Statics and Strength of Materials [2 ed.] 0028030672, 9780028030678

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RESULTANT OF COPLANAR FORCE SYSTEMS

2-1 INTRODUCTION Two systems of forces are said to be equivalent if they produce the same mechanical effect on a rigid body. A single force that is equivalent to a given force system is called the resultant of the force system. We shall first introduce the parallelogram law and use it to find the resultant of concurrent coplanar forces. Then the rectangular components of forces are discussed and used to find the resultants. As we Shall see later, any system of nonconcurrent coplanar forces can be replaced by a single resultant that is equivalent to the given force system. The location of the line of action of the resultant is not immediately known. To determine the line of action of the resultant of a nonconcurrent coplanar force system, we will introduce the concepts of the moment of a force first. The resultant of some simple types of distributed line loads will be discussed in this chapter also.

2-2 VECTOR REPRESENTATION Notations.

In this book, vector quantities will be distinguished from

scalar quantities through the use of boldface type, such as P. An italic type, such as P, will be used to denoted the magnitude of a vector. In long-hand writing, vectors may be represented by the notation P. Graphical Representation. A force F (or any vector quantity) is represented graphically by a line segment AB with an arrowhead at one end, as shown in Fig. 2-1. A is the point of application and x is a reference coordinate axis. The length of the line segment AB represents the magnitude of the force measured according to some convenient scale. The direction is indicated by the angle 6 from the reference axis.

39

40

CH. 2/RESULTANT OF COPLANAR FORCE SYSTEMS

A

a

FIGURE 2-1 Equal Vectors. Two vectors having the same magnitude and the same direction are said to be equal. Two equal vectors may or may not have the same line of action (Fig. 2-2). Equal vectors may be denoted by the same letter.

We Ae FIGURE 2-2 Negative Vector. The negative vector of a given vector P is defined as a vector having the same magnitude as P and a direction opposite to that of P (Fig. 2-3). The negative vector of P is denoted by —P. According to Newton's law of action and reaction presented in Section 1-6, the forces of action and reaction must always be equal in magnitude and opposite in direction. Thus, the forces of action and reaction may be represented by P and —P.

S/o

se

FIGURE 2-3

2-3 RESULTANT OF CONCURRENT FORCES Parallelogram Law. As mentioned in Section 1-3, vectors are added according to the parallelogram law. Figure 2-4 shows two vectors that are added according to this law. The two vectors P and Q are placed at the same point A and a parallelogram is constructed using P and Q as its two adjacent sides. The diagonal of the parallelogram from A to the opposite corner represents the sum of P and Q and is denoted by P + Q. Note that, in general, the magnitude of the vector sum P + Q is not equal to the algebraic sum of the magnitudes P and Q.

SEC. 2-3/RESULTANT OF CONCURRENT FORCES

41

FIGURE 2-4

Triangle Rule. The sum of two vectors can also be determined by constructing one-half of the parallelogram, or a triangle. This method is called the triangle rule. To find the vector sum P + Q, we first lay out P at A (Fig. 2-5a), then lay out Q from the tip of P in a tip-to-tail fashion. The closing side of the triangle, drawn from A to the tip of Q, represents the sum of the two vectors. Figure 2-5b shows that the same result is obtained if the vector Q is laid out first. Hence, the vector sum is not affected by the order in which the vectors are added; that is, vector addition is commutative:

P+Q=Q+P

(2-1)

FIGURE 2-5

Polygon Rule.

The sum of three or more concurrent coplanar vectors

may be carried out by adding two vectors successively. For example, the sum of three concurrent coplanar vectors, P, Q, and S (Fig. 2-6a) can be obtained by first finding P + Q, then adding S to P + Qto findP +Q+S, as shown in Fig. 2—-6b. Notice that the dotted line in the figure could be omitted, and the sum of the vectors can be obtained directly by laying out the given vectors in a tip-to-tail fashion to form the sides of a polygon. The closing side of the polygon, from the starting point to the final point, represents the sum of the vectors. This is known as the polygon rule for the addition of vectors. A polygon formed by forces is called a force polygon. Since the vector sum is commutative, the order in which the vectors are added is arbitrary. In Fig. 2-6c the vectors are added in the order of P, S, and Q. We see that, although the shape changes, the resultant obtained remains the same.

