable x to represent the number of customers ordering a meal with your flier, and variable y to represent the amount of money you make, we could write an equation to describe y: y = 2x + 6 (amount you make is $2 times the number of customers plus $6). If we choose a pair of numbers to substitute into this equation for x and y, the resulting equation is either true or false. For example, for x = 4 and y = 14, the equation is 14 = 2(4) + 6, which is a true statement. We call the pair (4, 14) a solution to the equation, and say that the pair of numbers satisfies the equation. But for x = 14 and y = 4, the equation is 4 = 2(14) + 6, which is not even close to true, so the pair (14, 4) is not a solution to the equation. Here’s a list of some pairs that satisfy the equation y = 2x + 6: (0, 6 ) ​ (1, 8 ) ​ (2, 10 ) ​​     ​  ​  ​  ​  ​  ​  ​  ​   ​​​ ​ (3, 12 ) ​ (4, 14 ) ​ If we plot the points corresponding to these pairs, something interesting happens (see Figure 6-4). Notice that all of the points appear to line up in a straight line pattern. This is not a coincidence. In fact, for this reason, equations like y  =  2x  +  6 are called linear equations. If we connect the points plotted with a line (Figure 6-5), the result is called the graph of the equation. This is one of the most important ideas in all of math, so it deserves a cool definition box:

y 16 14 12 10 8 6 4 2

–5 –4 –3 –2 –1

Figure 6-5

(4, 14) (3, 12) (2, 10) (1, 8) (0, 6)

1 2 3 4 5

x

Section 6-3  The Rectangular Coordinate System and Linear Equations in Two Variable     321

The graph of an equation is a way to geometrically represent every pair of numbers that is a solution to the equation. Each of those pairs corresponds to a point on the graph. When an equation can be written in the form ax + by = c, where a, b, and c are real numbers, it’s called a linear equation in two variables. (Our example can be rearranged to look like −2x + y = 6, so it fits that definition with a = −2, b = 1, and c = 6.) In Example 4, we illustrate the process of drawing the graph of a linear equation in two variables.

Example 4

Graphing a Linear Equation in Two Variables Graph x + 2y = 5.

SOLUTION Only two points are necessary to find the graph of a line, but it’s a good idea to find three to make sure that you haven’t made a mistake. To find pairs of numbers that make the equation true, we will choose some numbers to substitute in for x, then solve the resulting equation to find the associated y. I decided to choose x = −1, x = 1, and x = 5, but any three will do.

y x + 2y = 5 6 5 4

(–1, 3)

2 1

–6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6

(1, 2)

x + 2y = 5   x + 2y = 5   x + 2y = 5 − 1 + 2y = 5   1 + 2y = 5   5 + 2y = 5 ​​              ​  ​  ​​​ 2y = 6  2y = 4  2y = 0 y=3   y=2   y=0

(5, 0) 1 2 3 4 5

x

Three points on the graph are (−1, 3), (1, 2), and (5, 0). We plot those three points and draw a straight line through them. The arrows at each end of the graph are important! They indicate that the line continues indefinitely in each direction. The graph doesn’t suddenly end because we ran out of space on our drawing.

Try This One

4

Graph 2x − y = 10.

Using Technology: Making a Table of Values Both graphing calculators and spreadsheets can be used to make a table of values for graphing an equation. In each case you first have to solve the equation for y. In Example x 3, x + 2y = 5 becomes y​ = ___ ​ 5 −2 ​​.   

Graphing Calculator Step 1: Press , then enter the right side of this equation next to “y1 =.” Make sure you enclose the entire numerator in parentheses! Step 2: Press 2nd to get to the table setup window. Step 3: Use the arrow keys and enter to select “Ask” next to “Indpnt.” Step 4: Press 2nd to display the table. Key in any x value and press ENTER to ­calculate the associated y value.

Spreadsheet Step 1: Enter x values you’d like to input in column A. Step 2: In cell B1, enter “=” followed by the formula found for y, with A1 in place of x. In this case, we would enter “=(5–A1)/2.” This will put the y value in cell B1. Step 3: Copy the formula in B1 down to as many rows as you have x values for.

322   Chapter 6  Topics in Algebra

Intercepts When drawing the graph of an equation, the first thing you need to decide is how to scale the two axes. A good way to help decide is to find the key points where the graph crosses each axis. The point where a graph crosses the x axis is called the x intercept, and the point where a graph crosses the y axis is called the y intercept. Every point on the x axis has y coordinate zero, and every point on the y axis has x coordinate zero, so we get the following rules. Finding Intercepts To find the x intercept, substitute zero for y and solve the equation for x. To find the y intercept, substitute zero for x and solve the equation for y.

2. Graph linear equations.

Example 5

Finding Intercepts

Math Note

Find the intercepts of 2x − 3y = 6, and use them to draw the graph.

SOLUTION

Intercepts are points on the graph, not numbers. Make sure you give both coordinates, not just the result of solving the equation in the procedure.

To find the x intercept, let y = 0 and solve for x. 2x − 3y = 6 2x − 3(0 ) = 6​​​  ​​      2x = 6 x=3 The x intercept has the coordinates (3, 0). To find the y intercept, let x = 0 and solve for y. 2x − 3y = 6 2(0) − 3y = 6 ​​  ​  ​  ​      ​​ − 3y = 6

y

y = − 2

6 5 4 3 x intercept (3, 0) 2 1 –6 –5 –4 –3 –2 –1 –1 –2 2x – 3y = 6 –3 –4 –5 –6

1 2 3 4 5 6

x

The y intercept has the coordinates (0, −2). Now we plot the points (3, 0) and (0,  −2), and draw a straight line through them. (It would still be a good idea to find one additional point to check your work. If the three points don’t line up, there must be a mistake.)

y intercept (0, –2)

Try This One

5

For each equation, find the intercepts and use them to draw the graph. 3. Find the intercepts of a linear equation.

(a) x + 5y = 10   

(b) 4x − 3y = 12

Slope One of the key features of any line is its steepness. To describe the steepness of a line, we’ll use the term slope. The slope of a line on a graph is very much like the slope (or steepness) of a road. Look at the two roads in Figure 6-6. The “slope” of a road can be defined as the “rise” (vertical height) divided by the “run” (horizontal distance) or as the change in y compared to the change in x. In road A, we have 30 ft ​​ _________ ​ = 0.6​ 50 ft

Section 6-3  The Rectangular Coordinate System and Linear Equations in Two Variable     323 That is, for every 50 feet horizontally the road rises a height of 30 feet. Road B has a slope of

dA

a Ro

50 ft

Since the slope of road A is larger than the slope of road B, we say that road A is steeper than road B. On the Cartesian plane, slope is defined as follows:

50 ft (run) Slope = 0.6

50 ft (run) Slope = 0.2

10 ft _________ ​​   ​ = 0.2​

30 ft (rise)

Road B 10 ft (rise)

The slope of a line (designated by m) is ​y​  2​​ − ​y​  1​​ ​m = ​ ______  ​​  ​x​  2​​ − x​ ​  1​​

Figure 6-6

Change in y coordinate ​​ ​​ ______________    Change in x coordinate

where (x1, y1) and (x2, y2) are two points on the line.

©Arthur S. Aubry/Getty Images RF

In words, the slope of a line can be determined by subtracting the y coordinates (the vertical height) of two points and dividing that difference by the difference obtained from subtracting the x coordinates (the horizontal distance) of the same two points. See ­Figure 6-7. Example 5 shows the procedure for finding the  slope of a line when we know two points on that line.

The slope of this road is 0, because the rise is 0.

Example 6 Math Note If the line goes “uphill” from left to right, the slope will be positive. If a line goes “downhill” from left to right, the slope will be negative. The slope of a vertical line is undefined. The slope of a horizontal line is 0 (see sample graphs at the top of page 324).

y

m=

y2 – y1 x2 – x1

(x2, y2) (x1, y1) (x2 – x1) Run

(y2 – y1) Rise (x2, y1)

x

Figure 6-7

Finding the Slope of a Line Find the slope of a line passing through the points (2, 3) and (5, 8).

SOLUTION Designate the two points as follows:

Substitute into the formula:

(2,3) and (5, 8)    ​​     ↓​​ ​​ ​↓​ ​​​ ​  ​  ​ ​ ​  ↓​​ ​​ ​↓​​​​ ​ ​​ (​x​  1​​, ​y​  1​​) ​ (​x​  2​​, ​y​  2​​)

​y​  ​​ − ​y​  1​​ ____ 8 − 3 __ 5 ​m = ______ ​  2  ​  = ​   ​ = ​   ​​  ​x​  2​​ − ​x​  1​​ 5 − 2 3

The slope of the line is _​​ 53 ​ .​A line with that slope would rise 5 feet vertically for every 3 feet horizontally.

Try This One Find the slope of a line passing through the points (−1, 4) and (2, −8).

6

324   Chapter 6  Topics in Algebra

y

y

x

m>0

y

x

y

x

m0

a 0, it should open upward, and if a  1, the function increases as x increases, as in Figure 6-10a. 2. When 0  1 on the same set of axes. (c) What can you conclude about the differences between linear, quadratic, and exponential decline? 85. The richest family in town hires you to do 2 weeks of work landscaping their estate, but they’re a little bit eccentric. Instead of just making an offer, they give you three choices of payment. With choice 1, you start out at $10 the first day, then $10 is added to the payment for each additional day until the 14th day. For choice 2, you get one dollar the first day, and your pay in dollars for any day is the square of the number of days into the job. For choice 3, you get a dime the first day, 20 cents the second day, 40 cents the third, and your pay doubles every day until the 14th. (a) You have 10 seconds to decide or the job goes to someone else. Which do you choose? (b) Calculate how much money you would make in those 14 days with each option. How well did you choose? (c) Which type of growth is illustrated by each option? 86. A new resort in Costa Rica is managed by a retired math professor with a strange sense of humor. Patrons can choose one of three payment options. In each case, the cost decreases each day, and you have to leave when it reaches zero. Option 1: per-person cost is $200 the first day, then decreases by $25 every day. Option 2: perperson cost is $300 the first day, and in each successive day, you subtract 10 times the square of the number of days you’ve been there from the previous day’s cost. Option 3: per-person cost is $500 the first day and $250 the second day; in general, every day is half as much as the previous day. (a) Which option is the best if your primary goal is to spend the least amount of money? How much would you spend and for how many days? (b) Which option is best if your primary goal is to stay the longest? How long would you end up staying? (c) Which type of decline is illustrated by each option?

CHAPTER

Section

6 Summary Important Terms

Important Ideas

Variable

Many problems in our world can be solved by writing an equation that describes the situation in the problem and then evaluating or solving that equation to find desired quantities. The first step in doing so is usually identifying quantities that can change (or vary) and represent them with letters. In many cases, a statement within the problem can be translated into an equation.

Ratio Proportion Cross multiply Direct variation Inverse variation

The most efficient way to compare the sizes of two quantities is to divide them, forming a ratio. A proportion is a statement that two ratios are equal; proportions come in handy in solving a ton of problems in our world.

Rectangular coordinate system Cartesian plane x axis y axis Origin Quadrants Coordinates Linear equation in two variables Graph of an equation x intercept y intercept Slope Slope-intercept form

The rectangular coordinate system (also called the Cartesian plane) consists of two number lines: one vertical axis called the y axis and one horizontal axis called the x axis. The axes divide the plane into four quadrants. A point is located on the plane by its coordinates, which consist of an ordered pair of numbers (x,y). The graph of an equation is the set of all points in the plane whose coordinates make the equation true when substituted in for the variable. The graph of a linear equation in two variables, which is an equation that can be written in the form ax + by = c, is a straight line. The slope of a line is a measure of how steep it is, and is defined to be the change in y coordinate divided by the change in x coordinate. The slope describes the rate at which the y coordinate is changing with respect to the x coordinate. A horizontal line has slope zero, while a v­ ertical line has undefined slope. When an equation is written in the form y = mx + b, m is the slope, and the point (0, b) is the y intercept. Linear equations can be used to effectively model situations where the rate of change of some quantity is either constant, or very close to constant.

Relation Function Function notation Evaluate Domain Range Vertical line test

A relation is a rule matching up two sets of objects. Relations can be represented by sets of ordered pairs, graphs, or equations relating two variables. A relation is called a function if every element in the domain (the set of first coordinates) corresponds to exactly one element in the range (the set of second coordinates). Function notation is used to name functions: f (x) represents a function named f whose variable is x, and is read “f of x.” A linear function has the form f (x) = ax + b. The graph is a straight line with slope a and y intercept (0, b). Linear functions are used to model data with a rate of change that is constant, or close to constant.

Linear function Quadratic function Parabola Vertex Axis of symmetry Exponential function Asymptote

A quadratic function has the form f (x) = ax2 + bx + c. The graph is a parabola that opens up if a > 0 and down if a I (discussed shortly). The bending stresses just discussed are developed in response to the applied external moment and, in the examples presented, act horizontally and do nothing to balance the external shear forces, which are balanced by shear stresses that act on the face of the section. [See Figure 5(c).] These stresses are distributed in a complex manner that is covered later in the chapter, where it is demonstrated that they are maximum at the neutral axis of the cross section and decrease nonlinearly toward the outer faces. The important point, however, is that the shear stresses acting over the face of the cross section produce a resultant vertical shear force that equilibrates the applied external shear force. Shear stresses of another type also are developed in beams. These stresses act horizontally rather than vertically. (See Section 3.3.)

Beams

FIGure 4 Bending stresses in a beam. Parts (a) and (b) illustrate the general deformations produced by the external loading. Part (b) illustrates the deformations present at a cross section, and part (c) illustrates the bending stress at the same cross section.

strains

Copyright © 2013. Pearson Education Limited. All rights reserved.

deformations in the beam

neutral axis

It is important to know how bending and shear stresses vary along the length of the beam. Because the stresses at a cross section depend on the magnitude and sense of the external shears and on the moments at that cross section, the distribution of stresses along the length of the beam can be found by studying the distribution of shears and moments in a beam in general.

Beams

FIGure 5 Basic load-carrying mechanisms in a beam.

Copyright © 2013. Pearson Education Limited. All rights reserved.

, which

Note that when the equilibrium of an elementary portion of a member is considered, the effect of the set of external forces (including reactions) produces a net rotational effect (the external moment ME) about the cut section considered and a net vertical translatory force (the external shear force VE). These quantities are balanced by an equilibrating set of shears and moments 1VR and MR 2 developed internally in the structure 1i.e., VR = VE and MR = ME 2. The magnitudes of these shears and moments depend on the extent of the elemental section and the forces acting on it. Thus, the distribution of such shears and moments can be found by considering in turn the equilibrium of different elemental portions of the structure and calculating the shear and moment sets for each elemental portion. To help visualize the distribution of the shears and moments, the values thus found can be plotted graphically to produce shear and moment diagrams.

Beams

FIGure 6 Distribution of forces and stresses in a beam.

depends directly on the

depends directly on the

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 6 shows the relationship between the shear and moment distribution in a structure and the shear and bending stresses developed. In addition to bending stresses and horizontal and vertical shear stresses, other important considerations in beam analysis include bearing stresses, torsional stresses, combined stresses, the shear center of a beam, principal stresses, and a beam’s deflection characteristics. These topics are discussed later in the chapter.

3

analysIs oF Beams

In this section, we review briefly topics addressed in detail in textbooks devoted exclusively to studying the strength of materials.2 The following topics, which are the basic issues involved in analyzing beams made of linearly elastic materials, are addressed: (1) bending stresses, (2) shearing stresses, (3) bearing stresses, (4) combined stresses, (5) torsional stresses, (6) shear centers, (7) principal stresses, and (8) deflections. The examples used in these section are based on Allowable Strength Design (ASD) methods. Other design methods specific to different materials will be covered in Sections 4.2–4.6.

3.1

Bending stresses

Introduction. As noted in the introductory discussion on the general behavior of beams, bending stresses that vary linearly at a cross section are developed in response to the action of the external bending moment in the beam at that point. 2

See, for example, Ferdinand L. Singer, Strength of Materials, 2nd ed., New York: Harper & Row, 1962.

Beams

FIGure 7 The actual bending stresses at any point in a beam of any cross section are given by fy = My>I. The maximum bending stress present is given by: fb = Mc>I.

linearly

d

Copyright © 2013. Pearson Education Limited. All rights reserved.

Mc

Before looking at how to determine these bending stresses in a quantitative way, it is useful to review the general factors on which the bending stresses in a beam depend. Consider the beam illustrated in Figure 7. The stress distribution is based on the assumption that stresses are linearly dependent on deformations. The magnitude of the actual bending stresses 1fy 2 that are present at a point directly depends on the magnitude of the external moment 1M2 that is present at the section. The magnitude of fy must also depend directly on the distance 1y2, which defines the location of the point considered with respect to the neutral axis of the beam. It is quite reasonable to expect that the stress 1fy 2 will be inversely dependent on some measure defining the size and shape of the beam. Increasing the size of the beam should decrease the stress in the beam for a given moment. Let us temporarily designate this measure as I (typically called the moment of inertia), without as yet defining it further. The bending stress can be dependent on no other parameters. The actual stresses developed, for example, are not influenced by the material used, in the same way that the forces in a truss member are not dependent on the material of which the member was made. Ascertaining whether a member is safe under a given loading requires considering the strength characteristics of the material, but this problem is different from the one addressed here. In this text, a lowercase f denotes the actual stress that is expected to develop in a member. As previously discussed, it is a variable quantity that depends on the type and magnitude of the external force and moment system causing the stress, the size and shape of the beam cross section, and the location of the point considered in a member. The stresses developed in a member that can be calculated should not

= C.d = T.d

Beams be confused with allowable yield or failure stresses in the material that makes up the beam. When using Allowable Strength Design (ASD) design methods, safety factors are used to reduce permissible stress levels below yield or failure stresses. When using Load and Resistance Factor Design (LRFD), on the other hand, safety factors are primarily applied by enhancing the loads. The coverage in this section refers to ASD approaches. (See also Section 3.2.) Yield or failure stresses, denoted by an uppercase F, are a function of the type of material and are experimentally determined. Thus, fy is the actual stress in bending that exists at a point y measured relative to the neutral axis of the member, fb is the actual bending stress developed at the extreme outer fiber of the beam, Fy is the yield stress of the material of which the beam is made, and Fb is the permissible or allowable stress that the material can carry 1Fb = fy , safety factor when using ASD methods2. The actual and the allowable, or yield, values are compared with one another to determine how safe the beam is. For example, in a steel beam, when the calculated expected stress fb exceeds the allowable stress Fb, the beam is considered unsafe, and if Fb exceeds the yield stress Fy, failure could occur. Because the bending stress 1fy 2 depends directly on the moment 1M2 and the locational parameter 1y2 and inversely on the measure of the properties of the beam section 1I2, the variables generally relate in the following way: 1 fy 1 a M, y, b I

M increases y increases As c y increases fy increases s fy increases fy decreases

Thus, fy 1

My I

It is interesting to note that if the units of the actual stress fy are lb>in.2, the units of M are in.-lb, and the units of y are in., then the unit of I (the term characterizing the nature of the cross section) must be the unique in.4, so that the expression My>I is dimensionally correct 3i.e., lb>in.2 = [email protected]>in.4 4. In a beam of any cross-sectional shape, the maximum bending stress fb normally occurs at the outer fibers of the beam where y = c: fb = Mc>I

Copyright © 2013. Pearson Education Limited. All rights reserved.

A following discussion notes that for the specific case of a rectangular beam with a width b and a depth h, I = bh3 >12, and the maximum bending stress developed in the beam occurs at its outer faces where y = h>2. For a rectangular beam (only), we then have the following for the maximum bending stress present: f = Mc>I =

M1h>22 1bh3 >122

= M> 1bh2 >62

Later, we see that the denominator in this expression represents a property of a rectangular cross section—the section modulus—defined as S = 1bh2 >62. We have not yet formally derived either the general expression for bending 1f = My>I2 applicable to any beam or f = M> 1bh2 >62 for a rectangular cross section. We do so in the next section. First, however, it is useful to see an example of how these expressions are used in practice, using Allowable Strength Design(ASD) methods. The example also illustrates how to determine whether a beam constructed of a given material carrying a certain load is adequately sized with respect to bending. This involves calculating the maximum critical bending stress developed in the beam (which occurs at the section of maximum moment along the length of the beam and at the fiber that is farthest removed from the neutral axis of the beam) and comparing this stress with a level that is predetermined as acceptable for the material used (the allowable stress when using ASD methods, or the stress level close to failure stresses when using LRFD methods where critical moments are calculated based on factored loads). As discussed in Section 2.6, this allowable stress contains a specified factor of safety. If the actual stress that is present at a point

Beams exceeds the maximum stress for the respective design method (ASD or LRFD), the beam is overstressed at that point and is not acceptable. examPle A simply supported rectangular beam 15 ft long supports a uniform loading of 400 lbs>ft. It has a cross-sectional width of 6 in. and a depth of 12 in. What is the maximum bending stress developed in the beam? Ignore the dead load of the beam. Recall that the maximum bending moment present for a loading of this type is given by M = wL2 >8 and occurs at the midspan of the member. Hence, this bending moment is used in the stress calculations. solution: Maximum bending moment: M = wL2 >8 = 1400 lb>ft2115 ft2 2 >8 = 11,250 ft@lbs = 112 in.>ft2111,250 ft@lb2 = 135,000 in.@lb

Maximum bending stress for a rectangular beam: fb =

M 1h>22 1135,000 in.@lb2 Mc M = = = = 937.5 lb>in.2 I 1bh3 >122 1bh2 >62 6 in. * 112 in.2 2 >6

These maximum stresses occur at midspan where the bending moment M is maximum and at the extreme outer fibers of the beam (top and bottom faces). These are the maximum bending stresses present in the beam and are located where the bending moments are maximum. If the material used is wood with an allowable stress (see Section 2.5) in bending of Fb = 1600 lb>in.2, then the member is adequately sized with respect to bending because the stress of 937.5 lb>in.2 is less than the allowable stress of 1600 lb>in.2

In the preceding example, note that the expression fb = M> 1bh2 2 >6 can also be used to quickly understand how stresses vary with changes in beam width or depth. Doubling the beam depth (from h to 2h) decreases bending stresses by a factor of 4 (the depth is squared and in the denominator). Doubling the beam width decreases stresses only by a factor of 2. It is more efficient to increase beam depths rather than widths to reduce stresses.

Copyright © 2013. Pearson Education Limited. All rights reserved.

examPle A simply supported beam carries a concentrated load of P at midspan. Assume that the dimensions of the beam are as follows: b = 5 in. 1127 mm2, h = 10 in. 1254 mm2, and L = 10 ft 13.048 m2. Assume also that P = 4000 lb 117.792 kN2. If the allowable stress in bending for the specific type of timber used Fb = 1500 lb>in.2 110.34 N>mm2 or 10.34 MPa2, is the beam adequately sized with respect to bending? Note that the maximum bending moment developed for the loading condition described is M = PL>4. Use the fb = Mc>I expression directly for a rectangular beam instead of fb = M> 1bh2 >62. Note that c = h>2 for a rectangular beam and I = bh3 >12. solution:

Maximum bending moment: 14000 lb2110 ft2 PL = = 10,000 ft@lb = 120,000 in.@lb 4 4 117.792 kN213.048 m2 = 13.56 kN # m = 13.56 * 106 N # mm = 4

M =

Moment of inertia: I =

15 in.2110 in.2 3 1127 mm21254 mm2 3 bh3 = = 416.7 in.4 = = 173.4 * 106 mm4 12 12 12

Beams Maximum bending stress = fb at y = c: fb = =

1120,000 in.@lb2110 in.>22 Mc = = 1440 lb>in.2 I 416.7 in.4 113.56 * 106 N # mm21254 mm>22 1173.4 * 106 mm4 2

= 9.93 N>mm2 = 9.93 MPa

Copyright © 2013. Pearson Education Limited. All rights reserved.

Note that the calculated stress level is independent of the type of material used in the beam (i.e., the same stress would be present whether the beam was made of timber, steel, or plastic). If the material used has the allowable stress value originally noted, however, the wood beam is adequately sized with respect to bending because the actual stress developed is less than the allowable stress—that is, 1 fb = 1440 lb>in.2 2 … 1Fb = 1500 lb>in.2 2 or 19.93 N>mm2 2 … 110.34 N>mm2 2.

General theory. The basic method described here for finding the stresses in a beam can be applied to beams of any cross-sectional shape. When the shape of the cross section is complex, however, the solution is involved. To simplify calculations, the problem has been solved in general terms and the results presented in the form fy = My>I. In this formulation, a general expression for I appears that can be evaluated for any different shape of cross section. To derive the expression fy = My>I, operations are performed with respect to an elemental area within the cross section. Thus, the moment resistance in a beam of any cross-sectional shape is generated by stresses acting over areas to produce forces, which in turn act through a moment arm to produce a resisting moment. In the case of an elementary area dA, the force on the element is directly proportional to y 3i.e., force = fy dA = fb 1y>c2dA4, and its moment is the force acting over the distance y 3i.e., moment = y fy dA = 1fb >c2y2dA4. Thus, the term y2dA represents the resistance to bending associated with the elemental area dA and its location, defined by y, in the beam. The total moment resistance of the beam is the sum of the contributions of all the element areas, or M = 1A 1fb >c21y2dA2. If I is defined as 1A y2dA, then M = fb I>c, or fb = Mc>I, as before. The term 1A y2dA represents the total resistance to bending associated with the sum of all elemental areas in the beam. This expression is encountered as the second moment of an area in mathematics and is discussed in basic calculus textbooks, where the term is evaluated for different shapes. The expression is called the moment of inertia in the structural engineering field. The moment of inertia for a rectangular beam of width b and depth h, for example, is demonstrated to be I = bh3 >12. The expression fb = Mc>I can be applied to find the stresses at the extreme fiber of any beam, located by the distance c. For the stresses at any point located by the distance y, the expression becomes fy = My>I. The distance y, like the distance c, is measured from the neutral axis of the member, so it is necessary to know exactly where this axis lies. The neutral axis usually coincides with the centroid of the beam section. (For prestressed or pretensioned beams, this is not the case; nor is it the case with sections made of multiple materials.) The centroid of a cross-sectional area, defined by 1A y dA = 0, can be visualized as the point at which the geometric figure defining the area balances. The location of the centroid of a figure also determines the location of the plane of zero deformations and bending stresses. In cross sections that are symmetrical about their horizontal axes, such as a rectangle, a square, or a circle, the centroid (and hence neutral axis) is at the midheight of the section. Deformations and bending stresses vary linearly in the member and are proportional to the distance from the neutral axis of the member. [See Figure 4(c).]

Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

centroids and moments of Inertia. In the preceding discussion, the centroid and moment of inertia of a cross-sectional shape emerged as crucial descriptors of the beam cross section. In simple terms, the centroidal location of a geometric shape can be thought of as the point at which the cross-sectional shape can be balanced. For a circular shape, this would be the center point of the circle. For a rectangle, it is the midpoint. When only one axis is considered, the centroid can be considered a line. For nonsymmetric or built-up shapes, the location must be found by using the expression 1 ydA = 0. This expression means that the A moments (ydA) of each part of the area of the shape on either side of the centroid are equal (thus, the shape balances about the centroid—see Figure 46). The centroid of a triangle, for example, lies at a point one-third the height of the triangle from a reference base. The centroid for the T-shaped section lies nearer the top of the T than at half-height, but its exact location depends on the relative sizes of the horizontal and vertical parts of the T. The moment of inertia (I) of a cross section can be considered a measure of the stiffness, and hence resistance to bending. Bending stresses decrease when members with a larger moment of inertia are used. Moment of inertia values can be obtained in relation to x-x and y-y axes of a shape. As noted earlier, the moment of inertia of a cross section is defined by I = 1A y2dA. This expression can be evaluated for different cross-sectional shapes. The process for finding the I value of a typical rectangular shape, for example, is illustrated in the following problem, where I = bd3 >12 is found for a rectangular beam. For a triangular shape, I = bd3 >36 and for a circular shape, I = pd4 >64. Typically, I values depend on a dimension of the cross section raised to some power; therefore, small increases in these dimensions yield large increases in I values. Many cross sections are not simple geometric figures but are built from configurations of rectangular, circles, triangles, and so forth. They might also have holes within a solid shape. In these cases, more complex analysis methods are needed to determine the locations of centroids and moment of inertia values. The “parallel-axis theorem,” for example, is used to determine I values for built-up cross sections, such as T shapes. If shapes have complex curved boundaries, the reader must resort to first principle analyses using I = 1 y 2dA. A For common shapes, the centroidal locations and moment of inertia values can be easily tabulated. The method for determining these values is straightforward, many computer programs are available that can be used to quickly and easily find values for even complex shapes. Using such a program, cross-sectional shapes may be directly drawn and values quickly found. examPle A cantilever beam that is rectangular in cross section and of length L carries a concentrated load of P at its free end. Determine the maximum bending stress that is present in the member. Assume that the dimensions of the beam are as follows: b = 4 in. 1101.6 mm2, h = 6 in. 1152.4 mm2, and L = 8 ft 12.44 m2. Assume also that P = 500 lb 12224 N2. The maximum bending stress occurs where the moment is a maximum and on the outer fibers of the beam at the same section. Note that the maximum bending moment for a beam with a concentrated load at its end is given by M = P * L (see Figure 48). Also note that for a rectangular beam, I = 1 y2dA = bh3 >12. A

solution:

Maximum bending moment: PL = 500 lb * 8 ft = 4000 ft@lb = 48,000 in.@lb

= 2224 N * 2.44 m = 5426.6 N # m = 5.426 * 106 N # mm

Beams Maximum bending stress = fb at y = c: fb = = =

M1h>22 Mc M = = 2 I bh3 >12 bh >6

48,000 in.@lb = 2000 lb>in.2 4 in. * 16 in.2 2 >6

5.426 * 106 N # mm = 13.79 N>mm2 = 13.79 MPa 101.6 mm * 1152.4 mm2 2 >6

The stresses are predicted to exist within the cross section of the beam, regardless of the material present, and say nothing about whether the member is safe or unsafe in bending. If the material used has a low allowable stress level of 1200 lb>in.2, as might exist for timber, the member is unsafe in bending and could fail before reaching the predicted stress level of 2000 lb>in.2. If the material has a higher allowable stress level, say, 24,000 lb>in.2, as might exist for steel, then the member is safe in bending and the predicted stress level would be reached.

examPle A cantilever bean that is 10 ft long with a 5-in.-by-10-in. rectangular cross section carries a concentrated load of P = 10,000 lb at its end. What maximum bending stresses are developed at the base of the cantilever? While the problem could be solved, as in previous examples, for a rectangular beam, use the more general theoretical expressions presented earlier to find the bending stresses, including determining the moment of inertia for the cross section. Note that the maximum bending stresses present occur at the outer top and bottom faces of the beam or at y = c = 10 in.>2 or 5 in. What bending stresses would be present at the centroid of the beam cross section (or at y = 0)? In this problem, the centroid and neutral axis have the same location. solution: Maximum bending moment: M = PL = 5000 lb110 ft2 = 10,000 ft@lb = 120,000 in.@lb Moment of inertia I =

L

+h>2 2

y dA =

A

L

y2 1b dy2 = bc y3 d

-h>2

3

+h>2 -h>2

= bh3 >12

I = bh >12 = 15 in.2110 in.3 2 >12 = 416.6 in.4 Copyright © 2013. Pearson Education Limited. All rights reserved.

Maximum bending stresses: fb at y = c = h>2 = 10 in.>2 = 5 in. fb = Mc>I = My>I = 1120,000 in.@lb215 in.2 >416.7 in.4 = 1440 lb>in. in.2

Bending stresses at the centroid of the beam: fb at y = 0 1neutral axis location2 fb = Mc>I = My>I = 1120,000 in.@lb2102 >416.7 in.4 = 0

examPle For the same span and loading condition in the previous example where M = 120,000 in.@lb, what stresses would develop in a beam with a circular cross section having a diameter of 15 in.? solution: Moment of inertia I =

L A

y2dA = pd4 >64 = p115 in.2 4 >64 = 2483.8 in.4

Beams Bending stresses at y = c: fb = Mc>I = 1120,000 [email protected] in.2>2483 in.4 = 362.3 lb>in. in.2

comparing cross sections. The moment of inertia I may be evaluated for circular, triangular, and other geometric shapes. The discussion that follows compares beams with similar areas but different shapes. Note the importance of the I and the c values in determining stress levels. In evaluating beams with two different c distances (e.g., the triangular shape), the critical stress determining the beam capacity is usually associated with the largest c distance.

examPle A simply supported beam spans 25 ft and carries a uniform loading of w = 600 lb>ft (Figure 8). As shown, a maximum bending moment of M = wL2 >8 = 1600 lb>ft2125 ft2 2 >8 = 46,875 ft@lb = 562,500 in.@lb is developed at midspan. Three beams of equal cross-sectional area, but different shapes, are considered for use. Which beam has the lowest maximum bending stresses? It is critical to use the correct distance c in calculating bending stress. The triangular cross section has a different c for its upper and lower extremities, implying that stress magnitudes vary accordingly. solution: Rectangular Beam: Section properties: I = bh3 >12 = 18 in.2115 in.2 3 >12 = 2250 in.4

cmax = 7.5 in. Maximum bending stresses:

fmax = Mc>I = 1562,500 [email protected] in.2 >2250 in.4 = 1875 lb>in.2

Circular Beam:

1top or bottom2

Section properties: I = pd4 >64 = 12 * 6.2 in.2 4 >64 = 1160 in.4

Copyright © 2013. Pearson Education Limited. All rights reserved.

cmax = 6.7 in. Maximum bending stresses:

fmax = Mc>I = 1562,500 [email protected] in.2 >1160 in.4 = 3006 lb>in.2

Triangular Beam:

1top or bottom2

Section properties: I = bd3 >36 = 112 in.2120 in.2 3 >36 = 2667 in.4

ctop = 13.3 in. and cbot = 6.7 in. Maximum bending stresses:

ftop = Mc>I = 1562,500 [email protected] in.2 >2667 in.4 = 2805 lb>in.2

1top2

fbot = Mc>I = 1562,500 [email protected] in.2 >2667 in.4 = 1403 lb>in.2

1bottom2

FIGure 8 Beam using either a rectangular, circular, or triangular section. The maximum moment present M = 562,500 in.@lb.

Beams

FIGure 9 Comparison of bending stresses in different beam section with identical areas subjected to the same bending moment of 562,500 in.> lb.

Comparison: The smallest stresses are developed in the rectangular beam (which could carry the greatest external loading before becoming overstressed). Note that the triangular section has a higher I value than any of the sections but develops higher stresses than the rectangular beam because of its larger c distance. The circular cross section has a lower I value than the other two because more material is nearer the neutral axis (Figure 9).

Copyright © 2013. Pearson Education Limited. All rights reserved.

examPle For the same loading condition shown in the previous example 1M = 562,500 in.@lb2, assume that a symmetrical box section was used (Figure 10). What is the maximum bending stress present? [The crux of this problem is calculating the I value for the section. In this example, the hole is treated as a negative area.] The approach shown is

FIGure 10 Box beam.

Beams valid only for symmetric cross sections. Otherwise, the location of the centroid must be found and I calculated by I = 1A y2dA. solution: Section properties: Inet = Igross - Ihole = b1 h31 >12 - b2 h32 >12

= 11221252 3 >12 - 1821102 3 >12 = 14,958 in.4

cmax = 12.5 in. Maximum bending stress:

f = Mc>Inet = 1562,5002112.52 >14,958 = 470 lb>in.2

The maximum stress present is therefore lower than in any of the three sections in the previous example. Box beam shapes are efficient because they move material away from the neutral axis and toward the outer fibers of the beam. (See Section 4.1.)

Built-up sections. The general principles discussed previously are applicable to nonsymmetrical as well as symmetrical sections. The critical difference is that, in a nonsymmetrical section, such as a T beam or triangular shape, the centroid location is no longer obvious and is rarely at the midheight of the section. Calculating the moment of inertia I for a section built of simpler shapes presents special issues. The parallel-axis theorem must be used to find moment-of-inertia values. In the T beam illustrated in Figure 11, the centroid is located near the top flange of the member. Deformations and bending stresses in the member vary linearly and are proportional to the distance from the member’s neutral axis. This implies that the stress levels at the top and bottom of the beam are no longer equal as they typically are in symmetrical sections. The stresses are greater at the bottom of the beam than at the top because of the larger y distance. A difference in stress levels between the top and bottom surfaces is characteristic of nonsymmetrical sections. This point has enormous design implications, as discussed in Section 4. examPle

Copyright © 2013. Pearson Education Limited. All rights reserved.

Determine whether the T beam illustrated in Figure 11 is adequately sized to carry an external bending moment of 120,000 in.-lb. Assume that the allowable stress in bending of the material used is Fb = 1200 lb>in. solution: To find the actual bending stresses developed in the T beam, fy = My>I can be used, as before. The real task is to evaluate I and ymax, which is more complicated than before because of the nonsymmetrical nature of the cross section. First, locate the centroid of the cross section and then use the parallel-axis theorem to evaluate the moment of inertia about the centroidal axis. Location of centroid: Consider the shape to be made up of two rectangular figures (A1 = b1 * h1 and A2 = b2 * h2), as illustrated in Figure 12(c). Assume a reference axis as illustrated. Let yT represent the distance from the base of the figure to the centroid of the whole figure, and let y1 and y2 be the distances to the centroids of the two rectangular figures from this same base. Thus, yT =

a Ai yi i

a Ai i

=

= 1A1 y1 + A2 y2 2 > 1A1 + A2 2

12 in. * 10 in.2113 in.2 + 12 in. * 12 in.216 in.2 12 in. * 10 in.2 + 12 in. * 12 in.2

= 9.18 in.

Beams

FIGure 11 Bending-stress distribution in a T beam. Because the cross section is nonsymmetric, the centroid of the section is nonsymmetrically located. The bending-stress distribution is still linear and passes through zero at the centroid (or neutral axis). This results in a difference in stress levels on the upper and lower faces of the beam.

C

Copyright © 2013. Pearson Education Limited. All rights reserved.

T

Moment of inertia: Let I T represent the moment of inertia of the whole figure about its centroidal axis. Let I1 and I2 represent the moments of inertia of the two rectangular figures about their own centroidal axes and d1 and d2 be the locations of these axes with respect to the centroidal axis of the

Beams

whole figure. I T can be found by using the parallel-axis theorem thus:

b1 h31 b1 h32 ≤ + 1b1 h1 21d1 2 2 R + J ¢ ≤ + 1b2 h2 21d2 2 2 R IT = a 1 Ii + Ai d2i 2 = J ¢ 12 12 i = J

110 in.212 in.2 3 12

+ J

+ 110 in.212 in.213.822 2 R

12 in.2112 in.2 3 12

+ 12 in.2112 in.213.182 2 R

= 16.7 + 291.92 in.4 + 1288.0 + 242.72 in.4 = 829.3 in.4

Bending stresses: The general type of bending-stress distribution present is that of the T beam, as illustrated in Figures 11(a) and (b). The stresses at the top face of the beam are defined by fb = My>I, where y = 4.82 in. (the distance from the top face to the centroidal axis of the figure). Thus, fb =

My I

=

1120,000 [email protected] in.2 829.3 in.4

= 697.5 lb>in.2

The stresses that occur at the bottom face of the beam are found similarly, except that y = 9.18 in. Hence, fb =

My I

=

1120,000 [email protected] in.2 829.3 in.4

= 1328.3 lb>in.2

The stresses at the bottom face of the beam are much larger than those at the top face. This is due to the nonsymmetrical nature of the cross section. If the allowable stress of the material used in the beam is given by Fb = 1200 lb>in.2, then the beam is overstressed on its lower face (i.e., 1328.3 lb>in.2 7 1200 lb>in.2). Note that stresses on the upper face are within the acceptable range. The beam is still inadequately sized, however, because of the overstress occurring on the lower face. We have the following:

Copyright © 2013. Pearson Education Limited. All rights reserved.

y T = 233.17 mm

I T = 345.15 * 106 mm4

fb = 4.82 N>mm2

fb = 9.16 N>mm2

composites and sandwiches. The term composite is used to describe members made of multiple materials that appear within the same cross section and work together to carry bending moments and other forces. Many composite structures are sandwich constructions involving two or more layered materials. A typical sandwich construction consists of thin layers of high-strength materials on the outer extremities of the cross section and a thicker inner core of shear-resistant materials (e.g., dense foams, honeycombs). Usually, they are made as large flat or curved panels that are thin relative to their overall dimensions. Many different outer layer and inner core materials can be used. Plywood sandwich panels, for example, are common. Here, a dense form core material is used that has plywood sheets bonded to its faces. Such sandwich constructions type can be good at carrying normal bending moments and forces associated with distributed floor and wall loadings. They are often used to satisfy structural and enclosure surface needs. Several other material combinations are possible, including various fiberglass or carbon fiber layers bonded to different types of honeycomb cores. The outer layers in a sandwich are responsible for carrying the stresses associated with bending moments and are located where bending stresses are the highest. The core carries shear forces and accompanying shear stresses. Both the outer layers and cores must be designed to carry actual stresses. Care must also be taken to prevent common failures in sandwich panels, including delaminating or debounding between outer layers and cores. The latter can occur due to horizontal shearing stresses. Localized buckling of thin outer layers can accompany

Beams delamination failures. When mixed materials are used in a cross section, the structural analysis procedure is more complicated than those previously described. A transformed area concept is often used that has an equivalent cross section approach. Often a dominant base material is selected, and the equivalent are of the other materials is computed based on their relative mechanical properties. Exact procedures are beyond the scope of this text. Other kinds of composite structures are possible. The classic reinforced concrete beam described later in this chapter is a composite structure because steel and concrete are integrally used in the same cross section to carry bending moments and forces. Other kinds of composites can occur at the material level, where a higher-strength material (typically more costly) is distributed within a lowerstrength matrix. section modulus and symmetrical Beam sizing. It is a straightforward process to determine the required cross-sectional dimensions of a simple symmetrical beam to carry a given bending moment safely. First, a material is selected and allowable stresses in bending are defined. Required section properties are found next, on the basis of making the bending stress level in the beam equal to or less than the allowable stress level in bending. This process is based on using the bending stress relationship fy = My >I. The maximum stress occurs when y is maximum (or ymax = c). Thus, to make the actual bending stress at ymax = c equal to the allowable stress, we substitute Fb (the allowable stress for the material) for fy, c for ymax, and solve for the other terms in the relationship fy = My>I. Hence, for any beam shape, I M Srequired = a b = . c req’d Fb

The I>c measure is the section modulus S of the beam. The problem is then to find an S or I>c value for a beam equal to or greater than the value of M>Fb. For a rectangular beam 1b = width, h = depth2, Sreq’d = I>c = 1bh3>122 > 1h>22 =bh2 >6. Thus, 1bh2 >62 req’d = M>Fb, or 1bh2 2 req’d = 6M>Fb. Any rectangular beam with a combination of b and h dimensions that yields bh2 = 6M>Fb is acceptable with respect to bending. Further beam design considerations are discussed in Section 4.

Copyright © 2013. Pearson Education Limited. All rights reserved.

examPle A simply supported rectangular timber beam that is 12 ft long carries a concentrated load of 4000 lb at midspan. What is the required depth of the member if the width b is 2.0 in.? If b is 4.0 in.? Assume that the allowable stress in bending is FB = 1600 lb>in.2 solution: Bending moment at midspan: Required section modulus: Section size:

bh2req’d

M = PL>4 = 14000 lb2112 ft2 >4 = 12,000 ft@lb = 144,000 in.@lb

S = M>FB; 6 bh2 >6 = 144,000 in.@lb>1600 lb>in.2

= 540 in.3

Depth: If b = 2 in., h = 16.4 in.; if b = 4 in., h = 11.6 in. Other dimensions could also provide a section modulus of 540 in.3

Examining these results indicates that the bending stresses developed in a  beam are extremely sensitive to the depth of the beam. For a given applied moment, doubling the depth of a rectangular beam while holding its width constant reduces bending stresses by a factor of 4. Alternatively, doubling the width of such a beam while holding its depth constant reduces bending stresses only by a factor of 2. These observations follow from manipulations with the expression f = Mc>I = M1h>22>1bh3 >122 = M>1bh2 >62.

Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 12 Stress variations.

distribution of Bending stresses. The previous examples considered the bending stresses at only one cross section of a beam (where the maximum stresses exist). Because bending stresses are directly dependent on bending moments, it follows that bending stress magnitudes vary along the length of a beam as described by the moment diagram in Figure 12. When checking to see if a beam with no variation in depth or width along its length is adequately large, take care to use the maximum bending moment anywhere in the structure. When sizing a beam with no variation in width or depth along its length, using the maximum bending moment value to determine needed depth and width dimensions means that the beam may be oversized where moments are smaller. Beam depths can also be made to vary along the length of a beam because, stress levels are not constant throughout the beam. (See Section 4.1.)

3.2

lateral Buckling of Beams

Consider the thin, deep beam illustrated in Figure 13. Applying a load may cause lateral buckling in the beam, and failure will occur before the strength of the section can be utilized. The phenomenon of lateral buckling in beams is similar to that found in trusses. An instability in the lateral direction occurs because of the compressive forces developed in the upper region of the beam, coupled with insufficient rigidity of the beam in that direction. In the examples discussed thus far, it was assumed that this type of failure does not occur. Depending on the proportions of the beam cross section, lateral buckling can occur at relatively low stress levels. Lateral buckling can be prevented in two primary ways: (1) by using transverse bracing and (2) by making the beam stiff in the lateral direction. When a beam is used to support a roof deck or a secondary framing system, these elements automatically

FIGure 13 Lateral buckIing in beams.

Beams

FIGure 14 guides only.

Lateral bracing required for timber beams. These proportions are rough

2:1

3:1

6:1

5:1

7:1

provide transverse bracing. If a beam is used in a situation where this type of bracing is not possible, the beam can be made sufficiently stiff in the lateral direction by increasing the transverse dimension of the top of the beam. In a rectangular beam, the basic proportions of the cross section can be controlled to accomplish the same end. Figure 14 illustrates when lateral bracing is not required for timber beams and suggests types of bracing when it is. Different rules apply for steel plates. examPle A cantilever beam 10 ft in length supports a concentrated load of 1000 lb at its free end. The maximum moment developed in the beam is, therefore, M = PL = 110 ft * 12 in.>ft2 11000 lb2 = 120,000 in.@lb. Assume that the allowable stress in bending is Fb = 1200 lb>in.2 Determine the required member size. solution

Copyright © 2013. Pearson Education Limited. All rights reserved.

f =

Mc M M = or Sreq’d = I S Fb

1120,000 in.@lb2 bh2 M = = 100 = 6 Fb 1200 lb>in.2

6 bh2 = 600 in.3

Any beam with a bh2 value of 600 in.3 is satisfactory with respect to bending. The following table indicates acceptable beams (in all cases, bh2 = 600):

trIal Beam sIze (In.) cross-sectIonal dePth/ area (In.2) b h WIdth ratIo

lateral BracInG

0

5.5

110

0.27:1

None required

10

7.7

77

0.77:1

None required

5

11.0

55

2.2:1

None required

3.5

13.1

46

3.7:1

Bracing required

2

17.2

34.4

8.6:1

Bracing required

Beams As evident from the preceding example, many different beam cross sections can be used in a common situation. Beams that are relatively wide and shallow require much more material to support a given load than do beams that are relatively thin and deep. Which type to use is up to the designer. Design trade-offs are invariably involved. If relatively efficient, thin, deep beams are used, the designer must assure that lateral bracing of the appropriate type is provided. Providing such bracing usually requires additional material and is an added cost item. If there is no way to provide the necessary lateral bracing, or if the designer chooses not to provide it, the proportions of the beam must be selected so that the member itself provides sufficient inherent resistance to lateral buckling. Note that in the common situation of a beam’s supporting floor decking, inspecting the bracing requirements indicates that using beams which are relatively shallow (e.g., with 2:1 or 3:1 depth-width ratios) is inefficient because, the decking inherently provides lateral bracing and it is not necessary to have a beam capable of doing so independently. Consequently, much thinner and deeper beams are used in such instances. (Beams with depth-width ratios of between 5:1 and 7:1 are found in common house construction—so much for the exposed beams using 2:1 or 3:1 depth–width ratios that are typically found in brand-new suburban chalets.)

Copyright © 2013. Pearson Education Limited. All rights reserved.

3.3

shear stresses

As noted earlier, an elemental portion of a beam remains in equilibrium with respect to vertical shear forces through the development of vertical shearing stresses in the beam. The resultant force VR = 1A fv dA that is equivalent to these stresses is equal in magnitude, but opposite in sense, to the external shear force VE. To understand the magnitude and distribution of the stresses involved, it is more convenient to look first at the horizontal shear stresses that also exist in the beam. The presence of horizontal shear stresses may be visualized easily by considering two beams, one made of a series of unconnected planes (similar to a deck of cards) and the other of a solid piece of material. (See Figure 15.) Under the action of a load, a slippage occurs between each of the planes in the former case as the whole assembly deforms. The structure is behaving as a series of thin superimposed beams. The second structure behaves in a composite way, is much stiffer than the first, and can carry a much higher load. If the surfaces of the planes on the first structure were bonded together, the resulting structure would also behave as a composite whole. If this were done, stresses acting in the bond parallel to the adjacent surfaces would be developed because, the planes would still have a tendency to slide. These stresses are called horizontal shearing stresses. As long as the material can carry such stresses, adjacent planes in a beam remain in contact and the structure behaves as a composite whole. When a beam fails due to horizontal shear, slippage between planes occurs. Some materials, such as timber, are particularly weak with respect to stresses of this kind, and failures are not uncommon. [See Figure 15(d).] In the example in Figure 16, the magnitude of the horizontal shear stresses can be found by considering the equilibrium, in the horizontal direction, of the upper-left portion of the beam. (Remember that any elemental portion of a structure must be in equilibrium, so any portion can be selected for analysis.) With respect to equilibrium in the horizontal direction, the bending stresses acting on the right face of the element produce a horizontal force acting to the left. For equilibrium to obtain, the horizontal force due to bending must be balanced by other internal forces acting in the opposite direction, which implies the existence of horizontal shear stresses. The equilibrating force is provided by shear stresses acting in the horizontal direction over the horizontal face of the beam. Other horizontal planes in the beam also have shearing stresses but of varying magnitudes. In Figure 16(c), a plane near the top of the beam is shown. Again, the bending stresses produce a force in the horizontal direction that causes shearing stresses

FIGure 15 Horizontal shear stresses in a beam.

Beams

FIGure 16 Horizontal shear stresses in a beam. In a rectangular beam, the horizontal shear stresses vary parabolically from a maximum at the neutral axis of the member to zero at the top and bottom faces. The distribution of stresses along the length of a horizontal plane depends directly on the variation in the external shear force along the same length. Where the external shear force is high, shear stresses will be high, and vice versa.

because they

Copyright © 2013. Pearson Education Limited. All rights reserved.

that equilibrate this force are conse-

to develop on the horizontal plane. These forces and stresses, however, are smaller than those at the middle section of the beam because, the bending stresses act over a smaller area and thus produce a smaller horizontal force. At the top layer of the beam, no forces or shear stresses can exist. A close look at the magnitudes of these shearing stresses indicates that they vary parabolically from a maximum at the neutral axis of the beam to zero at free surfaces. The horizontal stresses also vary along the length of the beam in most situations (in particular, when the bending stress varies along the length of the beam). An exact expression, based on concepts similar to those just discussed, can be determined for the horizontal shearing stress in a beam. The horizontal shear stress at a layer located a distance y from the neutral axis can be calculated by the expression fv = VQ>Ib, where V is the vertical shear force at the vertical section under consideration, b is the width of the beam at the horizontal layer under consideration, I is the moment of inertia at the section, and Q is the first moment of the area (about the neutral axis) above the horizontal layer 1Q = 1y dA 2. Note that the expression describes not only the horizontal shear stresses at a point but also the vertical shear stresses because it can be demonstrated that the magnitudes of the horizontal and vertical shear stresses acting at a point are always equal. (See Figure 17.) Thus, the vertical shear stresses also vary parabolically, with a maximum value occurring at the neutral axis.

Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 17 The numerical magnitudes of the horizontal and vertical shear stresses at any point in a beam are always equal.

The general shear-stress equation fv = VQ>Ib can be evaluated for the specific case of a rectangular beam. (See the next example.) In a rectangular beam, the maximum shear stresses occur at the neutral axis of the member (at midheight) and are given by fv = 32 1V>bh2 = 32 1V>A2, where b and h are the dimensions of the cross section. Hence, the maximum shear stress in a rectangular beam is 1.5 times the value of the average shear stress in a rectangular beam. [See Figure 18(a).] In beams constructed of thin-walled sections with a lot of material in flange areas, the shear stresses are more uniform because of how the material is distributed throughout the cross section. The shear stress in such sections, such as a wide-flange beam, can be approximated by fv = V>Aweb = V>td with relatively little error. [See Figure 18(b).]

Beams

FIGure 18 Shearing-stress distributions in a rectangular beam and in a wide-flange beam.

examPle Is the timber beam illustrated in Figure 19(a) adequately sized with respect to shear? Assume that the allowable shear stress for the material used is Fv = 150 lb>in. 11.03 N>mm2 2. solution: The maximum shear stress occurs at the neutral axis of the beam at the cross section where the shear force V is a maximum. The maximum shear stress is given by the following:

Copyright © 2013. Pearson Education Limited. All rights reserved.

fv =

V1A′y′2 V1bh>221h>42 VQ 3 V 3 2000 lb = 60 lb>in.2 = = = = = Ib Ib 2 bh 2 15 in.2110 in.2 1bh3 >122b

This is the maximum vertical or horizontal shear stress that occurs in the beam. Because 60 lb>in.2 … 150 lb>in.2, the beam is adequately sized with respect to shear. The following is an alternative using metric units: fv =

3 V 3 8896 N = 0.414 N>mm2 6 1.03 N>mm2 = 2 bh 2 1127 mm21254 mm2

examPle For the same T beam previously analyzed and again illustrated in Figure 19(b), determine the shear stress present at the interface between the top flange element and the web element if V = 2000 lb. If glue were used to bond the elements together, what would be the stress on the glue? If nails spaced at 3.0 in. on center were used as connectors, what would be the shear force on each nail? What is the maximum shear stress present in the cross section? solution: The expression fv = VQ>Ib will be used again, with Q = A′ y′ evaluated for the area of the beam above the interface between the top flange and web element. We have the following: Q = A′ y′ = 1b1 * h1 21y′2 = 110 in. * 2 in.213.82 in.2 = 76.3 in.3 I = 829.3 in.

1See previous example.2

Beams b = 2 in. V = 2000 lb VQ fv = = 12000 lb2176.3 in.3 2 > 1829.3 in.4 212 in.2 = 92.0 lb>in.2 Ib

The quantity fv is the horizontal or vertical shear stress that occurs at the interface. If the two rectangular elements were glued together at this interface, the glue must be able to withstand 92.0 lb>in2 of stress. If some other connector, such as nails, were used, the connector must to be able to carry 2 in. * 92.0 lb>in., or 184 lb>in. If nails were placed 3.0 in. on center, each nail must carry 3 in. * 184 lb>in., or 552 lb of shear. This value could be compared with experimentally derived allowable forces on nails to determine if the nail spacing were adequate. The maximum shear stress in the section occurs at the neutral axis. In evaluating this maximum stress, it is more convenient to consider the area below the neutral axis: Q = A′ y′ = 12 in. * 9.18 in.219.18 in.>22 = 84.3 in. fv =

12000 lb2184.3 in.2 VQ = = 101.7 lb>in.2 Ib 1829.3 in.4 212 in.2

The shear-stress distribution for the cross section is illustrated in Figure 19(b). The discontinuity at the interface occurs because of the change from b = 2 in. to b = 10 in. Because shear stresses are inversely dependent on the width of the beam, actual shear stresses are significantly reduced in the wide-top flange.

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 19 Shearing stresses in beams.

Beams

3.4

Bearing stresses

The stresses developed at the point of contact between two loaded members are called bearing stresses. Such stresses, for example, are developed at the ends of a simply supported beam where it rests on end supports having certain dimensions. The magnitude of the stresses developed depends on the magnitude of the force transmitted through the point of contact and the surface area of the contact between the two elements. The smaller the contact area, the greater are the bearing stresses. The stresses produce deformations in both elements at the point of contact. These deformations typically extend only a small distance into the elements. The magnitude of the bearing stress at a point is equal to the load transmitted, divided by the area of contact, or fbg = P>A. This equation assumes that the bearing stresses are uniformly distributed over the contact area, an assumption that is not quite correct, but reasonably so. Many materials, such as timber, are particularly susceptible to bearing-stress failures. When a compressive load is transmitted, bearing-stress failures are not catastrophic and are typically manifested by an appearance of crushing in a material. The failure is generally localized but should be avoided.

examPle Assume that the timber beam illustrated in Figure 20 carries a uniformly distributed load of 500 lb>ft 17300 N>m2 over a span of 16 ft (4.87 m). Assume also that the contact area between the beam and the column on which it rests is 4 in. * 4 in.1101.6 mm * 101.6 mm2. If the allowable stress in bearing of the timber in the beam is Fbg = 400 lb>in.2 12.76 N>mm2 2, is the area of contact sufficiently large? Determine the actual bearing stress that exists at the reaction. The reaction is 4000 lb (17,792 N) and the contact area is 4 in. * 4 in. = 16 in.2 110,322.6 mm2 2.

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 20 Bearing stresses at the end of the beam.

Beams solution: Actual bearing stress = fbg = =

P 4000 lb = = 250 lb>in.2 A 16 in.2

117,792 N2

110,322.6 mm2 2

= 1.72 N>mm2

Because the actual bearing stress is less than the allowable bearing stress (250 lb>in.2 6 400 lb>in.2, or 1.72 N>mm2 6 2.76 N>mm2), there is sufficient contact area between the two members.

3.5

torsion

Torsion is a twisting. Torsional forces develop in a member through the direct application of a torque or twisting moment MT, or they may develop indirectly because of an off-balanced load or force application. For example, framing one member into the side of another can cause twisting because of the off-balanced load application. Figure 21(a) illustrates a round member subjected to torsion. The analysis of the stresses produced by this type of force is not particularly complex but is not covered here in detail. Figure 21(b) illustrates the type of stresses generated in the member by the applied torque. The resultant forces of the stresses developed produce a couple that equilibrates the applied twisting moment. The stress at a point, t, is dependent on the magnitude of the applied twisting moment MT , the location of the point is defined by the distance r from the centroid, and the properties of the cross section are designated by the symbol J. The relation between these parameters is given by t = Mr>J. This expression is analogous to that for bending stresses 2 1 fy = Mc>I2. J is analogous to I and is again given by 1Ar dA, except that polar coordinates are now used and J becomes the polar moment of inertia.

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 21 Torsion in beams. Tensional stresses are developed in members from applying a twisting force to the member.

Beams

FIGure 22 Closed sections are better for resisting torsion than are open sections. Many bridge configurations are open sections.

In instances where the section considered is not circular, the analysis of torsional stresses is fairly complex. Figure 21(c) illustrates a rectangular beam subjected to torsional stresses. Twisting is induced rigidly into the beam by its framing. The torsional stresses in such a beam are again dependent on the twisting moment MT, the location of the point considered, and the properties of the rectangular section and are given by an expression of the form t = aMT >b2d, where a depends on the relative proportions of b and d. For d>b = 1, a = 0.208; for d>b = 3.0, a = 0.267; and for d>b = 10, a = 0.312. Thin-walled tubes of thickness t may be analyzed in terms of shear flows: q = tt. In general, MT = 1r qds, where ds is along the perimeter and r is the distance to the center of twisting. Thus, q ds contributes an element to the torque resistance (e.g., for a thin circular tube, MT = q12pr2r, or q = MT >2pr2). More generally, for any thin-walled tube, MT = 1r qds becomes MT = 2Aq and q = MT >2A, where A is the area enclosed by the center line of the tube; then t = q>t. In design, closed forms such as tubes, pipes, or box beams are usually better for carrying torsional forces than are open forms such as plates or channels. The latter twist badly. Box forms are frequently used for bridge cross sections and other situations in which torsion is problematic. (See Figure 22.)

3.6

shear center

Copyright © 2013. Pearson Education Canada. All rights reserved.

It is important to note the phenomenon of twisting under the action of transverse loads when using members that are not symmetrical about the vertical axes and that are composed of thin-walled sections (e.g., a channel section). A symmetrical section, such as a rectangular beam, as used in Figure 23(a), simply deflects downward when loaded. A channel section carrying the same load would twist as indicated in Figure 23(b) because the line of action of the load does not pass through what is called the shear center of the member. This twisting leads to the development of torsional stresses in the member. The twisting can be predicted by a close investigation of the action of the stresses present in the section and can be described in FIGure 23 Shear center in beams. In nonsymmetric members, applying a load directly to the member causes the member to twist. Applying the load at the shear center of the beam causes the member to deflect downward without twisting. The shear center of many nonsymmetric members often lies outside the member.

Beams terms of shear flows. The presence of the twisting complicates but does not exclude the use of such members as beam elements because, a section can still be designed to carry external forces safely. If it were desired to use such a member to carry transverse forces without twisting, it is possible to do so by locating the load so that it passes through the shear center of the beam. In Figure 23(c), applying the load as indicated causes the twisting in the directions shown. Therefore, some point must exist to the right of the section where the load could be applied and cause no twisting. That point is the shear center. Other references describe how to locate the point.3 Note that, in the example illustrated, the shear center lies outside the section, thus complicating the use of that location as a point of application of the load.

3.7

deflections

Consider the beam in the upper left in Figure 24(a). The deflection ∆ at a particular point in this beam, or in any other, depends directly on the load P or w, directly on the length L of the beam, inversely on the stiffness of the beam, which depends

FIGure 24 Deflections in beams.

Copyright © 2013. Pearson Education Canada. All rights reserved.

—

—

—

—

—

3 Stephen Crandall and Norman Dhal, An Introduction to the Mechanics of Solids, New York: McGraw-Hill Book Company, 1959.

—

Beams on the amount and way material is distributed in the cross section (as characterized by the moment of inertia, I), and inversely on the stiffness of the beam, which depends on the stress-deformation characteristics of the material used in the beam (as characterized by the modulus of elasticity, E). Thus, ∆ depends on P or w, L, 1>I. In other words,

Copyright © 2013. Pearson Education Canada. All rights reserved.

w increases, ∆ increases L increases, ∆ increases d t I increases, ∆ decreases E increases, ∆ decreases

∆ = C1 1wl4 >El2 ∆ = C2 1PL3 >El2

The values of L must be raised to the powers noted for the expressions to work out dimensionally. For uniform loads, in. = 1lb>in.21in.a 2 > 1lb>in.2 21in.4 2, so the exponent a of L must be 4. For point loads, in. = 1lb21in.b 2 > 1lb>in.2 21in.4 2, so the exponent b of L must be 3. Support conditions also influence the amount of deflection present: C1 and C2 are constants that depend on support conditions. For a simply supported beam with a uniformly distributed load, C1 becomes 5>384, so ∆ = 5wL4 >384EI. Deflections are highly sensitive to the beam length. Also, it is interesting to note that, for a beam that is similar in all respects except that the member ends are fixed rather than simply supported, the deflection is given by ∆ = wL4 >384EI. Thus, fixing the ends of a simply supported beam carrying a uniformly distributed load reduces the midspan deflection by a factor of 5. Note that the form of both expressions is the same, but there is a difference in the modifying constant—which reflects the different boundary conditions. The increased rigidity associated with fixed-ended beams, as well as continuous beams, is a primary reason such members are used extensively. Deflections for other loading and support conditions are shown in Figure 24. In designing beams, controlling the magnitude of deflections is always a major problem. It is difficult to even establish what constitutes an allowable deflection in a beam. Still, some objective criteria can be established. If a beam deflects such that it interferes with or impairs the functioning of another building element, for example, the allowable deflection could be based on acceptable tolerances for other systems. In many cases, however, the problem is more subjective. For instance, many beams are safe from a strength viewpoint but are said to visually. Likewise, when people walk across floors that feel bouncy or springy, the supporting beams are said to deflect too much. With respect to the latter point, attempts are made to control the bounce of a floor by limiting deflections. This is, however, a misconception because people do not feel deflection; instead, they sense the accelerations associated with deflections. This is an important point because accelerations could be controlled in other ways than by artificially limiting deflections. The problem of what determines excessive deflections is tough. Empirical guidelines are often used. A common empirical criterion used to control visual sag and bounce problems is that the deflection of a floor should not exceed 1360 of its span. (The criterion is usually expressed as ∆ allowable = L>360, where L is the span of the member.) Usually, the L>360 criterion for floors applies to deflections caused by live loads only. Dead-plus-live load deflections are typically limited to L>240 of the span. Steel members or longer glued laminated timber beams are often cambered upward an amount equal to the dead-load deflection so that the live-load deflection occurs with respect to a horizontal member. Roof deflections are typically limited to L>240 for live loads and L>180 for live-plus-dead loads. If a member deflects more than is listed in these guidelines, usually, it is not considered acceptable, and a member of increased stiffness (i.e., increased I) must be used, no matter how low the stress level in the member might be. Evidently, the

Beams size of such a member can be found by equating the appropriate deflection expression to the maximum allowable deflection and solving for the required stiffness (e.g., ∆ allowable = L>240 = 5wL4 >384EI, or Ireq’d = 512402wL4 >384EL for a uniformly loaded beam). Many designers feel that, the criteria noted previously and both L>240 and L>360 are conservative, in view of their origins. The criteria have roots in historical building traditions in which rules were used largely to prevent plaster affixed to the underside of a floor (or on the ceiling of the room beneath) from cracking. The criteria have since been widely used for many other applications. Floors designed to these limitations are usually perceived by occupants as comfortable and not excessively saggy. It is interesting to note that our perceptions of what constitutes an acceptable level of both visual sag and floor bounciness are probably derived from our cultural conditioning in accepting prior experiences as a measure of correctness. These experiences are in turn based largely on an antiquated plaster-cracking criterion. examPle A simply supported floor beam that is 20 ft (6.1 m) long carries a uniformly distributed live load of 200 lb>ft 12920 N>m2. The beam has a rectangular cross section with dimensions b = 8 in. 1203.2 mm2 and d = 16 in. 1406.4 mm2. The modulus of elasticity of the timber is E = 1.6 * 106 lb>in.2 111,032 N>mm2 2. Assume that the beam is initially cambered upward so that it is level under dead loads. What is the maximum live-load deflection at midspan, and is it excessive? solution: Moment of inertia: I = =

18 in.2116 in.2 3 bd3 = = 2730 in.4 12 12 1203.2 mm21406.4 mm2 3 12

= 1136.6 * 106 mm4

Deflection due to live loading: ∆ =

Copyright © 2013. Pearson Education Canada. All rights reserved.

