Graphical Analysis: A Text Book on Graphic Statics

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GRAPHICAL ANALYSIS

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HkQraw-Tlill Book PUBLISHERS OF BOOKS

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World ~ Engineering News-Record Power v Engineering and Mining Journal-Press Chemical and Metallurgical Engineering v Coal Age Electric Railway Journal American Machinist Ingenieria Internacional Electrical Merchandising v BusTransportation Journal of Electricity and Western Industry Industrial Engineer

Electrical

GRAPHICAL ANALYSIS A TEXT BOOK ON

GRAPHIC STATICS

BY

WILLIAM

S.

WOLFE,

M.S. &

Head

Grills. Formerly of Structural Department, Smith, Hinchman Instructor in Architectural Engineering, University of Illinois.

Associate Member American Society of Ciril Engineers. Associate Member Society of Naval Architects

and Marine Engineers

First Edition

Fourth Impression

McGRAW-HILL BOOK COMPANY, Inc. NEW YORK: 370 SEVENTH AVENUE LONDON: 6 & 8 BOUVERIE

1921

ST., E. C. 4

re no W64-

•2.

Copyright

1921,

by the

McGkaw-Hill Book Company,

Inc.

PRINTED IN THE UNITED STATES OF AMERICA

PREFACE This book has been developed from notes and blue prints prepared by the writer and used in his classes at the University of Illinois. Certain additional material has been added, a part of

which has been

tributed

by the

to above were modeled after a set several years before, to

Many

number of articles conThe notes referred prepared by Dr. N. C. Ricker,

briefly presented in a

writer to the technical press.

whom the writer is greatly indebted.

additional problems might have been added

discussion greatly extended.

This

is

and the

especially true of Chapters

X

However, it has been thought best inclusive. to to keep the book to smaller proportions, and the writer believes that a thorough mastery of the constructions and solutions

IV and VII

here presented will give the student an excellent grasp of the subject.

The object has been to deal with the analysis of stresses rather than with design or the computation of loads. Nevertheless it has seemed desirable to give some attention to the determination of loads, and Chapter IX has been devoted almost exclusively to Material and ideas have, of course, been drawn from design.

many

sources.

writer is especially indebted to Prof. C. R. Clark for assistance in connection with the early part of the work and to Prof.

The

L. H. Pro vine for

encouragement and constructive

criticism.

w. Detroit, Mich. October, 1921.

s.

w.

..

1

.

TABLE OF CONTENTS PAGE v

Preface Notation

xiii

CHAPTER

I

GENERAL METHODS ART. 1

Introduction

1

2.

Definitions

2

3.

Graphical Construction and Measurement of an Angle

3

4.

Representation of Forces

5.

Composition of Forces

4 6

6.

Resolution of Forces

7.

Concurrent-Coplanar Forces, Conditions for Equilibrium

8.

Unknowns

11

9.

13

12.

Non-Concurrent Coplanar Forces Funicular or Equilibrium Polygon Resultant a Couple Non-Concurrent Coplanar Forces, Conditions

13.

Parallel Forces

14.

Resultant of a

10. 1 1

15.

Reactions for

16.

Reactions for

17.

Reactions for

8

14 18 for Equilibrium

Reactions for a Truss, Vertical Loads Reactions for a Truss with Vertical and Inclined Loads 20. Horizontal Component Divided Equally between the Two Reactions 21 Reactions of a Rafter Supported on Purlins 19.

.

Members of a Frame a Bicycle Frame

22. Stresses in the

24. Passing

19 19

Number of Parallel Forces a Beam a Beam with a Distributed Load a Beam with Inclined Loads

18.

23. Stresses in

10

a Funicular Polygon through Three Given Points

CHAPTER

.

