The Elements of Graphic Statics by Jeroen van den Bovenkamp

THE ELEMENTS OF GRAPHIC STATICS

THE BROADWAY SERIES OF ENGINEERING HANDBOOKS

VOLUME XXVI

THE ELEMENTS GRAPHIC STATICS OF

A TEXT-BOOK FOR STUDENTS,

ENGINEERS

AND ARCHITECTS

ERNEST

SPRAGUE,

ASSISTANT AT UNIVERSITY COLLEGE, LONDON

A.M.lNST.C.E. ;

LECTURER ON HYDRAULICS

AND STRENGTH OP MATERIALS; LECTURER AT THE WESTMINSTER TECHNICAL INSTITUTE; FORMERLY PROFESSOR OF ENGINEERING AT THE IMPERIAL CHINESE RAILWAY COLLEGE, SHAN-HAI-KUAN

WITH WORKED EXAMPLES AND ONE HUNDRED AND SIXTY-THREE ILLUSTRATIONS

LONDON

SCOTT, GREENWOOD. (E.

GREENWOOD)

E.C.

SON

PREFACE THE graphical treatment of problems in Engineering and allied subjects is often much less laborious and much more lucid than the corresponding mathematical analysis. Large mistakes are less likely to be overlooked, whilst the accuracy obtained is all that is required in most cases. Also, the more difficult the problem the more necessary becomes the use of graphical methods, the procedure in the more difficult cases being usually similar to that in the more simple ones, and therefore readily understood though perhaps complicated in appearance when whilst in the mathematical treatment the use of the higher branches of Mathematics is frequently necessary even in comparatively simple cases, and the labour involved and the consequent liability to error is so great as to finished

;

render a solution in many cases impracticable. Experience in the use of graphical methods and comparison of the results with those obtained by calculation will soon convince the sceptical that the accuracy obtained with reasonable care in draughtsmanship is not only sufficient, but is greater than might be anticipated. The Author believes that the graphical anof Engineering problems, which is of comparative recent development will grow in favour as the advantages of the method are alysis

370661

PREFACE

better appreciated, and hopes that this elementary exposition of the fundamental principles as applied to Statical problems may be of assistFurther applications ance to the beginner. will be found in Vol. xvii., "The Stability of " " Vol. xx., The Stability Masonry Structures of Arches"; and in "Moving Loads by Influence Lines and other Methods," to which the present book may be regarded as an introduction. ;

ERNEST SPRAGUE, UNIVKKSITY COLL HI IB, LONDON, March, 1917.

A.M.lNST.C.E.

CONTENTS

...

PREFACE

CHAPTER

I'AGES

v-vi

COMPOSITION AND RESOLUTION OP FORCES

1-15

Scalar and Vector Definitions Vector Addition Quantities Application to the Composition and Resolution of Forces Equilibrium of Concurrent Forces

Advantages

Graphic Statics

Equilibrium of Three Forces Triangle and Polygon of Forces Parallelogram of Forces Practical Illustrations Examples.

CHAPTER

COMPOSITION AND RESOLUTION OF FORCES MATHEMATICALLY CONSIDERED 16-25

Composition and Resolution of Forces Rectangular Components-^Resultant of Two Forces Resultant of a System of Concurrent Coplanar Forces Equations of Equilibrium Three Forces in Equilibrium Illustrations and Examples.

CHAPTER

III

FRAMED STRUCTURES

26-14

Assumptions Bow's Notation Sense of the Internal Forces Application to Roof Trusses, Framed Girders, etc. Structures in Equilibrium under the Action of Three External Forces Roof under Wind-pressure Resolution of a Force into Three Non-concurrent Components Illustrations and Examples. b

vii

CONTENTS

Vlll

I'AGKS

CHAPTER IV SUPERPOSED AND SPACE FRAMES

man

Truss

Lattice Girder

45-58

Perfect, Imperfect, and Redundant Frames ditions for a Perfect Frame Finck Truss

ConBoll-

Braced Tower

Sheer Legs Space Frames Belfry Tower Resultant of Concurrent Tripod Examples Forces in Space of Three Dimensions Resolution in Space of Three Dimensions.

CHAPTER V MOMENTS

59-77

Definition

Resultant

brium

Examples

Moment Method

Equations of Equiliof Section-

Appli-

cation to French Roof Truss

Graphical TreatComposition of Couples Resultant of Non-Concurrent Forces Link-polygon Graphical Conditions of Equilibrium Parallel Forces Graphical Treatment of Moments Application to a System of Parallel

ment

Moments

Couples

Forces.

CHAPTER

ROOFS

78-94

Dead and Live Loads

Weight of Roof Coverings, Rafters, etc. Slopes for Roof Coverings Weight of Snow Wind-Pressure Duchemin's Formula for Construction Duchemin's Graphical Increase of Formula Wind-pressure with Height Wind-pressure on Various Surfaces Distribution of Load on Roof French Roof Truss Station Roof Crescent Roof.

Tru^s with Fixed Supports

CHAPTER

VII

SHEARING FORCE AND BENDING MOMENT DIAGRAMS

95-111

Shearing Force and Bending Moment Diagrams for Point-Loading and for Uniformly Distributed Loading Shearing Force at a Point Special Cases of Shearing Force and Bending Moment Graphical Construction of Parabola (i) by Points, Combined Diagrams Indirect (ii) by Tangents Loading Determination of Forces in a Framed

CONTENTS

IX PAGES

Girder from the Diagrams of Shearing Force and Parallel and Non-parallel Bending Moment

Booms.

CHAPTER

VIII

RELATION BETWEEN THE CURVES OP LOAD, SHEAR, AND BENDING MOMENT 112-119 ... .

Load-Curve Sum-Curve or Curve of Integration Shear-Curve as the Sum-Curve of the Load-Curve Bending Moment-Curve as the Sum-Curve of the Shear-Curve Scales Application to a Concrete Raft.

CHAPTER IX SHEARING FORCE AND BENDING MOMENT FOU

MOVING

LOAD

120-185

Direct Loading

Single Concentrated

Two

Load

Concentrated Loads Train of Concentrated Loads Maximum Values of Shearing Force and Bending Moment Schlotke's Method Indirect Loading Maximum Values of Shearing Force and Bending Moment for a Train of Concentrated Loads Distributed Loading.

CHAPTER X MOMENTS OF AREAS, ETC

136-162

Graphical* Determination of the First and Second Moments of a System of Parallel Forces Centres of Gravity Two Point Loads System of Point Loads Triangle Trapezium Radius of Gyration Equimomental Systems Triangle Rectangle Relation between Second Moments about Parallel Axes Polar Second Moment Circular Areas Graphical Treatment of Moments of Inertia of Areas Momental Ellipse Symmetric and Asymmetric Sections Unequal-legged Angle Formulae for Centroids of Lines, Areas, and Solids Formulas for Moments of Inertia of Com-

mon

Cross-Sections.

CHAPTER XI STRESS DISTRIBUTION ON CROSS-SECTIONS

Load Point

Axis of Flotation

163-183

Neutral Line and Resultant Pressure on Sub-

or Centre of Pressure

CONTENTS PAGES

merged Area Relation between Neutral Axis and Load Point Position of the Load Point for a ExRectangular Section Submerged Area amples Stress Intensity at any Point -Examples Core of a Section Rule of Critical Distance the Middle-Third Asymmetric Sections When the Material is Incapable of Resisting Tension

Mohr's Method

Numerical Example.

CHAPTER

XII

THE LINE OF PRESSURE

184-191

Thrust, Shear, and Bending Moment on a Structure Crane Structure Three-Hinged Roof or Arch

Diagrams of Shear and Bending Moment Stability of a Masonry Structure Conditions of Stability Flying Buttress.

INDEX

192-19G

CHAPTER

COMPOSITION AND RESOLUTION OF FORCES. 1.

Statics is that

branch of Mechanics which deals

with bodies in equilibrium, and since Structures are usually in equilibrium as a whole, or in the case of a moving structure the parts which compose it are in equilibrium

amongst themselves, a knowledge

the principles of Statics

essential to the science of

structural engineering. Also, as will soon appear, the action of a system of forces is usually much

more conveniently investigated by graphical than by purely mathematical processes, and therefore the application of drawing to statical problems under the of Graphic Statics has come to be regarded as

name

an important branch

The engineering study. more than sufficiently

results so obtained are usually

accurate to satisfy error

all

practical requirements,

careful

the

usually draughtsmanship within about 1 per cent of the truth, and therefore within the limits of accuracy warranted by the data

being

;

whilst the

economy

effected in time

and the ease with

which mistakes exhibit themselves, recommend

its

use not only to the less mathematical designer, but also to those who are skilful in analytical methods whilst the one process may often be used* as a valu;

able check

upon the

other.

i5LEMENtS O^ GRAPHIC STATICS

The student cannot be too strongly advised to make himself familiar with the fundamental principles

based. ledge,

upon which the graphical constructions are Many mistakes arise from lack of this knowand it is impossible for anyone to think a way

out of a difficulty unless the principles involved are understood.

The mechanical

solution of a problem without an

intelligent appreciation of the

only burdens the

memory

modus operandi, not

with numberless variations

of construction, but leads to the adoption of

wrong

The most

methods.

profitable course of study will always be found to consist in understanding the reason of every process used. The fundamental principles

are few- in number, and their application to new problems not only develops an interest in the work,

but also a constant improvement in a knowledge of the theory of the subject. Festina lente (make haste slowly) is therefore a motto, which, though always to be recommended, deserves particular attention in the present subject, where the applications of a few important constructions are so varied and extensive.

The quantities 2, Scalar and Vector Quantities, with which we have to deal in everyday calculations is often assumed, amenable to the same mathematical laws, and it will be a great simplifica-

are not, as

tion

we may be divided

at the outset

that they quantities

Scalars

;

realize the important fact into (2)

two groups, Vector

(1)

Scalar

quantities

Vectors.

Scalar quantities are such as may be represented completely by numbers, such as intervals of time, lengths, speeds, temperatures

sums

money, and

COMPOSITION AND RESOLUTION OF FOKCES so on.

have

All these quantities

size

which can be

numerically expressed as in ordinary arithmetical but since size may vary by becoming operations ;

larger or smaller, it is desirable to indicate the sense of growth by the algebraic device of affixing a positive or negative sign, according as the size is increas-

Scalar quantities may, of course,

ing or decreasing.

be represented graphically by lines drawn to scale, and their sense may be indicated by affixing an arrow-

head

In this case the addition or sub-

to the line.

traction

of such quantities

be performed by

will

joining thes'j lines end to end in their proper sense and in the same straight line. Thus if a person have

^c *

AT-

*T| K--2--M?

3---*** FIG.

assets of

these amounts

5 say, and debts of 4 and be represented by lines drawn

and

may

any convenient scale, the assets being distinguished from the debts by arrowheads pointing in opposite to

senses.

and

plot

we choose

AB =

a scale of 1 inch

2",

BC =

3",

and

represents the total assets and EF = li" in the opposite ;

sents the total debts, and

excess of assets over debts.

1 say,

AD = 4" DE plot then DF sense, repreCD = if

(fig. 1),

represents to scale the All this

quite simple, usually no advantage gained by performing the operations of addition and subtraction of scalar magnitudes in this way, and the

but at the same time there

graphical

method

special cases, such as

therefore of little use except in its application to the Slide-Eule,

THE ELEMENTS OF GRAPHIC STATICS

where lengths representing the logarithms bers are added and subtracted mechanically, to

num-

in order

perform the operations of multiplication, division, Vector (see "Mathematics for Engineers").

etc.

on the other hand, not only have magnitude and sense, but also direction, and it is because the direction of a line may be so readily expressed on the quantities,

drawing-board, that the composition or addition of vectors may be so much more conveniently effected by graphical methods for whilst a line may repre;

sent

magnitude by

its

length,

and sense by an

arrowhead, the slope of the line will represent its direction, whereas in expressing this property as an angle in degrees we introduce the processes of trigonometry and the comparative difficulties attending it.

Any

vector magnitude therefore, such as a dis-

placement, a velocity, an acceleration, force, etc., in fact any quantity which has magnitude, sense, and direction,

may be

completely represented by a line or

in general no definite a a definite vector having position is called position a rotor or a localized vector.

Such vectors have

vector.

;

Vector magnitudes 3, Law of Vector Addition, do not obey the ordinary laws of arithmetic and If by the sum of two quantities we are to algebra. understand the result of their combination, then it is clear that in general this combination must obey a law which involves the idea of direction. Thus, for example, the sum of two forces will only be their arithmetic or algebraic sum if the forces have the same direction but when they are inclined to one ;

another they can no longer be added or subtracted in the ordinary way.

Suppose a body to move from

Then the

line joining

(fig.

2).

represents the displace.ment of the body, and this no matter what the path

along which it moved in passing from one point to the other. Let the body be further displaced from

to C, and again from C to D. Then if the lines AB, BC, CD represent to some scale the magnitude, sense, and direction of the several displacements,

these lines or vectors

when

joined end to entl in the

same

sense, that is to say, with the arrowheads all pointing the same way, will represent the combination of the displacements,

and the

line

which

joins the beginning of the first to the end of the last will represent the total or resultant displacement and since this law is not confined to displacements, ;

but evidently applies equally to all vector quantities, it is known as the law of vector addition. It is obvious that this law

quite different to the law

which

THE ELEMENTS OF GKAPHIC STATICS

governs the addition of scalar magnitudes, except in the special case when the vectors have one and the

same

direction.

Order of Vector Addition Immaterial. Since a body which moves through given distances in given directions and in given senses will arrive at the same spot in whatever order these displacements occur, it is evident that the vectors which represent them may be added or compounded in any order.

Thus

vectors in resultant, s

represent the magnitudes of the s represent their vector sum or

and

fig. 2,

we may

write

o.-H-&-ff-cors

where the symbol

-ff

c 4f 6 or again s = b -ff a -jf c used to denote the operation

-[[-

of vector addition.

Law

4. Application of the

the

Composition

of Vector Addition Concurrent Forces.

being affected by direction, are vector magnitudes, and must therefore obey the law of vector addition enunciated above. Consequently

Forces,

the vector

sum

or resultant of

any given system

acting along given directions which through a common point, may be found by pounding the vectors which represent them.

forces

For

Pj,

?<>,

fig. 3,

pass

com-

be three given forces

acting on a body at the points A, B, and C, in such directions that their lines of action when produced

pass through a

any one

common

them

point O, then the effect of

will be the

same,

we suppose

This is often spoken of as applied at the point O. the "Principle of the Transmissibility of Force," according to which a force may be supposed to act '

any point

in its line of action.

Hence P lt P 2 and ,

COMPOSITION AND RESOLUTION OF FOBCES

P 3 may

O instead of at A, B, in without respectively, any way affecting their influence upon the body as a whole. Consebe supposed to act at

and C

quently, as there is no tendency to produce rotation about 0, the force which is equivalent to the given

must also pass through O, only necessary therefore to find its magnitude, sense, and direction in order to determine it completely. This is accomplished at once by adding forces, or their resultant,

and

it is

the vectors 01, 12,

23 which

represent the forces, which closes the

as .shown; then the vector 03,

vector figure, represents the required resultant and a line drawn through O parallel to it will represent the line along which it acts. ;

5, Equilibrium of a System of Concurrent Forces in a Plane (Coplanar Forces). Let P p P^, P 3 (fig. 3) be any given coplanar forces whose directions

common

pass through a

point O.

These we have

shown above may be reduced to an equivalent force E = 3 acting as shown in the figure. Now the only single force that can keep another in equilibrium is one which is equal and opposite to it in the same

Hence a

straight line. to E, will

keep

force E, equal

in equilibrium,

and

and opposite

will, of course,

3 reversed, that is by be represented by the vector Such a force is known as the equilibrating 3 0. force, or Equilibrant,

at rest,

will also

and

since it keeps the resultant keep at rest the forces P lf P 2 ,

P 3 of which E is the equivalent. The forces P P 2 P 3 E will therefore be a system in equilibrium. Hence we see, that when a system of concurx,

rent forces

the

is in

equilibrium, their vectors added in and con-

same sense form a closed polygon

;

THE ELEMENTS OF GRAPHIC STATICS

versely,

system

when the of forces

vectors form a closed polygon the

they represent, if concurrent, will

be in equilibrium.

FIG.

6, Equilibrium of Three Non-parallel Forces. Let Pj, P._, (fig. 4) be any two non -parallel forces whose

FIG.

Then the line 2 which closes vectors are 01, 12. the figure in the same, sense represents their equili-

COMPOSITION AND RESOLUTION OF FORCES

it is equal and opposite to the vector 2 which represents their resultant. Hence, when three forces are in equilibrium, (1) these forces are

brant, because

represented by the sides of a triangle drawn parallel them and taken in order, and (2) the forces must

be concurrent, because the resultant of P l and P 2 must pass through the point 0, and the only force which can keep it in equilibrium is one equal and opposite to

Hence

one

and acting

FIG.

can be found. (fig.

5) to pass

same straight line. known, the other two

in the

of the forces is

For, suppose three forces Pj, P.2 P3 through O and to be in equilibrium. ,

any one of them, say P x is known, and a 1 be drawn to represent it to scale, lines 1 2, 2 parallel to the other two forces will represent them also, and not only in magnitude, but in sense

Then

vector

also,

by the arrowheads.

These indicate whether

the forces are acting towards or away from the point O, and therefore whether the forces are tensile or

compressive. It is of

no consequence on which side

vector the triangle

constructed, for

of the given

we draw

1 3,

THE ELEMENTS OF GRAPHIC STATICS

3 parallel to

P 2 P3 ,

respectively to

meet on the

opposite side of 01, it will be seen that we get vectors having the same magnitude and sense as in the

first case,

The vector

monly spoken will readily

Polygon

and

this is

sometimes convenient.

figure obtained in this way is comof as the " Triangle of Forces," and it

be seen to be only a special case of the " Forces which we obtain as the vector

figure when more than three forces are concerned. Also the Triangle of Forces is identical in principle with the "Parallelogram of Forces," which states .

that

two

forces are represented

FIG.

by the sides

triangle, their resultant is represented

For

(fig.

of a

by a diagonal. P p and

6) represent a given force

OP., a given force

P 2 and we

complete the parallelo-

gram, the diagonal OR will represent the resultant R for it will be seen that if we set off OP 2 to repre;

P 2 and add to it a vector PR 2 = OPj, the line which joins O to R is the resultant, by the triangle of forces ; and in fact it is not necessary to refer at all to the parallelogram of forces which is only another way of obtaining the same result as sent

that derived by the triangle of forces. 7. Application of the Triangle of Forces,

Let

7) be a given load supported by cords passing and B, and let Pj, P 2 be the tenover pulleys at (fig.

COMPOSITION AND KESOLUTION OF FORCES sions in the cords.

down

Set

1 to represent the

load

012

for the forces acting at C.

to scale,

and construct the triangle

of forces

Then the

vectors

1 2, 2

respectively represent the parallel to CB, tensions in these cords, a fact which may be tested

experimentally with flexible cords and freely moving pulleys.

the point

If the tensions

will rise to

angle of forces will

now

P l and P 2 are increased, some point C', and the tribe

1 2'

;

so that as

FIG.

more nearly do the cords and the the longer become the horizontal, approach vectors which represent the tensions, so that it is 1 may produce other evident that a force such as increase the tensions, the

forces

much

greater than

itself.

This fact

indeed

utilized in mechanical engineering in the form of the " toggle-joint," used for hand-brakes, stonecrushers, etc., a small force being used to develop

very large forces nearly in

line.

when

the arms of the toggle are

THE ELEMENTS OF GRAPHIC STATICS

As, however, the development of large forces is, in structural work, to be avoided as far as possible,

members of a structure should as far as conmake fairly large angles with one another. Thus in the roof truss (fig. 8) the force in the bar the

venient

will increase as the angle

CAD

gets smaller, as

may be seen by the vector triangle 012, and will decrease as the angle gets larger, the smallest stress

FIG.

feasible, occurring when the bar perpendicular to AC, as shown by AD', in which case 103 would be the triangle of forces for

possible,

were

the forces at A, arid the force in AD' would then be that represented by the vector 1 3. When AD,

them is represented by drawn parallel to AB.

are horizontal, the tension in

the length 1

be seen from the foregoing that when not more than two unknown forces act at a point, It will

these forces can always be found, for however many may be, so long as these are known, a

others there

COMPOSITION AND EESOLUTION OF FOBCES

vector figure can be drawn for them and their resultant determined. This resultant being in equilibrium

with the two unknown forces, the latter can be found by constructing the vector triangle. Thus if P lf P.,, P3 (fig. 9) be three known forces acting at O, and X, Y be two unknown forces, the vector figure 0123 being drawn for the known forces, 3 is their resultant K, and if we construct on it a triangle

043

having

its

sides parallel to X, Y, then the forces

FIG.

04, 43, which are

in equilibrium with K, are also in forces of which E is the rewith the equilibrium sultant. Bufr it will be seen that there is no necessity to draw in the line result is obtained by

3 at

all,

because the same

drawing lines from the extremities 0, 3 of the vector figure which represents the known forces, parallel to X and Y, to close the polygon.

Examples. Find the resultant of two equal forces inclined to one another at angles of 0, 30", 1.

60",

Ib.

90,

THE ELEMENTS OF GRAPHIC STATICS

120, 150, and 180. Plot the results on a base of degrees and construct a curve to show the resultant for

any other angle. Ans. 20; 19'3

;

17'3

14-1; 10; 5-18;

;

Ih.

2. Find the resultant of four forces of 8, 12, 15, and 20 lb., making angles of 30, 70, 120, and 155 with a fixed line and check the result by compound;

ing the forces in a different order. Ans. 39-5 lb. at 111 46' to the line. 3.

weight of 6 lb. is suspended by two strings, and 12 ft. from a horizontal beam at

of lengths 9

points 15

ft.

Find the tensions

apart.

in the strings.

Ans. 48 and 36 4.

bar 10

ft.

long, hinged at

its

lb.

lower end

against a wall, and attached to the wall by a chain 6 ft. long at a point 8 ft. above the hinge, carries Find the forces in the bar and a load of 2000 lb.

Ans. 2500 and 1500

chain. 5.

at B,

Two

rafters

AB,

BC of

equal length are hinged

and loaded there with 50

lb. vertically.

the compressive force in each bar

when

ABC

(iv).150

is (i)

30;

(ii)

90;

(iii)

Ans. 96-59 6.

A body weighing

11).

120; ;

lb.

70'71

;

rests

Find

the angle

;

25'88

lb.

on a smooth

plane inclined at an angle of 30. It is acted on by Find a pull inclined upward at 20 from the plane. the magnitude of the pull necessary to keep the body at rest. 7.

ends

Ans. 13-4, 25, 50, 186-5

lb.