= =

5wL4 384EI 51200 lb>ft>12 lb>in.2120 ft * 12 in.2 4 38411.6 * 106 lb>in.2 212730 in.4 2

= 0.165 in.

512.920 N>mm216100 mm2 4

384111,032 N>mm2 211136.6 * 106 mm4 2

= 4.2 mm

∆ = 0.165 in. 6 L>360 = 121202 >360 = 0.67 in.

The deflection is not excessive.

3.8

Principal stresses

One of the most interesting aspects of beam analysis is how bending and shearing stresses interact. Stresses that act in different directions cannot be added algebraically, but their resultant interaction can be found in much the same way that a vector resultant force can be found to represent the combined action of several different forces acting at a point. In a beam, the interaction between bending and shear stresses produces a set of resultant tensile and compressive stresses, typically called principal stresses, that act in different directions from either the bending or shear stresses individually.

Beams In a cantilever beam, the stresses acting on several elements are illustrated in Figure 25. A set of equivalent principal tension and compressive stresses could be found for each element. Note that, for an element at the neutral axis of the beam, where bending stresses are zero, only shear stresses exist. As diagrammed in Figure 25, these stresses can be resolved into equivalent principal tensile and compressive stresses acting at 45° angles to the neutral axis. At the extreme surfaces of the beam, an element carries only bending stresses because shear stresses are zero. Thus, the principal stresses in tension become aligned with the bending stresses in tension and have the directions indicated in the figure. For an intermediate element subject to both shear and bending stresses, the principal stresses have an inclination that depends on the relative magnitudes of the shear and bending stresses. By considering elements at other sections in the beam, stress trajectories can be drawn as illustrated. It is important to note that the lines are not lines of constant stress but are lines of principal stress direction. Stress intensity, therefore, can vary, depending on the relative magnitudes and distributions of shear and bending stresses in the beam.

FIGure 25 Principle stresses in a cantilever beam. Principle stresses result from the interaction of bending stresses and shearing stresses. The lines shown are often called stress trajectories and depict the direction of the principle stresses in the member. They are not lines of constant stress.

Copyright © 2013. Pearson Education Canada. All rights reserved.

,

Beams

FIGure 26 Lines of principle stresses: implications on general load-carrying mechanisms present in beams.

Figure 26 illustrates the stress trajectories present in a simply supported beam. Calculating the magnitudes of principal stresses is straightforward. In general, it can be shown that maximum and minimum principal tensile and compressive stresses occur on planes of zero shearing stress, and maximum shearing stresses occur on planes at 45° angles to the planes of principal tensile and compressive stresses.

Copyright © 2013. Pearson Education Canada. All rights reserved.

3.9

Finite-element analyses

The types of stress and deformation analyses presented thus far are the basic tools common for analyzing typical structures. Many structures have complex geometries or loading conditions, however, that demand more sophisticated treatments. One widely used approach is computer-based finite-element methods. A continuous structure is replaced by a conforming meshed network of interconnected, discrete pieces of varying shapes and sizes. Forces are applied at nodal points. Energy-based structural analysis techniques and various compatibility requirements are then used to predict forces and displacements. Outputs include various stress analyses (typically, three principal stresses; see Section 3.8), strain distributions, and displacement analyses. In a typical complex volumetric solid structure, magnitudes and distributions for three principal stresses are obtained. Problems involving thin-shelled structures can be beneficially limited and two principal stress components typically found. The basic results for a cantilevered structure (a complexly shaped structure with interior voids) analyzed by finite-element methods are shown in Figure 27. Many different automatic computer-based mesh generators are available. A variety of mesh types and elements are also available. Sizes may be varied as well, with smaller sizes used where more detailed information is sought. Restraints and external forces are specified next. The solution of the structural analysis algorithms demands the solution of multiple simultaneous equations (hence the need for a computer environment). Detailed results for all stresses and deformations are obtained in tabular form and can be displayed graphically. One set of results, the first principal stresses, is shown in Figures 25c and 27. The analysis that is performed indicates the power of finite-element techniques. A novice analyzing this structure might be tempted to treat it as a single cantilever beam with a unique cross section, and might seek to find stresses via f = My>I approaches. In that case, the presence of the large voids simply cannot be incorporated. Stresses and deformation patterns vary considerably once the openings are considered, and maximum stresses are much higher. Once principal stresses are found, a variety of more complex failure criteria can be used to determine an element’s safety. Rather than using simple allowable

Beams

FIGure 27 Structure in bending analyzed by finite-element method. 3RLQWORDG

D &DQWLOHYHUHGEHDP

E )LQLWHHOHPHQWPHVK

/RFDOVWUHVVFRQFHQWUDWLRQVUHVXOWIURP RSHQLQJVLQWKHEHDP

VWSULQFLSDOVWUHVVHV 7HQVLRQ 

&RPSUHVVLRQ

F 'HIRUPHGVKDSHZLWKVWSULQFLSDOVWUHVVHV

stress measures individually associated with just bending or shear, as was previously done for simple beams, stress interactions are considered. Typical failure criteria include the following: the maximum von Mies stress criterion, the Mohr-Coulomb stress criterion, the maximum normal stress criterion, and the maximum shearstress criterion. The choice of failure criteria depends on many factors, including the loading environment (e.g., rate of load application), material characteristics, and so forth.

Copyright © 2013. Pearson Education Canada. All rights reserved.

4 4.1

desIGn oF Beams General design Principles

approaches. Primary variables in designing beams include spans, member spacings, loading types and magnitudes, types of materials, cross-sectional sizing and shaping (including variations along member lengths), and assembly or fabrication techniques. The more a design situation is constrained, the easier it is to design a specific member. The easiest condition is when everything is specified but a member’s size, in which case the process can be almost deterministic. As more and more variables are brought into play, as is the case in a true design situation, the process becomes more difficult and less deterministic. Iterative approaches are common in which several beam designs for different sets of variables are developed and compared according to prespecified criteria (e.g., economy of material, depth, cost, appearance). This section focuses on design situations with well-defined constraints. Any beam design must meet specified strength and stiffness criteria for safety and serviceability. Design approaches to meeting these criteria depend on the materials selected. Timber, steel, and reinforced-concrete beams are discussed extensively in the following sections. Even within a specified material context, however, broadly differing attitudes are taken to member design: (1) A member’s size and shape may

Beams be determined on the basis of the most critical force state anywhere in the beam and this same size and shape used throughout the length of the member (even if force levels decrease). The member is thus worked to its maximum capacity at one section only. This strategy is used widely for ease and expediency of construction. (2) An attempt may be made to vary the size and shape of a member along its length in response to the nature and magnitude of the forces present at specific locations, with the intent to equally stress the beam along its length and with commensurate advantages of economy of material. Shaping a beam along its length, however, may prove difficult, depending on the construction approach selected. The following sections explore these and other design approaches conceptually, without getting involved in specific and quantitative design methods.

Copyright © 2013. Pearson Education Canada. All rights reserved.

strength and stiffness control. Beams must be sized and shaped so that they are sufficiently strong to carry applied loadings without undue material distress or deformations. When design is limited to finding a beam size and shape for a given span, loading, and material context, the problem is reduced to one of finding section properties for a member such that the actual stresses generated in the beam at any point are limited to predetermined safe levels, depending on the properties of the materials used. Typically, shear and moment diagrams are drawn first. The maximum bending moment 1Mmax 2 is determined. For simple steel or timber members with symmetrical cross sections, initial member size estimates are made, using the concept of a required section modulus 1S2, as described in Section 3.1. Thus, Sreq’d = Mmax >Fb. A member with a section modulus equal to or greater than this value is selected as a trial size. Remaining stresses and deflections are checked. Note that, in this procedure, we started with a member size based on bending. Depending on the situation, a trial member size may be determined on the basis of deflections or other types of stress. These procedures are illustrated for timber beams in Section 4.2 and for steel beams in Section 4.3. Reinforced-concrete members are treated in Section 4.4. cross-sectional shapes. The moment of inertia 1I2 and the section modulus 1S2 are of primary importance in beam design. A common design objective is to provide the required I or S for a beam carrying a given loading with a cross-sectional configuration that has the smallest possible area. The total volume of material required to support the load in space would then be reduced. Alternatively, the objective could be stated in terms of taking a given cross-sectional area and organizing the material so that the maximum I or S value is obtained, thus allowing the beam to support the maximum possible external moment (and hence loading). The basic principle for maximizing the moment of inertia obtainable from a given area is in the definition of the moment of inertia, which is I = y2dA. The contribution of a given element of area 1dA2 to the total moment of inertia of a section depends on the square of the distance of this elemental area from the neutral axis of the section. The sensitivity to the square of the distance is important: It would lead one to expect that, for a given amount of material, the best way to organize it in space is to remove it as far as practically possible from the neutral axis of the section (i.e., make the section deep, with most of the material at the extremities). Consequently, beams with high depth–width ratios are usually more efficient than beams with shallower proportions. A highly economic cross section also based on the preceding principle is illustrated in Figure 28(a), where a thin web member connects two widely separated thick flanges. The question arises of how thin the web can be. Recall that maximum shearing stresses always occur at the neutral axis of a beam. Thus, it is necessary that the web be a certain minimum thickness to carry these shear stresses safely. In addition, lines of principal stresses cross the beam in the middle. The principal compressive stresses could make the web buckle locally if it were too thin. Beams of the general proportions indicated in Figure 28(a) result after those factors

Beams

FIGure 28 Efficient beam cross-sectional shapes. Material is removed from the neutral axis to maximize the moment of inertia of the cross section and hence its resistance to bending.

x-x axis

Copyright © 2013. Pearson Education Canada. All rights reserved.

are considered. Sections of this type made of steel, called wide-flange beams, are commonly used in building construction, where bending stresses are typically a more important consideration than shearing stresses. When shear forces are high, beams with thicker webs should be used. An analogous shape in timber is shown in Figure 28(b). The discussion thus far has focused on ways of increasing the moment of inertia 1I2 of a section. Remember, however, that, in the design of a beam, the measure of primary importance is the I>ymax value of a section. In cases where the section is symmetric about the neutral axis, increasing the I value automatically increases the section modulus S. In nonsymmetric sections, such as T beams, the design must be based on the condition that the maximum bending stress at any point on the beam is limited to the allowable stresses. At a section, this point is defined by ymax and occurs on one face of the section. The other face, with a value of y less than ymax is therefore understressed. Hence, the use of such a section is not advantageous in beams made of homogeneous materials (e.g., steel). When a composite material such as reinforced concrete is used, however, T beams can have definite advantages. (See Section 4.4.) material Property Variations. In addition to changing how material is distributed at a cross section, the types of material used within a given cross section can be varied to achieve a better match between the stress state and the characteristics of the material. This principle is manifested in beams as diverse as those made of laminated wood (Section 4.2) and reinforced concrete (Section 4.4). See the discussion on composites in Section 3.1. shaping a Beam along Its length. The previous section considered ways to design beams of different materials at a single cross section. Now consider the shaping of a beam along its axis to improve the overall efficiency of the beam. Varying the shape of a member along its axis can provide a better fit between the characteristics of a beam and the shears and moments, which typically vary along the length of a beam. Shaping beams in responses to the shears and moments present in them has a long history in the field of structures. In the early nineteenth century, investigators were already determining appropriate responses for cantilever beams. Consider the cantilever illustrated in Figure 29(a), which carries a concentrated load at its end. Let us attempt to find an appropriate beam configuration based on the criterion that the bending stresses developed on the top and bottom surfaces of the beam should be constant along the length of the beam. This is done by setting up an expression for the dimensions of the beam as a function of the external moment present in the beam. An expression for the moment in the beam with respect to an arbitrary reference point is also needed.

Beams

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 29 Shaping a beam in response to its bending moment distribution.

For a rectangular section, the beam dimensions as a function of M are S = bd2 >6 = M>Fb, or bd2 = 6M>Fb. The external moment as a function of x is M = Px; hence, bd2 = 6Px>Fb = x6P>Fb. If the beam depth d is assumed constant, and the last expression is rewritten as b = k1 x, then the width b varies directly with x because the quantity k1 = 6P>d2 Fb is a constant. Consequently, a beam shaped as in Figure 29(b) would result. If the width b is assumed constant, the depth d depends on 1x [i.e., d = 1x1 26P>Fbb2 = k2 1x]. Consequently, a beam shaped as illustrated would result. If it is assumed that the beam section should always be square 1b = d2, then 3 d3 = x16P>Fb 2, or d varies as 1 x. For comparison purposes, repeat this exercise with a cantilever carrying a uniformly distributed load. For a rectangular cross section, the beam dimensions as a function of M are S = bd2 >6 = M>Fb, or bd2 = 6M>Fb. The external moment as a function of x is M = wx2 >2; hence, bd2 = x26w>2Fb, or b = k3 x and d = k4 x. If the depth d is assumed constant, then the width b varies with x2 because k3 = 6w>2Fb d2 is a constant. If b is assumed constant, then d varies directly with x: d = x 26w>2Fb b = k4 x. The shape in Figure 29(f) results. The shapes just derived are not intended to represent practical structural responses to the loadings indicated. Only bending stresses have been considered, to the exclusion of other considerations (e.g., shear stresses, deflections) that might influence the final shapes found for the loadings. They are idealized responses to the loadings shown, and their value stems from such idealization. They are a useful tool for visualizing whether a structural response is appropriate for a given situation. The shapes found are not necessarily generalizable to other beams carrying similar loads but supported differently. The moment equations differ for different types of beams, so it can be expected that shapes would, too. Students are encouraged to explore beams other than the cantilevers illustrated and determine appropriate shapes. Cross sections other than rectangular can also be used to determine idealized responses. Figure 30 illustrates a steel wide-flange beam designed in response to the moment present in a simply supported beam with a uniformly distributed loading, assuming that the beam remains a constant depth and that the flange width is the only variable. Varying the width of the flange causes the section modulus S to

Beams

FIGure 30 Varying the sectional properties of beams in response to the bending moment distribution.

and bending moment

Copyright © 2013. Pearson Education Canada. All rights reserved.

The section

change. This variation could be coupled with the moment distribution present in the beam to obtain a configuration in which the bending stress level in the flanges remains constant. In a wide-flange shape, the beam depth and flange width could be assumed constant and the flange thickness allowed to vary. Figure 29(e) illustrates such a beam where, for simplicity of construction, the variation in flange thickness is accomplished with horizontal layers of thin steel plates bonded (i.e., welded or bolted) together. Again, the properties of this beam at each section can be coupled with the amount of moment present to create a situation of constant bending stresses in the beam flanges. Stresses cannot be held purely constant because discrete changes occur in the flange thickness due to the layering used. Often, standard structural steel wide-flange shapes have additional steel plates (commonly called cover plates) added at regions of high moment. Other shapes based on radically different cross sections may also be optimized. (See Figure 30.) Again, dimensions are varied, so that Sx depends on Mx . Varying support locations and Boundary conditions. Manipulating support conditions can lead to major economies in material use. Such manipulations

Beams are intended to reduce the magnitudes of the design moments that are present or to alter their distribution. A classic way to reduce design moments is using cantilever overhangs on beams. Cantilevering one end of a simply supported beam with uniform loads causes a reduction in the positive moment, while a negative moment develops at the base of the cantilever over the support. The greater the cantilever, the higher the negative moment becomes and the lower the positive moment becomes. The cantilever can be extended until the negative moment even exceeds the positive moment. Of interest here is that a certain overhang must exist such that the numerical value of the positive moment is equal to that of the negative moment. Figure 31(b) illustrates a simple beam with variable-length cantilevers on both ends. If no cantilevers were present 1x = 02, the critical design moment would clearly be M = 0.125wL2 (the maximum positive moment at midspan). If the ends were cantilevered to a specific extent, the structure could be designed for a moment of M = 0.021wL2 at midspan and M = -0.021wL2 at either support. The moment can be reduced no further than this. Reducing the design moment this way leads to considerable economies when member sizes are determined. The exact extent of the overhang on either end that results in the positive and negative moments being equal can be found either by trial-and-error or by setting up one equation in terms of the variable x for the negative moment at a support and another for the positive moment at midspan and equating them. The optimum overhang is approximately one-fifth of the overall span. The same approach can be taken when the location of only one support, instead of both, can be varied. The optimum overhang is about one-third of the span. Considerations of the type discussed in the previous paragraph are one way to answer the question of how far a beam can be cantilevered. Rules like the preceding, however, should be tempered with other considerations. In many cases, for example,

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 31 Locating supports to minimize design moments in beams.

Beams the amount of deflection at the end of the cantilever, rather than criteria based on moment considerations, controls how far one can extend the member. The type of construction also is important. The measures are not applicable to prestressed or posttensioned reinforced concrete members, for example, without special considerations. By and large, the one-fifth and one-third measures are best applied to short- or moderate-span members made of homogeneous materials.

4.2

design of timber Beams

Copyright © 2013. Pearson Education Canada. All rights reserved.

The design of timber beams has long followed the principles of allowable strength design (ASD) using working loads and permitting members’ stresses to reach allowable levels set well below the failure stresses. More recently, load and resistance factor design (LRFD) methods have become broadly accepted as an alternative approach. Here, amplified or factored loads are used to find forces and moments, and the associated stresses can be close to average failure stresses. Typical load combinations and load factors in LRFD approaches are 1.4 D and 1.2 D + 1.6 L. The design process for both methods involves checking the actual stresses f against adjusted design stress values f ’. Adjusted stresses for both design approaches take into account a plethora of similar end-use related aspects such as moisture content or service temperature, to name a few. We return to a discussion of these factors later and focus on the core difference between ASD and LRFD first. LRFD increases the allowable stresses with a format conversion factor KF. For bending stresses, for example, the factor is 2.54, thus more than doubling the allowable stresses to a value close to the failure stresses. (See Figure 32.) Multiplying allowable stresses with KF obtains the so-called nominal design value. Next, a resistance factor Φ is introduced, slightly reducing stress values in recognition of uncertainties associated with specific structural actions. The relevant resistance factors for beams are 0.85 for bending stresses and 0.75 for shearing. The third factor unique to LRFD is the time effect factor l, with 0.6 for the load case 1.4 D and 0.8 for the load case 1.2 D + 1.6 L. Its equivalent in ASD approaches is the load duration factor CD. The first example illustrates the process for determining the required size of a simple timber beam within a highly defined span and loading context. A rectangular beam cross section constant throughout the length of the beam is assumed. The only variables are the width and depth of the member. The size of the member is estimated first, on the basis of controlling bending stresses, and is then checked for adequacy with respect to shear stresses, bearing stresses, and deflections. The example does not include adjustment factors beyond those crucial to understand the conceptual difference between LRFD and ASD design methods.

FIGure 32 Floor joist system.

Beams examPle Determine the required size of one of the rectangular floor beams illustrated in Figure 32. Assume the following: live load = 50 lb>ft2, dead load = 15 lb>ft2 (includes weight of floor deck, flooring, and estimate of beam weight), allowable stress in bending = Fb = 1200 lb>in.2, allowable shear stress = Fv = 150 lb>in.2, allowable bearing stress = Fbg = 400 lb>in.2, and allowable deflection = ∆ = L>360 for live loads and L>240 for live plus dead loads. Assume also that E = 1.6 * 106 lb>in.2, L = 16.0 ft, and a = 16 in. solution: ASD

LRFD

Loads: w = 115 lb>ft2 + 50 lb>ft2 2116 in. >12 in.>ft2 = 86.7 lb>ft

wu = 1.2 D + 16. L = 1.2115 lb>ft2 2 + 1.6150 lb>ft2 2 = 98 lb>ft2

wu = 198 lb>ft2 2116 in.>12 in.>ft2 = 130.7 lb>ft

Design for Bending Moments: M =

186.7 lb>ft2116 ft2 2 wL2 = 8 8

= 2773 ft@lb = 33,280 in.@lb Bending Stresses:

Mu =

1130.7 lb>ft2116 ft2 2 wuL2 = 8 8

= 4181 ft@lb = 50,176 in.@lb F′b = fbKF Φl = 11200 psi212.542 10.85210.62 = 2072.6 psi

Sreq′d =

33,280 psi M = 27.73 in.3 = fb 1200 psi

bd2 6 bd2 = 127.73 in.3 2 6 = 166.4 in.3

Sreq′d =

50,176 psi Mu = 24.2 in.3 = f ′b 2072.6 psi

Copyright © 2013. Pearson Education Canada. All rights reserved.

S for a rectangular beam: S =

bd2 = 124.2 in.3 26 = 145.3 in.3

Any beam with a bd2 = 166.4 in.3 for ASD, and 145.3 in.3 for LRFD will be adequately sized with respect to bending. The difference in results is partially due to the simplified approach that does not take several adjustment factors into account. Assume a beam width of 1.5 in.: Assume b = 1.5 in. 11.5 in.2 d2 = 166.4 in.3 S d = 10.5 in.

Assume b = 1.5 in. 11.5 in.2 d2 = 145.3 in.3 S d = 9.8 in.

The nearest stock timber size is a 2 * 12 (actual dimensions, nominal size 1.5 in. * 11.25 in.). This beam would be slightly oversized, but is reasonably close and will be used. Lateral buckling: The depth-to-width ratio of the trial beam size is 11.5:1.5, or between 7:1 and 8:1. Deep, narrow beams like this are sensitive to lateral buckling. Bridging should be used, or the proportions of the beam must be changed (Section 3.2). Assume that bridging is used.

Beams Shear stresses: The maximum shear force occurs at either end of the beam: ASD

LRFD

Shear Stresses: V =

Vu =

186.7 lb>ft2116 ft2 wL = 2 2 = 694 lb

V 3 fv = a b a b 2 bd

694 lb 3 b = a ba 2 11.5 in.2111.25 in.2 = 61.7 psi

61.7 psi″ fv, allowable = 150 psi

1130.7 lb>ft2116 ft2 wuL = = 1045.6 lb 2 2

3 Vu fv = a b a b 2 bd

1045.6 lb 3 b = 92.9 psi = a ba 2 11.5 in.2111.25 in.2

Adjusted shearing stress F′v

F′v = 1 fv, allowable 2KF Φ l = 1150 psi212.88210.75210.62 = 194.4 psi 92.9 psi … F′v = 194.4 psi

Both design methods show the beam is adequate in shearing. The LRFD method uses an adjusted shearing stress versus the allowable shearing stress used in ASD. Bearing stresses: Assume a bearing area of 1.5 in. * 2 in. fbg =

Pu P 694 lb 1045.6 lb = = 231 psi fbg = = = 348.5 psi A 11.5 in.2111.25 in.2 A 11.5 in.212 in.2

231 psi … 400 psi

Find adjusted bearing stress F′bg :

F′bg = fbg KF Φ l = 1400 psi212.08210.9210.62 = 449.3 psi

384.5 psi … 449.3 psi

The actual bearing stress is less than the allowable bearing stress (ADS) and less than the adusted bearing stress (LRFD). The beam is not overstressed at the supports.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Live-load deflections: Deflections are checked for working loads, and the calculation is identical for ASD and LRFD methods. Live load = 150 lb>ft2 2116 in.>12 in.>ft2 = 66.5 lb>ft ∆ = =

5wL4 384EI

5 3166.5 lb>ft2 > 112 in.>ft24116 ft * 12 in.>ft2 4

38411.6 * 106 lb>in.2 231.5 in. * 111.5 in.2 3 >124

= 0.32 in.

The beam deflection under live loads is 0.32 in. The allowable deflection for the beam under live loads only is given by ∆ = L>360 = 116 ft * 12 in.>ft2 >360 = 0.53 in. Because 0.32 in. 6 0.53 in., the beam does not deflect excessively under the design live loads.

Live- and dead-load deflections:

v = 86.5 lb>ft

6 ∆ = 0.42 in.

The actual beam deflection under combined dead and live loads is 0.42 in. The allowable deflection for this combined loading condition is given by ∆ = L>240 = 116 ft * 12 in.>ft2 >240 = 0.80 in. Because 0.42 in. 6 0.80 in., the beam does not deflect excessively under combined dead and live loads. The beam thus meets all the deflection criteria.

Beams The same general procedure is followed when SI units are used. The following example uses the ASD method: load = 86.7 lb>ft = 1265.3 N>m unit length

bending moment = 2773 ft@lb = 3759.7 N # m allowable stress in bending = Fb = 1200 lb>in.2 = 8.27 N>mm2 beam size = Sreq’d =

M 3759.7 * 103 N # mm = Fb 8.27 N>mm2

bd2 = Sreq’d = 454,617 mm3 6

Copyright © 2013. Pearson Education Canada. All rights reserved.

If b = 38.1 mm 11.5 in.2, then d = 267.6 mm 110.54 in.2. Shear stresses, bearing stresses, and deflections are checked similarly.

Note that constant-depth rectangular beams are not efficient. In the rectilinear beam illustrated in Figure 32, at only two spots on the entire beam (points A and B at section M 9 N) are the beam fibers stressed to the maximum extent. At all other points in the beam, the material is understressed. Consequently, the maximum potential of the material is not fully utilized. The reason for this lies in the nature of bending at a cross section, where deformations and bending stresses vary from zero at the neutral axis to a maximum at the extreme fibers. Thus, the stresses doing the most to generate an internal resisting moment to equilibrate the external moment are those near the outer fibers. Those near the neutral axis are of little consequence in this respect. Accordingly, from a design viewpoint, the beam material could be used more efficiently by moving it away from the neutral axis and toward the extremities of the beam. A second reason the beam material in the example illustrated in Figure 32 is underutilized is that the size of the section was determined in response to the maximum external bending moment in the beams (at section M 9 N). At other points in the beam, the external bending moment is less. (See the moment diagram.) Thus, using a section designed in response to the maximum moment results in inefficiencies at other points. From a theoretical viewpoint, it is possible to vary the size of the beam in response to the internal forces present. A beam with variable dimensions along its length would result. (See Section 4.1.) The practical design of timber beams is influenced by many other factors that respond to particular characteristics of the material used or the type of loading. The modulus of elasticity 1E2 and permissible stress values depend not only on the design method employed but also on the type, grade, and use of the wood employed. Wet-use conditions, for example, necessitate reductions in the value of E. Other use factors also are important. For example, timber has an intrinsic ability to handle high stresses for short periods of time. Higher allowable stresses are thus permitted for short- as opposed to long-term loadings. Permanent loads require that allowable stresses be reduced by a factor of 0.9, while a loading factor of 1.33 may be used for wind loadings. Normal loadings have a factor of 1.0. Many other specific adjustment factors also apply to the design of timber beams. In U.S. practice, most factors take the form CA, where A describes the phenomenon addressed. Most of these factors increase or reduce allowable stresses (ASD), nominal design stresses (LRFD), or other material properties. The load duration factors Cd was mentioned earlier; in ASD it replaces the time effect factor l (LRFD only). Both factors recognize that timber can be more highly stressed for loads of short duration. For wind loads, for example, allowable stresses in ASD can be multiplied by 1.6. Other environmental conditions are taken into account

Beams through the temperature factor Ct and the wet service factor Cm. Both reduce permissible stresses either for elevated use temperatures (generally larger than 100 F) or high moisture content (above 19 percent for sawn lumber and 16 percent for gluelam). The stability factor CL decreases stresses for beams that are not continually braced on the compression side. The empirically determined correlation between the beam size and its stress distribution is taken into account through the size factor CF (sawn lumber) and the volume factor CV (for gluelam). Yet other factors consider the repetitive use of joists or the effects of chemical treatment. Various texts examine timber design in greater detail and cover these factors.4 The next example briefly illustrates the use of several of these factors. examPle A cantilevering, sawn timber beam 6 in. by 16 in. is intended to carry a uniform dead load of 20 lb>in.2 and a live load from snow as a point load at its end of 2000 lb. Assume allowable bending stresses of fall = 1500 lb>in.2. Dry-use conditions prevail. Is the member adequate in bending? solution: First, the adjusted design stress values are determined. Some factors only apply to ASD or to LRFD methods. Wet use factor CM: Temperature factor Ct: Size factor CF : Beam stability factor CL: Repetition factor CR: Load duration factor CD: Format conversion factor KF : Resistance Factor Φ: Time Effect Factor l:

Dry conditions 1 Service temperatures are below 100 F. 1 For sawn lumber deeper than 12 in., CF is found using CF = (12>d)1/9 = (12>16)1/9 = 0.968 No reduction is necessary because the beam is continually braced on the compression side. 1 ASD only. The beam is not part of a repetitive joist system. 1 ASD only, for snow loads with a duration of no more than 2 months 1.15 LRFD only 2.54 LRFD only 0.85 LRFD only, dependent on load combination 0.8

Other factors that incorporate flat use or chemical treatment do not apply. ASD

LRFD

Copyright © 2013. Pearson Education Canada. All rights reserved.

Adjusted Design Stresses: F b= = 1500 psi [KFΦl] CM Ct CF CL

Fb = 1500 psi CD CM Ct CF Cr CL = (1500 psi)(1.15)(0.968)

= (1500 psi) [(2.54) (0.85) (0.8)] (0.968)

= 1669.8 psi

= 2507.9 psi

Actual stresses: M = a

(20 lb>ft)(10 ft)2 2

b + ((2000 lb)(10 ft))

= 21,000 ft@lb = 252,000 in.@lb

(6)(16)2 bd2 = = 256 in.3 6 6 M fb = S 252,000 in.@lb = = 984 psi 256 in.3

M= a

(20 lb>ft)(1.2)(10 ft)2 2

b + ((2000 lb)(1.6)(10 ft))

= 33,200 ft@lb = 398,400 in.@lb

S =

4

Mu S 398,400 in.@lb = = 1556.3 psi 256 in.3

fb =

See, for example, Breyer et. al., Design of Wood Structures, New York: McGraw-Hill., 2007., or Ambrose, James, Tripeny, Patrick. Simplified Design of Wood Structures New York: John Wiley and Sons. 2009.

Beams For both methods, actual stresses are well below adjusted design stresses: 984 psi 6 1669.8 psi (ASD) and 1,556.3 psi 6 2,507.9 psi. Thus, the member is adequate in bending.

laminated timber. A simple rectangular beam shape may also be made of laminated wood rather than solid-sawn shapes. A laminated beam is made of thin layers of wood bonded together to form a composite whole. Large sections and curved beams are possible. Permissable stress values in bending and shear are normally higher than those for solid-sawn timber sections. For example, an allowable stress in bending for a laminated member might be F = 2500 lb>in., rather than 1200–1600 lb>in. that is common for normal sections. Efficient long-span members are consequently possible. Most laminated members use high-quality wood throughout. Still, laminated members allow for the possibility of varying material properties at a cross section so that a better match is obtained between actual stress levels and material characteristics. Timber comes in different grades having different allowable stresses. Grades with high allowable stresses are typically made from high-quality, and hence costly, wood. Lower stress grades are cheaper. This is the basis for achieving an economic section. The typical bending-stress distribution in Figure 33(a) shows that it is not necessary to use wood layers made of the same quality wood throughout the beam. Layers using wood having high allowable stresses could be used at the extremities of the beam (away from the neutral axis). Successive layers toward the neutral axis could be made of less expensive wood having lower allowable stresses. This technique of putting the stronger material on the outer faces of a beam and weaker material toward the middle can result in economic sections. The same principle can apply to making more effective use of timber in T beams.

4.3

steel Beams

This section introduces the two alternative methods for the design of steel beams, allowable strength design (ASD) and load and resistance factor design (LRFD). It is important to recall that these methods cannot be mixed because ASD is based on working loads and LRFD is based on factored loads. The resulting member sizes are similar, but LRFD methods may yield slightly more efficient steel beams in certain cases.

Copyright © 2013. Pearson Education Canada. All rights reserved.

allowable strength design. Steel members can be designed on the basis of service loads and allowable stresses using principles of elasticity. Members used may FIGure 33 Varying the materials in beams to match the bending stresses present.

Grades of timber vary to match actual bending stresses.

Beams be either rolled sections (wide flanges, channels) or sections built up of individual elements (plates, angles). The design of rolled sections is straightforward. Only a set number of section types and dimensions are rolled, so their geometric properties are easily tabulated and used to selecting them for use in a structural context. Normally, the bending present is calculated first, after which a required section modulus value is determined 1Sreq’d = M>Fb 2, and the least-weight beam with a section modulus value equal to or greater than the required one is found. This member is then checked for shear, deflections, and other phenomena. The practice of determining required sizes of steel beams is facilitated by the a priori definition of cross-sectional properties for commonly available steel members. A small sampling of this tabulation is shown in Figure 34. In U.S. practice, the first number (W) in the shape designation is the nominal depth of the wideflange beam sections, and the second is its weight per linear foot. Other common listings are for WT sections (T shapes made from wide-flange beams), C or channel sections, L or angle sections, and several other common cross sections, including pipes, double angles, square tubes, and so forth. These listings are available in several sources and are included in libraries of computer-based structural analysis programs. Note that even within a single shape—a wide flange—numerous specific shapes are available that have very different section modulus (S) or moment of inertia (I) values. Values are normally given for both x-x and y-y axes. Some sections have high Sx and Ix values, but relatively low Sy and Iy values, and are thus primarily useful for bending about the x-x axis. Other sections have S and I values about each axis that are relatively close to one another and could thus be useful for beams where bending occurs about both x-x and y-y axes. FIGure 34 Properties of structural steel shape. Example: W27 * 194 Y

1.34 in.

28.11 in. X

X

examPle A simply supported steel wide-flange beam made from A992 steel spans 25 ft and supports uniformly distributed loads. The dead load of 200 lb>ft, and thenlive load equals 400 lb/ft. We are assuming an allowable stress in bending of fB = 33,000 lb>in.2 Determine the required size of the wide flange, based on bending stress considerations. Next, check shear stresses and deflections. Assume an allowable stress in shear of fv = 20,000 lb>in.2 Because the beam is used for a floor, live-load deflections are limited by L>360. Assume also that the beam is cambered upward so that it is level under dead-load conditions; hence, deflections need to be checked for live loads only.

Copyright © 2013. Pearson Education Canada. All rights reserved.

solution: Maximum bending moment: 0.75 in.

M = wL2 >8 = 1600 lb>ft2125 ft2 2 >8 = 46,875 ft@lb

Required section modulus:

14 in. Y Cross-sectional area: A = 57.0 in.2 Stiffness Properties: Axis X-X Ix = 7820 in.4 Sx = 556 in.3 rx = 11.7 in. Stiffness Properties: Axis Y-Y IY = 618 in.4 SY = 88.1 in.3 rY = 3.29 in.

Sreq’d = M>fb = (46,875 lb>ft)(12 in.>ft)>33,000 lb>in.2 = 17.05 in.3 Beam size based on bending stresses: Note that either of the following will work: W 8 * 21 1S = 18.2 in.3) or W 10 * 19 (S = 18.8 in.3) W 8 * 31 1S = 27.4 in.3 2 W 12 * 27 1S = 34.1 in.3 2. The beam having the least weight is most economical. The W 10 * 19 would therefore be selected, even if its S value is slightly higher than needed 1Sactual = 18.8 in.3 7 Sreq=d = 17.05 in.3 2. Check shear stresses:

Shear force: Shear stress: Actual: The value is acceptable.

V = wL>2 = 1600 lb>ft2125 ft2 >2 = 7500 lb

fv = V>td = 7500 lb> 10.25 in.2110.2 in.2 = 2941 in.2

f v = 2941 in.2 6 allowable fv = 20,000 lb>in.2

Beams Check deflections: Assume that of the 600@lb>ft load, 400 lb>ft is due to the live loads present. ∆ actual = 5wL4 >384EI = 51400>122125 * 122 4 >384129.3 * 106 2196.32 = 1.25 in. ∆ actual = 1.25 in. 7 L>360 = 125 * 122 >360 = 0.83 in.

The value is not acceptable, so we choose the heavier W 10 * 30. The deflection of this beam is 0.71, an acceptable value.

Copyright © 2013. Pearson Education Canada. All rights reserved.

For exceptionally large loads or spans, girder sections that are built up with angles and plates are often used. Welding has replaced older riveted technology. A trial size is selected, its moment of inertia is calculated via the parallel-axis theorem, and the member is then checked for bending, shear, deflections, and so on. Practical steel design encompasses many considerations, including the need to check for compact sections to prevent buckling of flanges and many other phenomena. Most steel cross-sectional shapes are open profiles. In a wide-flange beam, for example, the edges of the top and bottom flanges stick out. High compressive stresses can cause these protruding flange elements to buckle locally, which would in turn cause the whole beam to fail. The designer must prevent the flanges from buckling. This is often done by limiting stress levels to comparatively low values when the beam’s proportions are slender. The simple approach noted earlier is therefore suitable for rough estimation purposes only. Plastic Behavior of steel Beams. The analyses presented thus far have been based on the assumption that the beam material is linearly elastic (i.e., that stresses are proportional to deformations). Steel is linearly elastic only up to a certain point, when the material begins to deform massively under a relatively constant stress level. Only after significant deformation has occurred does the material actually rupture. These are plastic deformations. A simplified stress-strain curve depicting this behavior is shown in Figure 35. An appreciation of plastic deformation is important in understanding how steel beams actually fail. Failure modes are important to understand in the context of load and resistance factor design methods covered in the following section. Consider the rectangular beam shown in Figure 36. When external loads on the beam are low, the material in the beam is in the elastic range and bending stresses are linearly distributed across the cross section. As loads increase, the bending stresses increase, until the material at the outer fibers reaches a point, Fy, at which it begins to yield and enter the plastic range. As indicated in Figure 36(d), this corresponds to a resisting moment in the beam of My = Fy 1bh2 >62. (See Section 3.1.) Continuing to increase loads causes increased deformations in the beam fibers.

FIGure 35 Idealized stress-strain curve for ductile material. At a certain stress level Fy, the material begins to undergo increased deformations without any additional increase in stress level. The material does not fail until relatively large deformations occur.

Beams (A linear variation in strain still occurs because this is a function of the gross geometry of deformations in bending in the entire beam.) There is, however, no corresponding increase in the stress level in the material in the region where beam fibers are deformed into the plastic region. At this stage, some fibers near the neutral axis are still below Fy. In Figure 36(e), the beam is still capable of carrying a load. (The outer fibers have yielded but not ruptured.) Increasing the external load causes increased deformations,

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 36 Plastic behavior in beams. As the material in a beam begins to yield under load, the stress distribution in the beam begins to change. The beam can continue to carry moment until all the fibers at a cross section have yielded.

Beams until all the fibers in the cross section begin to yield. The load that finally causes the fibers nearest the neutral axis to yield is the maximum that the beam can carry. This load corresponds to a maximum resisting moment of Mp = Fy 1bh2 >42. Up to this point, the beam was always able to provide an increased moment resistance capacity to balance the applied moment associated with the external load. After that point, however, the beam has no further capacity to resist the external moment, and continuing to apply loads simply causes additional deformations until the beam ruptures and collapses. A plastic hinge is said to develop when all fibers are fully yielded at a cross section. For a rectangular beam, the ratio between the moment associated with the formation of a plastic hinge and that associated with the initial yielding is Mp >My = Fy 1bh2 >42 >Fy 1bh2 >62 = 1.5. The term bh2 >4 is often called the plastic section modulus, Z. In a simple beam that is continuously braced against buckling of the flanges, the 1.5 ratio implies that the load Pf required to cause the beam to fail is 1.5 times the load Py that would cause initial yielding in the member. Other shapes have different ratios of this type. The ratio is often called the shape factor for a beam because it is identical to the ratio of the plastic section modulus 1Z2 to the elastic section modulus 1S2 (i.e., shape factor = Z>S). For a wide-flange beam, shape factors vary but average around 1.14. Other factors are 1.7 for a round bar and 2.0 for a diamond. Using the concept of the shape factor, the plastic moment capacity of a section can be found at MP = FYZ. This equation is extensively used when designing beams for bending using the LRFD methods that are introduced in the following section. In statically indeterminate structures, formation of a single plastic hinge need not lead directly to beam collapse. Several hinges must form until a collapse mechanism is created. Consider the fixed-ended beam shown in Figure 37 and

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 37 Formation of plastic hinges in a fixed-ended beam. Plastic hinges develop initially where moments are the highest. The beam does not fail, however, until a sufficient number of hinges have formed to cause a collapse mechanism to develop in the beam. Beams of this type have a large measure of reserve strength.

s

Beams

Copyright © 2013. Pearson Education Canada. All rights reserved.

the moments that are developed in the structure. As the load w is increased, the beam yields first at its ends, where moments are maximal. As the load intensity is increased, plastic hinges begin forming at these points. The moment at the center of the structure is also increased. Note that the formation of plastic hinges at the ends of the beam does not cause the beam to collapse because the structure still has moment-carrying capacity at midspan. Further beam loading would eventually cause a plastic hinge to develop at midspan. A collapse mechanism would then exist, and the beam could carry no additional load. (Total failure would be imminent.) The failure load, ww, of the beam is greater than would be implied by considering only the shape factor of the beam. It is shown elsewhere that, for a restrained rectangular beam, the load ww required to cause failure is double that of the load wy required to initiate yielding. load and resistance Factor design methods. The preceding approaches have been developed into a formal design procedure called the load and resistance factor design (LRFD)method. Based on limit state concepts, load factors are specified to amplify service loads, and ultimate yield stresses are used to design members. Typical load factors for limit state design loads are 1.2D + 1.6L or 1.4D. The amplified loads are used to compute the factored moment MU . A resistance factor, f, is a strength reduction factor to modify the nominal resistance RN of a member and obtain its usable capacity RU . The resistance factor is similar to the capacity-reduction factor in ultimate strength design in concrete. The required strength RU thus equals RU = ΦRN . The values of f are 0.9 for a tension member, for flexure, and for a common compression member, and 1 for shear. In designing a beam for bending, it is often practical to determine a required plastic section modulus Zreq=d using Zreq=d = MU >0.9 fy. The equation, as was shown before, applies only to the limit state with yield stresses present throughout the cross section. It is also assumed that the compression side of the beam is continually braced laterally, and there is no risk of local buckling. Z values for standard shapes are provided in the literature. The moment MU is the factored moment calculated using factored loads. Methods for partially or wholly unbraced beams are complex. Checking a beam for shearing involves calculating the nominal shearing strength VN, which, only in the case of shearing, is equal to the required shearing strength VU. For most W, S, and T shapes, the equation VN = 0.6 FYAW provides a simple way to determine maximum shearing stresses in the web. The area of the web AW is determined using AW = tW h, with tW the web thickness and h the overall section depth. In addition to flexure and shearing, deflections would have to be determined. Here, working loads, thus without amplification factors, are used, and the process is identical to that in allowable strength design.

examPle The beam previously designed using ASD methods is now sized using an LRFD approach. The dead load of 200 lb>ft and the live load of 400 lb>ft need to be factored for the calculation of moments and shears. Loads: wD = 1.4 D = 1.4 (200 lb>ft2) = 280 lb>ft

wU = 1.2 D + 1.6 L = 1.2 (200 lb>ft2) + 1.6 (400 lb>ft2) = 880 lb>ft

The combined dead and live loads are larger than dead loads only, and the combined load will be used for the design process. Maximum factored bending moment: MU = (wU L2)>8 = (880 lb>ft)(25 ft)2 > 8 = 68,750 ft@lb

Beams Required plastic section modulus: Zreq’d = (MU)>(0.9 fY) = (68,750 lb>ft)(12 in.>ft) / (0.9) (50,000 lb>in.2) = 18.34 in.3 Beam size based on bending stresses: Note that either of the following will work: W 8 * 21 (Z = 20.4 in.3) or W 10 * 19 ( Z = 21.6 in.3) The more economical W 10 * 19 is selected. Check shear stresses: Shear force: VU = (880 lb>ft)(25 ft)>2 = 11,000 lb Nominal and required shear strength: VN = 0.6 fYAW = 0.6 (50,000 lb>in.2)(10.2 in.)(0.25 in.) = 76,500 lb The nominal shear strength is larger than the actual factored shear force present; the beam is adequate in shear. Procedures for checking deflections are identical to those shown in the ASD section, using working loads. The larger section must be chosen based on servicability criteria.

4.4

reinforced-concrete Beams: General Principles

Basic Behavior. When a beam is made of plain concrete, both tensile and compressive stresses normally associated with bending develop. Concrete is normally capable of carrying appreciable compressive stresses, but cracks develop under very low tensile stress levels. In a plain concrete beam, these cracks tend to propagate quickly (see Figure 38) and the beam would collapse virtually immediately. To make a concrete beam viable, it is therefore necessary to place a reinforcing material

Copyright © 2013. Pearson Education Canada. All rights reserved.

FIGure 38 Reinforced-concrete beams.

Beams (normally, steel) that can carry appreciable tension forces in the tension zone to intercept any crack that is beginning to propagate through the beam and then arrest its development. The tension stresses associated with the crack’s propagation are then transferred into the reinforcing steel, which in turn develops high tension forces. Sufficient steel must be used, or it, too, will pull apart. Cracks can result from bending stresses, shear stresses, or an interaction of the two and occur throughout the beam, necessitating the use of extensive patterns of steel. Longitudinal reinforcing steel is usually placed to counteract cracks caused by bending, whereas either stirrups (U-shaped bars) or diagonally bent bars are used to provide for cracks developed by shear stresses or principal stresses. Amounts of steel required usually vary in direct proportion to the magnitudes of the bending and shear forces. Maximal cross-sectional areas of longitudinal steel are necessary where bending moments are highest, and maximal areas of stirrup steel are necessary where shear forces are the highest. Steel must be designed precisely. Too little steel is insufficient to carry the internal forces developed, and the beam will begin yielding too soon and not carry required loads. If too much steel is used, there is a danger that the concrete will fail in compression long before the steel starts yielding in a ductile fashion. A compression failure is dangerous because it occurs without much prior visual warning. (The concrete seems to suddenly burst.) Ideally, sufficient steel is present to carry desired loads but not so much that it does not yield prior to the concrete’s failing. Welldesigned beams exhibit significant ductility and associated visual deflections prior to actual failure.

Copyright © 2013. Pearson Education Canada. All rights reserved.

4.5

FIGure 39 Reinforced-concrete T beam. The shape of the section causes the stresses in the top flange to be lower than stresses in the web of the member. Concrete is used at the top, and steel is used where the section is assumed to be cracked.

reinforced-concrete Beams: design and analysis Principles

Basic design approach. The addition of steel in the tension region to create reinforced-concrete beams ensures strength and ductility. Reinforced-concrete members inherit the properties of concrete and steel. Relationships, such as fy = My>I used previously to analyze and design beams, are based on the assumption that the material is not cracked and is linearly elastic. Neither of these assumptions is valid in the case of reinforced-concrete members. Rather, analysis and design procedures are based on a direct consideration of the idea that an internal resisting moment is generated in the member to balance the external moments that are present. As loads induce bending in a beam, a deformation pattern of the type illustrated in Figure 38 develops. The region below the neutral axis is in tension. The concrete is assumed to be cracked in this region and the forces present are completely taken up by the steel. The cracks extend upward until they terminate at the edge of the compressive region. An internal resisting moment generated in the beam balances the external moment at the same section by the couple formed by the tension force in the reinforcing steel and the resultant compressive force (composed of compressive stresses in the concrete acting over the uncracked part of the cross section) in the upper region of the beam (Figure 39). Hence, in a plain reinforced-concrete beam, only a portion of the concrete participates in carrying the load. The steel and concrete are reasonably assumed to bond to one another and to have the same strains at adjacent locations. For a given structural size, the area of steel required is proportional to the magnitude of the bending moment that is present 1A ∝ M2. An increase in the level of the applied bending moment demands an increase in the required area of steel. The designer must ensure, however, that the beam remains ductile, a condition that can fail to result when too much reinforcement is used. Beams must also be designed to resist shear stresses, which cause a different kind of cracking. Stirrups are often used to carry high shear stresses.

Beams

Copyright © 2013. Pearson Education Canada. All rights reserved.

ultimate strength design. In ultimate strength design (USD), the working loads are amplified to convert them into design failure loads. The underlying principle is similar to load factor and resistance design methods used for timber and steel. It is assumed that the reinforced-concrete structure starts to fail when these ultimate loads are reached, but not before, and that the member performs adequately until then. The material is controlled for ductility and a combined concrete-steel failure. For gravity loads, the amplification load factors in the United States are typically of the form U = 1.21dead load2 + 1.61live load2. In the presence of snow loads or lateral loads, these loads must be modified and the load factors become U = 1.2D + 1.0L + 1.0E, U = 0.9D + 1.0E, or U = 1.2D + 1.6L + 0.8W, where D and L are the working dead and live loads, and E and W represent earthquake and wind forces, respectively. The various coefficients and combinations reflect varying levels of uncertainty in load estimation or likelihoods of combinations. In ultimate strength design, safety factors are put on loads rather than on materials. Thus, a typical analysis uses service loads multiplied by some type of factor of safety and then uses actual failure stresses. ultimate strength design: design for Bending moments. In the United States, the ultimate strength design (USD) moment-carrying capacities of reinforced-concrete beams are given in standards developed by the American Concrete Institute (ACI 318), and which have a strong empirical basis. In these expressions, the width of a typical beam with only tension reinforcement is b, and the depth from the extreme compression fiber to the centroid of the reinforcing steel is d. (See Figure 40: The physical depth of the beam is not used because the cover of the steel below does not contribute to its bending strength.) M is the ultimate strength bending moment obtained from a moment diagram (with load factors applied). Moment-carrying values are given by expressions such as M = T1d - a2 2 or C1d - a2 2, where T is the total tension force developed in the reinforcing steel, C is the total resultant force equivalent to the compression stress field developed in the concrete, and d - a>2 is an internal moment arm, which is the distance between C and T (or the distance between the steel and the centroid of the compression stress field). A moment-resisting couple is thus formed that balances the applied ultimate bending moment (a behavior like a truss, except that the moment arm is not the same as the depth of the structure). In the expression d - a>2, a is the depth of the compression field stress block. Steps for determining the value of a are discussed shortly. The total tension force in the steel is given by T = A * Fy, where A is the cross-sectional area of the steel and Fy is the material’s yield stress. (Recall that ultimate loads and failure stresses are considered in this kind of analysis.) The moment capacity thus becomes M = T1d - a>22 = AFy 1d - a>22. We see that the required area of steel depends directly on the magnitude of the

FIGure 40 Ultimate strength design: strains and stresses.

Beams

Copyright © 2013. Pearson Education UK. All rights reserved.

ultimate bending moment 1M2, or Areq’d = M>Fy 1d - a>22. Note that the actual required area A varies with the bending moment: If M varies, so does the amount A of steel used. In U.S. practice, the moment capacity is called the nominal moment capacity MN of a beam. For design purposes, it is reduced by a factor Φ of typically 0.9, obtaining what is called the ultimate moment capacity MU = Φ MN. This reduction of the moment capacity takes variations in material strengths and other factors into account. It is important to recall that the ultimate moment capacity represents the maximum amount of moment a beam can carry without incorporating any significant safety factors. Any additional external moment would lead to a failure of the beam. The ultimate moment capacity therefore always must be used with the factored loads as described earlier. ductility. In USD, the failure mode of a beam is governed either by an initial failure of concrete or by an initial failure of the reinforcing steel. Depending on the amount of steel present, the failure can be sudden and brittle, or the member can fail in a ductile mode with the associated positive values of energy absorption and prior visual warning. Too little or too much steel can be problematic. Because ductility is provided by the reinforcing steel, a failure mode should begin with a steel yielding that can be sustained by the concrete until the steel reaches its ultimate strength and fails. Such a controlled failure mode suggests that the amount of reinforcing steel should be within a certain range. If the amount of steel is small, it will yield first, but it may reach its ultimate strength rapidly and with little energy absorption and little warning. This condition can be prevented by positioning more steel than a certain minimum amount. At the other extreme, if more than a certain maximum amount of steel is present, the steel will have an excess capacity for carrying the tension forces involved and will not yield before the concrete reaches its failure levels. In such a case, the concrete will crush suddenly in a brittle manner before the steel yields, with little energy absorption and little warning prior to collapse. This state of overreinforcing is a dangerous condition and requires special attention because it is associated with the placement of excessive amounts of steel—a condition that might normally be considered a conservative and safe move. In reality, the situation is quite the contrary. The condition is prevented by limiting the amount of steel used so that it yields before the concrete crushes. For any beam, there is a specific amount of reinforcing steel such that both steel and concrete will start failing at the same time. This is the balanced-steel condition, and the beam is said to be a balanced beam. An amount of steel equal to the balanced steel is not desirable because simultaneous failure of concrete and steel is a brittle, sudden failure. The maximum amount of steel allowed is usually 70–75 percent of the amount used in a balanced beam. Thus, if the amount of tension steel is between the minimum allowed and the maximum allowed, an appropriate ductility level in the beam is ensured.

4.6

reinforced-concrete Beams: General design Procedures

designing Beams for Bending. For designers, it is useful to understand the basic assumptions and mechanical models that underlie reinforced-concrete design. Initially estimating the size of a beam can be difficult, but the overall depth h of the beam can often be derived from the slenderness criteria that satisfy deflection (see Table 1). For a simply supported beam, for example, the overall depth is L>20 of the span L. From h, the effective depth d must be calculated by subtracting the distance of the steel centroid to the outermost part of the beam. For approximate calculations, it is often sufficient to subtract 3 in. from h to obtain d. In more accurate design procedures, the designer must estimate the amount of steel present in the beam. Conservative

Beams

taBle 1 Minimum overall depth h of beams and one-way slabs. memBer

sImPly suPPorted

Solid one-way slabs

L

Beams

L

>20 >16

one end contInuous L

L

>24

>18.5

Both ends contInuous

cantIleVer

L

L

L

L

>28 >21

>10 >8

guesses are better in the early design phases. If, for example, a single layer reinforcement of 1-in. diameter bars is assumed, 0.5 in. (the radius of the bars) plus the concrete cover and stirrup diameter must be subtracted from the depth h to derive d. Hence, for an internal beam, 0.5 in. + 1.5 in. + 0.5 in. = 2.5 in. in total subtraction. Once d has been determined, the width of the beam is estimated. Based on the earlier estimate of the steel area As, the depth of the stress block a must be derived with the equation a = (AsFy)>0.85 f c= b. Once a has been determined, the ultimate moment capacity MU = fMN = 0.9 AsFy 1d - a>22 can be found. U.S. codes require additional procedures, including procedures that ensure sufficient ductility and ultimately limit the amount of steel in a beam cross section. In cases where the maximum allowed amount of steel is insufficient to carry the factored loads, several measures can be taken. These include choosing higher material grades, having a deeper or wider section, adding compression steel (see next section), or changing the span or spacing of the beam to reduce the loads and moments.

Copyright © 2013. Pearson Education UK. All rights reserved.

compression-reinforcing steel. The bending capacity of a beam is governed by the size of the beam’s cross section, the strength of the materials of which the beam is composed, and the maximum allowable amount of tension steel in the beam. The addition of reinforcing steel in the compression zone alone has little effect on the overall bending capacity of the beam. However, the placement of proportional quantities of steel in both the tension and compression zones, in addition to the maximum allowable tension steel that is present, increases the bending capacity of the beam without violating the beam’s ductility characteristics. Notice that the compression steel may not reach its yielding stress because its strain depends on the surrounding concrete, while the tension steel always yields under ultimate loads. The forces in the compression steel and in the portion of the concrete subject to compression are always equal to those in the tension steel. The additional compression and tension steel behave the same way as a structure with a trusslike action between paired amounts of steel in the compression and tension zones, plus a normal reinforced concrete beam utilizing the remaining steel in the tension zone. (See Figure 41.) t and I section Beams. A T configuration beam under gravity loads provides a larger area of concrete in the compression zone, and at the same time removes concrete that is expected to crack from the tension zone. As long as the concrete provides a sufficient cover for the tension-reinforcing steel, a T section beam is lighter than, and can have the same strength as, a rectangular beam whose width would be equal to the width of the flange of the T section. However, in areas of negative moment, such as at the supports of a continuous beam, the compression zone is at the lower part of the beam, and the strength of the beam would be reduced significantly. I-beams may be more suitable in such cases. design of slabs. The design of slabs is similar to that of beams, except that it is typically done on a unit width basis. In addition, the minimum reinforcement for a slab is governed by shrinkage and temperature requirements, and the minimum steel should extend into both directions of a rectangular slab. The required concrete cover for the reinforcing steel in slabs is smaller than the required cover for beams with the same exposure.

FIGure 41 Compression steel in U.S. practice. The sum of the force vectors representing the stress block (Cc) and the compression steel (Cs) equilibrates the tension forces Ts in the steel at the bottom of the beam. Compressive stress block a

Cc Cs

Ts

Beams shearing stresses and diagonal cracking. Unreinforced concrete is a brittle material, and its failure criterion is based on the magnitude of the principal stresses. Thus, the shearing strength of concrete depends on the corresponding principal tensile strength because the compressive strength of concrete is significantly higher than the tensile strength. The principal stresses in tension result from the interaction of shearing and bending stresses. The lines of principal tensile stresses in a uniformly loaded beam are illustrated in Figure 26. For such a beam, the magnitude of shearing stresses is higher near the supports, with the result that cracks are often generated, as shown in Figure 42(a). These cracks are called diagonal tension cracks and are quite dangerous because they can lead to rapid failure of the whole beam. To prevent this, steel reinforcement is used to intersect such cracks. In the past, the steel reinforcement for diagonal tension was provided by bending the tension reinforcement bars toward the supports, as shown in Figure 42(a). The use of stirrups is now preferred because of uniformity and ease of construction and in the case of stresses resulting from horizontal loads, such as earthquakes, that might lead to a reversal of shear stresses. design for shear stresses. Shear stresses are usually higher near the supports, but they are checked at a distance d, the effective depth, away from the supports. Concrete has a shear strength that may be sufficient to take the shearing stresses without requiring steel reinforcement. The shear strength of concrete for members without axial or torsional loads is given in U.S. practice by the ACI 318

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 42

Shear and diagonal tension in reinforced-concrete beams.

Positive moment

Positive moment

Negative moment

Beams

code expressions, such as Vc = 21f c= bw d, where bw is the width of the web and d is the effective depth of the section. If the shear strength of the concrete is less than half the shear force at a section of a beam, steel reinforcing stirrups or other forms of reinforcement are required. The yielding strength of reinforcing steel in shear should not exceed 60,000 lb>in.2 The spacing of the stirrups depends on the magnitude of the shear force and the size of the stirrups but should not exceed d>2 (half the effective depth) or 24 in., so that potential cracks will definitely cross stirrups. Stirrups provide a shear strength Vs = Av fy d>s, where Av is the area of stirrups within a distance s. If Vs is more than 41f c= 1bw d2, the spacing between stirrups should not exceed d>4 or 12 in., and if Vs is more than 81f c= bw d, the section of the beam should be increased. Thus, for a given size of stirrups, their spacing along the beam may change to provide the required strength. Torsion reinforcement for beams consists of both longitudinal bars and closed stirrups, which are added to the required reinforcement for bending and shear.

4.7

Prestressing and Posttensioning

A way to make beams more efficient that is particularly suitable for concrete structures is by using prestressing or posttensioning. These techniques for permanently loading a beam in a controlled way are intended to build up stresses in the member opposite to those developed by the external service loads. First, consider an ordinary (not reinforced-concrete) beam, shown in Figure 43(a), that is subjected to both axial forces acting through the centroid

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 43 Combined stresses in beams. The final stress distribution in a beam carrying both axial and bending forces is a combination of the stresses associated with the axial forces and those associated with the bending moments.

,

Copyright © 2013. Pearson Education UK. All rights reserved.

Beams of the element and bending forces. The stress distribution associated with the axial forces is of uniform intensity and of magnitude fa = P>A. The stress distribution associated with bending is given by fy = My>I. The stress at any point on the cross section of a beam is the combination of these two stresses, added or subtracted according to whether the stresses acting at the point are similar, or not. Thus, either of the stress distributions shown in Figures 43(b) or (c) is possible, depending on the relative magnitudes of the axial and bending stresses. In members of this type, the neutral axis, or plane of zero stress, no longer corresponds to the centroidal axis of the section. By applying a large enough axial force to a beam, it is possible to develop compressive stresses of a magnitude sufficient to dominate the tensile bending stresses and thus put the entire cross section in compression. This is important with respect to concrete, which cannot withstand tension stresses and cracks when subjected to them. By putting the entire cross section in compression, the concrete can be used more effectively. It is important to note that the compressive stresses from bending and the applied axial force are additive. Thus, the advantages of applying this technique to a beam made of a material (e.g., steel) that is inherently capable of withstanding tension stresses is dubious because the increased level of the combined compressive stresses would govern the amount of load the member could carry. When the axial load is applied eccentrically, the types of stresses induced are particularly useful for offsetting the bending stresses associated with the service loads (Figure 44). Consider the first of the two most common ways to apply an axial force to a member, that of prestressing. In this process, which is typically done in an off-site factory, high-strength steel wires, or tendons, are stretched between two piers so that a predetermined tensile force is developed. Typical steel grades for prestressing tendons have yield strengths of over 200 ksi. Concrete is then cast in formwork placed around the wires and is cured. The wires are then cut. In effect, the tension force in the wires becomes equivalent to a compressive force applied to the member. (The tension stress in the wires is transferred to the concrete through bond stresses.) If the wires are placed eccentrically, cutting the wires induces stresses of the type illustrated in Figure 44(c) and causes the member to bow upward. When upward bowing commences, the dead load of the concrete member begins inducing stresses of the type illustrated. Neither of the stress patterns in Figure 40(d) occurs independently, and a combined stress pattern of the type shown at the right of the figure is present after the wires are cut. The exact amount of prestress force must be carefully controlled. Indeed, this stage of manufacture is when failure often occurs. Turning the beam upside down or on its side would lead to a rather dramatic failure because dead-load stresses would no longer offset tensile stresses developed by the eccentric load but instead would accentuate them. When the member is put in place carefully (in this case, by supporting it from its ends), the live load can be applied, which results in a final stress distribution of the type illustrated in Figure 44(e). The second most common method of applying a normal force to a member is that of posttensioning (a form of prestressing done at a later stage of construction. In this method, which is typically done on site, a tube, conduit, or equivalent containing unstressed steel tendons is set in place and concrete cast around the tube. After curing, the tendons are clamped on one end and jacked against the concrete on the other end until the required force is developed. The tendons are then anchored on the jacking end and the jacks removed. Next, the tendons are bonded in place by injecting grout into the tube. The net effect is that of the prestressing technique. Tendons can be draped in a funicular manner, if desired. Careful design can then yield a member in a state of compression only for the anticipated loading. Typical posttensioned beams are shown in Figure 45. In both of these approaches, several problems exist. An important one is the effect of creep in concrete (deformation with time in a constant-stress situation), which causes a loss in prestress or posttension forces. These forces must be

Beams

FIGure 44 Prestressed concrete members.

Copyright © 2013. Pearson Education UK. All rights reserved.

Because the strands are in the lower part of the

calculated carefully so that such losses with time are taken into account. Another consideration is that the stresses due to an eccentric force of the type illustrated are constant along the length of the beam, whereas the stresses associated with live and dead loads vary along the length of the beam. Indeed, with a constant eccentric force, critical prints in the beam are often found near the ends and are due to the prestress force because the offsetting effects of stresses due to the live and dead loads are not present. One advantage of posttensioning over prestressing is that the cable can be draped fairly easily so that the eccentricity of the cable force is a variable that can be adjusted to account for this phenomenon. Deforming prestressing steel in such a manner is more difficult. In situations where the beam undergoes a reverse curvature, special care must be taken with prestressed or posttensioned members. In small members (e.g.,

Beams

FIGure 45 Posttensioned double T beams. Cambers induced during posttensioning are visible on the bottom left. Right: typical construction detail. On site, a layer of poured-in-place concrete connects the precast beams into a one-way ribbed slab. 3UHFDVWSUHVWUHVVGRXEOH7EHDPV 3RXUHGLQSODFHFRQFUHWHOD\HU

Copyright © 2013. Pearson Education UK. All rights reserved.

&DPEHU

concrete planks), this is often handled by not placing wires eccentrically but instead putting them at the centroid of the member. Thus, reliance is placed only on the effect of the uniform stresses produced by the wires. In large structures, it is feasible to drape a posttensioned cable to reflect the anticipated reverse curvature. A final important point has to do with the ultimate load-carrying capacity of prestressed or posttensioned members. Usually, the primary criterion in designing such beams is their behavior under service or working loads. As loads increase beyond design parameters, the beam begins behaving much like a standard reinforced-concrete beam (i.e., cracks develop, etc.). The ultimate strength of a prestressed or posttensioned beam is not overly superior to that of a similarly proportioned plain reinforced-concrete beam. The primary value of the tensioning operation is thus to improve the performance of the beam at design or service loads. In particular, a highly significant reason for prestressing or posttensioning is to create a larger effective section to resist deflections. In plain reinforced-concrete beams, cracks develop in the tension region and propagate until they reach the compressive zone. This cracked section makes no significant contribution to the stiffness of the section. (The magnitude of I is based only on the extent of the compressive zone and on the amount of steel present.) In a prestressed or posttensioned beam, no tensile stresses are developed, and hence no cracks develop. Thus, the entire cross section of a beam contributes to the stiffness of the beam. Hence, for two beams of similar gross dimensions, a prestressed or posttensioned one can better resist deflections than its plain reinforced counterpart. In addition, the upward camber produced by eccentric prestressing forces is useful in controlling deflections. All members should be carefully designed, however, because prestressed members sometimes exhibit an undesirable springiness. Even so, such members are highly versatile and useful.