42

CH. 2/RESULTANT OF COPLANAR FORCE SYSTEMS

FIGURE 2-6 Resultant. A given system of concurrent coplanar forces acting on a rigid body may be replaced by a single force, called the resultant, equal to the vector sum of the given forces. The resultant will produce the same effect on the rigid body as the given force system.

——— EXAMPLE 2-1 Determine the resultant of two forces P

and Q acting on the hook in Fig. E2-1(1). Solution. Two methods are presented here.

P=650N

(a) The Graphical Method. A force triangle ABC is drawn as shown in Fig. E2-1(2). AB represents force P and BC represents force Q. The magnitude of each force is laid out by using a properly chosen linear scale. The direction of each force is measured by using a protractor. The closing side of

FIGURE E2-1(1)

the triangle, AC, is the resultant. The

magnitude and direction of the resultant are measured to be

P = 650N Force triangle (to scale)

FIGURE E2-1(2)

2

Scale

SEC. 2-3/RESULTANT OF CONCURRENT FORCES

43

R = 1090 N R=1090N

0 = 23° —


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CHAPTER §

STRESSES IN BEAMS

14-1 INTRODUCTION In this chapter, the stresses caused by bending moments and shear forces are considered. Normal stresses along the longitudinal direction are caused by bending moments, and shear stresses are caused by shear forces. The purposes of this chapter are to study the distribution of normal and shear stresses in a beam, and to relate these stresses to the internal

bending moment and shear force in the beam. Normal stresses in beams due to bending (also called flexural stresses) are discussed first, followed by a study of shear stress variation in beams. Inelastic bending of beams will also be studied in this chapter so that ultimate strength design of beams can be presented in the next chapter.

14-2 NORMAL STRESSES IN BEAMS DUE TO BENDING For a straight beam having a constant cross-sectional area with a vertical axis of symmetry, as shown in Fig. 14-1a, a line through the centroid of all cross-sections is referred to as the axis of the beam. Consider the two cross-sections, ab and cd, in the beam. Before the application of loads, the cross-sections are in the vertical direction and the beam axis is a straight horizontal line. Assume that the beam segment between the two sections is subjected to a positive bending moment +M. The beam bends and the cross-sections ab and cd tilt slightly. Note that lines ab and cd remain straight and perpendicular to the axis of the beam, as shown in Fig. 14-1b. Experimental results indicate that a plane section before bending remains a plane after bending.

487

488

CH. 14/STRESSES IN BEAMS

Vertical axis of symmetry Cross section

Beam axis

d — Elongation Initial length

(b)

FIGURE 14-1 Imagine that the beam is composed the longitudinal direction. The fibers along ac become shorter. Hence, the sion and those along ac are subjected

of an infinite number of fibers along along bd become longer and those fibers along bd are subjected to tento compression.

Neutral Surface and Neutral Axis. The fibers mn along the beam axis do not undergo any change of length due to bending; hence, these fibers are not stressed. The surface mn is called the neutral surface. The intersection of the neutral surface with a cross-section is called a neutral axis. For a beam subjected to pure bending (no axial force), the neutral axis is a horizontal line that passes through the centroid of the cross-sectional area.

Variation of Linear Strains. In Fig. 14—-1b, consider a typical fiber pq parallel to the neutral surface and located at a distance y from it. From point

n, draw a line ns parallel to mp; then

ps = mn = initial undeformed length of the fiber and the fiber pq elongates by an amount sq. From triangle nsq, we see that the elongation (or contraction) of a fiber varies linearly as the distance y to the neutral surface. Since the initial length of all fibers between the sections

is the same, it follows that the linear strains of the longitud inal fibers due to

bending vary linearly from zero at the neutral surface to the maximum value

at the outer fibers.