21 23 25 26 27 27 3

32 33 35 37

II

CENTROIDS Broken Line an Arc 27. Centroid of a Curve 28. Centroids of Areas 25. Centroid of a

40 42

26. Centroid of

44 45 vii

TABLE OF CONTENTS

viii

PAGE

ART. 29. Centroid of

a Triangle 30. Centroid of a Quadrilateral 31. Centroid of a Trapezoid 32. Given the Centroid, Area and Distance between the Two Parallel Sides of a Trapezoid to Find the Lengths of the Parallel Sides

45 46 47

34. Centroid of a Circular

Segment

48 49 50

35. Centroids of Irregular

Areas

51

33. Centroid of a Sector

Volumes

36. Centroids of

53

37. Centroid of a Triangular 38.

Pyramid

53

Complex Volumes which can be Divided Volumes

into a

Number

of

Regular

39. Irregular Volumes, Division into Slices 40. Centroid 41. Stress

Located by the Use of Sections and Area Curves

Volumes

54 57 59 61

CHAPTER

III

MOMENTS Moment of a Force about a Point Moment of a Number of Forces about a Point First Moment of an Area about a Given Axis Moment Diagram for a Beam

42. First 43. First 44. 45.

Second Moments Second Moment of a Number of Parallel Forces 48. Moment of Inertia of an Area Mohr's and Culman's Methods 19 50. Radius of Gyration Radius of Gyration of Rectangles, Parallelograms and Triangles 52. Moment of Inertia, Exact Method 53. Higher Moments 46. 1

7

.";1

,

54. Central Circle or Ellipse of [nertia 55.

Complicated Problems

CHAPTER

64 65 67 68 69 70 72 73 74 75 76 77 79 80

IV

BEAMS 56 57

Construction of the Elastic Simple Beams

(

lurve

61

antilever Beams Beams with an Overhanging End Beams with a Variable 1 Beams uiili tae End ix< d

62

B


s

'->

'

'.»1

94

96 99 K)2 107

TABLE OF CONTENTS

ix

CHAPTER V TRUSSES ART. 65. 66.

67. 68.

p AGE 113 114

Weight of Trusses Other Weights. Snow Load Wind Loads

115 115

Notation 70. Stresses Obtained Analytically 69.

116 116

by Analytical Moments Stresses by Graphical Moments Stresses Obtained Graphically by Joints Stresses Obtained by Stress Diagram Stress Diagram, Upper and Lower Chord Loads Wind Loads; Reactions and Stress Diagrams Stress Diagram Combined Loads Maximum and Minimum Stresses, Reversals

71. Stresses

117

72.

117 118

73.

74.

81.

120 122 124 126 128 134 137 141

86. Bridge Trusses of the

145 147 151 154 156

75. 76.

77. 78.

79. Cantilever Trusses

Bent Three-Hinge Arch 82. A Large Three-Hinge Arch 83. A Large Mill Bent 84. Cantilever Truss with Four Supports 85. Combination Truss, Three-Hinge Arch and Mill Bent 80. Mill

K Type

Dome

87.

Trussed

88.

Ring Dome, Dead Loads Ring Dome, Wind Loads

89.

158 159 163

CHAPTER

VI

MOVING LOADS Concentrated Moving Load Concentrated Moving Load and a Uniform Dead Load 92. Uniform Moving Load Longer than the Span 93. Moving Uniform Load Shorter, than the Span 94. Uniform Dead Load, and Uniform Moving Load Shorter than the 90. Single

168

91. Single

171

Span 95. 96. 97. 98.

99. 100.

Two

Concentrated Moving Loads Three Concentrated Moving Loads Four Concentrated Moving Loads and a LTniform Dead Load A Large Number of Concentrated Moving Loads Maximum Shears and Moments in a Turntable Moving Loads on Trusses

175 177 179 179 184

187 190 193 196

TABLE OF CONTENTS

x

CHAPTER

VII

MASONRY PAGE

ART. 101. Stresses in Rectangular Piers

201 204

Volumes

102. Stress

103. Problems, Rectangular Piers 104. Problems, Irregular Piers 105.

Kerns on the Edge of the Kern Analytically on Wall Footings Retaining Walls

106. Location of Points

107. Pressure 108.

109. Line of Pressure in a Pier 110.

Masonry Chimneys

an Arch Three-Hinged Arches 113. Three-Hinged Arch Symmetrically Loaded 114. Two-Hinged Arches 111. Line of Pressure in

231

112.