QuestIons 1. Two experimental methods based on the principle for determining the location of the centroid for any irregularly shaped cross section are illustrated in Figure 46. From a rigid sheet of cardboard, cut out several common geometric shapes; including a triangle, a T shape, and an L shape. Using one or both of the experimental methods illustrated in Figure 46, determine the location of the centroid of each shape and mark it on the cut-out shape. Discuss your results.

Beams

FIGure 46 Two experimental methods for determining the location of the centroid of an irregularly shaped figure. Pin

Pin

Centroid Balance edge

Centroidal axis

(a) Balance a rigid sheet representing the cross-section. The point of balance is the location of the centroid about that axis.

(b) From two or more pins, loosely suspend the rigid sheet representing the cross-section and draw vertical lines downward. The point of intersection of the lines is the centroid for the cross-section.

2. Determine the value of the moment of inertia, I, and the section modulus, S, for each of the following cross-sectional shapes: a. A 4 in. * 10 in. rectangle b. A 10 in. * 20 in. rectangle c. A 10 in. * 20 in. rectangle with a symmetrically located interior rectangular hole of 8 in. * 12 in. Consider values about the centroidal strong axis of each section only. Answer: (a) 333 in.4, 66.6 in.3; (b) 6666 in.4, 666 in.3; (c) 5514 in.4, 551.4 in.3 3. Determine the value of the moment of inertia about the centroidal axis of a triangular cross section having a base dimension of 5 in. and a height of 15 in. Answer: 468.75 in.4 4. Floor joists having cross-sectional dimensions of 112 in. * 912 in. are simply supported, span 12 ft, and carry a floor load of 50 lb>ft2. Compute the center-to-center spacing between joists to develop a maximum bending stress of 1200 lb>in.2 Compute what safe floor load could be carried if the center-line spacing were 16 in. Neglect dead loads and use allowable strength design methods.

Copyright © 2013. Pearson Education UK. All rights reserved.

5. Consider a cantilever beam that is 10 ft long and that supports a concentrated load of 833 lb at its free end. Assume that the beam is 2 in. * 10 in. in cross section and that lateral bracing is present. Draw the beam, showing the bending-stress distribution that is present at the base of the cantilever and at every 2.5-ft increment toward the free end of the beam (a total of five locations). Indicate numerical values for the maximum bending stresses at each section. Do a similar exercise for shearing stresses. What generally happens to the magnitudes of the bending and shearing stresses found when the width of the beam is doubled and its depth is held constant? What happens when the width of the beam is held constant and its depth is doubled? 6. Repeat Question 5, except assume that the beam is simply supported at either end and carries a uniformly distributed load of 200 lb>ft. 7. A simply supported beam 12 ft long carries a uniformly distributed load of 100 lb>ft. Assume that the beam is 112 in. * 912 in. in cross section and is laterally braced. Assume also that the beam is made of timber that has an allowable stress in bending of 1200 lb>in.2 and in shear of 150 lb>in.2 Using a simplified ASD approach, is the beam safe with respect to bending and shear stress considerations? Assume that E = 1.6 * 106 lb>in.2 What is the maximum deflection of the beam, and is it acceptable? Answer: 1fb = 9592 6 1Fb = 12002, 6 safe in bending; 1fv = 63.12 6 1Fv = 1502, 6 safe in shear; and 10.292 6 1L>240 = 0.62, 6 deflections are okay.

8. A simply supported steel beam will be used to span 30 ft and to support a uniformly distributed live load of 400 lb>ft. Assume that the yield stress in bending is 50,000 lb>in.2, and that allowable bending stress is 33,000 lb>in.2 Use both ASD and LRFD methods to determine the most efficient wide-flange shape to be used, based on a bendingstress analysis. Assume a factor of 1.6 for live loads when using LRFD methods. Ignore dead loads.

Beams 9. Assume that a laminated timber beam having cross-sectional dimensions of 8 in. * 20 in. is available. Based on bending-stress considerations only, how far could this beam span if it carried a uniformly distributed load of 250 lb>ft and was simply supported at either end? How far could it span if it carried the same load but was cantilevered? Assume that the allowable stress in bending is Fb = 2400 lb>in.2 and that the beams are all adequately laterally braced. Ignore dead loads. Answer: 58.4 ft if simply supported 10. Assume that a series of laminated timber beams will be used at 5 ft on center to span 25 ft and that the series carries a uniformly distributed floor live load of 40 lb>ft2 and dead load of 20 lb>ft2. Design an appropriate prototypical beam. Assume that the allowable stress in bending is Fb = 2200 lb>in.2, in shear is Fv = 400 lb>in.2, and in bearing is Fbg = 400 lb>in.2. Assume also that E = 1.8 * 106 lb>in.2 Make any assumptions that you must (e.g., about lateral bracing), but clearly state them or illustrate them with sketches.

Copyright © 2013. Pearson Education UK. All rights reserved.

11. With respect to bending-moment considerations only, sketch the shape variation present in a beam carrying two equal concentrated loads located at third points in the structure such that a constant bending-stress level is maintained on the top and bottom surfaces of the member. Do one sketch, assuming that the width of the beam is held constant and the depth varies. Then do another sketch, assuming that the depth of the beam is held constant and the width is allowed to vary.

Members in Compression: Columns

Copyright © 2013. Pearson Education UK. All rights reserved.

1

IntroductIon

Along with load-bearing walls, columns are the most common vertical support element. Strictly speaking, columns need not be only vertical. Rather, they are rigid linear elements that can be inclined in any direction, but to which loads are applied solely at member ends. Columns are not normally subject to bending that is directly induced by loads acting transverse to their axes. Columns can be categorized in terms of their length. Short columns tend to fail by crushing (a strength failure). Short members that ultimately exhibit a strength failure via a crushing action can be quite strong and carry high loads. Long columns tend to fail by buckling, which is an instability failure rather than a strength failure. Material rupture occurs only after the member has buckled. Long columns are members in which the length of the element is relatively great compared with its least lateral dimension. Pushing on the ends of a simple plastic ruler will convey a feeling of the buckling phenomenon. In this example, the buckling load causes the straight member to instantly bow. It does not take much force to buckle the member, and this is the maximum force the ruler can ever carry. Continuing to push on the ruler would ultimately cause a failure by bending, but by that time, the member has lost its load-carrying capacity. The same behavior is true of long columns in structures. The potential for buckling limits the load-carrying capacity of long members. Buckling can occur when compressive stress levels are quite low. While the phenomenon of buckling is usually described in connection with long columns, instability failures of this type also can occur in any member or structure with little transverse stiffness that is subjected to a compressive force. Localized buckling also can occur in parts of a cross section of a beam, such as in the flanges of a steel beam. Also, a thin shell subjected to compressive forces can also undergo localized or even snap-through buckling. In all these cases, the full load-carrying potential of the member cannot be reached because of the buckling phenomena, which can occur at low actual stress levels. The buckling phenomenon itself was recognized quite early as being of unique interest. Several investigators attempted analytical solutions that would predict exactly what load would make a slender member buckle. The problem was

From Chapter 7 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Members in Compression: Columns finally solved by Leonhard Euler (1707–1783), a mathematician born in Switzerland and related through training and association to the celebrated Bernoulli family, who were long recognized for their contributions to mathematics. Euler correctly analyzed the action of a long, slender member under an axial load while he was living in St. Petersburg, Russia, in 1759. The form he gave to the solution is still used today. It is one of the few early contributions to the structural engineering field that has survived unchanged.

Copyright © 2013. Pearson Education UK. All rights reserved.

2

General PrIncIPles

Short columns are members in which the least cross-sectional dimension present is appreciable relative to the length of the member. The load-carrying capacity of a short column is independent of the length of the member, and when excessively loaded, the short column typically fails by crushing. Consequently, its ultimate loadcarrying capacity depends primarily on the strength of the material used and its cross-sectional area. As a compressive member becomes longer and longer, the relative proportions of the member change to the extent that it can be described as a slender element, or long column. The behavior of a slender element under a compressive load differs dramatically from that of a short column. Consider the long compression member in Figure 1 and its behavior under increasing loads. When the applied load is small, the member maintains its linear shape and continues to do so as the load is increased. As a particular load level is reached, the member suddenly becomes unstable and deforms into the shape illustrated. This is the buckling phenomenon. When the member has buckled, it no longer can carry any additional load. Added loads immediately cause added deformation, which eventually causes the member to snap. This snapping, however, is regarded as a secondary failure because the maximum load-carrying capacity is that associated with initial buckling. A structure in a buckled mode is not serviceable. The phenomenon of buckling is curious. It is a failure mode due to an instability of the member, induced through the action of the load. Failures due to instability need not initially involve any material rupture or distress. Indeed, internal force levels can be quite low, and buckling can still occur if the member is long. The buckling phenomenon is associated with the stiffness of the member. A member with low stiffness will buckle before one with high stiffness. Increasing member length reduces stiffness. When a member initially becomes unstable, as does a column exactly at the buckling load, the member does not, and cannot, generate internal forces to restore the structure to its original linear configuration. Such restoring forces are developed below the buckling load. A column exactly at the buckling load is in neutral equilibrium. The system does not have restoring characteristics to reestablish the original configuration of the element once it is displaced. Numerous factors influence the buckling load, denoted Pcr, of a long compressive member. Length, of course, is critical; in general, the load-carrying capacity of a column varies inversely as the square of its length. In addition to the member’s length, the other factors that influence the load necessary to cause a member to buckle are associated with the stiffness characteristics of the member—both those that result from the inherent properties of the specific material used (as reflected by its modulus of elasticity) and those that result from the amount of material used in a cross section and how that material is distributed. The stiffness of the member is strongly influenced by the amount and distribution of the material present. In the thin rectangular member illustrated later in Figure 4(a), it can be experimentally demonstrated or theoretically proven that the member will always buckle in the direction indicated. The member is

Members in Compression: Columns

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 1 Behavior of columns under load.

less rigid about this axis than the other—that is, the member has much less ability to resist bending about this axis than about the other. The member will tend to buckle about the weak axis (the axis associated with the lesser ability to resist bending). The same member, however, can be sufficiently stiff about the other axis to resist buckling associated with that axis. The load-carrying capacity of a

Members in Compression: Columns compressive member thus depends not just on the amount of material present in the cross section (as is the case in tension members) but also on its distribution. A useful measure in this connection is the moment of inertia, I, which combines the amount of material present with how it is distributed into a single stiffness characteristic. Another factor of extreme importance in influencing the amount of load a member can carry is the nature of the end conditions of the member. If the ends of a column are free to rotate, the member can carry far less load than if the ends are restrained. Restraining the ends of a member increases its stiffness and thus its ability to resist buckling. Bracing the element in a direct way also increases stiffness. These important influences are explored in detail in Section 3.2. Figure 1(e) summarizes the general relationship between the load-carrying capacity and length of a member in direct compression and with simple pinned ends. As illustrated, the phenomenon of buckling reduces the load-carrying capabilities of compression members. The shape of the buckling curve illustrates that the load-carrying capacity of a long compressive member varies inversely with the square of the length of the member. The maximum load a column can carry, however, is associated with crushing (in the short-column range) rather than buckling. Long, slender columns fail at loads less than the crushing load. It follows that the actual stress level present when buckling occurs (the critical buckling stress) is less than the crushing or yield stress (i.e., fcr = pcr >A … fyield). Long columns can fail by buckling at stress levels that are considerably lower than those associated with yielding in the material. Remember that the actual axial stress level present in any symmetrically loaded compressive member, long or short, is always given by f = p>A. The stress level at which failure or buckling occurs, however, depends on whether the member is long or short.

3

Copyright © 2013. Pearson Education UK. All rights reserved.

3.1

analysIs oF comPressIon members short columns

axial loads. Compression members that fail primarily by a strength-related crushing action and whose ultimate load-carrying capacity is consequently independent of member lengths are easy to analyze. When the load is applied to the centroid of the cross section of the loaded element, uniform compressive stresses of a magnitude f = P>A are developed. Failure occurs when the actual direct stress exceeds the crushing stress of the material (i.e., fa Ú Fy). The crushing load is given by Py = AFy, where A is the cross-sectional area of the column and Fy is the yield, or crushing stress, of the material. Actual analysis methods must incorporate additional safety factors that are discussed in section 4.2. eccentric loads. When loads are applied eccentrically (i.e., not at the centroid of the cross section), the resultant stress distribution is not uniform. The effect of eccentric loads is to produce bending stresses in the member, which in turn interact with direct compressive stresses. If the load is considerably eccentrically applied, tensile rather than compressive stresses can even be developed at a cross section. Consider the member shown in Figure 2, which is subjected to an eccentric load P acting at a distance e from the centroidal axis of the member. Stresses produced by this load can be found by resolving the eccentric load into a statically equivalent axial force producing only uniform stresses fa and a couple (moment) producing only bending stresses fb. This resolution is illustrated in Figure 2. Final stresses are the combination of the two stress distributions. Uniform stresses = fa = P>A; bending stresses = fb = Mc>I, where M = Pe. Hence, fb = 1Pe2c>I. Finally, combined stresses = factual = fa + fb = {P>A { 1Pe2c>I.

Members in Compression: Columns

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 2 columns.

An inspection of the stress distributions in the figure reveals that the magnitude of the bending stresses is proportional to the eccentricity e of the load. In a situation of this type, the vertical load P can produce tensile stresses on one face of the element if the eccentricity is large (i.e., bending stresses fb dominate over axial or normal stresses). When e = 0, only compressive stresses fa exist. A limit to the eccentricity of the load must exist if the intention is to have only compressive stresses in the member. This limiting point can be found simply by equating the resultant stress to zero 1fa + fb = 02 and solving for the eccentricity, or P>A = 1Pe2c>I and e = I>Ac. For a rectangular cross section (see Figure 2), P>bd = Pe1d>22 > 1bd3 >122 and e = d>6. Thus, if the load were placed within this maximum value, the stresses produced would all be compressive. Placing the load exactly at the point such that e = d>6 produces zero stresses on the opposite face. Exceeding this eccentricity causes tensile stresses to develop on that face. This location is called the Kern point. Note that, for a load that can vary in either direction, Kern points exist on either side of the centroidal axis. These locations are third points on the face of the element. The existence of these points gives rise to the middle-third rule, often referred to in the design of masonry structures (particularly in a historical context), where the design intent is to keep the load inside the middle third to prevent tensile stresses, which masonry cannot withstand, from developing. If the third dimension is considered, a Kern area can be found, as indicated in Figure 2.

3.2

long columns

euler buckling. As previously mentioned, Leonhard Euler was the first investigator to formulate an expression for the critical buckling load of a column. The critical buckling load for a pin-ended column, termed the Euler buckling load, is Pcr =

p2EI L2

Eccentrically loaded

Members in Compression: Columns where E = modulus of elasticity L = length of column between pinned ends I = moment of inertia p = pi 1a constant2 ≈ 3.1416

The expression clearly shows that the load-carrying capacity of a column depends inversely on the square of the member’s length, directly on the value of the modulus of elasticity of the material used, and directly on the value of the moment of inertia of the cross section. The moment of inertia of concern is the minimum one about any axis of the cross section if the member is not braced. The Euler buckling expression predicts that, when a column becomes indefinitely long, the load required to cause the member to buckle begins to approach zero. (See Figure 3.) Conversely, when the length of a column begins to approach zero, the load required to cause the member to buckle becomes indefinitely large. What happens, of course, is that, as the member becomes short, the failure mode changes into that of crushing. Consequently, the Euler expression is not valid for short members because it predicts impossibly high values. The crushing strength becomes a cutoff for the applicability of the Euler expression.

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 3 Euler buckling in long columns.

Members in Compression: Columns The dependency of the buckling load on the inverse of the square of the length of the column is important. Doubling the length of a column reduces its load-carrying capacity by a factor of 4. Thus, if P1 = p2EI>L21 and L2 = 2L1, then P2 = p2EI1L2 2 2 or p2EI> 12L1 2 2 = P1 >4. Similarly, halving the length of the column increases its load-carrying capacity by a factor of 4 [i.e., P2 = p2EI> 1 1>2L1 2 2 = 4P1]. The buckling load of a column is thus extremely sensitive to changes in the length of the member. examPle Determine the critical buckling load for a 2@in. * 2@in. 150.8@mm * 50.8@mm2 steel column that is 180 in. (4572 mm) long and pin-ended. Assume that E = 29.6 * 106 lb>in.2 10.204 * 106 N>mm2 2. solution:

Moment of inertia, Ix or Iy: I =

2 in. 12 in.2 3 50.8 mm 150.8 mm2 3 bh3 = = 1.33 in.4 = = 555 * 103 mm4 12 12 12

Because Ix = Iy, the column is equally likely to buckle about either axis or about a diagonal because that I value also is equal to Ix or Iy. Critical buckling load: Pcr = =

p2 129.6 * 106 lb>in.2 211.33 in.4 2 p2EI = = 12,000 lb L2 1180 in.2 2 p2 1204,000 N>mm2 21555,000 mm4 2 14572 mm2 2

= 53,458 N = 53.5 kN

The actual stress level corresponding to this critical buckling load is fcr =

Pcr 12,000 lb 53,458 N = = 3000 lb>in.2 = = 20.7 N>mm2 A 2 in. * 2 in. 50.8 mm * 50.8 mm

Thus, the column buckles at a relatively low actual stress level, below the crushing stress for steel.

Copyright © 2013. Pearson Education UK. All rights reserved.

examPle At what length will the square column previously analyzed begin to crush rather than buckle (i.e., what is the transition length between short- and long-column behavior for this specific member)? Assume the yield stress of the steel to be Fy = 36,000 lb>in.2 1248 N>mm2 2 solution:

For failure by crushing: Pmax = Fy A = 136,000 lb>in.2 212 in. * 2 in.2 = 144,000 lb = 1248 N>mm2 2150.8 mm * 50.8 mm2 = 640 kN

Buckling length for Pmax:

Pcr = 144,000 lb =

p2EI = Pmax L2 p2 129.6 * 106 lb>in.2 211.33 in.4 2

= 640,000 N = 6 L = 1320 mm

6 L = 52 in. L2 p2 1204,000 N>mm2 21555,000 mm4 2 L2

Members in Compression: Columns Consequently, For L 6 52 in. 11320 mm2, the member will crush.

For L 7 52 in. 11320 mm2, the member will buckle.

For L = 52 in. 11320 mm2, either failure mode is likely.

Such an exactly defined transition point does not really exist. Rather, a more general transition occurs.

shape of cross section. When an unbraced member is nonsymmetrical, it is necessary to account for the different moments of inertia of the member (Figure 4). Members of this type generally buckle in the direction of their least dimension, or, more precisely, about their weaker axis. A typical rectangular column has two primary moments of inertia, Ix and Iy. A load is associated with each that will cause the member to buckle about each respective axis, Pcrx and Pcry. The load that causes the whole member to buckle is the smaller of these two values: Pcrx

p2EIy p2EIx = and Pcry = L2x L2y

examPle Determine the critical buckling load for a column with the same cross-sectional area as the column previously analyzed, but rectangular: b = 1 in. and d = 4 in. 125.4 mm * 101.6 mm2. Assume that L = 180 in. 14572 mm2. E = 29.6 * 106 lb>in.2 1204,000 N>mm2 2 as before.

solution:

Moment of inertia: 112143 2 125.4 mm21101.6 mm2 3 bd3 = = 5.33 in.4 = = 2.20 * 106 mm4 12 12 12 142113 2 1101.6 mm2125.4 mm2 3 db3 Iy = = = 0.33 in.4 = = 0.139 * 106 mm4 12 12 12 Ix =

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 4 Buckling of asymmetric cross sections.

Members in Compression: Columns Critical buckling loads: Load that causes buckling about the x-axis: Pcrx = =

p2EIx L2x

=

p2 129.6 * 106 215.332 11802 2

= 48,060 lb

p2 1204,000 N>mm2 212.20 * 106 mm4 2 14572 mm2 2

= 213.8 kN

Load that causes buckling about the y-axis: Pcry = =

p2EIy L2y

=

p2 129.6 * 106 210.332 11802 2

= 2975.5 lb

p2 1204,000 N>mm2 210.139 * 106 mm4 2 14572 mm2 2

= 13.3 kN

Because the load required to cause the member to buckle about the weaker y-axis is much less than the load associated with buckling about the stronger x-axis, the critical buckling load for the entire column is 2975.5 lb (13.3 kN). The member will buckle in the direction of the least dimension. Compared with the load-carrying capacity of a member having an equivalent area, but square in shape, the load-carrying capacity of the rectangular member analyzed is greatly reduced.

critical buckling stress. The Euler expression is often rewritten in a slightly different form that is more useful as a design tool. The critical buckling load for a column can be converted into a critical buckling stress fcr by dividing both sides of the Euler expression by the area A of the column. Thus, fcr = P>A = p2 EI>AL2. This expression contains two measures related to the dimensional properties of the column—I and A—that can be combined into a single measure called the radius of gyration, r, defined by r = 2I>A. Note that I = Ar2. Accepting this as a definition, we can rewrite the expression for the critical buckling stress of a column as

Copyright © 2013. Pearson Education UK. All rights reserved.

fcr =

P p 2E = A 1L>r2 2

The term L>r is called the slenderness ratio of the column. The critical buckling stress depends inversely on the square of the slenderness ratio—that is, the higher the slenderness ratio, the lower is the critical stress that causes buckling, and vice versa. The slenderness ratio (L>r) is an important way to think about columns because it is the single measurable parameter on which the buckling of a column depends. The radius of gyration, r, can be interpreted in the following way: The moment of inertia of the cross-sectional area of the column is equal to the product of the area and the square of the distance r, by definition (i.e., I = Ar2 2. Consequently, the radius of gyration of this area with respect to an axis is a distance such that if the total area were conceived of as concentrated at this point, its moment of inertia about the axis would be the same as the original distributed area about that axis. The higher the radius of gyration of a section, the more resistant the section is to buckling (although the true measure for resistance to buckling is the L>r ratio). Radius-of-gyration values for different sections are often tabulated in the same way as moments of inertia. examPle For the column previously analyzed (b = 1 in., d = 4 in., and L = 180 in.) what is the critical buckling stress, fcr, that is present? Iy = 0.33 in.4

Members in Compression: Columns solution: Determine ry = 2Iy >A = 20.33>4.0 = 0.29 in. Therefore, L>ry = 1180>0.292 = 620. The high L>ry value indicates that the column is extremely slender and prone to buckling.

Determine the critical buckling stress:

fcry = p2E> 1L>ry 2 2 = p2 129.6 * 106 2 >6202 = 750 lb>in.2

Note that this buckling stress level is far lower than that associated with material, yielding 1Fy = 36,000 lb>in.2 2.

end conditions and effective lengths. The discussion in the previous section focused on columns having pin-ended connections in which the ends of the members were free to rotate (but not translate) in any direction. This condition allows the member to deform as illustrated in Figure 5(a). Other end conditions are possible as well. Restraining the ends of a column from a free-rotation condition increases the load-carrying capacity of the column. Allowing translation as well as rotations at the ends of a column reduces its load-carrying capacity. This section considers the relative load-carrying capacity of four columns that are identical in all respects, except for their end conditions. The conditions are illustrated in Figure 5 and represent theoretical extremes because end conditions in practice are often combinations of these primary conditions. Column A represents the standard pin-ended column already discussed. Column B has both ends fixed (i.e., no rotations can occur). Column C has one end fixed and the other pinned. Column D has one end fixed and the other completely free to both rotate and translate (and is similar to a flagpole).

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 5 Effects of end conditions on critical buckling loads.

Copyright © 2013. Pearson Education UK. All rights reserved.

Members in Compression: Columns The theoretical buckling load for each of these columns can be computed in a way similar to that a pin-ended column. Different boundary conditions would be used. Another way to find the load-carrying capacity of the columns is to consider the effects of the end conditions on the deformed shape of the columns. Consider the deformed shape of the fixed-ended member (column B) illustrated in an exaggerated way in Figure 5(b). The shape of the curve can be sketched with a high degree of accuracy by noting that a fixed-end condition of the type illustrated causes the tangent to the member to remain vertical at each end. Under the buckling load, the member would begin bowing as indicated, with the curvature beginning immediately outside the connection. For the curve to be continuous, as it must be, the resultant curve must be similar to that illustrated. It is critically important that the member undergo a change in curvature. Points of inflection must exist at two locations where the sense of the curvature changes. The locations of these points can be estimated fairly accurately. In this case, they are L1 >4 from each end, where L1 is the actual length of the column. Note that the shape of the column between those two points of inflection is similar to that of a pin-ended column. This is sensible because pin-ended connections and points of inflection are analogous. It follows that this portion of the column behaves as if it were a pinended column of a length equal to the distance between the points of inflection, in this case, one-half the actual length of the column, or L1 >2. This distance is called the effective length (Le) of the fixed-ended column. This portion of the column controls the buckling load of the whole column. The buckling load of the portion is given by PB = p2EI>L2e , where Le is the distance between the points of inflection. Consequently, the buckling load of the column, in terms of its original length, is PB = p2EI> 1L1 >22 2 = 4p2EI>L1. Because the initial buckling load for a pin-ended column of the same actual length is PA = p2EI>L21, the effect of fixing both ends of the column is to increase its load-carrying capacity by a factor of 4. This substantial increase is equivalent to that caused by halving the length of the column. The foregoing process can be repeated for other end conditions. Deformed shapes are sketched and locations of inflection points are found. Effective lengths are determined next. Critical buckling loads are then given by P = p2EI>L2e . The effective length of the column with one end pinned and the other fixed is Lc = 0.7 L1. Its buckling load is consequently Pc = p2EI> 11>0.7 L2 2 = 2p2EI>L21, or twice that of a column with pins on both ends. For the flagpole column, the deformed shape of the actual column is one-half of the shape analogous to that of a pinended column. [See Figure 5(d).] Its effective length is thus 2L1. Consequently, PD = p2EI> 12 L1 2 2 = 14 p2EI>L21, or one-fourth of that for a pin-ended column of the same length. The concept of effective length is useful in analyzing columns with different end conditions because it provides a shortcut for making predictions about their load-carrying capacities. The numerical value that modifies the actual length 1L2 is called the k factor of the column. Thus, k = 1.0 for a pin-ended column. Hence, Le = kL = 1.0L and Pc = p2EI>11.0L2 2. Also, k = 0.5 for a fixed-end column, so Le = kL = 0.5L and Pc = p2EI>10.5 L2 2 = 4p2EI>L2. Often a column has different end conditions with respect to one axis than another (e.g., it may be pin-ended with respect to one axis and fixed with respect to the other). Hence, care must be taken to couple the correct effective length with the appropriate moment of inertia or radius of gyration. Thus, Pcrx = fcrx =

p2EIy p2EIx and P = cr y 1kLx 2 2 1kLy 2 2

p 2E p 2E and f = cr y 1kLx >rx 2 2 1kLy >ry 2 2

As one of the examples previously discussed, the critical buckling load for the entire column is governed by the smaller of Pcrx and Pcry.

Members in Compression: Columns The effects of imperfect end conditions in field applications are often taken roughly into account by modifying the effective length of a member. If a column is intended to have fixed ends, but the actual field connection is such that only partial restraint is obtained and, for example, some rotations occur, the effective length of the column lies between 0.5 L and 1.0 L. The more nearly full restraint is obtained, the closer is the effective length to 0.5 L. The more the column rotates, the closer its effective length is to 1.0 L. Values such as 0.6 L to 0.7 L are often used. Often, in the case of frames, only partial restraint is obtained, due to the rotation of end joints. This end rotation results in an increase in the effective lengths of members. When the end rotation is coupled with translation of the joint in space, the effective member length can be longer than the physical length of the column and can constitute a serious design problem.

Copyright © 2013. Pearson Education UK. All rights reserved.

bracing. To reduce column lengths and increase their load-carrying capacities, columns are frequently braced at one or more points along their length. (See Figure 6.) The bracing can be part of the structural framework for the rest of the building, which also serves other functions. Figure 6(b) illustrates a pin-ended column braced at its midheight. When the member buckles, it deforms into an S shape, as illustrated. Because the shape of the member between the bracing and the point of inflection is analogous to that of a pin-ended column, the effective length of the column is equal to that distance. The length of the column shown has thus been halved, which increases its load-carrying capacity by a factor of 4: P = p2EI>L2e = p2EI> 1L1 >22 2 = 4p2EI>L21. If the bracing were placed two-thirds of the way up from the bottom, the member would again deform but this time into a modified S shape. Its effective length would then be 2>3 L or 1>3 L. The longer unbraced portion would buckle prior to the shorter portion. Consequently, the critical load for the column would be P = p2EI>L2e = p2EI>[12>32L21] = 1 9>4 21p2EI>L21 2. Bracing the member at this point is not as effective in increasing the load-carrying capacity of the column as is midheight bracing. Note that a column can be braced about one axis but not the other. The column will tend to buckle in the direction associated with the highest slenderness ratio. Any column must be checked for buckling about both axes (using appropriate

FIGure 6 Effects of lateral bracing on column buckling. Bracing a column changes its buckling mode and, consequently, its effective length. The more a column is braced, the shorter its effective length becomes and the greater is the load required to cause buckling. If bracing is used, it is usually more effective when placed symmetrically.

Members in Compression: Columns combinations of I and Le or kL values) by calculating Pcrx and Pcry values or comparable critical stresses. The smaller of the loads is the critical load at which the column initially buckles.

examPle Consider the column shown in Figure 7. Assume that the overall height of the column is L = 180 in. and the bracing is at midheight. (a) If the narrow dimension b = 1 in. and the deep dimension d = 4 in., what load causes the buckling? (b) Repeat part (a) for a column with an identical area with dimensions b = 1.75 in. and d = 2.28 in. solution: a. Determine Ix = bd3>12 = 112143 2>12 = 5.33 in.4 and Iy = db3>12 = 1421132>12 = 0.33 in.4 Note that Lex = L = 180 and Ley = 1L>22 = 180>2 = 90. Hence, Pcrx = p2EIx >L2ex = p2 129.6 * 106 215.332 >1802 = 48,060 lb and Pcry = p2EIy >L2ey = p2 129.6 * 106 2 10.332>1902 2 = 12,015 lb. Because 12,015 6 48,060, the column buckles at Pcry = 12,015 lb about its weak axis in the plane of the bracing and in an S-shaped buckling mode. It is in no danger of buckling out of the plane of the bracing. b. Determine Ix = bd3 >12 = 11.75212.283 212 = 1.72 in.4 and Iy = db3>12 = 12.282 * 11.753 2 >12 = 1.02 in.4 Note that Lex = L = 180 and Ley = 1L>22 = 90.

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 7 Column bracing in one plane only. When a column is braced in only one plane, it can buckle in two modes. The column will buckle in the mode associated with the higher slenderness ratio (L>r).

Members in Compression: Columns c. Determine critical buckling loads. Pcrx = p2EIx >L2ex = p2 129.6 * 106 211.722 >1802 = 15,500 lb and Pcry = p2EIy >L2ey = p2 129.6 * 11.022 > 1902 2 = 36,750 lb. Because 15,500 6 36,750, the column buckles at Pcrx = 15,500 lb about its strong axis, out of the plane of the bracing in a simple curve. It will not buckle in an S shape because that requires a higher buckling load than that associated with out-of-plane bracing.

actual Versus Ideal column strengths. When columns are physically tested, a difference is usually found between actual buckling loads and theoretical predictions. This is particularly true for columns of lengths near the transition between short- and long-column behavior. Reasons for the difference include such factors as minor accidental eccentricity in the column loading; initial crookedness in the member; initial stresses present before loading, due to the fabrication process; nonhomogeneity in the material; and others. Bending stresses not accounted for in the Euler expression often exist. The result is that buckling loads are often slightly lower than predicted, particularly near the transition zone between short and long columns, where failure is often partly elastic and partly inelastic (crushing). For that reason, several additional theoretical and empirical expressions have been developed to take this phenomenon into account. Expressions are available, for example, to more accurately predict behavior in the intermediate range. eccentric loads and moments. Many columns are subjected not only to axial loads but also to large bending moments. For short columns, how to take these moments into account was discussed in Section 3.1. The moment is expressed as M = Pe, and combined stresses are considered. For long columns, the Euler expression does not take into account such additional considerations. There are expressions that are available for similar situations.

4

Copyright © 2013. Pearson Education UK. All rights reserved.

4.1

desIGn oF comPressIon members General design Principles

sections. A general column design objective is to support the design loading by using the smallest amount of material or, alternatively, by supporting the greatest amount of load with a given amount of material. Whenever the buckling phenomenon enters in, it is evident by looking at the buckling curves discussed in the previous section that the full strength of the material in the compressive member is not being exploited to the maximum degree possible. As was previously noted, the primary factor of importance in connection with buckling is the slenderness ratio 1Le >r2 of a column. A low slenderness ratio means that the column is not prone to buckling. High ratios are not desirable. The most efficient use of material can be achieved by either minimizing the column length or maximizing the value of the radius of gyration for a given amount of material. Either or both will reduce the slenderness ratio of a member and hence increase its load-carrying capabilities for a given amount of material. Most design techniques hinge around one or the other of these operations. Determining the required cross-sectional shape for a column intended to carry a given load is a conceptually straightforward task. In most cases, the objective is to find a cross section that provides the necessary rx or ry values by using the smallest amount of material possible in the cross section. The problem is similar to that of designing beams, wherein the objective is to find a section that provides the greatest moment of inertia for the smallest amount of material. In column design, the task is more complicated because of the need to consider radii of gyration about different axes. In general, however, an efficient column section removes material as far from the centroidal

Members in Compression: Columns

FIGure 8 Cross-sectional shapes for columns.

Copyright © 2013. Pearson Education UK. All rights reserved.

axis of the column as is feasible. This, in turn, increases the moment of inertia, I, of the section for a given amount of material, A. Consequently, the radius of gyration, r, is increased because r = 2I>A. Increasing r decreases the slenderness ratio L>r for a column, thus making it less susceptible to buckling. The techniques for optimizing moments of inertia are also relevant in column design. Several cross sections based on this principle are illustrated in Figure 8. bracing. Column strengths also may be greatly improved by increasing the end restraints on the column. By changing from a pinned to a fixed connection, the load necessary to cause a member to buckle is greatly increased. It follows that if the intent is to support a specified load, a member with restrained ends generally requires less material volume to support the load than one with unrestrained ends. The end conditions associated with the flagpole column illustrated at the right of Figure 5 are the least desirable. The choice of whether to use restrained ends, however, should be tempered by other considerations. Fixed joints are more difficult and costly to make than pinned joints. (This is true particularly when materials such as timber are used.) If the joint were at the foundation, the foundation would have to be designed to provide restraint to rotation rather than simply receive axial forces. For relatively small columns, this usually does not require significant additional material or effort, but that may change as column sizes and loads increase. One of the most common tools available to the designer to increase the efficiency of compressive elements is bracing. Adding bracing decreases the effective length of columns. It must be done with care, however, because, quite often, if bracing is not added properly, no benefit is gained and even a loss is incurred due to the extra material and effort necessary to add the bracing. Consider the columns shown in Figure 9. In these cases, adding bracing does not increase the load-carrying capacity of any of the columns. As a result of either their symmetrical cross sections or relative Ix and Iy values, all would buckle in a mode associated with their actual, rather than braced, lengths because of their higher slenderness ratios with respect to the unbraced direction. Buckling loads would be the same as if no bracing were present. If a symmetrical member is to be braced effectively, then the bracing must reduce the slenderness ratios in all directions (by providing bracing in more than one plane). A highly effective bracing is illustrated in Figure 10(a). Here, a symmetrical member is frequently braced in all directions at several points along its length. The ideal number of brace points is easy to determine. As previously discussed, for a member of a certain size and material, there exists a certain length that marks

Members in Compression: Columns

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 9 Ineffective use of bracing. Columns always buckle in the mode associated with the highest slenderness ratio (L>r). The columns shown will buckle as illustrated. The corresponding buckling loads relate to an unbraced or unsupported column length of L and are the same as if the columns were not braced at all. The bracing patterns shown are ineffective in increasing the load-carrying capability of the columns.

the transition between short- and long-column behavior. By putting in a sufficient number of brace points, it is possible to eliminate the possibility of buckling anywhere in the member. Consequently, the member could be sized for direct stresses only 1i.e., A = P>Fallowable 2, and no premium (in terms of additional area) would be paid to prevent the possibility of buckling. The minimum number of brace points thus corresponds to making column lengths between brace points exactly equal to the transition length between short- and long-column behavior. Putting in additional brace points does not accomplish anything more in terms of reducing required areas because the column size is already governed by direct stresses. In many situations, it is not possible to use bracing in more than one plane. Columns are often used in walls, for example, where the wall can serve as lateral bracing in one plane but where no bracing can be provided in other planes for functional reasons. In such situations, nonsymmetrical members can often be employed wherein the strong axis of the member is organized so it is associated with the possible out-of-braced-plane buckling mode and the weak axis with inplane modes. When this is done, the column properties can be related exactly to the number and kind of brace points required to achieve a high level of efficiency. This is done by creating a set of conditions such that the critical load associated with buckling about one axis is equal to that associated with buckling about the other axis 1i.e., Pcrx = Pcry 2. Thus, p2EIx >L2ex = p2EIy >L2ey, so Ix >Iy = L2ex >L2ey. Note that, for a rectangular cross section, Ix >Iy = 1bd3 >122 > 1db3 >122 = L2ex >L2ey, or d>b = Lex >Ley. Hence, a column with a depth-to-width ratio of 3:1 should have two symmetrically placed brace points (so that Lex = 3Ley). When the proportion shown obtains, the member is equally likely to buckle in either direction. Evidently, either the member configuration (Ix and Iy) can be varied, or the effective member lengths 1Lex and Ley 2 can be changed by bracing to achieve this proportion; or both may be varied simultaneously.

Members in Compression: Columns

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 10 Effective use of bracing.

The general relationship, Ix >Iy = L2ex >L2ey, also can be used to determine the appropriate number of brace points for a given column in which Ix >Iy is fixed. In practice, instead of Ix >Iy ratios, rx >ry ratios are often used. The concepts are similar, except that now fcrx = fcry, or p2E>1Lx >rx 2 2 = p2E>1Ly >ry 2 2. Hence, rx >ry = Lx >Ly.

columns in a building context. The previous discussion tacitly assumed that no penalty was involved in using as many brace points as necessary. However, the brace points themselves must be built. This, in turn, requires material volume. While bracing members can be quite small, because prevention of buckling requires little force, construction effort is still required to put them in place. If column design

Members in Compression: Columns is placed in the context of a building, trade-offs must occur in which optimizing the complete column and bracing system, rather than the column alone, makes the most sense. Quite often, it may be preferable to use fewer brace points and a slightly larger column. Figure 11 illustrates a column in a wall of a simple industrial building. The wall is made by attaching vertical siding to horizontal elements, which in turn frame into the column. Either a greater or lesser number of horizontal elements (girts) could be used than the number illustrated. Using more horizontal elements implies that the span of the siding, which functions as a vertical beam carrying wind loading, is reduced (and thus a lighter gauge siding could be used). Similarly, the columns would be braced at more points in one plane. This, in turn, would influence the choice of the properties of the column itself. More horizontal elements are involved, however, and consequently, more construction difficulty. Using fewer horizontal elements increases the loads on the girts (and hence their sizes), decreases the number of bracing points, and increases the span of the siding, but it tends to make construction easier. Typically, the spacing of the horizontal elements is based on the optimum span for the most economic form of siding available. This, in turn, establishes where the column is braced. Column properties would be selected on the

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 11 building.

Small industrial

Members in Compression: Columns basis of these bracing points (i.e., an Lex >Ley ratio would be found and, from this, an rx >ry ratio). A column with this rx >ry ratio would then be sized to carry the axial load involved. Note that this example implies that it is preferable to optimize the siding system rather than the column itself. This is often true in smaller buildings but not necessarily always so. The larger the loads on the column and the longer it is, the more important it is to pay attention to optimizing the column itself. Trade-offs are involved, and each situation must be looked at individually.

4.2

column sizes

Column design follows similar strategies as were discussed earlier in the context of beams or tension elements. Safety factors can be incorporated by reducing the material crushing strength to a lower allowable stresses value as is common in allowable strength (ASD) methods. Alternatively, safety factors can be assigned primarily to loads and to a lesser degree to material stresses. Factored loads then permit columns to be designed for the limit state of failure. This approach is referred to as load and resistance factor design (LRFD) in the context of steel and timber columns or as ultimate strength design (USD) for concrete columns. All methods for column design are ultimately based on Euler expressions, but straightforward distinctions between long and short columns are rarely made. The maximum permissible stress values are partly derived on an empirical basis, acknowledging the extensive transition phase between long- and short-column behavior. Even columns that would be considered short columns in the previous discussion can be affected by slenderness and buckling effects. Design procedures for estimating column sizes are iterative because the permissible stress value cannot be known prior to selecting a column size. As a first guideline for selecting a column, the minimum required cross-sectional area of a short column is often easily determined. Based on this initial calculation, a structural shape can be chosen which is then evaluated with expressions that incorporate issues of buckling, end conditions, and the presence of eccentricities or bending moments as well as safety factors. During that step, the maximum permissible stress to which the column can be subjected can be obtained. The compressive capacity of the column is then found by multiplying the permissible stress with the cross-sectional area. The following sections highlight common analysis and design approaches. Reference is made to other sources for a full treatment.1

Copyright © 2013. Pearson Education Limited. All rights reserved.

4.3

timber columns

Timber columns must have a compression capacity equal to or larger than the actual service load (ASD methods) or factored load (LRFD methods). In U.S. practice, an adjusted compressive stress f ′c is determined by modifying the material strength values for different species with a range of adjustment factors. Slenderness ratios for timber columns should be limited to 50 or less. The adjusted compression capacity P′c can be found with P′c = 1 f ′c 21Area2. It must be equal to or larger than the applicable loads. Adjustment factors incorporate effects such as moisture content, exposure to high temperature, and load duration, to name a few. Factors take the form of CX, with X describing the phenomenon considered. CX factors vary for sawn lumber versus glulam elements. Specific factors are prescribed depending on whether elastic design methods (ASD) or limit state design methods (LRFD) are used. Code-compliant design approaches that take into account these many factors are described in a following section. Because they are cumbersome to use for preliminary studies, a highly simplified approach is presented first. 1 Hassoun, M. N., Al-Manaseer, A., Structural Concrete Theory and Design, Hoboken: John Wiley & Sons, 2005; Breyer, D. et. al., Design of Wood Structures, New York: McGraw-Hill, 2007; McCormac, J., Structural Steel Design, Upper Saddle River: Pearson Prentice Hall, 2008.

Members in Compression: Columns simplified methods for approximate studies: Code-compliant design methods are lengthy and often not suitable for initial studies that are geared toward gaining an approximate sense of column sizes. Simpler expressions can be used to find allowable buckling stresses (long column) and allowable material stress values for short columns. The column capacity is obtained by multiplying the cross-sectional area by the smaller of the two stress values. Approximate allowable buckling stresses FC,all are calculated using the expression FC,all = 0.27 E> 1Le >d2 2. The expression incorporates typical ASD safety factors and is based on the Euler buckling load expression. Allowable material stress values Fall are determined by reducing the published material strength value by 40 percent; hence, Fall = 1fc 210.62. The smaller of the two stress values Fall and FC,all is then used to find the capacity P′c of the column. The simplified approach is meant for use with ASD service loads. examPle Simplified Method. Is a nominal 4 in. * 4 in. (actual size 3.5 in. * 3.5 in.) pin-ended timber column 10 ft (120 in.) high adequate to support axial loads of 1000 lb (dead load) and 3500 lb (live load)? Assume that E = 1.62 * 106 lb>in.2 and fc = 1500 lb>in.2. The column is located on the interior (dry condition) and is not subject to elevated temperatures. solution: Load calculations: ASD Service Loads P = 1000 lb + 3500 lb = 4500 lb FC,all = 0.27 E> 1Le >d2 2

= 0.2711.62 * 106 lb>in.2 2 > 1120 in.>3.5 in.2 2 = 372 lb>in.2

Because FC,all = 372 lb>in.2 6Fall = 1 fc 210.62 = 11500 lb>in.2 210.62 = 900 lb>in.2 FC,all = 372 lb>in.2 is used to determine the capacity P′c of the column. P′c = F C,all 1Area2 = 372 lb>in.2 13.5 in. * 3.5 in.2 = 4557 lb

Copyright © 2013. Pearson Education Limited. All rights reserved.

Because P′c = 4557 lb 7 P = 4500 lb, the column is adequately sized based on the simplified analysis. A code-compliant approach described next would also indicate that the column is safe but would yield a higher P′c.

code-compliant methods. In code-compliant ASD and LFRD design methods, issues of buckling are incorporated through a so-called column stability factor CP. This approach is largely based on empirically determined transitions between shortand long-column behavior. The column stability factor is obtained through expressions that compare the critical buckling stress FCE for long-column behavior with maximum adjusted stresses F*C for a short column. F*C is used only to determine the all-important stability factor CP and is not to be confused with the adjusted compressive stress f ′c used to derive the load-carrying capacity of the column. F*C is derived by adjusting the species-related, tabulated compressive strength with all applicable adjustment factors CX. FCE, on the other hand, is obtained through the Euler expression FCE = 10.822 E′min 2 > 1Le >d2 2. Here, E′min is Young’s modulus adjusted for temperature, stability, and other effects. It should be noted that Emin values are generally lower than general E values for the same species. Emin is used primarily in column design, while E is needed to determine deflections in beams and in the simplified method of column analysis just presented. The slenderness ratio should be obtained for the weak axis, and for d, the minimum value would be used for sections that are not square. The ratio ac = FCE >F C* is then used to find CP using the following expression: CP = a

1 + ac 1 + ac 2 ac ac b d - c db c 2c C 2c

Members in Compression: Columns In this expression, values for c are 0.8 for sawn and 0.9 for glulam timber. Adjusted compressive strength values for the two respective design methods can be obtained as follows: ASD: f′c = 1 fc 21CP 21CX 2

where CX represents several other adjustment factors. Just considering CP will give good initial indicators of stress values for approximate studies. LRFD: f ′c = 1 fc 21CP 21CX 21KF Φ λ2 = 1 fc 21CP 21CX 211.7282

where Φ is the resistance factor 0.9 for a compression element. KF , the format conversion factor, can be found with 2.16>Φ, so the value for a column is 2.4. λ is the time effect factor; a typical value is 0.8. Considering all factors together, they increase the product of material strength and column stability factor by 1.728. The same 3.5 * 3.5 * 120 in. column carrying a 1000-lb dead load and 3500-lb live load just analyzed using a simplified technique also has been analyzed using these code-compliant measures and is again found to be adequately sized.

Copyright © 2013. Pearson Education Limited. All rights reserved.

4.4

steel columns

The design of steel columns determines the nominal compressive strength Pn of the column based on maximum stress values that reflect the slenderness ratio of the column. In LRFD methods, the nominal strength is reduced by 0.9 to obtain a design compressive strength that must be sufficient for carrying factored loads. ASD methods divide the nominal strength by 1.67 to find the allowable compressive column strength. In U.S. practice, the method for finding Pn is identical for both design methods. While ultimately based on the Euler expression, the calculation methods have strong empirical roots that reflect certain inelastic behaviors in steel members. The Euler formula does not predict values for low slenderness ratios well, but it is more appropriate for larger ones. The expressions used no longer differentiate in an obvious way between short and long columns. Instead, maximum permissible stress values are obtained through a series of expressions that must be selected, indirectly, depending on the actual slenderness ratio of the column. For typical steel columns, the elastic critical buckling stress fe is determined using the Euler expression fe = p2E>1Le >r2 2. The slenderness ratio Le >r is generally limited to 200 or less. End conditions are reflected through the k value; hence, Le = k L. The stress value fe is then compared with the yield strength fy of the material. For fe Ú 0.44 fy, the critical buckling stress is found using the expression fcr = [0.6581fy>fe2]fy. For all other cases, the expression fy = 0.877 fe is used. The nominal compressive strength Pn can then be obtained by multiplying the critical buckling stress fc with the cross-sectional area of the steel shape. In some cases, steel columns can be shaped to address buckling issues, providing larger cross sections in the middle and smaller ones toward the supports (Figure 12). examPle An unbraced pin-ended steel W 10 * 19 column 20 ft long carries a dead load of 15 kip and a live load of 25 kip. Assume A-992 steel with fy = 50 ksi, rx = 4.38 in., ry = 1.37 in., and A = 8.84 in.2 Is the member safe to carry the loads? solution: Load calculations: LRFD: Loads must be factored in order to incorporate safety factors : ASD:

P = 1.2 PD + 1.6 PL = 1.2 115 kip2 + 1.6 125 kip2 = 58 kip P = 15 kip + 25 kip = 40 kip

Members in Compression: Columns

FIGure 12 Columns can be shaped to address issues of buckling. Larger sections at the mid-point provide better stiffness, while the tapering toward the supports makes the element appear visually lighter.

Slenderness ratio: kL>r = 11.02120 * 122 >1.37 = 175. Because the column is unbraced, the minimum r is used and k = 1.0 because the column is pin-ended. The slenderness ratio of the column is smaller than 200. Elastic buckling stress: fe =

Copyright © 2013. Pearson Education Limited. All rights reserved.

Critical Buckling Stress:

3.142 * 29, 000 ksi p2E = = 9.33 ksi 2 1Le >r2 1752 9.33 ksi 6 10.442150 ksi2

fcr = 10.877219.33 ksi2 = 8.18 ksi

Nominal compressive strength:

PN = A18.18 ksi2 = 18.84 in.2 218.18 ksi2 = 72.3 kip

Check Load Capacity:

LRFD: Design Compressive Strength = 0.9 PN = 10.92172.3 kip2 = 65.07 kip 7 58 kip OK

ASD: Allowable Compressive Strength = PN >1.67 = 172.3 kip2 >1.67 = 43.3 kip 7 40 kip OK

The load capacity for both design methods is larger than the loads present, so the column is safe.

4.5

reinforced-concrete columns

Reinforced-concrete columns are difficult to analyze and design because of the composite nature of the material, the complex state of stresses as a result of axial and bending loading, and the compressive axial load that may lead to buckling.

Members in Compression: Columns Therefore, only general observations about such columns are made here. Two kinds of reinforced-concrete columns are of interest: those that are spirally reinforced and usually round, and those that are tied and usually rectangular in cross section. Both the spiral and the ties contain the longitudinal reinforcement and prevent separation and buckling of the reinforcement itself. Spirally reinforced columns have a more desirable behavior near failure and in lateral loads than those that are more simply tied, although the latter are usually cheaper and easier to construct. The different behaviors are reflected in the use of different f values and other factors in USD approaches used to size columns. ductility. To ensure ductile behavior, the longitudinal reinforcing steel of concrete columns should be not less than 1 percent and not more than 8 percent of the gross area of the cross section. In practice, however, an upper limit of 4 percent makes it easier to pour concrete because the reinforcing steel is less dense.

Copyright © 2013. Pearson Education Limited. All rights reserved.

strength. In an axially loaded column, the applied load is shared between the concrete and the steel, which are expected to deform the same amount because they are bonded together. The ultimate axial load-carrying capacity of a short column is normally taken to be Pult = f Pn, where Φ = 0.7 for columns with spiral reinforcement, and Φ = 0.65 for tied columns. These factors are lower than in beams, where the reduction factor is typically 0.9. This is because columns rely much more on the actual concrete strength than beams that rely equally on steel strength. Because concrete strength varies much more than that of industrially produced steel, the Φ factor for columns is smaller, thus adding an additional safety margin. Pn = 0.85 fc′Ac + Fy As where Ac is the area of the concrete and As is the area of the steel (a USD approach). The factor 0.85 reflects an uncertainty in the strength of concrete and the development of crushing zones. The expression for Pu does not reflect buckling phenomena or a bending moment. Reinforced-concrete columns are invariably subject to bending as well as axial loads, as a result of loading conditions and connections with other members. For a particular column, an interaction curve showing the relative capacity of the column for different axial load and moment combinations may be developed. (For a steel member, this curve would be a straight line because the material is linearly elastic.) Axial load capacity is highest when the applied moment is zero. It is interesting that moment capacity is highest when a small compressive load exists (because a smaller area of concrete is subject to tension). These curves depend on the precise layout of the cross section of a column. Tabular information provided by the American Concrete Institute or others is typically used to design columns in bending. long columns , buckling. The design of reinforced-concrete columns, as given in the preceding paragraph, is based on the strength of the materials. The design process must be modified to include design for buckling only if the reinforcedconcrete column is defined as a long column. Short columns are not expected to buckle. The criteria for determining whether a column is short or long include the slenderness ratio of the column, the type of its end supports, and its support against lateral movement. If elements in the structure other than the column can provide the lateral support, the column is braced against lateral movement. Such elements are shear walls, elevator shafts, stairwells, and so on. A column is unbraced if it and the other columns provide the lateral support for the building. Unbraced columns are more prone to buckling than braced columns, and this relationship is quantified by establishing the appropriate checks. The bending moments for long columns are amplified according to the applicable codes, and then the column is designed for strength criteria only.

Members in Compression: Columns

QuestIons 1. An unbraced pin-ended square steel column has cross-sectional dimensions of 1.5 in. * 1.5 in. and is 20 ft long. What is the critical buckling load for this column? Assume that Es = 29.6 * 106 lb>in.2 Answer: 2139 lb 2. An unbraced pin-ended steel column has rectangular cross-sectional dimensions of 3 in. * 2 in. and is 25 ft long. What is the critical buckling load for this column? Assume that E = 29.6 * 106 lb>in.2 3. A pin-ended cylindrical steel column has a diameter of 20 mm and a length of 5 m. What is the critical buckling load for this column? Assume that Es = 204, 000 N>mm2. Answer: 632.5 N 4. Assume that a pin-ended column of length L has a square cross section of dimensions d1 * d1 and has a critical buckling load of P1. What is the relative increase in loadcarrying capacity if the cross-sectional dimensions of the column are doubled? 5. An unbraced steel column of rectangular cross section 1.5 in. * 2 in. and pinned at each end is subjected to an axial force. Assume that Fy = 36, 000 lb>in.2 and E = 29.6 * 106 lb>in.2 Find the transition point between short- and long-column behavior. 6. Compare the relative load-carrying capacity of a square column and a round column of equal lengths, similar end conditions, and identical cross-sectional areas. Which can carry the greater load? What is the ratio of the critical buckling loads found? Explain your answer in qualitative terms. 7. A pin-ended column of length L is to be braced about one axis at quarter points (i.e., there are three equally spaced bracing points). There is no bracing about the other axis. If a rectangular section of b * d dimensions will be used, what is the most appropriate b>d ratio for maximizing the load-carrying capacity of the section? Assume that the dimension d corresponds to the strong axis of the section. 8. Using the same bracing and end conditions noted in Question 7, determine the most appropriate Ix >Iy ratio for a column having any cross-sectional shape. Also determine the most appropriate rx >ry ratio. Assume that Ix and rx correspond to the strong axis of the section.

9. A column is made from three equal pieces of timber, each measuring 2 in. * 6 in. in cross section. Both connections at the column ends are pinned. Please compare the critical buckling loads for the following two cases:

Copyright © 2013. Pearson Education Limited. All rights reserved.

Column A: The pieces of timber are glued together (shear resistant). The section acts as one structural section. Column B: The pieces of timber are not glued together. Individual pieces can slide with respect to adjacent pieces. The three pieces of timber work independently as three smaller sections.

Continuous Structures: Beams 1

IntroductIon

Copyright © 2013. Pearson Education Limited. All rights reserved.

This chapter explores the analysis and design of beams that are continuous over several supports or that have fixed ends (Figure 1). Members that have more than two points of support or multiple fixed ends are typically indeterminate structures. Mathematically, a statically indeterminate structure is a structure in which reactions, shears, and moments cannot be determined directly by applying the basic equations of statics ( g Fy = 0, g Fx = 0, and g Mo = 0). Usually, there are more unknowns than there are equations available for solution. Another way to view this class of structures—one that has more intrinsic importance from a design viewpoint—is that, in indeterminate structures, the values of all reactions, shears, and moments are dependent on the physical characteristics of the cross section and the specific material used in the structure (and any length variation in them), as well as on the span and loading. This dependency contrasts with what occurs in statically determinate structures, in which reactions, shears, and moments can be found through directly applying the basic equations of statics and, consequently, are independent of the shape of the cross section and the materials used (as well as any variations along the length of the member).

2

General PrIncIPles

Two indeterminate structures are illustrated in Figure 2. The first [Figure 2(a)] is a continuous beam over three simple supports. Reactive forces are developed at each support. The magnitudes of these reactive forces cannot be found by directly applying the basic equations of statics because there are more unknown forces (RA RB and RC) than there are independent equations ( g Fy = 0 and g Mo = 0) that can be used to solve for these unknown reactions. The same is true for the fixed-ended beam illustrated in Figure 2(e). In this case, four unknown reactions (RA RB M FA and M FB) and even fewer independent equations of statics ( g Fy = 0 and g M = 0) are available for use. Both of these beams are consequently termed statically indeterminate. An added layer of difficulty is involved in analyzing structures of this type. Despite added analytical difficulties, indeterminate beams are frequently used. One reason for this is that such structures tend to be more rigid under a given loading and span than are comparable determinate structures. Another reason for their extensive use is that the internal moments generated in the structure by the external From Chapter 8 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 1 Continuous versus simple beams and typical types of continuous beams.

loads are often smaller than in comparable determinate structures. Thus, member sizes can often be reduced. Smaller amounts of material also are more efficiently used. Disadvantages in structures of this type include their sensitivity to support settlements and thermal effects. Support settlements, for example, can cause undesirable bending moments to develop in continuous beams over several supports, while not necessarily affecting a comparable series of simply supported beams.

2.1

rigidity

The improved rigidity of beams having restrained ends or resting on several supports can be demonstrated through deflection studies. Using relevant techniques, for example, it is possible to determine that the midspan deflection for a simply supported beam carrying a concentrated load at midspan is given by ∆ = PL3 >48 EI. [See Figure 2(h).] If the ends of the same beam were restrained by changing the nature of the support conditions to create fixed ends, the deflection at midspan would be given by ∆ = PL3 >192EI. [See Figure 2(g).] Consequently, fixing the ends of the beam reduces maximum deflections by a factor of 4. This is an enormous difference. A careful look at the deflected shapes of the two members reveals the role of end fixity in increasing the rigidity of the structure under loadings. [See Figure 2(h).] Note that the deflection pattern associated with a fixed-ended beam can be obtained from that of a simply supported beam by applying a rotational moment at either end of the simply supported beam. The end of the beam can then be rotated until the tangent of the beam is exactly horizontal. Rotating the

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 2 Behavior of simply supported versus continuous beams.

ends of the beam would, of course, cause the center of the beam to rise upward, thus inducing the absolute midspan deflection of the member to decrease. The value of the moment corresponding to that needed to rotate the tangent of the beam to the horizontal corresponds to that naturally developed as a reaction—in this case, a fixed-ended moment—when the end of the member is initially restrained.

Continuous Structures: Beams A similar observation could be made concerning the differences in rigidity between a beam resting on multiple supports compared with a series of simply supported beams. The continuous structure is generally more rigid than its simply supported counterparts. In a design context, this means that the continuous beam is often favored.

Copyright © 2013. Pearson Education Limited. All rights reserved.

2.2

Force distributions

The second reason that continuous or fixed-ended beams are frequently preferred to simple beams is that the magnitudes of the internal shears and moments generated in the structure by the external force system are often smaller than those in simply supported beams. One way to visualize why this is true is to think in terms of effective, rather than actual, beam lengths. Referring to the fixed-ended beam shown in Figure 2(e), for example, note that the deflected shape of the beam between the two points of inflection (where the curvature of the beam changes sense) is similar to that of a simply supported beam like that in Figure 2(h). The portion of the fixed-ended beam that is between the two points of inflection, which also are points of zero moment, can consequently be considered simply supported. The distance between the two points of inflection is termed the effective length of the fixed-ended member. The magnitude of the bending moment developed in this segment of the beam depends on that length, as well as on the value of the loading. The shorter the effective length, the smaller is the magnitude of the bending moment present, and vice versa. In a simply supported beam, the effective length of the beam is identical to its actual length. In a fixed-ended beam of identical actual length, the effective length is considerably less than the actual length. Thus, the midspan moment is smaller in the fixed-ended beam than in a comparable simply supported beam. It is evident, however, that bending moments are present in parts of the fixed-ended beam—namely, at its ends—that are absent in the simply supported beam. This is evident in the deflected shapes of the two members and by recalling that moments and beam curvatures are related. Fixing the ends of the beam changes not only the magnitude of the moments present in the structure but also their distribution. Thinking in terms of effective lengths is useful in analyzing any type of indeterminate structure. By sketching the probable deflected shape of a structure, the location of points of inflection can be estimated. Effective beam lengths can then be estimated. In the beam over three supports in Figure 2(c), for example, points of inflection are developed as shown. The effective length of each span is less than the actual distance between support points. Consequently, moments in the middle portion of each span are expected to be less than those present in two comparable simply supported beams whose effective lengths are identical to their actual lengths. It is important to note that the points of inflection are induced by different types of loading. If compressive loads were applied to each end of the member over three supports illustrated in Figure 2(a), the member would not buckle into a shape having two points of inflection, as illustrated in Figure 2(c), but would rather assume the shape of an S.

3 3.1

analysIs oF IndetermInate Beams approximate Versus exact methods of analysis

The subsections that follow explore methods for determining reactions, shears, and moments in indeterminate structures. Over the years, several methods have been developed for analyzing indeterminate structures. Several elegant early techniques were based on least-work theorems, such as that developed by Alberto Castigliano in the nineteenth century. His approach was based on an analysis of the internal elastic energy stored in various parts of a structure under a loading. The internal work

Continuous Structures: Beams performed can be shown to be the least possible necessary to maintain equilibrium in supporting the loading. Deflection methods of analysis also were introduced, in which sufficient supports were first conceptually removed to make the structure determinate, and then critical deflections were calculated. Forces required to push the structure back into its original shape also were calculated. Another technique, called moment distribution, was introduced by Professor Hardy Cross in the 1930s. This technique, based on successive cycles of computation that drew nearer and nearer to exact results, was in wide use for many years. Most of these elegant approaches, however, have been replaced by matrix-displacement techniques or various forms of finite-element techniques. These newer approaches are particularly appropriate for use in a computer-based environment and have virtually replaced the older techniques noted here. Their black-box character, however, does little to help a student understand the characteristics of indeterminate structures and how better to design them. For this reason, the discussion that follows focuses more on approximate methods of analysis that convey a more physically based picture of how these complex structures work.

Copyright © 2013. Pearson Education Limited. All rights reserved.

3.2

approximate methods of analysis

While virtually all continuous-beam analyses are now done via computer, this section discusses a simplified method for determining the approximate values of reactions, shears, and moments in indeterminate structures as a way to understand basic structural behaviors. The method is based on the concept of effective member lengths discussed in the preceding section. The method involves sketching the deflected shape of the structure, noting the location of points of inflection, and then using the condition that bending moments are known to be zero at points of inflection as a device to reduce the number of unknown reactions. The accuracy of the method depends on the accuracy of the original sketch of the deflected shape of the structure. Experiments are useful. The method is a conceptual way to understand continuous beams, rather than a practical analysis tool. Consider the fixed-ended member illustrated in Figure 3(a). Through a careful sketch of the deflected shape, points of inflection can be estimated at the locations indicated. Because the internal moment is known to be zero at these points, the structure can be decomposed as illustrated in Figure 3(c). Each of these three pieces can now be treated as a statically determinate beam. The reactions for the center portion can be determined as indicated. These reactions then become loads on the cantilever elements shown. Vertical reactions and moments can then be found for those elements. Moment diagrams also can be drawn for each piece analyzed. The final moment diagram for the whole structure is a composite of individual diagrams. The maximum bending moment developed in the structure, M = wL2 >12, occurs not at midspan, but at the fixed ends. The midspan moment, M = wL2 >24, is only one-half of this value. The structure must be designed to carry these moments. A comparable simply supported beam has a midspan moment of M = wL2 >8. Consequently, fixing the ends of the member greatly reduces the maximum design moment compared with the simply supported case. A substantial savings in materials is thus possible because of these reduced design moments. It is interesting to observe that the sum of the absolute values of the positive and negative moments 1wL2 >12 + wL2 >242 is equal to wL2 >8 (the same moment present at midspan in a simply supported beam carrying a uniformly distributed load). The moments the structure must be designed to carry at the ends and middle, however, are still those illustrated in Figure 3(d) (wL2 >12 and wL2 >24, respectively), and it is these moments that the beam must be sized to carry. Figure 4 illustrates a similar analysis made for a beam with one end pinned and the other fixed. The approximate shape of the structure is sketched and the location of the point of inflection estimated. The structure is then decomposed into

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 3 Approximate analysis of a fixed-ended beam.

two determinate pieces, which are analyzed to determine the reactions, shears, and moments associated with them. The shear and moment diagrams for the whole structure are found by combining individual diagrams. Figure 5 illustrates an analysis of a continuous beam over three supports. By carefully sketching the deflected shape of the structure under the loading shown, the member can be seen to have two points of inflection. These points are symmetrically located approximately 0.25L from the center support. Next, the structure can be decomposed for analytical purposes in the manner indicated in Figure 5(c). Each

Continuous Structures: Beams

FIGure 4 Approximate analysis of a propped cantilever beam.

Copyright © 2013. Pearson Education Limited. All rights reserved.

piece can then be treated as a determinate beam and appropriate reactions, shears, and moments found. The shear and moment diagrams for the structure are found by combining individual diagrams. With practice, it is possible to sketch the moment diagrams of complex beams and, from the sketches, infer approximate values of moments.

