Elementary Graphic Statics

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ELEMENTARY

GRAPHIC

STATICS

Whittaker's ADAMS,

H."

Practical

ARNOLD

and

ATKINS,

E.

BAMFORD.

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EX.

ELEMENTARY GRAPHIC

STATICS

BY

T.

JOHN

Lecturir

Mtuhine

in

and

Medallist

and

{City

WITH

and

Design

Heriot'WoUt Honours

A.M.I.Mech.E.

WIGHT,

Prizeman

186

in

Mechanical

of London

Guilds

64

AND

66

Institute)

ILLUSTRATIONS

"

LONDON, AND

Engineering

CO. SQ.

ST., PATERNOSTER

HART

WHITE

Movers^

Collegey Edinburgh

WHITTAKER 2

Prime

FIFTH

E.G.

AVENUE,

NEW

YORK

o

"

"

"

"

"

"

I

""

W

"

*

^

""

t

*

\

a

"

1 V'

CONTENTS

I

CHAPTER

PAGE

1

Introduction

.....

of a Force Graphic Representation Specification Force Ooncnrrent of a Ooplanar Forces Forces and Non-concurrent Compositionand Resolution Resultant of Forces Equilibrant Units

Definition

and

"

"

"

"

"

"

"

CHAPTER Composition Forces

Resolution

and

in the

Acting

Two

Oonfturrent

Triangle Worked

Out

Tripod Chain

III .

Composition Funicular

Polygon

"

Resultant

Unlike

Parallel

"

Moment

Resultant

Forces "

of Non-concurrent Forces

Moments

"

Moment of

Forces

Parallel

Like

of

of Moments Forces

Lifting

IV

Non-concurrbnt

op

a

"

Crane

Warehouse

"

CHAPTER

"

Legs

"

Rotating Crane

,33

.

"

"

"

"

Examples

"

Triangular Frame-^Shear with Simple Crane Simple Crane

Platform

Loaded

of

Notation

Problems

Practical

.

Forces

of

"

CHAPTER Simple

11 .

Resultant

"

Parallelogram Polygon of Forces

"

Bow's

"

Foucbs

op

Straight Line

same

Forces

Forces

of

II

of

a

System

.

Forces

Resultant

of

tion Graphic Representaof System Non-parallel "

of

Couples "

"

46

.

"

Vll

282178

Parallel

Forces

"

viii

CONTENTS

CHAPTER

V PAGE

Bending

Moment

Definitions

Oonatruction

"

(Concentrated Supported

Shearing

and

Load)"

Force Parabola

of

Cantilever

Loads)

S.F.

and

Loaded

Supported

"

Beam

Beam

(Compound

Beam

Overhung

"

(Compound

"

(Concentrated hung Over-

Loading)"

VI

Single Concentrated

Load

Rolling Lpad

tributed Dis-

Uniformly

"

Dead

Combined

"

.89

.

.

Rolling

Beam

Beams

Loads

Rolling

with

"

Loading)

CHAPTER

Beams

Cantilever

"

Supported

"

"

of B.M.

63

(Dietribnted Load)

Beam

(Concentrated Load) (Distributed Load) Unsymmetrically Scales

Diagrams

and

Rolling

Loads

CHAPTER

Roofs

Symmetrical

"

of

Plain

Roofs"

Tie-rod

with Truss

Distribution

"

Rafters

Simple Swiss

Compound

"

99

Loads

Dead

"

Definitions"

Frames"

VII

of

King-rod "

Truss

"

Truss"

Truss

post Truss

"

with

Lights

Pillar"

Simple

Swiss

Swiss

Truss

(Single

Truss

Inclined

and

Simple Truss

QueenFrench

"

VIII

.

Roof"

Island

135

Loading

Unsymmetrioal

Northern

Rafters

Truss

CHAPTER

Roofs

Truss"

Queen-post

Compound

"

Mansard

"

English

Types

Plam

Compound

(Right-angled Struts)" Simple King-rod (Single Struts Struts)" King-rod Truss Ties)" Belgian

.

Leading"

Tie-rod"

without

Truss

.

Overhung

Station

Roof

Roof"

.

.

Overhung

Roof

CONTENTS

ix

IX

CHAPTER

PAGB

Roofs

Presburb

Wind

"

.144 .

Calculation

Wind

Maximum to

Wind

Presanre

Pressure

Comparison

"

Saw-tooth

"

Wind

of

Determination

"

of

.

Experimental

"

Methods

Northern

or

.

Roof

Lights

with

Free

due End

Station

Island

"

"

Stress

of

Roof

"

Resnlts

Roof

CHAPTER Beams

Braced

Stanchion) Truss

"

Trapezoidal

Truss

Girder

Lintille

Girders

"

Cantilever

"

Gravity

of

Figures

Gravity Centre

^Moments

"

of

Gravity

Parallelogram Funicular

Girder

"

Inertia

of

a

Centre

"

Gravity

of

of

of

a

Section

Modulus

"

Forces

"

of Ferro-concrete

of

of

a

"

Section

Particle

Mohr's

"

"

Method

Section

XII

Walls

Retaining

.218 Case

Definitions*~Simple Pressure

Earth "

Example"

"

of

Pressure"

Water

Rankine*s Rebhann's

"

.

,

.

Theory

"

Irregular Surfaces

Inertia

of

C.G.

Gravity

of

Beam

Figures

System

the

finding

of

CHAPTER

"

.189

"

Moment

"

"

Resistance

"

Method-^ast"iron

of Inertia

Moment

of

Truss

Pratt

"

Axis

Triangle

Resistance

of Inertia

Moment "

"

of Inertia

Moment

a

BoUman

"

XI

Method

Centre

Polygon Axis

Neutral "

of

Truss

(Douhle

Pier

of

Experimental

"

Beam

Fink

"

Neutral

"

,

Braced

"

CHAPTER Centre

.166

"

(Single Stanchion)

Warren

"

Lattice

Qirders

and

Beam

Braced

X

Theory Method

"

Example Coulomb's

PREFACE

In

lecturing

of

suitable

a

of

knowledge

of

solution

the

to

and

Engineering It

is not

but

much

in

him. science

the of

of

is

be

comparatively

There

forgetting

that

there

engineers

the

useful

methods.

The

simple

who

practical most

which

in

youDg

studying appeals

the

problems,

cumbrous

all

problems

of

means

and

can,

to

easily,

too

with

an

air

difficulty.

those

are

and

intricate

equations

of

ready

a

with

the

privilege

engineering

many

methods

met

why

presents

working

mathematically

interests

of

commended re-

practice.

be

to

the

be

a

problems

the

which

the

and

graphical

reason

denied

Graphics

mystery

of

one

no

solution

mathematical

of

Construction

every

this

circumventing

clothe

to

engineering

In

search

simpler

Building

should

engineer

the

given

brilliant,

in

application

the

could

which

to

felt

always

has

text-book

Statics

Graphic

Author

students

to

of

subject the

students,

first-year want

the

on

scoff

are

of

problems work

of

graphical

hundreds

are

who

at

such xi

of

daily

first-class

solving

everyday men

methods,

is alone

life a

of

some

by

such

sufficient

PREFACE

xii

of

justification their

the

application

It

lies

only

problem

set

carefully Mere

the

with

in

out

the

the

in

methods

graphical problems.

student

to

realize

following and

out

of

of

engineering

to

worked

reading

existence

that

understood.

is

matter

be

must

pages

thoroughly

subject

every

than

worse

useless. The

aH

student

the

have

in

the

much

works

conscientiously and

problems acquired

stead

who

which

examples that

performance

will

him

useful

in

Hbbiot-Watt

Edinbubgh,

Collkgb, October

1913.

good

engineering

work.

JOHN

will

follow,

stand

of

through

T.

WIGHT.

"

"

'

"

ELEMENTARY

"

"

'

"

'

.

GRAPHIC

STATICS CHAPTER

I

INTRODUCTION

It in

is unnecessary to tell those of science,that any branch

depends, c"*

accurate

restricted that

the

to

no

small

measurement. to

a

need

few, yet

who

scientific

extent,

on

Scientific it does not

for accurate

interested

are

discovery

measurement

"

discoveryis indeed follow necessarily

measurement

is likewise

in importance of it is exeipplified practically everything around us, and in no other so profession, perhaps,is its importance and utility of engineering. In in that as significant every of engineeringits importance is undisputed, branch and its application, in conjunctionwith geometric methods, to the solution of engineeringproblems, and interest. providesa study of extreme utility Accuracy, both in drawing and in measurement, be too stronglyemphasized,for this is the cannot in the solution of all problems in keynote of success Graphic Statics. The one objectionlevelled against this science

restricted; the

A

"

"

"

I

"

"

ELEMENTARY

GRAPHIC

STATICS

is the

introduced by inaccurate liabilityto error In its applicationto drawing and measurement. engineering problems such an objectionis hardly in the valid, for, by exercising moderate care attain to an different operationsinvolved, we can of 1 per

accuracy

cent.

"

accuracy

sufficient for

all

practical purposes. Its advantages are manifold, enablingus to solve, would problems which readily and quickly,many and intricate cumbrous otherwise require much mathematical investigation.In many engineering quantities, problemswe deal with forces and velocities, and surelyit is a great in themselves,purely abstract, mind to be able to repreadvantage to the practical sent in a comparative such things on paper, if even of a few way, greaterstill to be able,by the deft use simple drawing instruments, to discover the effects produced by such forces and velocities at any point in a given system. Definition and Specificationof a Force." Before graphicalpart of the proceedingto treat the more subject,it might be well to deal brieflywith "force and its specification" which producesor tends Force is any cause Force. or to produce motion change of motion in a body. This definition is quite general in its application, conveying to us only a very vague notion of a of the effect produced force through the medium thereby,and before we can put anything on paper "

regardingit,we must few pointsconcerning

have

fuller information

the force.

on

a

3

INTRODUCTION The

necessary

Such

must specification

Point

(a)

to

force?*'

a

things,viz.

four

embody

the

are

completely specify

points a

"What

arises:

question naturally

:

"

of Application.

(b) Direction, (c) Magnitvde. Sense.

(d)

Graphic consider

of

Representation

how

of these

each

a

Force.

items

Let

"

be

can

us

now

represented

graphically. Point

"

statics if

considered

have

we

be

must

we

In

of Application.

all

forces

accurate, how

point of application,since precludes any idea of area, and make

on

how

matter

no

give

this

up

small

mathematical

"

our

of

point have

we

may

some

area, must

we

point

a

of

point

a we

purpose

of

points ;

at

mark

any

in

represent

we

idea

our

idea

the

which

to

area

For

consider

can

necessity,cover

small?

connection, and a

of

must,

paper

acting

as

then

the

calculations

our

in

this

application as the

given

name

point." Take a

for train.

calling

of the

them

yet

which

accuracy

of

the will

of

"points force be

applicationby

small

so

end

speaking, the point in

an

engine pushing

stance, points of application,in this inbuffers, in themselves presenting an

appreciable area, area

of

case

The the

are

whole

the

illustration

of

of

comparison

train

acts,

as

to

the

justify our

to

Generally exact region

application."

applicationis

satisfied a

the

in

so

if

pencil dot

that we on

the the

demands

represent the

paper.

the

of

point

ELEMENTARY Direction.

at

see

point

next

force.

the

that

glance

a

oa

to

rotate

oa,

we

would

It does

about a

make that

of

the

force

a,

unless

of

makes

have

we

to

line from

measurements.

our

in

o

the

point which that

straight

line

measured drawn

providedwe be

measured.

immaterial,

to

state

the

of

The

absolute

provided that

the

always

in the direccon*

of

that

the

wise. counterclock-

or

the

force

attention.

is the

Anything

represented by suitable

length or

taken

being

to

clock

some

unit

measure-

our

adopted being

be

can

what

quite immaterial

our

this line

direction, the

magnitude demands

next

be

can

The

all care

trary a

horizontal

a

from

measure

tion

of

draw

Fig. 1, we

right

make

one

"

some

which

to

Magnitude.

angle

an

datum

ments,

bands

direction

or

we

the

force.

our

line ox^ and

of

any

base

to

rotation

we

predetermined

From

1

have

ideas

our

make

yj

1

imagine the line new position of

direction

new

clear to say

could

we

is the

Fig.

to

might

and, in each

not, however,

more

to be considered

force

the

o

have

STATICS

By referring

of directions, for

number

any

The

"

of

direction

GRAPHIC

we

scale

scale.

make

the

to which

length

of

length of

the the

It

a

is

line,

it is to line

is

line,the

6

INTRODXJCTION

quantity

long

example, 4

or

scale

our

of

the

more

must

1

be

2'5

of

choice

forces

with at

space

scale

a

inches, and

depends

which

we

disposal on

our

in.

3

length

the

not

fraction

a

of

possibleextent

our

while

the

size

larger

the

scale, the

other

of

the

drawing-paper. is

drawing

percentage

words, the

larger

on

The

is,that the maximum

drawings varies,

our

also

dealing, and

in

error

of

magnitude

the

on

are

in mind

important point to keep

in

be

accordingly,but, if lbs.,then

in. =20

at

line

the

of

force

a

less.

The

the

represent

to

whether

matter

not

fixed

line

or

the

wish

we

say

in.,if the scale be chosen be

ment measure-

harmony.

lbs. ; it does

50

the scale of

represents and

it

all in

are

For

that

the

constant,

a

less,or,

is the

error

the

scale

the

that

so

greater

the

accuracy.

Sense, is the

"

The

only point remaining

of

sense

represented

three

the

still be uncertain

from

the

push

a

"

arrowhead In

point but on

so

the

the

that

minus

or

towards

this the

to

in all cases,

60 lbs.,pullingon

indicated

with

the

sense

negative forces the

it that

for a

of the

we

may

acts

away

it is

pull or

a

by putting

an

force. of

a

force, it is

.that positive forces act away

unless

take

force

whether

"

line of action

assume

point. Thus,

be

can

point, while

sign,we

it

the

settled

settled and

have

may

whether

to

as

We

preceding points,but

dealing,however,

conventional

from

force.

the

be

to

force be

the

example,

point at

an

act

we

angle

it,

preceded by

force acts if

towards

had

away a

a

from

force of

of 45", it would

6

ELEMENTARY

be

GRAPHIC

written, 6O45lbs., but

pushing It

the

on

point

Pi80+a.

This

will

be

reference

to

Fig. 1.

As

as

pull

a

oa^

in

are

the

it oaca

force, and have

that

seen

force

is

the

To

show

the

foregoing

the

an

on

We

will

we

select we

we

proceed

the

point

We

further

This

only

the

off

mark

that must

we

But

we

"

ideas

in

embodied

the

consider

now

applicationo

direction, we

that

from

a

this to be

length force

the

put

of

on

oa

is

oa

1

line, must

we

in.

equal

=

it

angle,

datum

our

length

that

see

off this

mark

we

shown

as

lbs.,

30 2

to

in.

negative, and

an

arrowhead

one

definite

of

graphics

acting

point o.

diagram one,

the

assuming

notice

consequently towards

of

fix the

can

to

negative

a

o.

scale,and

a

we

the force, "GOgg lbs.,acting

counterclockwise

Before

written

definitions

Fig. 2 (a),and, as for makes an angle of 35^, so

ox.

be

this may

in

measuring

a

o

the

given

are

hence

oa'i80+a.

application of

given point

a

is

produced by either In oa'i80+o. oaa="

on

graphicalrepresentationof

oa

and

oa

and

oa^ is

that

so

"

by

oa

Now

along "

=

along

acts

a,

it

write effect

Po

is concerned,

angle

force

we

same,

the

at

P

line.

straight

general terms

more

6O45 lbs.

"

push, provided

a

were

understood

point o

force

the

the

hence

the

as

as

same

while

readily

far

oa^

positive force acting write

force

same

be written,

more

whether

along

or

the

noting, in passing, that

worth

it is immaterial

if

it would

be

may

STATICS

and

now one

represents

advantage

force

and

becomes

GRAPHIC

ELEMENTARY

8

and

Concurrent

of

if

they do not are applied

but

situated would

be

to

have

by

suitable find

so

of forces

force

Forces.

of

acting on

forces

a

of forces,and

the

as

single force, so

the

can,

forces, and would

forces

several

is known

process

we

acting alone,

the

as

When

"

body

a

these

which,

effect

same

Such

together.

points, so

point,such

common

Resolution

single

the

;

action, if produced,

constructions, combine

one

produce

and

number

a

a

concurrent

various

at

lines of

in

to be

point of application,

body

several

point

common

a

said

are

If, in

"

non-concurrent.

Composition we

have

common

a

the

to

intersect

not

said

are

have

the

that

forces

forces

application,such

but

Forces.

Non-concurrent

several

the

system,

any

STATICS

acting

tion composi-

found,

the

as

resultant Resultant. is that

the

The

"

resultant

of

single force tvhich,acting alone, the

effectas

same

several

of forces

system

any

would

produce

forces acting/

primary

together. When we

of

can,

have

by

suitable

acting that

of

a

of

number

any

when as

we

in

the

single force forces

unison,

single as

several

forces, so

found,

any

number

given

of

of

number to

the

noted, that

produce

force

is knowu

be

whose

as

the

components

the

the

values

effects, effect

same

Such

alone.

of forces,and It

components. resolution is

body

a

combined

acting

resolution

on

find

constructions,

process

however,

acting

capable

of of

a

a

the

should,

force into an

infinite

solutions, but, in general, definiteness

the

solution

by practical considerations

is "

9

INTRODXJCnON

either

the

forces

will be

magnitudes given.

Equilibrant.

equilibrant of

from

resultarU.

the

these

(a)

The

(6)

The

up,

point

of

applied

to

effect.

Fig. 3,

to

of the

lines

and

paring com-

specification

"

applicationis

the

same.

i80-^c^

equal,

^

oa^,

=

(c) The the

is

sense

same,

and

both

positive

being

acting from

away

the

Q^y

Poi^*^-

Fig.

3.

directions

(d) The

diametrically opposed,

are

of

tudes magniare

oa

the find

we

its

system

carefully distinguished

Referring

along

two

drawn

have

neutralizes be

equilibrant must

The

a

when

single force which,

or equilibrates

that system,

we

The

"

that

is

forces

of the

of certain

directions

or

resultant

being

a, while

the

of

angle

that

of

then

is in the

the

equilibrant

the

islSO-hct. The the so

only point

two

that

of difference

being diametrically opposed the

effect

of

the

of the

into

equilibrium.

In

the

general,when

force-effect,produced resultant

bring

so

represents

equilibrantrepresents

by that that

a

system

another,

one

equilibrant is

resultant, and

that

to

directions,

to neutralize

whole have

we

of

system a

forces, the

force-effect, while

single force

given

which

the must

10

ELEMENTARY

be

applied

GRAPHIC

the

to

STATICS

neutralize

to

system

force

that

effect. In

Units*

absolute

unit

the

measurements

of

force

"

which

is

used

problems, The

is

acting

on

acceleration

We

know

of

that,

earth's

if

equal hence

follows

it

acceleration,

feet

a

force

produce

in

produced

is

per

the

on

which,

pound

poundal

acting

pound.

force

1

second

the

the

the

at

mately approxiand

second,

unit

produces mass

it

second.

per

of

mass

per if

that,

the

second

acceleration

32

to

would

per

drop

we

the

surface,

foot

1

is

that

as

pound,

1

engineering

force

of

defined

of

mass

a

unit

be

may

In

poundal.

the

however,

poundal

an

the

be

must

32

poundals. It

use

is

this

as

This

parts small

The when

unit

our

value, of

as

force

attractive

be

in

force

a

but

surface, in

ton,

consisting

of

dealing

with

problems

2240

all

the

practical lbs., is

involving

that

we

problems.

slightly

varies

negligible

pound

mass

engineering

however, earth's

the to

of

on

at

different

variation

is

problems. frequently large

used forces.

so

CHAPTER

COMPOSITION

Forces

in

case

shown

be

apparent

in at

attempting

the

same

glance.

a

is

FORCES

Straigrht

Line"

of

method

the

4

No

of

obviously

is

such

algebraic

the

ABC

In

the

should

solution

advantage

solution

graphical

a

resultant

Fig.

OF

RESOLUTION

AND

Actingr

II

The

case.

a

of

sum

by

gained

six

the

tx^ C

"

"

1-

o 4.