115.

Two-Hinged Arch, Method of Least Work Method of Least Work

116. Hingeless Arches,

117. Hingeless Arches, General Discussion I

1

8

Solution of an Arch Using Theory of Least

Crown

Pressure

119. Investigation of a Gothic Vault

120.

A

Study

Domes

of

CHAPTER RE1 \

'< I

>R(

'ED

I

CONCRETE

Beams Beams

12A

I

For Rectangular

254 259

r-Beamfi of

237 238 238 242 242 245 245 250

VIII

121. Simple Rectangular

122

205 206 209 214 216 218 222 224 228

2(10

T-Beams

264

264

126

Double Reinforced Joncrete Beams Bending Stresses in lomplex Sections

i_'7

Combined

J7(>

128

Eccentrically

125

(

'

Stresses

Loaded Columns

kmcrete Ihimneys 130, Deflection uf Reinforced Concrete 129

Reinforced

J7

275

(

(

(II

-t>7

Beams

278

U'TKlt IX

DESIGN 131

D

ign of n of

'..ri mi 1.;

I

i.. 136

Beams Plate


'.

mtinued

until

all

of the

loads

are

laid

off.

A

pole

/>

may

Chap.

GENERAL METHODS

I]

25

now be chosen, and the funicular polygon drawn in Fig. 38. Then parallel to the closing string p-X, a line is drawn in Fig. 39 from p, and the intersection X located. The string d-p is a component

and

of the force c-d

also of Ro, therefore in Fig. 39 the

vector for Ro will have one end at is

up, the direction from

X-A, and its sense

is

to

X

D and the other at X. being up.

also up, the direction

Reactions for a

16.

D

Beam

from

X

beam shown

carries a distributed load, as indicated

Its sense

vector for Ri to

A

with a Distributed Load.

required to find the reactions of the

The

The

is

being up.

— Let

in Fig.

it be 40 which

by the shaded area above. any section repre-

vertical dimension of the shaded area at

some scale, the intensity of the loading at that section in pounds per lineal foot or in some such unit. Also the resultant

sents, to

Fig. 41.

load on any length of

beam 4-3

is

given by the

mean

vertical

dimension of the shaded area above, measured to scale, times the length 4-3; or in other words the area 1-2 — 3-4 measured to the proper scale.

The slices

load.

first

step

is

to divide the distributed load into a

and, for each It will usually

slice,

number

of

substitute an equivalent concentrated

be found convenient to make

all

the slices of

equal width, and obtain the magnitude of the equivalent concentrated loads by scaling the mean ordinates and multiplying by the width.

The

action line of the concentrated

load should pass each case, but, if the width of the slices is small, the error in approximating the location of the centroid will be very small.

through the centroid of the

The concentrated slices of

slice in

loads that were substituted for the various

the distributed load are

now

laid off in Fig. 41.

A

pole

GRAPHICAL ANALYSIS

26

p

[Art. 17

chosen and the funicular polygon in Fig. 40 drawn. From p a drawn parallel to p-X, locating X, which determines Ri

is

line is

and Rz-

Beam

Reactions for a

17.

with Inclined Loads.

thus far considered have carried vertical loads only. carries inclined loads, as

shown

in Fig. 42,

is

In this particular case, R2

ficult

problem.

tical,

thus simplifying things a

The

little.

— The beams

A beam which

a somewhat more is

vectors for the loads

are laid off in Fig. 43 in the usual way, and the pole

Fig.

Since R2

is

vertical its action line

actiOD line of thai

it

is

A'i

not

All that

through the point

passes

in Fig.

The

is

is

12.

and constructed drawn and parallel to

in

known about

/.,

is is

the usual way.

a line

it

of the

funicular polygon

The

0.

therefore started at closing string

p chosen.

known, bul the direction

is

known,

dif-

assumed to be ver-

is

The

drawn from

/>

I:;.

string

/'

of the vector for

I

IS /•':

a \B

COmponenl

of

therefore at

lor Locating the centroid.