A string ABCD hangs in a vertical plane, the A and D being fixed. A weight of 1Q lb, is

COMPOSITION AND EESOLUTION OF FORCES

hung from B and an unknown weight from the point The middle portion BC is horizontal and the portions AB and CD are inclined at 30 and 45 to the horizontal respectively. Determine the unknown C.

weight and the tensions in the three portions of the string.

(I.C.E.

Ans. Weight

Exam.)

17-32

Ib.

;

AB, 20

17-32

;

in BG, CD, 24-49 Ib. ;

FIG. 10. 8.

weight

shown in and BC.

fig.

of 10.

tons hangs from a chain as Find the forces in the bars AC Ans. 15 and 8'66 tons.

The above examples may with advantage be worked analytically also as explained in the following chapter.

CHAPTER

II.

COMPOSITION AND RESOLUTION OF FORCES MATHEMATICALLY CONSIDERED. Although many classes of problems are very conveniently solved by graphical methods, it is equally true that others are more adapted to mathematical treatment, and it is essential to a clear grasp 8.

understand both methods, the one being often not only useful as a check upon the of the subject to

to it. The following traverse the ground of the previous chapter from the mathematical standpoint.

but

other,

supplementary

chapter will therefore

slight acquaintance with trigonometry is necessary, this is desirable in any case. 1

and

Composition of Two Forces at Right Angles, P p P 2 (fig. 11) be two such forces, whose lines

Let

See " Mathematics for Engineers," (16)

vol. xxi. iu this Scries.

COMPOSITION AND RESOLUTION OF FORCES

Then

of action intersect in O.

represent

and add

we draw Oa

triangle

right-angled

Ob we have shown P.2

the line

represents their resultant K, as in the previous chapter. Consequently

Oab

since

the

we have

R = PS + 2

k= vpTTiy.

or Also,

6 be

its

inclination to the force

from which the magnitude

;

tan

of the angle 6

can be found.

Example Then and 10,

Let Pj

Ib.

and

P2 =

Ib.

E = ^/fdOTWd = 18-03 Ib. = = 1-5, whence = 56 18'. tan Resolution of a Given Force into Two -}-J

Rect-

angular Components, Let P (fig. 12) be any given force, which it is desired to resolve into components

FIG. 12.

along two given axes

OX,

perpendicular to one

represent the given force P to scale, its projections Oa, ab parallel to the axes OX, will be the required components, by reversal of another.

Then

THE ELEMENTS OF GRAPHIC STATICS

If we call these Px, PY we the above reasoning. have P x = P cos 0, and P v == P sin respectively.

Example 2. Let P = 100 Ib. and = 30. P x = 100 cos 30 = 86'6 Ib. Then

PY =

and

Two

to (fig.

dicular axes

501b.

13) intersect in

Forces Inclined at any Let fche given forces P l Take two perpenO.

OX, OY, and

let

One Another.

Angle and P 2

100 sin 30

Composition of

11.

(for

Then

simplicity)

FIG. 18.

coincide with the direction of

re-

and place the force P 2 by its components P., cos P., sin 0, we have a total force of P, + P., cos

axis, and a force P.. sin acting along the along and since these are at right angles to the Y axis ;

one another they

may

ceding paragraph, and

R-

be compounded as in the pre\va get

(?!

PJ-

cos 0)2P,Po cos

2?^ cos

-f

P22 +

2 t

P.,

sin 2

since .-.

(P.,

+ +

sin 0) 2 2

P,,

cos 2

(cos

+ P. 2 sin 2 + sin 2 0)

^PjPo cos + cos 2 0=1.

K = yP/2 +

P, 2

+ 2P P 2 T

cos

COMPOSITION AND RESOLUTION OF FORCES and

inclination a to the

its

axis

given by the

formula

_?2 sin B

tan a

\ + P 2 cos & = Let P! 10 lb., P 2 = 201b., and = 35.

Example 3. Then E = ^/lOO + 400 + 400

cos 35

v/827'68

= 28-77 lb.

and

tan a

.-.

12,

sin

11-472

56'.

To Find the Resultant of a Given System

of Concurrent Forces,

Let P,, P,,

etc. (fig. 14)

the given forces which pass through O and make Let each angles O lt 6.,, etc., with the axes OX, OY. of the forces be resolved into

components along

THE ELEMENTS OF GRAPHIC STATICS

and OY. Then taking into account the sense in which the forces act, and denoting by 2x an(l 2y the algebraic sum of their components along the X and

axes,

2x = PI 2v = PI 1

get cos ~M #,

sin O l +.

whence we

get

P.> Z

cos

sin

- P., cos 3 - P 3 sin

0., 2. #._>

V^x + Sv

will be given

0., .1

+ P4 - P4

cos 0,1\ sin

2 M

and the direction

20 0,

lb.,

P4 =

= 48,

P =

Let

Example.

lb.,

= 50,

and 4

lb.,

by tan

P2 =

lb.,

the angles be B l

let

= -

P3 = = 32,

= 35.

FIG. 15.

Then 10 cos 32 - 15 cos 48 - 20 2X '

cos 50

10 cos 35

48 - 20 sin 50 - 10 sin 35 J = 10-0368-480 12-856+ 8-192 = - 6'221 .-. ^ x 15-32 - 5-736= -4-61J 11-146=5-299 and S Y + vv

= 10

sin 32

.-.

+ 15

and

76-222 tan

.-.

sin

4-61 2

4-61

-- -

32'

7-743 '7411.

(fig.

15).

lb.

COMPOSITION AND RESOLUTION OF FORCES

13, Equations of Equilibrium for a System of Concurrent Coplanar Forces, When a system of forces is in equilibrium their vector figure must close, or in other words the resultant must be

zero.

Therefore

+ 2y2 =

and

can

this

and 5y = 0, whether only occur when both 2x = we consider the matter from the graphical or the mathematical standpoint.

Whenever a system

of non-parallel

forces

FIG. 16. it is always possible to form two independent equations by resolving the forces into components along a pair of axes and equating

equilibrium therefore,

the

sums

in each case to zero.

Consequently

all

the forces except two are known in magnitude these two can be found, provided their directions are given.

Example

a point O, and

Let the forces shown in let

P P3 x

unknown

fig.

16 act at

in magnitude.

THE ELEMENTS OF GRAPHIC STATICS

Then

and

for equilibrium

2x = 2Y = -

+ P! cos 40 0. cos 65

100

PI sin 40 + P3 sin 65 = + -7660 P! -

100

150

sin

cos 30

= P3 =

129-9

- -4226 P 3

- -9063

6428 P!

and

- 150

0\ 1

7660 P! - -4226 P a = 29'9 ) 6428 P, - -9063 P 3 = - 75 j - -552 P 3 = 39-0 ) P! - 1-41 P = - 116-5)

whence 14,

P.,

Special

181-2

Ib.

Case,

and P, - lOO'O Three

Forces

Ib.

Equili-

brium.

Lanic's Theorem.

Although, of course, the

general

method explained

ab'ove will apply to this

FIG.

17.

particular case also, the following method is more when only three forces are concerned.

convenient

Let

unknown

P and

17) be a known force and Q, S two forces acting at the given angles a, y with

(fig.

equilibrium with

it.

COMPOSITION AND RESOLUTION OF FORCES

Then

P Q S :

if :

the vector triangle abc be drawn cb ba ac by the triangle of forces :

sin (180

sin

ft)

sin a

sin (180

6).

(

sin (180

sin y, 1

that

is to say, the forces are proportional to the sines of the angles opposite them. "Example 6. Two cords attached to a beam carry

a vertical load of 100 at

The cords

Ib.

120 and 1651 to the load.

them

are inclined

Find the tensions

(fig. 18).

FIG. 18.

Here

p^ =-^ 100

sin

sin .-.

Also

sin

sin 75

P = sin

26-8

12Q

100 /.

lf>r>

2588 9659*

Ib.

sin

sin 75

Q =

Ib.

89-5

8660

Example 1. A. body of weight 20 Ib. rests on a smooth plane inclined at 25 to the horizontal, being supported by a force P acting upward at an angle See " Mathematics for Engineers,"

p. 54,

Form

(8).

THE ELEMENTS OF GRAPHIC STATICS

24 of

Find the value

to the plane.

pressure on the plane If,

for convenience,

(fig.

2X = P 2v = N + P

and

and the

19).

take the

the plane and the.Y axis normal to

then

cos a sin a

axis parallel to

it,

W sin 6 W cos

(I)

(2)

FIG. 19.

and since

Pcos

equilibrium these must

for

Wsin0orP =

W COS W cos 6

1x3

zero,

get

and ,.

and

25 cos d5

8in

p. 20 cos 25

- P

sin a.

18-1

- 5-9

8-45 -819

10-3 sin 35

12-2 Ib.

A body whose weight is 130 Ib. is 8. from a horizontal beam by strings whose suspended 2 ft. and 4 ft., the strings being attached are lengths Find the tensions. to the beam at points 5 ft. apart. Ans. 88-9 and 126-6 Ib. Example

Example span and 3

9. ft.

simple triangular roof truss 24

deep

supported

at its

ft.

ends and

COMPOSITION AND RESOLUTION OF FORCES

carries a load of 3 tons at its apex. Find the forces and in the tie-rod. Ans. 6'2 and 6 tons.

in the rafters

Example 10

ft.

10.

long, hinged

wall crane consists of a strut at the

bottom end and supported

by a horizontal rod 6 ft. long attached to its upper end, and carries a load of 2000 Ib. Find the forces Ans. 1500

in the bars.

Ib.

;

2500

Ib.

Example 11. Five bars of a structure meet at a The bars make angles of 0, 30, 90, 120, joint. and 210 with the horizontal measured contra-clockJfr

FIG.

120.

and the forces in them are 10, 8, 6 tons, Q, the first two being tensile, the third compressive, and the last two unknown. Find the values Ans. P = 10-22 Q = 13'65 tons. P and Q. wise,

P and

;

A cord supported at A and B of 2 T at C,

(fig.

20) carries a load Find the at D.

and an unknown load tensions in the cord and the value is

horizontal.

Ans.

AC = 2-83; GD = 2; DB = 4;

W when CD W = 3-46

tons.

The above examples may with advantage be worked graphically also, as explained in .the preceding chapter.

CHAPTER

IIL

FRAMED STRUCTURES. 15. In dealing with the forces which occur in framed structures, it is usual to suppose that the loads which come upon the frame are transmitted to the joints.

ject to

When

this

not so, the

memher

hending and the stresses induced

will he considerable unless the loads are

will

he sub-

in this

way

very small

In any case it is desirable to avoid bending Moreover, it is usual to neglect the rigidity of the joints and to assume that these are free to turn. Actually of course this is

or act near the joints.

stresses.

frequently not the case, the principals of a roof truss being continuous from end to end, and the bars being frequently, and in fact usually, riveted at their extremities both in roof trusses and girders but it is necessary to make the assumption of pin-joints for ;

the sake of simplifying the problem.

Bow's Notation, The system of notation commonly known as Bow's is very convenient in 16,

If P lf P 2 P3 dealing with forces and their vectors. a forces at acting point O, let figures (fig. 21) be any in the spaces bebe written or letters, e.g. a, 6, c, ,

tween them, and let the vector which represents the force lying between the spaces a and b be denoted by the same letters placed at its extremities, and ('26)

FRAMED STRUCTURES

similarly for the other forces and their vectors. Then since the equilibrant is represented by the closing vector da, this force P 4 must be parallel to da, and

have the magnitude and sense which this denotes, a and d lying on either side of it. The convenience of this notation will best be

understood in

its

applica-

tion.

Example 1. Suppose a load (fig. 22) to act at the vertex of a triangular frame ABC, pin-jointed at the apices. Then if the direction of one of the reactions at

or'B be known, the values

of these re-

actions can be found, as well as the forces acting in For first, since the external forces at A, the frame. B, and C are in equilibrium, if the direction of the reaction at

A, say, be given, in a point

vertical force

;

will

intersect the

and since three forces

in equilibrium are concurrent the direction of the must pass through D. A triangle of reaction at

forces abd for these forces which intersect then determine their magnitudes.

will

THE ELEMENTS OF GRAPHIC STATICS

It is to

be noticed that

of the reactions at

the direction of neither

known, the problem

finding these reactions is impossible, because there are any number of solutions the only condition for ;

equilibrium being that their directions shall intersect on the line of action of W, and for each pair of forces satisfying this different.

condition

the magnitudes will be

the bearing surfaces are horizontal then

FIG. 22.

the reactions will be vertical for vertical loading, and in this case their values can be found.

Lettering the spaces of the frame and the vectors by Bow's notation, as explained above, the reaction

action

A will be given by the KB at B by the vector

vector oc, and the recb.

Also

we draw

cd in the vector figure parallel to AB, the forces at A will be represented by the sides of the triangle acd, the force cd in the frame being represented by cd in the triangle both in magnitude and sense, and as the sense is away from the point A, it indicates

FRAMED STRUCTURES

pull. Similarly the vector triangle cbd represents the forces acting at the sense of dc in the vector ;

now

opposite to what it was before, and a force indicating acting away from B, and therefore a pull ; which checks with what we got when configure being

sidering the force in the same bar acting on A. It will be seen therefore that the vectors referring to the bars of the

acting in the bars,

frame represent the internal forces which balance the external forces ;

in fact they represent the elastic forces material exerts to prevent being deformed,

act in both forces

which the and which

Thus when considering the

senses.

at C, the force in

acts towards C, but

when

considering the forces at A, it acts towards A, the elastic forces at every section being equal and opposite on either side of the section. For this reason it is convenient to indicate the sense of the internal force

by an arrowhead

in the

frame diagram

directed towards the point whose equilibrium is under consideration, when the force is a thrust ; and away

from

it,

when

the force

a pull.

It will

be under-

stood therefore that after having found the force in any bar AC by consideration of the forces at A say,

we must its

reverse the arrowhead

when

considering

action at the opposite end C.

Let fig. 23 represent a roof truss Then if the loads as shown. loaded symmetrically on the joints are set down along a vector line from

Example

e to e'

sected in

not shown in the figure), and ee be biThe resultant ke are the reactions.

k, e'k,

vertical force at

which

will

supports

may

therefore ke

kf,

from

be seen that any load acting at the be disregarded, so far as any effect

THE ELEMENTS OF GRAPHIC STATICS

on the forces

the

being known, and

The

frame are concerned.

resultant reaction at A, viz. kf, being triangle of forces kfa may be drawn.

known, the

Then fa

the polygon of forces whose sides are a/, fq in the vector figure may be closed by drawing gb ba parallel to the corre/</

also,

sponding bars

in the frame.

forces in the bars

Next, having found the

ka and ah, the polygon

ab can

be closed by drawing he, At', and so on. It is only to oneconstruct half of the force diagram, necessary because the other half will be a figure symmetrical

with the

first half

Example a

maximum

pivot at A.

about the horizontal line ak.

Let

fig.

24 represent a crane carrying

W, and let the crane turn about a Find the forces in the bars and the

load

counterbalance

weight required at

maintain

FRAMED STRUCTURES

equilibrium with W, neglecting the weight of the Set down ab to represent and construct the triangle of forces abf for the forces at the apex. Then if there are no loads acting at the joints, the

crane.

stresses in the bracing bars will be zero apart from Hence in negligible forces which prevent buckling.

the vector figure

k, j,

and /

all coincide.

Fin. 24.

Draw

ae, fe in

the vector figure parallel to the corre-

sponding bars in the crane. Next ed and bd. A horizontal through c should meet them in the same point d and this affords a check on the accuracy of the drawing, be will represent the magnitude of the counterbalance weight, and ca will represent the reaction of the pivot A.

large variety of roof frames

and other triangu-

THE ELEMENTS OF GRAPHIC STATICS

lated structures

might be considered, but as the priniii involved ciples constructing the force diagrams is the same in most cases, and special difficulties will be considered in due course,

to multiply

examples which

appears unnecessary

illustrate

no new

prin-

ciple.

When

a girder is supported at the ends, and carries loads both on the top and bottom booms the vectors must be arranged in order, as

Example

shown

fig.

25.

Thus we

Fia.

plot in

order ab,

2/i.

de, ef (equal to the reaction at B),

ia.

The

way

as shown.

force

diagram

When

be,

hi, /</, yh then drawn in the usual

the load

the reactions must of course be

not symmetrical,

first

found either by

calculation or graphically, as explained in chapter x., but when the load is symmetrical this is unnecessary,

because the reactions are the total load.

known

to be

each half

FEAMED STRUCTURES 17,

Structures

Equilibrium under the Action

of three External Forces which do not pass through

Common

Point.

The determination

of the forces involved

principle already enunciated sections 6 and 15, that when a rigid

upon the

depends and applied in body is at rest

under the action of three forces only, the lines of must intersect in one and the

action of these forces

same

point.

Thus

tig.

26 represent a flap-door

FIG. 26.

hinged at A and kept in position by a rope attached at B, as shown, then if the resultant weight of the flap act at C, it will be in equilibrium under the action of the pull P in the rope, reaction of the hinge at A.

its

own

Now

weight, and the the lines of action

two forces intersect in D, and therefore A must have such a direction as to D Hence if 1 represent W, and also. pass through 1 2 be drawn for the three forces the vector figure of the first

the reaction at

THE ELEMENTS OF GEAPHIC STATICS

intersecting in D, the magnitudes of P and be given by the vectors 2 0, 1 2 respectively.

Further Ilhistrations.

Let

tig.

will

27 represent a

crane structure, which turns in a footstep bearing at E and is kept up by a horizontal thrust at ED, a load

FIG. 27.

W being applied

at A.

Then the

structure being in

equilibrium under the action of three external forces, W, ED and EE, of which the first two intersect in G, the third force EE must also pass through G, and the detriangle of forces 012 being drawn, its sides will termine the values of the two forces EE and ED in

terms of

Again,

require to find the thrust

FRAMED STRUCTURES along CD, the bar

BA may

be regarded as a rigid

body in equilibrium under the load W, the thrust along DC, and the force at B, unknown both in magnitude and direction. Since the first two intersect in H, the third force at B must have its direction such as to pass through H also and the triangle ;

FIG. 28.

013

being drawn for these forces, 3'0 will give the thrust along DC, and 1-3 will give the pull at B. the resultLet fig. 28 represent a roof truss and

ant wind pressure on one side. at B is known to be vertical, as

Then if the reaction when it is supported

will intersect roller-bearings, this reaction and A will in D, and the reaction at pass through this

point.

known andB.

The vector force

triangle

012

drawn on the

1 will then give the reactions at

THE ELEMENTS OF GRAPHIC STATICS

If, however, both ends are pin-jointed to the supports the problem of finding the reactions is indeterminate for although these are often found by ;

assuming that they act parallel to the resultant external load, this supposed solution is only one of any number of possible solutions, because the only condition essential for equilibrium is that the directions of the two reactions shall intersect in a point lying on the line of action of the resultant external load. 18.

Three

Resolution of a Force into Components in Unknown Directions.

This problem

only capable of solution

when

the

forces

Thus

concerned are non-concurrent.

29) be a known force and X, Y, Z be three unknown forces acting at along the given directions, it will be seen that any number of vector figures may (fig.

drawn having

directions, is

e.g.

their sides parallel

0123

and

1 2

therefore indeterminate in this

to the given

The problem But if the case.

3'.

forces are non-concurrent, as indicated in

fig.

30, the

FRAMED STRUCTURES resultant of

any two

them, such as

P and

act through their point of intersection A ; larly the resultant of the remaining two,

must

act

But

through B.

the resultant of the

since there

first

Z, must and simiX and Y,

equilibrium,

pair must be equal and

apposite to the resultant of the second pair, both If therefore acting along the line joining A and B. 1 represent the known force P, and we draw 1,

1 2 parallel to P and represents respectively, 2 the magnitude and sense of the force Z, and 1 2 those of the force acting along AB necessary to keep P at rest. But this force reversed is in equilibrium

with the forces

and Y.

Consequently

the

vector triangle 213 for the forces at B be drawn, 1 3 and 3 2 and Y in magnitude and sense. give

Illustration

Suppose

fig.

portion of a braced rib supported

R, and suppose

31 to represent a by a given reaction

require the forces in the bars X,

THE ELEMENTS OF GRAPHIC STATICS

Produce X and R to meet in A. Also produce Y and Z to meet in B, and join AB. Then if we draw 1 to represent R and construct the Y, and Z.

1 2 for the forces at A, 1 2 will represent triangle that along X. the force along AB, and 2 Again if on 1 2 reversed we draw the triangle 123 for the

forces at B,

will represent the force along Z,

and 3 2 that along Y.

FIG. 31.

Illustration

32 suppose

In the braced girder shown in fig. known reaction at B, and we

to be the

The require the forces in the bars X, Y, and Z. force along Z meets R in B, and the forces along

Join AB, and setting up a vector 1 to represent the value of R, draw the triangle Then 2 of forces 012. gives the tensile force

and

meet

in A.

along Z, and if further we construct a triangle of forces 213 with its sides 1 3, 3 2 parallel to X and Y respectively, then these sides will determine

FBAMED STBUCTTJRES

the magnitude and sense of the forces along

and Y.

Examples. 1.

ABCD

(fig.

33)

a four-bar gate 10

action at

ft.

by 4

ft.

E and F, the reBE = CF = 6 ins. A

swung upon hinges

F being horizontal.

FIG. 33.

man rail.

weighing 140 Ib. sits on the middle of the top Find the forces on the hinges and the forces in

the bars due to the man's weight. = 70 DC = 172 = 187 Ans.

;

EF =

233

;

BE =

270

140

;

Ib.

THE ELEMENTS OF GRAPHIC STATICS

trap-door of uniform thickness, 5

ft.

long by

wide, weighing 5 cwt., is kept open at an angle of 35 with the horizontal by means of a chain, one 3

ft.

end

which

fastened to a hook 6

ft.

above the

Find the tension in the chain and the forces hinge. Ans. 2-65 cwt. 2 '5 cwt. on the hinges. ;

crane of the form sketched

(fig.

34) carries a

FIG. 34.

Find the

load of 2 tons at A.

and the reaction at

at B.

force in the strut

EC, assuming the reaction its magnitude and the

Also,

to be horizontal, find

reaction at the foot F.

Ans.

EC =

4-67

;

RB =

7-1

;

BE =

17-3

;

20 tons. 4. A Warren girder (fig. 35), of the form shown in sketch, projects from a wall and carries a load of 2 tons at its extremity. Find the forces in the

bars.

FRAMED STRUCTURES

Ans.

122334

Bars:

7'8

Diagonals.

Forces: 1-16 3-47 5-77 -2-31 -4-62 -6-94 tons.

roof truss of the

a span of 40 and

ft.