Fig.

and

forces,

(P

+

Q

R)

+

(A

or

with

case

have

to

Such

the

first

at

which find

second

have

admits aid

the

of

the

with

Forces" the

deal

is that

of

two

concurrent

the

two

Possibly important

most

to

of

in

of

which

we

forces.

methods

of

Parallelogram

aid

as

greater.

time,

same

resultant

the

with

the

we

problem

a

is the

left, according

or

Concurrent

Two

simplest and,

the

right

to

B+C)

+

of

Resultant

the

either

acts

solution

:

of Forces^

position

and

force

diagrams. First

method,

Method, we

Parallelogram

"

make of

In

solving

use

of

a

Forces.

problem

well-known This

II

the

method

by

theorem, of

solution

this the is

12

GRAPHIC

ELEMENTARY

not

in

yet

there

are

problems

many

readilyapplied,and it

'point he

the

it may

In

sides

their

be worth

Fig. 5

of

in

be very

can

considering

point 0. OB

have

we

00

Then

direction This

off

equal the

to

forces,P

two

along

P and

diagonal

resultant

these

to

in

Q, acting

at

a

two

lines distances

OBCA

scale

and

the

OA scale.

some

join 00.

magnitude

and

resultant.

about

the

proofs have

been

The the

and

Q respectively,to

represents

Newton

probably

the

5.

parallelogram

of the

various

by

direction

represents the

important proposition was

Sir Isaac

and

on

direction.

Mark

Complete

forces acting

two

parallelograniythen

a

Fig.

is

it

while

magnitvde

intersection

and

magnitude

time

practicalproblems,

in which

Forces."//

of

represented

adjacent

through

and

of

little in detail.

a

Parallelogram a

solution

in the

general use

STATICS

year

first enunciated

1687, and

given by

different

following proof, due

best

known.

since

to

by that

maticians. mathe-

Duchayla,

RESOLUTION

Let

AB, and

ACD.

be

and

act

forces

two

direction C

P

force

a

OP

Q

The

be

can

of the

T, acting

and

its

can

to

act

at

to scale.

and

T

at

points

the

now

and

CE

AE.

These

point

of

the

E.

EG

C,

W

acting along P

and

tion applica-

This

and

have

two

of

force

forces

at

we

some

its line

in

application be

of

point

be

to

replaced by

resultant

the

the

Complete

point

any

directions

of

the

along

(Fig. 5a.) line

the

be

now

Acting

assume

A

point of application be

let their

F.

on

act

Q is assumed

and

assumed

E, may

at

the direction

along

ECDF.

replaced by

acting along

we

W

Q, acting in the

and

further

P

force

and

of P

force

Let

action.

C

also

(say T) acting along

forces

to P

W

CD

represented by

resultant

force

and

Assume

parallelograms ACEB

at A

body

a

on

13

FORCES

equal

EF, and

removed

two

forces

CD.

W

to "

Again

to be

some

(say S) acting along the line CF, and by taking point of application as F, we thereby apply all the

force its

forces effect

at

one

Hence

point F

is

without a

point

altering their on

the

combined

line of action

of

14

GRAPHIC

ELEMENTARY

the

resultant.

Therefore

STATICS

direction

is the

AF

of

the

resultant. For

P

we

and

(nxf) general

the

must

system

force

know

the

that

OC, hence

OD

line, and

if

OC

O

is in

will

and

FA

equals DO, The

of

the

line

aid

known

of

of

in

which

two

of cord

be

the

A

spring this

the

Complete

the

the

ant result-

the

point forces

three be

in the

is therefore

equals OC.

FA

a

But

OD. the

magnitude

inclination

the

of

the

hooks is

B

two

are

fixed

attached.

are

of the

hung

a

being

apparatus

and

cord,

and

balances

balances

the

quite simply by

demonstrated

spring

Fig. 6.

in

of

must

represents Q,

same

of

OF

and

equals

and

along straight

acts

since

action

and

its direction.

can

two

connects

point O,

OC

P

and

to the

Q.

Now

figure OFAC

The

magnitude,

shown

OB

therefore

resultant

OF.

the

of

equal Q

OD.

consequently

OC

theorem

the

that

hence

representing

This the

length

or

magnitude

the

join

under

E, it follows

parallelogram,and

direction

and

the

in

along OF,

act

straight line.

same

the

by OD,

equal

must

equilibrium

P, Q and

in

apply

of P and

be

will

represents OC

P

Fig.-5, we

to

resultant

parallelogram ODFA, and

the

effects of P

OC

and

resultant, then

of E

Hence

magnitude

E, represented combined

the

opposite to We

the

consider

Referring back

a

(W+Q),

resultant.

now

resultant.

holds.

diagonal represents

of the

line of action

for

and,

(mxf)

theorem

same

the

case

We

write

may

a

weight

of

set

as

points A

balances, and

weight

up

of

to

length at

the

known

ELEMENTARY

16

We

by drawing

start

P, and

of

GRAPHIC

such

line ah

a

the

direction

h

line

he

parallel and

0

in

the

draw

next

we

Join

aCy

draw

and line

a

a

through R

parallelto

PositioD

of

line

ac

resultant, while the

being

sense

To

show

that

Fig.

6

to

equal

Q.

to

position diagram

we

Force

7.

Diagram.

now

the

represents line OR

indicated this is

the

the

(6)

the

its

of

the

direction,

arrowhead.

actually the

(a) Magnitude.

magnitude

represents

by

from

of view

letteringa

ac.

resultant

the

consider

scale

of the force ; from

sense

Diagram.

This

to

represents

magnitude of P, the with being in accordance

the

the force

parallelto

ah

that

length

a

STATICS

case,

we

will

now

followingthree points

"

Direction,

(c) Sense.

RESOLUTION

Dealing first with

will show

magnitude, we

that

Produce cb to e, making represents the value of R. equal to c6, complete the parallelogram abed and

ac

he

join

bd.

or

as

hence

force he acts

the

equal

are

in

eb from

of ab

of ab and

resultant

and

point 6, it does a as pull from to

e

not

6 to

the

6, since

two

Pl80+a)

(Po="

magnitude

resultant

the

the

now

push along

a

forces

the

Considering

whether

matter c,

the

17

FORCES

OF

eb will be

the

and as

same

be.

In the

that the diagonal previous proof we showed bd equalled the resultant in magnitude and direction, and therefore, if ae is the resultant of ab and 6c,then it must

paralleland equal to bd. Compare the trianglesd"a and bed.

bd

be

eb is

equal

and

parallelto eb,

ba is

equal

and

parallelto ed,

Angle abe is equal to the angle deb. Hence the trianglesare equal in every respect,and is therefore paralleland equal to ac. Therefore

and

represents the

ac

resultant

in

magnitude

direction.

It is not

but

diagram, show

which

We

is from

be noticed

opposite It is

of

know

that way

to

c

the

in the

resultant

force

this direction round

important

to

added

been

this

in

to note

the that

act

diagram,

other this

as

and

rule

ab

to

of at

the

R,

it will

arrowhead forces

force

case

sense

must

of the

the

on

investigating the

that

a

put arrowheads

to

have

they

method

the

force.

advisable

is the and

always

be.

holds

good. Rule.

"

B

In

the

force diagram

the resvltant

always

STATICS

GRAPHIC

ELEMENTARY

18

points the oppositeway round to the other forces^while round. the equilibrantpoints the same way The Triangfle of Forces."// three forces acting on a point be representedin magnitude and direction by the three sides

of in

three forces are The

meaning by

a

these

in order, then

triangle taken

a

equilibrium. will be

of this theorem

reference

to

Her6

Fig. 8.

stood under-

better

three

have

we

S,

forces, P, Q and

acting on a point 0 in equilibrium. To

construct

force

our

diagram

we

begin by selecting of

one

P, and

say

line

a

Fig.

line

finallya

if the

Now, a

of the

line a

ca

line

8.

paralleland forces

three

will coincide

ca

aby for,if they

closing

of the

evidence know

from

being

in

The

line

did

for

not

our

existence

the

of

6

we

to

it;

draw

a

line be

paralleland

equal

to

and

Q,

S.

to

equilibrium,the

end

of

the

the we

end

require

which

resultant, and not

a

would

diagram, a

parallel

from

coincide,

force

ab

equal

with

hypothesis, does

is

such,

an we

exist, the forces

equilibrium.

triangle dbc

directions

in

drawing

and

equal

are

forces,

the

of

the

represents

three

the

forces, P, Q

magnitudes and

S.

and

RElSOLtfTIOIlJ The

taken

phrase

in order

indicating the in sequence

other

is worth

it

which

that

noting

is coincident

follows

round

It is forces must

be

that

their

forces

on

must

equilibrantof

Any one force is the The foregoing theorem of

case

The

a

on

be

point,

a

by

the sides

forces are

in

If

the

known

since

as

other a

the

two.

particular

"

number

any

of forces, and

represented in magnitude

of a polygon

in order, then

taken

polygon does from the starting

the

equilibrium; if

close,then the closing line drawn

not

and

only

is,however,

general theorem,

more

direction

these

equilibrantof

Polygron of Forces."

acting

through O,

pass

equilibrium the force S must form the P and Q, and must also pass through 0.

in

are

that, if three

fact

the

equilibrium,their directions Considering P and Q, we know

concurrent.

resultant

lettering, forces,also

in

body

a

of

order.

notice

to

each

triangle,and

of the

sense

triangle in

the

of the

direction

the

heads arrow-

forces follow

sides

the

with

important act

the

round

the

that

means

of the

sense

Id

FOftCHS

OF

the

and

stopping point represents in magnitude direction the resultant of that system of forces.

to

Consider

the

we

then

force

diagram

equal

to

from

Q,

and

b

P.

we so

is reached

with

The

draw with

in the

a

each line

a

force

next

line of

ef.

at

force P,

the

by drawing

now on

Fig. 9, where a

find the

required to

are

Starting

in

Q, S, T, W, acting

five forces, P, that

shown

case

be

the

point 0,

have ing assum-

resultant. we

line in

we

commence

our

ab

paralleland rotation is Q, and

parallel and several

forces

equal until

to

W

20

ELEMENTARY

Should end

of

a

GRAPHIC

the

end

the

line a6,

figure,and

the

of

/

five

STATICS

line

the

ef

would

we

coincide

then

with

have

forces,P, Q, S, T, W,

the

closed

a

be

would

in

equilibrium. If,however,/

through

O

does

in

not

fall

on

position diagram

our

join af

we

a,

draw

and

line

a

parallelto af. This

line

represents the

now

of action

liue

of

the

Fig. 9.

R,

resultant its

Note, are

taken,

several this

force

as

can

polygons

is the

system of habit

diagram.

is immaterial

It

"

thus when

the

length af represents its magnitude, being found by following the forces

sense

the

round

while

case,

be and

formed

what

order

the

forces

easily proved by drawing out comparing results,but, although

it is

orderliness

in

advisable

in

cultivate

tackling such

will obviate

dealingwith

to

more

much

problems worry

intricate

and

some

; the fusion con-

problems.

RESOLUTION

Resolution

of Forces."

given provide the various

is

solutions in

to

volving in-

forces.

reversed

in the

The

tion applica-

problems dealing Avith

to deal

necessary

reduce

We

cases.

into

two

reduced

to

resolving

have

following: (a) Direction (6) Direction

imposed

problems, in general,to

the

of forces,we

number

fully with

very

restrictions

the

simple geometry. In general, whether

any

problems

concurrent

simply

subject,as

the

all

just

cases

of forces.

hardly

of

part

21

general

two

solving

of

constructions

resolution

It

of

composition

these

FORCES

The

means

a

operations are

of the

OF

are

to

find

one

the

on

exercises

forces four

other

or

this

or

ticular par-

of

the

"

(c)

of

{d) Magnitudes This

force.

one

of the

magnitude

of two

forces.

expressed generally thus : forces acting at a point,then

of

other,

forces.

be

can

number

and

one

of two

Directions

of

magnitude

and

If to

ti

be

the

specify

know must 2n facts about point we It is the forces, e,g.y n magnitudes and n directions. only possible to solve the problem if (2n" 2) facts about the given forces. known are of dealing with such problems Probably the method of will be most by an examination easilyunderstood the

a

forces

all the

Examples 1. A

and

worked

particularcases

few

with

at the

oz,

of

make

to scale

in the

and

dealing

previous

pages.

by

cords, oy

out.

worked

weight which

mentioned

cases

out

500

lbs. is

angles of

hung 50" and

150"

two

respectively,

22

ELEMENTARY

with

the

each

of the

datum

Produce

GRAPHIC

line

the

ob

equal

ba

parallelto

line of

to 2 in.

Find

shown.

as

Scale

cords.

STATICS

of the

action

load

be

b draw

From

lbs.

1 in. =250

tension

the

in

(Fig. 10.) and

mark

parallelto

oz

off and

oy.

SCO

Fig.

Lbs

10.

"

Then

equals

oc

the

2. Two

resistance the

tension

equals

the

tension

in the

in

cord

the

cord, oy and

oz.

oc

=

175

in.

=

1-75x250

=

437-5

lbs.

oa

=

1-32

in.

=

132

=

3300

lbs.

forces to

directions

are

motion

of the

x

250

applied is 400

oa

to

a

point 0, whose

lbs.,acting

applied

forces

be

at

270".

60"

and

If 135'

24

to

ELEMENTARY

Pj,

then

reached the

in

be the

STATICS

GRAPHIC

parallel to Pg, and line

magnitude

of

ef.

Join

pull

the

so

until

on

af represents

a/, and exerted

tackle,

the

by

while

is

Pg

direction,

its

centre-line.

the

a/=2-09in.

2 09x

=

lbs.

60=125

of R=

Direction

derrick

guyed by

dicates in-

R,

represented by

4. A

as

83".

pole of

five

four

of

means

tension

rods,

which,

P, Q, R,

shown

in

known

that

S,

are

It

plan. the

forces

is

is

zontal hori-

exerted

on

top of the pole by

the these and

1|

2, 3, 2 J

are

guys

respectively.

tons

(Fig. 13.) Find

the direction of

magnitude force the

Fig. 12.

fifth

in. =2

Beginning parallelto P, is reached

draw

a

in

line T

horizontal

with and the

P, so

draw

we

on

line

with de.

parallelto

force

exerted

represents its direction.

ea.

by

a

the Join Then the

the

and zontal hori-

exerted

by

Scale

guy.

1

tons.

ab,

line other

ea

fifth

long,

until

S

through

0

forces

and

ea,

1 in.

represents

the

while

T

guy,

ea

l'38

=

in.=l-38

5. The

from

point

0

is the

Fig.

of four the

of

them

tensions

regarding inclined extent

the

in

each

other

at 180", while of

30

lbs.

of

by

be

telegraph pole,

a

20

the

The

positions

other in

lines, P, Q, R, S, It

lbs.

lines, that

the

If O

301".

13.

being two

tons.

radiate.

indicated

are

=

plan

of wire

six lines

which

2=2-76

x

of guy

Direction

25

FORCES

OF

RESOLUTION

one

is in

is of

tension

known, them to

is the

equilibrium,determine

26

ELEMENTARY

the full

particularsof

lbs.

20

=

in.

T

The

y

add

latter

the

is reached

it

of 30

a

this force and

must

force

each

length).

(of indefinite have

we

unknown

ab

yet

to

direction,

u.

close the

a/, equal

radius

line

a

with

on

lbs. of

Fig.

and

draw

we

so

in the line ey

force

a

and

P,

is still unclosed, but

polygon to

1 in.

Scale

lines.

two

before, with

as

long, parallel to P,

until

STATICS

(Fig. 14.)

Beginning, 1

GRAPHIC

to

With

polygon. in.,

1*5

cut

centre

in

ey

the

point/. Then

the

information

ef

lines

two

regarding e/=l-30

the

in.=

6. A

T

=

in

us

the

unknown

two

1-3x20

26i8oolbs. and

barge, stuck

fa give

=

26

required

lines

:

"

lbs.

of a/ =22".

Direction .-.

and

a

canal

W

=

3022olbs.

bank, is

to

be

hauled

RESOLUTION

off

by

of two

means

bank.

barge off, and of

a

lbs.

600

900

are

lbs. the

just pull the capable of exerting

hauling to

the

through

bank

the

Find

relative

ropes

Scale

is stuck.

barge

opposite

lbs. to

respectively.

perpendicular the

the

on

1

to

the

in.=

(Fig. 15.) O

Assume the

point at

the

barge

to be eOO

which

Lbs.

4

W^

is stuck. O

Through

draw

perpendicular

a

Ooj.

The

ance resist-

of the is assumed

barge to act

this

along

line,

therefore

and

we

draw,

parallel to

Oa;,

line ah

in.

a

2*0

long. With

and

radius

the

the

900-lb.

the

of

b

(represent-

ing centre

6

centre

a

1*8 in.

at

of

directions

which

at

1000

winches

and

line drawn

point

the

600

limiting

placed

requires a pull of

It

pulls

winches

27

FORCES

OF

pull

a

pull

Fig. 16.

of

winch) and of

describe

radius

a

the

of

1*2

then

arc,

an

in.

600-lb.

winch)

through

0

with

(representing cut

the

arc

c.

Join

ac

parallelto

and ac

6c, and and

be

draw

respectively.

OW^

and

OWg

n

28

ELEMENTARY

These ropes

lines

STATldS

GRAPHIC

represent

now

the

directions

of the

"

Rope of small winch at 61" Rope of largewinch at 36" /3. the previousquestion,the positions of =

rt.

=

7. If,in winches Ox

at

have

such

were

anglesof been

the

that

30^ and

50"

ropes

were

inclined to

what respectively,

would

minimum

to give a necessary Scale 1 in. = 500 lbs.

Set

the

the

of the winches power pull of 1000 lbs. on the barge?

to Ox; long, parallel through b draw be parallelto Wj, and through a draw dc parallelto Wg. Then ac and be represent the pullsexerted by the winches Wg and Wj respectively. out

ac

=

6c Therefore

afi,20

line

a

=

in.

500=525

l-05 in. =

l-05x

l-57 iD. =

l-57x 500

=

lbs.

785 lbs.

be

capable of exerting pullsof 785 and 525 lbs. respectively. In nearly all our Bow's Notation. problems we kinds of diagrams,viz.,positiondiadeal with two grams and forcediagrams. is drawn In the positiondiagram,which to some space scale,the lines only indicate the directions of the other hand, in the force the forces,while, on to some force scale,the diagram, which is drawn the magnitudes of the forcesactingat lines represent the corresponding pointsin the positiondiagram. In passingthus from one diagram to the other,it winches

must

"

is essential common

to

that

we

both, yet

some

method

distinct

in each.

have

of

The

lettering method

RESOLUTION which

is

known

probably

OF

letteringis extremely in equilibrium, as in various

but

it is not

problems

acting Instead

S, T,

we

of

the

roof

be

point

naming a

the

Bow*s

use

"

0

the

with

in

an

forces in

the

girder,

solution

the

of

method

being on

we

of

ing preced-

explaining

actual

case.

forces, P, Q, S and

forces, as

capital letter

braced

it in

consider

with

forces

or

of

method

the

truss

four

the

(sometimes

dealing of

best to

O

in

dealt

the

of

case

the

put

to

29

This

case

have

we

would

the on

useful

a

Possibly

notation

Consider

as

is

notation.

advisable

such

pages. the

of

members

best

the

Henrici*s)

as

FORCES

T,

in

equilibrium.

have

done, P, Q,

either

side

of the

30

ELEMENTARTf

line of

action

to the

of the

of the

T

between

the

forces

to indicate

the

Q becomes

Passing

the

a

line

following point this the

line.

so

is

indicated

BC,

that

in

used

The

are

indicated

polygon

examination

in the

of

by

the

is finished

Fig.

17

force

begin by

line

the in

will

the serve

the

regarding

ab,

the

letteringa forces

and

sense

to

6

/

cated indi-

are

force diagram

our

corresponding in

and

AB,

figuring it

of

direction

position diagram

our

wish

the

we

force

the

represented by the

and

on.

carefully noted

are

P we

as

the

17.

parallel to be

call

now

of it

so

the

direction

A, between

diagram,

by capital letters,while they

we

and

force

italics

by

Thus

speak

we

force

should

The

force AB

being

ab

the

Suppose

Fig.

drawing

shown,

space

on.

P,

the

to

on

P so

force

the

as

amounts

between

spaces

clockwise.

and

B, and

space

do, what

we

the

letter

lettering being

space

AB,

force, or

the

thing, we

same

of action

lines

Q

of

StATlCS

GRAMIC

usual to

italics. way,

and

convince

an

the

ELEMENTARY

32

of

the

The

horse

per

of

140

ton

if

7.

lbs.

steel

A

south,

of

angle with

a

is

34"

ft. with

taking

the

north of

long.

and

intensity

the

30

lbs.

the

pull

in What

load.

the east

equal

the

guys, to

40

ft.

lbs.

per

slope

at

projected ing assum-

be

the

north-easterly sq.

an

blows

guy,

would

a

attached

of

north

with

running

wind

north

in.

30

are

and

sq.

a

motion

and

ropes

ground

per

bank.

tons.

ropes

A

the

exerts

high

The

bank

to

wire

ground.

the

all

four

west. from

25

ft.

20

the

horse

resistance

weighs

by

of

edge The

the

is

and

of

force

it

ft.

from

ft.

34 the

barge

up

Determine

area.

from

guyed

east

10

strap

a

ft.