3.3

computer-Based methods of analysis

As noted earlier, computer-based structural analysis techniques are used almost exclusively today to analyze indeterminate structures such as continuous beams. Most are based on the matrix-displacement or finite element techniques. Many analysis packages of this type are available today; some are free to academic users. Moreover, users must input the structural geometry to be analyzed. Users also must specify support or boundary conditions and specify real or default member types or properties. Live- and dead-loading conditions must be specified. Analysis results take the form of graphical and tabular displays of bending moments, shear forces, and torsional forces throughout the structure as well as accompanying stress states. Values may normally be obtained for any point in the structure. Shear and moment diagrams are readily obtained. Deflections and overall deflected shape patterns also are obtained (the latter are particularly useful in understanding a structure’s behavior). Another great advantage of a computer-based tool is that it allows different load combinations to be quickly explored.

Continuous Structures: Beams

FIGure 5 Approximate analysis of a beam continuous over three supports.

Note that member sizes cannot be left unspecified because analytical results in a statically indeterminate structure depend on what member sizes are initially chosen. If sizes are not known, default values can be used. Based on the results, improved size estimates can be made. Doing so also allows improved dead-load estimates. The analysis process is thus iterative.

Copyright © 2013. Pearson Education Limited. All rights reserved.

3.4

effects of Variations in member stiffness

The beginning of this chapter noted that one interpretation of why reactions, shears, and moments cannot be found directly by applying the basic equations of statics is that these values depend on the exact geometrical cross section of the member used, its material characteristics, and any variation in either along the length of the structure. The analyses in the previous section were based on the assumption that E and I of the member studied were constant along its length. To study the effects of a member not having a constant E and I, it is useful to first consider two extreme cases of the fixed-ended beam analyzed in the previous section. The extreme cases involve assuming dramatic differences in the cross-sectional properties at different points in the beam. With respect to Figure 6(a), assume that at the ends of the member, either E or I (or both), begin to be very small or approach zero. E and I are measures of the stiffness of the member. The modulus of elasticity, E, is a measure of stiffness as related to material. The smaller the value of E, the more flexible is the material (e.g., steel has a high E and rubber has a low E). The moment of inertia, I, is a measure of the amount and distribution of material at a cross section. The higher the I value, the stiffer is the section. The overall stiffness of a member at a cross section (i.e., its ability to resist deformations or rotations induced by the external load) depends on both parameters (E and I). If either or both are low at a cross section, the stiffness of the member is low at that cross section. When the whole member is loaded, the beam is more likely to undergo internal rotations at cross sections of lower stiffness than elsewhere. The capacity to resist rotation is then smaller. If E and I approach zero at certain locations, the structure begins behaving rotationally as if a pin connection were present. With respect to the

Continuous Structures: Beams

FIGure 6 Effects of variations in member stiffness.

Copyright © 2013. Pearson Education Limited. All rights reserved.

m.

fixed-ended beam, instead of points of inflection occurring at 0.2L from each end, as they did for a member with a constant E and I, the points of inflection occur at the locations of the drastically reduced E and I values. The deflected shape of the structure is influenced accordingly. By applying the same type of analysis as was done for the beam with a uniform cross section, the moment diagram illustrated in Figure 6(c) is obtained. (This is the same diagram as that for a simply supported member.)

Continuous Structures: Beams Alternatively, consider the behavior of the structure when it is assumed that the I or E value of the member becomes very small or approaches zero at the midpoint of the structure. Again, the structure’s capacity to resist rotation is reduced at this point, and the structure begins behaving as if a pin connection were present at that location. The resulting deflected shape and associated moment diagram are illustrated in Figures 6(e) and (f), respectively. Again, note that the total moment present is wL2 >8. Now, instead of extreme cases, consider the behavior of the structure if the ends are made stiffer than the central portion of the beam (often called haunching the member). In sketching the deflected shape of the member, it is evident that, by making the beam stiffer at the ends, the member is better able to resist rotation at these locations. The net effect causes the points of inflection to move more toward the center of the beam relative to their location in a beam with a uniform cross section throughout. The exact location of these points can only be estimated at this stage. The net effect, however, of the points moving inward is to increase the magnitude of the moment present at the ends of the member and to reduce that present at the central portion of the member. These examples generally illustrate why the internal moments in an indeterminate structure depend on the properties of the cross section and their variation along the length of the member. Note that the net effect of increasing the size of the member at a particular location (relative to other parts of the structure) was to increase the magnitude of the moment present at that point. Conversely, reducing the capacity of a member at a particular point resulted in a reduction in the moment present at the point. The moment was, so to speak, attracted by the stiffer portions of the member. This is a point of fundamental interest from a design viewpoint. In all the variations just discussed, the sum of the positive and negative moments was always wL2 >8. The appearance of a constant total is of interest. The effect of end fixity and variations in E and I was to alter the way this total moment was distributed and carried by the beam, but the total moment associated with a simply supported beam remained the same. The change in distribution, however, was advantageous because the actual moments the member should be sized to carry, of course, are the positive and negative moments themselves (each less than wL2 >8), which make up the total of wL2 >8.

Copyright © 2013. Pearson Education Limited. All rights reserved.

3.5

effects of support settlements

Continuous or fixed-ended beams are sensitive to support settlements, which can occur for a variety of reasons, the most common being consolidation of the soil beneath a support. The larger the load on the soil, the more likely consolidation is to occur. Rarely is the amount of settlement the same beneath all support points on a structure. If it were, the whole structure would translate vertically downward and cause no distress in the structure. The more usual case of differential settlement, however, can cause undesirable bending moments to develop in continuous and fixed-ended structures. Consider the fixed-ended beam illustrated in Figure 7. The beam is not loaded with any external force. Assume that a differential settlement of an amount Δ occurs. When one support simply translates with respect to the other, the fixed-ended nature of the support still restrains the end of the beam from rotating. The net effect is that curvatures develop in the beam. Associated with these curvatures are internal bending moments. The greater the differential settlement, the greater are the internal bending moments developed in the structure. This behavior contrasts with that of a simply supported beam. As differential settlement occurs under a simply supported beam, the member follows along. Because the ends of the member are unrestrained, the whole member rotates when settlement occurs. The differential settlement does not cause curvature, and hence bending moments, to develop in simply supported members. For this reason, simple support conditions are often preferable to rigid ones when problems with support settlement are anticipated.

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 7

Effects of support settlements.

Settlements also affect beams that are continuous over several supports. Consider the beam illustrated in Figure 7(c). If such differential settlements occurred, curvature and associated bending moments could be induced in the beam. If settlements occur when the beam is fully loaded, the support settlement causes a change in the moment distribution present in the structure. The general effect of the center support settling relative to the two end supports, for example, causes the points of inflection to move inward toward the center support. This movement, in turn, increases the effective lengths of the end spans, thus causing an increase in the

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 8 Cable-suspended beams. Top: Bowstring configuration. Bottom: Simple system resembling an inverted king-post truss.

positive moment present in the member. If the member is not sufficiently sized to carry the increased moment, the member could fail or become seriously overstressed.

3.6

cable-supported Beams

A variation of the continuous beam over several spans is the cable-supported beam. Here, a trusslike system of compression struts and cables creates the intermediate supports. The cables are designed such that a change in angle occurs at the end of each compression strut, thus generating a lifting force that creates compression in the strut, which in turn generates the intermediate reaction force for the beam element. Simple cable-supported beams superficially resemble the inverted king-post or queen-post truss. A major difference is that the beam element of cable-supported beams is continuous and not subdivided into pin-connected portions as is the case in a pure truss. Cable-supported beams are statically indeterminate. Bowstring configurations feature a polygonal cable that supports a series of compression struts, while simple cable-supported beams may have only a single strut (Figure 8). Cable-supported beams behave similarly to the continuous beams discussed in the previous section. The intermediate supports do not provide the same degree of resistance to loads that a rigid column or foundation would generate because the cable supports show displacements upon loading. The effect on the moment distribution is generally that positive moments increase while the magnitude of negative moments over the intermediate supports decreases—much as seen in the previous section and in Figure 7. Figure 9 shows the study of a cable-supported beam for a pedestrian bridge. As can be seen, the displacement of the intermediate support can be altered by modifying the cross section of the supporting cables. Depending on the stiffness of the intermediate support, the moment diagram takes on the characteristics of a continuous beam (very stiff support) or more that of a simply supported beam (lack of stiffness in the support). Other factors that impact the resistance of the intermediate compression strut to the loads of the beam include the depth of the strut and with it the angle between cable and strut. Deeper compression struts generally create a stiffer support condition and vice versa. Cable-supported beams are statically indeterminate, so varying cross sections will affect the distribution of bending moments. The cable and strut support system of any cable-supported beam creates an internal compression force in the beam element not unlike those familiar from normal trusses. Designers must pay attention to issues of local member buckling as compressive stresses from bending and those from internal compression induced by the cables combine. Lateral stability and buckling of the compression strut or struts also must be considered in the design process.

3.7

effects of Partial-loading conditions

One of the more interesting aspects of the behavior of indeterminate structures under load is illustrated at the right of Figure 10. A typical indeterminate beam over four supports is shown carrying three different sets of loads. The structure is portrayed under full-loading conditions in which all three spans of the structure are similarly loaded. Two different partial-loading conditions are illustrated in which one span is not loaded at all. These different conditions reflect different load patterns that might exist on the structure at one time or another. When different locations on the structure are considered (e.g., at midspan to the left and over the first support), it is evident that each loading condition produces a different moment at each of the respective locations. Inspecting the moments

Continuous Structures: Beams

FIGure 9 Continuous steel girder for a pedestrian bridge. Comparative study of different structural models and their effects on the inner forces and moments. In each, case loads are assumed to be uniformly distributed and self-weight is ignored. Pedestrian Bridge in Graz, Austria Architect: Domenig Eisenköck Structural Engineer: H. Egger Completed 1992

A continuous welded steel girder, triangular in cross section, is supported at midpoint by a system of compression strut and cables. The deck is a triangulated folded plate system. System Diagrams Alternative structural system models

C

(a)

Bending Moments

Axial Forces C C T

Copyright © 2013. Pearson Education Limited. All rights reserved.

Statically indeterminate, continuous beam with compression strut and large (very stiff) cables. The cables help to lift up the compression strut that provides intermediate support for the beam. (b)

T

Relatively large axial forces indicate that the load-carrying mechanism largely depends on the triangulation of the beam/cable/strut system. C

Reduced positive moments at the intermediate support. The dent in the moment diagram indicates that the strut/cable system is equivalent to a spring support.

C C

T

Same system as in (a), but the cables are much smaller in cross section and less stiff. Their ability to resist the downward force of the compression strut is reduced.

T

Smaller axial forces reflect the lack in stiffness of the cables. The load-carrying mechanism depends less on the triangulation than on bending of the beam. C

(c)

The smaller cable stretches and cannot effectively resist the downward deformation of the strut. The moment diagram resembles that of a simply supported beam.

C C

T

Statically determinate truss with pinned connection in the center, linking the two beam elements and the compression strut. This model studies the effect of a very small beam stiffness in the center.

T

The truss system shows the largest axial forces. The truss carries the loads primarily through axial compression and tension.

The beam elements are now the upper chords of a truss, and they show smaller bending moments between their pinnedend connections.

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 10 Shears and bending moments in common continuous structures for different loading conditions.

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 11 Critical loading conditions in continuous beams.

associated with the three loadings shown reveals some curious results. The maximum midspan positive moment in the first span occurs not when the structure is fully loaded but rather under a partial-loading condition. Nor does the maximum negative moment over the first support occur when the structure is fully loaded; it, too, occurs under a partial-loading condition. In no case does the maximum moment at a point result from a full-loading condition. A partial-loading condition always produces the critical design moment. Similarly, no single partialloading condition simultaneously produces the critical maximum moment at all locations. The curvature of the member, and hence the bending moment present, at a location is affected by loads placed anywhere on the structure. With respect to positive moments in the first span, for example, a load on span 1–2 causes a positive curvature and bending moment to develop. If span 3–4 also is loaded, the effect is to further increase the curvature and, consequently, the positive moment in the first span. A more critical moment is thus produced than when the first span alone is loaded. [See Figure 11(a).] Note that loading the middle span, however, causes the left span to rise upward. The effect of the rise is to decrease the curvature and bending moment in the first span. This loading condition thus produces a less critical moment than the partial-loading condition previously considered. Similar arguments could be made about other locations and loading conditions. Loadings that produce maximum moments of the type just discussed are called critical loading conditions. Several formal techniques can be used to establish which loading conditions are critical on a structure.

4 4.1

desIGn oF IndetermInate Beams Introduction

The process of designing a continuous beam is similar to designing a simple beam. Once the value of the maximum moment that can be present under any loading is known, determining the required member size at that point is straightforward. Principles concerning how to distribute material optimally at a cross section are similarly applicable. This section, therefore, addresses only general issues that are unique to the design of continuous beams.

Continuous Structures: Beams

4.2

design moments

Of particular importance in the design of continuous beams is the assurance that the member is sized to account for the moments associated with all possible loading conditions. The discussion in Section 3.6 is particularly relevant here. There, it was noted that the maximum moment that could be developed at any point in the structure rarely, if at all, resulted when the structure was fully loaded but typically occurred when the structure was only partially loaded. Note, however, that both maximum positive and maximum negative moments occur with a full load on the span under consideration. The question of the effects of partial loads relates primarily to loading on adjacent spans. From a design viewpoint, it is therefore necessary to consider all the possible variants of the loading that might exist on a structure and to determine the moments produced by each of these loadings at all points in the structure. Some loadings produce higher moments at certain locations than others. The size of the structure at any specific point is based on the critical loading, which produces the maximum possible moment at that point. The size of the structure at other points is based on maximum moments associated with other critical loads for those points. Consequently, under any one loading condition, the structure is somewhat oversized everywhere, except where the critical moment associated with that loading condition exists. In complex situations, it can be tedious to determine appropriate design moments for the different critical points in a structure. For commonly encountered situations, appropriate design moments have been tabulated. For two or more approximately equal spans (the larger of the two adjacent spans not exceeding the smaller by more than 20 percent) carrying uniformly distributed loads in which the live load does not exceed the dead load by more than a factor of 3, the shears and moments listed in Table 1 are reasonable for rough design purposes. More rigorous analyses must be made for unusual situations. While important, the discussion that follows does not further consider the effects of partial loadings in detail or the need to design a structure for a variety

taBle 1 Typical Design Moments Positive moment End spans If discontinuous end is unrestrained

Copyright © 2013. Pearson Education Limited. All rights reserved.

If discontinuous end is integral with the support Interior spans Negative moment at exterior face of first interior support Two spans More than two spans Negative moment at other faces of interior supports Negative moment at face of all supports for (a) slabs with spans not exceeding 10 ft and (b) beams and girders such that the ratio of the sum of column stiffness to beam stiffness exceeds 8 at each end of the span

1

>11vL2

1

>14vL2

1

>16 vL2

1

>9 vL2

1

>10 vL2

1

>11vL2

1

> 2 vL2

1

> 24 vL2

Negative moment at interior faces of exterior supports for members built integrally with their supports Where the support is a spandrel beam or girder Where the support is a column Shear in end members at first interior support Shear at all other supports

1

> 16 vL2 vL 1.15 2 vL 2

Continuous Structures: Beams of loading conditions. Rather, the discussion focuses on shaping the structure in response to its primary design loading condition. Quite often, the effects of partial loadings in structures used in buildings are taken into account after the structure is initially shaped for its primary loading.

4.3

shaping continuous Beams

The size of the cross section at a point of a continuous beam depends on the magnitude of the moment present at that point. There is wide variation in how moments are distributed in a continuous structure. The structure could be sized for the absolute maximum moment present anywhere in the structure and the same size simply used throughout [Figure 12(c)]. This is often done in building structures as a matter of convenience in construction. Alternatively, the size of the structure could be varied along its length in response to the moments present in the structure. Techniques for doing this have already been discussed in connection with simple beams.

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 12 Alternative strategies for designing a continuous structure.

Continuous Structures: Beams Shaping members is fairly frequently done in connection with highway bridge design when the possible material savings overshadow the added construction difficulties. The issues involved in shaping a beam along its length are relatively straightforward, with one major exception. Consider the continuous beam illustrated in Figure 12. If the member depth were designed to be dependent on the magnitude of the moment present at a point, and no deviations from this relation were allowed, a structure having roughly the configuration illustrated in Figure 12(d) would result. Clearly, where points of inflection exist, the structure has no moment and its depth consequently can approach zero (if shear forces are ignored). Actually doing this, of course, is absurd: The resultant structure is a configuration that is not stable under any loading other than the exact loading illustrated. Any slight deviation in the loading (which is, of course, bound to occur) would cause immediate collapse. A reasonable alternative, however, is to design the structure to reflect maximum positive and negative moment values and ignore the points of inflection. When this is done, and the structure is also designed for the shear forces present, a configuration of the type illustrated in Figure 12(e) results. This configuration is frequently used as a practical structure, particularly in steel highway bridges. Even if the effects of partial loadings are considered, the shape illustrated in Figure 12(e) is reasonably appropriate, unless the live-load–dead-load ratio is very high. Figure 13 shows several examples of shaped beam and truss structures. The exact shape of a member as it varies from points of maximum positive and negative moments depends on the choice of structure used. The exact optimum shape for a truss structure for a given variation in moment differs from the optimum shape for a solid rectangular beam. Other beam cross sections (e.g., wideflange shapes) would result in different variations of depth with moment. For this reason, the sketched shapes showing possible variations of depth with moment are for illustration purposes only.

4.4

use of Pinned Joints: Gerber Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

Because of construction difficulties, it is often difficult to make a long continuous member from one piece and it is desirable to introduce pinned or different types of construction joints. Figure 14 illustrates using construction joints in a beam that is continuous over three spans but is constructed from several pieces. To simplify making the joint, and thus the connection between discrete pieces that make up the

FIGure 13 Common shapes for continuous structures in which the depth of the structure is partly shaped to reflect the bending moment present.

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 14 Use of construction joints in continuous members. Construction joints often facilitate construction. Creating a condition of zero moment by design at points of inflection models the behavior of a continuous member by a series of statically determinate members.

total span, the joints are placed at or near points of inflection. Consequently, the joints need carry no moment and are sometimes designed as simple pinned connections. This type of connection is much easier to design and fabricate than are rigid joints that can carry moment. The beam that is illustrated shows four points of inflection. Pinned construction joints, however, cannot be placed at each point of inflection because that could result in a structure that would be unstable if the loading changed. Pinned construction joints can be used only when the resulting structural configuration is stable. For the beam illustrated in Figure 14(a), there are two basic ways of doing this. The

Continuous Structures: Beams

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 15 A typical Gerber beam with a pinned connection to control moment distributions.

&RQQHFWLRQORFDWLRQ

first is by putting joints in the end spans, leaving a center span with two cantilever ends. The structure would be built by putting this center span in place first and then adding the end pieces. Alternatively, the joints could be placed in the middle span, leaving two stable beams with single cantilevers. This structure would be built by first putting the two end members in place and then adding the center piece. In either case, the structure is converted into an assembly of statically determinate structures that function together in a way that reflects the behavior of a continuous member. The advantage of using pinned construction joints occurs when the behavior of a continuous member can be reflected exactly. Pinned construction joints are most effectively used when the design loading—in this case, a uniformly distributed load over the full structure—is expected to remain the primary loading. If loading conditions change from primary design loadings, these structures would still behave as assemblies of determinate structures but would not reflect the behavior of a continuous member and thus would not possess the implied advantages. For this reason, rigid, rather than pinned, construction joints are often used. The technique illustrated in Figure 14(e) is frequently coupled with a shaping of the structure in response to the extreme positive and negative moments that are present [Figure 14(f)]. Many large bridge structures reflect this general approach. In structures of this type, the dead load typically is the primary design load and far exceeds the variable live load in magnitude. These types of beams have long been in use. The 19th-century German engineer Heinrich Gerber extensively developed structures using internal pinned joints. Structures of this type have since been known as Gerber beams. They were initially used extensively in long-span timber construction where making fully continuous structures was difficult due to limited available member lengths. Today, they are used extensively in laminated timber construction. They can be employed in steel or reinforced-concrete systems but are not widely used. A typical steel Gerber beam is shown in Figure 15. Note that the primary beam is shaped according to bending moment variations. The pinned connection itself at the point of infection is a simple bolt and a bearing plate, but surrounding parts of the beam are specially strengthened and stiffened with welded plates. High shear forces are still transmitted through the connection and the steel webs could locally buckle if not strengthened. Note that in the beam itself over the support to the right, other stiffening plates are used, vertically over the reaction and horizontally where compression stresses are high. Often other pinned construction joints are initially used to facilitate construction by reducing member lengths and subsequently are converted to a fully rigid moment-carrying connection (e.g., by adding steel plates on top and bottom flanges in steel construction). This is done when the structure might potentially become unstable during different loading conditions if too many pins are used [Figure 14(c)]. The advantages of fully rigid connections are otherwise desired.

4.5

3LQQHGFRQQHFWLRQWKHEHDP IODQJHVDUHQRWFRQWLQXRXV

controlling moment distributions

The moments developed in a continuous and fixed-ended member can be affected significantly by the designer’s decisions. This can occur in several ways. One way is through paying careful attention to the spans and loadings involved. Often, by not using identical span lengths, the moment distribution is affected in an advantageous way. Moreover, using cantilevers on the ends of beams is frequently desirable, particularly for reducing positive moments on end spans. Deflections also are reduced by using end cantilevers. Another way to control moment distributions advantageously is by using construction joints. In the preceding section, using such joints at the locations of point

Continuous Structures: Beams of inflection was discussed. Pinned construction joints could be put anywhere and automatically cause a point of zero bending moment to develop at that point in the structure rather than where they would naturally occur. By locating such pin connections carefully, it is possible to reduce design moments. Consider the fixed-ended beam previously analyzed (Figure 3). The moments naturally developed are wL2 >12 at the beam ends and wL2 >24 at the beam midspan. Points of inflection occur naturally at 0 . 21L from each end. While the maximum design moment of wL2 >12 is considerably less than the wL2 >8 associated with a comparable simply supported span, it is possible to reduce the design moment to an even smaller value by inserting pin connections at points nearer the ends than where the inflection points naturally develop. Doing this increases the effective span of the midsection (increasing the positive moment) and decreases the effective length of the end cantilever portions (decreasing the negative moment). Indeed, the maximum positive moment can be made equal to the maximum negative moment by placing the pin connections at 0 . 15L. As before, the total moment (the absolute sum of maximum positive and negative moments) remains wL2 >8. By forcing the positive and negative moments to become equal, design moments are reduced to wL2 >16 for each. This maximum design moment is considerably less than that associated with the unaltered fixed-ended member. Member sizes are reduced. Inserting pins at locations other than inflection points does, however, have other consequences. One is that there is no longer a smooth curve between regions of positive and negative moments, as occurs at a natural point of inflection or when a pin is located exactly at such a point. Rather, a discontinuity is developed due to different end rotations of the midspan and cantilever portions of the member. If the magnitude of this discontinuity is large, it could cause problems if the beam is used in a building context. If a roof area is directly over the beam, for example, the sharp discontinuity could cause cracking, and hence leaking, in the roof. This is not a problem when such discontinuities are not present and the reversal in curvature is smooth. Another example of locating pins to control moments is illustrated in Figure  16(c). By carefully locating pins, positive and negative moments can be made approximately equal, thus minimizing the maximum design moment present. Again, discontinuities in curvatures occur at pin locations when the pins are moved from the natural points of inflection.

Copyright © 2013. Pearson Education Limited. All rights reserved.

4.6

continuous Beams made of reinforced concrete

Reinforced concrete is a particularly suitable medium with which to construct continuous beams. Continuity is achieved by how the reinforcement is arranged. Figure 17 illustrates the reinforcing pattern for a two-span continuous beam. Reinforcing steel is put into regions where tension stresses normally develop. The amount of steel used at a particular location depends on the magnitude of the moment that is present. Continuous beams can also be posttensioned or prestressed. Figure 17(d) illustrates how a posttensioning cable is draped to be effective for the type of moments that are present. Posttensioning is done frequently, but it is difficult to prestress a continuous beam. The nature of the prestressing process is such that getting reverse curvatures in the prestressing strands is not easy. Because prestressing is most normally (although not necessarily) done in a factory circumstance, transportation is an added difficulty (i.e., a beam to be used over three supports must be transported with similar support conditions to keep moments in undesirable locations from developing). Prestressed beams are more normally used under simple support conditions.

13.15 Steelpans

299

Copyright © 2013. Pearson Education Limited. All rights reserved.

around the edges by rapid hammer work known as piening, which prepares the note areas for tuning. The actual tuning process begins by lightly tapping the underside of the pan to create a small at to convex portion of the note area. The grooves between the note areas on the playing surface are retightened with a small hammer, so that when the player strikes the note area, the vibrations are restricted mainly to that note area. The fundamental frequency of a note area can be lowered by glancing blows across the top of the area, and it can be raised by increasing the height at the center or by tapping down at the corners of the area. A skilled pan-maker also tunes at least one overtone of each note area to a harmonic of the area s fundamental frequency. The rst overtone is nearly always the octave, and if the note area is large enough, a second overtone is tuned to the third or fourth harmonic. Tuning the third mode to the musical twelfth (third harmonic) gives the note a more mellow tone, and tuning it to the double octave (fourth harmonic) gives it a bright tone. Tuning the harmonics is more dif cult than tuning the fundamental frequency. In general, tapping the underside of the pan along or just outside a boundary of a note area that runs parallel to the nodal line for that mode will lower a harmonic, whereas tapping down or outward on the playing surface just inside this boundary will raise the harmonic frequency. The layout of the notes on a typical tenor pan is shown in Fig. 13.26. The outer ring has 12 more or less trapezoidal note areas tuned from D4 (294 Hz) to C#5 (554 Hz). The middle ring has 12 more or less elliptical note areas tuned from D5 to C#6 and the inner ring has four to six near-circular note areas tuned from D6 to F#6 . Note that the notes in the outer and middle rings are arranged in circles of fths: moving counterclockwise, one goes up a perfect fth or down a perfect fourth (which is equivalent to going up a fth and down an octave) on the musical scale. Each note in the middle ring is an octave higher than the corresponding note on the outside ring. The inside ring adds a third octave to ve notes (six notes when an optional F#6 is included). Thus, each note has several harmonics in common with its nearest neighbors, which leads to the strong interaction between notes characteristic of steel pans. (Another note arrangement, sometimes called the invaders layout, is also used for tenor pans by some pan-makers.) Holographic interferograms of the tenor pan driven at several frequencies that excite modes of vibration in the C5 note area (at 2 o clock, viewing the pan as a clock face)

FIGURE 13.26 Layout of note areas on a typical tenor pan.

300

Chapter 13 Percussion Instruments

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGURE 13.27 Vibrations of the tenor pan at frequencies that excite modes in the C5 note area: (a) 522 Hz; (b) 1050 Hz; (c) 1421 Hz; (d) 2064 Hz; (e) 2184 Hz (Rossing, Hansen, and Hampton 2000).

are shown in Fig. 13.27. Observe that several other note areas show appreciable vibration, especially at the higher frequencies. At the lowest frequency the active (roughly elliptical) portion of the note area moves in a single phase, while at the frequency of the second harmonic, there is a nodal line parallel to the rim dividing the note area into halves. At 1421 Hz (2 72 f 1 ) there is a radial node; at 2064 Hz (3 95 f 1 ) there are two nodal lines parallel to the rim, and at 2184 Hz there are two nodal lines perpendicular to each other. Besides ve modes of the C5 note area, many sympathetic vibrations of other note areas are apparent in Fig. 13.27. The various note areas can be identi ed by comparing Fig. 13.27 with Fig. 13.26. Holographic interferograms of several modes observed in the D4 , D5 , and D6 note areas of the same tenor pan are shown in Fig. 13.28. These are designated by m n , where m is the number of circumferential nodal lines and n is the number of radial nodal lines. 13.16

SOUND SPECTRA The sound spectra of steelpans are rich in harmonic overtones. These appear to have three different physical origins: 1. Radiation from higher modes of vibration of a given note area, tuned harmonically by the tuner;

13.16 Sound Spectra

301

FIGURE 13.28 Holographic interferograms of several modes in the D4 , D5 , and D6 note areas of the tenor pan. Modes are designated by m n , where m and n are the numbers of circumferential and radial nodes, respectively. Frequency ratios to the fundamentals are given (Rossing, Hansen, and Hampton 2000).

2. Radiation from nearby notes whose frequencies are harmonically related to the struck note; 3. Nonsinusoidal motion of the note area, vibrating at its fundamental frequency.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Sound spectra of two notes on a double-second steelpan are shown in Fig. 13.29. Note that harmonics as high as the ninth harmonic are detected. The rst three or four harmonics result from higher modes harmonically tuned, whereas the radiation from harmonically related nearby notes excited by sympathetic vibration could contribute to harmonics such

FIGURE 13.29 Sound spectra from two notes on a double-second steelpan.

302

Chapter 13 Percussion Instruments

as two, four, six, and possibly eight. The balance of the harmonics are attributed to nonsinusoidal motion of the note area. 13.17

BELLS AND CARILLONS

Copyright © 2013. Pearson Education Limited. All rights reserved.

Bells have been a part of nearly every culture in history. Bells existed in the Near East before 1000 B . C ., and a number of Chinese bells from the time of the Shang dynasty (1600 1100 B . C .) are found in museums throughout the world. A set of tuned bells from the fth century B . C . was recently discovered in the Chinese province of Hubei. Bells developed as Western musical instruments in the seventeenth century when bell founders discovered how to tune their partials harmonically. The founders in the Low Countries, especially the Hemony brothers (Franc‚ ois and Pieter) and Jacob van Eyck, took the lead in tuning bells, and many of their ne bells are found in carillons today. The carillon also developed in the Low Countries. Chiming bells by pulling ropes attached to the clappers had been practiced for some time before the idea of attaching these

FIGURE 13.30 The rst eight vibrational modes of a tuned bell. Dotted lines indicate the approximate locations of nodes. Frequencies relative to the strike note are given. The notes that correspond to these in a C4 bell are shown on a musical staff. In some bells only ve modes are tuned harmonically.

13.17 Bells and Carillons

303

Copyright © 2013. Pearson Education Limited. All rights reserved.

ropes to a keyboard or handclavier occurred to bell ringers in the sixteenth century. Many mechanical improvements during the seventeenth and eighteenth centuries, including the breached wire system and the addition of foot pedals for playing the larger bells, led to development of the modern carillon. Today, the term carillon is reserved for an instrument of 23 (two octaves) or more tuned bells played from a clavier (smaller sets are called chimes). The largest carillon in existence is the 74-bell (6-octave) carillon at Riverside Church in New York, whose largest bell is more than 18,000 kg (20 tons). Perhaps it is stretching the imagination a bit to think of a bell as being a plate, but the general principles of its vibrational behavior are similar. Although the mathematical description of the vibrations of a bell are understandably complex, the principal modes, at least, can be described by specifying the number of circular nodes and meridian nodes. The lowest mode of vibration, (called the hum tone), for example, has four meridian nodes, so that alternate quarters of the bell essentially move inward and outward. Figure 13.30 shows the principal vibrational modes for a carillon bell. The mode called the third is tuned a minor third above the strike note, whereas the upper third is usually a major third above the octave. The strike note is determined by the octave, the twelfth, and the upper octave, whose frequencies have the ratios 2 3 4, just as in chimes (see Section 13.5). Unlike chimes, however, carillon bells have a mode called the prime or fundamental with a frequency at or near the strike note. Careful studies have shown, however, that the pitch of the strike note is determined by the three modes mentioned above rather than by the prime. Vibrational frequencies of groups 0 X in a church bell with a D5 strike note are shown in Fig. 13.31. Horizontal lines on the axis at the right indicate relative strengths of several partials in the bell sound (these vary with location, of course). Arrows denote the three partials (octave or nominal, fth (twelfth), and upper octave) that determine the strike note.

FIGURE 13.31 Vibrational frequencies of groups 0 X in a D5 church bell. Also shown on the right are the relative strengths at impact of several partials in the bell sound. Arrows denote the three partials in group I that determine the strike note. (From Rossing and Perrin 1987.)

304

Chapter 13 Percussion Instruments

A new type of carillon bell has been developed at the Royal Eijsbouts Bellfoundry in The Netherlands. The new bell replaces the strong minor-third partial with a majorthird partial, thus changing the tonal character of the bell sound from minor to major. This requires an entirely new bell pro le. The new bell design evolved partly from the use of a technique for structural optimization using nite element methods on a digital computer. This technique allows a designer to make changes in the pro le of an existing structure, and then to compute the resulting changes in the vibrational modes (Lehr 1987). 13.18

HANDBELLS

Copyright © 2013. Pearson Education Limited. All rights reserved.

Handbells also date back at least several centuries B . C ., although tuned handbells of the present-day type were developed in England in the eighteenth century. One early use of handbells was to provide tower bellringers with a convenient means to practice change ringing. In more recent years, handbell choirs have become popular in schools and churches some 40,000 choirs are reported in the United States.

FIGURE 13.32 1984.)

Time-average hologram interferograms of inextensional modes in a C5 handbell. (From Rossing et al.

13.19 Summary

305

Although they are cast from the same bronze material and cover roughly the same range of pitch, the sounds of church bells, carillon bells, and handbells have distinctly different timbres. In a handbell, only two modes of vibration are tuned (although there are three harmonic partials in the sound), whereas in a church bell or carillon bell at least ve modes are tuned harmonically. A church bell or carillon bell is struck by a heavy metal clapper in order to radiate a sound that can be heard at a great distance, whereas the gentle sound of a handbell requires a relatively soft clapper. In the so-called English tuning of handbells, followed by most handbell makers in England and the United States, the 3 0 mode is tuned to three times the frequency of the 2 0 mode. The fundamental 2 0 mode radiates a rather strong second harmonic partial, however, so that the sound spectrum has prominent partials at the rst three harmonics (Rossing and Sathoff 1980). Some Dutch founders aim at tuning the 3 0 mode in handbells to 2.4 times the frequency of the fundamental, giving their handbell sound a minorthird character somewhat like a church bell. Such bells are usually thicker and heavier than bells with the English-type tuning. Hologram interferograms of a number of the modes are shown in Fig. 13.32. The bull s eyes locate the antinodes. Note that the upper half of the bell moves very little in the 7 1 mode; the same is true in m 1 modes when m 7.

Copyright © 2013. Pearson Education Limited. All rights reserved.

13.19

SUMMARY Percussion instruments have experienced a surge of interest, especially in contemporary music. Bars, membranes and plates vibrate in modes that are not harmonically related to the fundamental, and this gives percussion instruments a distinctive timbre. The lower bars of marimbas, xylophones, and vibes have arches cut on the underside in order to tune the rst overtone to a harmonic of the fundamental (fourth harmonic, in the case of the marimba and vibes; third harmonic, in the case of the xylophone). These three instruments have a tubular resonator placed under each bar to increase the loudness of the sound. In vibes, these resonators can be opened and closed rapidly to generate a type of vibrato. Chimes have no mode of vibration at the frequency of the strike note, which is a subjective tone. Drums have membranes of animal skin or synthetic material stretched over some type of air enclosure. The modes of vibration are shifted in frequency away from those of an ideal membrane by the air loading. Timpani convey a strong sense of pitch; the rst overtone is tuned to a fth above the principal tone. Other drums, such as the bass drum, have an inde nite pitch. The Indian tabla has several overtones that are harmonics of the fundamental. Examples of percussion instruments that are vibrating plates are cymbals, gongs, and tamtams. The sound spectrum of cymbals are characterized by a buildup and subsequent decay of sound in the range 3 to 10 kHz. The spectra of Caribbean steel drums are surprisingly rich in harmonic overtones. Tuned carillon bells and handbells may also be described as platelike. Carillon bells have ve to eight modes whose frequencies are harmonics of a fundamental. In a small handbell, only two modes are usually tuned harmonically.

306

Chapter 13 Percussion Instruments

REFERENCES AND SUGGESTED READINGS Bork, I., and J. Meyer (1982). Zur klanglichen Bewertung von Xylophonen, Das Musikinstrument 31(8): 1076. English translation in Percussive Notes 23(6): 48 (1985). Brindle, R. S. (1970). Contemporary Percussion. London: Oxford University Press. Chaigne, A., and V. Doutaut (1997). Numerical simulations of xylophones. I. Time-domain modeling of the vibrating bars. J. Acoust. Soc. Am. 101: 539. Fletcher, H., and I. G. Bassett (1978). Some Experiments with the Bass Drum, J. Acoust. Soc. Am. 64: 1570. Fletcher, N. H., and T. D. Rossing (1998). The Physics of Musical Instruments, 2nd ed. New York: Springer-Verlag. Chapters 18 20. Flugge, W. (1962). Statik und Dynamik der Schalen. New York: Springer-Verlag. Hardy, H. C., and J. F. Ancell (1963). Comparison of the Acoustical Performance of Calfskin and Plastic Drumheads, J. Acoust. Soc. Am. 33: 1391. Lehr, A. (1987). From Theory to Practice, Music Perception 4: 267. Raman, C. V. (1934). The Indian Musical Drums, Proc. Indian Acad. Sci. A1: 179. Lord Rayleigh (J. W. Strutt) (1894). The Theory of Sound, Vol. 1, 2nd ed. London: Macmillan. Reprinted by Dover, New York, 1945. Rossing, T. D. (1976). Acoustics of Percussion Instruments, Part I, Phys. Teach. 14: 546. Rossing, T. D. (1977). Acoustics of Percussion Instruments, Part II, Phys. Teach. 15: 278.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Rossing, T. D. (1982a). The Physics of Kettledrums, Sci. Am. 247(5): 172. Rossing, T. D. (1982b). Chladni s Law for Vibrating Plates, Am. J. Phys. 50: 271. Rossing, T. D. (1984a). The Acoustics of Bells. Stroudsburg, PA: Van Nostrand Reinhold. Rossing, T. D. (1984b). The Acoustics of Bells, Am. Scientist 72: 440.

Rossing, T. D. (1987). Acoustical Behavior of a Bass Drum, J. Acoust. Soc. Am. 82: S69. Rossing, T. D., I. Bork, H. Zhao, and D. Fystrom (1992). Acoustics of Snare Drums, J. Acoust. Soc. Am. 92: 84. Rossing, T. D., and N. H. Fletcher (1982). Acoustics of a Tamtam, Bull. Australian Acoust. Soc. 10(1): 21. Rossing, T. D., and N. H. Fletcher (1983). Nonlinear Vibrations in Plates and Gongs, J. Acoust. Soc. Am. 73: 345. Rossing, T. D., U. J. Hansen, and D. S. Hampton (1996). Music from Oil Drums: The Acoustics of the Steel Pan, Physics Today 49(3): 24. Rossing, T. D., U. J. Hansen, and D. S. Hampton (2000). Vibrational Mode Shapes in Caribbean Steelpans. I. Tenor and Double Second, J. Acoust. Soc. Am. 108: 803. Rossing, T. D., and G. Kvistad (1976). Acoustics of Timpani: Preliminary Studies, The Percussionist 13: 90. Rossing, T. D., and R. Perrin (1987). Vibrations of Bells, Applied Acoustics 20: 41. Rossing, T. D., R. Perrin, H. J. Sathoff, and R. W. Peterson (1984). Vibrational Modes of a Tuned Handbell, J. Acoust. Soc. Am. 76: 1263. Rossing, T. D., and R. W. Peterson (1982). Vibrations of Plates, Gongs, and Cymbals, Percussive Notes 19(3): 31. Rossing, T. D., and D. A. Russell (1990). Laboratory Observation of Elastic Waves in Solids, Am. J. Phys. 58: 1153. Rossing, T. D., and H. J. Sathoff (1980). Modes of Vibration and Sound Radiation from Tuned Handbells, J. Acoust. Soc. Am. 68: 1600. Rossing, T. D., and R. B. Shepherd (1983). Acoustics of Cymbals, Proc. 11th Intl. Congress on Acoustics (Paris): 329. Rossing, T. D., and W. A. Sykes (1982). Acoustics of Indian Drums, Percussive Notes 19(3): 58. Taylor, H. W. (1964). The Art and Science of the Timpani. London: Baker. Wilbur, C., and T. D. Rossing (1997). Subharmonic Generation in Cymbals at Large Amplitude, J. Acoust. Soc. Am. 101: 3144 (abstract).

GLOSSARY harmonics A series of partials with frequencies that are simple multiples of a fundamental frequency. (In a harmonic series, the rst harmonic would be the fundamental, the second harmonic the rst overtone.)

inharmonic partials Overtones or partials that are not harmonics of the fundamental. nodes Points or lines that do not move when a body vibrates in one of its modes.

Questions for Thought and Discussion overtones Upper partials or all components of a tone except the fundamental. radius of gyration A measure of the dif culty of rotating a body of a given mass. strike note The subjective tone that determines the pitch of a bell or chime; in most tuned bells it corresponds closely to one of the partials, but in chimes it does not. tension The force applied to the two ends of a string, or around the periphery of a membrane, that provides a restoring force during vibration. torsional mode An oscillatory motion that involves twisting of the vibrating member.

307

vibrato Frequency modulation (FM) that may or may not have amplitude modulation (AM) associated with it. Some musicians speak of intensity vibrato, pitch vibrato, and timbre vibrato as separate entities; others understand vibrato to include all three. Sometimes the term tremolo is used to describe AM, but this is not recommended because tremolo is used to describe other things, such as a rapid reiteration of a note or even a trill. Young’s modulus The ratio of stress to strain; also called modulus of elasticity.

REVIEW QUESTIONS 1. How much does the fundamental bending mode frequency change when the length of a bar is cut in half? 2. Where should a bar be supported in order to best excite its fundamental mode? 3. In a marimba bar, what is the frequency ratio of the rst overtone to the fundamental? 4. In a xylophone bar, what is the frequency ratio of the rst overtone to the fundamental? 5. Which instrument has motor-driven discs at the tops of the resonators? 6. Which modes determine the strike note of an orchestra chime (tubular bell)?

Copyright © 2013. Pearson Education Limited. All rights reserved.

7. Why is the 0 1 mode frequency in a kettledrum considerably higher than the corresponding mode in a membrane at the same tension without the kettle?

8. Which of the following drums produce a sound with a de nite pitch: timpani, bass drum, snare drum, conga, tabla? 9. Describe the motion of the two heads in the lowest vibrational mode of a snare drum. 10. Explain the delayed sound, or shimmer, in a cymbal. 11. How are the modes of a tabla tuned harmonically? 12. What is the difference between gongs and tamtams? 13. Arrange the following steelpans in order from highest to lowest frequency: cello, guitar, tenor, double tenor, bass, quadrophonics, double second. 14. What are three physical origins of harmonic overtones in steelpans? 15. What is the frequency ratio of the rst two vibrational modes in a tuned handbell? 16. Which modes in a carillon bell determine its strike note?

QUESTIONS FOR THOUGHT AND DISCUSSION 1. What is wrong with the statement The resonators of a marimba prolong the sound ? 2. The lowest mode of vibration of the triangle shown in Fig. 13.8 has four nodes. Make a sketch of the type in Figs. 13.2 or 13.3 to illustrate how it vibrates. 3. Explain why a microphone placed some distance above the center of a kettledrum picks up very little of the

principal tone (except, of course, by re ection from the walls, etc.). What modes would be picked up best by a microphone in such a location? 4. Compare the frequencies of the modes of the kettledrum given in Table 13.3 with those given by Rayleigh (Section 13.8).

308

Chapter 13 Percussion Instruments

EXERCISES 1. Compare the ratios of the frequencies of transverse vibrations in the glockenspiel bar in Fig. 13.2 with the theoretical ratios for a thin rectangular bar given in Section 13.1. Can you account for the difference? (Hint: Compare Fig. 13.6.) 2. Write an expression for the frequency ( f 1 ) of the lowest mode of vibration of a rectangular bar in terms of its length L and the speed of sound L . 3. Write an expression for the frequency ratio of the lowest transverse mode in a bar to the lowest longitudinal mode. Find this ratio for a glockenspiel bar 21.4 cm long and 0.90 cm thick (K t 3 46). Compare this ratio with the ratio of the corresponding values given in Fig. 13.2. 4. What are the actual ratios of the numbers 81, 121, and 169 (discussed in Section 13.5)? How close are they to the ratios 2 3 4? 5. Using the formulas for longitudinal and transverse vibrations in a bar (Table 13.1), show that the lowest longitudinal and transverse modes will have the same frequency when L 9 4 K . Find the ratio of length to diameter for a bar (rod) of circular cross section having the

frequency for its lowest transverse and longitudinal vibrations. 6. Determine the frequencies of the (41), (51), and (61) modes of the bell in Fig. 13.32. Show that they are nearly in the ratios 2 3 4, and that they will produce a strike note (virtual pitch) corresponding to D5 (see Section 7.4). Is there a vibrational mode with this frequency? 7. In large bells, the (61), (71), (81), and (91) modes create a secondary strike note. Determine the frequencies of these modes in Fig. 13.31, and estimate their virtual pitch (see Section 7.4). What note on the scale is this nearest (see Table 9.2)? 8. If a bell does not have perfect circular symmetry (perfection is seldom achieved in practice), one or more modes of vibration will show doublet behavior: that is, there will be two modes of vibration with slightly different frequencies. Suppose that such a doublet exists, having components with frequencies of 440 and 442 Hz. At what rate will the bell warble ? How is it possible to minimize this warble?

EXPERIMENTS FOR HOME, LABORATORY, AND CLASSROOM DEMONSTRATION

Home and Classroom Demonstration

Copyright © 2013. Pearson Education Limited. All rights reserved.

1. Modes of a rectangular bar Sound the rst three modes of a steel or aluminum bar (Fig. 13.2 suggests where to hold it and where to strike it for each mode). Note the musical intervals between the modes (the ratio f 2 f 1 2 56, for example, is slightly less than an octave plus a fourth). Repeat with a bar of hardwood (preferably rosewood) and note the difference. 2. Modes of a tuned bar Sound the rst three modes of a tuned marimba, vibraphone, and/or xylophone bar from the lower part of the scale. Note the musical intervals between the modes. 3. Vibraphone resonators Demonstrate the difference in loudness and timbre of a vibraphone (or vibraharp) when the resonators are open and closed. Demonstrate the intensity vibrato obtained by rotating the resonator discs at different rates. 4. Chimes Locate and mark the points where holding a chime rmly will preferentially excite vibrational modes 2 through 6. (They are approximately at 0 5L, 0 36L, 0 28L, 0 22L, and 0 17L, where L is its length.) Then show that modes 4, 5, and 6 are nearly in a harmonic 2 3 4 fre-

quency ratio, with the audible strike note being one octave below mode 4 and two octaves below mode 6. 5. Kettledrum Contrast the difference in the tone obtained by striking a kettledrum at its center (which emphasizes the symmetric modes) and at the normal playing spot (about onefourth of the way from edge to center); see Fig. 13.11. 6. Kettledrum Stretch your ngers as far as possible to touch points on the nodal lines of the 2 1 mode (see Fig. 13.9). Show that the 2 1 mode is a fth above the fundamental 1 1 mode. 7. Kettledrum Show how the pedal, by changing the tension, raises the frequency of all modes nearly proportionately and maintains the tuning of the overtones quite well (but not exactly). 8. Cymbals Strike a cymbal with a wooden drum stick to show how the timbre (especially the way the aftersound develops) varies with strike point and the strength of the blow.

Experiments for Home, Laboratory, and Classroom Demonstration 9. Steelpans By touching a note area in different places as it is struck, show how the harmonic modes are shaped (see Fig. 13.28, for example). 10. Chinese opera gongs Small Chinese opera gongs show an interesting pitch-glide when struck (Fletcher and Rossing 1998, Section 20.3.1). The larger gong has a downward pitch-

309

glide, whereas the smaller one glides upward. 11. Handbells Several individual modes of vibration can be excited by touching the bell in the vicinity of a node (see Fig. 13.32) and striking it in the vicinity of an antinode. Note the harmonic tuning of the 2 1 and 3 1 modes. 12. Chladni patterns See Demonstration 8 in Chapter 2.

Laboratory Experiments

Copyright © 2013. Pearson Education Limited. All rights reserved.

Acoustics of mallet instruments (Experiment 20 in Acoustics Laboratory Experiments)

Vibrations of membranes and drums (Experiment 19 in Acoustics Laboratory Experiments)

CHAPTER

14

Keyboard Instruments

This chapter features three different types of keyboard instruments. In the piano, strings are set into vibration by striking them with hammers; in the harpsichord, the strings are plucked; and in the pipe organ, sound is produced by blowing air into tuned pipes. In this chapter you should learn: About piano action and how the hammers interact with the strings; How the piano soundboard vibrates and radiates sound; About clavichords and harpsichords; About pipe organs.

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.1

FIGURE 14.1 A simplified diagram of the piano. When a key is depressed, the damper is raised, and the hammer is thrown against the string. Vibrations of the string are transmitted to the soundboard by the bridge.

CONSTRUCTION OF THE PIANO Invented in 1709 by Bartolomeo Cristofori in Florence, the piano has become the most popular and versatile of all instruments. It has a range of more than seven octaves (A0 to C8 ) and a wide dynamic range. Pianos vary in size from small home uprights or spinets to large concert grands. The main parts of the piano are the keyboard, the action, the strings, the soundboard, and the frame. A simplified diagram of a piano is shown in Fig. 14.1. More than 200 strings extend from the pin block or wrest plank across the bridge to the hitch-pin rail at the far end. When a key is depressed, the damper is raised, and the hammer is “thrown” against the string, setting it into vibration. Vibrations of the string are transmitted to the soundboard by the bridge. A typical concert grand piano has 243 strings, varying in length from about 2 m at the bass end to about 5 cm at the treble end. Included are 8 single strings wrapped with one or two layers of wire, 5 pairs of strings also wrapped, 7 sets of three wrapped strings, and 68 sets of three unwrapped strings. Smaller pianos may have fewer strings but they play the

From Chapter 14 of The Science of Sound, Third Edition. Thomas D. Rossing, Richard F. Moore, Paul A. Wheeler. Copyright © 2002 by Pearson Education, Inc. All rights reserved.

14.1 Construction of the Piano

311

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGURE 14.2 The top view of a studio grand piano showing the cast-iron frame, the overlapping strings, hammers, and dampers. The cutaway portion at the treble end shows the tuning pin block. (Courtesy of Baldwin Piano and Organ Co.)

same number of notes: 88. A small grand piano with 226 strings is shown in Fig. 14.2. Note that the bass strings overlap the middle strings, which allows them to act nearer the center of the soundboard. The acoustical advantages of wrapped strings and multiple strings for most notes will be discussed in Section 14.2. The soundboardis nearly always made of spruce, about 1 cm thick, with its grain running the length of the piano. Ribs on the underside of the soundboard stiffen it in the cross-grain direction. The soundboard is the main source of radiated sound, just as is the top plate of a violin or cello. To obtain the desired loudness, piano strings are held at high tensions which may exceed 1000 N (220 lb). The total force of all the strings in a concert grand is over 20 tons! In order to withstand this force and maintain stability of tuning, grand pianos have sturdy frames of cast iron.

312

Chapter 14 Keyboard Instruments

The rather complicated action of a grand piano is shown in Fig. 14.3. When a key is pressed down, the damper is raised. At the same time, the capstan sets the whippen into rotation around its pivot. The rotating whippen causes the jack to push against the hammer, knuckle or roller, starting the hammer on its journey toward the string. Just before the hammer strikes the string, the lower end of the jack strikes the jack regulator and rotates away from the knuckle. The freely rotating hammer now strikes the string, rebounds immediately (in order not to damp the string), and falls back to the repetition level. The back check prevents the hammer from bouncing back to strike the string a second time. When the key is lifted slightly, a spring pulls the jack back under the knuckle, so that pressing the key a second time repeats the note. Most upright pianos do not have this feature; thus a note cannot be repeated unless the key returns to its starting position.

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGURE 14.3 The action of a grand piano.

The upright piano developed about the middle of the nineteenth century. The nearly rectangular soundboard and the strings in upright pianos are vertical; the hammers travel horizontally. Thus the action is different from that of the grand piano in Fig. 14.3. In fullsize upright pianos, which stand 130 to 150 cm (4 12 to 5 ft) in height, the striking mechanism or action is located some distance above the keys and connected to them mechanically by stickers. In studio uprights or console pianos, which stand about 100 to 130 cm (3 14 to 4 14 ft) in height, the action is mounted directly over the keys without stickers. In small spinet pianos (less than 100 cm in height), the action is below the keys and drop stickers transmit key motion to the action. Pianos may have two or three pedals. The right pedal is called the sustainingpedal. It raises all the dampers, which allows the struck strings to continue vibrating after the keys are released. The left pedal is some type of expression pedal. In most grand pianos, it shifts the entire action sideways, causing the treble hammers to strike only two of their three strings. This shifting type of pedal is called the una cordapedal. In vertical pianos, and a few grands, the left pedal is a soft pedal, which moves the hammers closer to the strings, decreasing their travel and thus their striking force.

313

14.2 Piano Strings

Many pianos have a third pedal. On most grands and a few uprights, the center pedal is a sostenutopedal, which sustains only those notes which are depressed prior to depressing the pedal, and does not sustain subsequent notes. On a few pianos, the center pedal is a bass sustaining pedal, which lifts only the bass dampers. On a few uprights, the center pedal is a practicepedal, which lowers a piece of felt between the hammers and the strings, muffling the tone. 14.2

PIANO STRINGS The strings are the heart of the piano. They convert some of the kinetic energy of the moving hammers into vibrational energy and pass it on to the bridges and soundboard in a manner that determines the sound quality of the piano. Piano strings make use of high-strength steel wire. Efficiency of sound production calls for the highest string tension possible, while at the same time minimizing inharmonicity calls for using the smallest string diameter (core diameter in a wrapped string) possible. This results in tensile stresses of around 1000 N/mm2 , which is about half the yield strength of steel wire. For steel with an elastic modulus of 2 1011 N/m2 , this results in an elongation of about 12 % when the string is under tension. Fortunately, when strings break, it is usually near the keyboard end, so that the broken strings recoil away from the pianist. An ideal string vibrates in a series of modes that are harmonics of a fundamental (see Section 4.3). Actual strings have some stiffness, which provides a restoring force (in addition to the tension), slightly raising the frequency of all the modes. The additional restoring force is greater in the case of the higher modes because the string makes more bends. Thus the modes are spread apart in frequency and are no longer exact harmonics of a fundamental. In other words, a real string with stiffness is partly stringlike and partly barlike. The inharmonicity of strings (i.e., the amount by which the actual mode frequencies differ from a harmonic series) is found to vary with the square of the partial number (Fletcher 1964). Thus the second harmonic is shifted four times as much as the fundamental. The formula may be written

Copyright © 2013. Pearson Education Limited. All rights reserved.

fn

n f1 1

n2

1 A

where f n the frequency of the nth harmonic and f 1 fundamental. For a solid wire without wrapping, A

the frequency of the

3r 4 E

8T L2

where r the radius of the string, E Young’s modulus, T the tension, and L the length of the string. Thus, the inharmonicity is smallest for long, thin wires under great tension (large L and T , small r ). Because the stiffness of a string increases sharply with its radius (the factor A in the box above increases as r 4 ), inharmonicity is especially noticeable in the case of the large bass

314

Chapter 14 Keyboard Instruments

Copyright © 2013. Pearson Education Limited. All rights reserved.

strings. Because wrapped strings are more flexible than are solid strings of the same diameter, the inharmonicity of the bass strings is reduced substantially by the use of wrapped rather than solid strings of the same weight. (The lower strings on a guitar and violin are wrapped rather than solid, for the same reason.) A small amount of inharmonicity of string partials is considered desirable in pianos. One study of synthesized piano sounds demonstrated the preference of listeners, both musicians and nonmusicians, for tones with inharmonic partials. Tones synthesized with harmonic partials were described as “lacking warmth” and sounding much less like piano tones than those synthesized from slightly inharmonic partials (Fletcher, Blackham, and Stratton 1962). Inharmonicity may also help disguise small tuning errors in the same way that vibrato serves the players of other instruments. The inharmonicity in strings is the main reason why pianos are “stretch-tuned” (see Section 9.6). If they were not, the upper partials of a note would be slightly sharp with respect to the notes in the upper octaves to which they correspond, and undesirable beats would result when chords are played. Because pianos are usually tuned by minimizing beats between notes, a stretched scale will automatically be the result. Figure 14.4 shows the deviations from equal temperament that resulted when a spinet piano was aurally tuned by a fine-tuner at a piano factory. Also shown is the Railsback stretch , which is an average from 16 different pianos measured by O. L. Railsback. The aural tuning results follow the Railsback curve generally, but with a few deviations that are probably attributable to soundboard resonances. A jury of listeners showed little preference between tuning done electronically according to the Railsback curve and tuning done aurally by the skilled tuner (Martin and Ward 1961). A large grand piano with long bass strings will show less stretch in the bass than does the small spinet piano in Fig. 14.4. However, the stretch in the upper octaves will not be much different, since the upper strings are not appreciably different in a spinet and a concert grand.

FIGURE 14.4 Deviations from equal temperament in a small piano. (From Martin and Ward 1961. Reprinted by permission of the Amer. Inst. of Physics.)

14.4 Hammer-String Interaction

14.3

315

THE TUNING OF UNISONS Over most of its playing range, the piano has three strings for each note. Studies have shown that the best piano sound results from tuning these strings one to two cents different from each other (Kirk 1959). If the strings are tuned to exactly the same frequency, the transfer of energy from the strings to the soundboard takes place rapidly, and the decay time of the sound is too small. If the unison strings are tuned too far apart, prominent beats are heard, and what we commonly call a “barroom piano” sound is the result. When the unison strings are tuned to be a few cents different, the decay curve takes on two different slopes. The hammer sets all three strings into vibration with the same phase, and energy is rapidly transferred to the soundboard initially. Since there are small differences in their frequencies, however, the strings soon get out of phase, and the rate of sound decay slows down, leading to a second slope in the decay curve (aftersound). If the frequencies differ by 2 cents, for example, about 400 vibrations would be required for the strings to fall out of phase. The actual coupling between the strings depends, in a somewhat complicated way, on the mechanical properties of the bridge (Weinreich 1977), and the uneven decay of coupled strings is an important characteristic of piano sound. Fine tuning of the unisons is a means of regulating the amount of aftersound. Hammer irregularities can also affect the aftersound, and a skilled piano tuner can probably compensate for these to some extent by adjusting the unisons.

14.4

HAMMER-STRING INTERACTION

Copyright © 2013. Pearson Education Limited. All rights reserved.

The dynamics of the hammer-string interaction has been the subject of considerable research, beginning with Helmholz (1877). The problem drew the attention of a number of Indian researchers, including Nobel laureate C. V. Raman, in the 1920s and 1930s, and it has recently been taken up by Hall (1986, 1987). When the hammer has less mass than the string, it will most likely be thrown clear of the string by the first reflected pulse. The theoretical spectrum is missing harmonics

FIGURE 14.5 Hammer velocity and hammer-string contact time at various dynamic levels for the C4 note on a grand piano. (Ashkenfelt and Jansson 1988).

316

Chapter 14 Keyboard Instruments

numbered n (where is the fraction of the string length at which the hammer strikes). If the hammer mass is not too small, the spectrum envelope falls off as 1 n (6 dB/octave) above a certain mode number. A heavier hammer is less easily stopped and thrown back by the string. It may remain in contact with the string during the arrival of several reflected pulses. Analytical models of hammer behavior are virtually impossible to construct, but computer simulations can be of value. Hall (1987) has considered the cases of a hard, narrow hammer and a soft, narrow hammer. In the case of the hard hammer, the mode spectrum envelope takes on a slope of 6 dB/octave at high frequencies. For the treble strings, where the hammer mass exceeds the string mass, the spectrum envelope may have a slope as steep as 12 dB/octave at high frequency. Figure 14.5 shows a rather clear relationship between hammer-string contact time and hammer velocity at different dynamic levels. Striking the key with greater force increases the hammer velocity and decreases the contact time.

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.5

THE SOUNDBOARD A piano soundboard performs both structural and acoustical functions. Structurally, it opposes the vertical force components of the strings, which are in the range of 10 to 20 N per string, or a total of about 900 to 1800 N (200 to 400 lb). Acoustically, the soundboard is the main radiating member in the piano, transforming some of the mechanical energy of the strings and bridges into acoustical energy. Nearly all piano soundboards are made by gluing strips of spruce together and then adding ribs at right angles to the grain of the spruce. These ribs are designed to add enough cross-grain stiffness to equal the natural stiffness of the wood along the grain, which is typically about 20 times greater than across the grain. Laminated wood (plywood) soundboards have occasionally been used in low-cost pianos, but these tend to have lower acoustical efficiency and, particularly, less bass response than those of solid spruce. The unloaded soundboard is generally not a flat panel, but it has a crown of 1 to 2 mm on the side that holds the bridges. When the strings are brought up to tension, the downward force of the bridge causes a slight dip in the crown, and in old pianos the downward bridge force may have permanently distorted the soundboard. Soundboards are often tapered to be thicker near the center and thinner near the edges. Modern pianos generally have two bridges: a main or treble bridge and a shorter bass bridge. The bridges couple the strings to the soundboard. Piano bridges have an important effect on the tone production. By changing the bridge design, the piano designer can change the loudness, the duration, and the quality of the tone. The low-frequency vibrational modes of an upright-piano soundboard are shown in Fig. 14.6. In the lowest 0 0 mode, all vibrating parts of the soundboard move in phase, whereas in the 1 0 mode, a nodal line divides the soundboard into two parts moving in opposite phase. The 0 1 mode has a nodal line running lengthwise, and the 2 0 mode has two transverse nodal lines dividing the soundboard into thirds. The observed mode frequencies lie about midway between those calculated for fixed and simply supported (hinged) edges (Nakamura 1983).

14.7 The Clavichord

317

FIGURE 14.6 Chladni patterns of an upright-piano soundboard (Nakamura 1983).

Grand-piano soundboards have a somewhat more complicated modal structure, starting with the lowest mode at around 50 Hz. Mode shapes for both a 6-ft and a large 9-ft grand are shown in Fletcher and Rossing (1998), where there is further discussion of soundboard design. The best source of piano design, including soundboards, is a series of three papers by Harold Conklin (1996a, b, c).

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.6

PIANO SOUND Production of sound by a piano is a rather complicated process. The strings are set into vibration by the hammer, and they in turn act on the bridge to set the soundboard into vibration. Vibrational waves travel in many directions on the soundboard, and this leads to a complicated pattern of sound radiation. Eventually, the various components of the sound, which come from different parts of the soundboard, reach our ears; we process them and identify them as coming from a piano. The various partials in a piano sound build up quite rapidly (typically in 3 ms) and decay at widely different rates. Thus the sound spectrum of the piano is constantly changing with time, as in the case of the percussion instruments, which we discussed in Chapter 13. Many of the clues we use to identify a sound as coming from a piano are contained in the initial interval. This can be demonstrated by playing backward a tape recording of a piano. Now the sounds, which still have the same overall spectrum as before, build up slowly and decay rapidly, and they are more suggestive of a reed organ than a piano (Demo. 29, Houtsma et al. 1987). Figure 14.7 shows how the spectrum changes over the wide range of the piano (Fletcher, Blackham, and Stratton 1962). In the lowest octave, as many as 45 partials can be detected; at the highest notes, only two or three partials can be detected.

14.7

THE CLAVICHORD The clavichord, like the piano, depends on struck strings for its sound; but there the similarity ends. A clavichord is a portable keyboard instrument with a soft, delicate sound, well

318

Chapter 14 Keyboard Instruments

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGURE 14.7 The spectrum of high and low piano notes. (After Fletcher, Blackham, and Stratton 1962.)

suited for small living rooms but not for the concert hall. Its great virtue is its sensitivity; the tone can be varied in loudness and even given a vibrato by varying the force on the key. The action of a clavichord is shown in Fig. 14.8. A small square piece of brass, called a tangent, is attached to the end of each key. When a key is depressed, the tangent strikes a string (or pair of strings) and causes the portion between the tangent and the bridge to vibrate. A damper prevents the other portion from vibrating and also damps the vibration of the entire string when the tangent is released from the string. The tangent will normally oscillate up and down at a frequency of a few hertz (determined by the mass of the string, the key, and the player’s finger; the tension of the string; and the force applied to the key). This up-and-down motion varies the string tension and generates a vibrato. The playing range of the clavichord is typically four octaves, from C2 to C6 . The soundboard of a typical clavichord is a nearly rectangular slab of spruce 2 to 3 mm thick. The first two resonances, at about 140 to 330 Hz in one clavichord, are due to coupling between the fundamental modes of the soundboard and the enclosed air, as in a guitar or violin (Thwaites and Fletcher 1981). The next two resonances, around 470 and 570 Hz, are due to coupling between the second modes of the soundboard and the enclosed air. The

FIGURE 14.8 Clavichord action (simplified).

14.8 The Harpsichord

319

clavichord is found to have a fairly uniform sound output except in the lowest octave where it is understandably weak. 14.8

THE HARPSICHORD

Copyright © 2013. Pearson Education Limited. All rights reserved.

Another keyboard instrument that was very popular in the Baroque period is the harpsichord. Like a number of instruments from that period, it has experienced a revival of interest in recent years. Many modern harpsichords are patterned after the best of the early instruments, such as those built by the Ruckers family early in the seventeenth century. The action of a harpsichord is shown in Fig. 14.9. Pressing the key raises the jack so that the plectrum plucks the string. Plectra were originally carved from birds’ quills, but plastic plectra are being used in many modern instruments. The plectrum is attached to a hinged tongue, so that when the jack is lowered, the plectrum contacts the string only briefly before swinging back out of the way. Lowering the jack (by releasing the key) also causes a damper felt to touch the string and damp its vibrations. The construction of a harpsichord is not unlike that of a grand piano, except that everything is smaller and lighter. The soundboard of a harpsichord is typically 2.5 to 3 mm thick, compared to 10 mm in a typical piano. Strings have about one-third the diameter and about one-tenth the tension of the corresponding piano strings. The thinner strings of the harpsichord have much less inharmonicity (recall from Section 14.2 that inharmonicity is proportional to the fourth power of the radius), even though its bass strings are solid wire. The string dimensions of harpsichords are usually scaled in some regular fashion from bass to treble. In one modern harpsichord design, for example, the string lengths are proportional to 1 f over the top half of the range, with the rate of increase of length decreasing toward the bass end. The bass strings are brass, and their diameters are proportional to 1 f ; the treble strings are steel, and their diameter varies as 1 f 0 3 . The strings are plucked at a distinct from the nut, where is a fraction of the total length L given by f 0 6 at the treble end. At the lower end L 0 13. In general, the harmonic content of string vibration increases as the plucking point moves away from the center (Section 10.2), so that the harmonic development of the harpsichord increases in the lower notes (Fletcher 1977a). In order to vary the loudness and timbre, most harpsichords add one to three additional sets of strings to the original set. The selection of the strings to be played is made by means

FIGURE 14.9 Harpsichord action (simplified).