Other methods may be used for Locating the centroid of a trapel>iii they arc not of enough importance to make it desirable

zoid,

to give

I

hem

here.

Given the Centroid, Area, and Distance between the Two Parallel Sides of a Trapezoid to Find the Lengths of the Parallel Sides. A problem of this nature is sometime- encountered in connect ion with footings. In r\£. 83 column 2 has a larger load In than column I, the resultanl of the two loads falling at g. 32.

\

the footing

ie

allowed to extend only

b

Bhorl distance

beyond

Chap.

CENTROIDS

II]

49

each column, a trapezoid footing with its centroid at g may be Our problem is to find the length of A-D and of B-C. desirable. divided by the allowable soil pressure gives the loading The total area of the footing, and this area divided by e-f gives the mean width, or one-half A-D plus one-half B-C. From g draw the line Also draw m-n and m'-n' parallel to A-D g-h parallel to A-D.

and passing through the third points of the line e-f. This means that m-n passes through the centroid of triangle A-C-D, and that Along m-n m'-n' passes through the centroid of triangle A-C-B. lay off o-s equal to the value already obtained for one-half plus one-half C-B, draw the line o-p and connect s and p. line s-p gives the intersection

r.

Through

r

draw

A-D The

t-v parallel to

measured to the proper scale gives one-half C-B, and v-p measured to the same scale one-half A-D. This must be true because the areas of triangles A-B-C and A-C-D are proIf s-t and v-p were not portional to B-C and A-D respectively. t-v and s-p would not triangles, these of areas the proportional to the similarity between note should student The g-h. intersect on

o-p.

Then

s-t

In Fig. 29 the lower part of Fig. 83 and Fig. 29 in Chapter I. resultant the them from and given are forces two of Chapter I the necessary is and it known is resultant the 83 In Fig. located. to break

it

into

two components.

One

is,

in

a sense, the reverse of

the other. If the trapezoid had been given and the location in of its centroid required, a construction similar to that shown m'-n' and m-n drawing After used. been have Fig. 83 might

make s-t equal to one-half B-C and p-v equal one-half A-D, Then draw v-t and s-p. Their intersection locates r which in turn The intersection of this line and e-f locates locates the line h-g .

trapezoid because r divides the distance g, the centroid of the into parts inversely proportional to the m'-n' and m-n between areas of the two triangles.

Sector.— Consider the sector shown in Fig. divided into small triangles somewhat as were sector the If 84. Fig. 84, the centroid of each would of left the shown at Therefore, as far as r from 0. two-thirds distance at a be found 33. Centroid of a

the location of the centroid is concerned, the area of the sector may be considered as concentrated along the arc m-s-n, which is a The centroid of this arc is located by the line of uniform density.

method

illustrated in Fig. 73.

the sector.

The length

s-t is

The line Y-Y is an axis bisecting made equal to the length of the arc

GRAPHICAL ANALYSIS

50 s-n,

are connected and from n a line

and

t

is

[Art. 34

drawn

parallel to

From the intersection of this line with t-0 a line is drawn parallel to X-X. The intersection g locates the centroid of the arc

O-Y.

m-s-n and

also of the sector.

Fig. 83.

I'i...

M

34. Centroid of a Circular find the centroid of the )'

)

to

Y

of

tin

Y,

is

Segment.

segmenl shown

drawn, and through

l

B

<

is

located

al

ii

be required to

The

Fig. 85.

the axis

Then, using the construction

sector

Lei

in

l

o\.

X

A"

:ii

bisecting

righl

angles

Fig. 84, the centroid

This Bector

may

be

Chap.