Determine

chimney It

north,

64

loaded

the

diameter.

wind

is

5

STATICS

is

barge

is

tow-rope,

pull

in

The

tow-rope.

a

while

to

GRAPHIC

ft.

?

pull

CHAPTER

III

PRACTICAL

Having

discussed

now

solution

the

Graphic the

of

Loaded

held

in

tie-rods this

FP,

case

we

L=the the

=

W

We

of

HP

in

cut

FP

on

the

by

three

W, at

c

of

centre

the

the the

ft.

in

ft.

in

lbs.

in

this

W

in We

the have

33

it in

and

that

act

Bisect to

is acted

equilibrium,

tie-rods seen

to

vertical

a

platform

the

keep

tension

raise

formly uni-

taken

platform.

K

Now

O.

be

be

to

case

can

the

through

which

hinge.

lbs.

of

area

and

point

forces

In

platform.

the

H,

of two

means

platform

LxBxti?

=

hence

K

at

carried.

and

in

reaction

load

distributed,

point

formly uni-

a

hinged

is

platform

ft. in

load

the

load

sq.

the

the

through

is illustrated

of

the

assumed

have

simple

few

a

let

breadth

W

consider

to

practice.

18

end

either

per

Then

to

position by

of the

total

=

in

which

HP,

length

w^loB,d

problems

in

Fig.

horizontal

suppose

B

In

"

at

general proceed

with

met

one

in

principles

platform its

involved

now

these

Platform. loaded

and

will

commonly

principles

simpler

we

of

application

the

the

Statics,

problems

PROBLEMS

R, when

viz. the we

^

have in

acted

body

a

these

tension also

T

in

meet

through

pass

O,

O

get

we

l^his

completes

simple

force

gram dia-

the

determine

scale, a line

the

forces

tie-rods

in the

to

W

The

at the

triangle

we

"

supported over

Determine the

G

hinge.

at A a

the

by

a

pulley

"

BW

A

hinge D.

pull in

Produce

until it cuts

each

Frame.

Triangular

passes

in

Tension

the

At the

tie-rod

=

scale

can

ac =

-^

B

at a

by

load and

rope

line of

in O ; draw

C

action a

abc off

lbs.

-^

triangular frame and

and

hinge. T

Note.

the

b draw

; from

parallelto R,

which

and

represent

to

T.

from

Set

down,

be

diagram

the

forces.

load

force

of

required ab

gives the

is to

magnitude

the

parallel to

it

matter

and

ac

of

line

which

obtain

draw

to

and

H

the

now

from

a

line

positiondiagram,

the

from

must

of the reaction,

action

a

the

and

point. By drawing a through pass

this

pass

reaction

hinge

the

hence

must

W

load

The

point.

common

a

forces

three

it

keep

forces, which

three

by

on

equilibrium, then

through

STATICS

GRAPHIC

ELEMENTARY

84

a

W the

ABC

rope

is

which

is attached. reaction

of the

line to pass

rope

at at

through

PRACTICAL A

and

W,

O.

Produce

scale.

to

Oa

the

reaction

Oc

gives

in the

mark

ba

85

oflf06

to

parallelto

represent be

00

and

as

shown

Then scale

to

of

magnitude

the

and

b draw

OA.

measured

gives

OB

From

parallel to

PROBLEMS

at

A, while

the

tension

rope.

Shear shear

Legrs.

The

"

shown

legs,

in

ing Fig. 20, as used for liftduces heavy loads, introtwo one or points of

The

interest

rangement ar-

consists

struts,

two

the

member

by

AC.

In

make the

ab

down

to

the

triangle of

The

gives

struts.

This

latter

between

the

how

this

this force first obtain the

arrow

W

and

line

ac

the

thrust

struts, and

is sustained.

be acts

with

H.

perpendicular; with

two

to

foot B

the

a

order

the in of

swing

the

two

centre-line

determin to

see

struts,

we

how must

direction

of

strut, raise

a

the the

Set

the back-

by to

now

the

triangle

pull in

along In

and

struts.

sustained

looking

B, the centre

the

we

determine

AC,

complete

respect

elevation

From

the

have

thrust

an

tie

acts

we

problem to

the

gives

thrust

the

forces

the

represent

be

tie,while

part

pull in sustained by

of

of forces abc.

of

first

is

which

back-tie,

a

the

magnitude

thrust

and

CB,

of

use

Fig. 19.

shown

as

by

of

round

the

strut

ELEMENTARY

36

BO

until F

point half

the

the

ground

spread E

along LG

LK.

and

seen

in the

acts

along

LK.

From

of the

feet

direction

of

set

off

LM

draw

MN

Then

LP

LK

LN

either

side

on

in GK

by

Join

in L.

F

elevation

LG

and

From

M

in

be.

as

force be

The

forces to

; project

20.

parallel to GL, and

out

H.

equal

Fig.

any

true

arrow

is resisted

Select

from

is the the

in E.

struts

vertical

and

L

set

of the

GLK

Then

LF

line and

the

cut

to

STATICS

perpendicular

it cuts on

the

GRAPHIC

MP

and the

represent

parallel to

KL.

in

and

thrusts

LG

respectively. Tripod.

liftingor

simple arrangement, either in Fig. 21. sustaining loads, is shown

arrangement

in

connected

their

hung. way

at

Draw

that

struts.

for

Another

"

OA

The

the

this

consists

case

top ends, from elevation

shows

struts,

the

of the true

being

all

of which

the

of

load

is

in

such

a

one

same

struts,

the

tripod

length of

three

The

of the

length,

^RACI'ICAL make

equal angles

hence

each

of the

third

with

ground

sustains

one

load

a

Mark

load.

total

37

ttlOBLEJitS and

and

centre*line

proportional to off

on

the

one-

line

centre

W a

length

Oa

OA

in 6.

of the

legs.

cut

Simple

equal

to

Then

06

Crane

a

in which a

tackle

point

required

is

BC

in

the

jib

in

the

tie AC Set

parallel

to

be

BC,

and

jib, while

the

pull in

a

(ic

Lift"

The

be

from

a

AC.

to

thrust

in

gives

the

with

Fig. of

the

load,

passes

over

back

another

pulley at

represent W

Single

23

running to

present re-

the tie-rod.

modification

which

to

draw

the

Simple Crane Chain

pull tion). nota-

ab b

It

thrust

(Bow's

gives

the

the

parallel

etc

Then

and

from

by

jib.

the

down

W, draw

from

of the

to find

in each

of crane,

suspended

outmost

thrust

shows

is lifted

load

the

the

pended Sus-

Fig.22

simple type

very

represents

with

Load."

project ab horizontallyto

;

-^

in

; from

case,

pulley

parallelto the

in

crane

this a

Fig.

21.

shows

the

top

of

b draw

at

the

previous

is lifted the

end

by of

tion. queschain

a

the

tie-rod and

passing

the

Set down

be

mast.

parallelto BC.

jib, over

ab

We

38

ELEMENTARY

have

the

pull in

the

tie-rod

CD

Concerning these followingfacts the pull "

W, its direction from

away

force in the CD.

We

having

a

a

a

forces

two

in

pull in

the

parallelto

the

know

has

the

chain

the

we

chain

only know

therefore

from

the

jib-end,namely,

tude, magni-

a

tie-rod, and

it

jib-end; while, regarding the

the

tie-rod, we

can

by drawing

is

the

and

DA.

acts

STATICS

forces acting at

two

now

GRAPHIC

continue

line ad

magnitude equal

that

The

the

along

diagram

force

our

parallelto

to W.

it acts

chain

force

and

diagram

be

d a line completed by drawing from parallelto DC cutting he in the point c. The student, in solving this problem, should use the same configuration and load as in the previous example, and note that the thrust in the jib is the can

now

in each

same on

the

case,

tie-rod

Simple

Crane

while

is reduced

with

in

the

second

case

the

pull

by W. Double

Chain

Lift

"

The

out-

40

ELEMENTARY

separately,or jib-pointdue pulls in the

we

find

can

the

to

STATICS

the

db "be

first set down

jib-pointcaused

by

direction

and

to

represent

the two

the

of

the

the

pulls

It will

falls.

rope

at

force separately,

each

Taking

ropes.

force

resultant

magnitude

various

the we

at

GRAPHIC

W

readily be draw

cd

parallelto CD,

point d, since in CD.

stress

and

ab=bc=".

that

seen

do

we

therefore

but

cannot,

we

know

not

know

of

pull

from

draw

a

the

line

a

ae

c

as

we

now

yet, fix

the

of

the

magnitude

the

We, however,

direction

the

From

both in

the

magnitude

and

AE,

parallelto

AE

we

can

and

of

W

magnitude equal rope). and the

so

jib

From

e

we

complete and

diagram.

to

"

tie-rod

force can

ed

diagram.

then

tension

constant

finallydraw

can

the

(the

be

in

the

parallelto ED, The

scaled

stresses

ofi* from

in

the

When the

the

magnitude

in the the

method

second

three

of the

is

adopted

point

of

pulls acting

jib. Complete

the

OM

that

parallelogram, remembering

first find

we

of the

(R)

resultant

at the

ropes

41

PROBLEMS

PRACTICAL

W

=

and

W ON

=

and

"

find the

R.

resultant

db

down

Set

=

,

R,

and

triangleof diagrams it will

complete the

two

Fig.

the

result is obtained

same

be

triangleabc on

method, with

a

the

traced

force

on

joining

be

found

in each

By

paring com-

exactly

that

and

paper

obtained

that the

If this force

case.

tracing

the

abc.

2B.

diagram

it will be found line

forces

the

by

superimposed previous

the

line oft will

points

c

and

e,

coincide

as

shown

dotted.

Botatingr Crane. crane

which

capable

of

"

is mounted

Fig. 26 on

shows a

rotating through

central a

a

simple type pivot

complete

at

0,

circle.

of

and

In

lifting the load equilibrium when weight W^ of to apply a balance is necessary O of Wi about magnitude that the moment

order

W,

maintain

to

it

such

STATICS

GRAPHIC

ELEMENTARY

42

a

The

O.

about of W just neutralize the moment of this problem should present no difficulty solution point in the following order" if the joints be taken

will

of

26.

Fig.

Notice

that

gives

m

the

DE.

member

lower

of

end

outer

of mast,

jib,head

of the

magnitude

balance

weight W^. Warehouse crane

from

the

fitted

as

point

a

CD,

and

fitted

extremity. force both

a

This on

vertical

shows

at

with

the

This

a

upper the

and

lower

crane

simple is

house ware-

extremity

bearing the

horizontal

exerts

lower

force

at

hung

capable

angle,being

simple bearing

post, while a

a

lifting tackle

a

considerable

a

pivot bearing

with

jib.

of the

swinging through on

Fig. 27

Crane."

mounted

of the the

only

post

upper a

pivot on

of

zontal horiexerts

its lower

PRACTICAL In

extremity. the that

for

solving of

magnitude

equilibrium

structure

crane

PROBLEMS

point; Rj

and

the

the

intersect

Fig.

reaction,

lower

the

Knowing W,

we

can

triangle of will

be

also

must

find

forces acb.

easily understood force

two

pass at

diagram.

mine first deter-

we

We

reactions.

forces

acting

through 0, and

a

know on

the

common

hence

R2,

the

27.

pass

directions

three

easily

problem

three

all

must

W

this

43

through and

this

point.

magnitude of R^ and Rg by drawing the The solution of this problem from a pleted study of the comthe

Elementary

44

GftAPiiic

sfATtcs

Examples. 1. A

draw-bridge,

supported

in

shown

Fig.

and

in

the

and

chains

bridge

is

A

2.

above

from

outer

end.

pull in

the

and

the

hinges, the taking

comer

in.

ft. 6

3

its

seating.

spread the

of

line

and the

boiler

and

A

ft.

of 4

load to

wall

tons

point

a

long,

the

to

reaction

the

at

3

is

ft.

hinge

ft. wide, is carried

6

the

at

bottom

vertical

load.

Determine

the

reactions

legs is legs

feet

makes the

tie

of

supported

is

is attached

the

one

all the

The

the

struts

load

rope.

set of shear

4. A

the

the

at

on

right-hand

The

hinges

magnitudes

hinges

if the

are

and

gate

lbs.

200

weighs

when

10

end

outer

hinge.

Find

lower

the

beam

a

tackle, attached

a

apart.

of

directions

in

of

the

gate, 4 ft. high

3. A two

The

end.

wire-rope,which

a

the

of

each

distributed

of

consists

crane

point 8 ft. lifted by means at

chains

in

hinge

the

at

manner

the

tension

uniformly

a

the

between

the

reaction

at its inner

steel

a

Find

in

sq. ft.

small

hinged by

the

carrying

lbs. per

120

is 56".

angle

ft. wide, is

20

position

The

18.

and

long

horizontal

a

beam

the

ft.

30

weighs

the 16

each

ft. 6 in.

angle

an

back

and

28

are

used

tie

is 54

thrust

of ft.

to

40

place ft.

a

boiler

on

and

the

long

The

plane containing

70"

with

long.

in each

of

the Find

the

ground the pull

legs

if the

tons.

tripod,consisting of three struts, each 30 ft. long, is used to lift a steam-engine cylinder weighing make struts The 2*4 tons. equal angles with the 5. A

PRACTICAL

and

ground, feet

their

A

6. the

thrust

in

in

(a)

Simple

(6)

With

chain

(c)

With

snatch-block,

case,

barrel

as

lift,

in

as

in

ft.

the

28

tie

of

has

ft.;

and

4

and

the

tons.

22.

Fig.

23.

in

Fig.

as

4

being

Fig.

12

22

Fig.

ft.; jib,

load

a

circle

a

in

10

pull

lifting

on

struts.

shown

Mast,

the

when

jib

lie

the

that

:

Determine

the

in

to

dimensions

ft.

21

thrust

the

45

extremities

similar

crane

following

tie,

lower

Find

diam.

PROBLEMS

from

24,

the

lifting

foot

the

of

the

has

the

mast.

7.

A

revolving dimensions:

following to

of

centre

balance

weight,

various

the

A

8.

27,

is

has

crane

R^

lifting

ft., and and

R2.

the

DE

of bales

=

7

ft.

the

ft.

18

in

load

the

Determine

of

10

also

in

tons.

in

Fig. of

if

members

proportions:

the

stresses

maximum

a

of

above

shown

type

mast

centre

the

a

to

up

stresses

following

the

lifting

of

to

mast

jib,

26,

centre

tabulate

and

crane,

for

of

point

when

Determine

cwts.

12

members

warehouse used

ft. ;

Determine

level.

ground

12

of

ft. ; centre

20

Fig.

ft.;

10

Mast,

load,

in

shown

as

crane,

AE=CD the

reactions

30 the

=

CHAPTER

COMPOSITION

The

forces

not

condition

the

the

case

point

for

necessary

shovM

forces,

non-concurrent

the

satisfied, namely,

shall

we

which

of

application. that

saw

equilibrium in

further

the that

was

with

dealing

condition

funicular

now

several

the

we

close, but, a

been

have

we

in

forces

concurrent

force polygon

far,

forces, but

common

with

dealing

one

a

FORCES

So

"

concurrent

investigate

to

have

In

Polygon.

with

dealing only proceed

NON-CONCURRENT

OF

Funicular

IV

be

must

polygon

also

must

close. Before be

well

The

"

assumed

In

by system

the

term

The

term now

forces

the

As

compression. a

funicular is retained a

Latin

a

funicular

or

be

would

subjected

is well and

polygon

46

in

are

known,

is

a

the

hence

principally

geometrical

in

acting

polygon

compression,

only

the

it

were

in

which

shape

might

polygon."

derivation

cord,** and

cord,

funicular

the

"funicular

its

the

it

condition,

to

the

forces.

cases

of

transmit

has

of

has "

endless

an

latter

term

a

represents

many

links of

funicular

polygon

given

"

funis," meaning

"

link

the

explain

to

word

word

the

investigating

various

the

string general

hardly for

the

cannot use

of

appropriate.

convenience,

significance.

nature

and

48

GRAPHIC

ELEMENTARY

We each

also

know

equal

members

again

that

OA,

00

OB,

find

we

lines

the

and

system

a

forming When

in the

such these

figurepqrs the

point 0

figurein

respectively,and

of radiating lines^

are

known

is known as

It is worth

the

the

as as

in

are

the here this

28.

converted

the

into

a

system

position diagram.

relationship exists

a

existing in

OD

force diagram

closed

a

od

forces

Fig. instance

06, oc and

oa,

the

parallel to

and

STATICS

between

two

grams, dia-

reciprocalfigures.

The

funicular polygon,

and

pole.

noting, however,

that

the

forces

AB,

NON-CONCURRENT

BC, CD and

while

DA

and

of

find

would

position of

that, for

this

would

would

o

may

polygon,

and

vary,

quently conse-

funicular

of

few

a

tion, direc-

and

force

one

number

any out

forces, the figureahcd the

only

drew

we

49

magnitude

links

the

have

if

in

have

can

may

polygons, and we

we

lengths we

fixed

are

hence

the

FORCES

different

cases

particular system

remain

the

both

vary,

while

same,

cular funi-

and

force

of

polygons always closing. A

little

will

thought

hold

will also have

we

and

to draw

under

that

good.

Suppose, for example, forces in equilibrium, shape taken up by any

non-concurrent

wish

we

cord

four

show

action

the

the

out

of

we

would

of the

We

would

polygon.

with

polygon lines

in

funicular

the

the

and

formed

closed

a

line

the

would

we

of

cular funi-

the

radiating find

that

figure,each

action

of

of when

one

of

a

system

applicationof

we possible,

will

Fig. 29, and magnitude P

it is and

and

required

direction,

direction

this construction

S, to

are

the

as

clear

following case.

acting

find

of

forces,and

concurrent

non-

consider

now

forces,P, Q

Three

of

latter make

we

as

of

the

make

This

Fopces."

just described, is the one finding the magnitude and

the resultant

as

parallel to

of Non-concuppent

construction,

to

in

diagram,

on

would

forces.

Resultant

use

draw

then

sides

fell

We

forces.

polygon, selectinga join the angular points

our

polygon

of which

corner

the

force

the

struction con-

converse

force

begin by drawing pole o, to which

out

these

the

their

as

shown

resultant

in

in

50

Draw

out

the

forces

not

be

first the

closed

a

in

represent the magnitude of the Select a pole o, and join oa, 06, oc

resultant

will

Fig.

p, any a

line pr

06, and pr

point

so

in the

ad, and

in the

parallelto on,

until

point

since

polygon abed, and,

force

equilibrium, this polygon will figure,and hence the closing line ad

not

are

STATICS

GRAPHIC

ELEMENTARY

r.

this line

the

;

from

represents

of the draw

p

last line

Through

ody then

from

29.

line of action oa

and

R.

r

sr

pq

parallelto

parallelto

draw

the

force P, draw

line

a

line of

od cuts

parallelto

action

of the

NON-CONCURRENT

resultant, the

being

by following

polygon, although, in unnecessary, Resultant

of

case

of the

the

resultant

of

Parallel

noted

forces a

round

the

proceeding

Fopces."

A

is

particular

the

finding of

parallelforces. AB, BO, CD, DE and EF.

that

force

the

in

polygon,

(It this

80.

straight line,and

a

arrowhead,

of like

Fig.

instance, is

the

application to

system

the forces

be

the

51

is obvious.

sense

is its a

by

this case, such

Like

above

Set down should

the

as

indicated

as

sense,

determined

FORCES

that

the

closing line

a/ represents the magnitude of the resultant.) Select a pole o, and join oa^ ob, oCy od, oe and of,and from then point on the line of action of AB, p, any draw, in the space in

draw,

until

on,

the

of

Produce draw

a

line

A,

B,

space

is reached

np

and

vu

line np

a a

parallelto

line pq, in the

parallelto a/.

;

from

parallelto o", and

line

to intersect

oa

uv,

in the

in r, and

space

through

p so

F. r

52

ELEMENTARY

Then the

line

R

of the

sum

unlike

parallel forces

an

examination

explanation.

Parallel

of

The

of

line

action

of

forces.

of Unlike

from

the

being repriesentedby af

downward

Resultant of

STATICS

represents

resultant,its magnitude

the

"

this

GRAPHIC

be

will

case

easily understood further

much

without

Fig. 31

The

Forces."

only point calling

for

any

par-

"K

Fig.

ticular force

is the

care

AB

is set

is measured measured force

EF

Note

of

out

is measured also

that

the

ofc,but

as as

downvxirds

be, the cd

as

upwards

point /

represents

dowv/wardSy and

the

doum

upwards

a/, which

No

setting

the

forces

in

the

polygon.