320

Chapter 14 Keyboard Instruments

FIGURE 14.10 A Flemish virginal built by Hans Ruckers, 1581. (Reprinted by permission. All rights reserved, The Metropolitan Museum of Art.)

of stops. Borrowing from organ terminology, we call a set of strings that play an octave higher a 4-ft set, and those that play an octave lower 16-ft; one or more sets of 8-ft strings play at the pitch of the key pressed. Some harpsichords add a second keyboard, and some of the largest instruments have a pedalboard as well. Couplers between the keyboards provide the harpsichordist with even more tonal options. Many harpsichords have a buff stopin which a piece of felt or soft leather is gently pressed against the string near its end to damp it and provide a short, distinctive sound. Sometimes a special set of jacks is provided to pluck the strings very close to the end in order to produce a nasal timbre (lute stop). The virginal or spinet is a small instrument like a harpsichord with strings running lengthwise in a rectangular case. The keyboard is built into one side of the case, making it a portable tabletop instrument. Virginals were particularly popular in Elizabethan England. An elaborately decorated virginal is shown in Fig. 14.10.

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.9

THE HARP Although it does not have a keyboard, the harp is related to the harpsichord and other string instruments described in this chapter, so a brief description of it is appropriate here. The modern harp, shown in Fig. 14.11, has 47 strings tuned to the notes of the diatonic scale, plus seven pedals by which each string can be raised or lowered a semitone in order to play the chromatics. With all the pedals in their middle position, the harp plays in the key of C major. Depressing the C-pedal raises all the C’s on the harp to C# . Figure 14.12 illustrates how the pedals raise and lower the pitch of the harp by means of two rotating discs attached to the neck of the instrument, which change the length of the vibrating part of the string. The strings are attached to a slanted sounding-board at the bottom of the instrument. The nominal range, from C1 to G7 , is almost as great as that of the piano. The most familiar sound of a harp is a sweeping glissando. The C-strings and F-strings are usually colored differently from the others in order to help the player locate the desired notes.

14.9 The Harp

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGURE 14.11 A harp.

FIGURE 14.12 Mechanism for tuning the harp. Depressing one of the tuning pedals increases the tension on all the strings with that note name by one (for naturals) or two (for sharps) units.

321

322

Chapter 14 Keyboard Instruments

(a)

(b)

FIGURE 14.13 Hazel Wright Organ in the Crystal Cathedral, Garden Grove, California. (a) The chancel organ contains the great organ, swell organ, solo organ, choir organ, positive organ, echo organ, trompetria organs, and pedal organ. (b) The south balcony organ consists of the gallery great organ, string organ, celestial organ, and gallery pedal organ.

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.10

THE PIPE ORGAN: ITS CONSTRUCTION The pipe organ has been called the “king of musical instruments.” No other instrument can match it in size, in range of tone, in loudness, or in complexity. The world’s largest organ, in the Convention Hall in Atlantic City, has more than 32,000 pipes of various sizes and shapes. No two pipe organs in the world are exactly alike. A large pipe organ in the Crystal Cathedral in Garden Grove, California is shown in Fig. 14.13. This organ, designed by organist Virgil Fox, has 273 ranks and nearly 16,000 pipes in several locations throughout the church. The modern pipe organ consists of a large variety of pipes arranged into divisions or organs. Each division is controlled by a separate keyboard or manual, including a pedalboard that is operated by the organist’s feet. The pipes in the swell division are usually enclosed behind a set of shutters that can be opened or closed (by means of the swell pedal) to change the loudness. The other divisions are not usually enclosed, and their loudness can only be changed by adding or subtracting pipes through the use of stops. The principal division of an organ is called the great organ, and it usually contains the most stops. Most organs have couplers that allow certain pipes in one division to be controlled from the manual of another. The windchests contain valves that can be opened to admit air into the pipes. The three main types of windchest actions used today are shown in Fig. 14.14. The oldest type is the tracker or mechanical action, in which the keyboards are connected to the chests by rather complicated mechanical linkages. With the second type, the direct electric, pressing the key energizes an electromagnet, which in turn opens the valve to allow air to flow into the pipe. The third type is called electropneumatic , because electromagnets are used to exhaust air from the bellows, which open the valve into the pipe. Full pneumatic actions, in which the key controls an air valve, are becoming rare.

323

14.11 Organ Pipes

FIGURE 14.14 Windchest actions: (1) tracker; (b) direct electric; (c) electropneumatic.

(a)

(b)

(c)

Organ pipes are organized into ranks of similar pipes. One rank of pipes will include one pipe for each note (61 in the divisions operated from a keyboard, 32 in the pedal division). Each stop on the organ usually corresponds to one rank of pipes, except in the case of mixture stops, which involve several ranks. On smaller organs, a rank of pipes may be included in more than one stop. 14.11

ORGAN PIPES

Copyright © 2013. Pearson Education Limited. All rights reserved.

There are two basic types of organ pipes: ßue(labial) pipes and reed(lingual) pipes. Flue pipes produce sound by means of a vibrating air jet, in a manner similar to the flute and the recorder (see Section 12.14). Reed pipes use a vibrating brass reed to modulate the air stream.

FIGURE 14.15 Organ pipes: (a) open flue pipe of metal; (b) stopped wood flue pipe; (c) reed pipe.

(a)

(b)

(c)

324

Chapter 14 Keyboard Instruments

The essential parts of a metal flue pipe are shown in Fig. 14.15(a). The air jet passes through a flue or windway, a narrow opening between the languid and the lower lip. Sound is produced when the air jet encounters the upper lip and oscillates back and forth, sometimes blowing into the pipe, sometimes blowing out through the mouth of the pipe. The large flue pipes usually have ears on either side of the mouth to guide the air jet. Sound Generation in Flue Organ Pipes

Copyright © 2013. Pearson Education Limited. All rights reserved.

When air is admitted at the foot of a flue organ pipe, it flows upward and forms a sheetlike jet as it emerges from the flue (see Fig. 14.15(a)). The jet flows across the mouth of the pipe and strikes the upper lip, where it interacts both with the lip itself and with the air in the pipe resonator. The physics of this interaction is discussed in Chapter 16 of Fletcher and Rossing (1998). Jets can be described as either laminar (arranged in layers or streamlines) or turbulent(characterized by eddies or vortices). Jets in organ pipes are almost completely turbulent, which makes it quite difficult to describe them mathematically. Somewhat paradoxically, however, a fully turbulent jet in an organ pipe is more stable than a laminar one, and organ builders go to some lengths to ensure fully turbulent jets by cutting fine nicks along the edge of the languid, for example. The air flowing during the attack transient in an organ pipe, although quite complicated, can be studied by means of Schlieren photographs. Figure 14.16 shows the

(a)

(b)

(c)

(d)

(e)

(f)

FIGURE 14.16 Schlieren photographs showing air flow at intervals of 0.1 s during the attack when a jet strikes a sharp edge such as the labium of an organ pipe. (Photos furnished by A. Hirschberg.)

14.11 Organ Pipes

325

jet at intervals of 0.1 s as the jet first strikes the edge, forming vortices. Eventually, this settles down to an acoustically driven jet oscillator.

Copyright © 2013. Pearson Education Limited. All rights reserved.

The stopped wooden flue pipe, shown in Fig. 14.15(b), uses a similar mechanism to produce sound. Since the resonator is now a closed pipe, it need be only half as long as an open pipe in order to sound the same pitch. Wooden pipes usually have a square cross section, and produce a sound with a flutelike quality. There are three families of flue pipes: (1) diapasons, (2) flutes, and (3) strings. Pipes of the flute family usually have the least overtone content, whereas the bright-sounding strings have the most. String pipes are generally slender cylinders, whereas diapasons are open cylinders of somewhat greater diameter. Flute pipes come in several sizes and shapes, are constructed of either wood or metal, and may be open or closed. Closed pipes sound mainly the odd-numbered harmonics of the fundamental. The reed pipe, shown in Fig. 14.15(c), has a vibrating reed or tongue, which modulates the flow of air passing through the shallot into the resonator. The reed is pressed against the open side of the shallot by a wire that can be adjusted up and down to tune the vibrating reed. The resonator is usually tuned to about the same frequency as the reed. Reed pipes are rich in harmonics. If the resonator is cylindrical, the odd-numbered harmonics will be favored, as in a clarinet. Conical resonators, however, will reinforce all harmonics, both odd and even. Figure 14.17 shows several different pipes and the names given to the corresponding organ stops. Resonators of several different shapes are shown, including open and closed cylinders, cones, inverted cones, rectangular cylinders, and closed pipes with chimneys. (The chimney is tuned to boost one of the upper harmonics, around the fifth or sixth.)

FIGURE 14.17 Organ pipes of various families with the names of the corresponding stops.

326

Chapter 14 Keyboard Instruments

Pipe Resonators Figure 14.18 compares the first four resonance frequencies of open and closed cylindrical, conical, and reverse conical pipes that can be used as organ pipe resonators. Note that the resonance frequencies of cones are essentially the same as those of an open cylinder, although the pressure amplitudes are quite different, as shown in Fig. 12.7. This is not the case with truncated cones and reverse cones, however, where the resonance frequencies are inharmonic except for the closed pipe r 2 r 1 1 .

FIGURE 14.18 First four resonance frequencies of open and closed cylindrical and conical pipes.

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.12

PIPE SCALING The scaleof a rank of pipes refers to the ratio of diameter to length for the pipe of lowest pitch. Large-scale (large-diameter) pipes tend to have a dominant fundamental and fewer harmonics, whereas small-scale pipes have more harmonics. The reason for this is related to a frequency dependence of the end correction of a pipe. For an open cylindrical pipe, the end correction at the open end is approximately 0.6 times the radius (see Section 4.5), whereas at the mouth of a flue pipe, one adds approximately 2.7 times the radius to calculate the effective length (Strong and Plitnik 1983). The end correction decreases with frequency, so the pipe effectively shortens for the higher partials; thus the pipe resonances are slightly less than an octave apart. However, the spectrum of the sound source (the oscillating jet) has exact harmonics. Thus in a large-scale pipe, only the first few harmonics in the source are reinforced by the natural frequencies of the pipe. For pipes of small scale, such as the strings, the end correction is much smaller to start with; thus the pipe resonances more nearly match the harmonics of the oscillating jet. Each rank of pipes is usually graduated in diameter according to a fixed relationship for that rank. For large-scale pipes, the diameter may reduce to one-half at the seventeenth note of the scale; for small-scale pipes, the diameter will reduce at a slower rate than this.

327

14.13 Sound Radiation From Flue Pipes TABLE 14.1 Pitch and harmonic number of pipe ranks Equivalent length (in feet)

32

8

4

2 23

2

1 35

1 13

1 17

1

1

2

3

4

5

6

7

8

C1

C2

C3

G3

C4

E4

G4

C

G

b1

C5

CC

g1

B4

CCC

c1

33

65

131

196

261

330

392

466

523

16

Harmonic number Lowest note

C0

“Organ” notation Lowest pitch (in Hz)

16

e1

c2

Normally, the mouth width, lip cut-up, and width of the flue opening follow the same scale as the pipe diameter. The mouth width, for example, may vary from 0.6 times the pipe diameter (large-scale flute) to 0.8 times the diameter (string or diapason). Similarly, the cut-up may vary from 0.25 times the mouth width (diapason) to 0.4 times the width (flute); the flue width is typically about 0.03 times the mouth width (Fletcher 1977b). The pitch of a rank of pipes is expressed as the approximate length of an open pipe having the same pitch as the lowest pipe in that rank. Thus the lowest pipe of an eight-foot 65 4 Hz), and this is sounded by the lowest key on the manual. The rank sounds C2 f octave from C4 (middle C) to C5 then occupies the same central position that it does on the piano keyboard. A four-foot stop plays an octave higher, and a sixteen-foot stop an octave lower. Table 14.1 lists the lowest note and the harmonic number of various ranks. Mixturesare generally ranks of diapason-scaled pipes designed to augment the normal 8-ft, 4-ft, and 2-ft diapason chorus. They generally have from 3 to as many as 10 ranks, all properly balanced and coupled together. In normal mixtures, only octaves and fifths are used, but the selection of pitches concentrates much of the sound energy over a broad band between 500 and 6000 Hz.

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.13

SOUND RADIATION FROM FLUE PIPES Because a stopped pipe radiates only from its mouth, the radiation pattern (at least for the fundamental and the lower harmonics, where the mouth is small compared to the wavelength) is nearly isotropic. For the higher harmonics, when the wavelength becomes less than a few times the width of the mouth, the radiation pattern becomes more concentrated

FIGURE 14.19 Calculated radiation patterns for three harmonics of an open flue pipe standing vertically (Fletcher and Rossing 1998).

328

Chapter 14 Keyboard Instruments

in front of the pipe, with a large angular spread in the vertical plane and a smaller spread in the horizontal plane (see Chapter 17 in Fletcher and Rossing 1998). An open pipe has two coherent sources, at the mouth and open end, which are in phase for odd harmonics and out of phase for even harmonics. Even though the mouth area is smaller than the area of the open end, the two sources have the same strength, because the power radiated depends only on the total acoustic flow and not on the aperture size, as has been confirmed experimentally by Coltman (1969). Typical radiation patterns for the first three harmonics of an open flue pipe are shown in Fig. 14.19. 14.14

REED PIPES In reed (lingual) pipes, a curved tongue (the reed) closes against a matching cavity, called the shallot, to modulate the flow of air into the resonator. The vibrating length of the reed is determined by a stiff wire (Stimmkr¨ucke) pressing it against the shallot. Resonators can take on a number of different shapes, including open and closed cylinders and cones. The pipe voicer generally adjusts the reed and the resonator to produce the best sound and then tunes the pipe by adjusting the tuning wire. Without the resonator attached, the reed frequency varies smoothly with reed length, as would be expected, but attaching the resonator lowers the frequency of the reed and also produces distinct regimes of sound, separated by discontinuities, as the system jumps from one regime to the next one (Rossing et al. 1998). Reed pipes tend to sound a rather large number of harmonics, which gives them a bright sound, and so they are often used as solo stops on an organ. A model of air flow in reed pipes has been developed by Hirschberg et al. (1990). Free-reed organ pipes were introduced at the end of the eighteenth century. In contrast to the more common striking reed pipes, the reed does not beat against the shallot but swings freely through a perforated oblong plate of brass similar to those found in reed organs or accordions. Free-reed pipes are characterized by a slower rise time than striking-reed pipes and a “mellow, round” sound (Braasch and Ahrens 1999).

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.15

TUNING AND VOICING ORGAN PIPES There are several ways of tuning different types of organ pipes. Reed pipes are tuned by moving the tuning wire up and down, thus allowing a longer or shorter length of the reed to vibrate. Flue pipes, on the other hand, are tuned by changing the effective length of the pipe in some way. In the case of a closed pipe, this is accomplished by moving the stopper up or down to change the pipe length. Many open pipes have a tuning sleeve that slides up and down; others have an adjustable slot near the open end. Open pipes without such a tuning device can be tuned by the use of tuning cones. To raise the pitch, the apex of the cone is inserted and tapped to widen the end of the pipe slightly. To lower the pitch, the cup end of the cone is placed over the pipe and tapped to close the end slightly. One of the most critical of all the steps in organ building is voicing the pipes, which means making coarse and fine adjustments in the various parts of the pipe so that it “speaks” properly. Much of the voicing can be done in the organ-builder’s shop, but the final voicing or finishing is done after the organ is installed, and it takes into account the

14.15 Tuning and Voicing Organ Pipes

329

Copyright © 2013. Pearson Education Limited. All rights reserved.

acoustics of the room, as well as the acoustics of the organ. One of the objectives of voicing is to achieve a uniformity of loudness and timbre within each rank of pipes. Another is to adjust the initial transient or attack of each pipe. The attack time varies substantially from one rank of pipes to another and also from large to small pipes within a rank. In general, the number of cycles necessary to build the tone to its steady state remains about the same within a given rank, so the attack time doubles in going one octave down the scale. Reed pipes have shorter attack times than do flue pipes of the same pitch. In many pipes, the attack time is shorter for the upper harmonics than for the fundamental, so the initial sound is higher in harmonic content than the steady state. (Figure 7.15 shows a diapason pipe in which the second harmonic builds up rapidly.) Octave or mixture ranks are often coupled to slow-speaking pedal flue pipes to decrease the attack time. Also important during the attacks are certain characteristic sounds, such as chiff. The main parameters that are adjusted during voicing of flue pipes are (Mercer 1951): 1. The size of the foot bore. An increase in bore diameter increases the air flow and produces greater power and greater harmonic content. 2. The condition of the bore. Rounding the edges of the bore can change the tone. 3. Nicking.Nicks on the languid or the lower lip tend to remove chiff and other extraneous sounds during the attack. 4. The width of the ßue. This affects both the attack and steady state in a way that is not well understood. 5. Obstruction near the mouth. A bar, or “roller beard,” placed between the ears is sometimes necessary to prevent a pipe of high pressure or small scale from jumping up an octave in pitch. 6. The height of the languid. This controls mainly the articulation or attack; lowering the languid speeds up the attack. 7. The height of the mouth, or cut-up. Increasing the cut-up reduces the harmonic content of the tone. 8. Setting of the upper lip. Moving the upper lip farther in increases the attack time. 9. The condition of the upper lip. Bevelling the upper lip increases the harmonic content of the tone, whereas rounding it decreases harmonic content. These voicing adjustments tend to be interrelated. For example, increasing the cut-up reduces the harmonic content, whereas increasing the bore diameter produces greater harmonic development and more power. Thus, by combining the two, it is possible to obtain approximately the same tone at increased power. Laboratory studies (Nolle 1979; Fletcher and Douglas 1980) of the effects of mouth height (cut-up) and mouth position (which relates to adjustments 6 and 8) show that: 1. A high cut-up causes the harmonic content to decrease, the ratio of high to low harmonics to increase, and the overall level to decrease. 2. A low cut-up causes the harmonic content to increase, the ratio of high to low harmonics to decrease, and parasitic oscillations due to the edgetone mechanism (see Section 30.4).

330

Chapter 14 Keyboard Instruments

3. Within the optimum operating range, slightly increasing the mouth height favors a “ping” transient, whereas decreasing the height favors a “chiff” transient. 4. A frequency jump to a higher mode is encouraged by a high cut-up or an inward mouth position. 5. Adding ears to a pipe of small scale (diameter to length ratio) increases the range of mouth height and position over which stable oscillation occurs. Voicing of reed pipes involves adjustments to the reed as well as to the resonator. One of the most important adjustments is in the curvature of the reed. A curved reed does not cut off the flow of air as abruptly as a straight reed does. Also, since tuning is done at the reed, changes in the resonator of a reed pipe will mainly adjust the loudness and timbre of the pipe independently of the pitch. Many organs have one rank of pipes that is deliberately tuned sharp so that it will produce beats when played with another similar rank. The stop associated with this is called the voix celeste . One or more windchests on an organ are usually equipped with a mechanical device called a tremolo or tremulant, which causes the air pressure to fluctuate at a regular rate and thus to produce a vibrato.

Copyright © 2013. Pearson Education Limited. All rights reserved.

14.16

SUMMARY The piano, with a playing range of over seven octaves and a wide dynamic range as well, has become the most versatile and popular of all musical instruments. The struck strings transfer vibrational energy through the bridge to the soundboard, which radiates most of the sound. Piano strings have “stretched” partials because of their stiffness; this makes stretch tuning desirable. Most notes on the piano have three strings, and their tuning in relation to each other affects the decay rate as well as the timbre of the note. The clavichord, which also uses struck strings, is a portable instrument with a soft, delicate tone. The strings are struck by tangents, which stay in contact with the strings after striking. The harpsichord, which plucks the strings, is experiencing a revival in popularity. It is strung to a much lower tension than the piano, and has a much lighter soundboard. Large instruments have several sets of strings or stops, which may play in different octaves. The strings of a harp are plucked by the player rather than by mechanical plectra (as in the harpsichord). Harp strings are tuned to the diatonic scale; to play sharps or flats, the pitch of a string is raised or lowered a semitone by pedals. The pipe organis the largest of all instruments and has an extremely wide range of tone and dynamics. Its many pipes are organized into ranks and divisions. The main families of pipes (in order of increasing harmonic content) are the flutes, diapasons, strings, and reeds. Flutes, diapasons, and strings use flue pipes with a sounding mechanism similar to that of the flute and the recorder. Reed pipes have a metal reed that modulates the air flow to produce sound. Organ pipe resonators may be cylindrical, conical, or rectangular, with open or closed ends, and made of metal or wood. The pitch of a rank of pipes is denoted by the length of an open pipe sounding the same pitch as the lowest pipe in the rank. Pitch designations range from 1 ft to 32 ft. Voicing of the individual pipes is a very important step in organ building; voicing determines both the initial attack and the steady-state timbre.

References and Suggested Readings

331

Copyright © 2013. Pearson Education Limited. All rights reserved.

REFERENCES AND SUGGESTED READINGS Ashkenfelt, A., and E. Jansson (1988). “From Touch to String Vibrations—the Initial Course of Piano Tone,” Report STL/QPSR 1/88, 31–109, Royal Inst. Technology (KTH), Stockholm. Blackham, E. D. (1965). “The Physics of the Piano,” Sci. Am. 99(6): 88. Braasch, J., and C. Ahrens (1999). “Attack Transients of Free Reed Organ Pipes,” J. Acoust. Soc. Am. 105: 1002 (abstract). Coltman, J. W. (1969). “Sound Radiation from the Mouth of an Organ Pipe,” J. Acoust. Soc. Am. 46: 477. Conklin, H. A., Jr. (1996a). “Design and Tone in the Mechanoacoustic Piano. Part I. Piano Hammers and Tonal Effects,” J. Acoust. Soc. Am. 99: 3286. Conklin, H. A., Jr. (1996b). “Design and Tone in the Mechanoacoustic Piano. Part II. Piano Structure,” J. Acoust. Soc. Am. 100: 695. Conklin, H. A., Jr. (1996c). “Design and Tone in the Mechanoacoustic Piano. Part III. Piano Strings and Scale Design,” J. Acoust. Soc. Am. 100: 1286. Fletcher, H. (1964). “Normal Vibration Frequencies of a Stiff Piano String,” J. Acoust. Soc. Am. 36: 203. Fletcher, H., E. D. Blackham, and R. Stratton (1962). “Quality of Piano Tones,” J. Acoust. Soc. Am. 34: 749. Fletcher, N. H. (1974). “Nonlinear Interactions in Organ Flue Pipes,” J. Acoust. Soc. Am. 56: 645. Fletcher, N. H. (1977a). “Analysis of the Design and Performance of Harpsichords,” Acustica37: 139. Fletcher, N. H. (1977b). “Scaling Rules for Organ Flue Pipe Ranks,” Acustica37: 133. Fletcher, N. H., and L. M. Douglas (1980). “Harmonic Generation in Organ Pipes, Recorders, and Flutes,” J. Acoust. Soc. Am.68: 767. Fletcher, N. H., and T. D. Rossing (1998). The Physics of Musical Instruments , 2nd ed. New York: Springer-Verlag. Chapters 11, 12, 16, and 17. Fletcher, N. H., and S. Thwaites (1983). “The Physics of Organ Pipes,” Sci. Am.248(1): 94. Hall, D. E. (1986). “Piano String Excitation in the Case of Small Hammer Mass,” J. Acoust. Soc. Am. 79: 141. Hall, D. E. (1987). “Piano String Excitation II: General Solution for a Hard Narrow Hammer, and III: General Solu-

tions for a Soft Narrow Hammer,” J. Acoust. Soc. Am. 81: 535 and 547. Helmholz, H. L. F. (1877). On the Sensations of Tone , 4th ed. Trans., A. J. Ellis. New York: Dover, 1954. Hirschberg, A., R. W. A. Van de Laar, J. P. Marrou-Mauri`eres, A. P. J. Wijnands, H. J. Dane, S. G. Kruijswijk, and A. J. M. Houtsma (1990). “A Quasi-stationary Model of Air Flow in the Reed Channel of Single-reed Woodwind Instruments,” Acustica70: 146. Houtsma, A. J. M., T. D. Rossing, and W. M. Wagenaars (1987). Auditory Demonstrations(Phillips Compact Disc #1126-061 and text). Woodbury, N.Y.: Acoustical Society of America. Kirk, R. E. (1959). “Tuning Preferences for Piano Unison Groups,” J. Acoust. Soc. Am. 31: 1644. Klotz, H. (1961). The Organ Handbook . St. Louis: Concordia. Martin, D. W., and W. D. Ward (1961). “Subjective Evaluation of Musical Scale Temperament in Pianos,” J. Acoust. Soc. Am.33: 582. Mercer, D. M. A. (1951). “The Voicing of Organ Flue Pipes,” J. Acoust. Soc. Am. 23: 45. Nakamura, I. (1983). “The Vibrational Character of the Piano Soundboard,” Proc. 11th ICA, Paris, 385. Nolle, A. W. (1979). “Some Voicing Adjustments of Flue Organ Pipes,” J. Acoust. Soc. Am. 66: 1612. Pollard, H. F. (1978). “Loudness of Pipe Organ Sounds,” I and II, Acustica41: 65 and 75. Rossing, T. D., J. Angster, and A. Mikl´os (1998). “Reed Vibration and Sound Generation in Lingual Organ Pipes,” Paper 2pMU6, 136th meeting, Acoust. Soc. Am., Norfolk, VA, Oct. 1998. Strong, W. J., and G. R. Plitnik (1983). Music, Speech and High Fidelity, 2nd ed. Provo, Utah: Soundprint. Chapters 35, 37, and 38. Thwaites, S., and N. H. Fletcher (1981). “Some Notes on the Clavichord,” J. Acoust. Soc. Am. 69: 1476. Weinreich, G. (1977). “Coupled Piano Strings,” J. Acoust. Soc. Am.62: 1474. Weinreich, G. (1979). “The Coupled Motions of Piano Strings,” Sci. Am.240(1): 118.

332

Chapter 14 Keyboard Instruments

GLOSSARY aftersound Second portion of a sound decay having a longer decay time. chiff A chirplike sound that occurs during attack, especially in flute pipes on an organ. clavichord A small portable instrument that produces a soft, delicate sound by means of struck strings. cut-up Height of the mouth opening in an organ pipe; distance from the lower lip to the upper lip. edgetone The sound produced when an air jet encounters a sharp edge or wedge and oscillates back and forth, first passing on one side, then the other. flue The narrow windway between the languid and lower lip in a flue pipe. flue pipe An organ pipe that produces sound by means of a jet of air passing through the flue and striking the upper lip. harpsichord A keyboard instrument in which strings are plucked by mechanical plectra. inharmonicity The departure of the frequencies of partials from those of a harmonic series. jack A device in piano and harpsichord actions that moves up and down when a key is pressed. In the piano, the jack sets the hammer in motion; in the harpsichord, it carries the plectrum. labial A flue pipe. laminar flow Fluid flow in which entire layers have the same velocity. languid A plate that partially blocks an organ pipe and forms one side of the flue.

lingual A reed pipe. plectrum The small tongue of quill, leather, or plastic that plucks the string of a harpsichord. soundboard The wooden plate that radiates much of the sound in string instruments. stretch tuning Tuning octaves slightly larger than a 2 ratio.

1

sustaining pedal Right hand pedal of a piano, which raises all the dampers, allowing the strings to continue vibrating after the keys are released. tangent The small metal square that strikes the string of a clavichord. tremolo, tremulant A device on an organ that produces a vibrato, usually by varying the air pressure. turbulent flow Fluid flow characterized by eddies and vortices; the flow velocity tends to vary randomly. una corda pedal Pedal on grand pianos which shifts the entire action sideways, causing the treble hammers to strike only two of the three unison strings. virginal A small plucked string instrument in which the strings run parallel to the keyboard. voicing Adjusting organ pipes to have the desired sound. voix celeste An organ stop that uses two ranks of pipes with slightly different tunings so that they produce beats. windchest The important part of the organ that distributes air to selected pipes to make them sound.

Copyright © 2013. Pearson Education Limited. All rights reserved.

REVIEW QUESTIONS 1. What is the approximate range of the piano (in octaves)? 2. Approximately how many strings does a grand piano have? 3. How does the contact time of a piano hammer on a string change with increasing hammer speed? 4. Why are the bass strings of a piano wrapped with wire? 5. What is the main reason for stretch-tuning a piano? 6. What is the effect on string decay time of having three strings for a single note? 7. What is the approximate thickness of a piano soundboard? 8. What mechanical action takes place when the sustaining pedal is depressed? 9. What is the function of a piano bridge?

10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

How are the strings in a clavichord excited? How are the strings in a harpsichord excited? What is a virginal? What is the purpose of the pedals on a concert harp? How many pipes does the world’s largest pipe organ have? What is a tracker organ? What is a lingual organ pipe? Is the jet in an organ pipe generally laminar or turbulent? What is meant by pipe scaling? Compare the sound power radiated by the mouth and the open end of an organ pipe. Which generally have more harmonics: flue pipes or reed pipes?

Experiments for Home, Laboratory, and Classroom Demonstration

333

QUESTIONS FOR THOUGHT AND DISCUSSION 1. When a chord is played on a piano while the sustaining pedal is depressed, the tone sounds richer than the same chord played without the sustaining pedal. Can you explain why? 2. If a piano key is depressed slowly enough, the hammer fails to contact the string at all. Explain why. 3. Will the two notes shown in Fig. 14.20 sound exactly the same when played on a piano? 4. Can you think of any advantages a tracker action might have over a direct electric action in an organ? any disadvantages? 5. In small pipe organs, the sixteen-foot pipes are almost always stopped wooden pipes. Explain why.

(a)

(b) FIGURE 14.20

EXERCISES

Copyright © 2013. Pearson Education Limited. All rights reserved.

1. Suppose that all the strings of a piano were of the same material and also had the same diameter and tension. If the longest string (A0 ) were 2 m in length, how long would the highest A-string (A7 ) have to be? Is this practical? 2. Show that two unison strings, tuned 2 cents (0.12%) different and initially in phase, will fall out of phase after about 400 vibrations. 3. If a particular string on a harpsichord has one-third the diameter and one-tenth the tension of the corresponding string on a piano, which string will be longer? What will the ratio of lengths be? (See Section 3.2 and 4.3; assume that both strings are steel.) 4. If the diameter of pipes in a particular rank is reduced by a factor of two every seventeen notes, show that the

diameter reduces by four over a range of about three octaves. 5. A piano tuner finds that two of the strings tuned to C4 give about one beat per second when sounded together. What is the ratio of their frequencies? Show that their pitches differ by about 7 cents. (One cent is 1 100 of a semitone and corresponds to a frequency ratio of approximately 1.0006.) 6. By noting the weak partials in the C1 spectrum in Fig. 14.7, estimate the fraction of the string length at which the hammer strikes. 7. Calculate the acoustical lengths of open organ pipes tuned to C0 , C1 , and C2 (frequencies are given in Table 9.2, also Table 14.1). Compare these to the equivalent lengths in Table 14.1.

EXPERIMENTS FOR HOME, LABORATORY, AND CLASSROOM DEMONSTRATION

Home and Classroom Demonstration 1. Mechanical model of piano actionMechanical models of piano actions can sometimes be obtained from piano dealers or manufacturers. 2. The pedals Depress each piano pedal to see and hear what it does. 3. Trichord Damp two of the three strings of one note (piano tuners wedges work the best) and determine the change in the sound.

4. Piano sound reversedDemonstration 29 on the Auditory Demonstrations CD. Piano tones, heard backwards, do not sound like piano tones, even though the spectrum remains unchanged, because the sound of hammer on string comes at the end of a note rather than at the beginning. 5. Organ pipes From an organ builder, obtain as many different organ pipes as possible. These can generally be sounded by blowing on them. Note the different sounds.

334

Chapter 14 Keyboard Instruments

6. Wide scale versus narrow scaleCompare pipes of wide scale (open flute), medium scale (principal or diapason), and narrow scale (string) having the same pitch.

7. Closed and open pipesCompare the sound spectrum of an open flute pipe to a closed flute pipe of the same pitch.

Laboratory Experiment

Copyright © 2013. Pearson Education Limited. All rights reserved.

Acoustics of organ pipes (Experiment 21 in Acoustics Laboratory Experiments )

PART

IV The Human Voice

Copyright © 2013. Pearson Education Limited. All rights reserved.

It is difÞcult to overstate the importance of the human voice. Of all the members of the animal kingdom, we alone have the power of articulate speech. Speech is our chief means of communication. In addition, the human voice is our oldest musical instrument. The human voice and the human ear are very well matched to each other. The ear has its maximum sensitivity in the frequency range from 1000 to 4000 Hz, and that is the range of frequency in which the resonances of the vocal tract occur. These resonances (called formants), we will learn, are the acoustical bases for all the vowel sounds and many of the consonants in speech and singing. Because speech and singing are such closely related functions of the human voice, it is recommended that all three chapters (15Ð17) be considered together. It is possible for a singer to begin with Chapter 17, but frequent reference will probably need to be made to Chapters 15 and 16.

From Chapter 15 of The Science of Sound, Third Edition. Thomas D. Rossing, Richard F. Moore, Paul A. Wheeler. Copyright © 2002 by Pearson Education, Inc. All rights reserved.

Copyright © 2013. Pearson Education Limited. All rights reserved.

This page intentionally left blank

CHAPTER

15

Speech Production

Throughout human history, the principal mode of communication has been the spoken word. The systems in the human body that send and receive oral messages are sophisticated in design and complex in function. Our understanding of both hearing and speech has progressed dramatically in recent years, largely because of new techniques for making acoustical as well as physiological measurements. The auditory system and its function were discussed in Chapter 5, and now it is appropriate to devote the same attention to the vocal organs. In this chapter you should learn: How the human vocal organ makes speech sounds; How speech sounds are the product of the source, the Þlter function, and the radiation efÞciency; About speech articulation by the different parts of the vocal tract; About formants as resonances of the vocal tract; How the glottis and the vocal tract are studied by speech acousticians. 15.1

THE VOCAL ORGANS

Copyright © 2013. Pearson Education Limited. All rights reserved.

The human vocal organs, as well as a representation of the main acoustical features, are shown in Fig. 15.1. The lungs serve as both a reservoir of air and an energy source. In

FIGURE 15.1 Human vocal organs and a representation of their main acoustical features. (After Flanagan 1965)

337

Continuous Structures: Rigid Frames

FIGure 21 Steel frame example. *LUD3URGXFWLRQ)DFLOLW\2IILFHV $UFKLWHFW,QJHQKRYHQ$UFKLWHNWHQ 6WUXFWXUDO(QJLQHHUV...,QJHQLHXUJHVHOOVFKDIWPE+:HUQHU6REHN,QJHQLHXUH &RPSOHWHG

6LPSOLILHGFURVVVHFWLRQ7KH VKDSHRIWKHH[WHUQDOVWHHOIUDPH VKRZVWKHGHHSHVWFURVVVHFWLRQ DWWKHFRUQHU

0RPHQWGLDJUDPIRUFRPELQHG JUDYLW\ORDGDQGODWHUDOORDGVIURP HLWKHUVLGH7KHVKDSHRIWKH IUDPHIROORZVWKHGLDJUDPFORVHO\

,PDJHFRXUWHV\*LUDZZZJLUDGH

7KH*LUDSURGXFWLRQDQGRIILFHEXLOGLQJFRQVLVWVRIWZRVHSDUDWHVWUXFWXUDOV\VWHPV$FRQFUHWHVWUXFWXUHIRUPVWKHEDVHPHQWDQGWKHULJLG IUDPHIRUWKHXSSHUIORRUZKLOHDVWHHOIUDPHIRUPVWKHVWUXFWXUDOHQYHORSHRIWKHEXLOGLQJ7KHVWHHOIUDPHLVDWZRKLQJHGIUDPHEXLOWXSIURP ZHOGHGSODWHVWHHO,WVSDQVPIW DQGLVDSSUR[LPDWHO\PIW KLJK)UDPHVDUHVSDFHGPIW DSDUW(DFKRIWKHWZR EXLOGLQJVFRQVLVWVRIVWUXFWXUDOED\V

Copyright © 2013. Pearson Education UK. All rights reserved.

7KHVKDSHRIWKHVWHHOIUDPHVLVGHULYHGIURPDVWXG\RIPRPHQWGLDJUDPV'LDJUDPVIRULQGLYLGXDOORDGFDVHVFDQEHDGGHGWRREWDLQ PRPHQWFXUYHVIRUFRPELQHGORDGFDVHV

6\VWHP DQGORDG GLDJUDPV

0RPHQW GLDJUDPV 8QLIRUPJUDYLW\ORDGV1HJDWLYH PRPHQWVLQWKHFRUQHUVSRVLWLYH PRPHQWDWPLGVSDQLQWKHEHDP HOHPHQW

8QLIRUPODWHUDOORDGV3RVLWLYH PRPHQWVLQWKHULJKWKDQG FRUQHUVXQWLOPLGVSDQQHJDWLYH PRPHQWDWWKHOHIWVLGHRIWKH IUDPH

&RPELQHGXQLIRUPJUDYLW\DQG ODWHUDOORDGV1HJDWLYHPRPHQWV LQWKHOHIWKDQGFRUQHUVDUH VLJQLILFDQWZKLOHSRVLWLYH PRPHQWVLQWKHYHUWLFDOPHPEHU RQWKHULJKWDUHUDWKHUVPDOO

Continuous Structures: Rigid Frames

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 22 Shaping frame members in response to internal moments in a multistory structure.

can be safely placed without interfering in building functions. Typical locations are around elevator shafts and other continuous vertical bays that are normally enclosed. The use of diagonal bracing greatly reduces the moments in the frame caused by lateral loads. Consequently, members can be reduced in size.

Continuous Structures: Rigid Frames FIGure 23 Typical types of rigid (moment-resisting) connections.

QuestIons 1. Find a frame building under construction in your local area. Draw a sketch of a typical beam-and-column connection. 2. Using an approximate method of analysis, analyze a single-bay frame of the type generally illustrated in Figure 4 that carries a horizontal load of 3000 lb acting at the upper-left joint. Assume that h = 12 ft and L = 22 ft. Draw shear and moment diagrams. Indicate numerical values. Answer: M max = 18,000 ft@lb and Vmax = 1636 lb 3. A single-bay frame of the type illustrated in Figure 5 carries a horizontal load of 5000 lb acting at the upper-left joint. Assume that h = 15 ft and L = 25 ft. Draw shear and moment diagrams. Indicate numerical values. Use an approximate method of analysis. 4. Draw a sketch that illustrates possible member-size variations (along the lines of the sketches in Figure 18) for the frame analyzed in Question 2. 5. Consider the two single-bay structures in Figure 18(b) and (c), respectively. Assuming that both structures are identical in all respects, except for member end conditions, and carry identical loads, which would you expect to deflect more horizontally? Why?

Copyright © 2013. Pearson Education UK. All rights reserved.

6. A single-bay frame like that illustrated in Figure 19 carries a horizontal load of 5000 lb acting to the right at the upper-left joint. A downward uniformly distributed loading of 200 lb/ft is also present on the horizontal beam of the frame. Assume a horizontal span of 25 ft, and a column height of 10 ft. What axial forces, shears, and bending moments should the beams and columns be designed to carry? Use an approximate method of analysis. 7. A fully fixed single-bay frame has a span of 50 ft and a height of 12 ft and carries a uniform loading of 1000 lb/ft on the horizontal beam. Using a computer-based structural analysis program available at your school, analyze the structure (shears and moments) for a situation where the moment of inertia of the beam is (a) equal to that of the columns, (b) twice that of the columns, and (c) three times that of the columns. Compare your results. (Assume any Ic value for the columns that you want, or use Ic = 1200 in.4) 8. Using a structural analysis program, conduct a comparative study of deflections of the four systems shown in Figure 18(a)–(d). Keep the span, height, and the load P constant and use the same member in all four systems. Comment on your findings.

Copyright © 2013. Pearson Education UK. All rights reserved.

Plate and Grid Structures

Copyright © 2013. Pearson Education UK. All rights reserved.

1

IntroductIon

Plates are rigid planar structures, typically made of monolithic material, whose depths are small with respect to their other dimensions. A multidirectional dispersal of applied loads characterizes how loads are carried to supports in plate structures. The advent of modern reinforced concrete has made the plate among the most common of building elements. Plates can be supported along their entire boundaries, only at selected points (e.g., columns), or with some mixture of continuous and point supports. Support conditions can be simple or fixed. The variety of support conditions possible makes the plate a versatile structural element. Although, strictly speaking, a plate is made of a relatively homogeneous solid material exhibiting similar properties in all directions, several other types of structures have a general structural behavior that is analogous to that of a plate (Figure 1). The space frame (actually a space truss), which is composed of short rigid elements triangulated in three dimensions and assembled to form a large rigid planar surface structure of relatively thin thickness, is one such structure. The grid structure is another type of structure that is essentially planar and is relatively thin. Plane grid structures are typically made of a series of intersecting long rigid linear elements such as beams or trusses, with parallel upper and lower chords. Joints at points of intersection are rigid. The basic shear and moment distributions in structures like these are not unlike those in comparably dimensioned monolithic plates. However, they do have many basic differences, which are explored later in this chapter.

2

GrId StructureS

Consider the simple crossed-beam system supported on four sides and shown in Figure 2(a). As long as both beams are identical, the load will be equally dispersed along them (i.e., each beam will pick up one-half of the total load and transfer it to its supports). If the beams are not identical, the stiffer member will carry a greater portion of the load. If, for example, the beams were of unequal length, the shorter member would carry a greater percentage of the load than the longer member because it is stiffer. Both members must deflect equally at their intersection because they are connected. For both beams to deflect equally, a greater force must be applied to the shorter beam than to the longer one. Therefore, the shorter element From Chapter 10 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Plate and Grid Structures

FIGure 1 Plate, grid, and space-frame structures. The general curvatures and external moments induced in plate, grid, and space-frame structures are similar if their loads and general dimensions are similar. The exact way each structure provides internal resisting moments and the specifics of behavior, however, is different.

picks up more of the applied load. The relative amount of load carried in mutually perpendicular directions in a grid system is dependent on the physical properties and dimensions of the grid elements. The key to analyzing a grid structure of this type is to recognize that a state of deflection compatibility must exist at each point of connection in a crossed-beam system. This compatibility requirement assumes that the beams are rigidly connected such that both undergo an identical deflection due to a load. By equating deflection expressions appropriate for each beam, it is possible to determine the relative percentage of the load carried by each element. Let PA be defined as the percentage of the total load (PT) carried by member A, and PB be the percentage of the total load carried by member B. By equating deflection expressions for each member, we obtain ∆ A = ∆ B so

PAL3A PBL3B = 48EAIA 48EBIB

PA LB 3 EAIA = a b a b PB LA EBIB

If the members are identical in all respects except their lengths, the latter expression becomes

Copyright © 2013. Pearson Education UK. All rights reserved.

PA LB 3 = a b PB LA

For members of equal length, it is evident that PA = PB = PT >2. If LB = 2LA, then PA >PB = 12LA >LA 2 3 = 8 and PA = 8PB. Thus, the shorter (more rigid) beam picks up eight times the load of the longer beam. Noting that PA + PB = PT , we see that PA = 8PT >9 and PB = PT >9. The associated beam moments are MA = PALA >4 = 18PT >92 >4 = 4PTLA >18 and MB = PBLB >4 = 1PT >9212LA 2 >4 = PTLA >18. Thus, moments in the short span are four times as large as those in the long member. The analysis of more complex grids with multiple crossed beams proceeds in a fashion similar to that just described. Deflections at the intersections of the various beams can be equated. An analytical difficulty arises, however, because deflections must be compatible at multiple points and the interactive reaction forces between members at one point contribute to the deflections at another. Invariably, for a complex grid, several equations are generated that must be solved simultaneously—a task for computer-based analysis systems. A study of the results of such analyses reveals what should be expected: If the cross members are different lengths, the shorter, more rigid members pick up the predominant share of the applied load. The more rectangular the grid becomes, the less load is carried by the longer members. For long, narrow bay dimensions, the longitudinal ribs can become nothing more than dead weight and have limited value as structural elements, except as stiffeners. (See Figure 3.) Other important considerations affecting the structural behavior of grids are their support conditions. Figure 4 shows the same horizontal grid structure of crossed beams supported in several different ways and with several different member end conditions. In Figure 4(a), members have pinned ends and the grid is supported on two sides only. The structure behaves more or less like a series of one-way beams. The cross beams do little. Moments are consequently like those in a one-way beam system and no efficiencies are gained by using a grid approach. In the second example shown in Figure 4(b), grid members have pinned ends and the whole grid is supported around its four edges by rigid frame structures. True twoway grid action is obtained, and bending moments in the grid are much smaller than those in Figure 4(a). Significant bending moments are developed, however, in the surrounding frame system that provides the grid support. This approach is

Plate and Grid Structures

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 3 Effects of bay proportions on behavior of two-way grids. Two-way grids operate most effectively when bays are square. When used on rectangular bays, the stiffer short-span members pick up the greatest portion of the applied loads and do the most work. Longer members are less stiff and do not contribute much to carrying the applied loads.

(a-1) Deflected shape (exaggerated) of a two-way grid on square bay, simply supported. The cross-beams have similar stiffnesses and hence similar deflected shapes.

(a-2) Bending moment diagrams. All four cross-beams share equally in carrying the loads and hence carry equal bending moments.

(b-1) Deflected shape (exaggerated) of a two-way grid on rectangular bay, simply supported. The stiffer short-span beams experience sharper curvatures (hence bending moments) than the more flexible long-span beam.

(b-2) The stiffer short-span beams carry a greater percentage of the loading—and hence have higher bending moments—than the long span member, which does very little.

commonly used because it decreases bending moments in the horizontal members but is still self-supporting for lateral loads via rigid frame action. The final example in Figure 4(c) has grid members turning down into columns around all four sides to form a series of crossed rigid frames. Bending moments here can be quite small and the whole system is quite efficient structurally. Deflections here would also be smaller than the structures shown in Figures 4(a) and (b). Architecturally, however, one has to contend with the surrounding column pattern, which may or may not be possible. Another aspect of the behavior of grids is evident in the one-way spanning system shown in Figure 5. As the center beam deflects downward under the action of the load, the transverse member transfers some of the load to the adjacent longitudinal elements. By looking at the geometry of the probable deflected shapes, it is evident that the center longitudinal member carries a greater portion of the load than do the outside members. (The center member is bent more, which means that the internal moments are higher, which in turn means that the percentage of the load supported is greater.) All the grid elements, however, share in carrying the load. In a simple beam system, only the member beneath the load would carry it. Another interesting aspect of the one-way grid shown is the twisting induced in the exterior members by the transverse member. As the transverse member deflects, its ends rotate. This tendency to rotate causes torsion to develop in the exterior members. At the same time, these members provide a torsional resistance to the end rotations of the transverse member, which is, in effect, stiffened by the torsional restraint offered by the exterior members and has end conditions somewhere between a fixed end and a simply supported end. The center longitudinal member

FIGure 2 Simple two-way grid structure.

Plate and Grid Structures

FIGure 4 This diagram compares one-way and several two-way systems, and shows that two-way systems have reduced bending moments and increased stiffness and reduced deflections. For comparison, each of the structures shown has the same spans, carries the same loading conditions, and horizontal members are all made out of identical members. Member sizes could be uniquely designed for each case so that the same stress and deflection criteria could be safely met. The approach shown in Figure 4(c) would then result in having the smallest members. 0.2 D

0.64 

1.0 

(a-1) Reference case—simple one-way beams (b-1) Two-way grid system supported with pinned connections by surrounding rigid with pinned ends. The maximum deflection frames. The deflection shown is relative to the present (at midspan) is defined as 1.0 . one-way system in (a).

1.0 M

(c-1) Two-way framed interconnected grid system. Columns and beams are rigidly connected with moment resisting joints. Deflections are reduced relative to (a).

0.45 M

0.3 M

MMAX 0.24 M

Copyright © 2013. Pearson Education UK. All rights reserved.

(a-2) Moment diagrams. The maximum bending moment at midspan is defined as 1.0 M.

(b-2) Moments relative to (a). Moments in spanning members are decreased. The maximum moment is in a surrounding edge beam.

(c-2) Movement diagrams. The rigid connections help reduce bending moments throughout the structure.

is also stiffened by the restraint offered by the transverse member, thus reducing the magnitude of its deflections. The net effect is that a fractional part of the load is eventually transferred to the supports by a twisting action. All the grid elements participate more in carrying the load because of this twisting action. The stiffness of the whole grid also is thereby increased. In a more complex grid, two-way action and twisting both occur. All elements participate in carrying the loads to the supports through a combination of bending and twisting. Note that if the beams were crossed and nonrigidly attached at intersection points, the bending rotation of one member would not cause twisting in the other. The consequent loss in overall rigidity due to the loss in torsional resistance associated with the twisting action would cause greater deflections to occur in a nonrigidly connected system than in a rigidly interconnected grid. In more complex interconnected grids, torsion also can develop in the transverse members. Figure 6 illustrates a two-way grid with pinned supports that carries uniform loadings on its members. A computer-based analysis of the grid yields the bending moment diagrams shown in Figure 6(c). Torsional moment diagrams are shown in Figure 6(d). Note that the bending moment diagram has a stepped shape [Figure 6(c)] that reflects the moments carried torsionally by transverse members. At any node and along either the x-x or y-y axis, the bending and torsional moments are in a state of equilibrium. Moments about the x-x axis at

Plate and Grid Structures

FIGure 5 Behavior of complex grid structures. Twisting is developed in all members as a consequence of how the structure deforms. The associated torsional resistance of members increases the stiffness of the grid.

z y

x

(a) Two-way grid with pinned supports. L = 60 ft. Uniform loading = w = 1.0 k /ft.





(b) Deflections (exaggerated)

(c) Moment (MB) diagrams

a typical node are shown in Figure 6(e). Similar moments would exist around the y-y axis. The torsional resistance provided by grid members contributes significantly to how the structure carries overall moments and significantly increases resistance to deflections. A stiff and efficient structure results.

3

Copyright © 2013. Pearson Education UK. All rights reserved.

3.1

Plate StructureS

(d) Torsion (MT) diagrams MB = 314 ft-k

MT = 40 ft-k

MB = 234 ft-k

MT = 40 ft-k

40 ft-k x

314 ft-k

one-Way Plate Structures

uniform loads. Plate structures behave in much the same way as the grid structures just discussed, except that the actions described take place continuously through the slabs rather than only at points of interconnection. Assume that a simply supported one-way plate spanning two walls carries a uniformly distributed load per unit area of w′. The plate can be imagined as a series of adjacent beam strips, each of unit width, that are interconnected along their lengths. The bending moments developed in the plate can then be expressed in terms of a moment per unit width (m) of plate corresponding to the strip width. This moment, m, is measured in ft-lb>ft or kN # m>m. Reactions are typically expressed for the strips in terms of a force per unit length. The plate bends into a more or less uniform curvature between the line supports at either end; hence, the moments developed in the different beam strips are similar, although moments in strips near midspan tend to be slightly higher than those near free edges. For a typical strip, m = w′L2 >8. Each unit width would be designed for this moment; in a reinforced-concrete slab, for example, an amount of steel would be designed for a moment per foot or per meter of slab width.

x

40 ft-k

234 ft-k  Mx = 0 234 ft-k + 40 ft-k + 40 ft-k = 314 ft-k (e) Equilibrium about x-axis

FIGure 6 Torsion in two-way grids. Torsional resistance helps carry loads and stiffens structure.

Plate and Grid Structures Point loads. Plates that carry concentrated loads are much more complex. They do, however, illustrate the positive load-carrying action of plates. Consider the simple one-way plate carrying a point load shown in Figure 7. The plate can be imagined as a series of adjacent beam strips of unit width interconnected along their lengths. As an applied load is picked up by one beam strip, it tends to deflect downward. The interconnected adjacent strips, however, offer resistance to this tendency, thereby picking up part of the applied load. A series of shear forces is developed at the interface between adjacent strips. Twisting related to these shears also is caused in the adjacent strips. At beam strips farther and farther from the strip under the load, twisting and shear forces are reduced because more of the load becomes transferred to the supports by the longitudinal action of the strips. Longitudinal curvatures also tend to decrease toward the edges of the plate. Internal moments must consequently decrease as well. Figure 7 illustrates how reactions and internal moments are distributed in the plate. The sum of the reactions must total the applied load acting in the vertical direction, and the sum of the internal resisting moments distributed across the plate at a section must equal the total external applied moment. These observations follow from basic equilibrium considerations. Assuming that the plate shown had a span of a1 and a width of a2, the internal moment per unit length in the plate at a transverse section at midspan would be the external moment divided by the plate width [m = 1Pa1 >42 >a2] if it were assumed that the resisting moments were uniformly distributed across that section. As argued previously, however, these moments cannot be uniform; instead, they vary from a maximum at the middle of the plate to a minimum at each edge. Still, thinking in terms of average moments is useful to establish an average reference line about which actual maximum moment values deviate to a greater or lesser degree. Note that using an average value for design purposes, however, is not conservative. ribbed Plates. The common ribbed plate is a combined beam-slab system (Figure 8). If the connecting slab is relatively stiff, the whole assembly functions as a one-way plate rather than a series of parallel beams. For design purposes, however, it is still common to treat the structure as a series of parallel T beams in the longitudinal direction. The transverse slab is treated as a one-way plate that is continuous over the beams. Negative moments are developed in the slab over the beams.

Copyright © 2013. Pearson Education UK. All rights reserved.

3.2

two-Way Plate Structures

Simple Plates on columns. Consider a plate that is simply supported on four columns. [See Figure 9(a).] The probable deflected shape of the structure is shown in Figure 9(b). The curvatures in the plate are highest in the plate strips nearest the free edges of the plate and become less toward the middle. This implies that the internal moments in the plate that are generated to balance the external moments due to the applied forces are greater at the edges (with respect to the bending behavior of the plate between the columns) than in the middle. Because the plate is deformed into a doubly curved shape by the load, it is evident that moments are developed in several directions rather than in only one. To find the magnitudes of the internal moments developed, consider the equilibrium of a section of the plate. Any segment of the plate may be contemplated because it is fundamental that any portion of any structure be in equilibrium. If the moments need to be known along a line D–E–F, a section is passed through that line and the equilibrium of the left or right plate segment considered. In doing this, it is convenient to view the plate from the side, as illustrated in Figure 9(c), in which the plate appears as a simply supported beam carrying a uniformly distributed load per unit length. If the width of the plate is denoted by a1, the span by a2, and the

Copyright © 2013. Pearson Education UK. All rights reserved.

Plate and Grid Structures

FIGure 7 One-way plate structure.

Plate and Grid Structures

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 8 One-way ribbed plates.

load per unit area acting on the plate surface by w′, then the uniformly distributed load per unit length that appears on the analogous simple beam is given by w′a1. The reactions of the analogous beam on either end are the reactive forces in the columns (which, by inspection, must be w′a1 a2 >4). The external bending moment at midspan for this analogous structure is given by the simple beam moment in the form wL2 >8, where w is expressed as a force per unit length. In the case of the plate, w′a1 is analogous to w and a2 to L. The total moment present is consequently MT = w′a1a22 >8, or MT = 0.125w′a1a22, where w′ is expressed as a force per unit area. This is the same result that would have been obtained by summing moments about the line D–E–F. If the plate is square, a1 = a2 and MT = 0.125w′a3. The preceding simplified analysis is useful because it is evident that the external applied moment along line D–E–F is, in total, equal to 0.125w′a3. The plate must therefore provide a total internal resisting moment of 0.125w′a3. In a beam, this moment would have been supplied by a single discrete couple formed by the compressive and tension stress fields internally developed in the beam in response to the external moment. In a plate, the total resisting moment is supplied by a continuous line of resisting moments in couples across the width of the plate. The net effect of these moments is to yield a total resisting moment of 0.125w′a3. The equivalent discrete moment in the analogous beam has been, so to speak, spread out across the width of the plate. If the total resisting moment that the plate provides is known, the next step is to find how this moment is distributed across the section. If the resisting moment were assumed to be uniformly distributed across the section, the resisting moment per unit of width (m) would be the total moment divided by the width of the section (i.e., m = 0.125w′a3>a = 0.125w′a2, where m is the internal moment per unit width across the plate width at the section considered). The argument made earlier, however, indicates that the total moment is not uniformly distributed, but is greater at the edges of the plate than in the middle. Quantitatively assessing the exact distribution is complex. It is obvious that maximum and minimum values vary about the average moment value of m = 0.125w′a2 discussed earlier. Figure 10(d) shows the results of a more exact analysis, where it is seen that the maximum moment per unit width occurs at the plate edge 1m = 0.15w′a2 2 and the minimum at the center of the plate 1m = 0.11w′a2 2. Similar results would be obtained for an analysis about line B–E–H because the structure is symmetrical. Note that the maximum moment developed in the plate occurs at the free edge about an axis perpendicular to that edge. The moment about an axis parallel to the free edge at the same point is zero. At the plate midpoint, however, a moment of m = 0.11w′a2 is developed about both axes. The analysis just presented highlights that maximum moments occur, not at the midpoint of the plate, where one might normally expect them by virtue of a beam analogy, but at the midspan of the edges. The midpoint of this plate is, for example, a good place to put a hole if it is necessary to accommodate another building element. Another interesting point is that this particular plate must provide the same total internal resisting moment as an analogous beam. When the plate is designed, it is evident that using a plate does not necessarily save material, compared with an analogous beam. The plate will, however, be shallower. The much-heralded advantages of two-way plate action in terms of material savings are associated with other types of support conditions, not the one described and analyzed herein. Plates Simply Supported on continuous edges. Consider a plate similar to the one previously analyzed, but supported continuously on all four edges by simple edge supports (e.g., walls), rather than being supported on columns.

Plate and Grid Structures

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 9 Square plate simply supported on four columns (uniformly distributed load w′).

Plate and Grid Structures

FIGure 10 Square plate simply supported along all edges (uniformly distributed load w′).

Copyright © 2013. Pearson Education UK. All rights reserved.

.

Plate and Grid Structures

Copyright © 2013. Pearson Education UK. All rights reserved.

An inspection of the probable deflected shape of the structure reveals a radically different behavior under load: Maximum curvatures occur at the midpoint of the plate and decrease toward its edges. Internal moments can be expected to vary accordingly. To find the magnitudes of these internal moments, a method of analysis similar to that used for the column-supported plate can be adopted. In this case, however, the first step in finding the plate reactions is not so easy. Considering the way the plate deflects, it can be seen that the reactions are not uniformly distributed but are at a maximum at the center of each line support and then decrease toward the corners. [See Figure 10(b).] A curious aspect of the plate reactions stems from the tendency of the corners of the plate to curl upward under the action of the vertical load. If the corners are to be restrained from curling upward, downward reactive forces must exist at the corners. Then the sum of the upward nonuniformly distributed reactions and the downward corner reactions must total the value of the externally applied load acting downward—a fact that is obvious from basic equilibrium considerations. The general form of the reactions is illustrated in Figure 6(b). By looking at the plate from the side and considering a beam analogy, it can be seen that the total external moment present along a midspan section is less than the moment existent in the previously analyzed plate on columns. The total moment in the plate on the columns was equivalent to the previously encountered wL2 >8 moment value, or, in terms of the plate dimensions and unit area loads, w′a3 >8. This value is associated with concentrated reactions at either end of the structure. By looking at the derivation of the expression [i.e., ML>2 = 1wL>221L>22 - 1wL>221L>42 = wL2 >8], where the first term represents the effects of the end reactions, it can be seen that the dominant moment-producing effect is caused by the moments associated with the reactions and not the external loads. In the plate resting on four continuous supports, the comparable moment caused by the reactions must be less than that associated with point supports because a greater percentage of the reaction forces are nearer the moment center. Consequently, the net external midspan moment must also be reduced because the moment from the load remains constant. With external moments reduced, internal resisting moments also are reduced—considerably from those present in the plate supported on four columns. A shallower plate can therefore be used. The maximum positive plate moment per unit width in a square plate can be shown to be m = +0.0479w′a2. Continuously supporting a plate is far more preferable to using point supports. When continuous supports are used, the true benefit of two-way action can be achieved. Plates with continuous Fixed-edge Supports. Figure 11 illustrates a plate similar to that just analyzed, except that the edges are now completely fixed. The moments in the plate are reduced even further by fixing the plate ends. This is a highly advantageous support condition for plates. For a square plate, the maximum moments per unit width can be shown to be m = +0.0231w′a2 and m = -0.0513w′a2. Bay Proportions: effects on Moments. In analyzing grids, we noted that the stiffer portion of a grid structure, which is often the short-span direction, usually carries the greatest percentage of the applied load. The same is true of rigid planar structures. Consider the (continuously supported) rectangular plate shown in Figure 11. Only the plate strips in the short direction exhibit significant curvatures. The longitudinal strips are bent less. The implication is that only the strips in the short direction

Plate and Grid Structures

FIGure 11 Effects of different boundary conditions and bay proportions.

Side ratio

Maximum positive or negative bending moments

Copyright © 2013. Pearson Education UK. All rights reserved.

provide internal moment resistance and participate significantly in carrying the applied load. The long-span strips almost ride along. The more rectangular a plate becomes, the more it behaves as a one-way, rather than a two-way, system. It is as if the plate were supported only along its long parallel edges. Consequently, none of the advantages associated with the two-way action of continuously supported plates are present. If the length–width bay dimensions of a plate exceed a ratio of about 1.5, the plate behaves as a one-way, rather than a two-way, system. design Moments. The effects of different support conditions and bay proportions for plates with uniform loadings (e.g., w′in lb>ft2) may be reflected directly by using bending moment expressions in the form ma = Cw′a2a and mb = Cw′a2b, where a and b are plate dimensions and C represents a constant reflecting the support conditions present. Several cases have been solved through more advanced analysis techniques, and results are presented in Figure 11. Plate on Beams Supported by columns. Figure 12(a) illustrates a variant of the first plate analyzed. The plate is supported by continuous beams, which are in turn supported by columns. This is an interesting structure that finds wide application in buildings. Assuming that the plate is simply supported on the beams gives the plate a set of end conditions somewhere between the situation in which the plate is supported only by columns and one in which the plate is supported by continuous edge supports. If the beams are extremely stiff, the support conditions of the plate approach those of continuous edge supports, and similar moments are developed in the plate. If the beams are flexible, however, little edge support is provided, and the plate behaves more as if it were simply resting on four columns. Higher plate moments are accordingly developed than occur in the previous case. The relative stiffnesses of the edge beams therefore crucially affect the magnitude of the moments developed in the plate.

Plate and Grid Structures

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 12 Two-way beam-and-slab system.

An interesting facet of this type of structure can be seen by passing a transverse section through the midspan. From basic equilibrium considerations, it can be seen that the plate-and-beam structure must provide an internal resisting moment that exactly balances the external moment at the section. As was the case before, when four point supports were used, the moment of interest is 0.125w′a3. If the beams were highly flexible 1Ib S 02, the sum of the internal moments developed in the plate would also equal 0.125w′a3, as described earlier. The beams therefore would not provide any of the moment resistance. If, however, it were somehow possible to have the plates be of minimum stiffness 1Ip S 02 and not provide any moment resistance, the beams would have to provide the entire resistance. Each of the two beams would then develop a moment of 0.125w′a3 >2. The real situation lies between these extremes. The point is that the structure must provide a fixed requisite moment resistance, but the exact way this occurs is dependent on the nature of the structure used. Moments are never dispensed with, but merely redistributed. The consequence of using stiffer beams is that plate moments are reduced and the plate can be made shallower. Alternatively, a stiffer plate can be used and the beams reduced in depth or eliminated. (The plate, however, must be made thicker than in the previous case.) other types of Plates. Of course, we have not yet considered a wide variety of other plates. Because of the importance of end conditions, each type must be treated separately. The reader is referred to more advanced texts on the subject. Yield line theory is a particularly interesting technique for reinforced concrete.

Plate and Grid Structures

FIGure 13 Unusual support conditions. The basic behavior of plates having unusual support conditions can be qualitatively understood by drawing and studying the probable deflected shape of the structure (i.e., noting regions of tension and compression).

A good way to get at least a general feeling for the way any plate behaves is to sketch its deflected shape, something that, with practice, can be done for any type of plate. (See Figure 13.) These sketches will at least identify regions of critical positive and negative moments and will be useful in many situations (such as the preliminary design of reinforced-concrete plates, in which it is necessary to know whether reinforcing steel goes on the top or bottom of the plate).

Copyright © 2013. Pearson Education UK. All rights reserved.

comparisons among Plates, Grids, and Space Frames. As mentioned earlier, the general curvatures and moments induced in plates, grids, or planar space frames of comparable dimensions and carrying equivalent loads are similar. The exact way each structure provides balancing internal moments, however, differs. In a plate structure, the internal moment is provided by a continuous line of couples formed by the compression and tension forces developed on the upper and lower surfaces of the plate. In a grid, the resisting couples are concentrated in the beams. In comparison to the moments developed in a plate (normally expressed in terms of a moment per unit length), the moment in a grid beam is approximately given by the beam spacing times the average moment per unit length at that location. (All the equivalent plate moments in a region are, so to speak, concentrated in a single beam.) This approximation, however, can lead to unconservative results in coarsegrained grids because grids and plates differ in many significant ways (particularly the relative contribution of the torsional stiffness of elements in carrying the applied loads). The approximation just described improves as the grid mesh becomes finer and finer (and begins to approach a continuous plate surface). Thus, in a grid structure with beam members spaced s1 apart, the average moments per unit plate width, m, present at the section may be considered collected into a moment M1 = m1s1 2. The grid beam would then be designed for this bending moment. A similar situation exists for space-frame structures, except that the resisting couples are provided in a trusslike, rather than a beamlike, fashion. Thus, in a space-frame structure of depth d1 and bar spacing s1, the average moments per unit plate width, m, present at the section may be collected into a moment M1 = m1s1 2, which is in turn balanced by a resisting moment formed by the tension and compression forces in the upper and lower bars acting over the depth of the structure. The bar forces themselves can be found by noting that T1 = C1 = M1 >d1. Because these relations exist, the general moment distributions discussed earlier in connection with plates also can be used to describe the approximate behavior of grids and planar space frames.