2-31 tons.

form sketched

has (fig. 36) the tie-bar horizontal high,

FIG. 36.

being 8 bars

ft.

Ans.

6. Fig.

Find the forces

below the vertex.

when loaded with

ton at each

Bars Ib

Forces 5-2

4-7

37 shows one

2 3

in the

joint.

1 2

2'35 0-9

a3 2-5 tons.

of the cantilevers of a bridge.

Construct a reciprocal figure for the forces in the members, indicating which are in tension and which in compression.

THE ELEMENTS OF GRAPHIC STATICS

42 Ans.

Bars: Forces

Diagonals: Verticals:

12345 6789

-76-4 -40-3 - 44

-93

-132

+ 40-2 +77-1 +95-3 +103-8 -16-9 -40-5 -44 -3'2 9'0 + 17'9 +20-0 + +95

tons.

Fio. 37. 7.

the

roof truss

stress

loaded as

diagram,

-10'

shown

- ->r* - I0f-

H*-

-/O'-

Draw

in fig. 38.

assuming the

if*

left

end

to he

10'-

FIQ. 38.

hinged and the right end resting on the stress in the members.

rollers.

Tabulate (B.Sc.)

FRAMED STRUCTURES

1234567

Ans.

Bars:

Forces: +9-1 +1-9 +1'9 +8-4 -5-2

-58

-8-4

Bars:

+ 5'62 -4-Q5 +2-Q -4-12 +2-2

Forces: -2-8

tons.

shown

the outline of a roof truss, fig. 28 ft. ft. and The full panel loads are 7 height span 8.

800

FIG. 39.

each 1600

member.

Ih.

Find graphically the

stress in each

Ans. Stresses marked on bars.

Part of a pin- jointed frame shown in loaded with a vertical dead load of 10,000 9.

(B.Sc.) fig.

Ib.

and a

FIG. 40.

normal wind pressure

of 15,000 Ib., both being taken

as uniformly distributed along

AB.

The supporting

THE ELEMENTS OF GRAPHIC STATICS

Find forces P, Q, and E are shown by dotted lines. these forces and the forces in the bars which meet

and

in C, indicating the struts

Ans.

P =

13,000

;

ties.

Q =

1950

;

R =

14,500.

Other forces are marked on bars.

CHAPTEB

IV.

SUPERPOSED AND SPACE FRAMES.

framed structure

said to be (1) Perfect, (2) Imperfect, (3) Bedundant, according as the members which compose it are just sufficient, less than or 19.

so,

keep the frame

simplest case of a perfect frame

The equilibrium. the triangle, which

indeformable by its geometrical property that the lengths of the three sides completely determine its is

shape.

Thus

ABC

(fig.

41)

a perfect frame.

FIG. 41.

another triangle

ACD be built upon

it,

two additional

bars will be necessary to fix the new joint D. Any less number of bars would render the frame unstable

and therefore imperfect, and any greater number would make it redundant in which case the forces in the bars would be statically indeterminate, that is ;

to say, their

magnitudes could not be found except (45)

THE ELEMENTS OF GRAPHIC STATICS

members. Since two and joint requires only two additional constitute the whole a perfect frame, we get

by considering the

elasticity of the

every new bars to

for a triangle

for a quadrilateral

for a five-sided

3 bars and 3 joints 4 ,, ,, ,,

frame 7

,,5

Hence if b be the number of bars and number of joints, the condition for a perfect frame is b = 2/ - 3. Thus in fig. 42, b = 8 and and so on.

j the

FIG. 42.

perfect,

- 3 = 9. The frame is therefore im.-. 2; and unless another bar be added, its stability

must depend upon the rigidity of its joints. In the case of the ventilator frame in fig. 43, b = 8 and j = 5.

Fiu.

- 3 2/ member.

.'.

43.

Hence the frame has a redundant

In the case of the Fink truss j

perfect.

(fig.

44), b

13,

- 3 == 13 and the frame is therefore 2j A framework of this description may be

.-.

described as a superposed frame, for

will be seen

SUPEEPOSED AND SPACE THAMES

that the bars overlap one another, and in this respect from those which form an open triangulation. In dealing with the 20, Superposed Frames,

differ

Fink truss at the

fig.

upper

44,

joints,

W W W 2,

we may

be loads acting

begin at joint

The

thrust in bf will be equal to Wj and may be repre1 in the vector figure. Construct the sented by Then 2 1 for 012 the of forces joint /. triangle

FIG. 44.

represents the tension in cf, and 23, 31 are its and horizontal components at c. Similarly

vertical

if 4 5 at d, and therefore the 3 represent the load thrust in dh, the line 4 6 will represent the tension

in ch

and 4

7 6 are

its

horizontal and vertical

The

load on the centre strut eg is and therefore if therefore equal to 2 3 + 7 6 + 2 we add these as shown and construct the triangle of

components

forces

for the joint

tft

viz.

268, the

lines 6 8, 8 2

represent the stresses in the bars ge and ga respectively.

21.

Bollman Truss,

The Bollman Truss, which

THE ELEMENTS OF GRAPHIC STATICS

48 is

common American

triangles as

shown

type, consists of superposed 45. Set down 01 = Wj,

fig.

and construct triangle of forces and draw triangle down 23 = 2

and so

line

on.

Then

9 8 will be the

for joint

Set

of forces for joint

these are set off as shown, the of the vertical components

sum

and and 9

of the tensions in the bars,

equal to the reaction at a,

will therefore

will be the

sum

FIG. 45.

of the horizontal

components

of the tensions

and

comwill therefore be equal and opposite to the the of chord the in force girder. upper

pressive

22, Lattice Girder, Girders of this type (fig. 46) are statically indeterminate in theory, but for practical purposes a quite sufficiently accurate result may be obtained by resolving the framework into its comthe forces which occur in ponent systems, finding each in the ordinary way, and then compounding

SUPERPOSED AND SPACE FRAMES

the results by adding the forces which occur in bars upper and lower chords which are common to

of the

Thus the compressive forces ab and cd must be added to get the force likewise the compressive forces in ab, cd, and

the separate systems.

which in

act in ;

ef to get the force in '

~~ /0/

DE, and f

''- gy*-

- |o'-

so on.

f -y-

- l0'-

f *r -

/0 '~

-4"^"*

FIG. 46.

Example.

The double Warren

girder

(fig.

47)

supported at the ends and carries the loads shown. Find the forces in the members. State the assumpAns. Forces shown on bars. tions made. (B.Sc.) is

In other cases an approximation to the stresses in the bars of a redundant frame may be obtained 4

THE ELEMENTS OF GRAPHIC STATICS

-50

by neglecting the effect of the compression members in cases where these are designed only to take

Thus

tension.

in the case of the trestle

members may be put to the direction according compression of the applied pull, but as the bars are usually weak

shown

fig.

48, the diagonal

in tension or

"b Fia. 48. to

resist

compression

they

may

be assumed to

buckle sufficiently under thrust to throw the load upon the tension members alone. In the present

SUPERPOSED AND SPACE FRAMES case, for example, the bars ab, be, bd, ce,

51 fy

may

be neglected, and then a reciprocal figure may readily be constructed for the remaining bars, from

which the forces

them may be found.

These are

shown on

the corresponding bars in the figure. 23. Space Frames. Space frames are those

whose members in the case of

these

may

act in space of three dimensions, as etc. The stresses in

domes, towers,

may be very complex, but a few simple cases be here considered. In resolving a force into

two components by means

of the vector triangle, the

FIG. 49.

force

and

force

components necessarily lie in a plane. however be resolved into definite commay its

ponents at a point along three given directions, provided that these directions are not in one plane.

Let

ABC

49) represent a sheer-legs carrying supported by a guy-rope AD. Take a plane containing the load and the guy-rope and let it cut the plane of the sheer-legs in AE. The a load

W may now be resolved into components along

load

(fig.

W, and

and

may way.

as before, and the

then be resolved along

component along AE and AC in the same

THE ELEMENTS OF GRAPHIC STATICS

24, Belfry Tower, Fig. 50 represents a belfry tower carrying a load of 10 tons at A, This load may be assumed to act half along a vertical plane containing AjH^Jj and half along a vertical plane containing AjIjLj. Having resolved the half load of

AB and AD, component along AB

5 tons along the

may

resolved

into

components along and BD, and the

may

BE last

be again resolved

in plan along

B^, BjC.,.

Similarly the force along may be resolved

along

and EF, and

the force along

into

components along EjF,, in plan, and so on.

The results obtained shown on the bars in

are

the

figure.

25.

case three

A more general would be that of bars

unequal

length meeting at a point. Such cases occur in the of framed Let domes. and towers

construction FIG. 50.

OA, OB, OC (fig. 51) rewhose plans are OA lf OB P

present three such bars OC 1 resting on the ground at a vertical load

Then

A B x,

C and l

carrying

this load be first resolved

SUPERPOSED AND SPACE FRAMES into

components along OC,

the force along

OD in the plane OCD, and

then resolved along

and

FIG. 51.

OB, the forces in the three legs will be found. In order to carry out this idea a plan of the points A,

FIG.

drawn

5U.

and the lengths the triangles OjAjBj, OjBjCj If we equal to the triangles OAB, OBC are drawn. B,

to scale, viz. AjBjCj,

of the legs being

known

THE ELEMENTS OF GRAPHIC STATICS

suppose these triangles turned about their bases AjBj,

BiCp the plans

of their vertices will

move

along lines perpendicular to these bases, and these which is the plan lines will intersect in a point

of the vertex of the tripod.

Hence FA P

FB FC lf

Produce will be the plan of the three legs. It is OD. of is the Then to Dj. plan X

only necessary to construct the true triangle

ODC, and

now

of the

readily be done, for if a perpendicularly to the base DjCj, this

may

FO 2 be drawn and the length of the leg

line

shape

CjF

be taken and turned

the triangle CjCXD, will x 2 be the true form of the triangle ODC. If therefore

about

to cut

we suppose

the load

W to act along

O._,F,

may

The first force will be resolved along O..C,, O 2 Dj. the compressive force along the leg OC, and the second may then be set off along the line OjD, and resolved along OjAj, O,Bj, by which forces along

obtain the

and OB.

Examples, pair of sheer-legs are 60 ft. high when upThe guy-rope is right, their feet being 30 ft. apart. in the legs when 90 ft. long. Find the forces acting 1.

lifting a load of

20 tons which overhangs 25

ft.

Ans. 17'9 tons in the guy-rope, and 17'5 tons in

each 2.

load of 7 tons

leg.

suspended from a tripod,

the legs of which are of equal length and inclined 60 to the horizontal. Find the thrust on the legs.

Ans. 2'75 tons. 3.

In an

unequal-legged tripod, the lengths of

SUPEKPOSED AND SPACE FEAMES the legs are OA = 11 ft., 10 ft., the length AB = 8 ft.,

Find the

BC =

ft.,

and

ft.,

CA =

OC = 10

ft.

when

carrying a load of 10 tons Ans. 4*5 tons 6*15 tons -87 tons.

stresses

at O. 4.

OB =

;

A rectangular

on castors and

frame 3

ft.

carries bars 7

;

x 8 ft.,

ft.

ft.,

supported ft., 4 ft.

long at its corners meeting in a point O. At O is hung a weight of 400 Ib. Find the forces in all the

members. Ans. In bars in 5.

weight of 1000

frame Ib.

192, 192, 110, 110 :

Ib.

;

25, 72, 94*5, 94-5 Ib.

carried

by three con-

vergent ropes 10, 10, and 15 ft. long, hanging from AB is points A, B, and C in a horizontal ceiling.

and BC, CA are each 20 ft. long. Find the on each rope. 881 881 Ib. Ans. 736'5

ft.,

pull

;

26, General Tripod,

When

;

Treatment of the Forces

a given force acts in any direction

FIG. 52.

in space,

may

be completely represented by the

Thus if P (fig. 52) its vector. of the a and-P' vector E, its elevation, plan represent the magnitude and direction of the vector may be plan and elevation of

THE ELEMENTS OF GRAPHIC STATICS

= AB, drawfound by setting off the plan along ing a vertical at B, and a horizontal through D to cut it in C. Then CA is the resultant vector and a

is its

inclination to the horizontal.

27, To Find the Resultant of a Given System of Concurrent Forces in Space. Let OP lf OP 2 OP3 (fig. 53) be the plans and 1} OP' 2 OP' 3 be the elevations of a system of forces

whose resultant

required.

Fia. 53.

Construct a vector figure with the given plans of Then the closing line ad will be the forces, viz. abc. the plan of the resultant. Similarly, if we construct a vector figure a'b'c with the given elevations of the forces, the closing line a'd' will represent the The plan and elevation elevation of the resultant. of the resultant being thus known, its actual magni-

tude and inclination can easily be found by 26 or we draw a'e' horizontal and set up de = d'e at right :

angles to ad, ae will represent the resultant and a will be its inclination to the horizontal.

SUPERPOSED AND SPACE FRAMES

28, Resolution of a Given Force into Components along Three Given Directions not in the same Plane,

Let

(fig.

54) be a given force kept in equilibrium

by three other forces acting along the C, whose projections are a, a' b, b' ;

lines A, B, ;

c, c',

and

p, p' be the projections of the given force P.

and let

FIG. 54.

two components, one acting along the line of one of the unknown forces, say A, and the other along the line in which the plane containing P and A intersects that containing the other two unknown forces B and C. First, if we join the feet of a and b we get the horizontal trace of the plane conNext find the horizontal taining the legs A and B. trace ; of the force P by producing p' to meet the line XX, and erecting a perpendicular to cut p profirst

resolve

into

THE ELEMENTS OF GKAPHIC STATICS

58 duced

Join this to the foot of

/,.

c, viz.

t.-,.

Then

the horizontal trace of the plane containing P t^j and the leg C, and the intersection of this plane with is

the plane containing the legs A and B whose projections are de and d'e'. elevation of P, viz. ab, a'b' are

the line DE, The plan and now drawn in another is

place for convenience, as shown, and these are resolved along ed, et z and along e'd', e't\, respectively, as the plan and elevation of the force be, b'c' as the plan and elevation of The first of them, viz. the force along the bar C.

giving ac k

along ac,

a'c'

ED, and

may

be easily resolved along directions parallel B and the second, viz.

to the plans of the legs A, a'c',

;

along directions parallel to the elevations of A, In this way we obtain the plans and elevations

of the required forces along the legs A, B,

and C,

from which the forces themselves are easily deduced, as in

$ 16.

CHAPTER

MOMENTS.

When the line

of action of a force does not pass tends to produce rotation about and this tendency increases both as the force in-

29.

through a point, it,

creases and also as

its

lever -arm increases. to

cause rotation

distance from the point or

its

In other words, the tendency jointly proportional to the

magni-

tude of the force and

its perpendicular distance from The value of this product is known as the point. Thus if the moment of the force about the point. Pj_ (fig.

55) be a force

and

a^ be its distance from a moment of the force

given point Q, then P l ci l is the about the point. Similarly, if

(59)

be another force

THE ELEMENTS OF GRAPHIC STATICS

whose distance from

a 2 then P,ao ,

moment

is its

about Q, and the resultant moment due to both forces will be the algebraic sum of the individual + P2 a2 Further, if no rotation moments, viz. takes place, then this resultant must be zero, and

therefore for equilibrium

+ P 2a 2 =

we or

get the equation

P 2a 2 = - P^,

is to say, the moments must be equal in magnitude and opposite in sign. This is the fundamental condition that there shall

that

be no rotation.

Suppose P! = 10 Ib. and a = 2 ft., and we rewhat force P 2 acting at li ft. from Q will maintain equilibrium. }

quire to find

Then

since

10 x 2

+ P2

x 1^

get

the negative sign indicating that the must be opposite to that of P,.

13J

moment

Ib.

have seen formerly that when a body is by a system of concurrent forces in equilibrium, two independent equations can be obtained, viz. 2x = 0(1) and 2y = 13), and (2) ( when the system of forces is non-concurrent we have the additional equation 2 (moments) = The (3). first two are the conditions that there shall be no translation, and the third is the condition that there shall be no rotation. These three equations are known as the Three Fundamental Equations of Equilibrium for forces in a plane, and it is there30.

acted upon

fore possible to determine three unknown quantities when such a system of forces is in equilibrium and

the forces are non-concurrent

;

a very

common

case

MOMENTS

being to find the magnitudes of the two reactions direction of one of them, when a system of

and the

on a structure supported at two given the direction of one reaction being given

forces acts points,

(see

Example

3, p. 40).

Examples, 1.

Suppose a load

of 2 cwt. to act

A and B

beam supported

whose weight

neglected.

(fig.

upon a horizontal shown, and

56), as

Then

as the

beam

is at

whatever point we choose to consider, the sum the moments about it must be zero, because there

rest,

of is

|2cwt Jh

B FIG. 56.

rotation.

To avoid unnecessary labour

therefore,

consider either the point A or B, because by doing so we get rid of one of the unknown forces. If e.g.

we consider B moments about

as our moment-centre, then it

where the clockwise sense

rotation has been

adopted as positive. Hence BA since KA + BB = 2 cwt. BB = horizontal pull of 200

taking

x7-2x4=0

we have BA

= ?

1? cwt.,

and

cwt. Ib.

applied to a

post which is supported by a stay, as shown in fig. 57. Find the pull in the stay if the foot of the pole is assumed free to turn. vertical

THE ELEMENTS OF GRAPHIC STATICS

Taking moments about the foot we have 200 x 15 + P x a = 0. Now a = 5-656

''*>.

ft.

Tof---**

POl

FIG. 57. 3.

loads

beam, whose weight

shown

actions at

Problems dealt with

this

kind

but

neglected, carries the

are

by graphical methods, as

FIG. later,

Find the values of the fig. and B, where B is a pin-joint. 58.

it is

culation also,

the loads into

re-

conveniently be shown

will

T>8.

desirable to be able to proceed by calin this case it is best to resolve

and

their horizontal

and

vertical

com-

ponents. This may of course be done graphically or mathematically.

MOMENTS Also

let

and

be the components of

Taking moments about B we get R A x 10 - 2 x8 - 3sin60 x 6 -

x 10

RA =

.-.

1-5 x 6

1 sin

-707 x 3

x3 = 0, 27-1.

2-71 tons.

Also since the horizontal forces must have a zero resultant,

get

XX=

and since the resultant,

.-.

- 3 cos 60 1-5

-707

forces

vertical

must

have a zero

get

- 1 sin 45 = 0. - 2-71 = 1-5 approx.

Y + R A - 2 - 3 sin Y = 2 + 1-5 + -707 R B == VX* +~^=

V7-12

.-.

and

tan .-.

1 cos 45

2 '207 tons,

Three loads

and 9

2-67' tons

|= ^=-678. =

of 2, 1,

8'.

and 3 cwt.

rest

upon a beam

respectively from the L.H. end. Find the vertical reactions. Ans. 2-4 and 3'6 cwt.

at 2, 5,

ft.

whose weight is 100 lb., is raised by a pull applied at an angle of 45 and at 6 ft. from the hinges. Find the pull required, if the resultant weight of the door acts at 3 ft. from the hinges. 5.

flap-door,

Ans. 70-7

lb.

100 lb., 100 lb. act at 45, 90, 60, and 90 at intervals of 2 ft. from the left along a beam 10 ft. long, the left end of which is hinged, the reaction at the other end being 6.

Loads

100

lb.,

150

lb.,

angles of

vertical.

Find the

reactions.

Ans. Vert, reaction 210

lb.

;

inclined reaction 233

lb.

THE ELEMENTS OF GRAPHIC STATICS

platform in the shape of an equilateral triangle, with sides 6 ft. long, supported at its three corners, carries a weight 3 ft. from one corner and 4 ft. from another. Find the fraction of the weight 7.

borne by each of the three supports. (I.C.E. Exam.) Ans. -232W -336W -432W. ;

8. fig.

;

Find the stresses in the frame sketched in members AB and CD being continuous.

59, the

Find the

maximum

show how

to calculate the direct stress in

Ans. Stress in

moment 31,

CD =

bending

2-5 tons

The Method

moment

- -416 T ft.

;

CD, and

it.

maximum

Stress in

EF=

(B.Sc.)

bending + 1-59 T

The ".Method

of Sections,

an application of the method of moments to framework structures which are statically determinate with the object of Sections," as

it is

usually called,

finding the internal force in any particular bar, when the dimensions of the structure and the loads acting

are

known.

This method consists in taking an

MOMENTS

imaginary section through the structure, so that the forces in all the bars cut by it are known except three at most. If then the force in one particular ^ar be required, and we take as moment-centre the

point in which the other two intersect, the equation

one unwhich can therefore be found. quantity,

of equilibrium so obtained will involve only

known

Example 1. In order to make this important method clear, consider the roof truss in fig. 60,

FIG. 60.

where we require to find the stress T in the tension bar. Suppose an imaginary section XX taken, as shown. Then the structure on the left of this section may be regarded as in equilibrium under the action of the

known

external forces together with the three

internal forces in the bars cut by the section, and which may now be regarded as external forces

applied along them at the points where the bars are If then we choose the point O as cut through. centre of moments,

1-1000 x 50 - 4000

(37-i 4-

get

12$)

+ T

x 30

THE ELEMENTS OF GRAPHIC STATICS

whence T

20000 lb., the negative sign indicating that the force acting on the extremity of the bar tends to produce contra-clockwise, rotation about O,

and is therefore a tensile force. As a further illustration, suppose we require the In this case the point O' in which the force S. other two bars meet must be taken as centre of moments, and we get for equilibrium 4000 ..

(12-|

300000 -

37^)

+ S

where

ft.

.'. S 15000 lb., the negative sign representing contra-clockwise rotation about O' and therefore a -

tensile force. if

Similarly,

the force

take

moments about we get

O'' to find

in the principal,

14000 x 29 - 4000(16-5 + 4) + 4000 x 8-5 + P x 10-8 = whence P = - 33200 lb., and contra -clock wise rotation about

Example

indicates a compressive force. Determine by the Method of Sections

the forces X, Y, in

fig.

in the three bars of the roof truss

61.

Ans.

32-2 tons comp.

Y =

;

Z =

11*7 tons tension

;

37-Q tons tension.

32. Graphical Treatment of Moments, Let P in a given direc(fig. 62) be any given force acting tion AB, and let Q be a given point at distance a

from AB.

Then P x a

about the point Q. the force

to scale,

distance from be

the

moment

of the force

Let be be a vector representing

let O be any pole whose Join Ob, Oc, and through any

and

MOMENTS

AB draw lines P6', PC' parallel to 06, Oc, an making intercept b'c' = i on a line through Q to AB. Then by similar triangles parallel P

point

- or be x a

p x

p .'.

x a

force, viz.