Thus,

that

31.

consequently

should difficulty

lines in their

the

proper

be

the

next

and as

next

force

BC

forces

two

de, while

are

the

last

ef.

is below

the

poiut

a,

so

resultant, is measured acts

as

encountered

places if

the

shown in rule

at

R.

drawing adopted

in in

NONCONCURRENT

FORCES be

previous question

the draw

in

space

B

the

If

Moments." effect of

P

on

tendency

to

cause

of P and to the

the

so

in

point O,

about

depends

on

length of

R

the

consider

the

that

the

there

point

things

two

is

0.

the

"

and

"

in

on.

see

we

this, viz., and

oa,

Fig. 32, and

rotation

the

is

a

This

tude magni-

numericallyequal

of P and

product

R, measured

parallel to

examine

we

effect

rotational

line

a

of

use

parallelto oh, aud

line

a

A

apace

made

53

in

suitable

units. The

tendency

tion to rota-

known

is

the

as

of the force P

momerU

^

0

about be

and

O,

in

measured

and

R

this

moment

in

P

if

pounds

inches, then PR

equals

lbs. in. If the

off,along

mark

we

of action

line

length AB, equal a

triangle whose of P

moment

equal

a

AB

suitable

Thus

about

to

P, O

x

00.

be P

0

in

a

OA

equal

to

is

equal the

the

we

x

00, that

is,it is

of P about

moment

of the

get The

^ABxCO.

to P

area

OB,

and

O

triangle AOB,

devised. which

moments,

force

join

and

Hence

be

scale

32.

a

is

area

about

quantities,may The

P,

represented by

be

may if

to

of

Fig.

the

are

represented by

product

would

clochmse

tend

to

direction

of

two

areas.

produce ;

such

a

rotation

is known

as

54

a

GRAPHIC

ELEMENTARY

negative Graphic

Let which

aid

the

it is

tending

the

also

can

of the

be

mxyment

causes

a

force

The

"

cally represented graphipolygon. (Fig. 33.)

be

funicular

given to

Moments.

and

O

the

point

about

rotation.

cause

*

Draw

ah

and

join

chosen inches.

AB,

force

a

AB

O

o,

of

Representation

of

by

positive

rotation.

counterclockwise

moment

A

moment

STATICS

draw

parallel and oa

that

From two

H

and

equal

oh^ the

position

is

preferably Q, any point in lines, QS

and

respectively. Draw

QW

an

of

O

QT

perpendicular

to

in

x

OD.

of

and

oa

line

a

so

of

of action

line

draw and

pole

number

even

the

a

being

o

QT, parallel to

respectively, and through parallelto AB, and cutting QS

oh

Select

to AB.

OD

and

y

ELEMENTARY

56

Moment

Fig. 34 and

of we

we

O,

given

required about

the resultant We

have

of

system

a

to find

is the

what

or,

STATICS In

of Non-parallel Forces."

System

a

are

are

about

GRAPHIC

the

nou-parallelforces, of the system

moment

thing, the

same

moment

of

perform

in

0. distinct

three

solving this problem.

operations

First, we

to

find

must

the

magni-

find

its line

Fig. 34.

tude of

action

the

resultant

of the

and

;

Draw

length

the ae a

pole

being

inches.

pole From

so

and

magnitude join

chosen p,

find

must

we

must

we

polygon abcde, and the

o,

second,

the

moment

of

O.

force

gives

Select o

third,

about

resultant

;

draw

ob,

oa,

that pr

H

is

of oCy an

join the od even

parallel to

The

ae.

resultant. and

oe,

the

number

of

and

pq

oa,

NON-CONOTJRRENT ob ; from

parallelto until the

on,

Through line will

q draw

in the

draw

r

Through

funicular

R

line

line

a

OW

cutting pr produced

in y and

the

resultant

of the

moment

{xy being the =

of

system

forces

lbs., then

n

equal

of be

only

Produce

0

with

the

such

be

moment

line

in

unlike forces

the

case,

about

of

resultant.

R, and

in

Then

x.

the

of

scale

in.

1

will

resultant

Parallel

Forces."

It

method

of

the

parallelforces.

in x, and

produce lU in both in r. Through r

nx

Then

ae.

the

point O it

then

as

to

of

moment

of the

resultant

R

and

are

P

them in

it is

position

a

coincide

R, and

if

that

would

have

of

moments

is

with

two

no

point.

particular case we

that

would

ae

resultant

the

seen

be in such

parallelto

dealing

parallel forces. P

it will be

Fig. 35

the

that A

"

which

be

a

indicate

in y and

through

between treatment

O W

line of action

Couples. one

to

Like

to

OW

cut

study

a

the

produced

of

of like

to cut

quite possiblefor that

of the

this

RxA=a:yxH.

=

From

and

ae,

parallel to

drawn

of

System

parallel to

line

a

about

to

tr cuts

lbs. in.

case

np

directions

be

necessary

in the

solution

draw

a

ir

moment

m,xnxE.

xy

Moment will

the

so

equals xyxB. lbs. in. pounds and H in inches). If

in

measured

polygon

parallelto

line of action

the

draw

O

oc, and

point r.

a

represent

67

parallelto

qs

last liue of the

first line pr

the

FORCES

the

In and be

Fig. the

K

36

let

equal

and

the

two

perpendicular

Adopting previous problems,

the

the we

tance dissame

begin

GRAPHIC

ELEMENTARY

58

by drawing upwards

to

the

to

the

point

join

coincide

will

ob and

Fig.

line np to

parallelto

ob, and It will

with hence

one

now

a, the

force

that

the

parallel to would

it follows

draw

p

that, since

same

our

and

draw

p

is

line pq

a

oc.

the

cides coin-

point c

closed

a

parallel

figure,and

equilibrium is satisfied,but

therefore that

a

parallelto

lines np

the

From

oc).

polygon

of

condition

another, and Hence

noticed

be

pole o,

a

35.

from

;

line qs

finallya

also notice drawn

oa

P, and,

equal, obviously

Select

a.

is also

(which

oa

with

be

then

force

paralleland

are

P,

force

downward

the

first ab

Draw

ahc.

upward

the

represent forces

two c

polygon

represent

downwards since

force

the

STATICS

and

qs, since

line, are never

funicular

they

parallel to meet

if

we are one

produced.

polygon

would

close, and

not

59

FORCES

NON-CONCURRENT the

consequently

is

system

not

in

equilibrium. If

such

motion

of

motion

be

system

a

being

body, obviously

a

place, the

resulting

translation

without

of rotation

one

to

take

will

kind

some

applied

in

direction.

any

Such R

is known

system

a

being

called

the

as

Fig. effect

its

the

on

is made

moment

Total

Total of the

arm

This

couphy the

clockwise.

the

total

=

P(^l+^2).

x

PxR. =

the

product

of

one

force and

the

couple.

while

according

that

see

we

parts. r^ + (P x r^.

(P

moment

product

sense

Considering

of two

=

=

.'.

couple.

distance

36.

given body, up

moment

the

of

arm

couple,the

a

of as

the the the

is

known

direction

couple, and rotation

as

of is is

the

of the

moment

rotation

is known

positive or

as

negative

counterclockwise

or

ELEMENTARY

60

The be

GRAPHIC

of

solution

problems dealing

readily obtained in the

adopted The in its

by

of

in the

couple

a

couples will

with

application of

the

solutions

moment

STATICS

methods

the

preceding examples.

is the

all

about

same

points

plane.

magnitude of its rotational tendency, and two couples are equal if they both produce the same turning effect on a body, but if we have two rotational tendencies couples,whose are equal in magnitude but opposite in sense, the total effect produced is zero, or, in other words, the the other. one couple neutralizes A

if

Hence, effect

of

measured

is

couple

wish

we

opposite in

apply

must

equivalerUcouple, that but

balance

to

couple, we

a

the

by

is

or

neutralize

to

the

system

couple equal

a

the

in

an

moment

sense.

Examples. 1. Draw

a

it clockwise Bisect

regular pentagon of ABODE, beginning A

sides

the

The

forces. tudes

:

AB,

13

on

DE,

18

the

perpendicular

the

apex

2.

P

=

lbs.; and

on

lbs. ;

12

EA,

lbs.

14

CD,

on

mine Deter-

of the resultant, its direction of

distance

forces, P, Q, S, T,

ABOD.

its line of

=

action

PAB=120";

act

at

the

The

401bs.; Q

lines of

BO,

on

and

from

action

point.

Four

square

point.

apex

perpendiculars to represent have the following magni-

lbs. ;

magnitude

the

at the

raise

forces

On

lbs. ;

20

and

Letter

1 in. side.

magnitudes of 30 lbs.; T lbs.; S

60

=

of the

a6q

=

forces

150";

are

such

of

corners

the =

25

that

forces

a

are

lbs.

The

the

angle

b6s=135"; c")T=120".

NON-CONCURRENT

Determine

AD

AD

and

side

a

in.

IJ

of

Forces

long.

RP, SQ

of these

three

State

forces.

a

off to

horizontal

Through

in.

represent forces P

20

=

magnitude

a

the in

tion its inclina-

lbs.; Q

in it mark

line and

right distances

0C=1

A, B, 0, C

D

35

and

lbs.; S

in. and

0B

=

in. and

0D

=

draw

verticals

to

following magnitudes:

the

having

=

OA=J

point O.

a

lbs.; W

lbs.; T=14

28

=

lbs.,all forces acting downwards.

=40

Determine

the

5. Three

and

vertical

body.

a

between

forces

to the

The

Q

and

and

Determine

the

force which

would

the

forces

three

6.

in

upwards

instead and

lbs.

and

P

is 4

Q

ft.,

of the

magnitudes

respectively.

the

relieve

magnitude body

of

a

of

the

of

forces

single

action

of

P, Q, S. the

Question 4,

magnitude 7. A

60

The

position and

Considering

given

S,act downwards

between

S, 3 ft.

its

O.

forces,P, Qand

distance

80, 45

are

point

and

resultant

of the

magnitude

position relative

on

its

P, and

from

left distances

the

to the

IJ in.,and 2J

in

PQ.

4. Draw Mark

of

magnitude

the

lbs.,its perpendicular distance to

lbs. act

4

respectively. Using

polygon, determine

resultant

and

10, 8

QR

and

each

continuously PQRS,

lettered

square,

the directions funicular

it cuts

at which

A,

if necessary.

produced 3. Draw

from

distance

61

clinatio resultant, its in-

of the

magnitude

the

to

FORCES

system

same

but

assuming P,

position of

lever, O A,

3

ft.

the

long,

W

determine

downwards,

of

S and

as

are

to

act

the

now

resultant. is

pivoted

.

at

0.

It

62

ELEMENTARY

carries

at

hoop

rim

108", of

end

the

at

and

210"

forces

the

ABOD

is

and

45

90,

60

diagonals Determine

9.

If

of in

outwards to

cause

what

system

points

the

A

corner

the the

of

lbs.

34

in.

3

B, of

0

to

28",

are

magnitudes respectively. O.

system

about

side.

Forces

of

direction

of

the

along A,

ciding coin-

hoop

tangentially

act

The

the

of

circular

inclinations

and

moment

previous

D

and the

of then the

question of

directions

the

rotation

about

74

a

80, the

respectively.

system

about

the

square.

along

would

93,

outwards

the

S

T,

the

respectively.

square

a

of

whose

moment

act

at

radii

of

80,

the

Q,

it,

to

centre

P,

295"

are

Determine 8.

the

Forces

A.

attached

rigidly

A,

diameter,

with the

end

its

ft.

3

STATICS

GRAPHIC

the

square been

have centre

of

the

the

forces

the

sides

in

the the

square

had

acted

(all tending

same

direction),

moment

of

?

the

ELEMENTARY

64

leaving

a

part ABCD

the

is fixed

which

GRAPHIC

into

same

position which

of the

of

part ABCD

the

wall. load

and, by the addition

keep

from

The

P

cantilever

We

can

carries

able

are

and

also

before

occupied

part

extremity,

parts, we

equilibrium

it

the

its outer

at

certain

in

section.

central

detached

the

single concentrated

STATICS

in

to

the

removal

the

investigate the

now

effects of these

ditional ad-

parts, and, in

doing

will

so,

that

aasume

thecantilever has

with

and

and

dealing

are

we

only

itself

weight,

no

that

P

we

the

load

effects

the

produced thereby. The

of

effect

first which

P,

we

notice, is that tends

we

apply

must

the

prevented by same,

point.

an

must

This

be noticed

and

is

that

the

as

force, whether

In

have

shown,

force and

the

been on

conditions

balancing

accomplished this

tendency,

material

the

keep

to our

would

of

resistance

apply

to

opposite force.

downwards

and, in order

AD, we

equal

ABCD

this

counteract

to

motion

originalbeam, section

order

in

the

bodily down-

move

wards, and,

cause

to

portion

Fig. 37.

it

at

the the this

it will

artificial or

real,

BENDING

with

together

is

moment

load

the

P

R, and

x

66

MOMENT

P, forms

whose

couple whose tendency is

a

rotational

clockwise. This

couple constitutes

We

have

of

that

seen

only by Suppose, now,

couple of equal and

a

ABCD

that effects

the

this

is

produced

and

the

other

distance

AA^,

and

these

two

in the

model

AAi

tie in the

nature

chain

pull

a

in the

is

a

distance

from

r

is p

moment

tendency is

X

r

equal

of

x

strut, and

a

force

The

form

strut

acting

couple

at

a a

whose

rotational

whose

These

at

acting

the

in

a

R, but

on

P is therefore

by

and

the

couples

two

a

the

compressive

AD.

AD

is

The

twofold, for

bending

effect of on

ward up-

x

ing applysection

any P

moment

is neutralized

the

represented by

R, which

pr, and

resistingmoment

shearing force, which

stresses

material, while

applied at

we

harve induced

is neutralized a

which

tensile

shearing resistance force

to P

neutralized

are

,

the

force

we

the

equilibrium. In the original beam, pull,constituting the couple p X f are fibres

AD

of

in

induced

a

increasing

forces

two

another

the

the

Thege

one

push and represented by the

in

the

effects

counterclockwise.

system

a

of

tension, that

or

compression.

or

an

E,

about

lessening

a

chain.

a

be neutralized

rotate

to

by introducing,at DD^, of

with,

oppositetendency.

starts

DD^

form

the

on

to deal

tendency tendency can

distance

push

moment

this rotational

now

have

we

one

bending

AD.

section

and

the

by

the

also

shearing

resistance.

Shearing E

Force.

"

The

shearing forceon any

section

66

GRAPHIC

ELEMENTARY

of

beam

a

is

acting either

equal to

have

in the of

in

then

those

for

the

the

in

would

the

the

shown

in

be

as

in

increase

reason,

length,and

was

material

the

top layers would

forces

S.F." stress

of

the

be

extensible,

nature

length, while,

bottom

final

and

above, the

ease

cantilever

obviously,if

opposite

decrease

regfarding B.M.

that, in the

top layers of

all

of

the section.

leftof

Use

just seen

pull,and

a

algebraic sum

the

right or

Conventions We

to

STATICS

would

layers the

of

shape

Fig. 38 (a),the

lever canti-

being

curve

upioards;

convex

this

ing bend-

the

case

is said

moment

^)

(01)

to

be

positive and

the

bending

moment

drawn

the

C*

(c^

above

datum

Should Fig.

the

upwards^ tivCj and datum

is

diagram to the

tends

the

force

is

(6), that

38

is

said

take

be

negor-

the

below

drawn

then

up

is, concave to

tends

neighbour, as

shearing force, if

to

move

in

upwards

move

right,as

shown

said

be

above

is drawn

section

to

to

section

shearing

diagram hand

moment

the

beam

adjacent

the

bending

regard

the

the

Fig.

beam,

line.

With of

in

shown

shape

line.

the

38.

however, the

of

diagram

is

in

to

the down

datum

any

section

relative in

to

Fig. 38 (c), the

positive^and line, while,

relative

an

to

its

if

right-

Fig. 38 (d),the shearing

force

BENDING

is

said

below

be

to

negative, and

datum

the

of

bending-moment this

the

the

E

and

draw

EO

length

that

EC

of

EC

to

divide

AB

corresponding

line,on

which

AB

D

in

equal

AE

in EC

As

raise

the

well, at of

DC

line

Bisect

AB

in

such

a

and

CD

D, making DB

and

them

of

the

then

be

the

the

intersections

Bisect

parabola.

plete Com-

AE

and

into

them

points 1, 2, 3, 4

so

given and

EC

as

in the

points 1, 2, 3, 4

the the

points

the

AB

to

equal parts, numbering

numbered

in

described.

divide

join

lines.

cross

perpendicular of

the Join

Trace

is to be

on

number

any

shown,

numbers.

division

of

EC.

=

into

as

point C. From perpendiculars to cut

the

the

required depth

the

correspondingly

AB

to

before, let AB

parabola

of

Join to

Through

the

rectangle AEFB,

indicated. line

"

draw

number

same

the

methods

be

described.

required depth

the

the the

the

and

to

with

as

be

might

Fig. 39,

in

tangential to

Method.

Second

known

approximate

is to be

represents

and

required curve

it

perpendicular

Produce

the

dealing

curve

that

AB,

equal parts, numbering

across

the

two

Let

parabola

BD

and

In

curve.

"

which

AD

often

so

Method.

parabola.

is drawn

diagram

Parabola."

a

indicate

stage, to

First

the

diagrams

appears

setting out

67

line.

Construction

parabola

MOMENT

in

lines from AE

as

obtained

the

shown

trace

the

required curve. Note on

the

number

that

the

number the

accuracy of

of either

subdivisions:

greater the

accuracy.

method the

depends

greater the

68

ELEMENTARY

GRAPHIC

Loaded

Symmetrically

problems, where

the

Beams."

loading

Frg.

formly distributed, diagrams once

presents

the maximum

the

STATICS

is

or

bending

beam

many or

uni-

and

S.F,

symmetrical

39.

drawing

little

In

no

of the

B.M.

and, difficulty,

moment

has

been

when

deter-

mined, the diagram and

then

proceed

difficult Case End-

This and

The

words, the B.M.

equals zero. a

line AC

diagram,

and

since the

B.M.

The

rectangle scale. above

The the

Let

us

Free

in this

case

force

at

whose

S.F. is

now

B.M.

B.M.X"

is X

when

a;

the free

at

set

=L,

A,

at

up,

scale ;

bending-moment line AB,

datum

the

positive. is

equal

S.F.

the DG

height

positive and

is

to

W, and

is the is

diagram

represents

W

a

to

consequently drawn

line DE.

datum

that

suppose

S.F. at any

the

above

wall.

the

to

in

or,

x=Oy

suitable

some

is the

sections, hence

perpendicular

then

more

at

B.M.

AB, and

to

it is drawn

DEFG

diagram

WL

triangle ABC

all

the

hence

line

represent

shearing for

same

and

datum

or

The

when

is

zero,

base to

CB.

value

greatest close Again equal to WL.

becomes

Select

join

B.M.

is then

end

If

of the

Load

Concentrated

:

its maximum

B.M.

S.F.

ing apply-

unsymmetrical loading.

has

the

and

of

method

solution

to the

polygon

special

any

problems first,

such

with

investigatethe

I. Cantilever

B.M.

other

a

to

without

Consideringa section of the beam at X, we see is W the bending moment at that section (h"x). is obviously an expression of the first degree, will therefore be representedby a straight line.

"

that

The

of

case

drawn

will deal

funicular

the

be

can

We

construction.

69

MOMENT

BENDING

section, say to cut

in

the

B.M.

be

such

to find

wish

we

X.

From

diagram

X in

we ca

the

B.M.

drop and

a

the

gd. scale

(nxca)

that

1

lbs. ft.,and

if

m.^n

the

lbs. ft., S.F.

scale

70

ELEMENTARY

be

(m

such

that

II.

Case

Load.

"

in.

1

lbs., then

m

=

STATICS

the

S.F.x

=

lbs.

gd)

X

GRAPHIC

In

Cantilever: this

case

Distributed

Uniformly

the

load

is

evenly

distributed

W

WL

Fig. 40.,

along assumed

the

beam, and to

weigh

each

w

lbs.

L

lbs.

foot The

length total

of

load

the W

load

J is

is therefore

Considering,as before, the B.M. at a section X, and dealing with the load to the that the total load see right of the section only, we equal

to

w;

x

This

contains

expression of

is therefore

second

the

squared

a

degree,

The

Select

a

to

line

base

scale, and

to

hence

the

curve

when

x=o,

WL

w

equal

is then

and

quantity

representedby a parabola. above expression is a maximum

will be

and

STATICS

GRAPHIC

ELEMENTARY

72

L^

x

"

AB

= --_

and

set

then, by method

AC

up of

two

equal

describe

Fig. 39,

paraboliccurve, having its apex at B. The shearing force at X equals w (L- x). is of the first degree,and is a maximum

to

the

a;

S.F.max=t^;xL.

therefore

=0,

x=h,

value

when

S.F.

diagram

in which

length

the

and

ca

respectivelyat X, Case and In so

IIL

Loaded this case,

that

each

represents wh

to their

simply a

shown

support

will take

an

the

Our

shown,

as

W

to

scale.

S.F.

and

respectivescales. at

Central

Fig. 42,

when

zero.