4

deSIGn oF tWo-Way SySteMS: General oBjectIveS For Plate, GrId, and SPace-FraMe StructureS

The process of sizing members for the shears and moments that are present in a two-way structure is the same as for any member in bending. However, several steps can be taken to initially minimize the bending. Other critical factors also affect the design of a plate structure and influence the appropriateness of a system selected. These factors are considered in the next section. Support conditions. A primary design objective is always to minimize the bending moments that are present in a structure. An effective way to do this in rigid planar structures is to operate on the support conditions. (As discussed in the previous section, moments in rigid planar structures are critically dependent on support conditions.)

Plate and Grid Structures

FIGure 14 Plates on a column grid. A continuous plate surface is preferable to a series of simply supported plates because design moments are reduced and rigidity is increased by the continuity.

Copyright © 2013. Pearson Education UK. All rights reserved.

Generally speaking, a design that provides as much continuous support to a rigid planar structure as possible is one in which moments in the structure are minimized. Structural depths can thereby be reduced. Thus, continuous edge supports are preferable to column supports. Fixed supports also result in smaller plate moments than simple supports provide. (See Figure 11.) Techniques such as using overhangs, described for reducing moments in beams, also are applicable to planar structures in bending. Cantilevering a portion of the structure tends to reduce the maximum design moment that is present. When the support system is basically repetitive in nature (e.g., a series of column bays), using a continuous rigid structure leads to lower design moments than using a series of discrete, simply supported plates. Using a continuous surface causes the portion of the plate over a single bay to behave more like a plate with fixed ends than does a simply supported surface. (See Figure 14.) Bay Proportions: effect on choice of Structure. As noted in the section on analyzing rigid planar structures, the less square, or more rectangular, a supporting bay becomes, the less the supported planar structure behaves like a two-way system and the more it behaves like a one-way system acting in the short-span direction. From a design viewpoint, the consequence of this phenomenon is that bays should be designed to be as dimensionally symmetrical as possible if two-way action is desired. Only if bay dimensions form a ratio between approximately 1:1 and 1:1.5 does two-way action obtain. Using what appears to be a two-way structure, such as a space frame, over a long rectangular bay does not accomplish much. The structure will behave like a one-way system in the short direction. Members in the short-span direction would carry most of the load. Longitudinal members would simply ride along, contributing little, except adding some stiffening. Therefore, a system that is deliberately designed and intended to be a one-way spanning system might as well be used. Less total material could be used to support the load in space. types of loads: effect on choice of Structure. A fundamental reason for using a rigid planar structure is often a functional one: These structures can provide usable horizontal floor planes. Accordingly, the loads they carry are usually uniformly distributed. When this is the case, either plates, grids with small meshes, or space frames can be used relatively efficiently. These same structures are, however, relatively inefficient when called on to support large concentrated loads. Surface

Plate and Grid Structures structures work best supporting surface loads. Large concentrated loads can, however, be effectively carried by coarse-grained grid structures in which the crossed beams intersect beneath the loads. An additional point worth noting is that many current, commercially available space-frame structures, such as those typically made of cold-formed light steel elements, are better suited to roof rather than to floor loads because roof loads tend to be lighter. These structures do not work particularly well for floor loads. This is a consequence, not of the basic structural approach, but of the specific way these structures are constructed. Span range. A wide variation in spans is possible. Rigid plates made of reinforced concrete can economically span anywhere from 15 to 60 ft (4.5 to 18 m) or so. Grids and space frames can span higher distances, depending on exactly how they are made. Commercially available space frames, for example, can easily span up to 100 ft (30 m), depending on the member sizes and support conditions used. Much higher spans are even possible with specially designed structures. For short spans of the type often encountered in buildings, for example, 15 to 30 ft (4.5 to 9 m), the construction complexities of grids and space frames often render them less attractive than simple reinforced-concrete plates.

Copyright © 2013. Pearson Education UK. All rights reserved.

5

deSIGn oF reInForced-concrete tWo-Way SySteMS

Plate Moments and Placement of reinforcing Steel. The thickness of a reinforced-concrete plate and the amount and location of reinforcing steel used in a constant-depth plate or slab are critically dependent on the magnitude and distribution of moments in the plate. Figure 15(a) illustrates the deformations present in a continuous plate resting on columns. Reinforcing steel must be placed in all regions of tension. This requirement results in a placement of reinforcing steel of the type illustrated in Figure 15(b). Because moments are continuous, reinforcing steel must be closely spaced. It is thus common to use a series of parallel bars. Because moments are multidirectional, mutually perpendicular sets of bars are used. The bars can be deformed as shown in the figure, so that the steel is properly located in regions of tension, or else discontinuous shorter linear pieces that are lapped can be similarly placed. While the design moments in the plate (on which plate thicknesses and steel sizes would depend) can be found from techniques already discussed, some characteristics peculiar to reinforced concrete mean a slightly different design approach should be taken. As noted previously, the total (positive plus negative) moments in a slab supported on columns is given by MT = w′a3 >8 = 0.125w′a3. It might, therefore, be thought that the total amount of steel needed would be in response to this value. If, however, a more detailed study of the plastic behavior of the plate is made, it can be seen that designing for the 0.125w′a3 moment is more conservative than necessary. Practically, continuous reinforced concrete plates are designed such that the total design moment MT (the sum of all positive and negative moments) is divided throughout the slab on a highly empirical basis. In U.S. practice, the slab is first divided into column and middle strips. (See Figure 16.) Moments per unit width are assumed constant across each strip, instead of varying continuously as theory suggests. The total moment is then defined as MT = wL2 L21 >8, where w is the loading in lb>ft2 1kN>m2 2, L1 is the direction considered, and L2 is the clear panel width

Plate and Grid Structures

FIGure 15 Reinforced concrete plate on columns.

Copyright © 2013. Pearson Education UK. All rights reserved.

.

at right angles. For a continuous-slab interior bay extending in both directions, this moment is empirically split into a total negative design moment of 10.652MT and a total positive design moment of 10.352MT . Then, 75 percent of the negative design moment is apportioned to the column strip and 25 percent to the middle strip. Also, 60 percent of the total positive design moment is apportioned to the column strip and 40 percent to the middle strip. The next example is intended only to convey the spirit of the approach used, not to represent actual design practice.

Plate and Grid Structures

FIGure 16 Column and middle strips. Column strips generally carry larger moments than middle strips carry and require greater amounts of steel.

exaMPle A 20 * 20-ft two-way interior-bay flat slab supports a live loading of 80 lb>ft2 and has a dead load of 90 lb>ft2. Determine the design moments. Use an ultimate strength design approach. Solution: Total moments: loading: w = 1.2DL + 1.6LL = 1.21902 + 1.61802 = 236 lb>ft2 total moment: MUT =

23612021202 2 wL2L21 = = 236,000 ft@lb 8 8

negative design moment = 0.65MT = 0.651236,0002 = 153,400 ft@lb positive design moment = 0.351236,0002 = 82,600 ft@lb Column strip: negative moment = 0.751153,4002 = 115,050 ft@lb positive moment = 0.60182,6002 = 49,560 ft@lb

Copyright © 2013. Pearson Education UK. All rights reserved.

Because the strip width is 20>2 = 10, the 115,050 = 11,505 ft@lb>ft 10 49,560 = 4956 ft@lb>ft design positive moment per unit width = 10

design negative moment per unit width =

Middle strip: negative moment = 0.251153,4002 = 38,350 ft@lb design negative moment per unit width =

42,575 = 3835 ft@lb>ft 10

positive moment = 0.40182,6002 = 32,040 ft@lb design positive moment per unit width =

32,040 = 3204 ft@lb>ft 10

The maximum design moments are negative and occur over the top of the columns 1m = 11,505 lb>ft2, while midspan positive moments are relatively small by contrast. Reinforcing steel would be designed on a unit-width basis in response to the column and middle strip moments just found.

Plate and Grid Structures effects of Span. For low-span ranges, for example, about 15 to 22 ft (5 to 7 m), common building loads result in moments that can be handled by relatively thin plates, for example, on the order of 5 to 10 in. (13 to 26 cm). As spans increase, moments become greater and plate thicknesses must increase. A consequence of using thicker plates, however, is that the dead weight of the structure also increases, often dramatically. For this reason, plates are often hollowed out to reduce the dead load, while at the same time maintaining an appreciable structural depth. One such system, called a waffle slab, is shown in Figure 17(d). A typical waffle slab is shown in Figure 18. Another approach used to increase spans is commonly referred to as a twoway beam-and-slab system. (See Figure 17.) Because the beams are relatively stiff and form a continuous edge support for the plate, the plate can be made relatively thin because of the favorable support conditions, which decrease the moments present in the plate. The waffle and two-way beam-and-slab systems can each span relatively long distances. A system exhibiting desirable properties of both systems is obtained by not hollowing out the waffle slab between columns. A surrounding stiff-beam system is thereby created. Spans longer than those associated with the basic waffle slab are possible. The systems can be posttensioned to increase their span capabilities even further.

Copyright © 2013. Pearson Education UK. All rights reserved.

Plate thicknesses. L>d ratios for estimating slab thicknesses are given approximately in Table 1. In current U.S. practice, a detailed calculation of actual deflections is usually not required if the span-to-depth ratios listed are maintained. effects of Shear Forces. Although the discussion has thus far dealt exclusively with bending in plates, shear also is present and is often the dominant factor influencing the design of a plate. Shear forces are highest in plates on columns or other discrete supports. A punch-through type of failure can occur in plates around such points because of the high shear stresses that are present. The entire reactive force in a column, for example, must be distributed in the form of shear forces in the plate around the interface between the column and the plate. The plate area resisting the external shear can be found by considering potential shear failure lines. A reinforced-concrete plate, for example, tends to fail in the manner illustrated in Figure 19. The diagonal crack pattern is caused by diagonal tension cracks associated with the shear stresses that are present. The area of plate providing the resistance to punch-through failure is therefore this same crack surface. Note that the extent of the surface depends critically on the thickness of the plate and the circumference of the column. Punch-through shear failures are most common in either thin plates or plates supported on pointed or small columns. The approximate magnitude of the shear stresses present is given by fv = V>Ap, where Ap is the plate area in shear. The punch-through failure that is possible because of high shear forces is often a crucial design issue. The magnitude of the shear stresses that are present depends directly on the magnitude of the shear force and the plate area in shear. The latter depends on the thickness of the plate and the length of failure plane possible (which in turn depends on the size of the support). If shear stresses are too high, the design options are to use special steel reinforcement in the overstressed regions (called shear heads) or to increase the plate area that is in shear. Increasing the plate area that is in shear can easily be done by increasing the plate thickness. This, however, may prove uneconomical to do for the whole plate if it is not needed for moment considerations. If so, the plate can be locally thickened at critical points by using drop panels. (See Figure 19.) Alternatively, the plate area that is in shear can be increased by increasing the support size, which can also be done locally by using column capitals. The larger the capital, the greater is the plate area in shear. Column capitals can be any shape, but, due to the natural

Plate and Grid Structures

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 17

Two-way beam-and-slab systems and waffle systems.

Plate and Grid Structures

taBle 1 Typical slab span-to-depth ratios tyPIcal SlaB SPan-to-dePth ratIoS Two-way flat plate

L>30–L>33 (with edge beam)

Two-way flat plate (with column capital and>or drop panel)

L>33–L>36 (with edge beam)

Two-way beam slab Two-way waffle slab One-way slab

L>33–L>44 L>33 L>20 (simply supported) L>24 (one end continuous) L>28 (both ends continuous) L>10 (cantilever)

Beam or ribbed one-way slab

L>16 (simply supported) L>18.5 (one end continuous) L>21 (both ends continuous) L>8 (cantilever)

Copyright © 2013. Pearson Education UK. All rights reserved.

tendency to fail in diagonal shear, capitals are often given a sloped shape (because material below the 45° line is inactive). Both drop panels and column capitals can be used simultaneously. The shear capacity of such a system is quite high; the system is thus used in special cases where loads are particularly high (e.g., warehouses). Reinforced concrete slabs that do not use column capitals are usually referred to as plates. Those using capitals and drop panels are referred to as flat slabs. Shear can also be a problem in waffle slabs but can be handled quite easily. The area adjacent to a column top is simply not hollowed out. A built-in column capital is thus created. (See Figure 17.) In a two-way beam-and-slab system, the beams framing into the columns pick up most of the shear. Because their areas are relatively large and reinforcement is easy, shear is usually not a great problem in this type of system. Shear forces may also be carried by various kinds of special steel reinforcing inserts around tops of columns. (See Figure 20.) These may assume any of several different forms, but all rely on placing steel where the shear crack is likely to form. effects of lateral loads. Plates or slabs that are cast monolithically with columns naturally tend to form frame structures capable of carrying lateral loads. Their capacity to carry lateral loads depends to a great extent on the thickness of the horizontal structure at the column interfaces. (Deep structures have larger possible moment arms for increased resistance.) A thin plate-and-column system can carry lateral loads, but its capacity is limited, and thus its use is restricted to low buildings. The use of column capitals or drop panels increases the system’s capacity. Waffles and two-way beam-and-slab systems can support substantial lateral loads. The twoway beam-and-slab system forms a natural frame system and is often used when lateral load-carrying capacity is important. Special Plates and Grids. To reflect variations in the internal force states within them, plate-and-grid structures may be shaped in a variety of ways, although doing so may or may not be economical. As idealized forms, however, such structures are worth attention. Figure 21(b) shows a simply supported plate structure resting on top of a surrounding simply supported beam structure. Both the plate and the beams have been shaped to better reflect the variation in bending moments present within them. (Depths are proportional to moments.) The structure is not stable

FIGure 18 A typical waffle slab with infills around columns to increase shear capacity.

Plate and Grid Structures

Copyright © 2013. Pearson Education UK. All rights reserved.

FIGure 19 Designing for shear in reinforced-concrete plates.

Plate and Grid Structures with respect to lateral forces. Figure 21(c) shows a variable-depth plate-and-beam structure that is continuous throughout and capable of carrying both vertical and lateral forces. Figure 21(d) shows a constant-depth grid structure in which the widths of grid elements are varied to match the bending moments that are present. Figure 21(e) shows a variable-depth structure similar to Figure 21(b) in which beam widths are held constant and depths allowed to vary. Figure 21(f) shows a continuous grid structure in which beam widths are held constant and depths allowed to vary to match the bending moments that are present. These different shapes only generally reflect variations in internal force states because compromises and judgments must be made to make the shapes viable or sensible. A final, special form of plate is shown in Figure 21(g). A detailed analysis of the elastic stress distributions in a plate reveals the presence of lines of principal stress. These lines, often called isostatics, are directions along which the torsional shear stresses are zero. Some designers have devised plates that are ribbed in a manner intended to reflect isostatic lines. These structures are, of course, expensive to construct, but their designers claim that the expense is not excessive and that material savings compensate for any added construction cost. Such structures are undoubtedly interesting but are curious from the viewpoint of classical theory, in that they represent the physical manifestation of a model of a stress distribution in an elastic material that is constructed from an inelastic material (concrete). That the ribs are lines of principal stress is not argued, but a self-fulfilling prophecy may be present in placing stiffer ribs along these lines.

Copyright © 2013. Pearson Education UK. All rights reserved.

6

FIGure 20 Construction approaches to avoiding punching shear failures in flat plates.

(a) Shear studs welded onto a steel flat

SPace-FraMe StructureS

approaches. Space-frame structures are typically made of rigid linear members normally arranged in repeating geometric or modular units to form a thin, horizontally spanning structure. Many types of modular units (e.g., tetrahedron organizations) are possible. When modular units are fully triangulated, these structures are better characterized as space trusses rather than space frames, although the latter name is commonly used. These structures are suitable for use with uniformly distributed loads and do not handle large, concentrated loads well. Members in triangulated units normally experience only axial tension or compression forces, and are thus often made with symmetric cross sections (e.g., pipes). Designing nodal connections to accommodate the complexities of how members meet at a point is problematic, and several interesting approaches have been developed. One approach is shown in Figure 22. Other approaches are possible. The advent of computer-aided design and manufacturing systems now allows structures of this type to be built without reliance on totally repetitive geometries. Geometric data can be easily extracted from digital models and used to automatically calculate and subsequently fabricate member lengths to exact dimensions as well as to calculate and fabricate nodal members. Forces in Members. Members’ forces generally depend on the loading magnitudes, spans, the depth of the structure, and the size of the modules. Member forces in shallow structures with coarse grids are normally higher than those in deeper structures with finer-grained grids. Magnitudes of member forces result from the behavior of the structure with respect to overall bending and shear in a way that reflects analogous distributions in a homogeneous plate structure, except that forces are concentrated in members rather than distributed in the plate. Forces vary throughout a structure in a way that is dependent on the support conditions present. Space frames may be supported in a variety of ways—along their edges, at corners, or with inset supports that allow the structure to cantilever outward. When few point supports are used, member forces in the immediate vicinity of the point support are quite high (a situation analogous to shearing forces causing punch-through failures in homogeneous plates or slabs).

(b) Crossbeam system made from steel W shapes. The photo shows the system without the upper steel layer installed.

Copyright © 2020. CRC Press LLC. All rights reserved.

The Shape of Space

Weeks, Jeffrey R.. The Shape of Space, CRC Press LLC, 2020. ProQuest Ebook Central, http://ebookcentral.proquest.com/lib/ybp-ebookcentral/detail.action?docID=6012538. Created from ybp-ebookcentral on 2020-04-02 16:08:32.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Shell Structures membranes or hanging fabric (the latter forms were then used in an inverted way similar to the Gaudi approach). In contrast, common free-form shapes—especially those with reverse surface curvatures and flattened portions—rarely exhibit membrane action, and inefficient bending is developed that necessitates the use of awkward structural approaches. (These types of structures are briefly addressed in Section 5.) In structures exhibiting membrane or shell action, loads applied to shell surfaces are carried to the ground by the development of compressive, tensile, and shear stresses acting in the in-plane direction of the surface. The thinness of the surface does not allow the development of appreciable bending resistance. Thin shell structures are uniquely suited to carrying distributed loads and find wide application as roofs of buildings. They are, however, unsuited to carrying concentrated loads. As a consequence of carrying loads by in-plane forces (primarily tension and compression), shell structures can be very thin and still span great distances. Span-thickness ratios of 400 or 500 are not uncommon; for example, 3 in. (8 cm) thicknesses are possible for domes spanning 100 to 125 ft (30 to 38 m). Shells of this thickness, however, are a recent structural innovation made possible by the development of new materials such as reinforced concrete, which is uniquely appropriate for shell surfaces. Older three-dimensional shapes, such as masonry domes, are considerably thicker relative to their span and cannot be exactly characterized as carrying loads by in-plane axial or shear stresses because more bending exists and final stresses are not uniform. However, an approximation of this type is good for conceptualizing the behavior of such structures. Terminating a shell surface is always problematic. In a dome, for example, the in-plane forces developed in the surface normally have outwardly directed components that must be absorbed by either a series of closely spaced buttresses or a tension ring. Three-dimensional forms also may be made of assemblies of short, rigid bars. These structures are not, strictly speaking, shell structures because they are not surface elements. Still, their structural behavior can be conceptualized as being similar to that of continuous surface shells in which the stresses that are normally present in a continuous surface are concentrated into individual members. Structures of this type were first used extensively in the nineteenth century. The Schwedler dome, consisting of an irregular mesh of hinged bars, for example, was introduced by Robert Schwedler in Berlin in 1863 when he designed a dome with a span of 132 ft (40 m). Other, newer structures have bars placed on curves generated by the medians and parallels of surfaces of revolution. Some of the largest domes in the world follow this latter scheme. To minimize the construction difficulties involved in having to use bars of different lengths to create the shell surface, some systems have been developed with the goal of using equal-length bars. The most widely publicized are the geodesic domes associated with Buckminster Fuller. Because the surface of a sphere is not developable, the number of identical repetitive patterns into which the surface can be divided is limited. The spherical icosahedron, for example, consists of 20 equilateral triangles, but the need to subdivide the surface further leads to bars of different lengths. The structural advantages often claimed for these forms are not necessarily greater than those for other reticulated domes. Shapes other than surfaces of revolution also can be made using bar elements. Examples are typical ribbed barrel roofs and the lamella roof, which is made of a  skew grid of arch-like forms composed of discrete elements. The latter form was developed in connection with wood as the building material, but concrete and steel have been used as well. Remarkably large spans can be obtained with lamella systems. The introduction of computer-aided design and manufacturing techniques has considerably reduced the importance of the historical imperative of minimizing the number of bars with unequal lengths. When free-form surface shapes are used, surfaces can be subdivided within the digital environment and exact bar lengths and

FIgure 2 Thin shells versus other structural forms. The membrane action of shells is highly efficient.

(a) Thin shell structure: Only in-plane tension or compression membrane forces are developed.

(b) Complex shape that does not exhibit membrane action. Large bending moments are developed. Increased thicknesses or supporting framing systems are consequently needed.

FIgure 3 Dome-like shell by F. Candela. The shell is composed of multiple hyperbolic paraboloid segments separated by glazed elements.

Shell Structures the geometry of related nodes easily extracted. Advanced manufacturing techniques can, in turn, fabricate these many different forms quite easily. Recall, however, that resulting free-form shapes might not all be structurally efficient.

2 2.1

SpherIcal Shell StructureS Introduction

This section considers a specific shell structure made from a portion of a spherical surface. Section 2.2 discusses in detail the in-plane forces developed within the surface of a spherical shell. Prior to this analysis, however, it is useful to look more broadly at how such a shell works and what kinds of forces are exerted on its supporting elements. This is best done by studying a simplified example: Il Palazzetto dello Sporto, built in 1957 by Pier Luigi Nervi. example Il Palazzetto dello Sporto The dome of this structure apparently floats on top of a series of Y-shaped buttresses. (See Figure 4.) The dome is composed of a series of highly reinforced, ferroconcrete elements used to make a closely spaced, ribbed surface supported by the Y-shaped buttresses, which transfer forces developed within the surface of the shell to a massive tension ring (see Section 2.8) buried in the ground. The ring contains the outward force components developed in the buttress and transmits the vertical force components into the ground. Determine the approximate force in a typical supporting buttress and in the surrounding tension ring. Use a simplified analysis that assumes a simple gravity loading condition and ignores variances induced by the open cupola on top of the structure or by the use of ribs. Assume that the combined dead load and vertical live load associated with the unit areas of the surface is given by wDL + LL = 80 lb>ft2.1 Solution: First, determine the approximate value of the total load 1W 2 acting downward (the combined weight of the shell and live load). This value can be found by multiplying the surface area of the dome 1A2 by the load per unit area that acts on it (wDL + LL = 80 lb>ft2). The surface area of a portion of a sphere from f 1 to f 2 is given in any reference book by A = 2pR2[cos f 1 - cos f 2]. Alternatively, this expression can be found by a simple area calculation, as follows (see Figure 4):

Copyright © 2013. Pearson Education Limited. All rights reserved.

A = 1 2pa df = 1 2p1R sin f2 df = 2pR2[cos f 1 - cos f 2] = 2p1159 ft2 2[cos 0° - cos 38°] = 33,673 ft2

W = A * wDL + LL = 133,673 ft2 2180 lb>ft2 2 = 2,693,840 lb

Thirty-six buttresses are present, so each buttress supports a downward load of 2,693,840 lb>36 = 74,829 lb. This is the vertical component of the force developed in each buttress. (See Figure 4.) The total inclined force in the buttress is given by F = 74,829 lb>sin 38° = 121,542 lb This is the total force present in the lower stem of the Y. (Forces in the upper arms of the Y could be found by considering their angular orientation.) Next, find the approximate force in the lower ring that keeps the buttress from spreading outward. (The ring serves a function analogous to that of a tie-rod in an arch.) As can be seen from Figure 4, the horizontal component of the force in the Y buttress acts outwardly. This force is resisted by other forces developed in the ring that have inwardly directed components. If the ring is modeled approximately as a 36-sided polygon, then it is evident that 2T sin u = 121,542 cos f, T = 549,455 lb

1

or

2T sin 5° = 121,542 cos 38°

This example is adapted from a classic older textbook, Structure: An Architect’s Approach, by H. Seymour Howard, Jr., New York: McGraw-Hill Book Company, 1966.

Shell Structures

FIgure 4 A highly approximate analysis of Il Palazzetto del Sporto by Pier Luigi Nervi. The dome was made of precast ferroconcrete elements supported by Y-shaped buttresses. A huge tension ring is buried in the ground.

 



Copyright © 2013. Pearson Education Limited. All rights reserved.



This force, developed in the ring as a consequence of the lateral spreading action of the buttresses, is massive! A segmentally posttensioned reinforced-concrete ring was used in the structure to carry this force. The posttensioning put the ring into an initial state of compression that counteracted the tension induced by the spreading action of the buttresses. The simplified analysis presented has not addressed the distributed forces developed within the surface of the shell itself. These forces consist of meridional and hoop forces or stresses that act in the plane of the surface. Meridional forces act in the longitudinal direction (from the apex to where the shell terminates) and are discussed in Section 2.4. Hoop forces act in the circumferential, or latitudinal, direction and are discussed in Section 2.5. In the latter case, the tops of the buttresses collect the outward- and downward-directed components of the meridional forces and transfer them down to the surrounding tension ring. The approach just discussed is a direct method for finding buttress and tension ring forces without first calculating distributed meridional forces.

Shell Structures

2.2

membrane action in Shell Surfaces

A good way to envision the behavior of any shell surface under the action of a load is to think of the surface as analogous to a membrane, a surface element so thin that only tension forces can be developed. A soap bubble and a thin sheet of rubber are examples of membranes. A membrane carrying a load normal to its surface deforms into a three-dimensional curve and carries the load by in-plane tension forces that are developed in the surface of the membrane. The load-carrying action is similar to that in a crossed-cable system. The basic loadcarrying mechanism of a rigid shell of a similar geometry is analogous to that produced in an inverted membrane. Of primary importance is the existence of two sets of internal forces on the surface of a membrane that act in perpendicular directions. Also important is the existence of a tangential shearing stress developed within the membrane surface (and that is associated with the twist normally present in the surface), which also helps carry the applied load.

Copyright © 2013. Pearson Education Limited. All rights reserved.

2.3

types of Forces in Spherical Shells

The existence of two sets of forces in separate directions within the surface of a spherical shell tends to make the shell act similarly to a two-way plate structure. The shear forces between adjacent plate strips in a planar plate structure that were shown to contribute to the load-carrying capacity of the plate are present in shell structures as well. These two characteristics—the development of shear forces and two sets of axial forces rather than one—characterize the difference between the structural behavior of a shell and that of a series of arch shapes of a similar general geometry that are rotated about a point to form a similar shape. In an arch, no bending is present if the arch is funicularly shaped for the applied full-loading condition. If the loading is changed, however, to partial loading, substantial bending is developed in the arch. In a shell, the in-plane meridional forces (see Figure 5) are induced under full loading and are not unlike those in the analogous arch. Under a partial-loading condition, however, the action of the shell differs remarkably from that of the analogous arch, in that no bending is developed because of the other forces that act in the hoop direction and which also are developed in the shell. These hoop forces act in the circumferential direction and perpendicularly to the meridional forces. The hoop forces restrain the out-of-plane movement of the meridional strips in the shell that is caused by the partial loading. (Bending under partial loading in an arch is accompanied by a movement of this type.) In a shell, the restraint offered by the hoop forces causes no bending to be developed in the meridional direction (or in the hoop direction, for that matter). As a consequence, a shell can carry variations in loads by the development of in-plane stresses only. The plate shears mentioned earlier also contribute to this capability. The variations in load patterns involved, however, must be gradual transitions (e.g., a gentle change from full to partial loads) for bending not to develop. Sharp discontinuities in load patterns (e.g., concentrated loads) cause local bending to occur. Thus, any unusual forces in an arch cause bending that is propagated throughout the entire arch, while the effects of analogous forces in shells remain more localized. The shell is thus a unique structure that can be said to act funicularly for many different types of loads, even if its shape is not exactly funicular. In the example just discussed, the funicular shape for an arch carrying a uniform load would be parabolic. A spherical shell with a surface form that is not parabolic also will carry loads by in-plane forces. In this case, however, hoop forces will be set up even in the fullloading case because the shape is not exactly funicular. The meridional forces in a shell under full vertical loading are always compressive (by analogy with the action of an arch). The hoop forces, however, may

Shell Structures

FIgure 5 Meridional and hoop forces in spherical shells.







Copyright © 2013. Pearson Education Limited. All rights reserved.











 

be in tension or compression, depending on their location in the shell. (See Figure  5.) In a semicircular shell or one with a high rise, lower meridional strips tend to deform in an outward direction. Hoop forces containing this movement are therefore in tension. Near the top of the same shell, the meridional strips tend to deform inwardly. Hoop forces resisting this movement are therefore in compression. The stresses associated with meridional and hoop forces are fairly small for a uniform loading condition. Point loads, however, can cause high stresses and should be avoided on shell surfaces.



Shell Structures Holes in shell surfaces of the type just mentioned are possible but should be avoided because they disrupt the continuity, and hence the efficiency, of the shell surface. If holes are used, the shell must be specially reinforced around the edges of the holes.

2.4

FIgure 6 Hoop forces in spherical shells.



meridional Forces in Spherical Shells

The internal forces and stresses in axisymmetric shells that are uniformly loaded can be found easily by applying the basic equations of equilibrium. We will analyze a dome in detail as an example. Consider the dome segment illustrated at the bottom of Figure 6. Assume that the loading is a uniform gravity load distributed on the surface of the shell (e.g., the shell’s own dead weight and the weight of insulative or protective coverings). If the total of all such loads acting downward is denoted as W and the meridional in-plane internal force per unit length present in the shell surface as Nf, a consideration of equilibrium for g FY = 0 yields W = 1Nf sin f212pa2, where f is the angle defining the shell cutoff and a is the instantaneous planar radius of the sphere at that point. The Nf forces in the shell are in-plane compressive forces developed in the shell at the horizontal section defined by f. The vertical components of these forces (assumed uniform around the periphery of the shell) are simply Nf sin f. Because the Nf forces are expressed as a force per unit length (e.g., lb>ft or kN>m) along the section, the total upward force associated with all the continuous Nf forces is the instantaneous circumference of the shell at that point (which is given by 2pa), multiplied by Nf sin f (i.e., a total length times a force per unit length yields a total force). This upward force must be of a magnitude that exactly balances the downward force—hence the expression W = Nf sin f12pa2. This expression can be rewritten in terms of the actual radius of the sphere by noting that a = R sin f. Thus, W = Nf sin f12pR sin f2. Solving for Nf, we obtain

Copyright © 2013. Pearson Education Limited. All rights reserved.

Nf =



W 2pR sin2 f

If the total load acting downward 1W2 is determined, the internal forces in the shell can be found directly. Because these internal forces are expressed in terms of a force per unit length (e.g., lb>ft or lb>in.), actual internal stresses expressed in terms of a force per unit area (e.g., lb>in.2 or kN>mm2) can be found by dividing by the shell thickness, or ff = Nf >t. [This expression, seemingly a force divided by a thickness, may look odd for a stress measure (always force/area), but one part of the area is already reflected in the lb>ft definition of Nf. For example, if Nf = 1200 lb>ft (or 100 lb>in.) and the shell thickness is 5 in., then ff = Nf >t = (100 lb>in.)>15 in.2, or 1100 lb2>15 in.2 2 = 20 lb>in.2, which is still force>area. Occasionally, the expression is written as ff = Nf >tL, where L is the unit length.] Note that Nf acts in the vertically oriented meridional direction, but the unit length is measured along the circumferential direction. An expression can be easily derived that directly incorporates the load acting downward. If the load per unit area of shell surface acting downward is denoted by w, equilibrium in the vertical direction yields g FY = 0: -

f2

Lf1

w12pR sin f2R df + Nf sin f12pR sin f2 = 0

where f 1 and f 2 define the segment of shell considered. The term on the left thus defines W. For f 1 = 0, Nf =

Rw 1 + cos f

Shell Structures This expression is identical to Nf = W>2pR sin 2 f. Either expression defines the meridional forces present at a horizontal section.

2.5

hoop Forces in Spherical Shells

Hoop forces that act in the circumferential, or latitudinal, direction are typically denoted as Nu, are expressed in terms of a force per unit length, and can be found by considering equilibrium in the transverse direction. Alternatively, use can be made of the results of the membrane analysis, where the in-plane forces in a membrane that act perpendicularly to one another are related by the general expression pr = T1 >r1 + T2 >r2. This expression is immensely valuable in the analysis of shells because, by using it, the hoop forces 1Nu 2 can be related to the meridional forces 1Nf 2 that act in the longitudinal direction. Because the load being studied acts downward rather than radially outward, however, the external force expression must be adjusted. The radial component of the downward load is given by pr = w cos f. The expression relating hoop to meridional forces then becomes w cos f = Nf >r1 + Nu >r2, or Nu = r2 1w cos f2 - (r2 >r1) Nf. In a sphere, r1 = r2 = R, and substituting the expression previously found for Nf, we have Nu = Rw a -

1 + cos f b 1 + cos f

The preceding is a simple expression for hoop forces in terms of the radius 1R2 of the sphere and the downward load 1w2. Figure 6 illustrates hoop forces in shells. Nu forces act in the circumferential direction (over a unit length in the meridional direction).

example Consider a dome having a spherical radius of 100 ft (30.48 m), a thickness of 4 in. (100 mm), and an aspect angle of 45°. Determine the meridional and hoop forces at the base of the shell for a loading of 100 lb>ft2 (4788 N>m2). The load includes all applicable loads. Solution:

Copyright © 2013. Pearson Education Limited. All rights reserved.

Meridional forces: Nf =

1100 ft2(100 lb>ft2) Rw = 5858 lb>ft in compression = 1 + cos f 1 + 0.707

=

130.48 m214788 N>m2 2 1.707

= 85,494 Nm

Meridional stresses: ff = =

Nf t

=

15858 lb>ft2 >12 in.>ft 4 in.

85,494 N>m

11000 mm>m21100 mm2

=

488 lb>in. 4 in.

= 122 lb>in.2 in compression

= 0.85 N>mm2

Hoop forces:

Nu = Rw a -

1 1 + cos f b = 1100 ft2 a100 lb>ft2 b c - a b + 0.707 d 1 + cos f 1.707

= 1212 lb>ft in compression

= 130.48 m214788 N>m2 2 c - a

1 b + 0.707d = 17,684 N>m 1.707

Shell Structures Hoop stresses: fu = =

1212 lb>ft Nu = = 25.3 lb>in.2 in compression t 112 in.>ft214 in.2 17,684 N>m

11000 mm>m21100 mm2

= 0.177 N>mm2

The stresses in the sphere at the point in question are thus extremely low, a characteristic of most shell structures.

2.6

distribution of Forces

The distribution of meridional and hoop forces can be found by plotting the equations for the two forces (Figure 7). As is evident, the meridional forces are always in compression, while the hoop forces undergo a transition at an angle of 51°49=, as measured from the perpendicular. Shells cut off above this angle develop compression stresses only in their surfaces, whereas deeper shells can develop tension stresses in the hoop direction. The magnitude of the stresses, however, always remains relatively low. An interesting way to look at the overall behavior of a dome-and-ring assembly is illustrated in Figure 8. As in other structures, the external moment at a section must be balanced by an internal resisting moment (in this case, provided by a couple formed between the hoop and ring forces). Thinking of the structure in these terms also helps explain the development of tension hoop stresses in a dome.

2.7

concentrated Forces

The reason concentrated loads should be avoided on shells is shown by analyzing the meridional forces present under such a loading. The general expression found before was Nf = W>2pR sin2 f, where W was the total load acting downward. For a shell carrying a concentrated load P, the expression becomes Nf = P>2pR sin2 f. If the load is applied at f = 0 (the crown of the shell), a situation arises such that, directly beneath the load, the stresses become indefinitely large in the shell 1i.e., as f S 0, sin f S 0 and Nf S ∞2. A failure would occur if the shell surface could offer no bending resistance and the load were characterized as a point force. In any event, such forces should be avoided on shell surfaces.

Copyright © 2013. Pearson Education Limited. All rights reserved.

2.8

Support conditions: tension and compression rings

A major design consideration in a shell of revolution is the nature of the boundary or support conditions. In much the same way that buttresses or tie-rods must be used to contain the horizontal thrusts of arches, some device must be used to FIgure 7 Stress distribution in a spherical dome carrying a uniformly distributed load along the surface of the sphere.

 

Shell Structures

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 8 Basic shell behavior.

absorb the horizontal thrusts associated with the meridional in-plane forces at the lower edge of the shell. In a dome, for example, a circular buttress system could be used. Alternatively, a planar circular ring, called a tension ring, could be used to encircle the base of the dome and contain the outward components of the meridional forces (Figure 9). Because the latter are always in compression, their horizontal components are always outwardly directed at the shell base. The containment ring is therefore always in tension. If a hole were cut out of the shell at its crown, however, the same meridional force components would be inwardly directed. A ring used to absorb these forces would thus be in compression. A tension ring is a planar ring against which the outward thrusts push, causing tension to develop in the ring. Consider the sum of the forces along any direction in a section cut through the ring as it sits on a horizontal plane. The horizontal components 1Nf cos f2 of the meridional forces act outwardly along the circumferential length of the ring and produce a total outward thrust that is in turn balanced by the internal forces developed in the tension ring. This outward thrust can be shown to be equal to the unit outward thrusts times the projected length over which they act (i.e., the diameter). Thus, 2T = 1Nf cos f22a,

Shell Structures

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 9 Spherical shell support conditions.

or T = 1Nf cos f2a, where a = R sin f is the radius of the tension ring. This expression follows from 2T = 1 1Nf cos f21a21df2. A tension ring absorbs all the horizontal thrusts involved. When resting on the ground, it also provides a continuous footing for transferring the vertical reaction components to the ground. Alternatively, the ring can be supported on other elements (e.g., columns), which then receive vertical loads only. The use of a tension ring does, however, induce bending in the shell surface where the ring and shell intersect. The bending moments generated are due largely to deformation incompatibilities that exist between the ring and shell. Because the ring is always in tension, it expands outward. The hoop deformations in the shell, however, may be compressive (thus, the shell edge deforms inward), depending on the Nf and Nu forces. In any event, it is unusual for the deformation tendencies to be the same. Because the elements must be joined, the edge beam restricts the free movement of the shell surface, so bending is induced in the shell’s edge. As

Shell Structures

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 10 Edge disturbances in spherical shells.

mentioned before, however, this bending rapidly dies out in shells, so the bulk of the shell’s surface is unaffected. The shell edge is locally stiffened and reinforced for bending [Figure 10(c)]. Ideally, the supports should be made so that they do not cause any bending to be developed in the shell surface. Fixed-edge conditions should be avoided if possible. One possible solution, illustrated in Figure 11(b), has the shell pinned along its peripheral edge. However, unlike the situation with the arch, the presence of hoop forces causes the shell to deform naturally in the out-of-plane direction. Restraining this deformation by a pinned connection would be equivalent to applying forces to the shell’s edge, thus inducing bending. A peripheral roller support, as illustrated in Figure 11(c), is preferable because no restraint exists in the out-of-plane direction. Such supports, however, are difficult to build for shells. In addition, slight angular changes due to movement still cause some bending to be induced (although less than in the case of fixed or pinned conditions). For reasons of practical construction, some bending is often allowed to develop at the shell edge for the sake of using easy-to-construct edge and foundation conditions. The shell is locally stiffened (usually by increasing its thickness) around the edge and specially reinforced for bending. The problem with deformation incompatibilities can generate design approaches intended to minimize undesirable consequences. One effective method is to utilize posttensioning to control deformations. A support ring that is normally in tension, for example, can be posttensioned so that compressive forces (and hence compressive deformations compatible with those in the shell in the hoop direction) initially exist in the ring. The outward thrusts of the dome would relax the compressive forces (and increase the tension in the posttensioning wires). If the initial amount of posttensioning is controlled, final ring deformations can be controlled and thus minimize deformation incompatibilities with the shell. The shell surface itself also can be posttensioned in the hoop direction for added control of shell forces and deformations.

FIgure 11 Shell support conditions.

Shell Structures

FIgure 12 Buckling in thin shells.

(a) (a) (a) (a) (a) (a) (a)

Snap-through buckling. The whole shell can inwardly buckle due to external loads. Shells with flat curvatures are sensitive to this type of buckling. Increasing shell curvature reduces the possibility of this type of buckling.

(b) (b) (b) (b) (b) (b)

Local buckling. Rather than the whole shell snapping through, a portion of the shell can buckle inward. Again, sharp shell curvatures decrease the possibility of this type of buckling.

2.9

other considerations

Several factors other than those already discussed must be taken into account in the design of shells. Critical among these additional factors is the need to ensure that a shell does not fail prematurely and disastrously in a buckling, or surface instability, mode. When the curvature of the shell’s surface is flat, snap-through or local buckling can be a severe problem. (See Figure 12.) As with the analogous phenomenon in long columns, instability can occur at low stress levels. It can be prevented by using surfaces of sharp curvature. The need to maintain sharp curvatures, however, hampers the use of large spans with low-profile shells in which little curvature is present. The problem also is severe in reticulated shells made of small, rigid linear elements (e.g., geodesic domes). More advanced books discuss the shell-buckling problem and propose measures for predicting whether instability is a problem in a given design context. Another major concern is that shells must withstand loads other than those acting vertically. Figure 13 illustrates the stress trajectories in a spherical dome due to wind forces. Usually, wind forces are not too critical in the design of shell structures. However, earthquake forces that also act laterally can pose serious design problems. When such loadings are possible, special care must be taken with the design of the support conditions.

3

cylIndrIcal ShellS

Copyright © 2013. Pearson Education Limited. All rights reserved.

The relative proportions of the shell and its support conditions are a critical influence on the behavior of structural forms that are defined by translational surfaces. Consider the cylindrical surface supported on walls, illustrated in Figure 14(a). In this structure, commonly called a vault, the surface behaves like a series of parallel arches, as long as the supporting walls can provide the necessary reactions. If the surface is rigid (e.g., made of reinforced concrete), the surface also exhibits plate action (i.e., shearing forces are developed between adjacent transverse strips), which is useful for carrying nonuniform loads. The same type of action occurs when such a surface is supported on stiff beams. The beams, in turn, transfer the loads to the supports by bending. The behavior of a very short shell may differ appreciably from that just described if transverse end stiffeners are used. Surface loads may then be transferred directly to the end stiffeners through longitudinal plate action. FIgure 13 Stress trajectories in a spherical dome due to wind forces.

Shell Structures As the shell structure becomes much longer relative to its transverse span, a different type of action begins, particularly if edge beams are not used or highly flexible ones are used. Any supporting edge beam, it should be noted, naturally tends to become more flexible as its length increases. The cylindrical surface will again tend to start behaving like an arch in the transverse direction. Flexible edge beams (or no-edge beams) do not, however, provide any appreciable resistance to the horizontal thrusts involved. Consequently, no arch-like action is ever exhibited in this direction. Indeed, if no-edge beams are present, the longitudinal free edges tend to deflect inward rather than outward under a full loading. A different type of load-carrying mechanism must therefore be present. Structures of this type are called barrel shells. The principal action in a long cylindrical barrel shell is in the longitudinal, rather than the transverse, direction. A type of longitudinal bending occurs that is analogous to that which occurs in either simple beams or folded plates. Compressive stresses in the longitudinal direction are developed near the crown of the curved surface, and tension stresses are developed in the lower part. The analogy with folded-plate structures is useful because many of the same design principles are present. Transverse stiffeners, for example, are useful in increasing the load-carrying capacity of a barrel shell. The more that stiffeners are used, or if the barrel shell considered is one of a series of adjacent shells, the beamlike behavior becomes more pronounced and beam analysis techniques yield more accurate results. Barrel shells whose lengths are three times (or more) their transverse spans exhibit this longitudinal type of behavior.

Copyright © 2013. Pearson Education Limited. All rights reserved.

4

hyperbolIc paraboloId ShellS

The behavior of shells having ruled surfaces may be envisioned by looking at the nature of the curvatures formed by the straight-line generators. If the edge conditions can offer restraint (i.e., foundations or stiff edge beams), an arch-like action will exist in regions of convex curvature and a cable-like action in regions of concave curvature. The presence of compression or tension forces in the surface depends on which action exists. When surfaces become flat due to reduced curvatures, a plate action in which bending dominates may be present (necessitating increased plate thicknesses). If shell edges are not supported, a beam behavior may be present. A ruled surface may be made by translating two ends of a straight line over two parallel, but twisted, straight lines. It is interesting that the shape formed also can be described as a translational surface generated by translating a concave parabola over a convex one. In this type of structure, an arch-like action will be present in the direction of convex curvature and a cable-like action in the concavely curved perpendicular direction. The stress field in the plate is thus compressive in one direction and tensile in the perpendicular direction, each of which is at 45° to the original straight-line generators. (See Figure 15.) Membrane forces are expressed on a unit width basis for the arch- and cablelike actions noted, which are at 45° to the edges. Under a uniform loading, the force at the top of any arch or cable strip is Cx = Tx = wsL2x >8dx, where Lx is the span of the strip, dx is the instantaneous height of the strip, and ws is the load carried by the cable or arch strip. Because arch and cable strips cross one another, the normal distributed load w present at a point is half-shared between crossing strips. Hence, ws = w>2 on a strip and Cx = Tx = wL2x >16dx. Because dx is small, this value also can be taken as the maximum force present in a strip of unit width. (Stresses can be found by noting that f = C>t, where C is a force per unit length and t is the plate thickness.) Thus, if C equals 6000 lb>ft, or 500 lb/in., and t is 4 in., then f = 1500 lb>in.2>4 in. = 125 lb>in.2 Both Lx and dx vary among adjacent strips because of the 45° strip orientation. By expressing dx in terms of Lx , and Lx in terms of sides a and b for a hyperbolic paraboloid surface of maximum height h, the forces become C = -wab>2h and T = wab>2h, where h is the maximum rise present.

FIgure 14

Cylindrical shells.

Supporting wall a) Vaults: Supported continuously along their longitudinal edges, loads are carried by an archlike action.

Stiff edge-beam carries loads to supports b) Short shell with stiff edge beam: The vault rests on a beam that replaces the wall support in (a), carrying loads in bending to the columns.

Compression Tension Neutral axis

c) Long cylindrical shell: Loads are carried in bending of the thin surface. Compressive and tensile bending stresses develop.

d) Long cylindrical shell. Principal stresses: Lines of tensile and compressive principal stress form a curved network over the shell surface.

Shell Structures Such an analysis also yields the curious result that C or T is a constant value in adjacent strips of any length throughout the shell. (Observe that when Lx is large, dx is large, and vice versa, so that Cx = Tx = wL2x >16dx remains constant.) Remarkably, the membrane force is more or less constant throughout the entire surface! When the envisioned strips terminate at the edges of the plate, they exert forces vertically and horizontally. An arch strip meets a cable strip at a 90° angle, 45° to the shell edge. Vertical components balance each other. The components of the horizontal thrusts perpendicular to the shell edges also balance one another. However, a resultant edge shear force is directed along the edge of the plate. Edge forces along a unit length are approximated by F = wab>2h. These forces accumulate along the length of the edge of the shell (a) against restraint points (e.g., a buttress) into large total edge forces: Ftotal = wa2b>2h. Depending on the location of free and constrained points, the total edge forces may be in either tension or compression. Large edge beams may be required to carry the edge forces, which in turn bear against buttresses or other support conditions and, in some cases, cause them to splay outward—as is the case in Figure 15(a). Tie-rods may be used effectively to carry the edge forces. Shell surfaces of many types may be made by aggregating simple hyperbolic shapes. Unusual shapes may be formed by cutting the surface in directions other than along the straight-line generators, as shown in Figure 15(d).

FIgure 15 Hyperbolic paraboloid. The general nature of the forces in many shells can be ascertained by studying the shell curvatures present. Concave curvatures indicate a cable-like action (a tension field) and convex curvatures an arch-like action (a compression field). W

Standing parabola— compression curve Hanging parabola— tension curve

Copyright © 2013. Pearson Education Limited. All rights reserved.

Umbrella shell

h b Edge beam transfers shell forces to the supports

Edge shears

Corner supported

a Reactions

Shapes cut from ruled surface

Shell Structures

5

Free-Form SurFaceS

Developments in computational form-finding have led architects and others into using a host of geometrical forms that we describe here as free-form shapes. These forms are not easy to characterize because they can be generated in a variety of ways within a computational environment. Multiple complex spine shapes can be developed, for example, and even more complex surfaces lofted over them. Techniques of this type result in surfaces with complex curvatures in all directions. Synclastic and anticlastic curvatures, for, example, may exist within the same surface. We noted in the opening of this chapter that not all structural surfaces carry loads by efficient membrane action. There must be a particular relationship between the geometry of the structure and the nature of the applied loads; otherwise, significant bending stresses develop (in addition to in-plane forces) that lead to structural inefficiencies. Unless a free-form surface is specifically shaped with the objective of obtaining membrane action, it is highly probable that the resulting surface will not exhibit membrane action. This does not mean, of course, that such forms should be avoided or not used—these shapes have become popular for several good reasons— however, nothing suggests they are intrinsically efficient in a structural sense. A typical example of a free-form surface created within an advanced digital modeling environment is shown in Figure 16. Note that the surface has reverse curvatures and some flat areas. Nothing intrinsic about this surface suggests it would carry loads by membrane action. The issue is further compounded by the underlying algorithms used within a digital modeling environment to generate the surface forms—most are formulated for general visualization purposes, not structural shaping purposes (some advanced systems do allow for funicular equations to be input that would provide this capability, but most systems are intended for general visualization purposes). In the shape shown, it is expected that high bending moments and associated bending stresses will be developed. In small-span structures, it might be possible to carry these internal stresses by thickening the surface. In longer-span structures, however, it is necessary to introduce a framing system of one type or

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 16 Free-form surfaces. These surfaces are not intrinsically structurally efficient because they do not exhibit membrane action, and bending moments are typically developed within them.

(a) Complex surface form not exhibiting membrane action. Consequently, high bending moments are developed as well as in-plane forces.

(b) Carrying moments and forces by surface action alone can require large thickness that normally are not feasible for large-span structures.

(c) Large spans normally require extensive primary and secondary support systems. In the approach shown, the supporting members are in bending and behave like beams.

(d) When the surface is intended to be transparent, additional surface subdivisions and supports are normally necessary.

Shell Structures

FIgure 17 Complexly shaped building envelope. Often, the curvatures present in complexly shaped surfaces do not allow for the surface to act as primary structural element. In the case of the BWM World in Munich (Architect: Coop Himmelb(l)au) the structural system in the roof is a deep steel space frame, while the twisted vertical element on the right works as a single-layer grid shell.

The steel spaceframe receives a nonstructural panel cladding.

(b) Space frame

The steel grid shell is filled in with glass panels.

Steel space frame

Grid shell

Copyright © 2013. Pearson Education Limited. All rights reserved.

(c) Section diagram

(a) The surface curvature is continuous over the space frame and the shell.

another as the primary load-carrying system. The surface itself ceases to become the primary load-carrying system and becomes nonstructural (except for transferring minor surface loads to supporting elements—rather in the fashion of curved decking). The framing system can have a geometry that reflects the surface shape, as is illustrated in Figure 16(c), or it can have some other geometry with transition elements used between the surface and the framing. Normally, the framing elements are in significant bending and are best thought of as either curved beams, frames, or space trusses. Consequently, structural depths are much larger than those associated with structures exhibiting true membrane or shell action, as described earlier. The advent of computer-aided modeling techniques has made creating these kinds of structures possible, but they nonetheless remain highly complex to build and analyze. An example of a complexly shaped structure is shown in Figure 17. The type of structural analysis most appropriate for a free-form shell is dependent on which of the structural alternatives just noted is used. In a free-form surface that uses a framing system to carry primary loads, the system is probably best thought of—and analyzed—as a curved beam, frame, or truss that is primarily in bending. If the continuous surface itself is anticipated to be the primary load-carrying system, then more-advanced finite-element techniques must be used to obtain bending moments and other forces and deformations. In addition, one should also note that the actual making of a complexly curved surface (whether structural or not) can be extremely difficult. It is frequently necessary to subdivide the surface into smaller units. Most subdivision approaches—say into patterns of quadrilaterals—result in individual units that are not intrinsically planar, and thus make the construction of the surface difficult.

Shell Structures

FIgure 18 Grid shells. A curved surface is resolved into linear or curvilinear members that are interconnected. The Mannheim Multihall (a) is a timber system with up to four layers of thin timber laths connected with rigid joints. The shape was derived from a hanging physical model. The steel gridshell in Bad Cannstadt (b), a vault-like shell, has glass infill panels and is stabilized by diagonal steel cables.

(a)

Copyright © 2013. Pearson Education Limited. All rights reserved.

6

(b)

grId ShellS

The grid shell is a rigid spatial structure formed by interconnected networks of curved or linear members to produce a patterned surface form with sufficient strength and rigidity to carry in-plane forces and some degree of bending. Members are either rigidly connected to prevent racking (parallelogramming of modules) of the whole surface in the in-plane direction, or crossed cables are used for the same purpose (particularly when roof panels are simple infills). Grid shells are most effectively used to create shapes that are structurally efficient, for example, funicular or spherical shapes, but also may be used to create free-form shapes that have some free-form qualities yet still have structural affordances (when spans are not long). Grid shells can appear quite thin, but many are fairly thick compared to true shell surfaces. Others may use out-of-surface cables for stabilization. The Mannheim Multihall grid shell illustrated in Figure 18(a) uses layers of rigidly interconnected curved and linear elements. The shape was derived from a hanging physical model and inherently has a positive structural action. It can accommodate bending due to partial loadings (it is relatively thick). Joint rigidity prevents racking of the whole surface in the in-plane direction. A more recent version of a grid shell is shown in Figure 18(b). The vault shape also is intrinsically efficient. Diagonal cables are used to prevent racking in the in-plane direction because the glass panels are only infills. The risk of buckling can be reduced by stabilizing cable systems. (See Figure 19.) Other grid shells have been developed, largely by the firm of Schlaich and Associates, that use a geometrical approach that yields surfaces that appear quite complex and have free-form qualities but which can still be subdivided into planar quadrilaterals. Surfaces are derived from the concentric expansion or contraction of parallel curves or by using particular translational surfaces that result from sliding one curve over another. A carefully designed roof sandwich construction involving transparent surfaces, supporting bar networks, and crossed cables then accompanies these special surfaces. Resulting surfaces can be thin and elegant, while at the same time allowing designers to explore interesting spatial shapes.

FIgure 19 Stabilizing cables. All cables join at a single point. The photograph shows the detailed cable connection of the grid shell in Figure 18(b).

Shell Structures

QueStIonS 1. What is the total weight of a portion of a spherical concrete shell that has a radius of 200 ft and is subtended by an angle of 45° and made using a roof surface that has a unit area weight of 65 lb>ft2? Answer: 4,787,786 2. A portion of a spherical shell has a radius of 150 ft and a subtended angle of 30°. Assume that the dead load and live load carried is wDL + LL = 70 lb>ft2. What hoop and meridional forces are developed at the base of the structure 10 = 30°2? Use the Nf and Nu expressions. Answers: Nf = 5627 lb>ft and Nu = 3465 lb>ft.

3. Assume that the same shell described in Question 2 has a structural thickness of 4 in. What stress levels correspond to the hoop and meridional forces present? Answer: ff = 117 lbs>in.2 and fu = 72 lbs>in.2 4. For the same shell described in Question 2, assume that a tension ring will be used at the base of the shell. Are the hoop forces found in Question 2 the same as or different from the force you would except to be present in this tension ring? Discuss. 5. A rigid shell having a spherical radius of 300 ft is cut off at an angle of f = 51°49=. Assume that the shell carries a distributed live load of 50 lb>ft2. What are the in-plane forces are in the shell at f = 0°? At f = 51°49=? What are the associated ff and fu stresses? Ignore the dead load of the shell itself. Assume t = 5 in. 6. A rigid reinforced-concrete shell having a spherical radius of 200 ft is cut off at an angle of f = 35°. The shell thickness is 3 in. Assume that the unit weight of the shell material is 150 lb>ft3 and that the shell carries a live load of 60 lb>ft2. Draw forcedistribution diagrams of the type illustrated in Figure 7. Indicate numerical values. 7. Assume that a tension ring is used in conjunction with the shell described in Question 6. What is the magnitude of the force developed in the ring? Answer: T = 1,007,283 lb 8. Assume that (1) a hyperbolic paraboloid surface is supported as shown in Figure 15 (left), (2) a = b = 50 ft, (3) h = 20, and (4) the combined live and dead loading is 60 lb>ft2. Find the membrane forces in the surface and the maximum edge forces present. Note the location of the maximum edge forces.

Copyright © 2013. Pearson Education Limited. All rights reserved.

9. By using library resources, review the work of Heinz Isler in relation to funicularly shaped shell structures. Try to replicate one of his physical models using hanging fabrics. Stiffen the hung fabric with a coating of plaster and try inverting it. Write about the problems you encountered when you tried to use a hanging fabric approach to determine a shell form.

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

1

IntroductIon

The process of designing the building structure is intimately linked to that of designing the overall building. Building design decisions both determine and are determined by decisions on the structural system level. Teams of architects and structural engineers approach the design process in many different ways. Some may be interested in structure as a way to organize, give scale to, and pattern the overall built volume (Figure 1). Structural elements may play a visually and spatially dominant role in defining the identity of the building (Figure 2.). Other approaches may seek to suppress a direct reading of structure and treat the structural system as a more neutral element in the overall assemblage of the building. The stage at which structural considerations enter the design process varies accordingly. It depends not only on the design philosophy of the architect and the structural engineer, but it also depends on the real structural challenges at hand. When designing long-span or tall structures, for example, design teams necessarily address structural needs early on because much of the essential design is that of the structure. Buildings with these extreme structural challenges often clearly communicate, through their forms and patterns, issues of structure. Programs that do not require large spans or slender forms often allow for more leeway, permitting architects to pursue other interests. Structural issues on the conceptual design stage are normally more manageable and can be handled in a slightly less aggressive way. This chapter deals with structural system decisions common to most design philosophies and building programs, with a focus on the design of the vertical load-resisting system. We first look at basic issues of selecting structural elements such as beams, trusses, arches, and so forth based on the loads and spans present (Section 2.1). The choice of these elements depends largely on the arrangement of the vertical supporting structure, an issue addressed in Section 2.2. The arrangement of these elements into the next larger unit—a single structural bay—is discussed next. Here, we differentiate between one-way and two-way systems as the most universal principles underlying the design of a typical structural bay that, once replicated and arrayed, can become the fundamental building block of the overall structural system (Sections 2.3 and 2.4). Expanding the viewpoint once more, Sections 3 through 5 discuss how these structural units are organized into common structural grids and patterns, horizontally as well as vertically, and how different grids merge, transition, and accommodate disruptions. Finally, the chapter From Chapter 13 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Structural Elements and Grids: General Design Strategies

FIgure 1 The shaped roof structure lends a more intimate scale to the vast interior space of the Madrid airport terminal.

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 2 Space and structure: Spatial characteristics of a simple rectangular space can be substantially modified through strategic choices and arrangements of structural system.

(a) Spacing of primary elements: Rigid frames, closely spaced, create a sense of enclosure.

(b) Spacing the primary frames further apart generates the need for a secondary beam system. The space feels more open.

(c) Increasing the spacing of primary trusses even further begins to suggest a subdivision of the enclosed space.

(d) Directional structure: Creating a pin-pin column on the left versus the rigid corner on the right suggests transparency to the left side.

(e) Arranging rigid frames in a zigzag pattern animates the space more dynamically.

(f) Randomized arrangements of beam and column members follow a more ornamental approach to structural envelops. Structural efficiencies are usually compromised.

Structural Elements and Grids: General Design Strategies reviews issues that relate, in a more direct way, to the overall spatial design process. The relation of structure to the building program and the space-forming and expressive characteristics of structural systems are discussed.

2 2.1

Structural element SelectIon and SyStem organIzatIon Horizontal Spans

Copyright © 2013. Pearson Education Limited. All rights reserved.

common Spans. Span length is a crucial factor in selecting a structural response for a given situation. Some structural systems are appropriate for certain span ranges yet not for others. To give a graphic sense of the scale and spanning capability of different systems, Figure 3 illustrates some typical economic span ranges for several one- and two-way systems. The maximum spans indicated are longer than typically encountered spans but do not represent maximum possible spans (most of the systems could be made to span farther). The minimum limitations are intended to represent a system’s lower feasibility range, based on construction or economic considerations. Nothing is sacred or fixed about the precise span figures shown; they are intended to give a feeling for relationships between structural systems and spans and nothing more. underlying Principles governing Span lengths. The importance of the structural span is evident from the observation that design moments for uniformly distributed loads are proportional to the square of the length of the span. Doubling a span length, for example, increases design moments by a factor of 4; quadrupling span lengths increases design moments by a factor of 16. Member sizes, of course, depend closely on the magnitude of the design moment present, even though deflections can be the governing structural design criteria for longer spans. Figure 4 illustrates how several different structural elements provide an internal resisting moment to balance the external applied moment. The basic mechanism is the same in all cases. A force couple is formed between compression and tension zones whose magnitude exactly equals the external applied moment. For a given applied moment, the magnitude of the internal forces developed in the compression and tension zones depends directly on the magnitude of the moment arm that is present. The deeper the structure, the greater is the moment arm and the smaller are the tension and compression forces present. The process of designing an appropriate structure for a given span range is directly linked to choosing a system with appropriate internal moment resistance. In low spans, the design moments are low, and any of the basic structural options shown in Figure 3 are possible. As spans lengthen, however, design moments increase so rapidly that some of the options become less feasible. Constant-depth members, such as beams, are relatively shallow; increases in span lengths quickly lead to large tensile and compressive stresses that provide the internal resisting moment. Because the depth of these members is inherently limited, the increases cannot entirely be compensated for by increasing the moment arm of the resisting couple or, indeed, by any other means (e.g., increasing the moment of inertia through increased flange areas). Therefore, members of this type are not capable of extremely large spans because a point is reached whereby the internal compression and tension forces become too large to be handled efficiently. Deflection control also becomes a governing consideration. long-Span Structures. As spans increase, fewer systems are available to resist the large external moments of long spans. As a result, it is often easier to choose a long spanning system compared to selecting short spanning elements where almost all systems can be made to work. A characteristic of appropriate long-span systems

Structural Elements and Grids: General Design Strategies

FIgure 3 Approximate span ranges of different horizontal spanning systems.

Copyright © 2013. Pearson Education Limited. All rights reserved.

300

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 4 Internal force couples that equilibrate external moments are developed in all structures. As the depth of the internal moment arm increases, the associated internal force increases, and vice versa.

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

is that their structural depths are large in comparison with their spans. The larger depth allows for a greater internal moment arm between tension and compression zones, thus keeping forces fairly small while still resisting significant external moments. Most long-spanning elements rely on axial forces rather than bending as a more efficient mechanism of load transfer. Shaped systems such as arches, cables, or shaped trusses are efficient for long spans because the distance between tension and compression zones responds to the varying external moment along the member. Cables, nets, or rigid shells have similar capabilities for long spans. Flat systems such as constant-depth trusses, space frames, and other structures also have been successfully used, but they do not normally match the efficiency of the shaped arch, cable, or truss elements. The geometry of common shaped structures limits their normal use to roofs of buildings or other situations where structural shapes can be allowed to vary. Rarely are shaped systems useful for providing the planar surfaces that are needed elsewhere in buildings—for obvious functional reasons. Intermediate- and low-Span Systems. When planar surfaces are a must, and as span or load-carrying requirements decrease in a building from the dramatic to the more commonplace, designers are faced with an increasing range of structurally viable options that are made from a wider range of materials. An inspection of Figure 3 reveals that, in the span ranges encountered in most buildings, several systems are potentially appropriate for use. For spans on the order of 16 to 20 ft (5 to 6 m), for example, either one- or two-way systems of all major types can be competitive. Which of the systems is most appropriate cannot be determined without a separate analysis consisting of designing alternatives for the specific span and load condition considered and without making a cost analysis. Usually, the lightest or least-involved construction type appropriate for a given span that is capable of carrying the design load is preferred. For example, in reinforced concrete, for a span of 20 ft (6 m) and for relatively light, uniformly distributed loads, the flat plate is often the preferable two-way concrete system. Other reinforced-concrete systems such as waffle slabs or beam/slab systems are possible, but their higher span and load capacities are not needed for short spans. From a structural viewpoint, there is no incentive to go to the trouble and expense of creating the special formwork required to construct these more complex systems when a simple flat plate would work just as well for the conditions at hand. With increases in span or load, however, the flat plate begins losing its viability, and these other systems become more appropriate. In low-span ranges, the deeper-shaped structures capable of long spans are still structurally possible, but the costs associated with the construction complexity of shaped structures do not offset possible material savings. Using a cable system, for example, to span a mere 15 ft (5 m), instead of wooden joists does not normally make sense—at least not from a structural point of view. The versatility of the truss in terms of its shape and how it is made make it one of the few structural elements that is useful for the full range of short, intermediate, and long spans. Specific characteristics and geometries of trusses for these extreme span ranges, of course, vary. Short- and intermediate-span trusses are typically closely spaced and highly repetitive—as is common with typical bar joist systems. Often, such elements are mass-produced. Long-span trusses are usually specially made and placed relatively far apart.