=p P x

;

a, is

say, the moment of the to the equal product of the polar

that

is to

FIG. 61.

FIG. 62.

distance measured in the scale of forces by the intercept i measured in the scale of lengths.

first

moment than

its

sight the

graphical representation of a

appear to be less convenient mathematical expression, but this disadvanin this

way

will

tage will disappear as the subject develops. 33, Couples, Two equal and opposite forces not

same

known

as a couple, and the measured by the product of one of the forces into the distance between them. For if P and - P (fig. 63) be two equal and opposite in the

moment

line are

of a couple is

THE ELEMENTS OF GEAPHIC STATICS

forces distant d from one another,

tance from the one force

P(x

and

we consider

moments about any given

their

then their resultant

- Px = Pd.

point Q whose disx and from the other

moment

HeDce,

about

the resultant

will be

moment

being independent of the value of x, is the same about all points whatever, being equal to P x d. 34. Effect of a Couple, If P (fig. 64) be any force acting through A, a pair of equal point

and if we suppose at any and opposite forces applied

along a line parallel to the original force P, then

FIG. 63.

FIG. 64.

these three forces produce the same effect as the original one, and may therefore be considered as replacing

force

it.

But these

acting at

in the

forces consist of a single same sense as the original

one, together with the couple P x d so that the force P acting at B is equivalent to the same force P actmg ;

at A, together to the

which tion

with a couple whose

of a couple is equivalent to

parallel to itself, so that the is

moment

equal

product of the force into the distance through it has been shifted. Conversely, the introduc-

equal to

tha.t of

shifting a force to its shift

moment due

the couple,

MOMENTS

Since couples have 35. Composition of Couples, and direction, they are vector magnitude, sense, and be magnitudes compounded by the law of may vector addition.

In order to

effect this the

moment

of a couple is represented

by a line perpendicular to its plane, the sense of its moment being indicated by the sense of the arrowhead attached to it. The vectors representing two or more couples may then be

compounded

way, and the resultant magnitude and sense the

in the usual

vector will represent by

its

FIGS. 66, 65A.

resultant

moment and

will be perpendicular to the

plane of the resultant couple.

Thus

P^, P a 2

are the moments of a system of forces 3 (fig. 65) to rotate a shaft AB, and if ab, be, cd tending fl

represent these moments, the resultant vector ad measure the resultant moment E, which is obvi-

will

same sense as that of the first two moments. 36, Resultant of a System of Non-Concurrent

ously in the

Forces. that

In the preceding paragraph we have seen P l (fig. 66) act upon a body at a point

a force

A, and another force

at a point B, the effect of

THE ELEMENTS OF GRAPHIC STATICS

the latter is the same as if it were to act at A, together with the couple whose moment is P 2 x d. But since a couple consists of two equal and opposite forces, having a zero resultant, the magnitude and direction of the resultant force cannot be affected

but only its position. Hence, when a system of non-concurrent forces acts upon a body, the magnitude and direction of the resultant may be found it,

exactly as but point ;

moment -of

the forces were

its

position will

all

transferred to one

depend upon the resultant

the couples due to this transference

all

FIG. 66.

The

of the forces. is

position of the resultant force

however more conveniently determined by means

of the following construction

Let

lf P.,,

current forces.

representing them, sense,

and

their resultant in magnitude,

direction, as

order to find

be any system of non-con(fig. 67) Then if 1, 1 2, 2 3 be the vectors

we have

its position,

just

shown.

take any convenient pole

Through any point a in xa parallel to OO, ab parallel to Ol, be parallel to O2, and cy parallel to O3. Then P! may be replaced by its two components acting along j'rt, ba and represented by 00, Ol in the vector join OO,

the force

Ol, O2, O3.

P draw 1

lines

MOMENTS figure.

Similarly,

P2 may

be replaced by

its

two

components acting along ab, cb and represented by 1O, O2 and again P 3 by forces along be, yc repre;

sented by 2O, O3. Now the forces along ab and also along be are equal and opposite forces, being represented by the same vectors 1O and 2O respec-

hence they cancel one tively in opposite senses another, and there remain only the forces along xa, ;

FIG. 67.

ya as the equivalent of the original system of forces P lf P2 and P3 The resultant of these forces along xa, ya must pass through the point of intersection d, and therefore if a line be drawn through d parallel ,

3, it will

sultant force

represent the line of action of the reE and since the same reasoning may ;

be extended to any number of forces, that if a polygon be drawn as above its links will intersect in a point

which

we conclude and last on the re-

first

lies

THE ELEMENTS OF GRAPHIC STATICS

sultant force

and therefore fixes the position of its Such a polygon is known as a Link-

line of action.

or Funicular- Polygon. 37. Further, if the first and last links xa, yc, along which act the two forces represented by OO and O3, coincide along the line ac, then OO and O3 will But as we have shown above these

coincide also.

two forces are equivalent to the original system, and they will now form a pair of equal and opposite forces acting along ac, from which it is evident that

when

the link-polygon is closed by the lines xa, yc produced being in coincidence, the given force system but if the first and last links are is in equilibrium ;

parallel, the

and

coincide,

polygon

will

along the

points

this figure not.

first

and

the

vector figure will but the link-

will close,

In this case the forces acting form a couple whose

last links

moment

is equal to that of either of the forces represented by its corresponding vector multiplied into the distance between the first and last links of the

Hence, link-polygon. 1. When the vector and link-polygons both close,

we have complete

equilibrium. the vector-polygon closes, but the linkpolygon does not, we have a couple. 3. When neither polygon closes, we have a definite 2.

When

resultant represented by the line

which

required

to close the vector-polygon and acting through the point in which the first and last links of the link-

polygon intersect. 38. Special Case, sists of vertical forces,

between the reactions.

When the load (a) When all This case

is of

system conthe loads

very

lie

common

MOMENTS

occurrence in practice, because the loads which act

on structures are usually

vertical forces.

The determination of the resultant of the given loads is exactly the same as for that of non -parallel loads explained above. Thus if lt 2 3 (fig. 68)

W W W ,

be any given vertical loads, and vectors which represent them, sultant vector. force take

any pole

1, 1 2,

2 3 be the

3 will be the re-

find the position of the resultant

and construct the link-polygon

FIG. 68.

abcde as before.

The

reactions are usually

vertical (though of course this really

assumed

depends upon

whether the supporting surfaces are horizontal or not), and therefore the first and last links cut them Join ae and draw through the pole O a in a and e. line

parallel to ae.

that the resultant load

Then

we may show / may be replaced

as before

B acting

by two forces along af and ef, and then again by forces along ae and Aa, and along ea and Be respectively, of which those along ae cancel out, 4 and 4 3 as the leaving forces represented by first

THE ELEMENTS OF GRAPHIC STATICS

components of R along the verticals through A and B. Consequently the line O4 drawn parallel to the closing line ae of the link-polygon divides the vector into two parts which represent the vertical

line

pressures on the supports, or the vertical reactions, reversed in sense. This method of finding graphic-

ally the

magnitude of the reactions at which

particularly useful for reasons

A and B will

appear

FIG. 69.

but if the reactions are alone required then they he obtained by calculation with less trouble than may 29. by drawing, as in later

;

(b)

When

actions.

1 2,

Let

all

the loads do not

the supports.

and 2 3

lie

between the

re-

W p Wo, W As

be the loads and 3 (fig. 69) before set down vectors 1,

to represent the loads,

O construct a

and with any

link-polygon abode, the being produced to cut the support 4 reactions in a and c.. Join ae, and draw a line

convenient pole

first

and

last links

MOMENTS Then 4

parallel to ae.

and

0, 3

4 are the reactions at

respectively, the reason being as in the previ-

ous case.

calculation, taking

RA ...

moments about B, we have

x9-2x6-4x2 + 2x4

K =

* ~ IS

= IJand R B =

0.-

9 - 1J

tons.

39, Graphical Determination of the Moment of any Given System of Forces about a Given Point,

FIG." 70.

P P 2 P3

be given forces and Q a given point. Let lf Construct the vector figure 0123 for the given ,

Then

3 is their resultant. With any pole a O draw link-polygon abcde. Then a line drawn through the point / in which the first and last links forces.

THE ELEMENTS OF GRAPHIC STATICS

meet, parallel to 0-3, is the line of action of the reNow the moment of this force, which

sultant force. will also

moment

be the

of the

given forces about Q,

can be found as described in

32, viz. by drawing through a point / on the line of action of the force E two lines parallel to the lines O O 3 in the vector ,

Now

these lines correspond to the first and last links of the link-polygon produced and they cut off an intercept ea = i, which measured in the linear figure.

scale

and multiplied by the polar distance p measured

in the force scale gives the

moment

about Q, and therefore also the

of the resultant

moment

of its

components.

\ \

F.G. 71.

a link-polygon be drawn for a given Hence, forces and its first and last links be of system cut line through the given point to a produced of the given forces, then to the resultant parallel the intercept on this line multiplied by the polar distance gives the moment of the forces about the given point. 40, Application to a System of Parallel Forces, Let E, Pj, P 2 be any given parallel forces whose if

MOMENTS moment about Q

0123

required.

Draw

77 the vector figure

and the link-polygon abode. Then if a line be drawn through Q parallel to the resultant 3, the intercept ae on this line between the first and last links produced to cut it measures the moment of the forces about Q.

CHAPTEE

VI.

ROOFS. 41. In calculating the probable maximum stresses that are liable to occur in the members of a roof truss, the first step is to estimate the

maximum

loads

which are likely to come upon the joints. These loads are due to the weight of 'the truss itself and of the roof covering and any other loads which the truss may support. These constitute the permanent or dead load, and in addition we have to consider the possibility of snow and wind. The following Tables I, II, and III will be useful in estimating the probable dead load that may be expected in any given case.

(78)

HOOFS

CO 00 O5

CO CO O5

'o 1

1 3

* 00 <N <M > CO 00 05 >0 n CO

<N U5 ~ C CO 00 T

3 00

IH IH

^ CO

O >0 ^ CO

>O 00

OOCOCO.

>OOOCO

^ t- 05 TH CD 4h CO 6>

<N CO

OOOOO

<jq

1C CO t~ 00 O5

i-J

THE ELEMENTS OF GRAPHIC STATICS

TABLE

II.

SLOPES FOB ROOF COVERINGS.

TABLE (CHIEFLY FROM

III.

" INSTRUCTION IN CONSTRUCTION

Weights of Roof Coverings,

".)

Lb. per Sq. Ft.

etc.

Lead

covering, including laps, but not boarding or rolls Zinc covering, including laps, 14 to 16 zinc

to 8

1$ to If

gauge Corrugated iron, galvanized, 16 W.G. . 18 . 20 Sheet iron, 16 W.G 20 Slating laid with a 3 in. lap, including nails, but not battens or iron laths Slating Doubles, 13 in. x 9 in. at 18 cwt. per 1200 Slating Ladies, 16 in. x 8 in. at 31-5 cwt. per 1200 Slating Countesses, 20 in. x 10 in. at 50 cwt. per 1200 Slating Duchesses, 24 in. x 12 in. at 77 cwb. per 1200 Tiles, plain, 11 in. x 7 in. laid with a 3 in. lap and pointed with mortar, including laths and absorbed rain .

2} 2

8J 8|

ROOFS Pan

Weights of Roof Coverings, etc. 13 in. x 9 in. laid with a 3

tiles,

Lb. per Sq. Ft. in.

. lap, etc., as above Italian tiles (ridge and furrow), not including .

the boarding Slate battens, 3 ,,

x 1

in.

in. for

Slate boarding, "

in.

thick

Doubles Countesses

,, .

..' '

Wrought-iron

;

. .

..'

1J 2|

Duchess

laths, angle bars for

slates

Wrought-iron

12 14

laths, angle bars for

Countess

slates in. plate glass Cast-iron plates, | in. thick Thatch, including battens Lath and Plaster Ceiling .

., .

....

Ceiling joists

Common

rafters

As regards the

42, Temporary or Live Load, snow,

30, the snow will

slide

off,

be blown away under the so that

greater than about and in any case it will

the slope of the roof

it is

maximum wind

hardly necessary to If it is desired to do

snow. used for the purpose. This table basis of 15 Ib. per sq. foot x cos for

slope of the roof.

TABLE

pressure,

make any allowance so, Table IV may be

IV.

WEIGHT OF SNOW.

calculated

where a

on a the

THE ELEMENTS OF GRAPHIC STATICS

82 43.

Wind

Pressure,

The pressure

wind

of the

often the greatest force which a structure has to withstand, but its action on a roof is a very difficult is

When

matter to determine.

a stream of air of small

cross-section impinges normally upon a plane surface much larger than itself, the change of momen-

tum per second per

sq. foot of air

pressure on the surface, or

mass per mass

the

.-.

P =

where

= mv, and where = -0807 Ib. say.

per second

sq. foot

of 1 cub.

P =

current will be the

ft.

of air

the

Ib./ft.-'

-0054

velocity in miles per hour. section of the air stream is

area on which

If,

2 ,

where

the

however, the crosslarger than the

much

impinges, the change of direction and moreover a partial

of the air is not complete,

vacuum ward

induced at certain points and on the leethe results of experiment in such

side, so that

cases will not agree with the simple theory suggested above. The gusty nature of wind suggests also that

the pressure is likely to be anything but uniform over a large area, arid, the pressures registered by small wind gauges may be greatly in excess of the

average value over a large surface. Sir B. Baker's experiments on large boards at the Firth of Forth 2 Bridge showed that 45 lb./ft. on small surfaces and 30 lb./ft. 2 on large ones is quite a sufficient allow-

ance in this country even in exposed positions, and this were not so, the large windows in many of our

churches and cathedrals would have been blown in long ago.

The action

of the

wind on

roofs is

moreover greatly

ROOFS

modified by the adjacent walls and surroundings. air current sweeping upward along the wall of

The

a building or roof is so deflected that the pressures are often negative at the top and bottom of a roof surface on the windward side, with a strong negative so that the common pressure on the leeward side the of pressure over a as to distribution assumptions roof are purely imaginary, and can only be regarded ;

comparative values for the purposes of

giving

design.

Many formulae have been suggested for calculating the normal pressure of a fluid stream on an, inclined surface, of which that of Duchemiri appears to give the best results with the least labour. This formula is

N =

5 + sm 2

where

the normal pres-

is the wind pressure when the is the perpendicular to the current, and inclination of the plane to the current. This formula

sure on the plane,

plane

gives

the following values for wind pressures of 2 and 56 lb./ft. 2 the latter value being the

lb./ft.

Board

Trade allowance.

Graphical Construction for Duchemin's Formula, Let AB (fig. 72) be the slope of the roof at any

Draw PC horizontal and PN perpendicular AB. Set off a unit length PD on any scale, and draw DE vertical, and DC parallel to AB. Make

point P. to

THE ELEMENTS OF GRAPHIC STATICS

and PG = PD. Also make PH = 2W on is the given wind any scale, where pressure. Join FH and draw GK parallel to It. Then PK is the normal wind pressure on the roof at P to the

CF = PE

same scale as PH = 2W. The Board of Trade figure, although monly adopted to allow for the effects

high, is comof impulsive

action and vibration as well as the uncertainty of the assumptions alluded to above.

It is provided in Section

of the L.C.C.

General

Powers Act that for the purpose of calculating the loads

adopted

roofs

the

following

figures

shall

Where

the pitch of the roof exceeds 20 28 lb./ft>' All other roofs, 56 lb./ft. 2 on a

of sloping surface.

This last condition for slightly for the possibility of a crowd is allow roofs to sloping of people collecting on the roof. horizontal plane.

At the same time the working stresses steel are given as follows

for iron

and

ROOFS

44. Increase of

For high

Wind

Pressure with Height.

structures, like factory

chimneys,

etc., it is

desirable to recognize the fact that the pressure and velocity of wind also increases very greatly as the

height above ground increases, as

lowing figures

shown by

the

fol-

These figures indicate that the pressure varies roughly as the square root of the height above ground. At the Firth of Forth Bridge, on 1-5 sq. ft. gauges, the pressure ranged from a maximum of 65 lb./ft. 2 at 378

elevatjion to

ft.

Wind

45,

lb./ft.

showing the ;

relative

taking a

unity Flat plate,

ft.

flat

are experimental coefficients pressures on various shaped plate normal to the wind as

Hexagon,

Pressure on Inclined and Curved Sur-

The following

faces,'

bodies

0*65.

Octagon, 0'75.

Cubs (normal

to face), 0'80.

THE ELEMENTS OF GRAPHIC STATICS

86 .

Cube

(parallel to diagonal), 0'66. Lattice girders, about O8.

Wedge

(vertex angle

Sphere,

0'3.

Elongated

90), 0'6 to O7.

projectile, 0'5.

Cylinder, -54 to '57.

diam.), 0'47 normal to axis. (vertex angle = 90), 0'69 to 0'72.

(height

Cone

( ,,

(height

60), 0-54. diam.), 0'38 parallel to base.

46, Distribution of Load on Roof, The loads are estimated on the assumption that each truss

FIG. 73.

carries half the load

on either side

it,

and that

this

transmitted to the joints of the frame by means of the rafters and purlins. Thus if w be the

load

load per sq. foot of roof surface due to the dead or live load, as the case may be, I be the length of the

and d the distance between the trusses, ivld on one principal such as DE and as each (fig. 73), joint, such as E or F, e.g., is likewise assumed to carry half the load on the panels to the right and left of it, the full panel loads will rafters,

will be the total load

on all the joints except those at the walls (in the case of the dead load), which carry one-half of a full

act

EOOFS panel load

;

but in the case of the wind pressure we it as being fully loaded on one side

have to consider

or the other in turn, and in this case we have full loads on the one side at all the joints except at the top and bottom. Special calculations must of course

made

for lanterns or other variations.

The end

trusses of the roof carry only half the load which comes on the intermediate trusses, but they are made to the

same dimensions

as the rest for the sake of

uniformity. 47. Example 1. French Roof Truss. One of the most economical types of truss, and therefore one

FIG. 74. of

very frequent occurrence,

shown truss.

in If

fig.

74, set

that particular type

and commonly known as the French

down

the joint-loads along a vector

THE ELEMENTS OF GRAPHIC STATICS

get ab as the reaction at one end, and if we proceed in the usual way we get the forces in the bars without difficulty up to the joints A and B. At line,

each of these points we are met with three unknown forces, and these cannot therefore be found unless

can determine one of the unknown forces

in-

directly. may be done in several ways I. By calculating the stress in the bar a7, as ex31, and plotting it on the force diagram. plained in

This

II.

the method of bar substitution, which conremoving the bars 4 5 and 5 6 and putting

sists in

in a

new

bar BC.

As the

force in the bar al

may

readily be seen by the reasoning of method I to be independent of the system of bracing adopted, this force

may

be found graphically from the

new

ar-

ordinary way, instead of by calcularangement The old bars are then replaced tion as in method I. in the

and the

force

diagram

may

be continued in the usual

way. III.

Suppose the bars 12, 23, 45, and 5 6 rehalf-load on the principal, viz. ce, con-

moved and the centrated

at A.

this force ce

then resolved

parallel and normal to the principal,' as shown, we get 3 4 as the force acting in the corresponding bar,

and when this is known, all difficulty and we can proceed in the usual way. 48.

Example

station roof of the

removed,

form shown

75 being given, find the stresses in the members, if the dead load works out at 200 Ib. per foot of rafter and the wind pressure at 300 Ib. per in

fig.

foot

normal

reaction

assuming that the support on the side opposite that on which Width 30 ft. height 10 ft. Base

to the roof,

vertical

the wind acts.

;

EOOFS and principals supported

at equal intervals.

the length of each principal on 1 truss = 2 x 18 x 200

ft.,

Since

the dead load

= 7200 lb., which gives 1200 lb. at each joint except the extreme ones, on which the load is 600 lb. Also the wind load on

FIG. 75.

one side

18 x 300

5400

intermediate joints and 900

or 1800 lb. on the on the extreme ones.

lb.,

lb.

Method I. A diagram may now be drawn for the dead load, and another for the live load, and the results

compounded. Method II. The dead load and the live load may be compounded at each joint, and one diagram drawn

THE ELEMENTS OF GRAPHIC STATICS

A link- and vector-polygon may then be drawn, by means of which the resultant of the loads may be found in magnitude and position or for the sake of greater accuracy the resultant of the for the whole.

;

whole dead load acting at the centre of the roof may be compounded with the resultant wind pressure This course has acting midway along the principal.

The two resultants intersect in drawn through C parallel to the re-

been adopted here.

and a

line

sultant of the total load, viz. ah, gives the line of action of this resultant, and this cuts the vertical

reaction at

in D.

Hence

will be the reaction

and the magnitudes of the reactions can be found by constructing ahj on all. We can now pro-

at B,

ceed to construct the force diagram in the usual way, In order to complete the investigation, as shown.

wind should also be assumed to act on the other and the stresses again determined. The maximum value in either case must then be taken as the load liable to come on any member. the

side of the roof

This will be illustrated in

49.

The

forces in the

bars will be found to be as follows, where

compression and

Truss with Fixed Ends, that

signifies

tension.

say pin-jointed at

When the truss is fixed, its

ends (because

ROOFS

always assumed that the joints have no

problems

of finding the forces in

the

rigidity), the

members

indeterminate, and the method often given in the textbooks of resolving the resultant load into two

components through the points of support and may easily be made to give contraAs no determinate solution is posdictory results. sible, the best course to pursue is to make an assumption which will introduce the worst conditions likely to arise, and this will be to suppose that one of the supports takes all the horizon(i) tal thrust and (ii) that the other takes it all, the wind acting on the same side in both cases. This of course means that we assume each of the reactions parallel is

incorrect

in turn to be vertical, the other being regarded as a pin-joint, exactly as is done in the case when one of is supported on rollers. 49, Crescent Roof. In order

the ends

above,

now

will

illustrate

the

consider a crescent roof with

The span was taken as from the chord AB to the and a depth of 5 ft. from the upper

fixed supports (fig. 76). 50 ft., with a rise of 8-J ft.

top of the roof, chord members, the. outer radius being 30 ft. and the inner 41'3 ft. The resultant wind pres-

to the lower

sures at the joints make angles of 56, 42, 28, and 14, with the tangents at those points, and the wind 2 pressure being taken at 56 lb./ft. normal to its direction, the values of the normal pressures by calcula-

tion or graphically will be found to be 55, 51*8, 43*2, 25*6 Ib. respectively. The lengths of the panels

measures

and assuming the distance between ft., the wind pressure on the be as follows 2410, 4540, 3780, 2240 Ib.

7'3

ft.,

the trusses to be 12 joints will

THE ELEMENTS OF GRAPHIC STATICS

FIG. 76.