B.M.

supported

Concentrated in

or

the

load

equal share

L

the

Ends, Load."

is central

of the load.

B.M.x=B,x^i^-a:) W =

/L

pression ex-

its minimum

equal to triangle FDE,

fd represent

with as

DF

a

measured

Beam

has

is then

is therefore

before,

As

and

It

This

\

2-n2-^)

=^(L-2a:)

;

BENBINO This the

is

expression

maximum

value

centre, while

the

whena;=-,

the

MOMENT

?d

represented by

occurring

minimum value

when

value

then

straight line,

a x

ends

at the

occurs

being

at the i.e.,

o,

=

The

zero.

maxi-

WL value

mum

a

=

The

-".

having triangle,

it is drawn

below is

diagram

equals

negative.

S.F.

at

obviously R^ beam.

its apex

or

Hence

and "

depth

in this of

line, since on

we

over

as

for

pass now

from

half, is

this

for

diagram,

estimating our We

have

S.F. in this half

the

loads

two

S.F.,we

take

must

acting upwards

"

and

their

that

This

the

S.F. is

value

is

equal

to

"

in

algebraicsum.

acting downwards,

W

W -

W=

-

"

.

from

constant

right-hand support, and

+

soon

with, and,

W so

as

S.F. alters,

of the

deal

to

positive.

find that,

right,we

centre, the value

the

is

a

above

is drawn

and (to scale),

"

left to

have

we

is

of the

in this half

is constant

S.F.

our

Passing

the

half

left-hand

the

in

W

datum

;

span

,

rectangle of height the

is'

scale.

section

any

of the

centre

maximum

some

j-

in this case,

lioe,since the B.M.

The

to

"

diagram,

at the

the datum

instance

The

B.M.

the

is therefore

centre

to

the

represented by

a

W

rectangleof depth", St

datum

line.

but

this

time

drawn

belUyw the

J

It

will

from

be

noticed, therefore, that

positive to

a

through

passes

a

a zero

of

point

the maximum

Case

IV.

of

zero

the

negative value, and, value, and

Fig. that

STATICS

GRAPHIC

ELEMENTARY

74

S.F.

S.F. in

so

passes

doing, noting

it is worth

42.

coincides

with

the

supported

at

the

poiot

B.M.

Beam

simply

Ends

Carpying this

in

a

the

across

lb.,

w

Each W

is therefore

reaction

equal

be

may

equal

is

or

half

the

taken

as

lbs.

wxh

to

^^,

to

to

case,

where ,

-"

tt7xL.

=

from

the

loads

the

to

load

at

Rg

whose {"n'~^\

distance

-k-

weight ( "

-

we

have

two

section, viz.,

the

and

X, of

piece

-

of

distant

X

an

at

acting counterclockwise

=

a

that

find

right

""

consisting of

section

any

we

(s* *) from

distance

a

B.M.

the

centre-line,

acting

upward

at

load

previous

equal

each

foot-run

per

total

the

hence

Considering X

load

The

load.

total

the

in

as

R^, are

reactions, B^^ and

the

and,

span,

tributed dis-

uniformly

is

Fig. 43,

in

shown

as

load

The

Load"

Distributed

Uniformly

case,

75

MOMENT

BENDING

the

isw

(

from a:)

total

load

-a:],acting

"

load

downward

a

of

length

clockwise

X.

={^x(i-)}-mh")(f-")}

.{^(l.-2"l)}-{H(l-^")(l-2. =

4a;L '^"2L2 -

-

-1(1.-4.=)

L2+

4a;L

-

4x^

)

16

fiLUMEtfTARY from

It appears is when

The

X

this

i^, hence

at the

its maximum

expression has the

at

centre, the

value

value

zero

B.M.

when

being

then

becomes

the

supports

gram dia-

B.M.

the

expression that parabola. The expression

a

==

STATICS

GHAtHlC

-"

is

a?

o,

=

or

-

zero.

i.e.,

"

"

o

o

The

of the

drawing

B.M,

diagram

should

present

now

difficulty.

no

S.F.

The

forces

two

at

X

is

left

the

acting to

of the

algebraic sum

to the

equal

(or right) of

the

section.

=u)x.

y Tlie and

S.F. is therefore

when

x^o,

The

centre.

a:

+-^, the

values

the

supports.

The

be

readily

=

The

been

of

then

S.F,

was

four

considering and

in view so

the

in

are

practice,the

practicalcases

being

of the

hardly

will

which

merit

of

are

the

term

of

B.M.

the

we

the and

diagrams,

the

quite justifiable

loadings. Examples

exception

or

at

adjoining figure.

arrangement more

diagram

setting out of the

the

the

calctdation

the

adopted

symmetry

nicely arranged

rule

before

occurring

-,

values

constructions of

and

"

at

when

occur

typical'cases

since

minimum

performed

still the

being

straightline, occurring

values

from

the

a

value

construction

"graphical solutions," maximum

this

maximum

understood

solutions

have

S.F.=o,

the

the

representedby

less

rather and

than

the

loadings

in

unsymmetrical.

BENDING

We

will

now

proceed to investigateone

unsymmetrical polygon

in

loading, making

constructing the

Fig.

Unsymmetpically first the

case

77

MOMENT

of

Loaded a

series of concentrated

simple loads

B.M.

use

two

or

of

the

cases

of

funicular

diagram.

43.

Beams." cantilever

Let

us

loaded

AB, BC, CD,

and

consider with

DE,

a ^

ELEMENTARY

78

Set down e

draw

to scale the

of

the

inches.

parallelto and

so

(w

in the

;

parallel to B.M.

the

A

space B

The

and

co

draw

with

an

In

do.

of the line

any

line qr

a

polygon

eo.

bo,

ao,

length

the lines of action

space

pq

parallelto 60, horizontal

the

figure pqrstu

now

sents repre-

diagram.

shearing force

The

projectingalong line

the

its

making

ae,

Join

closing the

on,

line tu

In

shown.

as

from

oft,be, cd, de, and

positiondiagram produce

loads

STATICS

loads

perpendicularto

eo

number

even

GRAPHIC

loads

the

constructed

is

diagram

from

shown

as

by load-

the

ae.

Note,

It is not

"

to the

that

necessary

load-line,but

be

eo

it

drawn

gives a

pendicula perB.M.

neater

diagram. Scales

scale

the

to which

drawn

drawn

been

is H

from

in.

in

ae), then

our

scale

example, I in.

lbs.,and

H

1 in.

10

=

=

=

4

L

ft.,and

that

been which

polar distance of

the

point

lbs. ft.

to

have

the

been

load-scale

drawn is 1 in.

to =

in.,then

(B.M. scale) =

10

o

scale

xnxR

suppose

to

the

further, that the

the

By

"

section.

has

the load-scale

lbs.,and

n

(N,B.

B.M.

1 in.=m

For

that

know

Suppose

cantilever

perpendicular distance

the

mean

is lin,

the

structed, con-

measured

S.F. at any :

so

we

be

follows

as

of

length

unless must

and

is derived

ft.,and

polar distance we

the B.M.

the

is 1 in.=m

has

ordinates

the

of B.M.

scale

scale

ae

which

to determine

in order The

by

diagrams,

worthless

less

or

more

are

The

S.F."

and

B.M.

of

X

60

X

4=

2000

lbs. ft.

a

50

80

GRAPHIC

ELEMENTARY

STATICS

y have,

section,

at that

is pq, and

therefore

moment

of the

puxB.j

pu

resultant

the

parallelto

of the

funicular

the

interceptis

last is tu^ hence

the

The

j"u.

first line

The

loads.

of the

line xy

a

about

system

representing

polygon line

the

this section

force

a

and

is

H

a

distaDce. be obvious

It will

things remaining also

definite

some

have

reduced

been

1 in.

ft.,it

m

=

scale, then

the

in

that

(pu x 12m) ins. lbs.,and consequently

would n

of this force

about

equal

to ^^^

This

may

The

depth

X

n

12m

x

x

H

written

in feet.

but

a

"

Let

Fig. 45, in

n)

diagram

length

lbs.

The

under

of

pu

moment

pux\2m

lbs.

ft.,which

the

section

study

us

which

often

of the

over

x

a

and

this

point,

will

come over-

arise.

the

now

expressed

inches

reasoning with

Beam

have

in

arises

above may

consider we

the result is

is measured

H

supported

Simply

is reduced

"

which difficulty

any

if

scale.

Confusion

careful

will

lbs. ft.

as

of the B.M.

of pu

is therefore

section

x

to

Hence,

diagram

puxl2mxnx"^

or

ft.,the distance

not

Loads.

m

length

full-size

would

reduced

ratio.

our

carefullythat, although

lbs.

as

be

B.M.

the Note

X

(pu

the

lbs. ins.,

xnxH is

is

pu

be

pu

other

Again, each inch represents of the force the magnitude

be

represented by

if L

same

times, and, therefore, the

12m

length

true

the

follows

the

that

so

doubled,

been

before, then

as

doubled,

been

have

if Lhad

that

Concentrated case

as

shown

simply supported

in beam

BENDING

carrying

conceDtrated

lines of

the

a

pole o,

the

majority

which

of

makes

aod

and

of

parallelto o", and

o

of inches.

(In

in

so

the

until

on

closed

for the

space

B,

space

the

a

is

point t

figure pqrst above

point A, a

any

p

draw, in the

q

the

diagram

draw

projecting

is the

beam.

oe

from

across

Considering the S.F. find

we

that

the

load-line

the

Rj

to AB

load-line. AB

between

Passing BC

and

is

Ri acting upwards

on

the

load-line

by

acting downwards. BC

is therefore

diagram The

for

remainder

a

and force The

eb

this

of

and

it is

be

can

Rj

or

ea

right,the S.F. represented by the sum

now

AB ea

nett

to

the

acting downwards, acting upwards S.F.

between

and AB

or

ah and

acting upwards, and hence our S.F. part is a rectangle of height eb. the

diagram requires no

explanation. Concentrated Beam with Overhung The following example is well worth F

beam,

is constant

to

the

on

indicated.

half of the

in the left-hand

S.F. from

as

R^, being also positive,so that height represented by a rectangle whose equal

of

Select

parallelto pty and the lines de and represent the reactions Eg and R^ respectively. The by shearing force diagram is constructed From

ea

from

oa;

pt, then

bending-moment

loads.

From

do.

R^ draw,

line

Join

down

set

best

60, co

ao,

parallelto

reached.

duce Pro-

positionfor o is that approximately an equilateral

line pq qr

CD.

and

the

number

even

the

line of action

the

forces

the

represent

an

cases

triangle.) Join in

is

and

AB, BC

of

cd to

H

that

so

loads

action

lines a", be and

the

81

MOMENT

further

Loads

oareful

"

cou-

GRAPHIC

ELEMENTARY

82

sideratiou, embodyiog, the and

Set

previous cases.

two

dCy select draw

space

A

arises

at

line

a

this

and

point

oa.

to which

as

have

a6, 6c,cd, de, and

have

ea.

the and

Now

point,a, ab,

or

of

represent R^

which

is the

the and

lines

AB,

that

oft,be, cd In

points. is the

the

force

close

the

line

the oa

meeting-point of

fa

and

loads

of

the

Difficulty usually

space

finallyto

notice

the

combination

space

A.

45.

round

we

a

the

parallelto

follow

we

down up

Fig.

If, however,

does,

join

pole o,

a

it

as

STATICS

and the

ab. line

These

polygon,

we

polygon

we

is drawn the

lines

latter

parallelto

to

oa

ea

lines must

line pq

crossed are

oa

in

be

position diagram.

the

is the line

lines should

other will

B.M.,

AB

parallelto

of the It

and

R^

connect

noticed

In

present this

in

of

in

constructing the

S.F.

points

the

sections

in which

of

diagram

to

for

be a

beam

diagram is

Beam: as

shown

adopted beam

the

S.F.

the

following case,

method

that

drawing

The

difficulty. have

we

a

with

when

a

46.

inflection

supported

Simply The

case

no

The

diagram. The points a:, jr, denoting zero indicate and called points of inflectioUy

curvature

that

required.

now

Fig.

change

83

MOMENT

BENDING

a

are

at

these

it is worth

points. noting

coincident,

with

maximum.

Compound in

Fig. 47,

Loading." indicates

constructing

compound

the

loading.

the

B.M.

84

ELEMENTARY

Draw

first the

out

loads

AB,

BC

CD

and

the

at

to

as

now

by

diagrams

combine

these

of

and

span

the and

equal parts, support,

this

beam.

the

Now,

this

at

gives the

us

point

one

The

shearing

loading

are

added

The

first to

form

example

another

employed without loaded

system

any

a

out

should

has

of

loading

be

with

the

each

and

the S.F.

Fig.

loading. from

apparent taken a

in beam

B.M.

From

manner

the

B.M., due

the mark

point

diagram.

for

explanation.

been

"

and

I

AH the

points obtained.

in

compound

beam

similar the

to

line a:y, we

B.M,

combined

shown

detailed

(ef+gh).

of

X

total

to

diagrams

the

as

of

case

in

drawn

The

gh. the

of

sidering Con-

due a

to

new

through

force

c/,and

to

(ef+gh), and

our

on

line xy

section

a

a

left-hand

the

B.M.,

a

cuts

to

treated

are

finallydrawn

curve

equal

kl, equal

ordinates

then

is

form

to

number

a

from

have

we

this ordinate

distance

a

X,

We

shown.

as

through

load, equal

then

point

into

ordinates

loads, equivalent

point kj where off

beam

passes

at

distributed

the

the

first ordinate

ordinate

concentrated to

of

taken

scale.

same

horizontal

a

B.M.

being

diagrams

Draw

raise

the

now

the

formly uni-

the

however,

two

B.M.

the

to

construct

to

gon poly-

bending

due

span,

care,

both

single diagram divide

uvw,

link

maximum

and

loading,

shown

construct

must

the

concentrated

the

by

the

of

centre

for the

shown

as

calculate

distributed

diagram

STATICS

B.M.diagram

Then

pqrst. moment

GRAPHIC

A

of

system

ordinates

are

diagram. 48

illustrates

The

tions construc-

the

diagrams

symmetrically

this

case,

but

of

this

type

any can

MOMENT

BENDING

be

quite readily

in the

employed It

should

load

on

treated

noted

beam

that

The

load

Fig. into

large

a

parts

of

acting should

the

at

work

should

of

number load

their out

are

the

be

is divided

beam

47.

treated of case

funicular

the

equal parts, and

centres

this

on

by

as

gravity. for

these

concentrated The

himself, a

readily accomplished further explanation. which

methods

distributed

uniformly

a

treated

be

can

method.

polygon

stages by the

in

preceding examples.

be

a

85

small loads

student

proceeding

without

any

ELEMENTARY

86

STATICS

GRAPHIC

Examples. 1. Draw describe

horizontal

a a

line AB,

4 ins.

long,and

its vertex

parabola having

it

on

ins. above

2J

AB. 2. A

is 14

cantilever

ft.

long

and

carries

free

end

2

a

its

load

of

Draw

tons.

mine deter-

and

by

ment measure-

B.M.

at

ft., 5

ft.

the 2

points

8 ft. from

and free

the

Check

end.

calculations.

by

3. If

lever canti-

the

in

above

the had

question

ried car-

uniformly

a

of

load

distributed

J

the

grams, S.F. dia-

and

B.M.

at

what per foot,

ton

been

have

would

the maximumB.M.? Fig.

Draw

48.

the

S.F.

by and

the

measurement

9 ft. from

the

B.M.

and

curves

and

checking

free end,

B.M.

S.F.

your

and termine de-

at 3 ft.

results

by

calculation. 4.

A

beam, 30 ft. span,

abutments.

of the

span.

It carries Draw

the

a

is

simply supported

load B.M.

of 3 tons and

S.F.

at the

on

two

centre

diagrams,

and

8S

ELtiMllNTARY

of

ft.

36

It

lbs.

200

weighing

beam,

A

9.

carries

1

left-hand What

is

the

maximum

What

is

the

B.M.

Assuming

10,

addition

in

graphically of and

the

beam

ft.

10

from

the

each the

diagrams. it

does

the where

the

beam.

B.M.

it

is

occur

S.F. the ?

of

160

loads

dead

and

What does

load the

to

Question

in

distributed

uniformly

run,

S.F.

where

of

from

S.F.

and

and

loads, ft

28

span

?

occur

right-hand

?

support

a

B.M., and

and

B.M.

the

a

load

3

to

14

8,

at

covers

distributed

addition

in

Draw

support.

foot,

per

uniformly

a

plfiW5ed

ton,

STAtiCS

lbs.

40

foot-run,

per

weighing

GRAPHtC

at

maximum

8

lbs.

given, the

centre

B.M.

to

carry foot-

per determine of and

length S.F.,

VI

CHAPTER

BEAMS

In

all

far,

the

fixed

the

known

dead

as

however, and

especially

position,

loads, free

are

loads

Such

beam.

the

girders,

crane

of

being

along

move

rolling^

as

are

practice,

instead

to

known

are

in

bridges,

in

loads

cases

many

centrated con-

distributed

Such

beam.

so

either

were

uniformly

or

In

structures,

in

loads

the

loads.

more

such

fixed

of

considered

have

the

points

length

LOADS

we

that

seen

at

along

which

cases

have

we

EOLLING

WITH

or

the live

loads.

Concentrated

Single examine

a

the

of

the

case

few

roll

from

the

B.M.

support,

we

Rg,

to a

Load"

concentrated

beam, shown

as

section

We

will

rolling loads, considering

supported

R^ at

of

cases

single

a

simply

Rolling

X

load

assuming in

Fig.

distant

x

49.

from

now

first

rolling along the

load

to

Considering left-hand

the

have "

RiXL p

.

"

=

"

W(L-a:) _W(L-a:)

^1

"

L

The

maximum

always

occurs

B.M., with

just

under

a

the

89

concentrated

load,

so

dead that

load, for

the

90

ELEMENTARY

given position of at

X,

and

STATICS

GRAPHIC

hence

the we

load

have

the

B.M.

maximum

occurs

"

B.M.x

=

KiXa: W(L-a;) XX

_W(Lic-a:2)

This

expression gives

the

for

B.M.

section

any

as

Fig. 49.

the

load

rolls

degree, the

across,

curve,

and, since

which

shows

it is of the

the

variation

second in B.M.

from

side

one

above

the

to

L, being then,

and

its maximum

values

equal

case,

when

to

when

a?=o

It has

zero.

and

x=-

B.M.Qentre

The

parabola.

a

its minimum

in each

value

be

other, must

has

expression

91

LOADS

ROLLING

WITH

BEAMS

=

T

_WL This

is the

obtained

for

as

horizontal

a

the

dead

treated

in

AB, and

set

which

B.M.

maximum

concentrated

a

supported beam, Draw

of

value

load

the

on

we

simply

a

previous chapter.

down,

at the

of

centre

WL the

span,

struct

B.M. its

parabola, having

a

below

draw

we

positions of of

for

diagram apex

if

perpendicular CD, equal

a

The

curve

and

is,

the

load,

out

the

B.M.

the

load,

on

The

that

account,

successive

that

the

apex

ADB,

curve

diagrams, of

spoken

the

as

'ofB.M,

^

+

\

when

The x=o,

the

left of X

is

obviously

"

degree, shows

straight line.

maximum

for

B.M.

the

often

shearing force just to

first

dotted, and

the

on

all

The

triangle,having

a

find

fall

con-

j-,

vertex.

shown

as

will

we

envelops

equal to +RiOr of the

is

and

"

its

as

diagrams

ADB

envelope curve

a

section

any

triangle will always

the

D

to

This, being an

that

shearing and

the

envelope

force

is then

expression

is

curve

obviously

equal

to

+W;

is a

92

ELEMENTARY a:=L

when

and

join

S.F.

the

line

GRAPHIC

and

EF, GF.

is

equal

E

at

The

set

the

that

the

then

equal

zero.

We

to

the negative the

As in

the

force

EH.

The

from

F

line

EH

from

rolls

B,^

left

the

EFH.

of

the

(or

the

load)

It

will

the

shearing

the

maximum

the

while

force

of

of

This

:r=:L, and

is the

is

is then

line FH,

equal

to

envelope of

is

be

force

shown

any

the

in

force

in

section it

triangle measuring take

must

we

be

the

to

the

by that

whether

S.F.,

shearing by the negative

shown

shearing

obvious at

is

variation

the

the

(or

load)

variation

the

Rg

to

force

positive shearing

shearing

"be

right

its value a

the

can

W^.