CENTROIDS

II]

51

divided into two parts, the given segment and the triangle O-A-C. The centroid of the sector has been located, and g~2 the centroid

on w-O, one-third of the distance from w to 0. assumed to act at gi with a magnitude proporNow, if tional to the area of the sector, and another one of opposite sense at g-2 proportional to the area of triangle A-C-O, and the resultant of the triangle lies

a force

is

found, the action line of this resultant will pass through the centroid of the segment. The resultant of these two forces, which are

Fig. 88.

Fig. 86.

Fig. 87.

but of opposite sense, is located by the construction shown Chapter I. The length of go-m is made proportional to the area of the sector, and gi-n is drawn parallel to it on the parallel

in Fig. 30,

same side of Y-Y and proportional to the area of the triangle A-C-O. Now m and n are connected and the line extended until the axis of symmetry Y-Y is intersected at g. The nt g is the centroid of the segment. 35. Centroids

of Irregular Areas.

ular area the centroid of which

most convenient method

is

is

—Fig.

desired.

86 shows an irregIn such a case the

to divide the area into small parallel

GRAPHICAL ANALYSIS

52

[Art. 35

full vertical lines. The centroid of each of now approximated, and from their approximate centroids lines are drawn parallel to the slices. The area of each one of these slices is determined approximately by scaling the mean length and multiplying it by the width. In Fig. 87 the

slices,

shown by the

these slices

is

vectors are laid off proportional to these areas, and from Fig. 87

the lower funicular polygon of Fig. 86

is

drawn.

Attention should be called to the fact that, in order to obtain satisfactory results, the slices should be reasonably small,

and yet

not so small that their number becomes excessive and thus the

The

chance of accumulation of small errors dangerous.

^-o_9-9-9-c>-

q.

\

\

\

of

/?x/s

vertical

\

i

i

i

i

'

'

'

i

"1-

symrnetrtf)

A



-4-

*

*-

/

s

^

/

/

-+7

Fig. 90.

Fig. 89.

line, or

more

st

rictly

Bpeaking, the line parallel to the

by the intersection of the

first

and

last

slices,

located

strings of the funicular

polygon, contains the centroid of the area, hut as ye1 the location along its length is unknown. The area is now

of the centroid

divided into another set of slices, in this case horizontal slices as shown by the dotted lines. Now the centroid of each slice is

approximated, the areas found, and the force polygon

drawn, using vectors proportional

force polygon, the funicular polygon at the right

tin is

drawn and

The

I

he horizontal line

m

of Fig.

to the a reus of the slices.

n located

of

88

Prom Fig.

by the intersection

86 o.

intersection g Locates the centroid of the area.

Figs.

89

the centroid

and of

B

90 illustrate the construction for locating sonicwhnt different shaped area. Since this

Chap.

CENTROIDS

II]

53

area has an axis of symmetry, only one force polygon and one funicular polygon are required to locate the centroid 36. Centroids of

Volumes.

g.

— The centroid of some volumes can

be located easily by means of planes of symmetry. There are others that have their centroids on some axis at a certain proportional distance from one end. For still others the centroid can be most conveniently located by cutting the volume into small slices and approximating the centroid and content of each slice.

Then the

may be located by the use of force There are some volumes for which it is

centroid of the whole

and funicular polygons.

convenient to find the areas for various sections, plot these areas, draw an area curve, and then locate the centroid of the area under

Then

the curve.

again

it

may

be possible to

split

a given volume

into a few smaller volumes, the centroid of each of

which can be and the centroid of the entire volume located by the force and funicular polygons or by some other convenient

easily located,

use of

method. It

is

evident that the centroid of a sphere

of three or

more planes

common

of

is

the

symmetry, the centroid

common

of a

cube

point

is

also

point of three or more planes of

point of a line connecting the centroids of

symmetry or the midtwo opposite sides. The

centroid of a rectangular parallelopiped

is

the

the

common

point of

three planes of symmetry, or the mid-point of a line connecting the

centroids of allel

bases

sides. The centroid of a prism with parthe mid-point of a line connecting the centroids of

two opposite

is

the two bases. 37. Centroid of a Triangular

pyramid shown triangular base

in Fig. 91.

Pyramid.