KOOFS

Also assuming the dead load of the roof and truss to be 10 Ib./ft.^, the dead loads on the end joints will amount to 438 Ib. each and on the intermediate joints to 876 Ib. each. As explained above we shall first assume the wind to act on one side and consider the whole of the

horizontal thrust to be taken alternately first by the one support and afterwards by the other. Taking

the wind to act from the

left,

the whole horizontal thrust

and assuming

(i)

that

taken by the left-hand support, then the right-hand reaction will be vertical. We now compound the live and dead loads on the joints,

and by means

of a link-

and vector-polygon

determined in magnitude, direcand position in the usual way. Producing the

the resultant tion,

line of action of this resultant to cut the vertical

support reaction at

in Q,

tion of the reaction at A.

as the direcget triangle of forces bb'a

may now be drawn which determines the reA and B. After this the force diagram in the usual way as shown in be constructed may (i)

actions at

(i),

and

may

their

magnitudes scaled

off,

whilst the sense

be found from the direction of the vectors in

as explained in These results 6. the be tabulated. whole of the Next, horizontal wind thrust will be assumed to be taken

diagram should

(i),

now

up by the right-hand support B.

In this case the

left-hand support will be vertical. Producing this to cut the resultant load E and joining to B, we get

the direction of the reaction at B, and a triangle of forces bb'a (ii) determines the magnitudes of the reactions.

A new

and the forces

force

diagram

now drawn

in the bars are again

found from

(ii)

THE ELEMENTS OF GRAPHIC STATICS

94 aiid

If we now compare the members which are symmetrically

entered in the table.

forces

the

and take the larger values, these will be forces which will occur whether the wind blow from the right or from the left. situated,

the

maximum

CHAPTEE

VII.

SHEARING FORCE AND BENDING MOMENT DIAGRAMS. The shearing force at any section the force parallel to the section which tends to cause sliding between it and the 50. Definition,

a structure

adjacent section.

Let

ABCD

(fig.

77) represent a cantilever

which

FIG. 77. is

BCEF

cut through at XX, the portion being kept by a spring-balance attached at F which

in position

prevents vertical motion, and by a wire FG attached to a spring balance at G, and a small strut XE, the tension in the wire together with the thrust in (95)

THE ELEMENTS OF GRAPHIC STATICS

The spring-balance the strut preventing rotation. will register the weight of the portion BCEF which tends to cause vertical movement, and which

therefore equal in magnitude to the shearing force be applied EF. If any loads x, 2 to the free portion, the spring-balance will register

W W

at the section

their

sum

At the as the additional shearing force. the moment of the couple due to the

same time

pull in the wire

and the push

of the strut will

be measured by the pull P in the spring-balance This multiplied by the arm a of the couple. moment is known as the Bending Moment at the section,

one side

Hence we see that the effect of a beam may be analysed into

beam.

any

being the moment of all the forces on the of the section which cause bending in the

section of a

load on a shear-

ing force tending to produce a slide and a couple

tending to produce rotation. The first is resisted by the shear stress in the material, and the second by the moment of resistance of the tensile and compressive stresses induced.

In order that a beam or girder

enough

be strong

may

at all sections to resist the joint action of the

shearing forces and bending moments,

it is

important

to be able to construct

tudes of these at

all

diagrams showing the magniSo far as the shearing sections.

concerned, this merely amounts to finding the algebraic sum of all the forces parallel to the force

section on either side of

it,

because at the section

these form a pair of equal and opposite forces which tend to produce a slide either upward on the one side or

the other.

Thus

78) the shearing force

on the

downward on

(fig.

any section

left of it will

SHEARING FORCE AND BENDING MOMENT be equal to the reaction of it will be equal to

KA upward, and on

the right since

W - E B downward, and

- RB = RA we see that the shearing forces are equal and opposite on either side of the section. This we indicate by plotting their values on opposite sides of the base line AjB^ and for the sake of uniformity force

will

make

which tends

move upward

the convention that a shearing make the left-hand segment

relatively to the right shall be con-

sidered positive, whilst one which tends to

make the

right-hand segment move upward relatively to the left shall be considered negative, the first being

FIG. 78.

plotted above

below

the base-line

and the second

it.

51. Diagrams of Shearing Force and Bending Moment, It will now be shown that the dia-

grams

of shearing' force

and bending moment may

be easily drawn at one and the same time by the same construction which has been explained in the last

chapter for finding the reactions. For suppose 2, 3, and 4 tons to act as shown in fig. 79

loads of

on a beam simply supported at A and B. Then if construct the link- and vector-polygons, and draw

O a line O4 parallel to the closing as get 3 4 as the reaction at A, and 4 Draw a horizontal 4C through 4. the reaction at B. 7 through the pole

line bg,

THE ELEMENTS OF GEAPHIC STATICS

Then

project the point

the ordinate of

represents the constant positive shearing force at any section of the beam between

the rectangle

CD 2

CDDjD 2

this shearing force being equal to the reaction

&t the left-hand support. After passing D.> however, the resultant of the forces on the left of a section

becomes equal

to the reaction 4

the load

01 =

Hence we

project the point 1 as shown to EEj. on Similarly, passing the next load the resultant

force

becomes negative and equal

to 4 2,

and

finally

FIG. 79.

on passing the last load

becomes 4 3 downward,

being equal in magnitude but opposite in sign to the reaction at the right-hand support. The diagram

thus obtained

shows at

any

shearing

force

diagram, and

at a glance the values of the shearing forces section.

difficulty frequently occurs

to the student

contemplating such a shearing force diagram, in that we get apparently two. different values for the shearing force at one and the same section. Thus at D 2 we get D 2 E and D 2 D This ambiguity arises from the assumption tacitly made that a load may .

SHEAEING FOEGE AND BENDING MOMENT This

act at a point.

is,

of course, impossible,

the shearing stress which

area

because

would then be

be a load this Thus if (fig. 80) must rest upon a base of some appreciable If we suppose this width to width, however small. be CD, then the shearing force on the left of C will be represented by AA 1? and the shearing force on the right of D will be represented by BB 1? and the value will change more or less uniformly from the one value to the other over the length CD on which the infinitely great.

load

load rests.

therefore

we assume

a load to be

6 PIG. 80.

concentrated at a point, as is often done for convenience, the base CD shrinks to a point and the It must, however, be line EF becomes- vertical.

understood that this is only a conventional representation of the facts. Actually the upper ordinate will represent the shearing force on the left of the load, and the lower ordinate will represent that on the right of it, however small the base may be, and the

two values do not occur at the same

section, although

so represented for convenience.

Not

only, however,

do we

in this

way

obtain a

shearing forces, but at the same time the figure abcde by which the reactions were found is a

diagram

THE ELEMENTS OF GRAPHIC STATICS

100

diagram

beading moments.

For if we

refer to

fig.

that the bending moment represented by the intercept be-

we have shown

71, p. 76,

about any point

last links of the link-polygon on the vertical through Q, and if we suppose Q to be on the axis of the beam, then it will be seen that the

tween the

first

and

fig. 79 corresponds exactly to the interae in cept fig. 71, which we have shown to represent the moment about Q. Hence, the bending moment

intercept ae in

of the beam under vertical loading be represented by the vertical intercept of the link-polygon which is made by the section produced,

any section

will

this intercept being

measured

in the linear scale

and

multiplied by the polar distance in the load scale. 52. When some of the Loads Lie Outside the

Supports.

The procedure

in this case

similar to

FIG. 81.

'that in the previous case. along a vector line (fig. 81) is .

The loads are set down and a link-polygon ab f .

drawn with any pole O for the given forces Wj the first and last links being produced back 4 .

SHEARING FORCE ANb BENDING 'MOMENT

101

a and /. The line af then the closing line and the hatched figure is the bending moment diagram. In constructing the to cut the support verticals at is

= 1 is first shearing force diagram the load x set down below the base line of shear drawn through 5. is set up At D the vertical reaction E A = 5

The points 2 and 3 are then pro2 when across as before, until we come to jected the vertical reaction RB rnust be set up from to from Dj to

G!-

The

figure so found

the shearing force dia-

gram, and it will be seen that it merely represents the algebraic addition of the loads from left to right. In constructing the bending moment diaNote. it is advisable to use a gram polar distance which is a round number of units, in order to simplify the numerical evaluation of the bending moments from the diagrams. 53, Shearing Force

grams

and Bending Moment Dia-

for Special Cases,

Case

Cantilever with

concentrated load.

acts at the end of cantiWhen a load (fig. 82) lever the shearing force at all points is constant and

THK ELEMENTS. OP GRAPHIC STATICS

IQ'J

equal to

Hence

the diagram will be a rectangle

of constant height representing

The bending moment

at a distance

x from the

W#.

This represents a value which to a increases uniformly from zero when x = maximum of WJ when x = I as indicated in the

end

free

will be

figure.

Case

run

II.

(fig.

Cantilever with uniform load

per foot

83).

In this case

consider a section at any distance

FIG. 83.

on the right of it and therefore the shearing force = wx that is to say, it will be represented by a straight line BC which increases from zero to a maximum at the fixed end, where its value is the whole load = wl. Also the bending moment r at x will be

x from the

free end, the total load

will be wx,

;

the load

x the distance of

from the section, .'.

centre of gravity

viz. \x.

M. x

= wx

the values of

values of x

its

x \x

= ^wx 2

be calculated for different

get a curve

whose ordinates vary

SHEARING FORCE AND BEMDlNG MOMENT

103

as the squares of the abscissae, viz. a parabola, the

maximum

moment

bending

= JWZ where

being -^-

the total load.

Case III.

Beam

with concentrated load

(fig.

84).

Taking moments about B, we get

and similarly

E A = W.^ a

FIG. 84.

Hence the shearing in

force

diagram

will be as

shown

84.

fig.

Also the bending moment will increase uniformly from the supports up to the load, where its value will be

EA x a =

b .

-j

= I

maximum

bending moment of segments = load x product span the load is central, then a = b =

or the

bending

-j-/

moment

W7 at the centre is -j-

and the

THE ELEMENTS OF GRAPHIC STATICS

104

Case IV. (fig.

Beam

W = wl

with distributed load

85).

this case the reaction at

being ^

the shear-

ing force will have this value at A and will decrease at a uniform rate, until it becomes zero at the midspan, and then increases to

at B.

Also the

FIG. 85.

bending

moment

a section x from the

centre

will be

Here again the bending moments depend upon the square of x and the curve is therefore of parabolic form, the maximum ordinate being obviously at the centre

when x =

and

its

value

wl-

or half the value if the same load were concentrated at the centre. (Compare with Case III.)

54.

It will

be seen that for uniform loading the the load is always

bending moment diagram under

SHEARING FORCE AND BENDING MOMENT

105

and therefore it is necessary to be able to construct a parabola for the following conditions of parabolic form,

Case

When

the base, height, and axes of the

parabola are given.

Let

(fig.

86) be the given base and

the

a c Fia. 8G.

and height.

AD, and draw any vertical Project c to d and join dD, cutting the line ab in e. Then e is a point on the parabola, and other points can be found in the same way. axis

ab to cut

Case

II.

given and

When its

Join

end

the tangents to the parabola are points.

FIG. 87.

Let AC, BC be the given tangents and let A and Divide AC into any be the points of contact. number of equal parts, and BC into the same

number

of equal parts as

shown

fig.

87.

Join 1 1

THE ELEMENTS OF GRAPHIC STATICS

106

22, 33. These lines will be tangents to the parabola and will sufficiently define it, if enough tangents are drawn. Construct combined diagrams of 55, Example 1. shearing force and bending moment for a beam and a distributed load w carrying a point load

per

ft.

Set up

CC = W^X

any

scale,

and

join

Fro. 88. j,

BC r

Then ACjB

Calculate the

maximum

the bending

moment

bending moment

dia-

WJ 8

it up at DDj (fig. 88) to the same scale as before and construct the parabola ADjB. Then the whole figure represents by its ordinates the combined diagram of bending moment.

Set

SHEARING FORCE AND BENDING MOMENT Next draw a vector to

AC GB

OK = W, and

i.e.

horizontal, will give

shearing force diagram via:

EFGHKP.

and

OO,

Then O

in O.

meeting the vector diagram, and a 1

line

107

parallel

the pole of

AB, The

parallel to

the two reactions.

may

we now

then be drawn as usual, plot

and

PM =

%wl

join get the shearing force diagram for the distributed load, and the whole figure will repre-

sent the combination of the two diagrams on a base

FIG. 89.

areas which are positive in one diagram and negative in another cancelling out. Example 2. A beam AB carries the load shown in fig. 89. The distributed load is first con2

LM, any

sidered as a concentrated load acting through centre of gravity, and a link-polygon

AKHLC

drawn

as usual

;

and the closing

to this determines the reactions at

however, the load on

AC. O 4 A and B.

line

its is

parallel

Since,

uniformly distributed instead of concentrated, as assumed, the link-polygon between F and G should be a parabolic one. is

108

THE ELEMENTS OF fJRAPHTC STATICS

56, Indirect Loading,

When

a load does not

bear directly upon a beam or girder supporting it, but is transmitted to it through the medium of cross-girders or otherwise, so that the load bears on the beam at intervals, the loading is said to be is the commonest case in practice, large girders, whether triangulated or plate girders, the load is usually transmitted to the main girders through the cross-girders. In this

This

indirect.

because in

all

case the bending moment will be modified, as will be seen from the following example :

W W

be any system of point Let Wj, Wo, 4 3 on a girder, as shown in fig. 90, the loads ,

loads

on the platform, which is carried by the Construct the bendcross-girders at C, D, E and P. resting

ing

moment diagram in

the usual way, viz.

aHJKLB.

Then

neglecting the continuity of the longitudinal members, if we join cd, d\\, K/', the new diagram

obtained

this

way

will

be the actual bending

moment

diagram, the parts cHd, dJeK, KL/ being the bending moment diagrams for the panels CD,

DE,

for the

loads upon them,

and

only

SHEARING FORCE AND BENDING MOMENT therefore

moment

109

the panel joints that the bending the same as for direct loading, being less

at is

between the on the main

joints

than

the load rested directly

girder.

Similarly, for a uniformly distributed load, if we draw the parabolic diagram in the usual way, and drop verticals

from the joints to cut

it,

the chords connect-

ing the joints will be the true bending moment diagram that is to say, the true bending moment ;

diagram is a polygon inscribed in the parabola. 57. Determination of the Forces in the Bars of a Braced Structure from the Shearing Force and

Bending Moment Diagrams, Let Case I. Parallel Booms,

ACDEFGB

91) be a triangulated girder, carrying loads

(fig.

Wp W

FIG. 91.

at its joints, as

shown, and

the shearing force and be constructed in the

let

bending moment diagrams

THE ELEMENTS OF GRAPHIC STATICS

110

Then

usual way.

the force in a diagonal CD is be found at once from the

required, this force can shearing force diagram

and

of the shearing force will be equal

to the vertical fore

chord members

the

have no vertical components, the

these will then

whole

when

are horizontal, because as the forces in

of the force in

component

we take

a vertical section

and opposite

CD.

There-

XX cutting the shear-

ing force diagram in a and b and draw cd parallel to CD, cd will represent the force in CD because it is ;

evident that the vertical component of cd is cc and is therefore equal to the shearing force ab. Similarly, the stress in the bar

drawn

and the

parallel to

it,

will be given

In order to find the

members such

a section

Now

AD.

stress in

by the line by gh.

one of the boom

force in

CE, we suppose the

girder cut by through the three bars CE, CD, and will be evident that if the bar CE be as

supposed cut, the girder will collapse due to the bending moment about D, and it is the force acting

whose moment is in equilibrium with this Hence the force in EC x the depth of the girder d must equal the bending moment about D. But this may be measured off the bendalong

bending moment.

ing

moment diagram,

ate

Kl x

or the force in

equal to

EC =

DF may

= moment 58. Case

then force in

scale of loads.

bar

being represented by the ordinin EC x d= Kl x p

and therefore the force x ^,

and

EC =

In the same

p had been drawn

K/ measured

way

in the

the force in the

be found by writing force in E = mn x p, and so on.

x d

about II.

Non-parallel

Booms,

Let

ACB

SHEARING FORCE AND BENDING MOMENT (fig. 92.)

111

be a braced 'girder, with curved upper boom, let AjLBj be the bending mo-

loaded as shown, and

ment diagram.

CD may by

plied

Then

the force in any bar such as

be found from the fact that this force multidistance from

its

is in

equilibrium with

FIG. 92.

the bending

moment about

O, which

given by the

ordinate EF.

For

force in .

force in

x d

= FL.

FL = CD = ^

moment at O bending ~'

~d~

we set off cd = force in CD and draw de parallel to DO, then de will be the force in the diagonal DO, If

because the vertical components of cd and dc together must balance the shearing force at the section.

CHAPTER

VIII.

RELATION BETWEEN THE CURVES OF LOAD, SHEAR, AND BENDING MOMENT.

When

a load is supposed to be represented by a straight but when the load is distributed, it will be

59, Load-Curve, act at a point line,

may

Actually as we have exshould always be represented in this way, because the load is distri-

represented by an area.

_ af\^ [

plained

buted over some area, however small, If the load is uniform and in all cases.

w per unit length over any given distance d, then the total load over that distance will be wd and equal to

t i

FIG. 93.

will be represented

by a rectangle

When the load base d and height w. is variable, the average intensity w lt

over a short length rf,, may be taken, and the load or graphically on d l will then be represented by the area of a rectangle whose base is cl in the

scale of lengths,

and whose height

of load intensity

(fig.-

93).

in the scale }

Similarly, the load

d 2 will be represented by a rectangle of height w<>, value of the load intensity representing the mean over

do,

and so on

;

and the sum

of these areas will

which represent the total load on the length (112)

the

RELATION BETWEEN CURVES OF LOAD, ETC.

sum

of their bases.

113

The smaller the lengths d lt d z more numerous are the rect-

etc., are taken, the

angles, and the more nearly do they represent the actual load distribution AB, but in practice unless the mathematical expression for the load-curve is

known, we are compelled to proceed graphically by dividing the load area into small strips and to assume that these areas are equal to their widths multiplied by their middle ordinates. The summation of the

load

may

then be effected graphically as follows

Fio. 94.

(fig.

94).

Suppose

load upon a

be the load-curve for the

beam AB, the to

scale

representing Let the area under

ordinates of the curve

the varying load intensity. be divided into vertical strips

by the ordinates aa lt bb lt cc l9 etc., and let a^j, %<$<>, etc., be the mid-ordinates. Project yv y 2 etc., upon Take a pole O on a vertical axis OY to y\, y' 2 etc. ,

produced and join Oy' ly Oy' 2 etc. Next, startfrom O draw a line Oa 2 parallel to Qy\, and from ing a 2 draw a.2 b>2 parallel to Oy'^ and so on. Then if we consider the area of any one of the strips, say 8 ,

THE ELEMENTS OF GRAPHIC STATICS

114

that on the base

be,

and compare the similar triangles

~ 6 263 b. c. 3 2

.-.

Hence we

p =

b2 bs x

# 3 2/ 3

base x

mean

height.

see that the increase in the ordinate in

going from b to c, viz. 6 3 c 2 multiplied by the polar distance p, measures the increase in area. Similarly,

P represents the area on the base cd, and so so that the ordinate of the curve at any point

C -A x

on x

;

For as a

represents the sum of the areas up to that point. this reason the curve Aa.2 b.2c.2 d.2 etc. is known

Sum-Curve

60,

or curve of integration.

The foregoing construction draw a curve of shearing force for

Shear-Curve,

will enable us to

Frc.

<>f>.

Thus, suppose the sloping line to irregular loading. be the load-curve for a beam AB, and let Aa e be the sum-curve of this load-curve drawn with a polar .

EELATION BETWEEN CURVES OF LOAD, ETC. Then

distance p.

115

since the increments of ordinate

sum-curve represent the increments of area, if we project the points a, 6, etc., to a 2 6 2 etc., the lengths Ba 2 a 2 6 2 etc., will represent the areas or loads on the bases Aa 1? a-J)^ etc. These loads are supposed to act along their mid-ordinates, and a linkof the

polygon ghk

.pis drawn whose closing line is gp. O draw Or parallel to gp. Then

the pole

Through

Br represents the reaction at A, and re that at B. Draw a horizontal rs through r. Then rs is the base-line of shear

and the diagram Asre

a diagram

of shearing force. For consider any section, say b lt of the beam the shearing force at this section will be the difference between the upward reaction at A and ;

on A6 X acting downward. Now represents the first and bb-^ represents the second, so that bt represents their difference ; that is to say, the ordinate at any point between the line the

sum

of the load

b^ = As

and the sum-curve Aa

& gives the shearing force

at that point.

61. at

any

Bending Moment Curve, The shearing force beam is by definition the summation

section of a

of the loads is it

on one side or the other

of the section,

and

represented by the area of the load-curve ; and may be shown that the bending moment at any sec-

tion

is in

way represented by the area of the For let the shearing force diagram for Then the bending fig. 96 be drawn.

a similar

shear-curve.

the loads in

any section XX will be equal to E A x A X X - x x D 2 X - a x E 2X - DD x D.N - EE; x = AC x

moment

= area

AjX

AjCDD^EEjPX

area of the shearing force diagram.

116

THE ELEMENTS OF ftRAPHIC STATICS

Hence

in the above

problem

the bending

60)

moment diagram might

be obtained by constructing a sum-curve of the shear-curve on the base sr, but as

the figure gh

diagram, to

moment

already a bending

unnecessary

in this case, at

any

rate,

draw a second sum-curve. 62, Scales,

In order that the numerical values

of the shearing forces

known,

it is

and bending moments

may

necessary to find the scales of the dia-

grams. If the scale of load intensity be 1 inch = a tons or pounds per unit length, then the ordinates of the shearing force diagram will have a scale of 1 in. = p^ where p is the polar distance used, measured in the l

linear scale of the drawing. Thus if the linear scale be 1 in. = I feet and p l is in inches, the scale of forces will be 1 in.

}

al,

and

shearing force diagram. the shear-curve be p.,

in.

Again,

measured

the sum-curve of

drawn with a polar distance

inches, then the scale of bending

distance if

this will be the scale of the

moment

in the force scale multiplied

in the linear scale or 1 in.

the polar distances are taken the

will

be 1

by the polar

p^al x

same

p.2 l,

each

KELATION BETWEEN CUKVES OF LOAD, ETC. case,

and equal to

as the case

may

lb./ft.-

then

tons/ft.

ap~l

be.

FIG

63, Concrete Raft,

1 in.

117

97.

Numerical Example.

Fig.