-

when

x=o

W,

to

shearing force.

triangle EGF,

a

down

set

load

to

right

when

it

the

to

=

value

equal

way

W

-

zontal hori-

a

envelope of

same

^(^^^)

to

W, and

"

join

is the the

its maximum

now

W, and

-

In

line EG,

a

up

Draw

zero.

shearing force, just

load, is equal

expression has

to

GF

line

positiveshearing force. shown

STATICS

positive

or

egative. It should say

Xj,

that, if

be noted

the

shearing

positive if

W

approaches

from

is

force

consider

we

at

that

section,

any

will

section

approaching from Rg, while, if R^, the shearing force will then

be W be

negative. Distributed

Uniformly case

have

per

which a

we

shall

uniformly

foot-run, and

Rollingr Load."

consider

distributed

of

length

L

is that

in

The which

load, weighing

moving

across

next

a

w

we

lbs.

simply

BEAMS

supported see

that

beam

of

LOADS

ROLLING

WITH

L.

span

ExamiDing

93

Fig. 50,

we

"

B.M.x and

R9

.-.

X

=R2(L-a;) L

=ti7a:

X

"

B.M.x=^(L-x).

Fig. 60.

This

expression,of

the

second

degree, has

its mini-

when

mum

values

case,

equal The

right. the

to

The

zero.

load

B.M.

the

the

at

fully loaded,

B.M.

the

left

from

moves

occurs

is

span

of

magnitude

the

as

maximum

when

span,

L, being then, in each

and

x=o

increases

gradually

STATICS

GRAPHIC

ELEMENTARY

94

of

centre

and

to

is then

WL

equal

; the

to -3-

parabolic

being

shown,

as

curve,

0

the

of

envelope

of

maximum

the

and

beam,

B.M.

hence

supported When

distance

a

obviously equal

for

curve

T"

Now

simply

a

will

travelled

we

the

to

on

S.F. at X

The

is

Rg.

to

XT

is

S.F.,

has

(see lower figure).

B.M.

load.

the

only

a

distributed

consider

load a

of

curve

uniformly

a

for

simply supported

a

B.M.

the

to

the

on

value

is the

obtained

we

envelope

with

come

that

load

to

beam

we

assume

which

the

similar

exactly

This

B.M.

distributed

uniformly

beam

maximum

^^^

"

T

RoXL

W?aX

=

V.

7r=

^

-rr-

2

2

"^"^ p

^^

. "

The of

a

and

greatest words, load

value

the

shows

the

X

beam

will

be

right

up

and

is

=

^rY

increases.

,

envelope

curve

It

equal to

becomes

a

at

S.F.x

a

as

when

S.F.

fills the

"21:

Rg obviously depends

increases

maximum sion

of

value

"

on

the will

to

the

have

its

x, or, in other

maximum

a

length

when section.

the The

negative. Theexpresof

negative

S.F.

to be

96

ELEMENTARY

diagrams, pqra B.M.

scale.

and

In

the

STATICS

GRAPHIC

uvWy

order

be

must

drawn

to obviate

maximum

the

same

culating cal-

necessityof

the

bending

to

in

moment

the

case

Fig. 51.

of the with

rollingload, the graphical advantage,

Set down value Draw

of the

fo

fo equal ut

a

the to

be

applied in

this

line de to represent

rollingload

which

perpendicular H.

Join

parallelto do, and

from

w

w

L, the maximum

x

come

bisector

on

of

wt

the

span.

de, making

From

eo.

draw

might,

case.

can

and

do

construction

u

draw

to parallel

oe.

BEAMS Draw

ROLLING

perpendicular

tx

Using

WITH

ut and

parabola

wt

and

uw

construction

as

having

uvw,

to

97

LOADS

it

bisect

lines, construct

its vertex

at

in

v.

the

v.

i a

nil////.

I..

11

1

"I

tint.

h

""

t"

.

f}}))})n}^

Fig.

Draw

a

new

base

of ordinates, the combined G

52.

line yz^ and

as

diagram

on

this,by

the

tion addi-

previously explained,construct for the dead

and

rollingloads.

98

GRAPHIC

ELEMENTARY

The

B.M.

greatest

which

STATICS

at

occur

can

section,

any

,

is

X,

say

-The

lbs. of

shape the

on

first

then

the

the

direction

and

addiug

By the

the

left.

then

case

lower

diagrams, when

the

the

diagram

load

load.

loads, load,

rolling first

span

ordinates

we

Fig.

in

from it

moves

can

the

one

construct

the

shows

52

diagram

upper moves

when

dead

the

the

for

pend de-

rolling

the

for

diagram. the

Will

other.

corresponding S.F.

of

the

across

in

scale.

B.M.

the

diagram

diagram

diagram

move

combined

required the

to

force

motion

S.F.

the

out

construct

it

of

by

feet.

shearing

direction

draw

assuming

multiplied

gh,

to

=g'AxmxwxH

B.M.x

We

equal

representing left

to

from

right, right

and

to

CHAPTER

ROOF

Framed

TRUSSES"

of

the

roof

and

trusses

external

In of

induced

stresses

loads,

have

DEAD

Structures."

practical application

VII

LOADS

ONLY

with

dealing

Graphics in

the

to

the

more

determination of

members

various

bridge girders by

the

of

application

the

we

consider

to

y////^y////////A what

known

are

framed

as

A

tures, struc-

is

Frame

ing consist-

structure

of bars

a

several

jointed

their

at

by

ends

gether to-

pins, 53.

Fig.

free

in

motion

composing Fig. and

53

it

is

Such

a

frame

shows

an

are

a

frame that

obvious

adapted

readily

their

round

plane

one

of

allow

which

to

carry

arrangement

known

members.

as

of

composed such a

load

is, 99

a

bars

several

The

centres.

members,

four

contrivance W

across

however,

could the

open

space to

be S.

this

100

ELEMENTARY

GRAPHIC

objection,that, should due

to

would

lateral

any

BD

tend

pins

about

to rotate

would

render

for

suitable

this motion,

frame

the

frame, owing

conditions

of Frames.

Types into

used

distinct

three

To these of

means

mation this defor-

some

roofs

in

and

way

and

bridges do

Firm

known

are

a

trtisses.

as

be divided

may

of

of the parts,

arrangement

Frames

classes

(a)

and

all loads.

in this way

"

D

theoreticallyrequired

different

to the

frames

hence

used

AC

under

some

in

and

load.

prevent

joints

under

frames

the practice, fulfil the

and

stable

adopt

must

we

the by stiffening,

making

bottom

its work

therefore

must

we

preventing

their

consequently displace the

frame

the

conditions

not

force, for instance

wind, be applied to it,the members

C, and

In

STATICS

roughly

"

Frames.

(6) Deficient Frames. (c) Redundant Frames.

Firm shown

in

frame

possesses

An

"

example

(a).

Fig. 54

just

Frames.

It

of

will be noticed

suflScient

under appreciable deformation plane of the frame, provided the

be or

beyond

noticed

that

any

shortened

limit

their one

without,

of

to

members

member

may

in

way,

is

such

a

prevent

load

any

in are

It should

safety.

any

that

members

any

stressed

frame

firm

a

the not

also

lengthened

be

affecting the

'

stresses

Such

in the a

breaking

frame

members.

two

is stable

for

all

loads

within

the

load.

Deficient frame

other

Frames.

is shown

in

"

An

example

Fig."4 (6). SuqH

of

a

a

frame

deficient is not

^ v#

^

ROOF

TRUSSES"

DEAD'

^

,

O

'

LO'AfoS ONti'S

lOI

to prevent deformation possessedof sufficient members the applicationof a load. If we refer to our on remarks will readily the funicular on polygon we that there will be, at least,one see system of forces which will keep the members in the of the frame given shape,but should one of these forces be altered, the effect would be such that the frame, as presently outlined by the members, would no longer coincide with the link or funicular polygon, and hence an alteration in shape would occur. Such a frame would

cb)

(OL)

CO

Fig. 54. therefore

not

be

so

as

made

by

the

dotted.

shown

it into

be stable

under

all

loads, but

applicationof Such

an

another

addition

would

it could

member, convert

frame, possessingstabilityunder any system of loading, in the plane of the frame, and a

firm

possessingthe further advantage that any one member yet be lengthenedor shortened without may in any way the stresses in the others. affecting An Redundant Frames. example of this type is frame in Fig. 54 shown (c). A redundant may also

"

possess for

one

or

more

but stability,

members

more

it is stable under

than all

is necessary

loads,although

"

""

"

"

' '

'

162'

*'

"

C

t

"

"

"

E'LEMEi"rTAEY

possessedof the length of

disadvantage,that

this

a

STATICS alteration

any

aflfectsthe stresses

member

one

Such

others.

GRAPHIC

frame

would

therefore

in

in all the

be aflfected by

workmanship, and carelessness in marking oif the result in initial would lengthsof the various members stressingwhen the frame was fitted together. Such frames are sometimes spoken of as self-strained frames. If ri= number of joints in a frame, then a perfect frame should have {2nr 3) members. Although roof and bridge trusses do not always bad

"

the

meet

is found

requirements of a frame, yet it that, in the generadvantageous to assume ality theoretical

of cases,

the

theoretical

conditions

are

satisfied.

Fig.55, for example, the rafters all in one and CE AC are piece,instead of being the tie-rod, while jointed at B and D respectively, of two of being formed instead separate members and FE, is all in one AF length. The joints B, C, D In the truss

shown

in

securelybolted or riveted, but, for exist at these that pin-joints assume we our purpose, and that AB, BC and such parts are all separate points, Such assumptionsadmit of the frame adjustmembers. and

F

would

be

ELEMEllfTARY

104

from

the

material

It

used

covering.

in

Distribution

of

on

In

"

elevation

in

truss,

the

of

weights

minimum

with

general,

joints, and of

the

the

loads

at

considered. shown

as

shown

As

56.

in

are,

the

be

now

the dead

dealing with

magnitude

roof

the

Fig.

piece

a

and

plan

in the

by

principal,carries

or

of the

nature

table

these

truss

will

points

consider

us

each

roof

a

computing

various

Let

A

acting vertically at

as

these

the

on

later.

of Loadtngr.

acting

method

the

depends

roofing materials, together

pitches,is given

taken'

STATICS

horizontal.

of various

loads

GRAPHIC

shaded

area

of

roof

the

P

extending

if the

P, and total

(2L

P)

X

obviously (2L indicate

distributed of

section We

have

x

sq. P

one

be

L

rafter

the

ft.,then

Now

total

if this

roof,

over

the

the

rafter

gets

four

rafter

members,

we

principalis ft.,then

rooting

the

for

the

to

carried

is

convenience,

take

can

be

material

load

weight

be

we

uniformly each

it that of

W.

R^, Rg, R3, R4,

and

equal

an

line,so

centre

principal will

one

w?) lbs.,which,

X

W.

as

by

Assuming

lbs. per

w

carried

supported by

ft.

sq.

the

of each

length

of roof

area

weigh

can

of roof

the total width

that

of

side

either

on

"

share

W each

hence

carries

one

either

assume

on

of "r-

rise

lbs.,which

its centre

along

R^, gives

say

load

through

distributed

uniformly acting

acts

a

of

the

member.

to

reactions

we

gravity This at

1

can

or

is

load, and

2

W

each

equal

.

to

"

;

similarly,the

load

on

Rg gives rise

TRUSSES"

ROOF

LOADS

DEAD

106

ONLY

W to

reactions

2 and

at

3, each

equal

to

and

---,

bo

on

8

It will

right round.

observed

be

Fig.

the

abutments

mediate

can

be Let

only get

points get stated N W

in

two

number

=

total

one

on

joints over

while part, i-rrjy

parts, (2x

of rafter load

the

56.

general form,

more

=

that

) or

"

thus

members

on

.

"

or

parts.

truss.

joints over

abutments

=

-^.

W Load

on

intermediate

This

"

W

Load

inter-

joints=r|^.

106

ELEMENTARY

In not

dealing be

with

that

An

of

the

Wi

roof

any

on

roof

a

itself may

load, and

add

hence

of

it must

truss

it is

very sary neces-

approximating

truss

before

formula

is

the

given

design below

is

to

got

"

feet,

pitch of principalsin feet,

=

approximate weight

=

following

foot of different minimum

in

STATICS

truss

means

some

W,

may

dead

Z=8pan P

The

the

approximate Let

The

loads

that

have

we

weight

oufc.

the

forgotten

considerably to the

GRAPHIC

=

in lbs.

|fP(l+i).

table

gives

the

weight

per

roofing materials, together

pitch at

which

they

may

square

with

the

be laid.

out, following simple example, fully worked lead to a better understanding of the foregoing.

Example. 55, has

a

"

span

A

roof truss, of

the

type shovm

the raftersbeing of 30 /^.,

16

in

Fig,

ft long.

TRUSSES"

R(X)P

DEAD

LOADS

107

ONLY .

pitch of

The

is covered

with

lbs, per

7 0

each

the

principals

|

ft.,and

10

boarding,

in,

and

the

slates

what

Determine

ft.

sq,

is

roof

weighing

load

comes

on

joint.

Our the

first

is to find

above

the

this

itself,and

truss

given

step

of

approximate weight

is

given by

the

formula

"

30\

=

Jx30xl0

=

fx

We

must

material

per

in. thick,

70

weigh

of

weight

roofing

weighs

lbs. per

2*5

lbs. per

sq.

purlins

1'5 lbs.

sq.

ft.,and

of

roofing material

ft.,

sq. ft. Hence

ft.

the total

sq. ft.

Boarding, f slates

lbs.

determine

now

per

30x10x4

900

=

(l+?^

the total

weight

70 + 2-5 + 1-5

equals

W

Now

Total

" .

.

=

load

is one

now

w;x2LxP 11x2x16x10

=

3520

weight one

divided

part

sq.

lbs.

=

by

This

11

=

per

each

Iba

carried

=

truss

=

900

=

4420

up to A

Wi+

+3520 lbs.

the

among

and

W

E,

and

joints,apportioni two

parts

each

B, C

to

D

and

be divided

must

STATICS

GRAPHIC

ELEMENTARY

108

;

other

part=

"

lbs.

=553

-

"

the

words,

eight parts, hence

into

Each

in

or,

load

"

(say)

8

Load

"

Load

"

E

lbs. each.

553

=

perhaps, of

plain rafter

the

which

have

we

their

together at

Fig. 57

in

shown

type

lbs. each.

1106

Roof.

of

Form

Simplest

with

and

B, C, D=

at

.

.

the

at A

.

.

With

"

used

as

to deal.

first

The

scale, the

the

the

three

these

the

a

are

beset

the

stand to with-

suitable

some

then,

side, and

to

problems adopt

joint

at

form

such

some

various

round

go

so

polygon for

each

vention con-

joints, joint in

time, starting

a

the

polygon of

joint.

with

unknown

three

with

of the

is to

direction, each

for each

on,

to

are

on

able

force

dealing

adopted

Starting then once

down,

the treatment

left-hand

the

forces

to set

unclosed

In

forces.

method

clockwise

from

be

rafters.

above, it is advisable

one

the

abutments

of the

realitythe

regarding and

ends

wards joint-loadsacting verticallydownand roof AB, BC CD, in the line abcdy

is in

which

fastened

are

bottom

to

form

three

on

as

operation is

simplest

rafters

The and

exception, lean-to roof,

a

the

spreading by prevented from supporting walls, so designed as thrust

in

is about

top ends,

the lateral

the

at the a

abutment,

inasmuch difficulty,

in either to

left-hand

the

forces, of which

magnitude apex one

as

or

point,

we

is known

we

the

are

reactions

direction. find in

at

there

ing Passare

magnitude

TRUSSES"

ROOF

and

DEAD

direction,while

directions

the

ONLY

LOADS

of

the

109 other

two

known.

are

Going with

point clockwise, find force BC, we

round

this

known

our

and in

starting our

force

next

force

Fig. 57.

line

polygon

the

in order

is CE,

parallelto EB, and to EB

the

draw

yet, the pointe

representing

and

from

member

therefore

(we

be

we

from

is not

c

CE,

we

in

this

draw

now

while

draw, from b

it ; the

the

line

a

third

ce

force

is

6, a line be parallel instance, because,

fixed). Join

ea

and

ed.

as

Then

eb and

the rejireseiit

ec

while and

direction

shows

CD

instead

frame have and

of

of

by

the

the

of roof

no

the

lateral

modified

rafters

is taken

this

on

noticing.Consideringany

that

the one

by

a

the

tie-rod

thereby making

In this case,

components, they

effect of

so

58

58.

abutments,

self-contained.

Fig,

Single Tie rod"

with

Truss

thrust

spectively, re-

reactions.

Fig.

outward

EC

and

EB

represent the magnitude

ed

previous type

the

in

stresses

and

ea

of the

Roof

Simple

STATICS

GRAPHIC

ELEMENTARY

110

the

since the reactions

will be

perpendicular,

loads p^ and of these

p^, ^^ worth

points,we

have

ELEMENTARY

112

GRAPHIC

complicated structures,

adopted whereby and

of the

in any

stress

if the

we

At

the

determine

can

members

be

three

the

be

of

the

nature

the structure,

composing of

must

be

procedure

adopted,

encountered.

point

apex

the

along

special method

some

following method

should diflSculty

no

STATICS

have

we

three

forces

given directions, and

acting

forming

a

simple system in equilibrium. The three forces will force diagram by a trianglea5tZ, be represented in our sides taken

whose

of the

nature

only

the

at

will

forces.

several

force

one

in order

give We

hence

(b)

to

left,or

in the direction

Similarly

Hg.

of

direction

pushing that

were

We

reactions.

each

on

Fig. 59 (a),and the

first

point

at

indicated

also

have

therefore

the

sight it

the

pointed outwards,

as

member

at

BD and

from

and

is

a

(a),we

the

in

are

find the

heads arrow-

as

in

shown

compressive

convergent,

seem

arrow

points at

that

6

right

would

we

6,

to

DA

divergent to

is in

a

from

it acts

the

on

two

stress

be

would

Both

similar

member, hence

DA,

pushing

in

a

the dotted

by

apex,

each

is

order

acts

in

force

the

in

in BD

Hj.

arrow

diagram

from

down

then

force

the

arrowheads

Fig. 59 (6),then At

with

the the

on

they

Should

the

point,

direction, so that

pass

and

d, indicating that

the

lettering a6

we

of

is dovm-

force

our

downward

Fig. 59.

to

in

the

to

sense

that

towards

the ^

the

apex-point, and

/Q^i

clue

a

know

mards

-*

us

one.

shown

in

tension.

that, if the arrowheads had

a

decided

indica-

ROOF

of tension, but

tion

be

DEAD

TRUSSES"

but

itself acts

member If

therefore

then

these

arrowheads

is well

to

of

modification

and

example, the

the to

apex

the

only duty

whose

horizontal The

is

and

order

in

following method, good

as

as

as

any.

60.

This

of

is

that

it to to

truss

of

length be

the a

assist

general,about

and, since the rafters

points, this H

ing indicat-

convergent,

the

stress

horizontal

in this

redundant in

a

king-

member,

supporting

the

tie.

rise is, in

20

and

simple shown in the preceding truss being a light king-rod from

found

showing

is zero,

joints,

two

the member,

system

some

the

centre

It will be

tie-rod. rod

addition

its extremities. on

on

will be

Truss."

the

the

compression.

Fig.

King-rod

heads arrow-

in which

outwards

by Fig. 60, is,perhaps,

Simple

the

expected.

cultivate

tabulating results, indicated

CO

on

is to be

tension, as It

in

it will be

will

the member,

manner

inwards

act

that

joints at

acts

jointsmust

consequently The

member

a

113

matter

acting on

of the

the

on

the

in mind

the forces

indication

an

as

serve

bear

we

indicate

do not

ONLY

in diflSculty

any if

easilyovercome

LOADS

are

construction feet,

one-fifth

unsupported is unsuited

of the

span,

at intermediate

for

spans

ceeding ex-

GRAPHIC

ELEMENTARY

lU

Roof

Swiss

Truss."

above, allowing

from rod to

the is

fg

longer

no

in

BC, CD

the

in

has

(i^.B." AB Now,

since

will

reactions

CD

and

be

the

will

divide

give

the

ad

in

magnitude

the

equal

seen

kingforce

loads

AB,

neglected) in

be

loading

each

Fig.

we

may

the

construct

down

be

equivalent

amount

To

the

centre

in the

stress

an

diagram.

will

It

62.

the

begin by setting

we

abed.

that, if

Fig.

that

but

zero,

force

the

shown

in the

headroom

more

of

modification

further

diagram

stress

diagram line

little

a

building,is

of the

A

STATICS

to

is

the

symmetrical, another,

one

so

61.

e, the

of the

two two

lines total

ae

and

ed

reactions.

the left-hand abutment, and going Considering now have round the point clockwise, we ea up, paralleland EA; then, following this, we equal to the reaction is the one force now have ab down next acting ; the 6 in the force diagram we draw along BF, and from The 6/ parallel to BF, in the position diagram. only remaining force acting at the point is that line ef parallel draw from c a we along FE, and to

and

EF

to

follow

cut

the

6/

in

/.

forces

Take round

now

the

apex-point

clockwise, and

lastly

TRUSSES"

ROOF

DEAD

the

right-hand abutment,

be

experienced

in

ONLY

LOADS

and

difficultyshould

no

completing

115

the

force

diagram

A

Fig. 62.

as

The

shown.

stresses

can

then

be

spans

up

scaled

off

and

tabulated. This rise

is

type

is suitable

usually

about

for

one-fifth

of

to

the

ft.