— Consider the triangular

First locate the centroid d of the

A-B-C by drawing

lines

from the vertices A and B If the pyramid were

to the mid-points of the opposite sides.

divided into very thin slices parallel to the triangular base, the connecting d and V would contain the centroid of each of

line

them.

Therefore the centroid of the pyramid must be on the line locate the centroid e of the side A-C-V by drawing lines from the vertices A and V to the mid-points of the opposite

V-d.

Now

Connect e and B. The centroid of the pyramid must lie B-e for the same reason that it is on the line d-V, since any side of a triangular pyramid may be considered the base. Therefore the intersection of B-e and d-V must be the centroid of the pyramid. Now it can be proved that this centroid g is at the sides.

on

this line

GRAPHICAL ANALYSIS

54

d-V

fourth point of

The

triangle

92.

In this

or of e-B, that

is

[Art. 38

d-g equals one-fourth d-V.

Y-f-B with the lines e-B and d-V is shown in Fig. Since e-f equals figure c-h is drawn parallel to f-B.

one-third f—V, k-d equals one-third d-V. Now triangles e-k-g and g-B-d are similar and e-k equals two-thirds/—d alsof-d equals

Then

one-half B-d.

e-k equals one-third B-d.

It follows that

k-g equals one-third d-g and k-d equals one and one-third d-g equals

four-thirds

d-g

equals

d-V.

one-third

Therefore

d-g

equals one-fourth d-V. It is also true that the centroid of any pyramid is on the line connecting the centroid of the base with the vertex, and one-fourth

up from the

the distance

For

base.

if

the pyramid

is

divided into

thin slices parallel to the base the line connecting the centroid of

the base with the vertex will contain the centroid of each of them,

and therefore the centroid of the pyramid. Now if the base is divided into triangles, and the pyramid into triangular pyramids, lie centroid of each one of these pyramids will be up from the base t

one-fourth the distance to the vertex. the large pyramid must be up one-fouri h to the vertex wit

li

t

and on the

line

Therefore the centroid of t

he distance from the baso

connecting the centroid of the base

he vertex.

Complex Volumes which can be divided into a Number The elevation of a rather simple complex volume lb shown in Fig. 93, and s plan of the same volume in to divide this volume into two parts is convenient It Pig. 94. 38.

of

v

Regular Volumes.

and

a

.

in

plan

Q

II

T&ndK M X

0.

Now

the portion

hownns.l B E F in elevation and Q T in plan, has the B E F constant. A plane parplane of symand half the way back will be to .1 B E II

dimension norma] to the Bide A allel

metry and

/•'

will

therefore contain

;i

the centroid.

This plane of

Chap.

CENTROIDS

II]

symmetry

is

shown

55

plan by the line r-s.

in

The

side

A-B-E-F

by the construction is already explained for the trapezoid. The volume (a) may be cut A into thin slices all parallel to this side and directly back of it. line projected directly back from g\ would contain the centroid of each of these slices, and therefore the centroid of volume 0. The point g\ in Fig. 93 is the elevation of the centroid of volume (a). By projecting down from g\ until r-s in Fig. 94 is interIn a simsected, gi', the centroid of volume (a) in plan, is located. ilar way the centroid of volume (b) is located in elevation as g-2 and in plan as go'. Now the centroid of the entire volume must lie and ®, on the line connecting the centroids of the two volumes which line is shown in elevation as gi~g2 and in plan as gi'-g2 r In Fig. 95 vectors are drawn proportional to the two volumes (a) and @, a pole is chosen, and the funicular polygon in the a trapezoid and

centroid g\

its

located

is

©

-

lower part of Fig. 94

is

drawn.

which contains the centroid

The

this plane, u-v, intersects g\'-g2

and

therefore the centroid of the entire

as g and in plan as Figs. 96

which

is

volume.

g\-gi. at g'

volume

is

The

trace of

and g respectively, shown in elevation

g'

and 97 show an

desired.

intersection v locates a plane

of the large

The

first

interesting volume, the centroid of

step

is

to divide the given

volume into

smaller volumes, the centroid of each of which can be easily

determined or approximated.