97 represents a concrete rafft or distributing the load due to a building or other structure over a soft foundation. its

The

weight

was designed to be 3 ft. thick, and be taken as 125 lb./ft. 3 the super-

raft

may

posed load consisting of the weights transmitted to the three piers, which are estimated at 14 tons, 12

and 14 tons per foot length of walls respectively, symmetrically situated, and which may be taken to

tons,

THE ELEMENTS OF GEAPHIC STATICS

118

where they

act at the centres of the walls

the

rest

upon

raft.

Total weight of concrete raft per foot length

= Load on

raft

44-5 x 3 x 125

lb./ft.

7'45 tons.

per foot length

The load on the

soil

Total load

47 '45

therefore

44-5

1-067

ton per sq. foot. This load was plotted

= ba and a horizontal line a the reaction of the soil. The through represents 125 x 3 = 0-167 ton weight of the concrete =

-g-

per

was deducted, as represented by the dotted line. The wall whose weight is 14 tons causes a pressure

foot

14 of ^TK

5'6 tons per foot, leaving a resultant

down-

ward pressure of 4'7 tons per foot, which was plotted In this way we get as shown below the base line. a load diagram abcdefghij. A sum-curve of this diagram was next drawn with a polar distance of

5 ft. This diagram is bklmnop, and represents the shearing force to a scale of 1 in. = 2 tons per ft. x 5 ft. = 10 tons. Next a sum-curve of the shear1 in.

ing force diagram

same polar bending x 5

ft.

was drawn, as shown, with the

distance.

found to measure 1-3

This will therefore be the

moment diagram to a scale of 1 in. = 10 tons = 50 tons/ft. The maximum ordinate was

780 x 2240

lb,/in.

in.

on the drawing

tons/ft.

RELATION BETWEEN CURVES OF LOAD, ETC. ..

119

the stress per sq. inch

780 x 2240 x 6

-isirssr

=18>71b -/ m

2 -

The concrete may

this

working

will

stress,

be required.

be relied upon to withstand and therefore no reinforcement

CHAPTER

IX.

SHEARING FOECE AND BENDING MOMENT FOR A MOVING LOAD. 64,

Single Concentrated Load Crossing a Maximum Shearing Force. Let 98) be the beam, and let C be any section

Simple Beam,

(fig.

distant

Then

x from B.

as long as the load

lies

FIG. 98.

on the right of C, the shearing force pend iipon the value of the reaction this will increase as

shearing force at

at A,

will de-

and as

W approaches C, the maximum will

occur

(120)

when

the load has

SHEARING FORCE AND BENDING MOMENT

121

C from the right. Now in this position can be found from the equation

just reached

EA =

W -y I

which shows that E A increases uniformly with and attains its greatest value when x = I, when

has the value

and

Hence

AjB, the triangle

AA^B

set

up AA = X

x, it

will be a

diagram of maximum shearing force for every section, and if we agree to call the shearing force positive which join

tends to cause the left-hand segment to move upward relatively to the right, then this diagram will represent the sections.

all

BBj =

W and

maximum positive shearing force at On the other hand, if we set down join

the triangle

AB B X

will

maximum negative shearing force. So example, when the load stands at C the CC will represent the maximum positive

a diagram of that, for

drdinate

shearing force at the section, that is, on the left of the load, and the ordinate CC 2 will represent the maximum negative shearing force on the right of the load.

At any other section such as D,

will represent the

maximum

positive

DD X and and nega-

tive values.

Maximum Bending Moment.

Since the bending

moment at a given section D (fig. BA (I x), and since EA increases

99) is equal to as the load ap-

proaches the section, the bending moment will be greatest when the load reaches D, and since EA is then equal to

the bending

x a

moment

THE ELEMENTS OF GRAPHIC STATICS

122

which being an expression of the second degree " Mathematics in x represents a parabola (see Engineers "), whose central ordinate will be found by making x = iJ, when we get for the for

central

bending

up CC

set

WZ moment MC = -j-. =

{

and construct the parabola will be the

ACjB, the diagram so obtained of

maximum

therefore

bending moments

diagram

for all sections of

FIG. 99.

the beam.

of all

ADjB which

This parabola the triangles such as

the locus of the vertices are the bend-

ing moment diagrams for the loads at various points of the span.

65, II. Two Concentrated Loads, Distant d from One Another, Let (fig. 100) be the leading load and \V 2 the following load, and supSet up pose them to move from right to left. and A Then B. AAj = Wj and AA 2 = join 2 AjB, 2

as in Case

maximum

Now when force at

A2B

will

be the diagrams of

shearing force for the separate loads. Wj stands at C, the maximum shearing is

and as \V 2 has not yet come

SHEARING FORCE AND BENDING MOMENT

123

the diagram of maximum When for the segment BC. positive shearing force reaches A, then AA X is the shearing the load x stands at At this instant 2 force at A due to it.

upon the beam,

A is re= DD P E A we make presented by DD r If therefore we if join EC^ AE will be the total shear at A, and the diagram ECjB will be the diagram of maximum D, and the additional pressure

causes at X

The diagram

for both loads. positive shearing force

?r ^ B maximum

negative shearing force will be a similar reversed below the base-line AB. diagram Maximum Bending Moment Diagram. Set up

CCj = _14

(fig.

as base.

101) and construct a parabola on

Similarly, set

struct another parabola.

mum

bending

loads.

Then

maximum

up CC 2

-^-

and con-

These represent the maxi-

moment diagrams

for the individual

as in the case of shearing force, the

bending

moment

for the

segment

BD will

THE ELEMENTS OF GRAPHIC STATICS

124

be represented by the curve

because the load

has only just reached B. If we now advance 2 the front load to E, EEj will be the bending moment is at D But at this instant that it causes at E. 2

and causes bending moments represented by the triangle AD 2 B. Consequently its effect at E is repreEE 4 Then EE Make sented by EE 4 will be the maximum bending moment at E, and .

Fio. 101.

In this

so on.

66.

Wj,

Ill,

AC.X^B

Maxi-

Suppose a train of loads cross a simple beam AB to 102) Then as long as the leading load

Shearing Force.

etc. (fig.

from right

get the diagram

maximum

bending moment. Loads. Concentrated Train of

the diagram of

mum

way we

lies to

force at

left.

the right of a given section C, the shearing will be the same as the reaction at A and

go on increasing as the loads approach Let the loads be set up from A in order to scale,

will therefore

as shown, and let these loads be also set off in re-

versed order with the leading load at B. Using x B as a pole, construct a link-polygon ABDEFG.

Then the ordinate y between the

first

link

and

SHEARING FORCE AND BENDING MOMENT

125

the link cut by the.ordinate at any section of the beam will represent the moment at that section for

Thus the assumed position of the loads reversed. is loads reversed of the C the moment

at the point

measured by the product of y l in the length scale, by / in the force scale. But this is the same as the

FIG. 102.

sum

of the

which

moments

of the actual

RA x /. x RA I = y x 1

Hence we reaction at

and when

RA =

C measures

A when

the

the leading load stands at C the leading load is not small relatively to

maximum

when

see that the ordinate at

the loads which follow the

loads about

equal to

;

it,

this value of

will be

shearing force at the section the leading load is relatively small, this

;

but

may

not be the case, because the increase in the reaction

THE ELEMENTS OF GRAPHIC STATICS

126

due to advancing the loads may be greater than the decrease due to subtracting the front load which has In order to test this, let the crossed the section C.

C and

second load be advanced to

sponding position of r Then new reaction at ; but since

of 2/ 2

the

section

so that

now on

shearing

we make

be the corre-

MH will be the

.y.2

the actual

let

force

the will

W MK lf

left

be will

value of the shearing force at C, and we can therefore now see whether the shearing force is be the

new

greater than before or not by comparing the values of

and MK. T/J Maximum Bending Moment.

be the

Let AB (fig. 103) the etc., given train of 2 This case is the most difficult to

beam and W,,

loads crossing

it.

Fio. 103.

find a direct solution for,[ because, in the first place we do not know which combination of loads will

cause the

maximum

bending moment, neither do load it occurs or at which

we know under which section of the beam.

The usual procedure

therefore

SHEARING FORCE AND BENDING MOMENT

127

draw link- and vector-polygons for the given loads, as shown in fig. 103, and to extend the first and last links indefinitely. Now if the beam AB be brought into any position as indicated, and verticals be drawn through A and B to cut the linkpolygon in Aj and B p the closing line A. 1 B 1 will is

evidently give the bending moment diagram for the loads which are on the beam, and the maximum

moment

bending be scaled

maximum

for

position

bending moment, however,

necessary to shift the

and

the assumed

which

to find

may

In order to arrive at the absolute

off.

beam

will

into various positions,

of these gives the greatest vertical

This method

is usually carried out by of curves bending moment for selected plotting sections of the beam, as it is moved by equal in-

ordinate.

under the loads. The highest point of each then the maximum bending moment for the If this is done for a section to which it relates. series of sections, a curve drawn through the maxitervals

curve

mum

values at each section will be a curve of ab-

solute

maximum

Schlotke

bending moment. Lehrbuch der graphischen Statik

has given an ingenious solution for the case when the span is longer than the load, so that no loads enter upon or leave the beam whilst these are shifted a little to one side or the other of their

maximum given

01, 12,

Let \V t

position.

loads etc.

be any

(fig. 104) represented by the vectors With any convenient polar O construct

a link-polygon abc g and produce the first and last Then a vertical through Q relinks to meet in Q. Draw in the of the resultant. presents the position .

128

THE ELEMENTS OF GRAPHIC STATICS

beam in any position A B, and at to the total load = 05 and join

set

AC=

cutting the in the stepped figure or

resultant in D, and draw load-line E6FGHIJKLMNP,

where CF, GH,

etc.,

FIG. 104. etc., set down in order, and the horizontal through D cuts from b where starting the first load. Bisect A S, B P in X and Y and join

are the loads 01, 12,

XY. occur

Then when

or points in

the the

maximum beam

which the

bending

moment

will

drawn through the point line

cuts the vertical

SHEARING FORCE AND BENDING MOMENT In the present case

steps of the load-line.

the steps the position

moment.

and

the

it is

cuts

of these gives the

shown

that the area

TLV

of the triangle

moment

bending

which

tell

value,

A B or A 2 B 2 is maximum bending

Therefore

beaih for

In order to

maximum

absolute

MN.

129

represents the diminution in which occurs in moving the beam

from the position AjB x to a position coinciding with the base LV, after which the bending moment increases again, until when it reaches the position it has increased 2 again by the area of the tri-

A B2

VMU.

The position AjBj or A 2 2 will thereangle fore be the position of absolute maximum bending

TLV

moment according as the triangle is greater or less than the triangle VMU. In this case therefore A^j is the position required, and if we now

A lt B to cut the link-polygon A 3 B 3 will be the closing line which

erect verticals through in

A3 B3 ,

the

line*

determines the bending moment diagram, and the value will be given by the ordinate which

maximum

passes through T. 67, Indirect Loading,

loading has been

.direct,

usually indirect, that

In the previous cases the but in practice it is more to say, the load acts on the

main girders through the medium of cross-girders. The load on any panel is then distributed so that it acts on the cross-girders which support the ends of the panel. I.

Maximum Shearing

Let

Loads,

Wj, W W W 2,

from right to

left.

direct loading,

Force for a Train of

girder (fig. 105) and let be a train of point loads crossing it

be an

66 for Proceeding exactly as in reverse the loads and construct a 9

130

THE ELEMENTS OF GEAPHIC STATICS

link-polygon with the span for polar distance. Suppose now we investigate the maximum shearing force in the second panel from the left. When the leading load arrives at the panel-point C, the reaction at A will be C^CX, which will therefore be the maximum

shearing force so

far,

and

only remains to investi-

gate whether it will become greater than this value. As the maximum shearing force will always be greatest when a load arrives at the panel-point, let

Wa FIG. 105.

be moved up to

;

then

will be at

Dj and

From this, however, DjDjj will be the reaction at A. must be subtracted the downward pressure at E due to the load

and

C^.

join

Then

Set up EjE 2 = x the ordinates of this line

standing at

represent the pressure at E as the load Wj rolls across the panel CE, and therefore represents the pressure at E when the load is at Subr

D^g

tracting this

viz.

DjDg, the

the shearing force at C when the 2 D 3 is has reached this point, and since 2

remainder load

from the reaction at is 3

SHEARING FORCE AND BENDING MOMENT found to be

maximum draw C 2 F maximum the

less

than

C^,

the latter value

moving

load.

which

gonals due

to the

the

due

maximum positive way the diagonals may be deter-

In this

and negative stresses in mined. The maximum

occur in

will

the

shearing force in the panel, and parallel to the diagonal, then C.2 F tensile force

131

the

tensile stresses in the dia-

dead load

be found in the

may

ordinary way by calculation or by means of a force diagram. The forces due to the live and dead loads

should then be tabulated, and

the compressive or

due to the live load exceeds the tensile stress due to the dead load in any diagonal, then the panel containing that diagonal must be counterbraced, because it is assumed that the diagonal negative stress

members

are incapable of resisting a compressive compression occurs therefore in the

When

force.

one diagonal the other deformation. '

in tension

Maximum Bending Moment.

II.

tem

of point loads

W W W

and prevents Let any sys-

lf 2 3 (fig. 106) say cross a span /, the loading being applied to cross-girders or otherwise transmitted at intervals to the main

girders.

In the present case

let 1, 2, 3, 4,

5 be the

Then we

require to know first of all cross-girders. the position of the loads which will cause maximum bending moment at each of the points, 1, 2, etc.

This see

may "

series)

be very easily found as follows

Moving Loads by Influence Lines

(for "

proof

in this

Set up at B vectors Jt represent the loads

1, 1 2,

W W W

through the points

1, 2, etc.,

2 3

draw

as shown, to

Join AO,

and

lines la, 26, etc.,

132

THE ELEMENTS OF GBAPHIC STATICS

parallel to AO.

Then the

loads which are cut by must act at the

these lines are the loads which

corresponding points to cause maximum bending moment there. Thus the lines from 1 and 2 cut the load Wj.

Hence Wj

will stand

maximum bending moment

both at 1 and 2

when

occurs at those points.

FIG. 100.

Similarly,

will stand at 3

bending moment

maximum

and so on.

In order

when

at that point,

there

to find the numerical values of these

maxima, place

the girder first in position AjBj so as to bring the point With any convenient pole O 2 under the load W,.

(and making the polar distance an even number of tons on the force scale for the purpose of simplifying the arithmetic), construct

a link-polygon for the

SHEAEING FORCE AND BENDING MOMENT

133

given loads. Project A p B t to a lt b r Then a-J)\ will determine the bending moment diagram and the ordinate y l in linear units multiplied by the polar distance in force units will be the maximum

moment at the point 1. Similarly, if we the place girder in position A 2 B 2 , so that the point 2* lies beneath load 1 and drop perpendiculars A 2 a.,, bending

6 from its ends to cut the link-polygon in a 2 6 2 2 2 the line a 2 b.2 will determine the bending moment ,

diagram, and the ordinate y., will give the maximum bending moment which can occur at the point 2. Also

2/ 3

the

same way

bending moment

under load

and so

will give the

when

at 3

the point 3

maximum is

brought

on.

The determishearing force and

68, Uniformly Distributed Load. nation of the

maximum values of is much more readily

bending moment

effected for a

uniformly distributed load, and for this reason it has been a common practice in this country to work out the values of distributed loads for various spans and

assumed axle loads which will produce maxishearing forces and bending moments at least as great as those which the concentrated loads will cause. There seems to be no particular reason why this indirect and more or less indefinite procedure

various

mum

should be adopted in these days when the use of lines leads to a reasonably simple and

influence

direct solution, but is

has been widely adopted and

simple in application, especially for girders with

parallel booms.

Maximum

Shearing

Force.

In the case

girder divided into equal panel lengths, it may be shown (see "Moving; Loads by Influence Lines")

134

THE ELEMENTS OF GRAPHIC STATICS

that the position of the load

which causes

maximum

shearing force in any panel can be at once found by dividing the span into a number of equal parts one less

than the number of panels.

Thus

fig.

107

where there are given equal intervals we divide the span into four equal parts at a, 6, and c. Then when the head of the load reaches a maximum shear will occur in that panel.

When

reaches

this position

FIG. 107.

of the load will cause

maximum

sponding panel, and so on.

shear in the corre-

Then

in order to avoid

the trouble of calculating the correct value of the maximum shearing force, the following graphical

method may be used, the proof book referred

to above.

Let

which

given in the

AB be the girder

and

let

w be the load per foot run which passes over it. On a base-line A B set up AjA = \wl, that is to say, one1

half of the load

and

join

AoB r

on the whole span when covered, Divide the into n - 1 span

equal

SHEARING FORCE AND BENDING MOMENT parts where n is the the span is divided.

number of Here n =

135

which n - 1 = 4.

intervals into 5,

therefore

Suppose now, for example, we require the maximum shearing force in the interval 1 2. Drop a perpen-

dicular from a, cutting in C, and project C Join A 3 B X cutting a perpendicular horizontally to A 3 from the point 2 in D, then is the maximum 1 .

possible shearing force which will occur in the interA similar construction will give the maxival 1 2.

mum

values for the other intervals, and also the In the latter case set negative values.

maximum down

B B2 = X

through a in

$wl and join AjBg cutting the vertical Project b to

and

Then a perpendicular through the point line in

and the ordinate

join AjBg. 1 cuts this

measures the maxi-

mum

negative value of the shearing force in the interval 1 2, and similarly for the other intervals.

Maximum Bending Moment. As regards the maximum bending moment nothing need be said in addition to what has already been said for a uniformly distributed load covering the span, because the maximum bending moment will occur at all sections of the span when the load completely covers it, and therefore apart from considerations of dyna-

mic

action, there

difference

between the case

moving load and that of a static load. is therefore dealt with as in 57. a

This case

For further treatment of moving loads, the volume on " Moving Loads by Influence Lines and Other Methods," already mentioned, may be referred to with advantage, where the subject is more fully dealt with.

CHAPTEK

MOMENTS OF AREAS,

ETC.

69. Graphical Determination of the First and

Second Moments of a System of

Parallel Forces, In later developments of the subject it will be necessary to find the first and second moments of a of quantities about an axis, and occasionally also the third moment where by the first, second, or

system

moment

meant the product of the quantity first, second, or third power of its distance from the axis. Thus suppose, for example, that a small area or mass be supposed to act at its centroid, and that a number of such quantities whose magnithird

into the

tudes are ap a 2 tant y lt

?/ 2

etc.,

etc., lie at

the points

from a given (136)

axis

A A2 lf

XX (fig.

etc., dis-

108).

Let

MOMENTS OF AREAS, vectors

1 2, 2 3, etc., be

and a pole

ETC.

drawn

137

parallel to

be taken with polar distance

XX, Also

'p.

be drawn through A x A 2 A 3 etc., parallel to XX, and a link-polygon abode constructed, and its in b lt c lt d^ sides produced to cut the axis Then let lines

b^ on the axis between the first and last points measured in the scale of lengths and multiplied by the polar distance p in the scale of the magnitudes of a lt a 2 etc., will give the resultant of the intercept

the

moments

first

of these quantities about the axis.

For the resultant moment

Proof.

etc

Now c.b, -V*

in the similar triangles bc b l9

Vi c^

.-.

Similarly,

x x

P = 1 p = 1.2

x yL

a, x

= =

a2 x

y.,

y.2

= 2 3 x y.3 a 3 x y z + c^i + d^Jp = a' y + a.2 y z + a z y 3 = sum of the moments about XX.

and .-.

001,

1 -

dfa x p

(bfr

This result was proved also in chapter

somewhat

in a

manner.

different

Further, the second moment of the system of loaded points about the axis will be represented by the area edcbb 1 between the first and last links multiplied by twice the polar distance.

For the

Proof.

second,

moment

of the quantities

is 2

i2/i

Now 6 1c 1

2 2

a 32/3 2

area of the triangle 66 1 c 1

etc

%b l c l x y l and

cldl

7/ 2

Also the area of the triangle

where

cldl

^- and ,

cc^

so on, so that

THE ELEMENTS OF GEAPHIC STATICS

138

the whole area

= bb^ +

cc l d l

+ dd^ 2

2p x

.'.

= the a,,

a lM

a.,v. = 4. flit/, 321-4. i -ML P P P whole area = a-flf 4- a 2 2/ 2 2 + a a 2/3 2

sum

of the

etc.,

second moments of the quantities XX, the area being

about the axis

measured in linear units on the and the polar distance p in the

scale of the scale of the

drawing magni-

tudes of the quantities.

The second moment of a system of areas or masses about any axis is commonly called their moment of inertia about the axis, but the expression is not a good one. If we find the distance from the an area or mass which is equal to the sum of the areas or masses, so that the second moment of this area or mass about the axis is identically the axis, of

same

as the resultant of the given areas or masses, is commonly known as the radius

then this distance

In other words, if A of the system. the sum of all the represent quantities a, and k x be the value of their radius of gyration, then of gyration

k? =

Hence

The

a!?//-'

a.2 y.r

+ a3 ?/ 3 - + 2[<

etc.

= S[ay2 ].

radius of gyration of a system of quantities

same purpose in regard to second moments that the centroid or centre of gravity does for first moments, by enabling the whole system of serves the

points to be regarded as concentrated at a single point, and so simplifying the subsequent treatment,

but the distance of the centroid of a system of areas

MOMENTS OF ABEAS, or masses from an axis

always

139

ETC.

less

than the radius

of gyration.

Example.

find the centroid distance

and the

radius of gyration of a system of points loaded with quantities whose magnitudes are 2, 3, and 4 and lying at distances of from a given axis.

and 2 units

re'spectively

Then since the moment, of the resultant = the sum of the moments of its components, (2

4) x their centroid distance

-2x1+3x3+4x2 = 2

.-.

the centroid distance

N.B.

+ 8 = is

19.

be observed that

It will

moments about is

the axis

2 if

units

the

sum

2 -11. of the

zero, the centroid distance

zero also, or the axis passes through the centroid. + 3 + 4) x &,2

Further, since (2

2 x

fc,s

.45_

+ 4 x 2* = 45, ^ = v/5 = 2-24.

3 x 3* 5

Let 70. Graphically, let 2, 3, 4 be the loads.

and

the axis (fig. 109) be Set off vectors parallel

to the axis to represent these loads

and take any

= sum of the polar distance (but preferably a length Then the Construct link-polygon aocde. vectors). moment about XX. But the = sum of the loads x the centroid distance x from the axis. Therefore sum of loads x x = ae x p. But if we take p = sum of the loads, then x = ae where x is the centroid distance = 2'11 as ae x

p is the moment about

before.

first

THE ELEMENTS OF GRAPHIC STATICS

140

= second moment of the 2j> sum of the loads x A;,2 sum of the loads, k,- = 2 area abode.