The

span,

and

20

the

of

rise

the

STATICS

GRAPHIC

ELEMENTARY

116

tie-rod

one-thirtieth

about

of

the

span.

Compound

Swiss

For

Truss."

larger

spans,

the

R.

Fig. 63. outline

in

the

previous truss

Fig. 63, Swiss compound

shown the

of

should

be

the

truss

truss.

apparent from

the

may now

The

be

modified,

being method

known

of

as as

solution

figure. If,however,

the

found

will be

the

that

since

that

point

readily

a

have

three

magnitudes

unknown

be

indeterminate,

is

solution

the

will

It

determined.

be

cannot

e

that

seen

we

R^, it just after the point over arises,due to the fact difficulty

taken

apex-point be.

117

ONLY

LOADS

TRUSSES"DEAD

ROOF

deal

to

with. be got over can by treating the point difficulty R^ and by so doing getting the force triangle

The X

after

makes

this

cde;

possible

it

the

construct

to

force

polygon abfed for the apex-point. It should be noted that, in completing the force diagram, the line c/ forms line,for the points c and / are fixed, a check coincide with and a consequently the line c/ must line through c parallelto CF. Swiss

Compound The

Struts." at their

known

as

that

on

the

case,

this

force

the

is

the

rafters the

right angles to

is sometimes

truss truss.

diagram

difficultyexperienced

solution

Right-angled

support

at

are

account

out

the

but

to

right-angled strut

drawing with

met

added

are

mid-points. They

rafter,and In

struts

with

Truss

possible by

we

in

are

the

first

again

previous

treating

the

point a;,before passing on to the apex-point Another solution,though not strictlygraphical,might easily be

know

be

We

adopted. the

got Let

very T

=

W

the

readily in tension

point h in HE, tension

the

=

=

as

soon

this

and :

we can

"

in HE,

8pan, load

as

following manner

perpendicular distance

y=

2S

of

value

fix the

can

at each

joint.

from

apex

to

HE,

GRAPHIC

ELEMENTARY

118

Taking

about

moments

apex

STATICS

we

have

"

R,xS=(Wx|)+(Txy), y

This

tension

can

now

be

marked

off in

the

force

I

f

Fig. 64.

diagram manner.

in

the line eA, the

point h being fixed

in this

The

of the

addition be used

to

GRAPfllC

ELEMENTARY

120

In

the taken

again Bj

and

^^^^

^^^

represented,in

are

and

The

de.

points

k

in

equal

figure,the and

and

that

in

CD,

the

member

round

going

the

common,

are

check

a

ea

the

that

are

under

joint

forms

KH

and

the lines

diagram, by

only points to be noticed / in the force diagram

and

reactions

the

hence

is

loading

J(AB+BC+CD),

to

force

the

king-rod truss

the

ft.

the

symmetrical,

as

Rg

30

to

up

shown

case

allows

struts

for spans

STATICS

line

the

load

in

force

polygon. Truss

King-rod In

order

of the

to

gain

a

with

Struts

little

more

headroom

floor,the previous type of

modified, and

The

horizontal. followed

from

the

k and

this

type of roof

/

longer for

is used

fit additional

to

centre

is sometimes

truss

inclined,in place of being

diagram, no

are

in the

will

of solution

method

that

usual

ties

the lower

Ties."

Inclined

and

and

be

it should

readily

be

noted

When

points.

common

ft.,it

30

over

spans

suspension rods,

is

shown

as

dotted.

Swiss

Truss

67

shows

Fig.

with

Two

further

a

or

Truss."

Belgian of

modification

of two

addition

truss, by the

Struts

struts

to

Swiss

the

support

the

rafter. In three

equal parts.

settled, and drawn

to

point X

the

tion

the

setting out the cut

truss, the The

height

perpendicular

the

struts

of the rafters.

line are

of

the

drawn The

rafter

is divided

of the

tie-rod

bisector

of

tie-rod to

drawing

the

is then

the

in x;

points

of the force

into

from

rafter this

of trisec-

diagram

TRUSSES"

ROOF

and difficalty,

presents

no

from

diagram.

the

This

DEAD

type

of truss

is

should

and

NO

of this of

are

tie-rod, and vertical.

type, in which

wood,

was

be

ONLY

the

case

a

simple

and

rafters

first introduced

without

any

66.

in such A

121

readily followed

frequently made

Fig.

rise in the

LOADS

by

the

EL

struts

inexpensive

and

struts

Mr

EL

P.

are

roof

made

Holt, C.E.

ELEMENTARY

122

This 20 10

type

to 50

ft.,and

be

can

ft.

GRAPHIC

standard

The the

designed

STATICS

for

spans

pitch

of

tied

principalsare

ranging

the

from

principals is

together by

the

Fig. 67.

purlins,which, being notched

on

English truss

with

three

of

a

substantial

section,are

rafters.

to the

Truss.

wood

"

The

English

struts, and

truss

is used

is

for

a

king-rod

spans

up

to

ROOF

TRUSSES"

LOADS

DEAD

123

ONLY

drawing of the stress diagram will present but, in drawing out the frame diagram, difficulty, The

70 ft. no

the

KL

members

and

WX

O

might

with,

as

stress

no

loads, their sections

Simple

sole

of the

lower

Queen-post

dispensed

68.

by

in them

is induced purpose

be

e

7

Fig.

well

being

to

external

the

support

the

long

tie-rod. Roof.

"Fig.

69

shows

a

type of

truss

X

known

known

being

wood, this forms up

to 35

ft.

superppsed on

a

being

noticed, from

types

of frames,

deficientframe

a

Such

frames.

a

under

would,

be deformed,

and

the

but

In be

cannot

two

portion

wings firm of

loading,

crossbeam

69.

massive

tions propor-

sufficient iron

obtained

side

wooden

tie, is usually of such

the

composed

centre

system

practicethe

in

to aflFord

as

the

unsymmetrical

an

is

diagram

theoreticallyconsidered,

structure,

Fig.

forming

the

for spans

line

the

the structure

the truss, that

in

constructed

of roof

good type

very

It will be

distinct

of two

As

queen-posts.

as

X,

members

queen-post truss, the

the

as

STATICS

GRAPHIC

ELEMENTARY

124

from

mation. rigidityto prevent defortruss, however, this rigidity the

iron

tie-rod,and

hence

of

counterbracing the central portion The be adopted. must simple queen-post truss, as in Fig. 70, shows shown modification of this truss one to adapt it to the requirements of iron construction, means

some

for spans

to

30

ft.

difficultywith

The when

up

we

come

to

indeterminate

consider

the

forces

occurs

apex-point, but

this

ROOF

can difficulty

marked

x

treating the solution

be

got

after

that

point

will be

LOADS

DEAD

TRUSSES"

125

ONLY

by considering the joint load the under AB, or by

over

The

counterclockwise. from

easilyfollowed

the

complete

diagram.

Fig. 70.

Queen-post.

Compound

Figs. 68 a a

little be

study,

truss

to the

the

the

members

comparison

a

that

this truss

The are

opposite order

uprights

ties,while of

of has

English trusSy but, by

following important

readily noted.

the inclined

From

71, it will be noticed

close resemblance

very

will

and

"

things

are

in holds

distinctions

struts, and the

English

good,

and

ELEMENTARY

126

also

inclined

the

slope

from

English

GRAPHIC

the

members

the

the

truss

queen-post

inwards, while

bottom

slope from

truss

in

STATICS

bottom

those

in

the

outwards.

^

71.

Fig. force

The

diagram,

clue to the

ready

of this solution

French

worked

complete

type exist,but

along Roof

the

lines

Truss."

out

alongside,gives

solution. all lend

Many themselves

tions modificato easy

indicated.

Fig. 72

shows

a

very

common

a

ELEMENTARY

128

ances,

the

forces

acting

and

solution

of

know In

(a). We OP

to

thrte diflSculty,

this

magnitude

the

four,

other

methods

tion of solu*

**

us.

that

assume

can

six

the

"*

the is

only

MN

in

stresses

equal. (Thip,however,

are

the

of

only two, while,

overcoming open

know

point,we

their directions.

only

are

STATICS

indeterminate, for, of

is the

on

direction

we

GRAPHIC

and the

when

true

we loading is symmetrical.) By this means the point p, by the aid of simple geometry, further in that half difficultywill be met

fix

can

and

no

of

the

frame. We

(6)

fix the

can

point r, by calculatingthe

tie-rod KR,

in the horizontal

example, by taking

explained in

as

about

moments

considering the equilibrium of (c) We This

make

can

The

most

first two

PQ,

in

shown are

member,

point,we

by

previous and

apex

of the

substitution

truss.

frame.

Barr,

Professor

is

by

method. satisfactory

explanation,but study. As

the

one-half

the

first introduced

method,

far the

of

use

a

stress

methods

do

the

is

Fig.

third

72

omitted, and, XY, have

worthy

(a),the in

two

their

is introduced. 6c

call

not

If

for any

of

parallel to BC,

little further

members,

stead, we

a

now

from

further

OP

and

substitution

a

go c

we

round

the

draw

ex

be drawn parallelto parallel to CX, and xn can now XN, the point n having previously been fixed by going We round the joint at the foot of the strut LM. can now

force

go round

the

point under

polygon being cd, dy^ yx,

CD, the lines forming the and

finallyxc.

We

now

to the

pass

lower

replaced,and before

Vlll

in

usual

undtfr

those be

The

BC

encountered

OP

acting

substitution

of the

CD.

and

until

and

74, are

the

Fig.

73.

Fig.

74.

over

spans The

through in

a

60

are

now

joints are

now

joint

over

roof

truss,

when

this

as

type

FG

under

is

use

tions modifica-

The

shown

1

difficulty

by making

before.

as

adopted

PQ,

further

No

member

French

and

force

the

treating joint marked

way,

is got reached, but the difficulty of the

form

the

at

129

ONLY

1, and

members,

forces

the

the

LOADS

joint marked

polygon Icnxyrh, treated"

t)fiAD

TRUSSES"

ROOF

in

of roof

Figs.

is used

73

for

ft.

student all the

who

has

conscientiously

preceding examples

position to solve

the

above

cases

should

worked now

for himself.

be

Roof.

Mansard

type of truss, the

of

the

made

of

same

from

queen-post truss, this iron, although retaining

modified

wooden

the

of

timber, while

the

Mansard

Fig.

76 shows

members

of

arrangement

which

truss

shows

Fig. 75

name.

constructed

as

the

when

Like

its members

the

roof

"

outline, differs considerablyin the arrangement

same

bears

STATICS

GRAPHIC

ELEMENTARY

130

when

structed con-

of iron. The

difficultyof seemingly

forces

indeterminate

Fig. 75.

will

be

again

AB, but

such

overcoming in

these

to

the

joint

have

now

under

load

the

trouble

no

in

difficulties.

modifications

Figs. 77

at

should

student

the

Other

with

met

80, but

of

the

Mansard

it is left to the

roof

student

are

given

to solve

for himself.

Examples. 1. A

simple

horizontal the

roof truss

tie-rod

abutments.

consists

connecting The

span

the

is 18

of two ends ft. and

rafters which the

and rest

rise 4

a on

ft.

The

ROOF

TRUSSES"

roof

covering

DEAD

is

ONLY

LOADS

equivalent

to

a

dead

load

131

of 660

Pig. 76.

lbs. at the and

the

apex.

pull in

Determine the

tie-rod.

the

thrust

in the

rafters

ELEMENTARY

132

Swiss

2. A The

the

on

storage

slates carried ft.

24

the apex

The

various

compound

on

the

rise to the apex

in

ft.

The

Swiss

4. A

glass. use

a

small

The

square

is 6

workshop

compound

The

rise to the

The

pitch of

the

The

ft.,and

the

and is

span

tie-

lower

78.

The are

pitch of J

ft.

the

in. thick

principalsis and

slates

the

and

Determine

5

tabulate

members.

in the

span

late tabu-

and

Fig. 80.

roofing boards

stresses

abutments.

trusses.

Fig.

rise of 1 ft.

8 lbs. per

weigh the

a

tie-rods

boarding

with

Fig. 79.

has

lower

members.

Fig. 77.

rod

ft.

of 20

span

Determine

lbs.

is roofed

shed

a

the level of the

is 800

in the

for

ft. and

4*5

ins. above

stresses

3. A

is

STATICS

is used

truss

apex

point 9

a

load

The

roof

rise to the

rise to

GRAPHIC

is 30 Swiss

apex

is to be

ft.,and

it has with

truss

is 6*5

covered

ft. and

principalsis

6 ft.

in with been

plate-

decided

to

right-angledstruts. to the

tie-rod

1 ft.

Plate-glassweighs

TRUSSES"

ROOF

5 lbs. per

be taken in the

Belgian The

The the

used

tie-rod has

and

into the

be

can

in

a

shed

roof, the pitch of

before. lower

The

height

tie-rods

rose

the

the

weight

per

compound

7. A a

are

lbs. A

workshop The

trusses. 8 ft.

The

pitch

of

lbs. per

span

is No.

500

sq. ft.

The

the

the

same

is

The

loads

ft.

roof is 40

at

tabulate

the

carried

is

ft. and has

8

other

joints

French-roof

rise to the

rise ft.

the

over

stresses.

on

the a

the

roof

a

horizontal,

tie-rod

lbs.,and

and

G.

level

in

carries

truss

principals being 16

of

as

the

the

stresses

being

lower

is 10

tie-rod

lower the

The

apex

Determine

to

before.

as

ft.

each

ft.,and

14

the

queen-post

rise of the

abutments

material

of 45

span

the

8.

sq. foot

tinguishi dis-

adopted

were

ft. above

2

material

roofing

truss.

altered

was

was

apex

point

a

the

struts.

trusses

Determine

abutments.

members,

1000

to

lbs. per

principalsbeing kept

the the

to

The

members,

question

English

thick, is

IJ

of

various ties and

above

the

of 60 ft.,and

span

the

and

in

pitch

sq. ft.

as

8

apex

The

in.

weight

the

carefullybetween 6. The

taken

the

account

stresses

1

lbs. per

8

engine

an

ins.

Boarding,

weighing

fastenings

Take

Determine

over

can

stresses

rise to the

the

rise of 18

a

ft.

10

slates

carry

sq. ft.

of

the

roofing over

in

ft. and

is 40

span

principals is

purlins

for

glass

the

Determine

sq. ft.

is used

truss

lower

to

give

for

supports

members.

shed.

of

the

lbs. per

1|

as

5. A

ft.

ft.,and

sq.

133

ONLY

LOADS

DEAD

of

18

apex

ins., the

The

roofing

corrugated iron, weighing

purlins

and

bearers

can

be

3*5

taken

GRAPHIC

ELEMENTARY

134

1*5

as

the

per

ft.

sq.

into

Taking

determine

truss,

STATICS

the

the

account

in

stresses

of

weight the

various

members.

for

used

a

roof

roof

of

span

lbs.,

600

of

members

the

ft.

40

Fig.

in

shown

as

Determine

lbs..

1200

various

the

a

each

are

joints

other

truss,

having

abutments

the

in

Mansard

A

9.

The

loads

and

at

the

stresses

is

over

of

each

the

induced

due

truss

77,

to

these

loads. 10.

9, the in

determine

roof each

transferred

the

Assuming the

case

in

stresses

shown

trusses

can

from

loading

same

be

in

taken the

the

Figs. as

diagrams.

50

as

given

ft., and

79,

Question

members

various

78,

in

80.

the

The

of

span

proportions

the

GRAPHIC

ELEMENTARY

136

loads

and

finishingwith

construct

the

the

STATICS

funicular

closing line

sp.

polygon

From

o

pqra^

draw

ody

Fig. 81.

da and

cd

Rg, respectively.

The

parallelto R^

and

5p, then

represent

the

drawing

of

reactions, the

stress

137

ROOFS

will

diagram

present

now

from

readily followed

be

I

the

and difficulty,

no

should

diagram.

ra

Fig. 82.

For the

wider

long

rafter

appearing

spans with as

in

it becomes additional

Fig. 82.

stiffen

necessary

to

struts, one

tion modifica-

138

GRAPHIC

ELEMENTARY

before, the

As in

determining the

out

funicular

the

STATICS

polygon

reactions

R^

and

is made

will diagram, difficulty

stress

drawing

In

Rg.

of

use

experienced

be

jointsunder the loads BO and CD, the forces acting at these points being seemingly indeterminate. The be got over can difficulty by treating the joints at the

Fig. 83.

Fig. 84.

the

in and

following order, viz., R^, AB,

by going the

at

at difficulty

The

Rg.

round

also

can

be

overcome

joint counterclockwise,

time, that

same

horizontal

a

the

BC

x^, CD

BC,

x^,

through

the e,

h

point

since

eh

must

is

noting, fall

on

horizontal

a

member.

Figs. 83 truss.

for

and

The

himself.

84

student The

show

further

modifications

is left to work

finding

of

the

out

of

these

reactions

and

this cases

the

139

ROOFS

of

drawing

the

should

diagrams

stress

present

now

difficulty.

no

Roof

Overhung: shown the

of

type

one

or

pent In

joints.

upper

particularsof

the

Trass"

truss

carrying

this

in

the

resultant

pqrstuv, and intersect

in

Now

truss,

the

which

keep

R, Rj

and

Through

w. as

it in

a

the

w

in

R

draw

whole, is acted

meet

first

must

As

explained funicular gon poly-

first and

the

equilibrium ;

Rg, must

loads.

construct

produce

on

obtain

can

we

is

85.

of the external

Chapter IV., Fig. 30,

85

five loads

Bg, we

reactions, B^ and

Fig.

find

before

case,

Fig.

In

Pent

on

hence a

common

last

lines

to

parallel to af. by three forces, the

three

forces,

point.

The

140

point X

GRAPHIC

ELEMENTARY

and

to

common

y will

give

determine

now

drawing The

R

and the

the

Rg

line of

taken

x^, DE,

Overhungr modification

so

that

in the

of

Rg.

Roof

with

the

preceding example.

line

R^.

R^

and

through We

if difiBculty

the

86.

Pillar."

overhung In

can

Rg by

following order, viz.,AB, BC,

EF, and

of

a

of

action

magnitude

Fig.

CD,

is x^

triangleof forces, afg. diagram will present no

the

stress

jointsbe

STATICS

this

Fig.

roof,

case,

as

the

86

shown

shows in

roof, instead

x^,

a

the of

being supported entirelyat the wall, is further supported so that, in reality,it is only by a front pillar,

141

ROOFS

the The be

of

panel

extreme

will

reactions

be

It will

in the

of

nature

joint under

the

with

the

tension, while

BC,

load

the

in

stress

all the

and

there

the

rafter

member

can

funicular

the

that

the

overhung.

case,

of

aid

is

change

a

members

at

being in

BH

members

rafter

other

is

this

in

the

noticed

be

which

truss

vertical

easily determined

polygon.

the

in

are

Fig. 87.

On

compression. member

the while

AH

all the

members

is in

in

solution

will

from

diagram

the

Island

shows

type The

stations.

presents the

becomes such with

no

a

readily

The

in

followed

Fig. 86.

Roof." of case

roof of

Fig.

a

little

more

case

will

be

loads.

When

87 used

frequently

simple loading,

and difficulties,

diagram.

wind

horizontal

tension.

Station a

from

be

side,

compression,

lower

other

are

under

the

will

acted

be

on

in the

next

as

island

shown,

easily followed

by wind,

complicated, but given

for

the

the

case

solution

of

chapter dealing

ELEMENTARY

142

GRAPHIC

STATICS

Examples. 1. A

Fig.

saw-tooth

The

81.

angles the

FG

is the

30"

pitch of

has

truss

the

is 90"

angle

apex

60" and

and

The

roof

and

the

principalsis

weighing

side

is covered

lbs. per

12

is covered

with

Determine

and

with

the

member

longer

slates

and

glass weighing tabulate

The

the

side

shorter

in

stresses

rafter.

boarding

lbs. per

5

ft.,

is 22

of the

ft.,while

sq.

in

abutment

span

ft.

8

perpendicular bisector

longer

the

The

respectively.

shown

outline

ft.

sq.

various

the

members. 2. Draw

out

is 32

Assuming

shown

that

to

in

The 90". angles being 30", 60", and ft. and the pitch of the principals 10 ft. the same type of covering as in Question

1, determine

tabulate

and

members. and

similar

truss

the

Fig. 82, span

roof

a

Write

the

down

the

roof,

as

in the

stresses

of

value

various

reactions

the

R^

Rg. Draw

3.

out

a

proportions from

the

in the above

as

pitch of

shape

Fig. 84,

in

the

the

roof

ft.

span

of

shown

of

determine

principals10

Question in

determine

members,

and

3

Fig. 84,

Adopting

before.

as

With

the

roof

The

4.

the

the

cases,

Fig. 83, taking

diagram.

of

if the

members,

the

in

shown

determine

in the

ft. and

40

all other

write

stress

the

finallyaltered

was

outline

the

and

is

ing load-

same

ing things remainand

proportions the

down

also

to

the

stresses

value

of the

reactions. 5.

in

An

overhung

Fig. 85,

the

station

amount

of

roof

is

arranged,

overhang being

20

as

shown

ft.