B-D is

A

vertical plane passing through

pyramid at the left. The base of this pyramid a rectangle whose edge is shown in plan as B-D, and in elevation will cut off a

as f-k.

Now if a horizontal plane is passed through /,

a triangular

pyramid is cut off, the base of which is shown in plan by B-C-D and in elevation by f-^i. There remains a triangular prism shown The centroid of in plan by B-C-D and in elevation by f-i-J-k. each one of these small volumes can be easily located. Also it is evident that a vertical plane passed through

symmetry

A-C

is

a plane of

for the given volume, as well as for each

one of the

The base

pyramid at the left is a rectangle, shown in plan as o, and in elevation as s. From s a line is drawn to the vertex E, and g\ located at the quarter point. Projecting up, g\ is obtained on the line A-C. The point the plan of the centroid of this gri shows the elevation, and g\ pyramid. Now consider the triangular prism, shown in plan by the triangle B-C-D, which has its centroid at gs'. The prism

divisions.

the centroid of which

of the is

GRAPHICAL ANALYSIS

56

I

'I...

91

I

[Art. 38

i

-

.

99.

Chap.

may

CENTROWS

II]

be considered as

to the base

and

of the

57

made up of a lot of thin slices, all parallel same shape, the centroid of each being on a

through the centroid of the base. Therefore the must lie on this vertical line and half-way up. Now project down from gz and locate gz mid-way between f-i and k-J; gz is the elevation, and gz is the plan of the centroid of the vertical

line

centroid of the prism

prism.

The

triangular pyramid, above the prism just considered, has

its base shown in elevation by t, and in plan by Connect t with the vertex H and locate go at the quarter point. The point go is the elevation of the centroid of the pyramid, and go', obtained by projecting up from go, shows it in plan. Vertical lines are now drawn from gi, go, and gz, and in Fig. 98

the centroid of

gz'.

A

vectors are laid off proportional to the different volumes.

pole

chosen and the lower funicular polygon in Fig. 97 drawn. The intersection v locates a vertical plane which contains the

is

centroid of the given volume.

The

trace of this plane

u-v

is

which intersects the trace of the plane of symmetry at g'. Therefore g' shows the centroid of the given volume in plan and in

some point along the line w—x. HoriIn Fig. 99 vecgi, go, and #3tors are laid off proportional to the various volume divisions, and after choosing a pole, the funicular polygon shown at the left of Fig. 97 is drawn. The intersection z locates a horizontal plane which contains the centroid of the entire volume. The trace of this plane is z-y, and it intersects w-x at g. The centroid of the entire volume is therefore shown in elevation as g and in plan as g' elevation the centroid

zontal lines are

39. Irregular

is

at

now drawn from

Volumes.

Division

into

Slices.

— Let

it

be

required to find the centroid of the volume shown in Figs. 100 and 101, is

which

is

therefore

lie

above

this line.

The

may

line

be cut into a number of thin parallel

A -B

centroid will

The volume cannot be divided

small regular volumes the centroid of each of which it

The

a portion of a hollow circular cylinder.

the trace of a vertical plane of symmetry.

is

into

known, but

slices at right

angles to

A-B. If the slices are thin the centroid of each may be assumed to be mid-way between its sides. Through these approximate centroids vertical lines are drawn as shown in Fig. 101. The volume of each slice may be obtained with very small error by multiplying the thickness by the mean width times the mean height. The mean widths are obtained from Fig. 100, and the the plane

GRAPHICAL ANALYSIS

58

[Art. 39

mean heights from Fig. 101, each being approximated. The mean width of the fifth slice is e-d plus c-f and the mean height It is convenient to make the thickness of all the slices the is r-s. Fig. 100.

I

Fig.

Fia

L02.

io:j.

thus simplifying the computations for the volumes.

Baxne,

slice

....

loi', .

tli''

the volumes vectors are laid off proportional pole i> is chosen, and Hie funicular polygon, Pig. i