Further, area abode x loads about /. if

XX = the

Now the area

measured

2-5 sq. inch.

by calculation. 71, Centres of Gravity.

Case

.-.

Both Loads

Centroid of Two Loaded Points. be any two points at which (fig. 110)

Positive.

and

k* = 5 as

Let

we may

suppose areas or masses to be concentrated, and

let

LA, LR be the magnitude of these loads. Then the resultant of these must lie on the line joining A to B, and since the moment of their resultant acting at the centroid C is equal to the moments of its components,

take

moment

centre,

MOMENTS OF AKEAS,

LA since the

AC - L B

moment

141

ETC.

BC = O

of the resultant

FIG. 110.

inversely as

the loads.

Graphical Construction. Set up AD = LA and BE = LB where AD and BE are parallel lines. Join DE cutting AB in C. AD = BE

Then by

= '-gn

similar triangles -r^

Hence

divided in

inversely as the loads,

and

therefore their centroid.

FIG. 111.

Case tive.

II.

One Load Positive and

In this case, referring

the Other

to fig. Ill,

Nega-

THE ELEMENTS OF GRAPHIC STATICS

142

LA .-..

-^ LA

^TH BC i

AC + L B

BC =

and since the moments about C

are of opposite sign, the point C must lie on AB pro= LB and duced, as shown in the figure. Set up

BE =

parallel to 'one another.

produce to cut as required.

produced in C.

Hence the

Join

Then

resultant lies

and

AC = LR *

pro-

duced, and on the side of the greater load. 72, Centroid of a System of Loaded Points on a Straight Line, This is identically the same

FIG. 112. of finding the position of the reHence if we sultant of a system of parallel forces.

problem as that

MOMENTS OF AREAS,

ETC.

143

construct a link- and vector-polygon for the loads, first and last links will de-

the intersection of the

termine a

line passing

through the centroid.

Thus

ABCD

with two circular (fig. 112) be a rectangle areas cut out of it, the actual area may be regarded as consisting of a positive part ABCD and two

negative parts, each acting through their centres. Hence if we set off 1, 1 2, 2 3 to represent these areas in magnitude and sense, and construct a linkpolygon with pole O, then the intersection of the

and last links fixes the position of the resultant of the area, load passing through the centroid and as the centroid must lie on the axis of symfirst

metry, its position is determined. 73. Centroid of a System of Loaded Points

not Lying

a Straight Line,

Let

L L 2 L 3 L4 ,

FIG. 113. 1, 1 2, 2 113) be the loads, and Construct the vectors. link-polygon ab if

we produce ab

meet dc

(fig.

in /, a line

their

Then

through

144

THE ELEMENTS OF GRAPHIC STATICS

parallel to the direction of the loads will fix the

L and L2 and act to at g lt the L., Similarly, Lj centroid of L lf L 2 and L 3 must lie on the line ^L 3 and it must lie also on the line through k in which the first and fourth links of the link-polygon inter-

point

f/ 1

L^L.,

which

the centroid of

we suppose ,

Hence

sect.

lies at g<2 .

and producing the the line through gf

L4 2

<7 3

which

first

and

Again, joining fifth links to

<7 2

meet

in ra,

parallel to the loads intersects

then the oentroid of the whole

system of loads. 74, Centroids of Areas,

When

an area has an

symmetry, the centroid will always lie on that axis, and when there are two axes of symmetry

axis of

the centroid will

lie

at their point of intersection, so

that the centroids of figures like circles, rectangles, equilateral triangles, etc., can always be found by inspection.

Centroid of a Triangle,

ABC (fig.

we suppose a triangle number of thin

114) divided into an infinite

Fio. 114.

BC, the centroid of each of centre, and therefore all of them

strips parallel to its base

these will

lie at its

MOMENTS OF AEEAS, will lie

base.

on the median

line

By similar reasoning we also lie on the median BE,

must and by geometry

triangle intersect

ETC.

145

which

bisects the

may show

that

which bisects AC, it is known that the medians of a in a point which lies one-third of

the

way along any median. This fixes its position. 75, Centroid of a Trapezium, First, the cen-

EF (fig. 115) which joins and F of the parallel sides. Draw CH parallel to BA, and let G p G 2 be the centroids of the parallelogram and triangle so formed. Join G 1} G 2 and produce to cut the sides CB and AD

troid

must

lie

on the axis

the mid-points

produced

and L.

G AL X

are congruent = x and write

Then

(i.e.

since the

As GjCK,

equal in every respect),

DL = y, x + BC = y + AD (1), and since .the As G ML, G KC are similar, and ~ AH G M = G C, x + BC = 2ML = + y\ we

AD - AH + 2y (2). + AD = AD - AH + 2y. =

by (1) and

(2)

= BC, and /. Hence make DL = BC and .:.

(1)

BK =

= AD.

AD.

Join

KL,

and the point G in which it cuts EF is the centroid. The above construction is, however, sometimes 10

146

THE ELEMENTS OF GRAPHIC STATICS

inconvenient, because the points K or L required may In this case lie outside the limits of the drawing.

the following method

may be used (fig. 116) Make CC^ = J(AD - BC) and draw C^A, parallel to the diagonal C A, cutting the median line EFinG. Then

the centroid.

Also by taking moments about the base it may be shown that the distance of the centroid from the

FIG. 116.

/i/B

base

equal to

where h

the distance

between the parallel sides, and B, b are their lengths. Also when one side is vertical, as in fig. 117, the distance of the centroid from the vertical side

given by the formula

Third Method. Draw in the median Join AE and CF, and make EG X

118).

FG = 2

JCF.

Join

G^, cutting EF in

(fig.

- AE,

and

line

Then

the centroid required.

Polygons bounded by straight lines

may

be split

MOMENTS OF AEEAS, up

into triangles,

and the areas

147

ETC.

of these being sup-

posed concentrated at their centroids, the centroid of the whole may be found by the method for a

system of loaded points (see

73).

FIG. 118.

76, Radius of Gyration for an Area. Equimomental Points. Definition. If a system of points can be found such that when an area is divided up and concentrated at these points, the moments of these loads about

area

itself,

any axis is equal to that of the then the points are said to form an equi-

momental system.

Equimomental System for a Triangle, Thus if BCD (fig. 119) whose area is A be supposed

a triangle

FIG. 119.

divided into three equal areas

and -^,

supposed concentrated at the mid-points

these are

of the sides

THE ELEMENTS OF GRAPHIC STATICS

148

F and G, then these points an equimomental system, and the

of the triangle, viz. at E,

so loaded form first,

second,

etc.,

moments of the loads concentrated same as for the area itself

at these points will be the

about

all axes.

Thus the second moment

of the triangle

about

bh*

its

base will be

A and

A if

k be the radius of gyration about the base,

Equimomental System be the area of the rectangle divides

into

- r. for a Rectangle, (fig.

120) then

two equal triangles each

its

diagonal

^ of area

and

Fia. 120. if

one-third of

of their sides,

in the figure.

be concentrated at the mid-points

we get the equimomental system shown Consequently the second moment of

MOMENTS OF AKEAS, the rectangle about the axis

2 x

//A

Also about

through

tas

Afe

its

149

ETC.

its

centroid

J^^bh =

XX, the second moment 2A //A 2

base

* *

T *(*)' AW

bh*

-5-

Any figure bounded by straight lines may in this way be split up into a number of triangles, for each which an equimomental system can be formed, and therefore a system for the whole area is found also, after which the moments for the loaded points of

may

be dealt with as already explained in

69.

Other simple cases of common occurrence however may be deduced at once from the above results. Thus for a hollow rectangle (fig. 121) the second

H FIG. 121.

moment about

llT

second

the axis

bh s 12*

an(^

moment

^ for

through the centroid

same formula

will also give the

I section (fig. 122),

which may

THE ELEMENTS OF GRAPHIC STATICS

150

between two recand depths H, h. 77. Relation between the Second Moment of an Area or Mass about any Axis and its Second Moment about a Parallel Axis through its be considered as the .difference

tangles of breadths B, b

Centroid,

Let

and

let

be an axis through the centroid (fig. 123), be any parallel axis distant d from it.

Let a be a very small element of the area or mass Then its second y from the axis CC.

distant

FIG. 123.

moment about XX is a(y + d)'2 and the second moment of all the elements about XX will be the sum of an infinite number of infinitely small elements which we may represent by 2a(?/ + d}1 = ,

Now

2a7/-

denotes the summation of

moments about

the axis

summation of the ments about CC. But

the

of the

moments

resultant,

and

this

all

the second

CC =

first

Also 2a?/ denotes TCmoments of all the ele-

this is zero, because the

equal

the

moment

sum

of their

by definition passes through the

MOMENTS OF AREAS, centroid,

and

fore zero.

moment about

its

the axis

Finally the last term the whole area.

151

ETC.

equal

there-

Ad'2

2a = Hence Ix = Ic + Ad 2 and since Ix = Afcx 2 and

since

78.

Example. To find the second moment of a an axis through its centroid parallel to

triangle about its base.

Since the second

shown

to be Ix

of is

moment about ,

we have

(fig.

the base has been

124)

79, Polar Second Moment, The second moment an area about any axis perpendicular to its plane

known

axis.

Thus

its if

polar second

fig.

moment about

that

125 represent any plane area and

an axis be taken through any point O perpendicular to the plane, the second moment of the area about this axis is 2a2 2 where z is the distance of an ele-

THE ELEMENTS OF GBAPHIC STATICS

152

ment from the

axis.

Let OX,

- x + >2

if,

2oz2

Then

rectangular axes through O.

be any pair of

since

= 2ox 2 + 2a#2 = Iy +

where IY and Ix are the second moments about the OX and OY respectively.

axes

80. Example. To find the polar second moment of a circle about an axis through its centroid.

FIG. 125.

Suppose the infinite

number

circle

(fig.

of triangles.

divided

126)

Then

into

we suppose

one-third of the area of each of these triangles concentrated at the mid-points of its sides, we get ulti-

mately one-third of the whole area of the circle concentrated along its circumference, and two-thirds along a concentric circle whose radius the

half that of

first.

Hence the polar second moment IP

Ar*

MOMENTS OF AREAS, From moment

153

ETC.

this we can easily deduce the second of a circle about a diameter ; for if XX,

be two diameters at right angles, IP = I x + IY. But I Y = Ix. or I x = JIp. /. 2I X = Ip

Y FIG. 126.

Hence the second moment diameter

one-half

polar axis through

From

its

~?p>

the above

its

centre,

a circle about

second

moment about

its

the

and since

~~ ~A

we deduce

once that

for

annulus

where

and d are the external and internal

dia-

meters.

81. Graphical Determination of the Moment of Fig. 127 represents a cross-section asym-

Inertia,

154

THE ELEMENTS OF GRAPHIC STATICS

metric about

its

NN.

neutral axis

It consists of

two

channel bars (BSC 26) 12 ins. x 4 ins., area = 10-727 in. 2 The whole area is divided into convenient parts which are regarded as forces acting through their centroids, and the areas of these parts are set off

along a vector line from

10 as shown.

pole

'I'

FIG. 127.

then taken with alpolar distance p (which for may be made one-half the length O'lO).

convenience

The

intersection of the first

and

last links of the link-

polygon drawn with this polar distance determines the line NN through the centroid of the cross-section,

which is therefore the neutral axis. The area abc between the link-polygon and its first and last links

MOMENTS OF AEEAS, was then measured and found

155

ETC.

to represent 21 '5 in. 2 to

the scale of the drawing. Now we have shown, 69, that this area multiplied by twice the polar distance in the scale of areas will be the second moment of

the whole area about the axis

NN.

But twice the

whole area polar distance was made equal to O10 of the cross-section, which was in the present case 47'8 in. 2 Hence the value of the second moment of the cross-section about

21-5

in.

= 1030

ins.

The Momental

82,

its

neutral axis

47*8

in. 2

Ellipse.

The moment

always greatest and least about two axes at right angles to one another, and these are known as the "Principal Axes of Inertia," and the principal radii of gyration can be found by dividinertia of

any area

ing the moments of inertia about the principal axes by the area of the section and taking the square root. If an ellipse be now constructed of which

these radii are the semi-major and semi-minor axes, the radius of gyration about any central axis may be

found by drawing a tangent to the

ellipse parallel to

that axis, the perpendicular distance between the This lines being the radius of gyration required. ellipse

known Thus

whose

centre is the centroid of the area is " " Momental for the area. Ellipse of an area and (fig. 128) be the centroid

as the

UU, VV

are the principal axes of inertia, an ellipse k\, k\j for semi-axes is the momental

ABA'B' having

ellipse, and if it be required to find the radius of inclined gyration about any other central axis to at an angle a to UU, draw xx parallel to

touch the

ellipse.

Then the perpendicular distance

THE ELEMENTS OF GRAPHIC STATICS

156

kx between

about

;

A*2 and

XX and YY

fcu

fcu is

the radius of gyration

sin 2

cos2

a ( fc u -

a(kf

fcv

)

(1)

2 )

(2)

being perpendicular to one another.

addition & x*

By that

and xx

or by calculation

2 ,

Y or I x

to say, the

= 2&u 2 - (^u 2 - &v 2 = + IY = Iu + Iv (3)

)

;

sum

of the

moments

of inertia

FIG. 128.

about any two axes at right angles is constant and equal to their sum about the principal axes. 83,

When

metry, as

the Section has an Axis of Symthe Case of a Channel-iron, for

example, this Axis

Axes

always one of the Principal

of Inertia,

Example. which

Given the channel-section

fig.

129

for

Iu Iv

= =

26-03 ins. 4

3-82 ins. 4

= 2'33 ins. = O89 in.

A =

4-?9 ins. 2

MOMENTS OP AEEAS, momental

to construct the

the

moment

ETC.

and

ellipse

157

to determine

about an axis bisecting the

of inertia

principal axes.

Since Ix

14-925

IY,

we have I x =

and since by symmetry

Iv,

i(I\j

Iv)

J(26-03

3-82)

in. 4 ,

14-925

and

3-11 ins. 2

4-79"

1-76

in.

FIG. 129.

84,

Asymmetric Sections,

When

the section

has no axis of symmetry, the position of the principal axes must be obtained by finding the second moments about three non-parallel axes. This is best done by finding the second

moments

Ix, IY

(fig.

130) about

any pair of rectangular axes XX, YY, and also the second moment Iw about a third axis bisecting Then the angle a the angle between the other two. which the semi-minor axis of the ellipse makes with

XX is

given by the formula tan 2a

IY IY -

2Iw (4).

158

THE ELEMENTS OF GRAPHIC STATICS

Iv and

I\-

can then be found from the formulae Iv

and

= ilx +

+ :== 5?

(

)

Iv -

(6).

Angle-iron, 10 ins. x 4 ins.

Example.

(fig.

130).

-f~

FIG. 130.

A =

9-00

Ix = x =

ins.-

3-20

T\y

\vaa

ins. 4

92-11

ins.

8'77 ins. 4

-987 ins.

moment

found by constructing a second

area to be 68-5 ins. 4 .-.

Then tan 2a .-.

2a =

k\y

92-11

= 2-70 ins. 2 x + 8-77 ^. u H ?7

25',

90-81)

- -207.

+ 8-77+

..Iu^-i(92-ll 4(100-88

- -433.

IT 42V.

and a =

.-.tan u =

68-5

95-85

92-11 - 8-77

ins.

^230^ 4

.-.

3-26 ins.

MOMENTS OF AREAS, Also

(100-88 .-.

90-81)

'748

159

ETC. 5-035.

in.

Centroids of Lines, Areas, and Solids.-

Trapezium:.-

|- J(*-^)

131).

(fig.

k- o-

FIG. 131.

Circular arc:

OG =

Semi-circular arc

r sin a

180

OG =

re s

-6366r.

Circular sector:

OG =

(fig.

132).

FIG. 132.

Semi-circular area

Circular segment

OG = OG =

4 Q O

r .

r~-

-4244r.

x area

(fig.

132).

160

THE ELEMENTS OF GRAPHIC STATICS o

Parabolic segment

--X

;

(fig.

133).

FIG. 133.

Pyramid or cone

;

at J height from base.

Segment

sphere at

from base, where h

Moments

^2r

/tV~

the height of the segment.

of Inertia of Areas,

bh 3 Rectangle: I

(fig.

134).

i..

FIG. 134.

Square about diagonal Triangle about base

I,,

bh* ^c"(fig-

(

n 8-

MOMENTS OF AREAS,

ETC.

161

ISft Fia. 135.

FIG. 136.

Ellipse

Semicircle about T

diameter:

I,,

nab*

4 -

(TT

(fig.

137).

8 \

(

138 >

Sir)

FIG. 137.

FlG

Parabolic segment T I about base

175**'

-i-

(

g- 139).

-*,

FIG. 139.

Hexagon about diagonals :

_ 5 >/3r>4 _ TfiT

(fig.

140).

THE ELEMENTS OF GEAPHIC STATICS

162

= diameter

-638B*(fig. 141).

FIG. 141.

Corrugated Plate, BJl 2

where

(fig

142).

H--B-H I

FIG. 142.

Flat Corrugated Plate,

where

/i,

/t,

= i(H + = |(H -

)

FIG. 143.

J(B + 2-6Q

- i(B -

2-60-

CHAPTBB

XI.

STRESS DISTRIBUTION ON CROSS-SECTIONS. 85. Load Point or Centre of Pressure, By the Load Point or Centre of Pressure of a section is to be understood the point at which the resultant of a given system of forces acting upon it cuts the plane

When a load is uniformly distributed over an area, the resultant of that load will pass through the centroid of the area, and therefore in

of the section.

this case the centre of pressure will coincide

the centroid.

with

When, however,

as often occurs, the such that the intensity

distribution of pressure is follows a straight-line law, that

is to say, it varies as the distance from a given line, the position of the load point will be determined by the rate at

which the intensity of pressure varies, and the line along which the pressure is zero is known as the Neutral Line or Neutral Axis. Thus when a plane area

submerged

in a liquid, the intensity of pres-

sure increases with the vertical depth below the surface, being zero along the line in which the

plane produced cuts the surface of the liquid. This line is therefore the Neutral Line, or as it is usually called in this connexion, the Line of

This line is strictly analogous to the Neutral Axis in the case of beams or other struc-

Flotation.

(163)

THE ELEMENTS OF GRAPHIC STATICS

164 tures,

which the stress intensity varies as the

distance from a line, and the relation which

are

about to deduce between the load point and the line of flotation is equally true for the load

point

and the neutral axis of a structure. 86, Total Pressure on a Submerged Area, If we consider the pressure on a horizontal plane immersed in a liquid whose weight is w per unit volume, the pressure per unit area at any depth y (fig. 144) below the surface being equal to the weight of the column of liquid above it will be wy,

FIG. 144.

and since the pressure tions,

of a liquid is equal in all direc-

will be the pressure per unit area in all

directions at a depth

then

consider an

infinitely small area a at a vertical depth y below the surface, the pressure on it will be wya, and the total pressure on the whole surface, whether plane

or not, will be

the

first

moment

^[wya] w^,[ay] where 2[a7/] is of all the elements of area about

the plane of the water surface. Hence P = 10 x first moment of the area,

= wAy depth

(1)

where A

of its centre of gravity,

the area and y

the

165

STKESS DISTKIBtJTION

Example

mersed so that

sphere of 2

its

centre

diameter

ft.

im-

below the surface, per cub. foot. Find

ft.

in fresh water weighing 62'4 Ib. the total pressure on the surface.

Here P = ivky =

= 2.

2358

Ib.

rectangular area 4

Example mersed in salt water weighing 64 with

its

62-4 x 4jrr2 x 3

centre of gravity 5

ft.

Ib.

approximately. ft.

x 3

ft. is

im-

per cub. foot, below the surface. Ib.

Find the pressure on one side. P - 64 x 12 x 5 = 3840 Here 87,

748-8*

Ib.

Position of the Load Point on a Plane

Submerged Area,

Let

(fig.

145) be the plane

FIG. 145.

inclined at

any angle

and

let

its

centroid

The plane produced cuts the the load point. water surface in a line through X which is the axis of notation, and the moments of the pressures on every element of area about this axis must be and

moment of the resultant pressure acting L. Consider a small element of area a at through Then the pressure on it is distance y from X. equal to the

way

sin

flotation

6, is

and the moment about the axis

way

sin 6 x y.

THE ELEMENTS OF GRAPHIC STATICS

166

x y2 = w sin 6 2[ay 2 ] moment about the axis.

Hence

But .*.

But

= Ay =

7/2

wAt/j sin

= w sin

x second

ne second moment about the

moment about the about axis second moment _ first moment about axis

the

first

axis.

Let Special Case, Rectangular Cross-Section, be a (fig. 146) rectangular cross-section with

FIG. 146.

its

Then the pressure

base in the water surface.

per unit width will be tion be vertical

P =

ivh

moment moment

second -first

P = wABy, and x i/t = -=-. h* =

the sec-

2; '

load point for a rectangle, whether vertical 2 or inclined, lies therefore at = of its length from the o

The

water surface, the resultant pressure being normal to the surface.

STEESS DISTRIBUTION 88.

16?

Relation Between the Neutral Axis and

the Load Point Since y 2

(fig.

= LN

147).first

NC + CL = .-.

CL =

moment about axis moment about axis

second

+ or

but

NO =

NC CL = .

Fio. 147.

mathematical language states that the radius of gyration is a geometric mean between NC and CL. Now if the form of the area is known, its radius

which

of gyration

ko about the axis through

the. centroid

THE ELEMENTS ox

168

STATICS

can be determined, as well as the parallel to position of its centroid C, and therefore we have a definite relation between the load point and the neutral axis, from which either can be found when

known. Thus if we make CD = ko, and draw DL perpendicular to ND, cutting

the other join

produced in L,

NC CL = .

2 o by the well-known " mean fc

construction for a geometric 1 p. 23), and therefore the point

(see

Geometry,"

so obtained

the

or being given the load point, the corresponding position of the neutral axis can be found by reversing the construction.

load point

;

Example diameter

A circular pipe (fig. 148) 4 ft. 1. closed at the end and is subjected to

FIG. 148.

water pressure, the level of the water being 3 ft. above the centre of the pipe. Find the total. pressure and the position at which the resultant acts. First.

Total pressure

P = wAy =

62-4 x

= 2260 1

x 3 Ib.

Geometry for Technical Students," E. Sprague. Messrs. Lockwood & Son.

Crosby,

STRESS DISTRIBUTION Second,

ko 1

CL =

CD = to

169

ft.,

join

ft.,

ft.

or by construction set off

DN, and draw

perpendicular

it.

89.