2.2

Basic Strategies

Support Strategies for optimized element design. Maximum moments in beams and frames can be reduced by strategically placing supports moved in from the ends of the horizontal elements,

Structural Elements and Grids: General Design Strategies

FIgure 5 Optimizing support locations. In many cases, a mutually beneficial fit between the vertical support pattern and the building program can be achieved.

.

Copyright © 2013. Pearson Education Limited. All rights reserved.

effectively producing cantilevering conditions. These manipulations, while potentially enabling the design of more material-efficient structural elements, have strong implications for the structural patterning used and may be a contributing factor to the design of overall structural grid. In the process of selecting a structural pattern in a building, it might be possible to introduce cantilevers into the building, but it is difficult to do so after the basic structural pattern has been formed. Cantilevers can serve well as overhangs covering circulation routes (Figure 5). The architectural implications of using cantilevers also include the visual suppression of the columns along the façade—the façade composition does not have to make reservations for the vertical support structure. These principles apply not only to simply supported beams but also to continuous beams and frames. density of structural elements. The discussion of primary spans in the previous section relates closely to the density of the structural system—its horizontal elements and the number and spacing of supports. Is it better to use what may best be described as concentrated support structures or to use distributed support structures? The descriptive terms refer to alternative design strategies for a given loading condition. The first approach is characterized by designing a few large members to carry the load, the second by designing a greater number of smaller elements. The question not only relates to the design of structural elements but also has wide-ranging repercussions for the organization of the overall structural grid discussed in Section 3. The support density will inevitably and significantly shape the appearance of the building as a whole and most likely impact the characteristic of the interior space. On the level of the individual structural elements, the question may be whether it is more efficient to carry a given load with one large element or with several smaller ones. How will choosing one large beam versus two parallel smaller beams affect structural behavior and the space the structure is forming? Is it preferable to use a space frame with many small members dispersed throughout the system or to use a series of more widely spaced large trusses? (See Figure 6.) The basic choice of concentrated versus distributed structural strategies appears again and again in many situations. Soil conditions and foundation design can, at times, play a significant role in guiding these decisions because highly concentrated supports with their larger reaction forces and moments could require special foundations which may or may not be advisable for given soil conditions.

Structural Elements and Grids: General Design Strategies

FIgure 6 Structural system choices often affect the characteristics of the internal space that they form. Space frames tend to create uniform, nondirectional spaces, whereas long-spanning trusses create a layered, directional space.

Copyright © 2013. Pearson Education Limited. All rights reserved.

2.3

one-Way Systems

Structures enclosing volumes are typically conceived in terms of vertical support systems and horizontal spanning systems. Many horizontal spanning systems, particularly those in steel and timber, are made up of hierarchical assemblies of different kinds of members. Common post-and-beam systems, for example, are made up of a series of beam, decking, and column elements. Alternatively, trusses or frames can be used in lieu of beams to support the decking elements. It is helpful to define members according to different locations in a hierarchical arrangement (Figure 7). Lower-level members gather loads from higher-level elements, and lower-level members define the necessary location of vertical supports. If the vertical support system is selected first, its type and geometry suggest something about the nature of the best hierarchical arrangement and appropriate horizontal spanning system. Usually, neither one dictates the design of the other, but the two are interactively designed together. Several simple hierarchies are illustrated in Figure 7. A simple one-level hierarchy can be made directly with basic surface-forming decking elements. Because reactions are line forming, load-bearing wall support systems are uniquely appropriate for this hierarchy. A two-level hierarchy may be made to achieve longer spans. Individual structural elements are arranged so that decking elements of shorter span are carried by periodically spaced longer spanning elements. These collector elements (they gather loads from surface-forming elements) can be articulated as beams, trusses, or other elements, sized and shaped for the span and loads present. The collectors deliver concentrated forces to the vertical support system. If the collector members are widely spaced and have long spans, the concentrated forces are quite high, and a columnar support system is normally dictated. If the collector members are closely spaced and spans are low, and if forces at the ends of members decrease accordingly, a load-bearing wall system or pilaster system may be a reasonable choice for the vertical support system. A three-level hierarchy involves introducing members at even lower levels. Lowest-level members are called primary collectors; middle ones are called secondary collectors. In systems of this type, the primary collectors end up picking up large forces from the secondary system and delivering highly concentrated forces to the vertical support system. A columnar support system is invariably required. Hierarchies with greater than three levels are possible, of course, but are rarely used, except with very long spans or in other unique circumstances.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Structural Elements and Grids: General Design Strategies

FIgure 7 Typical one-way horizontal spanning systems.

Structural Elements and Grids: General Design Strategies Short spans often may be made directly with one-level systems, whereas larger spans use two- or three-level systems. This principle is, of course, not universally applicable because several special elements (such as prestressed concrete tees) can efficiently span large distances directly. The principle is more applicable to systems that use minor surface-forming elements (e.g., planks, decking) initially to pick up a uniformly distributed loading. One-level systems tend to span distances with smaller overall structural depth when compared to multilevel systems, in turn impacting clear heights and floor-to floor heights. When minor surface-forming members are used, they are typically ubiquitous, mass-produced elements. In a three-level system in steel, for example, the effective span of available decking usually determines the spacing of the secondary collectors. Secondary members are, in turn, off-the-shelf wide-flange elements. They are sized exactly and are selected from a variety that is available for the specific spans and loadings involved. The primary collectors may, in turn, be off-the-shelf elements as well, if spans and loads are small, but often they are specially designed. For longer spans and loadings, a specially shaped truss, plate girder, or cable system would make sense. The spacing of these specially shaped forms is normally dictated by larger building design issues (relation to functional spaces, etc.). The spacing of the secondary collectors, which is influenced in turn by the decking, often dictates the spacing of subelements in specially designed primary collectors. For example, the spacing of truss members (so that loads may come in at nodal points) or the locations of stiffeners in plate girders may well be dictated by the spacing of secondary collectors. It is interesting that the design of a three-level system normally proceeds with a simultaneous consideration of all three levels. Each affects the other. Working out an exact relationship can be difficult because of the number of variables involved, particularly if some larger structural objective, such as minimizing total volume, is assumed. While it is common to think in terms of hierarchies of orthogonally placed elements, the general ideas are applicable to other patterns as well. Just because a bay is square does not mean that the horizontal spanning system has to be organized in a parallel way. The grid could be twisted or otherwise manipulated. Framing elements are predominantly linear, straight elements. Using beams that are curved in plan is highly unusual and potentially problematic from a structural viewpoint. Detrimental torsional (or twisting) effects are induced in the curved members by vertically acting loads, and spans are limited to small-scale structures.

Copyright © 2013. Pearson Education Limited. All rights reserved.

2.4

two-Way Systems

The intrinsic nature of two-way systems and the extensive use of reinforced concrete plates or slabs with multidirectional load-dispersal paths make the notion of structural hierarchies less obvious for two-way systems than for one-way systems. In short-span systems, flat reinforced-concrete plates have no beams. As spans increase, however, the need arises for two-way beam or ribbed systems to support flat-plate elements. A critical characteristic of two-way systems is the invariable presence of square or nearly square support geometries, rather than rectilinear patterns, in both beams and vertical support systems (e.g., in the column or wall grids), so as to obtain true two-way action. Figure 8 illustrates some typical hierarchies for two-way systems. When the surface-forming elements are reinforced-concrete plates, the extent to which a flat plate may span is limited. Spans may be increased by using column capitals to form a flat-slab system, by a peripheral wall support, or by introducing beams around the periphery of a plate to form a two-way beam-and-slab system. Using a two-way ribbed waffle slab further extends spanning capabilities. Extremely large spans between columns may be obtained by introducing a secondary beam system beneath a supported plate. This system, which is often posttensioned, would run between primary beams that span from column to column [Figure 8(f)].

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 8 Typical two-way horizontal spanning systems.

Two-way grid systems may be composed of beams of either reinforced concrete or steel and may in turn support other surface-forming substructures (flat or shaped). Reinforced concrete is a natural material for forming two-way grid structures. Steel plate-girder grids are possible but are somewhat more difficult because of the need to make rigid intersecting joints (Figure 9). Similar grid systems may be made using steel trusses or steel Vierendeel frames. The spacing of grid members depends largely on the nature of the supported substructure, but usually

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 9 The roof of Mies van der Rohe’s National Gallery, Berlin, Germany, is a welded two-way steel grid structure supported on eight cantilevering columns. By avoiding the corners for placing the columns, the impression of a floating roof is communicated.

it is fairly generous to reduce the number of difficult grid joints. The choice of the surface-forming elements logically leans toward two-way systems that load all supporting primary grid elements equally. One-way decking or boards also can be used, but it is not intrinsically obvious which way to orient the spanning directions of the modules. Vertical support systems for horizontal grid structures may be either walls or columns, depending on functional needs and the spacing of grid elements. When grid elements are closely spaced—for example, less than 5 ft (1.5 m)—walls are commonly used. In large structures with generous grid spacings—for example, over 20 ft (6 m)—columnar supports are suggested, with a column placed under each major horizontal grid element (a one-on-one relationship). In intermediate spacings, ring beams supported on periodically placed columns can be a good solution. Grids may effectively cantilever beyond inset wall or column lines to reduce the bending moments that are present. Space-frame systems are among the more commonly used two-way horizontal structures, typically used for longer spans. They are inherently fine-grained with closely spaced individual members, particularly when they are made of steel. Coarser grids may be obtained, however, simply by using larger members. Timber and even reinforced concrete may be used. (The latter must be carefully designed, and grid spacing becomes quite coarse.) Space-frame systems are quite effectively supported with walls, with a series of closely spaced columns, or on a beam system supported by columns. Cantilevers are useful in reducing design bending moments. Supporting space frames on four-corner columns is indeed possible and often done, but it is among the least-efficient support systems possible and leads to increased member sizes in the space frame itself. Branching columns that connect to the space frame at multiple points can mitigate these problems.

Structural Elements and Grids: General Design Strategies

3

tyPIcal HorIzontal grIdS

Structural grids and patterns are aggregations of individual structural units (or bays), which in turn consist of the elements just described. Many different structural patterns are possible. In most buildings, a repetitive geometrical pattern or grid governs the organization of the vertical support system and the horizontal spanning system. Vertical supports can be formed by load-bearing walls, columns, or a combination thereof. The sections that follow focus on buildings composed of repetitive structural bays or other forms of cellular units. Many buildings (e.g., gymnasiums, auditoriums) are actually single-cell structures. Most buildings, however, are composed of a large number of aggregated bays or units. Next, we review some basic characteristics of typical structural patterns.

3.1

orthogonal Systems

Perhaps the most commonly used structural patterns are orthogonal. Rectilinear building sites, ease of construction, and the functional logic of rectilinear spaces are among the most compelling reasons for their widespread use. Orthogonal arrangements allow for a wide range of structural systems to be applied. Figure 10 shows two examples of typical orthogonally arranged structural systems. Depending on the choice of the structure (e.g., one-way beam versus two-way plate system), interior spaces can have substantially different characteristics. Issues of construction and structural efficiencies weigh in heavily when deciding on system layouts. A typical example of closely knit orthogonal patterns is a multiunit housing development where unit size and structural unit are often identical (Figure 11). The structural pattern of the vertical support system itself consists primarily of a series of parallel lines, which in turn could consist of a series of load-bearing walls or a columnand-beam system arranged in the same pattern. The serial nature of the structural system leads to certain efficiencies in the engineering design. In early stages, the study of typical elements can quickly yield insights on issues pertinent for much of the structural system.

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 10 A given orthogonal column spacing can yield different spatial characteristics. One-way beam systems tend to generate directional spaces, while two-way plates are spatially more neutral.

Structural Elements and Grids: General Design Strategies

FIgure 11 Typical structural pattern in housing, Borneo-Sporenburg, Amsterdam, The Netherlands. Architect: West 8.

Orthogonal systems are widely used for office buildings, parking facilities, and institutional buildings, to name a few. A more complex tartan grid is illustrated in the plans of Louis Kahn’s community center in Figure 12. The building illustrates how the structural pattern of a building can be related to and reinforce

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 12 Trenton Community Center, Trenton, New Jersey. Architect: Louis Kahn.

Structural Elements and Grids: General Design Strategies a building’s functional zones. Special spaces in this building result in altering the individual structural units into larger spanning systems. Irregular site boundaries and building shapes often create an additional need to adjust otherwise regular orthogonal grids locally (Figure 13). The proportions of orthogonal grids vary widely. Repetitive column and beam systems such as those used in deep multistory office buildings are often close to square in proportion. Square or near-square column grids allow the use of efficient two-way spanning systems. Elongated spans in one direction usually lead to overall increased moments, and bay dimensions beyond 1:1.5 usually do not lend themselves to creating two-way action in the horizontal span. When using rectangular bays and one-way structural systems, a basic decision must determine the direction of both primary and secondary elements. Either the primary collector elements span in the longer direction, with secondary elements in the shorter direction, or vice versa (Figure 14). The decision usually involves issues beyond consideration of the immediate structural bays (lateral stability often being one), constructability, and the use of off-the shelf or custom-built structural elements. When spanning larger spaces, elongated orthogonal grids are common, with custom-built elements such as deep trusses, funicular systems, or girders bridging the primary span, and smaller elements spanning between them. The relative proportion of primary span to secondary spans remains an important factor and is driven by constructability and the absolute size of the span to be created. A starting point can be to place primary spanning elements at 1 > 4 to 1 > 3 of the primary span. Choices of secondary and other elements must be considered as well.

Copyright © 2013. Pearson Education Limited. All rights reserved.

3.2

triangulated Systems

Triangular structural grids are occasionally used to achieve particular spatial effects or because the overall building form lends itself most readily to a triangular subdivision (Figure 15). Another reason for using triangulated structural patterns is the need to create stiff horizontal diaphragms in response to lateral loads. (See Figure 16.) Triangular patterns are often created by offsetting parallel gridlines starting at the perimeter of the building. Compared to orthogonal systems, triangulating structural patterns tend to lead to geometrically more complex connection details, which in turn might impact cost and construction time. The horizontal spanning structure between primary triangulated beams is characterized by variations in span within a triangular bay. Compared to orthogonal systems, the span variations often lead to the deployment of a greater range of the structural shapes. The orientation of the secondary spanning structure is usually perpendicular to the longest side of the generating triangle, thus minimizing

FIgure 15 Triangular structural grids may be a logical consequence of special plan shapes. Twoway as well as one-way horizontal spanning systems may be possible. Two-way concrete system

One-way system using linear members

FIgure 13 Highly irregular building forms may lend themselves to the use of simple orthogonal structural grids. Additional column support along the edges may be necessary to avoid excessive cantilevers. Two-way concrete system

Additional columns are added to avoid excessive cantilevers.

FIgure 14 Bay proportions in orthogonal grids: One-way systems are usually preferred in square or near-square support systems. Strongly rectangular grids favor one-way systems with primary spans in either the long or the short direction.

Structural Elements and Grids: General Design Strategies

FIgure 16 Fuji Kindergarten, Tokyo, Japan: The oval building shape was constructed using a triangulated system of primary beams and column supports.

In the finished building, the beam system is hidden behind finishes.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Framing diagram: The uneven overall building form results in highly individualized beam spans.

The triangulated pattern of primary beams creates uneven spacing for decking and secondary beam systems.

the bending moments in the secondary spanning system. Span variations as a result of the triangular support geometry may lead to a differentiated sizing of the secondary structural elements. The constructional efficiency of using identical elements, on the other hand, may turn out to be more cost effective despite the increased use of materials.

3.3

radial and circular Systems

Radial and circular patterns are based on the same basic grid geometry. Radial grids arrange primary structural elements along the radiating lines, with secondary spanning systems such as slabs, plates, or secondary joist-and-beam combinations spanning concentrically around the radiating net. The spanning directions of primary and secondary systems are reversed in circular and near-circular approaches, where the primary elements are arranged concentrically, while the secondary span direction is aligned with the radial lines. Even though this distinction seems idiosyncratic at first, it turns out to be an effective way to identify structural opportunities for architectural design. Radial or circular systems have been used in cylindrical towers and low-rise buildings alike as well as in structures such as arenas for sports and public events. Patterns with radial primary elements systems are flexible in creating fanning geometries, staggered heights, or other variations on the basic pattern. When using

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 17 Tietgen Dormitory, Copenhagen, Denmark: Radially arranged bearing walls allow cantilevers to occur freely in the radial direction. A highly varied spatial pattern results despite the regular plan figure.

load-bearing walls as the radial structure, length variations can generate varied structural patterns even for multistory buildings (Figure 17). A challenge in radial networks is the ever-changing distance between primary structural elements and the associated difference in span of the secondary systems in place. When concrete slabs form the secondary system, the differences in span can result in varying degrees of reinforcement and slab depth. Spans above 10 m typically employ ribbed slabs and posttensioned systems. For discrete secondary systems, a typical response can be to adapt the sizing (mostly depth) of the secondary beam or truss to the span—even though this approach might affect clear interior heights. Patterns where primary structural elements are arranged along concentric circles limit the variations in member spans, especially for the secondary spanning system. Special circular arrangements include funicular or similar systems—singleor double-cable systems, membrane structures, radiating arches, or structural shells. These systems derive efficiencies from the interdependency of the radial and the circular structural layer (Figure 18). The horizontal thrust that is generated in the radial direction is usually contained within circular structural elements such as a tension or a compression ring. A tension ring, for example, can tie together the supports of radiating arches or a compression ring that resists the inward pull of a double-cable system. In these special cases, the circular structural pattern is required for efficient funicular load-transfer to take place.

4

multIStory grIdS

Structural grids should not be thought of only in plan. Their effect on floor-to-floor height and overall building height are of equal importance, especially when designing multistory buildings. Assigning floor-to-floor height is not only driven by structural concerns but also takes into account functional and architectural aspects such as the proportions of the interior spaces, the integration of mechanical services, or the build-up of floor finishes. Zoning requirements often limit the maximum permissible height of buildings, while clients might require a certain amount of usable floor space with a minimum required clear height. Clear heights also may be regulated by building codes, depending on the use for which an interior space is designed. Understanding the basic dependency of grid spacing and building height is an important question, especially in the early design phases.

Structural Elements and Grids: General Design Strategies

FIgure 18 Horizontal force containment: Circular system geometry can efficiently contain the horizontal forces generated in cable structures or membranes. Radially arranged beams or trusses do not derive particular efficiencies from the circular geometry.

FIgure 19 The spacing of column grids in plan often impacts building height, assuming that minimum clear heights remain constant for interior spaces. Services Zone h1 h1

(a) Radially arranged trusses. The structural grid impacts the structural system logic only minimally.

(b) Radially arranged double-cable system: Tensile forces are contained within the outer compression ring—the system logic hinges on the circular arrangment.

(c) Variation on the system shown under (b): The inner tension rings substitute the continuous radial cables.

(d) Circular pneumatic structure: The horizontal tensile forces of the inflated membrane are resisted in the outer compression ring.

Overall height

h2

a

(a) Closely spaced column grid: Small floor-to-floor height with flat concrete plate.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Services Zone h1 h1

Increased height

h2

2a

(b) Long-span column grid: Increased floor-to-floor height with beam/slab concrete system increases overall building height.

The plan grid spacing is closely related to floor-to-floor height. Coarser grids lead to larger bending moments that necessitate stiffer floor systems with more capacity to resist bending moments and deflections. A structurally efficient way to deal with these larger bending moments is to use deeper members as beams in steel systems and deeper plates or slab/beam systems in concrete systems. Deeper cross sections of the structural floor system have a larger moment of inertia that provides for the needed internal resistance to external loads and moments. Larger spans in concrete systems also can be accommodated by using prestressing. As a result of the deeper beam, the floor-to-floor height may have to be increased to preserve a required minimum clear height. (See Figure 19.) While larger grid spacing may be desirable for improved flexibility of a space, the increased structural depth needed for floor systems quickly translates into added building height. This is particularly true when horizontal layers for mechanical services remain below the structural beam layer. For longer-spanning grid systems, it is sometimes more efficient to design the mechanical system such that it occupies the

Structural Elements and Grids: General Design Strategies same horizontal layer as the structural floor system. Deep beams can be designed to accommodate regular openings that allow for mechanical services to communicate between adjacent structural bays. Multistory buildings with large grid spacing are usually heavier structures with larger beam-and-column elements that collect overall loads in fewer concentrated areas. Questions of soil conditions and foundation design can impact the basic layout of the grid because large, concentrated column loads at the base may require special attention to the design of foundations. Decisions on grid spacing and floor-to-floor height are crucial and must be made early in the design process. Changes at later phases are usually difficult to make because repercussions would be felt in almost all other areas and aspects of the buildings.

5 5.1

Irregular and dISruPted grIdS nonstandard Structural Patterns

Copyright © 2013. Pearson Education Limited. All rights reserved.

In situations where the pattern of the vertical support system is irregular, some structural systems are more attractive than others. The effective or economic use of systems involving several highly repetitive and typically prefabricated units (e.g., precast concrete elements) is usually problematic. Although recent developments in digital design and fabrication methods have facilitated the use of customized construction elements, significant economic incentives remain for the repetitive use of identical elements. Poured-in-place concrete systems are often preferable in cases involving irregular support patterns because anomalies can be more easily handled. Spans on the order of 16 to 20 ft (5 to 6 m) are preferable for flat concrete plates. For slightly larger spans, steel or timber beam systems can be designed to address irregular column patterns. Primary and secondary beam directions and spans can be directly derived from the column pattern, resulting in fairly complex structural and constructional issues. This approach also is feasible for long spans, albeit at the expense of very deep horizontal members. These have at times been treated as story-high structural elements, and are thus incorporated into the occupiable space itself. An alternative approach, albeit often limited to intermediate spans, employs an orthogonal or other regular primary beam grid. Columns either connect to grid intersections directly, or else an intermediary system of connection beams transfers loads between the irregular supports and the regular beam structure (Figure 20).

5.2

grid transitions

In many cases, more than one generalized structural pattern is used in a building. The reasons for this multiplicity of patterns vary. Quite often, variations in the programmatic requirements of the building lead to a variety of minimum clear span dimensions. One or more standard functional patterns may be adopted to respond

FIgure 20 Irregular column spacing is often easier to accommodate in concrete systems. Beam structures may require intermediate elements to allow columns to connect to the primary beam grid.

Structural Elements and Grids: General Design Strategies to this variation, instead of a single pattern that compromises the varying requirements present. In other cases, physical constraints, such as nonuniform foundation conditions or irregular site boundaries, may dictate using different structural grids in different areas. When more than one generalized structural pattern is used in a building, the way the patterns meet becomes a basic structural design issue. Intersection points always call for unique treatment or special elements. Structural issues that arise from grid transitions in the plan dimension are quite different from the problems associated with vertically transitioning grids. Constructional efficiency tends to be a primary concern for horizontally transitioning grids, while structural issues are crucial for transitions in the vertical, primary load-bearing direction.

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 21 Basic strategies for the meeting of generalized structural patterns.

Horizontal transitions. Figure 21 shows several strategies for dealing with the existence of two horizontally different grids. When grid systems intersect randomly, the structure in the region of the intersection will be atypical compared to regular grid elements. A beam located at the boundary line B, for example, would

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

be loaded in a way that is different from those present in either the grid to the left or the one to the right. It would therefore have to be designed differently. Additional columns also might be needed. An alternative approach illustrated in Figures 21(b) and (c) is to align grid elements in one direction, restraining differences in spaces to the other direction. This alignment reduces the misfit between the systems and does not require additional vertical supports. Another general approach is to use some sort of mediating system between the two general systems. One type of mediator would be a strip of space itself, in which case the whole problem is bypassed. [See Figure 21(d).] Another type of mediator would be a third structural system. This third system typically reflects the characteristics common to both of the general systems. A structural grid that is much smaller than either of the primary systems (but is a logical subdivision of each), for example, is often used to join primary systems. [See Figure 21(e).] The mediating system also can be articulated by using a different material from that used in the primary system. Two primary poured-place concrete systems, for example, could be joined by a smaller-grained light steel system. Mixing materials in this controlled way is usually acceptable, even in purely economic terms. Intermixing materials within primary systems, however, is more difficult. An advantage of using a small-grained mediating system is that the designer has greater freedom in placing the primary grids. Primary gridlines, for example, do not have to be aligned. Load-bearing walls also are often used as third-element mediators, particularly when the primary systems are not aligned, as illustrated in Figure 21(a). Other approaches to intersecting patterns also exist; for example, an interpenetration strategy of the type illustrated in Figure 21(f) could be adopted as a variant of the basic alignment approach. Figure 22 shows the application of several transition strategies in José-Luis Sert’s Science Center (Cambridge, Massachusetts). A large auditorium is separated from two different beam grids through a transition structure. The beam grids are based on the same overall grid spacing but vary between single- and double-beam elements that each feature distinctly different spans. corner conditions. Numerous local geometrical conditions could affect the choice of structure and whether a one- or two-way system is preferable. Whenever it is necessary to bend, distort, or do something else unusual to a structural grid because of an overall building geometry, there are certain to be structural implications. Even simply terminating a grid, for example, usually demands special end treatments. Consider the plan shown in Figure 23(a), in which the whole building literally turns a corner. Under some circumstances, structuring the corner region can pose problems. Assume that the functional modules in the building are basically rectilinear, as illustrated in Figure 23(b). This is a common pattern in many buildings, such as housing, in which each module represents a dwelling unit. As is evident, this basic pattern lends itself well to a structural system that uses parallel beams or loadbearing walls along the depth of the building and a planar one-way system that spans horizontally between these elements. The only difficulty in using such a system is at the corner, where the one-way system cannot be oriented in two directions at once. It would thus be necessary to adopt an approach like that shown in Figure  23(b). While such an approach is possible, it lacks elegance. The basic difficulty, of course, is not really a structural one but one associated with the rectilinear pattern and the corner-turning requirement. As a consequence, the structure that is most logical for the basic parts of the building becomes awkward at the corners. Now consider another type of functional pattern, which consists of aggregated squares. [See Figure 23(c).] In this case, the basic pattern lends itself well to a twoway structural system supported by columns at grid intersection lines. It is evident that no difficulty exists in structuring the corner region. The two-way system is naturally capable of turning corners because of its symmetry in both geometry and

Structural Elements and Grids: General Design Strategies

FIgure 22 Three different structural patterns are combined in Harvard University’s Science Center. A transition structure negotiates between radial and orthogonal systems.

Copyright © 2013. Pearson Education Limited. All rights reserved.

(c) The three structural patterns are clearly articulated on the exterior

structural behavior. It would consequently seem that, whenever L-shaped conditions exist in a building, square rather than rectangular functional and structural grids should be used. A rule of this sort is somewhat simplistic, however, because square grids may not accommodate building functional requirements as well as rectangular grids. (Housing is a case in point.) In addition, other structuring possibilities may prove effective. Several other ways a structural pattern can turn a corner are illustrated in Figures 23(d) and (e). The axes of the basic grid can simply be bent around the corner in a radial fashion. While possible in a great many cases, building functional requirements may preclude this approach in others. If a bent-axis system is adopted, one-way structural systems using prefabricated elements usually, but not always, prove difficult to use. Construction difficulties are involved in having to form a variable-length spanning surface in the bent region when materials such as steel or timber are used because member lengths are not constant. A poured-inplace, one-way reinforced-concrete system using radial beams and simple one-way slabs, however, is quite appropriate. However, span lengths are limited with this system. The two-way flat plate is another natural system for use in a bent-axis region. Again, though, spans are limited. In general, the bent-axis system is difficult to structure when span distances between gridlines are large. Another way to handle the corner condition is illustrated in Figure 23(e). In this case, the primary system is terminated at either boundary of the corner and

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 23 Structural patterns for building corners.

a transition structure is used. This transition structure has two-way characteristics and often has a relatively small grain. As discussed earlier in this section, particular strategies are common to negotiate the meeting of different structural patterns. The transition structure can be of the same materials as the parent structure, or it can be made of different materials altogether. This approach is somewhat of an overkill if the building is small and its absolute depth is shallow; then it is not worth the trouble. In large buildings that are relatively deep, the transition-structure approach works well. Vertical transitions. The spacing of vertical supports in a structural system does not always stay constant throughout the height of a multistory building. Stacking spaces with different programs vertically often leads to vertically shifting

Structural Elements and Grids: General Design Strategies

FIgure 24 Vertical grid transitions are normally accommodated through deep trusses or using transfer beams.

(a) Diagonal members transfer forces between framing grids via axial compression.

grids to meet different clear span requirements. When these functional reasons or other preferences prevent the vertical alignment of columns or walls, designers have to anticipate the incorporation of a transitional structural layer to merge otherwise disconnected vertical load paths. In typical office buildings with a relatively closely spaced regular square or rectangular column grid, for example, the column spacing for typical office levels is often incompatible with the larger clear spans needed for parking levels located in the basement. In these cases, a system of transfer beams is designed to redistribute the column loads into the new, more widely spaced basement columns. Transfer beams can require a significant depth. Alternative transfer structures include deep trusses that may be incorporated into the height of typical floors (Figure 24). Other, more extreme cases of grid transitions exist when buildings are bridging over infrastructure elements such as train tracks or roads, where column placements are dictated by traffic clearances. Yet, other issues arise when structural grids are deliberately shifted. Some recent buildings have exploited the principle of shifting grids as a primary mode of architectural expression (Figure 25).

5.3

Copyright © 2013. Pearson Education Limited. All rights reserved.

(b) A story-high truss distributes column loads to adjacent columns, creating larger column spacing on the ground floor.

(c) Deep transfer beams allow for typical upper floor grid spacing to permit uses such as parking in the lower levels.

accommodating large Spaces

Programmatic requirements of buildings often lead to the design of special spaces that are different in scale from the typical structural pattern used. For example, in a school with a large number of similar-sized small spaces (classrooms) and a few large spaces (gymnasiums, auditoriums, or cafeterias), handling both types of spaces with the same structural grid is often problematic. The resulting disruptions of the prevalent and regular grid are related to the discussion in the previous section and, in fact, are almost identical when thinking about disruptions of the horizontal plan grid in low buildings (Figure 26). Embedding large spaces in multistory buildings is more challenging. Normally, the larger space is treated as a local disruption to the otherwise regular fine-grain structural system. The resulting issues are thus different from considerations involved in grid transitions.

FIgure 25 New Museum, New York: The envelopes of the gallery volumes are shifted vertically, while a large vertical core penetrates the building vertically.

1

Core

6

7

5

4 3 2

Structural Elements and Grids: General Design Strategies

FIgure 26 Gund Hall, Harvard University, Cambridge, Massachusetts: The large lecture hall is embedded in a near-square column grid.

Auditorium

(a) Square column grid

Copyright © 2013. Pearson Education Limited. All rights reserved.

(b) Section through auditorium: A transfer beam allows for a a longer span.

(c) Exterior view shows the stepped studio space and the office/classroom wing.

Figure 27 illustrates several basic ways to handle these situations. Two situations are of interest here. In the first scenario, the large space is placed at a low level such that much of the gravity loads of the finer-grained system must be deviated around it. In the other case, the large space is placed above the finer-grained system. When the large space is embedded at a low level, several structural problems arise. One is to design large transfer members to pick up the loads from the closely spaced vertical supports above the open span space and to carry them to the vertical supports at the edges of the large space [Figure 27(a)]. Another option is to create a story-high transfer truss that, while requiring special accommodation in the design of spaces around it, does not infringe on the clear height of the space below it [Figure 27(b)]. Yet another strategy is to adopt the span of the larger space as the repetitively used dimension, thus using a series of large-span members consistently above the large space. In the first case, the transfer members must carry heavy loads and thus must be uniquely designed and constructed. Off-the-shelf members usually will not work. The loose-fit alternative with long spans everywhere is uneconomical: Spanning a large distance when a small one could be spanned instead usually does not lead to economical structures. Using special systems (e.g., precast

Structural Elements and Grids: General Design Strategies

FIgure 27 Strategies for accommodating large volumes in a fine-grain system.

(a) A locally arranged transfer beam deviates floor and column loads to adjacent columns.

and prestressed reinforced-concrete tee shapes) designed for long spans might offset this inherent disadvantage somewhat, but it remains an influential factor. Most of the difficulties mentioned here can be avoided by placing the largespan space on top of the smaller grid system. In that case, the long-span elements carry only roof loads, which are relatively light, and any of a variety of systems could work. The design of the finer-grained system is not appreciably affected by the presence of the large space above it. A difficulty with this approach is not a structural one, but a functional one, in that large, more public spaces are usually more appropriately placed on a lower level.

6 6.1 (b) Deep truss transfers loads without reducing the clear height in the higher volume below.

(c) Three large beams transfer floor loads to adjacent columns. The effect of the needed large span on the ground floor ripples through the entire system.

Copyright © 2013. Pearson Education Limited. All rights reserved.

(d) Locating the long span system on the roof eliminates the need to transfer floor loads. Roof loads tend to be lighter, making the design of the transfer structure easier.

ProgrammatIc and SPatIal ISSueS relation to Program and Functional zones

critical Programmatic dimensions. Associated with any building are certain critical dimensions that define the minimum clear span for the structural system used. These critical dimensions may stem either from functional necessities (e.g., minimum clear spans for basketball courts or some other programmatic use) or simply from more subjective design intents. We will assume that these critical programmatic dimensions define a minimum programmatic space and will denote them as a1 and a2. Many buildings will have different minimum clear spans for different spaces, but for simplicity’s sake, we will study the effect of a single such space. It also should be noted that critical program dimensions may have to anticipate possible future functions and thus may or may not be based on present-day needs. The length of either a1 or a2 must become the minimum clear span dimension for the structure, and vertical supports may not be placed any closer together than the minimum of these two distances. They can, however, be spaced farther apart because a1 and a2 define minimum, not maximum, clear spans. degree of Fit. A primary initial structural objective is to determine an appropriate degree of fit that is possible or preferred between the pattern formed by the vertical support systems and the critical dimension a1 or a2. The selection of a degree of fit determines the magnitude of the horizontal span that the structural system must provide; consequently, it has an important bearing on the final system selected for use. Either one-on-one or looser fits are possible. In a one-on-one fit, the repeatedly used structural bay corresponds directly to the space formed by a1 and a2. Alternatively, the structure selected may span over multiples of a1 or a2, and the span would thus be longer than is minimally necessary. In more complex programmatic arrangements, any number of vertical support patterns and related horizontal spanning systems could be fitted to a given arrangement. (See Figures 28 and 29.) Consider the hypothetical building plan shown in Figure 29(a). Many radically different approaches reflecting different degrees of fit and different structural systems are possible and feasible. By looking at this diagram only, nothing a priori suggests that one is preferable to another from a purely structural perspective, especially because critical information such as span magnitudes and loading conditions are not indicated. The resolution of which of the approaches diagrammed may prove most feasible depends on many of the issues discussed in this chapter. The relation between the span length and the appropriate type of structure is of primary structural concern, but in a design context, many other factors come to bear. Materials and available construction systems, for example, are important considerations that relate directly to possible primary spans. Choices at that level of the design development are often based on typical precedents and embrace what seems reasonable. Orderly, regular, and simple solutions with primary and secondary structural layers in consistent directions tend to be easier and more economical to construct. The solution indicated in Figure 29(h) would be

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 28 Typical clear spans and structural system options.

more difficult to construct and hence be more costly to build than some of the other solutions, at least for a low-rise building. By simply repeating more complex floor framing strategies over many floors in multistory buildings, however, even complex solutions can attain certain efficiencies. As an aside, it is interesting to note that when a building can be framed in many different ways with no one solution clearly better than others, the building probably will never prove to be problematical (from a structural point of view) as the design develops, no matter what option is selected.

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 29 Study of alternative systems for a given L-shaped plan configuration. Most buildings can be fitted with different types of structural patterns that correspond to different building configurations and functional needs.

Structural Elements and Grids: General Design Strategies Other factors that might affect the degree of fit used are the foundation conditions present at the site. Under some conditions, building loads are best distributed over the entire site so that the load concentrations are reduced. The structural implication is that a fine-grained grid system and one-on-one fits should be employed. If, on the other hand, some site locations can carry enormous loads and other locations virtually no loads, the structural pattern should respond to these conditions by having vertical elements concentrated at high-capacity points and by using long-span horizontal elements to bridge low-capacity points. A rough-grained pattern to the vertical support system often results. Examples of tight and loose relationships among the vertical support system, the horizontal support system, and critical functional dimensions are shown in Figure 30. In general, one-on-one fits are considered preferable when critical functional distances are fairly large, for example, a minimum of 15 to 30 ft (5 to 10 m), because minimizing bending is always a structural objective and this is not done by using spans of unnecessary lengths. If distances are quite small, for example, less than 10 ft (3 m) or so, one-on-one fits are usually desirable only when light timber systems are used, for example, wood joists. Looser fits are desirable when materials such as steel or concrete are used. The point is simply that the structural order may be equal to or greater than critical functional dimensions, depending on exact span lengths and system capabilities.

Copyright © 2013. Pearson Education Limited. All rights reserved.

6.2

Spatial characteristic of Structural Systems

Structure and Space. The design of a building goes hand in hand with the design of its structural system. The building design both determines and informs the structural system layout as much as the structural system can play an important role in defining spaces and architectural form. Even for a given overall form, the strategic choice of different structural systems can give rise to different architectural expressions. Efficiencies also will vary. Figure 31 explores various structural systems for the design of a horizontal spanning, single-story structure. The spaces formed by the bearing-wall system [Figure 31(a)] are one-directional (or linear) and have a strong planar quality that is imparted by the vertical and horizontal enclosure elements. The beam-and-column system [Figure 31(b)] also creates a primarily onedirectional space through the organization of the collector beams, but the secondary axis in the transverse direction is implied. These characteristics are significantly influenced by the shape, spacing, and orientation of the columns. Rectangular columns with a long axis in the direction of the beams would further emphasize the linearity of the space, whereas placing the long axis in the other direction would emphasize the secondary axis more. Round columns are a neutral statement, and square columns are bidirectional. The spaces formed by the flat-plate system [Figure 31(c)] are two-directional because neither of the axes is dominant. The spaces formed are relatively neutral. Round columns or square columns do not alter this two-way directionality. Rectangular columns start causing one or the other axis to become more dominant. The spaces formed by the two-way beam-and-slab system [Figure 31(d)] are less neutral, with the beams even further emphasizing the two-way directionality of the spaces. The space formed underneath a large, completely neutral surface (such as a flat plane) is not directional. This assumes that any directionality that the support system might impart does not influence the observer. The space frame illustrated in Figure 31(e) would provide this type of space. (The structural member spacing is sufficiently small as to impart little spatial influence.) While other structural arrangements may yield slightly different spatial properties, they are usually some variant or combination of the general types discussed previously.

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 30 Loose and tight fit between functional and structural units.

Structural Elements and Grids: General Design Strategies

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgure 31 Spatial characteristic of different structural approaches.

Structural Elements and Grids: General Design Strategies

FIgure 32 Relation of roof geometry and framing strategies for rectangular plan shapes.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Figure 2 at the beginning of this chapter visualizes the effect of different structural systems on the design of a simple rectangular space with a medium span. Simply assigning connection constraints, and then shaping members according to their internal forces and moments, will give a different spatial expression to the space that is being formed. Single-layer versus multilayered beam systems, even based on simple orthogonal framing plans, give a different sense of scale to the same space. Arrangements of criss-crossing beams tend to create a more dynamic feel on the interior when compared to straightforward orthogonal arrangements. These studies are not to convey a more or less efficient solution but rather suggest that the structural system can be activated directly to support architectural design intentions. roof Shape and Structure. Thus far, the discussion has focused on structures with flat horizontal spanning systems (e.g., floors, flat roofs). The basic idea of a structural hierarchy applies to other configurations as well. Roof shape, for example, is a primary determinant of the framing organization. The diagrams in Figures 32 and 33 illustrate several common framing strategies for different rectangular and square forms. With most roof forms, there are a number of different ways to structure the shape, so the diagrams are for illustration only. Usually, a basic decision involved in structuring most volumetric shapes includes whether to attempt to have the structure be only coincident with the external envelope of the roof shape (and not have any structural members exist within the internal volume of the space itself) or to allow the structure to exist within the internal volume. An example of a coincident system is shown in Figure 33(a) and 33(b), where sloped members are used in a simple pyramidal form. Members are tied together at the crown and around the base periphery. (The latter tie is needed to prevent outward spreading of the sloped members.) By contrast, a crossed-truss system forming the same roof shape passes through the interior space. This approach is often cumbersome and constructionally complex but can sometimes be used to advantage.

Structural Elements and Grids: General Design Strategies

FIgure 33 Relation of roof geometry and framing strategies for square plan shapes.

Copyright © 2013. Pearson Education Limited. All rights reserved.

When there is a change in the profile of a roof, a change of some sort occurs in the framing system at the same point (Figure 34). A change in the roof profile might generate a change in the framing system, or alternatively, a change in the framing system necessitated by other factors might lead to a change in the profile of the roof. Frequently, the profile change necessitates the introduction of a vertical support line (e.g., walls or a column line) at the point of change.

FIgure 34 Effects of changes in roof profile. A line of vertical support is often introduced at profile transition points.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Copyright © 2013. Pearson Education Limited. All rights reserved.

PrinciPles of structural Design structural design activities may be classified into several broad categories that reflect the sequence in which design decisions are encountered: First-order design issues deal with the relationships among structural fabric, the intent of the design, and contextual or programmatic dictates. these issues hinge around the type and organization of the structural system to be used in relation to the overall morphology of the building or other functional entity designed, including how specific spaces are configured and defined by built elements. of specific importance here is whether the structure to be designed is conceived of as a single large form or as an aggregation of smaller structural units or bays. Basic preconceived attitudes or images in relation to the general shaping and configuration of the structure also come in at this level (e.g., frameworks versus shell or cable structures). Second-order design issues deal with specific structural strategies in relation to the exact shape of the structural entity itself and how that shape is formed by specific constituent structural elements. choices of structural materials (steel, timber, reinforced concrete) are made at this level (if they are not already implicit in basic attitudes defined previously). intrinsically related choices of specific structural systems also are made at this level From Part III of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Principles of Structural Design

(e.g., reinforced-concrete flat plate versus two-way beam-andslab systems; or steel systems based on rigid-frame action and beams that span horizontally versus those using diagonal bracing and trusses as horizontal spanning elements). of importance here is the nature of the structural hierarchies present (levels, spacing of members), bay spans and proportions, and similar geometric characteristics, all of which are variable within certain limits imposed by the first-order decisions. Third-order design activities deal with specific member shaping and sizing. associated with the choice of a framing pattern for a spatial entity in a given context and the choice of materials are specific implications or even imperatives relating to the shapes and sizes of constituent structural elements. of initial importance here is the choice of member cross-sectional configuration: rectangular, square, round, hollow tube, open box, and so on. once the member cross section is selected, actual member dimensions can be determined. a detailed knowledge of element design is a prerequisite for engaging in first- and second-order design activities. While it is convenient to discuss design issues at three different levels, it is important to remember that strong interdependencies exist. certain structural systems may be structurally efficient and desirable when executed in certain materials, using certain member cross sections, and for certain ranges of spans. the same pattern might be impossible for other choices of span, materials, or cross sections. Hence, the structural design process is not really deterministic, nor does it begin with a consideration of an overall form in the abstract and eventually result in the sizing of each elemental member through a series of independent design decisions at each of the three design levels. nothing could be further from the truth! in reality, good designers deal with each of the three design levels at varying levels of engagement at all times as the overall design moves from a mental image to a fully developed design. the process is interactive and iterative. the structural system finally adopted is usually one that works well both on its own merits and with respect to other architectural objectives. structural design considerations arise during the early design phases of a building, when the designer is in the stages of establishing broad strategies and manipulating spaces within a building. Primary reference is in relation to vertical loadings. in addition, focus is on the effects of lateral loading conditions.

Structural Systems: Design for Lateral Loadings 1

Copyright © 2013. Pearson Education Canada. All rights reserved.

1.1

LateraL Forces: eFFects on the Design oF structures Basic Design issues

Basic issues. In designing structural systems, the way lateral stability is achieved is of fundamental importance in buildings of any shape and height. When designing a structure, a novice can easily lose sight of the need to address the resistance to lateral loads and instead focus on a system only in response to gravity loads. The way a structure resists lateral forces, however, influences not only the design of vertical elements but also the horizontal spanning system. While both needs must be addressed in every structure, it is true that dealing with lateral loads in singlestory or other low buildings is often straightforward. The opposite, however, is the case in slender, vertical structures such as towers and high-rise buildings. In such structures, the consideration of lateral stability is frequently at the very core of the structural design effort. Basic issues associated with the effects of lateral forces due to wind or earthquakes on structures are illustrated in Figure 1. Lateral forces cause structures to deform horizontally. They also can cause twisting or torsional deformations. If no adequate resistance mechanisms (e.g., shear walls, diagonal bracings, rigid frames) are present to resist these forces, complete collapse can occur. Shear walls, diagonal braces, or rigid frames can be used to prevent collapse. (See Figure 2.) Figure 3 illustrates these same effects and associated resistance mechanisms in more detail, with reference to a simple rectangular structure. The upper part of the diagram shows that a simple pin-connected series of columns would radically deform and possibly collapse in the lateral direction when a horizontal load is present. The columns in effect form a system that cannot transmit in-plane forces. When a shear wall replaces the columns, when they are connected through cross bracing, or when rigid-frame connections are used, the element becomes stiff and capable of transmitting in-plane forces. It is stable and helps prevent collapse. The term shear plane is often used for such planar stiff elements and also to describe any of these methods for providing in-plane stiffness and in-plane force transmittal capabilities. From Chapter 14 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Structural Systems: Design for Lateral Loadings

Figure 2 Approaches for creating rigid planes used to stabilize buildings with respect to lateral loads.

Figure 1 Typical effects of laterally acting wind or earthquake forces.

Shear wall (a) Basic structure without lateral stability devices

Cross bracing (truss action)

Rigid frame

(b) Major lateral deformations (racking) due to horizontally acting wind or earthquake loadings

(c) Torsional deformations

In the lower part of the figure, note that the roof (or floor) also plays a vital role. Laterally acting forces on the side of the building can cause a nonrigid roof plane to deform radically in the horizontal direction. The lack of stiffness would also prevent the roof from transferring laterally acting external forces to resisting vertical shear planes. For the latter to be brought into play, the roof (or some part of it) must be capable of carrying in-plane forces and act in the manner of a horizontal,

Figure 3 Rigid wall and roof or floor planes in a simply rectangular building.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Rigid frame with fixed supports

a) Alternative ways to create vertical stiff planes in a simple structure. Collapse would occur if only pin-ended columns are used (upper left). The needed stiff shear plane (gray) can be created with cross bracing, through a shear wall, or through a rigid frame.

b) Alternative ways to stiffen the roof plane of a simple structure. Lateral load would deform the roof even though it is held at the short ends through connection to a stiff plane. Cross bracing or a rigid surface can effectively prevent the roof deformation.

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

flat beam. Stiff planes of this type are often called horizontal shear planes or diaphragms. Alternatively, diaphragm action is said to be present. Diaphragm action can be imparted to roof or floor planes by similar mechanisms as were discussed in the context of vertical shear planes—stiff floor planes, cross bracing or truss action, or frame action. Typical buildings must have stiff shear planes in both walls and roof or floor planes. shear Planes and rigid Diaphragms. Before delving deeper into the organization of shear planes, it is useful to review the various strategies employed for constructing shear planes—the essential elements of lateral stability. Shear planes at their simplest level are stiff wall and floor or roof elements. Masonry walls work well for low- to mid-rise buildings. In timber construction, columns or studs sheathed with appropriate panels (e.g., plywood, oriented strand board) can provide the stiffness needed in a shear plane. Diaphragm action in simple timber beam-and-decking systems, for example, can be made to occur if the decking is substantially connected to the beams via shear connectors or welds along their interface length. By contrast, roofs made with large glass areas often do not form stiff horizontal planes, and alternative mechanisms (typically cross bracing) must be used to develop any needed stiffness. Poured-in-place reinforced concrete walls or slabs have sufficient in-plane stiffness to act as shear planes. In precast concrete construction, care must be taken to rigidly connect adjacent surface elements to prevent slippage. Triangulation is a second and widely used strategy to generate stiff shear planes. Used as cross bracing or configured as larger horizontal truss systems in roofs, triangulation is primarily used in the context of steel and timber construction. It can be employed for single bays or for complete vertical or horizontal surfaces. Individual member connections can be pinned, thus facilitating construction. With the exception of precast concrete systems, cross bracing is rarely used in concrete construction. Frame action is the third major strategy used to provide lateral load resistance through shear planes. Although less efficient than either shear walls or cross bracing, frame structures tend to be more flexible than walls or braces. Frames also induce severe bending moments in both beams and columns, and final member sizes are typically much larger than in pin-connected systems that rely on diagonal braces or shear walls for stability. Consequently, frames are not normally a preferred solution for larger buildings. However, they are spatially open and thus desirable in many architectural situations. Recall that the rigid-joint requirement means that beams and columns must be attached to one another so that no relative rotations occur between the attached members (although the joint may rotate as a unit). In steel members, this typically means that flanges are connected at tops and bottoms. In poured-in-place reinforced concrete, it is necessary to place continuous steel reinforcement bars in opposite faces of the member. In timber, knee braces or specialized steel connector systems must be used. Force transfer and Basic organization of shear Planes. The exact way a structure ultimately experiences wind or earthquake forces depends on many factors. Wind forces on the side of a simple building shown in Figure 4, for example, are picked up by surface members, which in turn transfer them to secondary framing elements. The exact pattern of forces acting on the primary structure depends in its turn on how the secondary framing elements are organized. Earthquake forces are concentrated at high-mass areas (especially horizontal roofs or floors). For this discussion, we assume that the effects of winds or earthquakes are represented by a series of forces acting laterally on the primary structure. As noted, the roof or floor plane plays the primary role of transferring these lateral forces to side shear walls, cross braces, or frames. Figure 4 illustrates two methods for picking up forces acting on transverse walls and transferring them to

Structural Systems: Design for Lateral Loadings

Figure 4 Force transference to rigid wall or roof–floor planes. The graphic convention of a gray tone with dotted X is used to depict a rigid plane, regardless of whether it is a wall, truss, or rigid frame. Loadings on Transverse Faces

Forces transferred through beams in compression Stiff beam

Rigid-plane (stiff plane, rigid frame, or in-plane truss)

(a) Forces from secondary framing are resisted by an edge beam with high lateral strength and stiffness and carried directly to side shear walls or diaphragms (typically small structures only).

(b) Forces are transferred through roof members to the roof diaphragm, which transfers loads to side diaphragms (roof members must be designed to carry compressive forces as well as normal bending from vertical loads).

Loadings on Longitudinal Faces Rigid roof diaphragm (stiff plane, in-plane truss, or rigid frame)

Rigid vertical plane (stiff plane or shear wall in-plane truss or rigid frame)

Copyright © 2013. Pearson Education Canada. All rights reserved.

(c) Forces from secondary framing are picked up by rigid roof plane or diaphragm and transmitted to rigid side planes.

(d) Horizontal and vertical rigid planes or diaphragms can be anywhere as long as loads can be transmitted to them.

side shear planes. The figure also illustrates how forces on the longitudinal face are transferred to transverse shear planes on each side. An edge-beam approach, as illustrated in Figure 4(a), demands that the edge beam be sized for both vertical and horizontal loads. This approach is often used for relatively small buildings. As building dimensions increase, a larger part of the roof plane is utilized for stiffness. An entire bay, for example, might be cross braced to provide sufficient in-plane stiffness for a large building. In these cases, it is critical that the whole system be organized such that the rigid horizontal shear plane both receives externally induced forces and transmits them to side shear planes. The rigid horizontal plane providing diaphragm action can normally be anywhere in the overall plane. If the horizontal diaphragm is removed from the exterior wall planes, however, roof or floor beams must be designed to transmit external forces to it. These beams then must carry normal bending due to vertical loads and act as columns in compression. As illustrated in Figure 5(a), a real building must be capable of resisting external loads acting in any direction. In a simple building whose shear planes are located in vertical or horizontal bays, this requirement usually means the use of at least three, and typically four, vertical shear planes and connecting horizontal shear planes. (The next section demonstrates that a minimum of three vertical planes is necessary when a fully rigid horizontal roof or floor plane is present.) Because imparting in-plane stiffness to a horizontal roof or floor plane is often easy to accomplish (or simply inherent in most in situ reinforced-concrete systems),

Structural Systems: Design for Lateral Loadings

Figure 5 Typical lateral-stability solutions for small rectangular buildings. Stiff roof planes (in-plane trusses or diaphragms) pick up loads immediately.

The whole roof is made into a diaphragm.

(See Figure 6.) (a) For many typical buildings, rigid planes can be located anywhere in the structural volume, as long as external loads are transferred to the planes.

(b) These two arrangements are frequently used in larger buildings (depending on the construction techniques used) because they exhibit excellent resistance to horizontal racking and torsional deformations.

good practice suggests making entire roofs or floors into diaphragms when possible. [See Figure 5 (b).] If doing so is not possible, converting the external edges of a roof or floor plane into a band of rigid planes also is good practice. (Note that interior roof beams then need not carry compressive forces.) The building shown in Figure 6 is fairly large and made primarily of laminated timber beams. Given the nature of the structure and the large glass area present, it would have been difficult to make the entire roof plane into a rigid diaphragm. Thus, a band of horizontal diagonal braces surrounding the entire periphery of the building was used. This band connects directly with diagonal braces in the vertical planes, yielding a structure fully capable of resisting all types of lateral loads.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Fundamental statics. The previous section broadly described basic principles and good practice vis-à-vis how simple structures carry lateral loads. This section looks briefly at the nature of the forces developed within shear planes. A simplified analysis is used, based on the fundamental premise that the entire horizontal plane is fully rigid and capable of transmitting through it any in-plane forces that act on it. Thus, it can, in effect, serve as the carrier of a moment arm in simple equilibrium calculations.

Figure 6 Lateral force resistance approaches: In-plane roof and wall bracing system. Structure over swimming pool Laracha, A Coruna Architects: C. Quintans, A. Raya, C. Crespo

Primary beams

Diagonal braces around periphery in roof plane provide rigid diaphragm action. (a) Framing plan

Braces in roof plane

Braces in wall plane

(b) Simplified drawing of wall and roof braces

Structural Systems: Design for Lateral Loadings

Figure 7 Basic equilibrium of rectangular structure with fully rigid roof diaphragms and three shear walls. Equivalent concentrated load for wind=P1 w=Distributed wind load (psf)

P2

P1 h a

P2

Plan view

b

(a) Basic structure with fully rigid roof plane (diaphragm) and three shear walls

(b) Typical wind forces acting on vertical faces: P1=w (a*h)/2 (assume one-half of force is picked up by the foundation) and P1=w (b*h)/2. P1=w (a*h)/2

(c) Basic equilibrium: ΣFy=0

R1+R2=w (a*h)/2

o

a/ 2

a/ 2

R1

R2

Symmetry R1=R2=w (a*h)/4 ΣMo=0

P2=w (b*h)/2 b

R3(b)+R2(a) – [w (a*h)/2] (a/2) – [w (b*h)/2] (b/2)=0 R3=w (b*h)/2

Copyright © 2013. Pearson Education Canada. All rights reserved.

ΣFx=0

R3-w (b*h)/2=0

or

P2

R3

R3=w (b*h)/2

The simple rectangular structure shown in Figure 7 has three vertical shear planes. A simplified static analysis is illustrated. Distributed wind loads of w psf acting on the sides of the building are converted into equivalent point loads. (Half of the forces on the wall would typically be picked up by the foundations.) Force equilibrium in the y direction and symmetry considerations yield forces in the two side transverse shear planes 1R1 = R2 = wah>42. Moment equilibrium about any point, in this case, point O, yields the force in the third shear plane 1R3 = wbh>22. A building with a = 50 ft, b = 20 feet, h = 10 feet, and w = 20 psf would thus develop forces in the transverse walls of R1 = R2 = wah>4 = 120 psf2150 ft2110 ft2>4 = 2500 lb and a force R3 = wbh>2 = 120 psf2120 ft2110 ft2>2 = 2000 lb. These forces are what the lateral-force-carrying mechanisms (shear walls, cross bracings, rigid frames) must then be designed to carry. This simplified analysis is useful in many ways. It suggests that at least three shear planes are needed for stability. (Thus, the analysis is similar to that of a member, such as a beam, requiring a minimum of three support restraints for stability.) The same analysis also can be used to understand how best to locate vertical shear planes in buildings with different plan configurations. Figure 8 briefly illustrates Figure 8 Effects of building proportions.

While stable, the forces developed can become extremely high, and undesirable deformations can develop.

Using symmetrical shear walls reduces forces in transverse walls.

R3=2 wah/ 2 R3=2 wah (a) Square b=a

(b) Long rectangle b=4a Problematic arrangement

(c) Long rectangle b=4a Preferred arrangement

Structural Systems: Design for Lateral Loadings

Figure 9 Structures with shear walls meeting at a point. Rotational equilibrium about point O cannot be satisfied when walls meet at a point; hence, the structure can potentially be unstable with respect to twisting actions.

0

Tower with these arrangements must be designed as single rigid members; torsional resistance, however, remains low.

Torsional resistance is high in this shear wall arrangement.

0

(a) Problematic wall arrangements for normal wall and floor or roof constructions

(b) Small structures (including towers)

(c) Preferred shear wall arrangements

the effects of building proportions. The organization shown in Figure 8(b) is poor due to the high forces developed in the short lower wall. Nonsymmetrical placements such as the one shown are possible, and stability can indeed be achieved, but a better practice is the placement shown in Figure 8(c), if possible. Figure 9 illustrates a particular problem with structures employing only three shear walls. If the walls meet at a point, a twisting instability can develop. (Rotational equilibrium about their point of meeting, O, cannot be satisfied.) Some tower structures do have these kinds of wall arrangements, but they are designed more like vertical beams with unique cross sections (all horizontal and vertical elements are rigidly interconnected along all of their interface lengths). Even here, however, twisting can be problematic. A better practice arrangement is shown in Figure 9 (c). Note that when more than three walls are present, the structure is statically indeterminate and approximate or computer-based approaches must be used to determine force distributions.

Copyright © 2013. Pearson Education Canada. All rights reserved.

1.2

Low- and Medium-rise Buildings

Fundamental strategies. In designing low- to medium-rise buildings, it is often adequate to note only the basic lateral-force-resistance strategy and to identify its pattern implications during the early design phase. It is not always necessary to decide immediately whether a frame, shear wall, or diagonal bracing action will carry the lateral loads because any one of these approaches will provide sufficient lateral resistance. Appropriate locations of planes for these elements, however, must be provided. Further decisions are made as design development occurs. Figure 10 shows several concepts applicable to stabilizing medium-rise buildings. A common rigid-frame system might inherently provide vertical plane stiffness throughout the entire grid present [Figure 10(b)]. Ideally, floor and roof elements also would be made into diaphragms in this system. Alternatively, as seen in Figure 10(c), a system of rigid walls might be designed to encircle a roof structure that cannot be made into a rigid horizontal diaphragm. Sometimes only end bays are stabilized, as shown in Figure 10(e), in which case it is absolutely necessary that the horizontal planes act as diaphragms. Figure 10(f) shows an arrangement that works well except for the protruding top bays. In some buildings, it is easy to provide enough stiff shear planes. In housing, for example, the cellular nature of the building lends itself well to using shear walls or diagonal bracing in the interface area between adjacent units. A frame could also be used in these locations, but frames are less efficient than shear walls or diagonal bracing as a lateral-load-carrying device. In housing, shear walls or diagonal bracing in the interfaces between units pose no functional problem and are preferred. In many other types of building, however, the barriers

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 10 The lateral stability of any structure under any type of loading must be assured by the correct placement of lateral-force-resisting mechanisms.

formed by shear walls or diagonal bracing systems create functional problems and cannot be freely used. Frames, a combination of frames and shear walls, or diagonal bracing are consequently used. Intermediate- and low-rise office buildings, for example, quite often employ a basic frame throughout the building that is further stiffened by placing shear walls or diagonal bracing at building ends or around service cores. Usually, placing shear walls or diagonal bracing in these locations poses no functional problems. In rectangular buildings, the greatest problem with lateral forces is in the short direction of the building, although stability must be assured in both directions. (See Figure 11.) Sometimes, one type of lateral-load-carrying mechanism is used in one direction and another in the other direction (primarily for functional reasons). In simple steel buildings, for example, the more efficient mechanisms (e.g., shear walls or diagonal bracing) are often used in the shorter direction. Stability in the long direction is achieved either by similar means or by frame action. (See Figure 12.) If shear planes are to be used effectively in conjunction with other vertical planes not having any significant ability to carry lateral loads, such as a pin-connected steel-beam-and-column system, floor planes must be designed to serve as rigid

Structural Systems: Design for Lateral Loadings

Figure 11 Organization of either shear planes or bearing walls, or both, in a low- to medium-rise structure.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 12 Possible lateral-load-resistance mechanisms in a simple steel structure.

horizontal diaphragms. Such diaphragms act as thin horizontal beam elements spanning between shear planes (Figure 13). This diaphragm action transfers lateral loads from interior non-load-carrying vertical planes to the load-carrying shear planes. When reinforced-concrete systems are used, making floors serve as rigid diaphragms is typically no problem. With steel, special care should be taken to ensure that diaphragm action does indeed occur. Depending on the spacing of the shear planes, it may be necessary to impart some lateral-load-carrying capacity to interior vertical planes as well.

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 13 Rigid floor diaphragms are used in framed buildings that rely on shear walls to carry horizontal forces.

Anytime a structure is specially stiffened with shear walls or diagonal bracing systems, care must be taken to place these elements symmetrically. Otherwise, highly undesirable torsional effects can develop (see Figure 14) because the center of rigidity of the building becomes noncoincident with the centroid of the applied lateral load. Placing elements symmetrically is particularly important in tall buildings or when earthquake hazards are present. In either case, lateral loads capable of inducing high torsional forces are possible. Structures can be designed to resist these torsional effects, but a premium is paid. stability approaches in relation to shape. Thus far, the discussion has centered on relatively simple building configurations. However, configurations are often complex, as in the series of diagrams in Figure 15, which are derived from

Figure 14 Use of shear planes in tall structures and effects of nonsymmetrical placement. When shear planes (walls or diagonally braced planes) are used to carry horizontal forces in a tall structure, the planes should be arranged symmetrically to avoid undesirable torsional effects.

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 15 Common approaches to providing lateral stability in real buildings.

examples of real buildings. Drawing such diagrams is a good way to assess the stability of real configurations consisting of complex assemblies of shear walls, rigid frames, and pin-connected elements. Several approaches to simple configurations are shown in Figures 15(a)–(c). Figure 15(a) shows a pin-connected structure with diagonal bracing in each of the end bays. Long, low buildings like this often need bracing only at the ends. When longer bay assemblies are used, intermediate braced bays also are utilized. A simple rigid-frame structure using moment-resisting joints is shown in Figure 15(b). Base connections are pinned because of the frequent constructional difficulty of tying moment-resisting base connections to foundations. In relatively low buildings,

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

structural performance is not unduly affected, even though some loss in stiffness occurs. Figure 15(d) shows a low-rise building made of stable and interconnected three-hinged arch structures. Figure 15(e) indicates how a horizontal spanning truss can be connected to the sides of columns to form a structure that, in an overall way, acts like a rigid frame in resisting lateral loads. The truss must be connected to each column face at two widely spaced points. (The truss must have significant depth at its ends.) This same approach is evident in Figure 15(i). Many ways can be used to stabilize a structure constructed on a slope. The structure in Figure 15(f) utilizes a stabilized central bay. Note that the large cantilevered space to the right is made by using pin-connected elements configured to form a simple trussed configuration. Special building features, such as large interior spaces, may require that adjacent structural assemblies each have their own stability mechanisms, as in the building shown in Figure 15(g). Features such as cantilevers, with their need for fixed or moment-resisting connections, may naturally suggest that the larger structural assembly utilize rigid-frame action as well. Note that not all connections must be designed to be moment resisting: Pin connections may be used judiciously. Figure 15(k) shows one of many different cable-stayed structural configurations. Here, the center tower, a combined rigid-frame–truss structure, also provides the primary lateral force resistance mechanism in the overall structure. In smaller buildings, the type and arrangement of stability devices is less critical than in larger buildings, although the earthquake design requirements discussed in Section 2 may impose strong requirements in some areas. As buildings get taller, it becomes increasingly necessary to have clearly defined lateral-load resistance mechanisms. As illustrated in Figures 15(h) and (i), a building’s core parts (e.g., around elevators or stairs) frequently provide an excellent location for these mechanisms. In buildings based on slender cores, as illustrated in Figure 15(i), the cores are designed to resist lateral forces. They must ultimately carry all wind and earthquake forces acting on the structure. The center truss supports the floors above it to provide a large void area on the ground level. Extremely tall multistory structures necessitate special consideration and are discussed separately in Section 1.2. general considerations: Member orientation. Adopting one or more of the lateral-stability strategies discussed in the preceding sections influences the design or selection of individual members and their connections. When employing frames as stiff shear planes, for example, members should be organized so that their maximum bending resistance corresponds to the axes about which maximum bending occurs. In buildings with narrow dimensions that rely on frame action for stability, this often suggests organizing their deepest dimensions parallel with the narrow direction so the strong axis resists bending. However, if frame action is used in the long direction and combined with diagonal bracing in the short direction (Figure 12), then, in low buildings, wide-flange elements are usually organized so they function as part of the frame in the longitudinal direction about their strong axis and as part of the diagonal bracing system in the short direction about the weak axes. Because the diagonal bracing acts like a truss, bending is minimum in a member in the short direction, so it is acceptable to have the weak axis oriented this way. In the long direction, however, lateral loads are carried by frame action involving high bending in the members. Consequently, it makes sense to organize the strong axis of the member in this direction. When the narrow dimension of a steel building is small relative to the building height (indicating a severe problem with lateral loads in the short direction) and the building is adequately braced in the long direction, a combination frame and diagonal bracing system can be used to carry loads in the short direction. In this event, wide-flange members are organized so their strong axes contribute to frame action in the short direction. The frame and diagonals supplement each other, yielding a total system of increased load-carrying capacity.