Variation of Flexural Stresses. For elastic bending of the beam, Hooke’s law applies; that is, stress is proportional to strain. For most materials, the

SEC. 14-3/THE FLEXURE FORMULA

489

moduli of elasticity in tension and in compression are equal. Under these conditions, we also conclude that the flexural stresses in a beam section vary linearly from zero at the neutral axis to the maximum value at the outer fibers. Figure 14-2 shows the normal stress distriCompression bution in a beam section. The stresses vary from zero at the neutral axis, to

a maximum

tensile stress at the bottom outer fiber and a maximum compressive stress at the top outer

fiber. Note that the resultant T of the tensile stresses below the neutral axis and the resul-

tant C of the compressive stress above the

Tension

FIGURE 14-2

neutral axis must be equal and opposite (in the absence of axial forces) and form a couple equal to the internal resisting moment M.

14-3 THE FLEXURE FORMULA Derivation of the Flexure Formula. Consider a beam segment subjected to a positive bending moment +W, as shown in Fig. 14-3a. At section m-m, the applied moment is resisted by normal stresses that vary linearly from the neutral axis. The maximum normal stresses occur at points on the bottom of the section that are most remote from the neutral axis. Denote the maximum normal stress by a, and the distance from the neutral axis to the bottom of the section by c. Then the normal stress o at the narrow strip of area AA located at distance y from the neutral axis is, by proportion,

ae

(14-1)

Centroid Neutral Axis

+M

(N.A.)

(a)

(b)

FIGURE 14-3 The force on the incremental area AA is oAA. The incremental moment of this force about the neutral axis is

AM = (aAA)y

490

CH. 14/STRESSES IN BEAMS

The total internal resisting moment developed by the normal stresses is the sum of the incremental moment over the entire section. This internal moment must be equal to the external moment M to satisfy the equilibrium condition. We write

M =

AM=%

(a AA)y

Substituting o = (/c)o,,,, from Equation 14-1, we have M= ac Gina aA)y Since g,,,,/C is a constant factor in each term, it can be factored out of the summation. Thus, Tnax

M=—~—

xy? AA

According to the definition given in Chapter 8, moment of inertia of the cross-sectional area axis. Its value is a constant and depends on the sectional area. Denote it by /, and the equation ye

the expression {y?AA is the with respect to the neutral size and shape of the crosscan be written as

O max Ae

From which we get

Me

a

Omax —

where

(14-2)

,,, = the maximum normal stress due to bending (also called the flexural stress) M=the internal consideration

resisting

moment

at

the

section

under

c = the distance from the neutral axis to the outermost fiber 7 =the moment neutral axis

of inertia of the cross-sectional area about the

Substituting in Equation 14-1, we have M

C=

where

o =

=:

,

(14-3)

the flexural stress at any point in the section

y = the distance from the neutral axis to the point where the flexural stress is desired Equations 14-2 and 14-3 are two forms of the flexure formula . The moment of inertia of simple geometric shapes can be computed by using the formulas listed in Table 8-1 in Chapter 8. (The table is also printed inside the front cover for easy reference.) The moments of inertia of composite areas or built-up steel sections can be determined by the method discussed in Sections 8-4 and 8-5.

SEC. 14-3/THE FLEXURE FORMULA

491

The sketches in Fig. 14-4a and b are helpful for determining whether a fiber is in tension or in compression due to a given bending moment.

Compression

) 4M

Tension —

Tension Neutral surface (or beam axis)

(a)

(b)

FIGURE 14-4 Section Modulus. Note that the moment of inertia / and the distance c are both geometric properties of the cross-section and their values are constant for a given section. Their ratio //c is also a geometric property and is constant for a given section. The quantity //c is called the section modulus. Denoting the section modulus by S, we write

me =