Example

Any Area Immersed

Let ACB (fig. 149) be Liquid, any area, which in the present case has been taken as a parabola

for the sake of

comparing with the calculated result. of 4-45 ft. and a height of

The parabola had a base 2-5

ft.,

the point

ft.

being 1

ft.

below the surface of

NN, and was drawn

to a scale of 1 in.

The area was divided

into five horizontal

the liquid

of equal width, the half-lengths of which were taken to represent their areas, and were set A pole O was to 5. off along a vector line from then taken and the vector link-polygon abc g was drawn, its first and last links intersecting in g.

strips

horizontal through g determines the centroid

THE ELEMENTS OF GEAPHIC STATICS

170

At the same time the area abc

of the area.

x 2p, where p is the polar distance, represents the second moment of the area about the axis NN. This area was measured with a planimeter and

found to be 5-92 load point the

formula OL =

intercept /.

OL =

first

moment about

5-92 x 20 7

measure 4-45

ins.,

and since

OL =

1 1

.04

given by

equal to the

bh*

was found

= 2-66

ft.

l.)

x 4-45 x 2-5*

3-18

ft.*

jfg

+ Ad 2 =

3-18

7 ; 7 4l"x~2 52

= 50'48 ft.*, = OG = 2-52 ft.

7-42 x 2-52* since

For a parabola

Check by Calculation,

moment about NN ...moment about NN

x the polar distance p.

the distance of the

second -first

But the

Now

sq. ins.

from the neutral axis

2 66 fi| exactlv a gKi n g '

with the graphical result. 90, Determination of the Stress Intensity at any Point of a Cross-Section when the Load on it is

known

Let

(fig.

Magnitude and Position, 150) be any section of a

structure,

whose centroid is C, and let P be the resultant load on it acting at L, and inclined to the section at an angle u. Then if P be resolved into components S and T parallel and normal to the section, the first will be the shearing force and the second will be the normal thrust on the section, and since CN CL = k 2 ( 88), .

STRESS DISTRIBUTION

get

CN =

the neutral axis.

171

which determines the position If

we now make

the usual assump-

tion that sections which were plane before the load

FIG. 150.

was applied remain plane after strains will

obey a straight

sented by the line

AjN

(fig.

strain, the stresses

line

151),

law,

and

being repre-

and S

S lf S 2

will

FIG. 151.

represent the stresses at the centroid edges A and B of the section.

C and

at the

Let S be the stress at any distance x from the

Then

since S

kx where

constant the total stress on a thin strip of area a kx a, and the total stress on the whole area

axis

(fig.

152).

which must be equal and opposite

2[kx a] normal thrust T. .

Hence

to the

THE ELEMENTS OF GRAPHIC STATICS

172

T =

k2[ax]

.-. -T-

k x

first

moment

about

the stress at the centroid.

Hence the stress at the centroid mean stress on the cross-section. If

NN =

therefore

set

up CC 1

(fig.

equal to the

151)

T-

FIG. 152.

C and

join

NCj, we get the distribution of stress

over the section. 91.

Formula for the Stress at any Point.

Since the stress varies as the distance from the neutral axis,

we have

But

CN =

152)

its

PC ON"

where

^np-

the load point or

(fig.

CN + PC ON

e is

the eccentricity of

distance from the centroid.

+ r|r

where y

be regarded as positive or negative, according as

STKESS DISTRIBUTION is

on the same

173

side of the centroid as the load point

or not.

At the extreme edges of the section writing y 2 respectively,* we obtain the y l and y =

corresponding values of the stresses.

Example 6

ins.

A column

diameter

Find the

= +

d*_

3 ins.,

(

1-77

= -)

and - 1'06

3 1

-354

and

edge with 10 tons.

"354 ton/in. 2

28-3 sq. ins. ins.,

circular section

its

on the edges.

stresses

Here

.-.

loaded on

9 4'

354 x 5 and - 3 2

tons/ins.

Example 2. A wall weighing 120 Ib. per cub. foot has the dimensions shown in fig. 153, and is subject

LEVEL.

FIG. 153.

to a water pressure to within 1 ft. of the Find the stresses on the edges of the base.

top.

174

THE ELEMENTS OF GKAPHIC STATICS

of wall per foot of length

Weight

60 cub.

x 120

ft.

7200

Ib.

Distance of centroid from vertical face

7 -

Resultant water pressure per

* _ acting at 3

2-63

ft.

62-4 x 121

~~2

of to

ID.,

from the base.

ft.

Now

= 355

;. LH = 3f x -525 = 1-925 = CL = LH - CH = 1-925'- 0-87

ft.

.-.

7200/ 7

l-05^x^5N _ ~

1955

49/12 2

)

and 98'6

lb./ft.

92, Critical Distance,

1-055

ft.

7200 7 2

lb./ft.

It is frequently a condi-

tion of stability for masonry and concrete structures that no tensile stress shall occur on any crosssection.

Now it will be

approaches the edge

seen that as the neutral axis

a section, the stress will

decrease in intensity until it changes sign when the neutral axis first cuts it so that if the section were in ;

compression the stress will become zero when the neutral axis touches tension.

Hence the

and

will

afterwards change to

condition that no tension shall

occur on a section subject to thrust is that the neutral The corresponding distance of axis shall not cut it. the load point from the centroid is known as the critical distance, because so soon as the load point lies

beyond

this limit,

tensile stresses

will

occur.

STEESS DISTRIBUTION

Now when

the neutral axis

175

touches the edge

first

(fig. 154), the corresponding position of the load point will be such as to satisfy the relation

CN CL = .

and when

CN =

- y 2

k*,

CL 2 =

2/2

CN =

and when

Example

CL,

2 .

find the critical distance e c in the

case of a circular cross-section.

Since

Hence the

and y 1

= yz =

critical distance is

D <

J radius.

If,

and

therefore,

radius

describe a circle with centre

JE, this

circle

within which the resultant must

defines the lie

limits

in order that the

not change sign. This area is known as the core of the section, and therefore the core of a stress

may

may be defined as the locus of the load point whilst the neutral axis turns about the edge of the section, keeping always in contact with it but nosection

where cutting

it,

THE ELEMENTS OF GRAPHIC STATICS

176

radius

Find the critical distance of a hollow whose outer radius is R and whose inner

Example

circular area

An,

is r.

Example cross-section

(fig.

=g=

find the core of a rectangular

155).

Since for a rectangle k

/j2

and y ^ LA

T i

K d

->}->; FIG. 155.

Consequently ad,

CLi =

the neutral axis touch the side

\h.

In the same way, side ab, CL 2 = h l .

the neutral axis touch the

Also

and

will

be the

the load point when the neutral axis Now it touches the sides be and cd respectively.

positions of

may

be shown that whenever the neutral axis turns

about a point, the load point travels along a straight line, so that if we suppose the neutral axis to turn

from the position da to the position ab by rotating about a, the load point will travel along the line Similarly, as the neutral axis joining L and L. turns about the other corners 6, c, and d, the load l

STEESS DISTRIBUTION

177

L L2 L3 L which

point will trace out the rhombus therefore the core of the section.

Example

Find the

critical distances

and the

core for a hollow rectangular cross-section, whose outer dimensions are B, H, and whose inner dimensions are

b, h.

BH s>

bh*

6H(BH -

;

bh)

- hb s

6B(HB On

93, Rule of the Middle-Third,

hb)'

account of

the frequent occurrence of the rectangular section in walls, dams, arches, etc., this particular form

has a special importance, and therefore it should be noted that the critical distance along either of the principal axes must not exceed one-sixth of the depth. In other words, the resultant must not outside the middle-third of the depth, in the direction of bending. This is known as the " Kule lie

of the Middle-Third," and is of frequent use in the treatment of masonry structures, the primary condi-

tion of stability usually being that the line of pressure shall lie within the middle-third limits for all

the cross-sections.

94.

Asymmetric Section,

Example

common

section of this type in practice is the T section, which occurs in the case of walls with

counterforts and elsewhere.

Let

156) C be the radius of

(fig.

the centroid of the section, and Jc gyration about the axis through

y\> e c

= -

-7-,

and when y

it.

Consequently the points Lj and from C, represent the critical 12

Then when ec

2 .

2/2

L2 at

these distances

limits, according as

THE ELEMENTS OF GRAPHIC STATICS

178

the bending moment about the axis XX.

clockwise or contra-clockwise

***J 1M--NL\

PIG. 15C.

95, Graphical Representation of the Stresses, when the Load Point and the Critical Distances

Known,

are

Let

the load point, and

L L

(fig. lf

L.,

157) be the position of be the critical positions.

FIG. 157.

Set up at C a perpendicular CCj = s 0t the mean Join CjLj, CjL.,, and erect stress over the section. a perpendicular at L, cutting these lines or the lines

Then LE, LE X will represent produced in E, Ej. the stresses on the edges A and B respectively to the same

scale that

}

represents the

mean

stress s (>

STRESS DISTRIBUTION

Proof.

and since

For since et

+ )

(

179

91),

*o(l

Now

by similar triangles ^^r-

^'GL;

:=SO

rr=^.

/CV-CL \ (

CL,

)

If, therefore, we project E and E x upon verticals drawn through A and B, we get the points F and H. Then the line FH represents the distribution of stress

over the section.

96. When the Material is Considered as Incapable of Resisting Tension. For the best brickwork in cement mortar a compressive stress of from 200 to 300 Ib./ins. 2 is probably permissible and a tensile stress of 35 Ib./ins.*

When, however,

the

tensile stress is neglected, the position of the neutral

no longer be the same as before. Suppose NN to be its position (fig. 158), and L the position of the load point. Then, if s is the stress at any distance y from the neutral axis, and S is the stress at unit distance from it, then s = s y, and if 8 A be a narrow strip of area parallel to NN, axis will

THE ELEMENTS OF GKAPHIC STATICS

180

the total stress on

sny

8A and therefore the

resultant load

P = 2*.8A = *2SA.

(1)

Fro. 158.

Again the moment to the

sum

strips,

and

.-.

of the

x x /. s

Example

about

moments

must be equal

of the stresses in all the

8A x y = s 2[8A y*] 2[8A y] x x = s 2[8A y*] 2nd moment of shaded area 2* y

(2)

1st

Rectangle

moment (fig.

159).

FIG. 159. 3

2nd moment about

NN of compression area

by = --'

STKESS DISTEIBUTION

1st

moment about .-.

Also the

I?/

NN f(a

mean stress =

compression area

x),

181

whence x =

= bip ~.

2a.

and the maximum

stress

2P ~ __2P

Hob'

by 97.

The determination

most cases best

the neutral axis

effected graphically.

Thus

is in

the load

point L (fig. 160) lie upon a principal axis, the whole area is divided into strips parallel to the neutral

and a link-polygon acb for these is drawn with A line LI/ is drawn perpendicular polar distance p. to AB to cut the first link produced in L', and the axis,

line Lib

then drawn by

trial so that

the triangle

acb between the polygon and its closing line ab, or so that the two shaded areas are equal. ablu'

This determines the position of the neutral axis. 69 the area acbh x 2p = 2nd moment, For by = 1st moment. and bh x p 2 area acbh x p - or area '----

but y

y x bh

the height of the triangle Lt'bh, and area y x bh = area ocbhi 2

and

.*.

area aL'c

area cdb.

This equality is most conveniently obtained by drawing an equalizing line by eye, and adjusting it until the required equality is obtained, the areas themselves being most conveniently checked with a

planimeter.

182

THE ELEMENTS OF GKAPHIC STATICS

98, Numerical Example, stack of external diameter 8

circular

chimney

and

inside diameter

lb.,

the eccentricity

ft.,

FIG. 1KO.

ft.

carries a load of 50,000

being 2

ft.

Find the

maximum

compressive stress

STRESS DISTRIBUTION on the assumptions being allowed

(i)

of tensile stress in the

(i)

Since

= tl4 +

D + 2

where

cement

tensile stress neglected.

(ii)

;

183

and

50000

x uy ^_ _

^ ^

100

5175

lb./ft.

and the position

compression .

160,

fig.

mean stress

CN =

17'84 sq.

f\_-i

2*88

ft.,

at the centroid

stress

r\r

and the area

ft.

50000

I^BT >

and maximum

lb./ft.

of the neutral axis is ****

(ii)

and - 636

Ib./ins.

= 2800

= 2800 lb.-ft.'2

^ = 5290

lb./ft.

36-7 Ib./ins. 2

Hence it will be seen that the load point may pass appreciably outside the critical limits before it causes anything but small tensile stresses.

CHAPTER

XII.

THE LINE OP PRESSURE. 99. We have seen that when the resultant load on any cross- section of a structure is known completely, the thrust, shear, and bending moment can be found, and the resulting stresses can then be In order that this resultant may be calculate^. for found any required cross-section, the most easily

convenient procedure is to construct the line of pressure for the external forces acting on the structure

;

where by the

understand the

line of pressure

are to

line of action of the resultant force

Thus let tig. 161 represent a crane as shown by means of structure carrying a load Set down a vector 1 a chain passing over pulleys. throughout.

to represent the load \V,

and

1 2 to represent the

pull in the chain between the pulleys A and B. This will be equal to the load for a single-sheaved

and

2 will be the resultant force acting of the pulley wheel at A. centre Also the through if we set down 1 3 to represent the pull in the chain between B and C, 2 3 will be the resultant of the

pulley,

tensions at B, which will act through the centre of

the pulley at B parallel to 2 3, and will intersect the Then 3 will be the new previous resultant in a. resultant,

which

will act

through

(184)

and

this again

THE LINE OF PEESSUEE will

meet the resultant

185

of the forces at

in b

;

that the final resultant lel

The

4 will pass through b paralline Aabcd will be the line of pressure

for the structure,

when

neglecting

its

own

has been drawn we have

this

weight, and all that is

required to determine the thrust, shear, and bending moment at any section. Suppose, e.g., that we require the values of these for any particular section

FIG. 161.

such as XX, which the point L.

The

cut by the line of pressure in

resultant force K' on this section

3, and represented by the vector and position are along ab. Hence

direction

its if

be the

inclination of E' to the plane of the cross-section, R' cos a = S will be the shearing force, and R' sin a

will be the

normal thrust on the plane, whilst

R' x r will be the bending

where r

moment

about the

axis,

the perpendicular distance of the resultant

186 from

THE ELEMENTS OF GRAPHIC STATICS it

;

or since K' x r

where

the

eccentricity of the load point L, the bending moment is equal to the product of the normal thrust into the eccentricity.

The

stress distribution

easily found as explained in

100, Line

Pressure

can then be

90.

Three-Hinged Let A, C,

Structure under Vertical Loading.

W W W W

4 2 3 lf 162) be the three joints, and let be any system of vertical loads on it, whose magnitudes

(fig.

FIG. 162.

are represented by the vectors 0, 1, 2, 3, any pole O construct a link-polygon ab .

af and draw reactions at

from

5 parallel to

A and

it.

Then

Draw vertical

reaction

With

4. .

Join

5 0, 4 5 are the

Also drop a perpendicular

to cut the polygon in g, 7 parallel to ag, fg. 6, at

and

Then

join ay, gf.

7 6

Through 6 and

the

draw

Then if Q be parallels to CA, CB meeting in Q. used as a new pole, a new link-polygon starting from A should pass through C and B, and this link-

THE LINE OF PRESSURE polygon

be the line of pressure, because

will

the reaction at A, and action.

187

Also

represents

QO compounded

with

its

line

gives

which acts along hk, and so on. The same construction will hold for a threeNote. pinned arch, where AC, BC are the chords of the

their resultant,

arch joining the pins.

To Construct Diagrams of Shearing Force and Bending Moment, If we resolve the force QO acting

Ah into components parallel and perpendicular AC, by constructing on QO a triangle of forces having its sides parallel and perpendicular to AC, is the thrust and Q8 is the shear. viz. Q8 0, then 8 These may be set up at AD, AE, and lines DF, EG drawn parallel to AC. Then the vertical ordinates along to

of the diagram so obtained will represent the shear and thrust respectively over that part of the rafter. Similarly, if we resolve the force Ql acting along hk into components parallel and perpendicular to AC, by constructing on Ql a triangle having its sides parallel and perpendicular to AC, viz. Q9 1, then 9 1 represents the thrust and Q9 represents the shear, and these may be plotted as before on the rafter as a base-line, and so on, as well as with the The thrust diagram is shown by the full rafter CB. line, and the shear diagram by the dotted line, while

the lower figure is the bending moment diagram. 101, Conditions of Stability in a Masonry

Structure,

Although the cement or mortar in a

masonry structure will ordinarily add to its stability quite appreciably, it has been the custom to neglect this

and

to require that the

shall be fulfilled

following

conditions

THE ELEMENTS OF GRAPHIC STATICS

188 1.

That there

shall be

tensile

stress

on any

cross-section. 2.

That the crushing strength

the

material

shall not be exceeded. 3.

That the resultant thrust on any bed- joint shall not be inclined to the normal on that joint at an angle greater than the friction angle.

The

first of

these conditions

one, and the circumstances that been considered in chapter xi.

Examples

usually the principal limit

have already

of the application of the line of pressure

to the determination of the stability of dams and retaining walls will be found in vol. xvii. of this series (" Stability of Masonry "), and of its application to arches in vol. xx. (" Stability of Arches "). The following example will show how it is applied to the case of a flying buttress, whose stability it desired to investigate. 102. Fig. 163 represents the nave wall of

cathedral,

which sustains the thrust

the arch

The buttresses acting on a corbel as indicated. are 3 ft. thick and the weight of wall bonded with a buttress, and acting with it, may be assumed as equivalent to 2 buttress.

ft.

of wall

The weight

taken as 145

Ib.

per cub.

either side of

the

masonry may be The thrust of the

ft.

main arch was found previously acting at

the horizontal.

the

be 30,000 Ib. This thrust is

assumed to act through the centre of the corbel on which it abuts. The thrust of the buttress arch which acts in the opposite direction is due to its own weight and the weight of the masonry resting on The springing joint was taken at AB, and the it.

THE LINE OF PRESSURE

FIG. 163.

189

THE ELEMENTS OF GRAPHIC STATICS

190

masonry was divided into four strips of equal The mid-ordinates of these strips were taken '

width. to re-

present their weights and were plotted along a vector to 4, the total length line from 4 representing the

whole weight

7610

The

Ib.

position of the result-

ant was found by taking a pole O and constructing the link-polygon shown, the first and last links of which intersect in C.

vertical through C was drawn to drawn through the upper middleof the arch-ring at the crown (see

cut a horizontal

third point

Stability of Arches ") at the point E, and this point was joined to the lower middle-third point

was then drawn

EGH

The

at the springing-joint.

triangle of forces for the resultant load and the

two and F, whose magnitudes were found from the figure to be 7000 and 10,200 Ib. respectively.

reactions at

The horizontal

thrust of the arch at

with the thrust from the main arch

compounded

by producing their lines and finding their resultant

action to

KM.

This

meet is

next com-

pounded with the weight of the wall above any selected section DD P by finding the point N in which their lines of action intersect, and compounding the previous resultant with the weight of this In portion of the wall (39,600 Ib.) at the point N. this

way we

get

NP as

the final resultant, and

when

produced backward to cut the section DD lf we get the load point L lf which is well within the middlethird of the wall.

next compound the weight

remaining part of the wall (97,500 Ib.) with the last resultant, and get PQ. This cuts the base

of the

section in third,

which

but not

lies

sufficient

little

outside the middle-

be at

all

dangerous.

THE LINE OF .PRESSURE Next proceeding

to the buttress wall

191

we compound

the thrust of the arch (10,200 Ib.) with the weight of the masonry above the section aa^ in order to test

whether the

line of pressure lies within the middle-

third at this section.

the section at

We get

and which

a resultant which cuts

well inside

it.

Simi-

at the section bb lt get the load point 4 L at cc section the finally r The normal thrust 5

larly,

and

at the base cc l

was found 10,000

Ib.,

and since the

21 sq. ft., we get for the average 10,000 Ib. 2 - = 477 This stress on the base lb./ft. Jl area of the base

was

set

at de to scale

third points (see these in / and/x

95).

and joined

vertical

which are the

of the wall.

gh,

to the middle-

through

cuts

on the edges g and h, and joining

stresses

Projecting these to get the diagram of stress distribution over the

on the one edge being 800 lb./ft. 2 and on the other 100 lb./ft. 2 The above investigation cannot of course be regarded as very exact, on base, the stress

account of the roughness of the assumptions made, but

sufficiently so for practical purposes.

[THE END.]

INDEX Areas,

moment

Asymmetric sections Axes of inertia .

...... ... '

PAGE 136-62 157, 178 .

Axis of flotation neutral

155 103 163

B Belfry tower

52 95 98, 100

Bending moment diagrams Bollman truss Bow's notation

47,48 26

Centre of gravity pressure

on submerged area Centroid of areas circular arc

sector

segment cone loaded points

- parabolic segment

- pyramid

...... ...... ....... ........

semi-circular area spherical segment

trapezium triangle

Composition of forces Concrete raft Concurrent forces

Conditions of stability

Coplanar forces Core of sections Corrugated iron, weight Couples composition of

140 163 165 144 159 159 159 160 142 160 160 159 160 145 144

1-25 117-19 6 187, 188 7

177

.80 67 69

(192)

INDEX

193 PAGE 68

Couples, effect of Crescent roof Critical distance

91-4

174

Dead load Diagrams of shearing

moment

bending

78 97-107 98-107 174 163-83 83 83

force

Distance, critical . Distribution of stress

Duchemin's formula

graphical construction for

E '

Eccentricity of load Ellipse,

momental

172 155 60

of equilibrium . Equilibrant Equilibrium, conditions of , equations of

Equations

60 72

...

of forces

147 148 147

Equimomental points for a rectangle a triangle

system

Finck truss

.....

Flotation, axis of Forces, composition

concurrent coplanar in space

and resolution .

space superposed

French

roof truss

Funicular polygon

% 4

. .

--.-.

...... ....

redundant

;

perfect

56, 57

transmissibility of triangle of Framed structures

Frames, imperfect

47 163

.... ....

10 26-57 45 45 45 51 47-51 81 72

G 109-11

Girder, braced

48,49

lattice

THE ELEMENTS OP GRAPHIC STATICS

194

Gravity, centre of See also Centroid. Gyration, radius of .

Imperfect frames Indirect loading Inertia, axes of

;

'./..'.

..... .'.-.

'..'

PAGE 140

138

.-'.-.

........

Lame's theorem

.... ...

Load-curve dead

84 184 72 81 112 78 172

.........81 ........ .81 ........ M ..... ....... .95 .61 ........ ...... .......

eccentricity of

- permanent point

48, 49

L'ive load

live

45 108 155

of inertia.

Lattice girder L.C.C. requirements for wind-pressure Line oi pressure

Link-polygon

.'_'.

Moment