Structural Systems: Design for Lateral Loadings general considerations: effects of Member characteristics. The choice of a method for achieving lateral stability does not always precede the choice of a horizontal spanning system. In cases where the horizontal span is extremely long or loads are unique, the nature of the horizontal element may dictate the lateral-stability approach used. If spans are 100 ft (33 m), for example, and precast single tees seem appropriate for the vertical loads, providing lateral stability through a technique other than joint rigidity is mandated. The type and scale of the horizontal element, and the difficulty of obtaining rigid connections, mean frame action is not feasible. In general, the longer the horizontal spans, the less likely will frame action be appropriate for achieving lateral stability (Figure 16). Often, the unique characteristics of one structural element influence the selection of another. When masonry load-bearing walls are used in low-rise buildings, the walls serve as shear planes in resisting forces in the lateral direction. It is best to organize them along the short dimension of the building. Lateral stability along the long dimension can be achieved (depending on the building height) through specially designed stair cores or interior walls placed transverse to bearing walls. The use of masonry walls implies that horizontal spanning elements should be simply supported because masonry walls, unless specially reinforced, cannot carry moments. Thus, attempting to use rigid connections at the end of horizontal elements (which would induce moments into the wall) would be counterproductive.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 16 Common methods of resisting lateral forces: implications on type of connection.

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

1.3

Multistory construction

strength Design. As buildings increase in height, responding to lateral forces caused by either wind or earthquakes becomes an increasingly important structural design consideration. Of the basic approaches to stability, frame action is probably the least efficient way to achieve lateral stability and is used only when lateral forces are not excessive—as in low- to medium-rise buildings. Typical steel-framed buildings that carry loads exclusively by frame action, for example, do so efficiently only up to about 10 stories. After that, a significant premium is paid in terms of excessive material used if a framed structure is utilized. Even in low- to medium-rise structures, where frame action may still be appropriate, the lateral-load-carrying capacities differ among the types of framed systems possible. Steel systems are unique in that they can be designed to be responsive to virtually any situation. Reinforced-concrete systems, by contrast, must be used more carefully. Systems such as the flat-plate-and-column assembly have a much smaller lateral-load-carrying capacity than other poured-in-place concrete systems. The flat-plate spanning element is uniquely suited for carrying the limited moments induced by relatively light floor loads but not the moments generated at the interfaces with columns by large lateral forces. The plate is too thin to carry such moments effectively. Systems with deeper horizontal members, such as the two-way beam-and-slab system or the waffle system, provide frame action more effectively in both directions. When flat plates are used, their lateral-load-carrying capacity is often supplemented by some other mechanism. In common rectangular apartment buildings, for example, which often use flat-plate construction, end walls are frequently turned into shear planes, which in turn largely carry the lateral loads. The flat-plate system can then be designed primarily in response to vertical loads. The enclosure around elevator cores also is often specially designed to serve this same function. As heights increase from the low to intermediate range, it is preferable either to begin supplementing any type of frame system used with additional lateral bracing mechanisms, such as diagonal braces around elevator cores, or to adopt a radically different structural approach. To get a feeling for the different structural approaches that might be possible for tall structures, it is useful to first briefly review some fundamental principles of how a tall structure carries lateral loads. Most high-rise structures are relatively tall and slender. Under the action of lateral forces, they act like vertical cantilever members. The lateral loads tend to produce an overturning moment, which must be balanced by an internal resisting moment provided by the structure. Couples formed between forces developed in vertical members typically provide this internal resisting moment. (See Figure 17.) If the building is very slender, the small moment arm present between the forces in the vertical members means that high forces must be developed to provide the internal resisting moment. Buildings of similar heights with wider bases and less slender proportions could provide the same internal resisting moment with smaller forces developed in vertical members because the internal moment arm is larger—an advantageous consequence. Viewing a high-rise building as a vertical cantilever is useful because, in addition to implying something about appropriate building proportions, it can help inform other design responses. The most efficient use of material in bending is obtained by locating the greatest amount of material as far as possible from the neutral axis of the section. The overall moment of inertia of the section for a given amount of material is increased by doing this, as is the resistance to bending. The same principle holds for truss design. In a high-rise building of a given proportion, applying this same principle means that the greatest amount of material should be located in the outer, rather than the inner, vertical elements. (See Figure 18.) Many efficient high-rise structures are conceived in this way. The

Structural Systems: Design for Lateral Loadings

Figure 17 Very tall buildings can be conceived of as vertical cantilever beams.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Wind

Compression

Tension

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 18 Typical structural approaches in high-rise construction: Exterior system elements carry the majority of all lateral loads, while interior columns carry mainly gravity loads. Rigid cores can be used for these systems to provide additional stiffness.

(a) Rigid moment frame

(b) Moment frame is further stiffened by a rigid core.

(c) Frame and core are connected with outrigger trusses for additional stiffness.

(d) Tube Structure. Exterior columns are closely spaced and rigidly connected to spandrel beams.

(e) Trussed frame. A rigid moment frame is further stiffened by a large truss element.

(f) Diagrid. Gravity and lateral forces are transferred through a triangulated column grid.

Structural Systems: Design for Lateral Loadings exterior column-and-beam assemblies are designed to provide a very stiff ring, or tube, capable of carrying lateral loads from any direction. Exterior columns are typically closely spaced to each other. Spandrel beams are usually rigidly connected to columns to ensure that the whole outer assembly acts in an integral way—like a stiff surface element. Although the outer frame assemblies can and do carry gravity loads and act like frames in the horizontal direction, their primary function is to carry forces generated by the overturning moments associated with lateral loads. Interior columns are designed to carry gravity loads only, so they are smaller than exterior columns. Cores, located at the interior, are often designed to complement lateral load-resisting frames and other systems located in the perimeter. Special attention must be paid to designing a floor system that makes the overall resultant structural assembly behave like a unit in carrying lateral loads. The stiffness of a tube like this can be increased even further by adding large cross bracing on the outside faces of the structure. Diagrids can be efficient systems to carry gravity loads and lateral loads. Here, columns are configured to generate stable triangles that often include horizontal floor beams (Figure 19).

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 19 Diagrid approach for a tall building: Diagonals can be visually emphasized over the horizontal members, as shown here at the Swiss Re Headquarters in London.

Structural Systems: Design for Lateral Loadings

Figure 20 Plan configurations of buildings with narrow dimensions can be shaped to provide overall structural configurations that are resistant to lateral forces.

The configurations of some types of tall buildings may ultimately involve a plan type that is long and narrow (e.g., as might be found in a residential high-rise building where units are grouped around central corridors). Narrow plan shapes are difficult to structure efficiently to carry lateral forces if those shapes are used in a simple linear way. By deforming the overall plan configuration, however, the overall stiffness of the structure and its related ability to carry lateral loads can be dramatically increased. (See Figure 20.)

Copyright © 2013. Pearson Education Canada. All rights reserved.

Deflection and Motion control. The structural responses of multistory buildings discussed thus far have all been strength oriented. Equally important are deflection and motion considerations associated with the dynamic effects of winds. A typical tall building, for example, sways under the buffeting action of damping devices. Such a device, often called a tuned mass damper, is placed in the upper floors of a tall building. As motion begins, the inertial tendency of the mass is to remain at rest. Hence, building movements cause the damping devices to accentuate and begin absorbing energy, thus damping out motions. (See Figure 21.) The use of damping devices just described is usually restricted to very large buildings. They have, however, been successfully used in other situations. More commonly, the stiffness of the structure is controlled. A rough rule of thumb often used to limit both accelerations and deflections when dynamic analyses are not appropriate or feasible is to use a static wind-load analysis and limit maximum steadystate deflections to a particular value, say, h/500, where h is the height of the building

Figure 21 Tuned mass damper in a high-rise building: A heavy mass or pendulum swings in a controlled way to counter the swinging movement induced by dynamic lateral loads.

(a) Initial position: The heavy mass is connected with tuned springs to the building. Its natural frequency is usually similar to that of the tower.

(b) Upon lateral movement of the tower top, the mass tends to stay in its original position. This delay causes the springs to deform.

(c) The mass oscillates in the exact opposite direction as does the building. The resulting forces dampen the tower movement.

Structural Systems: Design for Lateral Loadings in feet, and to vary the stiffness of the structure until this criterion is met. Such an approach is much easier than a dynamic analysis but is dubious at best. Still, it has historically proven to be useful, if not always a sure bet.

2 2.1

earthquake Design consiDerations general Principles

Earthquakes produce forces on structures that are primarily lateral in character, although forces in other directions can develop as well. Most of the principles discussed in the preceding sections are therefore as applicable to the design of structures for earthquake resistance as they are for wind resistance. The extremely pronounced dynamic character of earthquake forces, however, makes the design problem even more complex. This section briefly explores some of the primary issues involved and alternative strategies that are possible. Among the various hazards associated with earthquakes are surface fault ruptures, ground shaking, ground failures, and tsunamis (sea waves generated by earthquakes). It is useful to review the behavior of a structure subjected to an earthquake. Figure 22 illustrates characteristic vibration modes for

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 22 Earthquake-induced motions in a multistory building.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Structural Systems: Design for Lateral Loadings a tall building. The ground acceleration can cause all floors to move in the same direction, or it can cause different floors to accelerate in different directions in a whiplash type of movement. The whiplash movement can occur because of an elastic building’s tendency to spring back to the vertical after its base has been initially translated and accelerated. Because of inertial tendencies, the upper mass of the building would lag behind the base movement. As the upper mass springs back to the vertical after the translation has occurred, it begins gaining momentum. Because of the momentum gained, the upper mass can swing past the vertical. If, during this process, the ground movement is changed such that accelerations and translations are reversed, highly complex deformations can occur because of the inertial tendencies of the building masses to continue the movement already started, while at the same time countermovements begin. There is a general lag between the ground movements and their translation into displacements of the upper levels of the building. Associated with these complex accelerations are high inertial forces caused by nonconstant winds. Even when winds are steady state, a dynamic behavior is still present. As the wind blows against a building, the building bends and changes shape slightly. The exact magnitude and distribution of wind forces thus also change slightly. Increases or decreases in forces due to this phenomenon, coupled with any buffeting action of the wind that might be present, cause the building to oscillate. Typically, there is an average deflection in the direction of the wind force about which the building oscillates. The magnitude and frequency of the oscillations depend on the characteristics of the impinging wind forces and the stiffness and mass-distribution characteristics of the building. These deflections and oscillations are highly important. Excessive deflections can impair the functioning of other building elements (e.g., building service systems), even though the structure itself is unharmed. Oscillations can cause extreme discomfort to building occupants. Human beings do not sense absolute deflections if they occur slowly, but they do sense the accelerations associated with rapid oscillations. As a result, a type of motion sickness can occur. What constitutes an acceptable level of acceleration is not easy to establish rationally, although investigators have suggested some values. In addition, it is difficult to analyze a building to predict what motions might occur in a given circumstance. The most common procedure is to model the structure as an assembly of springs (having elasticities derived from a study of the load-deformation characteristics of the structure) and concentrated masses representing building weights. The vibrations a model like this undergoes when dynamically loaded can be used to predict the response of the real structure. Computer programs facilitate these analyses. Thinking about a multistory structure as a spring-mass assembly also is useful in formulating design responses to deal with excessive motions or deflections. One way to control deflections is to increase the stiffness of the structure (not necessarily its strength) until wind-induced oscillations are held to an acceptable level. Increasing member sizes, using diagonal braces or other shear planes, and redistributing the placement of material are all ways to increase stiffness. Another way to control motions, particularly accelerations, is to build physical damping mechanisms or friction connections into the building. (Dampers are conceptually similar to devices that cause a door to close slowly and evenly when opened and released.) Damping devices are typically installed at joints between beams and columns. As movements begin, the devices absorb energy and damp the motion. Excessive motions also can be controlled by installing a large mass of material mounted on rollers and attached to the building structure with dashpots, dampeners that slow down movement using viscous fluids. In addition to the high inertial forces expected, unique forces can be induced in a structure because of unusual

Structural Systems: Design for Lateral Loadings building proportions or features. Twisting, for example, can be induced because of the nonsymmetrical placement of building masses. These effects are explored in the next section.

Copyright © 2013. Pearson Education Canada. All rights reserved.

2.2

general Design and Planning considerations

Probably the single most important design principle with respect to ensuring that a building performs well during an earthquake is making certain that the general masses (e.g., floors and roofs) and the stiffening lateral-force-resisting mechanisms (e.g., shear walls and braced frames) present in the building are symmetrically located with respect to one another. Lateral forces associated with earthquakes are, of course, inertial in character and are thus related to the masses of different building elements. The nonsymmetrical placement of building masses and elements that resist the forces associated with these masses can lead to undesirable torsional effects in the building, which can be extremely destructive. This issue was generally addressed earlier in connection with wind forces (see Figure 14), but it has even more importance with respect to earthquake forces. The principle has significant implications for the overall building shape adopted because the mass distribution and the placement of lateral-force-resistance mechanisms are strongly influenced by the building’s shape. An L-shaped building, for example, has a nonsymmetrical mass distribution and, normally, a nonsymmetrical placement of stiffening elements. Destructive torsional forces develop in such shapes for the reasons just discussed. Another way to understand why the configuration is undesirable is to conceive of the building as being made of two separate masses (each leg of the L), each of which vibrates in its own natural frequency. Because the stiffnesses of the two units differ, their natural periods also differ—a condition that sets up a deflection incompatibility at the interface of the two masses. The consequence is that failures often occur at interface locations. The problem can be somewhat alleviated by separating the building into symmetrical units connected by a seismic joint, which allows free vibratory movement to occur independently in each unit. (The device is conceptually similar to an expansion joint.) From an earthquake-design viewpoint, preferred building shapes are those of simple plane geometry (e.g., squares and circles); L shapes, T shapes, H shapes, or other unusual shapes are difficult to structure when earthquake hazards are present. Seismic joints can be used to advantage in most such configurations. Difficulties can arise, however, even in buildings that seem symmetrical in plan, if stiffening elements are nonsymmetrically placed. This condition often appears when elevator cores or other natural stiffening elements are unusually located. [See Figure 23(b).] The structural U shapes of shear walls found in many common commercial buildings that seem symmetrical (rectangular) are subject to undesirable torsion effects. Such buildings should have their open faces, where glass is normally present, stiffened by the insertion of frames. Stiff roof diaphragms also should be used. Some completely symmetrical buildings can be susceptible to torsion because of nonsymmetrical placements of occupancy loads, a condition that frequently arises in warehouses. This situation can be prevented somewhat by careful design. The principle of imparting symmetrical stiffness and massing characteristics to a building also holds true for a building in its vertical dimensions as well as in plan. If the stiffness characteristics of different building elements change with building height, for example, nonsymmetries can develop that lead to destructive torsional effects at different levels of the building. Buildings with discontinuous shear walls can be particularly problematical. Other considerations can affect a building’s sensitivity to earthquake movements. Buildings that are extremely elongated, for example, should be avoided even if they are symmetrically organized (Figure 24). The longer a building is in plan, the greater is the possibility that opposite ends of the building will be subjected to

Structural Systems: Design for Lateral Loadings

Figure 23 General planning considerations: Symmetry versus nonsymmetry.

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 24 Elongated buildings are more susceptible to destructive forces associated with differences in ground movements along the length of the building than are more compact shapes. Long buildings can be subdivided by using seismic joints.

different relative ground movements that do not act in the same manner, a phenomenon that might cause the building to be torn apart. Designing to prevent such tearing apart is difficult. It is often necessary to separate extremely long buildings into series of adjacent volumes (through the use of seismic joints) that can move freely with respect to one another.

Structural Systems: Design for Lateral Loadings Another basic consideration is that tall, slender buildings are less able to resist the overturning moments associated with earthquake forces than are lower, wider buildings (Figure 25). Buildings that are constructed adjacent to one another should be adequately separated so that each can vibrate freely in its natural mode without touching the other. Otherwise, severe damage can occur because of pounding effects (Figure 26). While forces associated with lateral movements are of primary design importance, vertically acting accelerations due to earthquake motions also can cause trouble. Cantilevers, for example, are particularly susceptible to failure because of vertical movements. (See Figure 27.) In such events, failure often also follows in interior members because of the moment redistributions that occur after the cantilevers drop off. All the considerations just noted point to the fact that, from the viewpoint of designing for earthquake resistance, buildings that are symmetrically organized and have relatively compact proportions with few or small overhangs or other protrusions are preferable to nonsymmetrically organized buildings with exaggerated proportions. Other considerations, it should be noted, such as the specific site and soil conditions present, also may strongly affect the preferred building configuration.

Copyright © 2013. Pearson Education Canada. All rights reserved.

2.3

Figure 25 Relatively slender buildings are less able to resist the overturning moments caused by earthquakes than are shorter, more compact configurations.

general characteristics of earthquake-resistant structures

Structures that are continuous in nature and more or less uniformly distributed throughout a building generally perform well when subjected to earthquakes. The primary reason for this is that a structure’s earthquake resistance depends, to a large extent, on its ability to absorb the energy input associated with ground motions. Pin-connected structures, such as traditional post-and-beam assemblies, are far less capable of absorbing energy than are comparable continuous structures (e.g., frames with monolithic joints). The formation of plastic hinges in framed structures (which must precede their collapse) requires a significant energy input. Continuous structures, therefore, are effectively used in buildings in earthquake-hazard zones (Figure 28). Either steel or poured-in-place, reinforced-concrete-framed structures designed to be ductile under earthquake forces can be used. Precast-concrete structures, with their lack of continuity at connections, can be used, although working with them is more difficult. Even within the category of continuous structures, however, differences exist. Figure 29 illustrates the plans of two structures that derive their ability to carry lateral forces from frame action. The first, illustrated in Figure 29(a), with frames distributed throughout the structure, is preferable to that shown in Figure 29(b), which has frames around the periphery only. The first structure has a greater redundancy and consequently has greater reserve strength than the second structure. The failure of relatively few members in the outer plane of the structure shown in Figure 29(b) may lead to total collapse, whereas many more members must fail in the more redundant structure for it to collapse. Generally, structures with redundancy are preferable to those without redundancy. Another general characteristic of viable earthquake-resistant structures is that column-and-beam elements are generally coaxial. Offsets or nonaligned members often present extremely difficult design problems. Earthquake-resistant structures also typically have floor and roof planes designed as rigid diaphragms capable of transmitting inertial forces to lateral-loadresisting elements through beamlike action (Figure 30). Another aspect of earthquake-resistant structures is they are designed such that horizontal elements that fail due to earthquake motions do so before any vertical members fail (Figure 31). Failure should never occur in vertical members first because of lifesaving considerations. Horizontal elements in continuous structures (e.g., slabs or beams) rarely fall down completely, even after receiving extreme damage, and when they do, the collapse is fairly localized. (This is, in general, not true

Figure 26 Adjacent buildings should be separated so that buildings do not pound against each other during seismic events.

(a)

(b)

Structural Systems: Design for Lateral Loadings

Figure 27 Importance of the vertical components of earthquake ground motions.

Figure 29 Redundant versus nonredundant structures: Buildings with highly redundant structural systems in which multiple load paths exist for carrying earthquake forces to the ground usually perform better than less redundant structures during seismic events.

of pin-connected horizontal members.) When columns receive damage, complete collapse is imminent. The collapse of a column causes other portions of a structure to collapse as well (progressive collapse). The effects of a single column collapsing can be extensive. To ensure that horizontal elements fail first, care is exercised in the design and general proportioning of beam-and-column elements. Extremely deep beams (e.g., spandrels) on light columns are generally best avoided because experience indicates that such buildings often receive significant damage in the light columns, which must pick up all the laterally acting forces by shear and bending. In contrast, shear walls with a series of spaced, small openings perform somewhat better. Good engineering design, it should be noted, can make all types work adequately.

Copyright © 2013. Pearson Education Canada. All rights reserved.

2.4 Figure 28 Continuous structures are preferable to pin-connected ones because the plastic hinges that form in continuous structures before they collapse absorb large amounts of energy.

Materials

As noted before, fully continuous structures are desirable for use in earthquakeprone regions. Coupled with the idea of continuity is that of ductility. For energy absorption to take place, ductility is essential. Steel is a naturally ductile material, so

Figure 30 Importance of rigid floor and roof elements: For earthquake-induced inertial forces to be transferred to lateral-load-carrying elements, floor and roof elements must be capable of acting like rigid diaphragms.

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 31 Members should be designed such that failure occurs first in horizontal members rather than in vertical members (a strong-column, weak-beam strategy).

it is often used. Poured-in-place reinforced concrete also can be made to have a high degree of ductility by carefully controlling member proportions and the amount and placement of reinforcing steel. Reinforced-concrete buildings like this are frequently used in earthquake zones. Precast-concrete structures, however, can be difficult to design for safety in earthquake zones because of the problems involved in achieving a continuous, ductile structure. Such structures are also typically high-mass structures. The high mass contributes to the magnitude of the seismic forces developed and compounds the problem of their use. Timber can be an extremely good material for use in earthquake regions. It is lightweight and capable of absorbing large amounts of energy when deformed and before collapse. Low-rise wood-framed structures are highly earthquake resistant and perform well in earthquake regions. Unless highly reinforced with steel, materials such as masonry are not suitable for use in earthquake regions. Masonry structures are massive but have little ductility. Masonry buildings frequently crack when subjected to earthquake motions. Much of the damage that has occurred in cities during earthquakes has been because older, unreinforced masonry buildings failed. Masonry walls often fail because large diagonal tension cracks develop (recall that principal stresses in elements subjected to shear act only at 45° angles) and assume large X patterns. These cracks initially develop at corners of windows or other openings. Another common failure is the collapse of floor systems because of horizontal beams pulling off the masonry walls from movements induced by the ground shaking. Tying walls and beams together helps prevent this kind of failure.

2.5

stiffness issues

Of primary importance in seismic design is the natural period of vibration of the building under consideration. If the ground input motions have a similar frequency to that of the natural frequency of the building, then resonance can develop and high destructive forces can be generated. However, the forces that a building experiences are always directly related to its natural frequency even if resonance does not occur. Forces specified in many building codes, for example, often assume a general form, which implies that short- period (high-stiffness) buildings must be designed for larger forces than long-period

Structural Systems: Design for Lateral Loadings

Copyright © 2013. Pearson Education Canada. All rights reserved.

Figure 32 Different structural responses have widely varying natural periods of vibration, an important consideration in seismic design. The use of base-isolation systems and other devices can dramatically alter the structural response present. (See Section 2.7.)

(low-stiffness) buildings (Figure 32). As is discussed shortly, however, this is not the only factor that must be taken into account when choosing to opt for a relatively stiff building (e.g., one using reinforced-concrete shear walls) or a relatively flexible (low-stiffness) one (e.g., a steel frame). One important component of the decision has to do with the notion that, for ground motions that are close to the natural frequency of the building itself, the structure will receive maximum punishment because of the tendency toward resonance. If the frequencies are different, then lower seismic forces will affect the structure. When ground motions have a long period, a stiff building should undergo lower seismic forces than a flexible building, and vice versa. It, therefore, seems that the choice of relative stiffness should be made on the basis of the expected character of the ground motions that the building might receive. Unfortunately, the character of these motions is difficult to predict. The motions a building receives are strongly influenced by the nature of the soil conditions present beneath the building and the interaction of the building with the soil. Still, in certain circumstances, the general nature of expected ground motions can be predicted to a greater or lesser degree. No matter what the situation, many designers advocate designing fully flexible buildings in which a relatively flexible structure (e.g., a steel frame) is used. Such buildings also are designed so that no actual or potential stiffness is imparted by other building elements (e.g., partitions). These nonstructural elements are detailed to allow the structure to move freely (a difficult and sometimes unfeasible task). Briefly, the primary advantages of a full flexible approach are that such buildings are especially suitable for sites where ground-motion periods are expected to be short. Ductility and continuity also are easy to achieve because steel is used extensively. Disadvantages occur when long-period sites are present and resonance situations can develop. Nonstructural elements are also difficult to detail and often receive great earthquake damage it they are not designed properly. Stiff structures may have some advantages for long-period sites. Nonstructural elements are easy to detail in stiff buildings because structural movements are not large and damage to such elements from minor earthquakes is limited. Stiff structures, however, are not advantageous for short-period sites. For some materials, the notion of high stiffness also is not compatible with the notion of ductility, which is important from an energy absorption viewpoint. (The presence of some ductility is important even in stiff structures.) Still, using careful design, even stiff structures (such as those made of reinforced concrete) can be made highly ductile by incorporating appropriate amounts of steel in the proper locations.

Structural Systems: Design for Lateral Loadings On some occasions, mixed systems are advantageous. The mixed frame–shear wall system shown in Figure 32(d), for example, is often a good system because, with earthquakes of a low magnitude, the stiff shear wall takes most of the forces involved and the whole building responds stiffly, thus limiting nonstructural damage. By contrast, the more ductile frame provides a large measure of reserve capacity when high-magnitude earthquake forces are present.

2.6

Figure 33 elements.

Nonstructural

nonstructural elements

Many nonstructural elements can significantly influence the dynamic behavior of, and hence the seismic forces that are present in, a building. Depending on how such elements interact with the primary structure, they may alter the natural period of vibration of the structure, thus changing the forces present. They also can affect the distribution of lateral stiffness in a building, which can have significant consequences, too. One of two basic strategies is normally adopted with respect to nonstructural elements (Figure 33). One is to carefully analyze all elements and include those that contribute to the stiffness of the primary structure in the analysis and design of that structure. Contributing elements must be carefully detailed to ensure that they contribute to the building stiffness as expected. The second basic strategy, noted earlier, is to prevent any of the nonstructural elements from contributing to the stiffness of the structure. This is done by detailing connections such that gaps exist between the primary structure and the nonstructural elements. The structure can then deform freely. Which of these two approaches is assumed depends largely on the attitude the designer takes toward the overall stiffness of the structure and whether flexibility is desired or not.

Copyright © 2013. Pearson Education Canada. All rights reserved.

2.7

Base isolation systems and other techniques

There has been a spate of new developments in designing structures to perform well during seismic events. These approaches include using various base-isolation and damping systems, as well as careful manipulation of the stiffness of beam-andcolumn elements to achieve improved performance and reduce damage during a seismic occurrence (including so-called strong-column, weak-beam approaches; see Figure 31). These manipulations allow designers not only to control overall dynamic responses better but also to control more readily where and how damage might occur. Approaches include the use of one type of system or another for seismic base isolation and passive energy dissipation. Systems like this have been installed in buildings and bridges throughout the world. The idea is simple. The designer avoids transmitting the shaking action into the structure in the first place by inserting an isolating device between the base of the building or bridge and its foundations. One of the most widely used approaches is passive base isolation. A layer of low stiffness is placed between the structure and its foundation. (See Figure 34.) A widely used isolation system utilizes elastomeric bearings of either natural rubber or neoprene. Such bearings help decouple the building structure from the laterally acting components of an earthquake’s ground motion by introducing a layer with low horizontal stiffness between the structure and the foundation. This makes the fundamental frequency of the overall structure change so it is lower than its frequency without the isolation devices; often, it becomes lower than normal frequencies of the ground motion. In the first dynamic mode, a structure using an isolation system is more or less rigid above the isolation device, where most deformations occur. Higher modes can be problematic. Note that this basic system does not necessarily absorb the earthquake energy, but rather, it alters the dynamics of the whole configuration. Adding damping devices can be beneficial and reduce the potential for any kind of resonant action. Fluid viscous dampers are often used.

Figure 34 Base isolation. In a seismic event, the lateral movement of the soil is dampened.

Structural Systems: Design for Lateral Loadings Other isolation systems include the sliding system (in which the transfer of shear across the isolation interface is limited), friction-pendulum systems, and various active base-isolation systems. This whole field is in a rapid state of change and development.

questions 1. Identify at least one multistory building in your area or one that is documented in the literature that uses shear walls for lateral stability. Draw a plan of the building and indicate the locations of these elements. Do they form a pattern capable of resisting loads in all directions? If not, what other load-carrying mechanisms are present to ensure complete stability? 2. Repeat Question 1, using a framed building. 3. Repeat Question 1 for a diagonally braced building. 4. Review the literature and identify at least six different multistory buildings in different height ranges. The height range should vary from about 10 stories or fewerless to the heights of some very tall buildings (e.g., the Hancock Building or the Empire State Building). Identify the structural system used in each case. To the same scale, and emphasizing structural elements, do a schematic diagram of each building’s elevation. 5. Review the literature and identify a major building that was specially designed for earthquake forces. Identify all structural and nonstructural design devices that were incorporated to reduce earthquake hazards. Record your findings with a series of annotated diagrams.

Copyright © 2013. Pearson Education Limited. All rights reserved.

6. Review the literature and identify a major building that uses a tuned mass damper in its upper floors as a way to control lateral motions. (Consider, for example, the Taipei 101 Building.) Diagram how the device works.

Structural Systems: Constructional Approaches 1

IntroductIon

Copyright © 2013. Pearson Education Limited. All rights reserved.

This chapter provides a brief descriptive overview of many different structural systems in common use. A useful way to characterize these typical approaches is according to the nature of the primary material used in the structure. Three common materials— wood, reinforced concrete, and steel—are addressed here. Basic characteristics of systems are highlighted, grouped according to materials. The chapter also contains some rule-of-thumb information for determining approximate sizes of elements. Using rules of thumb is a time-honored way to initially size structures. At one time, to inform the design of new structures, it was the only way to utilize empirical knowledge gained from observing built structures that performed successfully. (See Figure 1.) Rules of thumb are still of interest for the same reason but should be used with care. Because the rules were derived by looking at previously built structures, their use in design propagates the past state of the art rather than looking to the future state of the art. Hence, they preclude innovative structural solutions based on an understanding and application of the theory of structures.

2

Wood constructIon

Primary systems. Many constructional systems using timber as the base material can be characterized as being composed of linear one-way spanning elements. Hierarchical arrangements are typical. Figure 2 illustrates a sampling of the different types of wood construction systems in common use. Many other systems also are used. Light Framing. The light-joist system illustrated in Figure 2(a) is ubiquitous in wood-framing systems used in current construction. The floor joist system is primarily useful for light occupancy loads that are uniformly distributed and for modest spans, conditions typically found in house construction. Joists are normally simply supported. Rarely are moment-resisting joists used because special connections are required. Usually, the transverse decking is not considered to act integrally with the joists, unless special care is taken with connections. From Chapter 15 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Structural Systems: Constructional Approaches

FIgurE 1 Early rules of thumb for sizing structures.

Copyright © 2013. Pearson Education Limited. All rights reserved.

The vertical support system is typically a load-bearing wall of either masonry or closely spaced wood elements (studs) sheathed in plywood. In the latter case, the lateral resistance of the whole structural assembly to horizontal forces is obtained by arranging plywood-sheathed walls to serve as shear planes. Structures like this are typically restricted in height to three or four stories, not so much because of their theoretical load-carrying capacity as for fire-safety requirements stipulated in building codes. Because each element is individually put in place on the job site, a great deal of flexibility is available in the use of the system and in how it is integrated with other building components. stressed-skin Elements. Stressed-skin members are related to the standard joist system. [See Figure 2(b).] In these elements, plywood sheathing is affixed to both sides of stringers in a way that assures that the sheathing acts integrally with the stringers in carrying bending. A form of plate is thus achieved. The rigidity of the complete system also is increased by this integral action. Structural depths are consequently less than in standard joist systems. Stressed-skin elements are typically made off-site and put in place as modular units on the site. They are useful when they can be used in a repetitive way. They also can be used in other ways, including being made into long-span foldedplate systems. Box Beams. The built-up nature of plywood box beams [see Figure 2(c)] allows great latitude in the range of span and load conditions that they can be designed to meet. Such beams are useful in long-span situations or when unique loading conditions are present. Box beams can efficiently span greater distances than can homogeneous or laminated beams. Heavy-timber construction. Heavy-timber beams with transverse planking historically preceded the lighter joist systems. [See Figure 2(e).] Laminated timber beams are now often used in lieu of homogeneous members. Such systems

Structural Systems: Constructional Approaches can have a much higher load-carrying capacity and span range than joist systems. With laminated beams, for example, relatively long spans are possible because member depths can be increased almost at will. Elements of this type are normally simply supported (see Figure 3), but rigid joints can be achieved through special connections. The vertical support system is typically either masonry walls or timber columns. The lateral resistance of the whole structural assembly to horizontal forces is obtained by using walls as shear planes. Knee braces also are utilized to obtain lateral stability when columns are used. Gaining lateral stability through momentresisting joints is possible in low structures but is not frequently done.

Copyright © 2013. Pearson Education Limited. All rights reserved.

(d)

FIgurE 2 Timber construction systems.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Structural Systems: Constructional Approaches

FIgurE 2 (Continued)

trusses. The timber truss is among the most versatile of all one-way spanning elements because a wide variation is possible in the configuration and member properties used. Special handcrafted trusses are particularly suitable for unique load or span conditions. Mass-produced timber trusses, however, also are available and are used in light-load and modest-span situations. The trussed rafter illustrated in Figure 2(g), for example, is used extensively in roofs of single-family detached housing. The system shown in Figure 2(h) is analogous to the steel open-web bar joist and is useful in long-span situations (particularly for roofs). Folded Plates and Arch Panels. A variety of special flat- or curved-plate structures, which are typically one-way spanning elements that are useful for roofs, can be constructed from wood. Most involve the use of plywood. Figures 2(j) and (k) illustrate two examples of this type of structure.

Structural Systems: Constructional Approaches Arches. Standard arch forms can be made from timber. Laminated members are most often used. Almost any shape of arch can be made from laminated wood. Relatively long spans can be obtained. These structures are typically useful as roofs only. Most are either two or three hinged and not fixed. Lamellas. Lamella construction is a way to make extensive singly or doubly curved surfaces from short pieces of wood. [See Figure 2(l).] This interesting construction can be used to make long-span cylindrical surfaces or long-span domed structures. The system is versatile and has found wide application in roof structures. Member sizes. Figure 4 illustrates approximate span ranges for different timber structures. As noted earlier, the “maximum” spans indicated on diagrams like this do not represent maximum possible spans, but instead represent spans that are not considered unusual even if they are longer than typically encountered. The minimum span limitations represent a system’s lower economic feasibility span. The figure also shows approximate depth ranges for the different spanning systems. One number represents the usual minimum-depth system, the other common maximums. An approximate depth of L>20 means, for example, that a member that spans 16 ft (4.9 m) should have a depth of about 16 ft 112 in.>ft2 >20 = 9.8, or 10 in. 14.9 m2 >20 = 0.24 m = 240 mm. Timber columns typically have thickness-to-height 1t>h2 ratios that range from 1:25 for relatively short and lightly loaded columns to around 1:10 for heavily loaded columns in multistory buildings. Walls made of timber elements have comparable t>h ratios, ranging from about 1:30 to 1:15. Tables of the type in Figure 4 and the t>h ratios noted here should always be used with care and never as the only means for selecting and sizing a structure. Rule-of-thumb information should be used only to develop a feeling for relationships among systems, spans, and depths, and nothing more.

Copyright © 2013. Pearson Education Limited. All rights reserved.

3

rEInForcEd-concrEtE constructIon

slabs and Beams. Among the simplest of reinforced-concrete spanning systems is the conventional one-way solid slab [Figure 5(a)]. The easy formwork is an attractive feature of this system. These constant-depth systems are particularly suitable for short spans. With longer spans, the dead weight of the solid slab becomes excessive and ribbed slabs are preferable [Figure 5(b)]. One-way beam systems with transverse one-way slabs can be used to span relatively long distances (particularly if the beams are posttensioned) and carry heavy loads. Such systems are relatively deep. Beam spacing is usually determined by what is most reasonable for the transverse slab. one-Way Pan Joist system. A one-way pan joist system consists of ribbed slabs constructed by pouring concrete around special forms made of steel or fiberglass [Figure 5(c)]. Transverse beams of any depth can easily be cast in place at the ends of pans so the system can adapt to a variety of column grids. Longitudinal beams that are more substantial than the normal ribs also can be easily cast in place by varying the pan spacing. The ribbed slab is more suitable for longer spans than a solid slab. With posttensioning, very long spans can be obtained. The pan joist system is too complex and uneconomical for short spans. The vertical support system can be either columns or load-bearing masonry walls. The ribbed slab-and-column system is capable of considerable lateral-load resistance because transverse and longitudinal beams are cast into the floor system monolithically with the columns. Frame action can thus be achieved in both directions.

FIgurE 3 Pinned timber lap joints. A steel plate is inserted into slots that have been cut in both members. Connections are made with bolts. Other connections are made via steel angles and bolts. Plate bolt Steel connecting angles

(a) Plate bolts Steel angles

(b) Elevation view

Structural Systems: Constructional Approaches

FIgurE 4 Approximate span ranges for timber systems. So that typical sizes of different timber members can be compared, the diagrams of the members are scaled to represent typical span lengths for each of the respective elements. The span lengths that are possible for each element are noted by the maximum and minimum span marks.

Copyright © 2013. Pearson Education Limited. All rights reserved.

feet

Flat-Plate construction. The flat plate is a two-way, constant-depth, reinforced-concrete slab system [Figure 5(d)]. Appropriate for use with light floor and roof loads and relatively short spans, the flat plate finds wide application in housing construction. Although a regular column grid is most appropriate, some flexibility is allowed. Indeed, flat plates are often used where the rigid orthogonality other systems demand on the layout of the vertical supports is not desirable or possible. Spans are, however, limited in comparison with ribbed or beamed systems. Lower floor-to-ceiling heights are more feasible with flat-plate construction than with many other systems. Relatively large amounts of reinforced steel, however, are required as a result of the thinness of the plates used. The governing design factor for flat plates is often the punch-through shear in the plate at the columns. Special steel reinforcement is often used at these points. At plate edges, columns also

Structural Systems: Constructional Approaches are moved in from the free edge to ensure that the interface area between the slab and column remains as large as possible. Lateral stability for the entire plate-and-column assembly can be a problem. Because the plate and columns are poured monolithically, joint rigidity is achieved that contributes to the lateral resistance of the structure and is sufficient for low buildings. Because of the thinness of the plate element, however, this resistance is limited. With tall structures, stability is often achieved through shear walls or stiff poured-in-place core elements in the building, as might be possible around elevators or stairways. The simple formwork involved is an undoubted virtue of this system. The planar nature of the lower surface also facilitates the design and placement of other building components. The system is often used for apartment buildings and dormitories composed of functional spaces that have limited span and are cellular in nature. Flat-slab construction. The flat slab is a two-way, reinforced-concrete system similar to the flat plate, except that the interface area between the plate and

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgurE 5 Reinforced-concrete construction systems.

Structural Systems: Constructional Approaches

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgurE 5 (Continued)

columns is increased by adding drop panels or column capitals [Figure 5(e)]. The drop panels or column capitals reduce the likelihood of punch-through shear failure in the slab. The system is particularly appropriate for relatively heavy loading conditions (such as those found in warehouses) and is suitable for larger spans than are possible with flat plates. The capitals and drop panels also help make the slab-and-column assembly more resistant to lateral loads than the flat-plate system. two-Way Beam-and-slab construction. A two-way, beam-and-slab system consists of a flat, reinforced-concrete plate with beams, monolithically cast in place, along the periphery of the plate [Figure 5(f)]. The system is good for medium spans and high loading conditions. Large, concentrated loads also can be supported if carried directly by the beams. Columns are invariably used for the vertical support system. Because beams and columns are cast monolithically, and a substantial interface is between these elements, the system naturally forms a frame in two directions. This provides an appreciable lateral-load-carrying capacity, so the system can be used for multistory construction with ease.

Structural Systems: Constructional Approaches the Waffle slab. The waffle slab is a two-way, constant-depth reinforcedconcrete system having ribs in two directions [Figure 5(g)]. The ribs are formed by the use of special domed pans made of steel or fiberglass. The voids formed by the pans reduce the dead-load weight of the structure. Waffle slabs are more useful than flat plates in longer span situations. These slabs can also be posttensioned to increase their spans. A thickness of concrete is usually left around column tops (by not using pans in these locations). This solid area serves the same function as drop panels or capitals in a flat slab. The possibility of shear failure is reduced and the momentresisting capacity of the system is increased (as is its lateral-load-carrying capacity). The span of the waffle system and its lateral-load-carrying capacity can be increased by casting in place beams spanning between columns. This is done by eliminating the pans along these lines (or spacing them farther apart), adding appropriate reinforcing, and casting a full depth of concrete. curved shapes. Any singly or doubly curved shape (e.g., a cylinder or dome) can be made from reinforced concrete. Reinforcing typically consists of a mesh of light steel rods throughout the shell, with special additional steel used in localized areas of high internal force. Posttensioning is commonly used for special elements (e.g., tension rings in domes).

Copyright © 2013. Pearson Education Limited. All rights reserved.

Precast concrete Elements. Precast concrete elements are fabricated offsite and transported to the job. They are one-way spanning elements that are most often pretensioned. A range of cross-sectional shapes is fabricated that are suitable for a wide variety of load and span conditions. Precast concrete elements are appropriate for uniformly distributed occupancy and roof loads and not for concentrated loads or unusually heavy distributed loads. These members are most often simply supported. Moment connections are made possible by using special steel connections but are difficult. Large cantilevers also are difficult and must be kept to a minimum. Precast elements are most successful when used in a repetitive way. Several different precast members are discussed next. (Many others also are available.) Precast Planks. Both short-span and long-span planks are available. Shortspan planks are used in much the same way as timber planking is used, although appropriate spans are slightly longer. A poured-in-place, concrete-wearing surface is placed on top of the planks, which are often used with precast reinforced-concrete beams or with steel open-web joists. Long-span planks are available that span between 16 and 34 ft (5 to 11 m), depending on the exact width and depth of the element. These long-span planks are usually prestressed and cored to reduce dead weights. A poured-in-place, concrete-wearing surface is placed on top of the planks. This concrete also forms a shear key between adjacent elements so the resultant structure behaves like a one-way plate. [See Figure 5(h).] Precast planks are most appropriate for light occupancy or roof loads. They are simply supported and often used with load-bearing walls as the vertical support system. (The walls must be either masonry or concrete, not wood stud.) Precast planks also are frequently used with steel or reinforced-concrete beams. channels and double tees. These ribbed, one-way, precast, prestressed elements are suitable for longer spans than planks [Figure 5(i)]. They are appropriate for occupancy and roof loads. A poured-in-place, concrete-wearing surface is placed on top of adjacent members. single tees. These typically large, precast, prestressed elements are most suitable for relatively long spans. They are rarely used with short spans because of the difficulties erecting them. Invariably simply supported, single tees are suitable

Structural Systems: Constructional Approaches for heavy occupancy and roof loads. They are, for example, often used in parking garages and other buildings having large spans and heavier-than-usual loads [Figure 5(j)]. special Building systems. Numerous systems are available to completely form the entire shell of a building [Figure 5(l)]. Systems specially designed for housing are common. Approaches used to date usually fall into one of two categories: (1) systems that employ off-site-produced linear or planar elements, such as walls or horizontal spanning systems that are assembled on-site (usually by a posttensioning system) to form volumes, and (2) systems in which volume-forming assemblies, such as complete boxes, are constructed off-site and then shipped to the site and aggregated. Member sizes. Figure 6 illustrates typical span ranges and depths for several reinforced-concrete spanning systems. Reinforced-concrete columns have thicknessto-height (t>h) ratios that range from 1:15 for short and lightly loaded columns to around 1:6 for heavily loaded columns in multistory buildings. Reinforced-concrete load-bearing walls have comparable t>h ratios, ranging from about 1:22 to 1:10.

4

stEEL constructIon

Primary systems. Most systems constructed of heavy steel are made of linear one-way spanning elements. Rolled steel sections (e.g., wide flange) are available in various sizes. This variety allows remarkable flexibility in the design of beamand-column elements. Although simple support connections are used whenever possible, for construction convenience, it is easy to make moment-resisting joints. Frames thus formed are capable of resisting lateral loads. Lateral stability also is often achieved by using shear walls or diagonal bracing elements.

Copyright © 2013. Pearson Education Limited. All rights reserved.

Beams. Wide-flange shapes are commonly used as horizontal spanning elements. [See Figure 7(a).] The possible span range is rather wide. Members are simply supported, unless frame action is needed for stability, in which case moment-resisting connections are used. Other shapes, such as channels, are utilized to carry bending but are limited to light loads and short spans. trusses and open-Web Joists. Endless truss configurations are, of course, possible. Potential load capacities and span ranges are enormous for uniquely designed or specially made trusses. Open-web joists, which are mass-produced [Figure 7(b)], can be used both for floor and roof systems. They are economical for intermediate- to long-span situations in which relatively light, uniformly distributed loads are involved. The bar joist, for example, is frequently used as a long-span roofing element. Open-web joists are simply supported (although it is possible to devise rigid connections) and thus make no direct contribution to the lateral resistance of the assembly. Often, wideflange and open-web joists are used in the same system, with the former having moment-resisting joints so that frame action capable of resisting lateral forces can be obtained. Special, heavier-gage open-web joists are available for very long spans. These joists are appropriate only for light, uniformly distributed roof loadings. composite construction. Many structural systems cannot be easily characterized according to material. The composite beam system illustrated in Figure 7(c) is common. In this structure, the steel is first put in place, and concrete is then cast

Structural Systems: Constructional Approaches

FIgurE 6 Approximate span ranges for reinforced-concrete systems. So that typical sizes of different members can be compared, the diagrams of the members are scaled to represent typical span lengths for each of the respective elements. The span lengths that are possible for each element are noted by the maximum and minimum span marks.

Copyright © 2013. Pearson Education Limited. All rights reserved.

feet

around the shear connectors at the top of the steel beam. The shear connectors make the steel and concrete act integrally. A T-section is thus formed, with steel in the tension zone and concrete in the compression zone. If a concrete-wearing deck is needed anyway, considerable economies can be achieved with this approach.

Structural Systems: Constructional Approaches

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgurE 7 Steel construction.

Plate girders. Plate girders are special beams. Their built-up nature [Figure 7(d)] allows plate girders great latitude in the range of span and load conditions they can be designed to meet. Plate girders are useful when very high loads must be carried over medium spans. They are often used, for example, as major load-transfer elements carrying column loads over clear spans. For typically encountered distributed floor loads and span ranges, however, plate girders are rarely used because wide-flange elements are more appropriate and economical. A girder support is shown in Figure 8. Arches. Rigid arches of any shape can be specially made from steel. Prefabricated arches are available for short-to-moderate spans. Specially designed arches have been used with long spans, for example, spans on the order of 300 ft (90 m) or more. Steel arches can be made of solid sections or open-web sections. space-Frame structures. Space-frame structures are frequently made of steel. Long spans can be easily obtained. Many different configurations are possible. [See Figures 7(h) and (i).]

Structural Systems: Constructional Approaches

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgurE 7 (Continued)

shells. A variety of shell forms are possible with steel. The primary problem in using steel to achieve doubly curved surfaces is how to create the shape by using line elements. In domes, for example, either ribbed or geodesic approaches are possible. Light steel decking of small dimensions is commonly used to form enclosure surfaces. In some small-span situations, curved steel surfaces can be made by specially pressing steel sheets in a way similar to that used to create singly or doubly curved steel forms for automobile bodies. cable structures. Steel is the only material that is used extensively to create cable structures. Cable structure forms vary enormously. Cables can be used to create permanent roofs whose enclosure surfaces are made either by rigid steel planar elements such as decking or by membrane surfaces. The former are usually singly curved shapes, whereas the latter are doubly curved. Movable roofs that use a cable system also are possible. [See Figure 7(l).] These movable roofs are usually temporary enclosures or semipermanent at best.

FIgurE 8 Typical pin connection beneath a steel bridge beam.

Structural Systems: Constructional Approaches

FIgurE 9 Approximate span ranges for steel systems. So that typical sizes of different members can be compared, the diagrams of the members are scaled to represent typical span lengths for each of the respective elements. The span lengths that are possible for each element are noted by the maximum and minimum span marks.

Copyright © 2013. Pearson Education Limited. All rights reserved.

feet

Member sizes. Figure 9 illustrates the span ranges and depth-to-span ratios for several common steel spanning systems. Structural steel columns typically have thickness-to-height (t>h) ratios that range from 1:24 to 1:9, depending on column heights and loads.

Structural Systems: Constructional Approaches

5

systEM IntEgrAtIon

No matter what material a structure consists of, it must respond to several special conditions that arise in it. These conditions represent particularized events related to building functions such as circulation (e.g., stairwells) or to building service systems (e.g., heating, ventilating, and air conditioning). Several basic strategies can be used to accommodate building service elements that run horizontally. One is to use a one-way spanning system that provides a space for parallel runs. (See Figure 10.) Except with trusses, service elements that run perpendicular to the spanning direction are more difficult to integrate, but it is possible to locally deform or penetrate the basic structural fabric to allow mechanical systems to pass. This is most often done when fewer and smaller elements must be accommodated. Most structural systems can be designed to accommodate minor penetrations in horizontal members. The penetration strategy also can be used to accommodate horizontal elements in two-way structural systems. [See Figure 10(b).]

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgurE 10 Accommodating horizontal service elements.

Structural Systems: Constructional Approaches Some structural systems, such as trusses, easily accommodate horizontal service elements because they pass between web members. In other systems, whenever penetrations become numerous or large, the basic structural fabric can be compromised to the extent that the structure is more correctly characterized as a series of ad hoc solutions rather than a system. When this occurs, it is preferable to switch to a different strategy to accommodate the special conditions. Another strategy to accommodate horizontal service elements in either oneor two-way systems is to pass them beneath the primary structural system (leaving it intact) and leave them exposed or enclose them with a dropped chase or ceiling. This approach is efficient from a construction viewpoint, and it also leaves flexibility for future installations. Total building height, however, is often increased. Most structural systems can be designed to accommodate minor penetrations for vertical service and functional elements without difficulty (Figure 11). Major penetrations, such as for stairways, must often be accompanied by special local framing. The orientation of openings for vertical service elements also must consider the direction of spans (e.g., one way, two way; see Figure 12). Still, the basic structural pattern is not affected to any great degree. One way to handle numerous vertical penetrations is to group the special functions into clusters and then design the structure to accommodate those clusters. This is often done for vertical elements: The elements are grouped into core units that fit into individual structural bays. The structure is eliminated or specially treated in each bay. The advantage of doing this is that the general structural fabric

FIgurE 11 Strategies for accommodating building service systems and special features.

(b) Clustering vertical service elements and special features: Vertical service elements are often clustered with circulation elements in so-called building cores.

Copyright © 2013. Pearson Education Limited. All rights reserved.

(a) Distributing vertical service elements: Minor vertical elements are often easily integrated into the structural fabric by locally penetrating floor structures.

(c) Interstitial installation zones: It is possible to provide regularly spaced zones for the installation of vertical as well as horizontal service elements. This strategy is appropriate in cases where relatively large and complex service elements are needed throughout the building.

Structural Systems: Constructional Approaches in other areas is undisturbed and can be designed to function efficiently on its own terms. This approach is common in many high-rise structures. Quite often, the cluster approach does not work well with the functional requirements of the service elements. In many instances, these elements must be distributed throughout the building, as in the case of some types of heating, ventilating, and air-conditioning systems or laboratory piping systems. Under some conditions, a structural system that provides spaces for locating the elements at fixed intervals may prove effective (Figure 11c). The interstitial spaces between the parallel beams and double columns are reserved for horizontal and vertical mechanical runs. The precast concrete structure of the Harvard University Science Center in Figure 13 has such interstitial spaces for mechanical service elements in the back of the building. A double beam-and-column system allows the integration of services for mechanically intensive physics and chemistry laboratories. (See also Figure 22.) A conceptual difficulty in using structural systems that provide fixed intervals for the provision of mechanical services is that nothing a priori suggests that the desired spacing and distribution of mechanical systems is the same as the structural system’s. Forcing the mechanical system into a rigid pattern based on the structural system could compromise the design of both the structural and the mechanical system! Whether this is true or not, however, depends on the specific building program involved as well as many other design variables. Other factors to consider are needed upgrades for mechanical systems and the inability to foresee future system needs. Because major upgrades to the mechanical system happen at shorter

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgurE 13 Science Center, Harvard University, Cambridge, Massachusetts. Designed and built as a precast-concrete structure, the laboratories use a double beam-and-column system. Mechanical services are integrated in the regularly spaced interstitial zones between the beams.

FIgurE 12 Slab penetrations should be carefully considered with respect to the primary span of slab and steel decking systems.

(a) Slab openings can be easily accomodated if they are aligned with the direction of the primary span.

(b) Special framing strategies may be necessary to allow for openings that are perpendicular to the primary span.

Structural Systems: Constructional Approaches intervals than expected changes to the structural system, predetermined connections between both may be problematic over the building’s life cycle. A more flexible approach to service integration is to create interstitial zones or service layers between floors. Service elements can be dropped from interstitial zones into floors below or can rise to upper floors. Changes and modifications can be made with minimum disruption to functional activities. A structural solution that takes advantage of the full depth of the interstitial zone is often employed. [See Figure 11(c).] In general, highly integrated and expensive approaches of the type described are justified only in cases when the mechanical system in a building is complex and extensive, as might be the case in a hospital, where such approaches can work well. In buildings of low mechanical intensity, like churches, such integrated approaches are not economically justifiable because distributed mechanical elements are few and small. The expensive interstitial spaces would not be utilized to the extent needed to justify their existence.

Copyright © 2013. Pearson Education Limited. All rights reserved.

6

LIFE sAFEty

Fire safety requirements. A strong determinant of the structural system used in a building is the fire safety requirements imposed by building regulations. Codes classify buildings according to type of construction, which can be roughly characterized as light, medium, or heavy, according to the construction’s degree of fire resistivity. Light construction is wood frame or unprotected metal framing. Medium construction uses masonry walls as load-bearing elements. Heavy construction typically includes reinforced concrete or protected steel framing. Building regulations mainly influence the choice of structure by placing restrictions on the type of construction allowed in accordance with the degree of fire hazard. High-hazard occupancies require substantial fire-resistant construction (necessitating reinforced concrete or protected steel structures), while less hazardous occupancies require less fire-resistant construction. In a similar vein, construction types are coupled with building heights. In low buildings, almost any type of construction is allowed, unless a high-hazard occupancy is present. As heights increase, the construction must have a greater degree of inherent fire safety. Buildings using masonry load-bearing walls are acceptable up to five or six stories, although much taller masonry structures have been built. Above that height, reinforced concrete or protected steel structures are typical. Any heavier type of construction suitable for more hazardous occupancies or taller buildings is allowable in less restrictive cases but, of course, not vice versa. Thus, a single-family detached house can be built of reinforced concrete, but a high-rise office building cannot be built of light wood framing. Using a heavier system when a lighter one is allowed, however, is not typically economical. Coupled with construction types and heights are limitations placed on the maximum floor areas allowed in buildings between specially designed fire division walls. These walls separate the building into compartments and restrain the spread of a fire beginning in one compartment and moving to another. In many cases— particularly, low- to medium-rise buildings—load-bearing walls are designed to be coincident in location with required fire divisions, so the same masonry walls serve both functions. This design has implications on where load-bearing walls are placed and often on the design and selection of horizontal spanning elements. Progressive collapse Prevention. Life safety is a foremost design objective for any structure. In some cases, this includes considering extreme loads such as those originating from explosions. Blast loads are unlikely to occur for most buildings, so a certain amount of permanent damage to the structural system is considered acceptable. Compromising life and personal physical safety, however,

Structural Systems: Constructional Approaches is not. A building damaged by a blast may show larger deformations and locally destroyed structural members, but a blast should not lead to what is called progressive collapse. The term describes the phenomenon whereby localized failure of a structural element leads to failure of neighboring elements that eventually ripples through large parts of or even the complete structure, ultimately leading to its collapse. An example would be a failed column on a ground floor that leads to the rapid downward movement of the columns it was supporting, ultimately causing all connecting beams or slabs to collapse progressively. Several strategies can be used to deal with this problem. Blast loads decrease exponentially with the distance from the explosion, so keeping a potential explosion as far away from the structure as possible is the most effective protection. Design of the structural frame also is affected. Column spacing is smaller to reduce the load on individual columns. Load transfer elements that gather loads from several vertical elements into one also should be avoided if the collector elements are within the zone of a possible explosion. Structurally redundant moment frames are preferred systems because they feature alternative load paths once local elements no longer can perform structurally. But these alternative load paths must be carefully designed and considered. Upon collapse of one or more columns, for example, horizontal spanning elements such as beams or slabs show extremely large deflections. A structural design objective would be to maintain enough strength in the horizontal elements to allow time for the evacuation of the structure and avoid total collapse. One strategy is to introduce tension ties and cables. A continuous beam or slab that sags because its intermediate column support has failed will act more like a funicular hanging cable than a member in bending. To do so, the beam or slab must have enough tensile strength to carry loads primarily as a cable system. For that purpose, cables are incorporated into slabs, and tension ties are added to the connections between beams and columns.

Copyright © 2013. Pearson Education Limited. All rights reserved.

7

FoundAtIons And rEtAInIng WALLs

Foundation structures transfer forces from the building structure into the ground. Among common foundations are discrete spread footings associated with single columns, continuous spread footings associated with load-bearing walls, raft foundations (similar to other spread footings, but larger and carrying multiple columns), various straight and battered bearing or friction pile systems, and different caissons. Spread footings distribute forces directly into the soil beneath them. The area of the footing depends on the magnitude of the loads carried and the allowable bearing pressure of the soil. Simple soil stress models assume that the pressure on the soil is evenly distributed. However, theoretical and empirical studies from the field of soil mechanics indicate that stress distributions in the soil are far more complex than simple models indicate. Settlement mechanisms are similarly complex. Simple models can be used in many situations, but many other situations warrant a more intensive study of the behavior of soil beneath a footing and the various structure– soil interactions that take place. Piles are long, column-like members that are driven into the soil and tied together at the top with a pile cap. Piles can carry loads by frictional forces developed during driving or by transferring the loads directly from the building structure to an underlying bearing stratum (e.g., a hard rock bed). In the latter cases, the piles simply pass through softer material. Caissons are larger versions of this general foundation class that can carry large loads. Caissons are made by a drilling, rather than driving, process. Concrete is typically poured into the drilled holes. Normally, caissons extend downward to a bearing stratum (a stiff clay or rock), where they are

Structural Systems: Constructional Approaches

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIgurE 14 Typical foundation conditions. (Adapted from Daniel L. Schodek, Structure in Sculpture, Cambridge, MA: M.I.T. Press, 1993. Used with permission.)

Structural Systems: Constructional Approaches belled out to distribute loads more evenly onto the stratum. Many versions of both pile and caisson foundations exist. Retaining walls are used to contain earth fill. Typical soils exert a lateral pressure on the interior face of a retaining wall that tends to cause the wall to either overturn or slide laterally. The amount of lateral force generated depends on the type and weight of the soil present, its cohesion characteristics, water content, and other factors. A triangular stress distribution is often assumed in which the total lateral pressure increases with the square of the height of the wall. With that distribution, a lateral force factor of Fx = 11>22KA ph2 is obtained. In this expression, KA is an empirically derived coefficient (with values ranging typically from 0.25 to 0.35) that depends on soil characteristics, p is the unit weight of the soil, and h is the height of the wall. The overturning moment is critically dependent on the height of the wall, h, which influences both the magnitude of the lateral force developed and the moment arm associated with overturning. The overturning force can be resisted via a large, heavy wall with a wide base. More sophisticated reinforced-concrete retaining structures use an extended base with a toe and heel. (See Figure 14.) The dead weight of the soil above the heel portion helps stabilize the wall to prevent overturning, and the toe helps extend the base width. The various forces cause tension and compression to develop at different points in the structure, which in turn dictate the location or placement of reinforcing steel. The wall and the flat base also must be reinforced in the tension zones. In addition to this type of wall, many other approaches (e.g., using counterforts—triangular walls placed perpendicularly to the retaining wall) are possible.

QuEstIons 1. Obtain a set of working drawings for a structural steel building, preferably one in your area that you can visit. Draw to-scale vignettes of the type illustrated in Figure 7. A scale of 3>4 in. = 1 ft is suggested for the connection detail drawing, and a scale of 1 >4 in. = 1 ft is suggested for the overview.

2. Repeat Question 1, except use a building made of poured-in-place reinforced concrete. 3. Repeat Question 1, except use a building made of precast reinforced-concrete elements. 4. Repeat Question 1, except use a building employing laminated timber elements.

Copyright © 2013. Pearson Education Limited. All rights reserved.

5. Visit four different buildings that each have a different structural system, and measure the dimensions of critical structural elements (e.g., beams, columns, and trusses). Convert this information into rule-of-thumb information (e.g., ratios).

Copyright © 2013. Pearson Education Limited. All rights reserved.

Structural Connections 1

IntroductIon

How structural members join or meet is often a critical design issue and one that, under certain circumstances, can influence the choice of the basic structural system itself, particularly its patterns and materials. The possible strategies in joining structural elements strongly depend on the physical properties and geometries of the elements to be joined. The next section highlights only basic design considerations. Building codes that govern structural engineering also define the rules of detailing connections. Design approaches for steel and timber connections follow the principles of load and resistance factor design or allowable strength design. This chapter provides basic example calculations, assuming an allowable strength design approach.

Copyright © 2013. Pearson Education Limited. All rights reserved.

2

BasIc JoInt GeometrIes

In the joining of simple linear rigid members, the most commonly used joints employ a strategy of either lapping the basic elements, deforming and interlocking them, or butting them. Monolithic joints, particularly in reinforced concrete, also are possible. Most joints using one of these basic strategies also use a third-element connector, of which bolts, nails, and welds are typical examples. Third-element connectors also may involve using additional pieces (e.g., cover plates). The basic function of such connecting elements or assemblies is to help transfer the total load present at the joint from one element to another. A bolt connecting two lapped pieces, for example, transfers forces from one member to another through shear forces developed in the bolt itself. As is discussed next, however, not all joints must employ thirdelement connectors to serve this function. When used, the connecting elements or devices are smaller in terms of absolute dimensions than are the primary elements and also have higher relative strength (e.g., bolts connecting two lapped-timber elements). When they have the same relative strength, connectors are often necessarily larger so that loads can be transferred safely. Figure 1(a) illustrates several butt joints. Third-element connectors are most often used with such joints, which can be made either pinned or rigid, as From Chapter 16 of Structures, Seventh Edition. Daniel L. Schodek, Martin Bechthold. Copyright © 2014 by Pearson Education, Inc. All rights reserved.

Structural Connections

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 1 Basic types of joints.

discussed previously. In joints like this, either the vertical or the horizontal element can be made continuous through the joint, but rarely both. This caveat has significant structural implications because, if continuity is desired both vertically and horizontally, special rigid joints must be used with at least some of the elements. In steel construction, for example, columns are typically continuous and beams are framed into column sides. If frame action is desired, rigid joints are then made.

Structural Connections Butt joints can be used advantageously for unusual horizontal framing patterns. Figure 1(a) illustrates how several horizontal members can frame into the same point in a column by using a third-element connector (a circular shelf). In such cases, the column is continuous and the beams are discontinuous, unless special rigid connections are made. Figure 1(b) illustrates several lapped joints. This approach is frequently used in timber construction to achieve continuity. The strategy lends itself well to oneway framing systems. Complex patterns of connected elements are difficult to handle with this type of joint. Figure 1(c) illustrates several joints in which members are deformed to make the connection. The strategy shown at the right in the figure is often used with easily moldable materials, such as reinforced concrete. Many precast-concrete building systems employ such joints. Timber elements also can be deformed to make connections. The ancient mortise-and-tenon joint is such a connection. By and large, these joints employ either simple or pinned connections between elements. It is possible, however, to make rigid connections. Joints made by deforming and interlocking basic elements sometimes do not need obvious third-element connectors, nor do those that are made by taking advantage of the moldability properties of certain materials (such as reinforced concrete). The latter is considered a subclass of a deformation strategy. Sometimes, using a deformed and interlocking strategy means that the member must be made larger at the joint to accommodate the deformations. In some cases, this increase implies that the remainder of the member is also made relatively larger so that a constant-size member can be used. Occasionally, third-element connectors are used internally in deformed joints (e.g., in poured-in-place concrete, special reinforcing steel is often used at joints).

Copyright © 2013. Pearson Education Limited. All rights reserved.

3

BasIc types oF connectors

General considerations. The third-element connectors commonly utilized in making joints can be characterized as either point connectors (e.g., bolts, nails, rivets), line connectors (e.g., welds), or surface connectors (e.g., glued surfaces). (See Figure 2.) The type of connector used depends on the specific physical nature and geometry of the elements to be joined. Joining large rigid surfaces, for example, when connecting a foam core to facing plates in a sandwich structure, calls for distributed connectors. Joining two sheets of plywood via a lapping approach with a single bolt, for example, is not feasible because large localized stresses would develop in the plywood around the bolt as it engages in transferring loads from one sheet to the other. Bearing failures would occur. A preferable approach would be to use a series of distributed smaller bolts such that the amount of load each bolt transfers is relatively small. Alternatively, a lapped and glued joint, or a combination of glued and mechanical connections, also is possible, depending on the loads. Using a number of smaller distributed-point connectors (or line or surface connectors) is a useful approach to consider for any situation. While it is true that single-point connectors can be used effectively in many situations (particularly those involving primary members with small dimensions), a distributed approach allows for reduced localized stress concentrations in the members that are joined, which is generally advantageous or necessary. rigid Versus pinned Joints. Single-point connectors are, of course, most appropriate for making pinned joints, while line and surface connectors lend themselves to rigid joints. Point connectors, however, can be used to create rigid joints if at least two are separated in space (e.g., at the top and the bottom of a member) so that a resisting internal moment can be developed.

FIGure 2 Basic lap joint connectors.

(a) Point connectors (e.g., bolts or rivets)

(b) Line connectors (e.g., welds)

(c) Surface connectors (e.g., glue)

Structural Connections Forces developing in the separated point connections act over the affected moment arm and provide the resisting moment. The greater the separation in space, the smaller are the forces developed in the connectors for a given moment resistance (Figure 3).

Copyright © 2013. Pearson Education Limited. All rights reserved.

FIGure 3 Selecting types of connectors.