The

CHAPTER

ROOF

TRUSSES

of

Calculation been

subjected which

be

induced

stresses

in

negligible Many

the

a

if

whose and

further

P t;

W ti;

this air

the

surface

Let

After

let

of

the

than

then

deduced

the

wind

however,

by

Let a

of

area

us

the sider con-

flat

surface,

the

stream,

perpendicular the

of

to

the

exerted

pressure

thus "

=

velocity

in

pressure of

air

weight

of air

weight

of

impinging

be

wind.

exerted

on

the

surface

be

air

=

impinge

to

the

can,

assumptions.

stream,

can

=

=

air

is greater

area

direction on

of

stream

a

We

force

for. the

almost

solution

mathematical

a

certain

make

we

by

which

pressure

the

cause

appear

caused

those

impossible.

is

case

expression

an

wind,

and

structure,

practical

obtain

the

to

pressure, to

as

cases,

loads

to

influence

on

some

dead

the

by

comparison

causes

exerts

in

great,

so

less,

or

more

wind

to

all

but

only, are,

due

loads

have

we

structures

on

loads

weather

the

to

far,

So

"

produced

dead

additional

to

may

effects

of

exposed

structures

PRESSURE

Pressure.

the

application

the

WIND

"

Wind

with

dealing

by

IX

on

1

lbs. in

per

ft.

per

delivered

cubic

the

ft. of

surface 144

sq.

ft.,

second, per air the

sq.ft.p. in

air

sec.

in

lbs.,

lbs. is assumed

to

ROOF have

TRUSSES"

WIND

velocitynormal

no

PRESSURE

145

surface.

to the

have, in

We

general "

of momentum

Change

impulse

=

"

W

g In

our

t=

unity ;

we

case,

dealing with

are

therefore

have

we

second, and

one

relationship

this

"

the pressure expressed in words ft. is equal to the change of momentum or

that

on

sq. ft. and

1

the

delivered

air

per

ft. per

v

sq. ft. per

second with

dealing

are

we

velocityof

per

sq.

second, hence

second

will

be

v

ft.

cubic .*.

air

an

Now

area.

lbs. per

in

"

occurring

hence

of air delivered

Weight

per

second

=ti;xv

lbs.

Wlb8.

=

W

Now

change of

momentum

p.

sec.

xv.

= "

9

g

9

in

Putting (

=

the

values

32"2)and converting

ft. per second

into

miles

( 0'08 lbs.)and g velocityof the wind from

of

the per

w

=

hour,

we

"08

Change

of momentum

p.

sec.

=

=

where

velocityin

V=air .-. K

P=0-0053V2

miles

X

get "

V2

X

5280

X

5280

x

3600

X

32-2

"^^eOO 00053V2, per hour.

lbs. per sq. ft.

U6

ELEMENTARY

Thus, with

wind

representing sq. ft. would the

above

at

be

gale,the

Experiments

These P=

by

00032V*

exposed it

show

experiments

lbs. per

found

was

and

that

for

that

of the

much

this

on

the

as

high.

too

what

For

wind.

of

sical Phy-

pressures,

matter

be

area

lattice-work much

was

pressure

structures

practice.

rectangular surfaces

ft.,no

the

in lbs. per

National

the

are

for

that

sq.

force

the

to

that

formula,

in

the

at

show

to

above

the

attained

Stanton

by Dr Laboratory go

hour,

per

lbs.,assuming

8*5

=

be

could

miles

40

pressure

00053x40x40

conditions

obtained

velocityof

a

moderate

a

STATICS

GRAPHIC

P

nature

greater, 000405V2

=

lbs. per. sq. ft. The

effect

important on

off

effect

the

wind,

suction

be

can

of

a

effect

on

the

effective

experiments of

the

wind

ground.

The

relative

to

very

wind.

does

suffers

leeward

which

effect the

of

of the wind.

an

exerts

is found

to

is

structure

Lattice-

innumerable

increased

has

wind

the

suction

length

side

work,

thin, fiat plates

pressures

due

this

to

effect.

slope the

this the

as

it

as

force

direction

in the

presenting

It

the

on

rapidly

increased

to

suction

structure, and

any

fall

of

taken roof

generally the

leeward

force at

the

shape other

materially

greater side, the

on

Forth

of

the

and

the

steeper

be

the

suction

hence

the

greater

side.

windward showed

the

height

above

Bridge surface

buildings

affect

the

will

with

increase

to

that

are

force

and

its

factors

exerted

the

The

effect the

position which

by

the

allowed

for

a

we

make

of

use

obtained,

taken

time

long

hurricane, which hour, per

find

we

the

means

sq.

which

we

exerted

by

miles

90

be

about

the

Tay

to

pressure

if

ft.,and

formula

velocity of

was

a

per

lbs.

41

sq. ft.

Following

efifect of

investigatethe bridges,and, as a B.O.T.

sudden

to

life and

limb

the

structure.

experiment

has

shown

than

40

allowance lbs. per

situations. force

It

stresses

In

Continental

the

wind

but

in

assumed

sq.

this

to

horizontal.

normally

to

angle of

country

the

roof

the In d

dependent

maximum

only

the

the

on

pressure

the

is

assumed

usually

lO'^-lS" to the of

surface, the

affects

roof

truss. that

horizontal,

the

the force

most

of

component

composing

termining

exposed

which

surface

direction e

and

design,however, hardly economical

it is

roof

the

members

an

wind,

that, only in very

ft.,and

practice it at

This

roof

greater

is, however,

in the

acts

a

of

gusts are

In that

for

normal

the

a

high, readily justifies is in an structure exposed

the

stabilityof to make

for

sq. ft.

lbs. per

56

of

designs

seemingly

where

when

so

of

railway tions, investiga-

made

be

should

pressure

situation, subjected more

in all

Bridge

Committee on

pressure Committee's that

allowance

figure, although cases

wind

recommended

wind

itself in

appointed a

of the

result

railway bridges maximum

Trade

of

to

the

after

inquiry

the

disaster,the Board

this

country

pressure

a

wind

the

this

lbs. per

theoretical

the

wind

maximum

in

40

as

calculate

and

The

structures

on

147

PRESSITRE

Pressupe."

Wind

Maximum pressure

WIND

TRUSSES"

ROOF

usual

wind exerted method

is

ELEMEl^TARY

148 is

make

to

follows

P

=

GRAPHIC Hutton*s

of

use

=

=

force

wind,

the

of

normal

to the

wind

Example.

blows

is

as

sq. ft.,on

of the

sq.

ft.,impinges

the

horizontal

surface

a

wind,

of surface, e

cos

-

1.

horizontally,tfis equal

to

the

with wiud, blowing horizontally,

velocity corresponding on

a

to

a

surface

lbs. per

40

inclination

whose

will be the

is 35", what

of

pressure

normal

to

pressure

sq. ft. ?

in lbs. per

P" Sin

sq. ft.,

lbs. per

direction

right angles

at

roof.

the

If the

"

surface

a

P" =P(sin. 0) 1*84

slope or pitch of

a

in

of inclination

angle

the

on

in lbs. per

pressure

Then

When

which

"

inclined tf

formula,

:

to its direction

Pn

STATICS

tf=sin

P

=

36"

(sin 0)1-84 and

0-57

=

^

cos

1.

-

35"

^=cos

cos

=

-82 ;

.".P"=40(0-57)a-84x-82)-i, 40 x(0-57)-5 40x756, =

=

=

It is not

usual

for snow, is not

wind The winds

in

on

likely to

sq. ft.

lbs. per

30

this

the

remain

assumption in

the

make

that

presence

ance allow-

any

the of

load

snow

the

mum maxi-

load.

following tables give and

to

country

the

normal

roofs for different

the

pressures

horizontal

velocityof on

various

variously inclined

pressures.

ROOF

TRUSSES"

Determination the

of

methods dead

treat

can

we

two

the

maximum

variation

As

By

due

we

have can

the

on we

former

In

to

either

of

combine

their

then

method

only get

we

members,

are

enabled

to

it may

be

necessary

the

get

members, to

the

combine

various

various

two

resultants,or

the

the

mining deter-

combined

option

separately,and

the

method,

make

for

while, tion variawhich

allowance,

considerable.

latter

understood,

structure

on

it be the

a

by finding

load

each

load

in

should

loads

loads

latter

the

to Wind."

We

solution.

effects.

the

by

loads,

wind

and

in

stresses

149

PRESSURE

due

of Stress

wind

and

dead

WIND

we

method

will

is

possibly the

consider

this

method

more

easily first and

150

ELEMENTARY

then

proceed

dead

and

The of

this may

Let

L=

wind

due

load

total

=

load

on

will

be

tion determinathe

on

truss,

ft.,

on

up

if N

number

be

the

in lbs.

truss

the

among

load

ft.,

lbs.

joints along

of spaces,

the joint (i.e., the

sq.

one

any

LxpxP"

=

lbs., while

"

wind,

to

W

end

each

in lbs. per

is divided

W

rafter, and

load

in

is the

"

pressure

Then

the

thus

of rafter

normal

=

This

combined

pitch of principalsin ft.,

=

P"

of

cases

acting

pressure

be found

length

p

few

a

considered

be

to

wind

total

the

out

STATICS

loads.

point

and

W

to work

wind

first

GRAPHIC

and

apex on

then

the

abutment)

intermediate

each

2N

joint

lbs.

be

will

-^

finding

In

Case*

First

consequently to

act

We

also

acts

through the the

over

Set

down

the

Select

a

nature

It

distinct

three

that

the

the

centre

of of

line of action abutments

draw

parallelto

poleo,

and

AB

join ao

roof

truss

tached at-

are

by

rigid fastenings, and

of

the

is, however,

parallel to the

assume

a6

of the

abutments

indeterminate. them

ends

^Both

"

the

to

have

we

with.

to deal

cases

reactions

the

is

reactions

admissible

resultant resultant

to

wind

quite

assume

pressure.

wind

pressure

duce Prolength of the rafter. W, and through the joints two

and

and

lines

parallel to

equal ho.

In

to

it.

it,to scale.

the

space A

152

GRAPHIC

ELEMENTARY

W

STATICS

E^ to intersect in 0, and to satisfy the conditions of equilibrium stated above, the line of action of Bg must pass through O as well as P, and Produce

hence

We

and

line of action have

B^

is determined.

determined

now

hence

magnitude,

one

of

it

directions

three

only

remains

fix

to

parallel

a

and

from

and

a

Then

and

and

Bg

Case.

Third

blowing

in

and

of construction case,

have

apply equally

working of

the

wind

to

a

employed

constructions the

out

roof, due loads

to

should

stresses a

dead

be

in the

on

load

treated

in

the and as

are

method

previous will

student if he

solution

ings fix-

previous

the

the

this, and the

difficultywith

no

carefully the In

well

the

truss

The

reasoning,as given

wind

the

89. case.

In

from

of the as

spectively. re-

"

the

type

left, while

Fig.

cb

magnitudes

the

this

line

a

ac

Bj

is

draw

AC.

parallelto

of

6

parallelto BC,

line

give

ab

equal

from

W,

B^^

Draw

Bg*

to

the

of

magnitudes and

and

studies

Fig. 90. various

bers mem-

wind

load,

acting

first

a

TRUSSES"

ROOF

from

and

right

the

values

maximum

of

of

we

comparison,

solving

the

the

the

Two

will

Methods.

problem by

the

Fig/

carried

be used

type. the The

boarding

wood

on

being of

The

For

"

in

the

actual

an

and

methods,

the

sake case, paring com-

results. is to be

engineering workshop

An

to

two

used

be

consider

now

the

left, and

the

should

found

thus

design. Comparison

from

then

153

PRESSURE

WIND

principals

is 8

the

J

the

loading

40 lbs.

the

is 24

pitch of are

truss

right-angledstrut-

as

ft.

Pitch

under

"

superficialft

ft. superficial

(horizontal) per

of

the roof is 30".

superficialft.

3 lbs. per

SlateSy 8 lbs. per

Windy

of

building

2 Jbs, per

Boarding^

purlins, the roof

and

ft, and

particularsof Purlins

the

slates

90.

steel and

of

span

with

roofed

sq.

ft.

GRAPHIC

ELEMENTARY

154

Approximate

weight

of truss

STATICS

"

w.=f?p(i+l) + ?|) fx24x8(l

=

489-6

=

of rafter

Length .-.

(from drawing)

of roof

Area

supported principal

each

by

lbs.

"

Dead

load

equals"

500

=

13'5

g x

ft.

13-6

8

X

)

sq.ft.

216

=

/.

lbs, say.

=

Purlins

216x2=

432

Boarding

216x3=

648

Slates

216x8

1728

=

2808 To 3000

for

allow

make

contingencies

Dead

.*.

Load

at

abutment

jointover

Load

="

^"

at intermediate

joints

acting horizontally

=

"

Pn

=

P

(sin ") 1-84

log Pn =(1-84 =

cos

/.

'P^

=

^-

1

"

lbs. sq. ft.

lbs. per

(^=30")

^-1) log sin ^+log

{(1-84X -86)- 1} log 0-5+

=("58x1-699)+ =

cos

-

880

40

=

lbs.

440

=

load

lbs. =W.

=3500

load, including truss

=

Wind

load

lbs.

.'.

.*.

dead

P

log 40

1-602

1-427 26*7 lbs. per sq. ft.,normal

to roof.

ROOF

Normal

/.

load

wind

of truss

side

one

i

on

Wind

load

2880

lbs.

W2

=

2880

joint over'i

at

.^

g

J =

.*.

155

PRESSURE

WIND

TRUSSES"

_

abutment

and

4~"

i

apex

.-.Wind

load

mediate

inter-)

at

2880

f

joint

2

the

set

now

can

jointsand

and

will

We

members.

is fixed

truss two

cases,

and

wind

load.

ah

line

forces

acting

of

direction

the

of the

direction

acts.

Now

resultant

through

of the the

the

roof

load and

of

loads

R^

which

the

dead hence

W) (i.e.,

roof.

(i.e., Wg)

is

the

No. the

on

and

a

",

acts

uniformly

Rg.

and

1,

truss

on

and

the

giving

we

also, the Before we

resultant is

load

can

we

external

R2,

the

the dead

combined

and, hence

resultant

of the

apex

that

inclination

truss, its

roof, and

dead

at

various

of all the

resultant

that

the

the

case,

reactions, R^ and

know

over

the total wind of

points

the

magnitudes

we

distributed

loads

point through

the

first find

the two

the

settle

can

the

on

diagram

diagram

to

external

the

represents

stress

in

the

first with

Join

ah.

load-line

loads

this

in

Referring

all the

wind

ends, and, in comparing

will deal

we

the stress

assume,

both

at

first set down the

maximum

the

and

to construct

proceed

determine

so

dead

these

out

lbs.

1440

=

We

lbs.

720

=

we

must

ah

uniformly draw

the

acting vertically

Again

we

normally distributed

know to

one

over

that side

that

Fig.

91.

HOOF

TRUSSES"

side, hence

therefore

point

a

We

resultant.

draw

it

with

o7

is

wind

been

of

the

of

load

will

and

x

of

the

a

line

x

to

join o^a

and

and

with

In

between

the

dealing

with

with

the

when

dealing

with

wind

load

the

at

use

forces

two

dead

the

becomes

abutment

wind

and

dead

making

along

the

now

bte scaled

the

space.

when

will

position diagram

same

letter

la

should

stresses

and

which

should

diagram

act

complete

and

equal Rg

stress

the

B, and

omit

again

of

assumed

omitted

Similarly

lines

action

polygon

economize

over

being

be

roof,

x,

through

dealing

omitted, thus,

B

AB.

In

to

of

pole o\

a

the

lettering,the

load, the letter

draw

separately, the used

been

has

in

li^ can

kl

tabulated. load

has

intersect

difficulty.The

no

off* and

The

line

the

closing line, parallel to

drawn;

equal R^. The drawing present

rafter.

funicular

dotted

now

the

the

on

Select

the

the

to

will

and

B^

parallel to ak. o'i;

normal

therefore

ak;

parallel to

loads

two

157

Wg,

of

mid-point

of these

action

draw

can

the

through is

we

PRESSURE

WIND

AC,

the load

wind loads

we

is

abutment

again AC. No.

Diagram loads

only.

diagram the

wind

No.

the

this

load

3, calls for

load

roof

alone) R3'

the

wind

The

is

be concentrated to

2 shows

a

surface.

line, and

little

The will

R^ it

single

now

diagram for

diagram,

acting alone,

in

and

a

stress

force

can

Wg

reactions therefore

remains

shown

as

acting act

to

it to

assume

(due

in

When

explanation.

we

dead

normal

to

wind

parallel

determine

to

the

GRAPHIC

ELEMENTARY

158

of

magnitudes o''a

join shown

Draw

o''g.

Select

funicular

the

line ; gl will

closing

equal R^

and

o"

pole

a

polygon

as

o"!

parallel

and

la

chain-dotted, finally drawing

the

to

and

R4.

and

R3

STATICS

will

equal R3. Diagram wind

load,

induced

2

3

and

it

in

stresses

II

No.

the

shows

will

noticed

be

diagram

be

that

seen

agreeing fairly closely,still These

drawing,

in

errors

tends

latter

bring

to

assumption

that

but

there

about

this

show

from

resultant

R^ new

and

reactions,

hi.

values

for

R^

and

obtained

diagrams

by

Rg

R^^and

Nos. 2 and

3.

positionof i,which

modified

stress

the

the

method

We

of of

not

This

The

the

the

find

coincide

with

give

a

slightly

in

a

slightly

result

a

the with

coincide

indeterminate to

that

as

of the reactions

would

would

each

were

quite easilyfind

would

which

the

nature

large extent, by

supporting

walls

and

the

the exact fastenings,makes of this makes the case impossible, and adopted, in diagram No. 1, as justifiableas

other.

action

do

by

reactions,

3, we

could

reactions, accentuated,

twisting

any

diagram.

yielding nature

solution

and

2

made

Rg

the

the combination

different

of

We

crepancie dis-

which

cause

difference.

found,

so

Rg parallelto

the values in

Nos.

diagrams

Nos.

for

accounted

parallel to Jca^ but, by combining found

Table

slight

some

be

of

the

diagrams

is another

directions

the

is

figures,although

the

might

shows

1, while

No.

the

stress

no

I

off from

scaled

stresses

that

Table

off from

for

diagram

stress

PQ.

It will

3.

the

member

scaled

as

and

the

shows

160

ELEMENTARY

abutment.

again

make

of

use

side

of the

down

all the

reactions

is

of

forces

STATICS

funicular

the

employed

windward

resultant set

finding the

method

the the

In

GBAPHIC

this case,

roof

acting

at

loads

we

polygon, although On

slightly different.

the

external

in

first

we

find

joint,and

each in the

line

.

a/.

the then

Select

^:^

a

pole o,

and

join

the

lines of action

the

funicular

point of the

arises

Rg, we

here

in

as

various

two

loads, and A

We

reactions

that the line of

very the

with

R^

and

action

construct

important

construction

closing Rg. Regarding

that

know

Produce

of.

and

oe

showu.

connection

polygon.

join the

know

of the

polygon

funicular

line must

oby oc^ od,

oa,

must

the

be vertical,

ROOF

while

the

TRUSSES"

the

through

pass

its direction line must we

from should The

and

and

Referring to the the line joining af all the If

external

we

and

to

a/, then

draw

will

loads

their

If

(R).

produced

first and

last

through o\ diagram for himself various

Saw-tooth

The

the

of

action

it will be

student

by drawing

wind

Fig. 93

"

dead

and

left,while

is fixed, the

of

can

the

and

A

line

a

of

of the

lines

parallel of the

line of action

the

o\

(R)

and

R^

found

Rg

that

R

complete the stress diagram

members.

Roof.

subjected to

intersection

in

that

seen

in the spaces

lines

give

to intersect

passes

from

truss.

lines

the

in the

resultant

the

it

good.

holds

it will be the

represent

acting on

through

polygon

easilydetermined

are

the

R^, hence

end, and

always

diagram,

this line will

resultant

for the

rule

the polygon (i.e.,

F)

be

force

produce

now

funicular

Rg

of

fixed

the

at

AB,

closing

funicular

our

this

that

R^ fg.

Our

line of action

commence

joint

noted

reactions

lines ga

to

abutment

be

the

on

it must load

the

still undetermined.

being

terminate

the

under

joint

abutment

compelled

are

R^ is,that

about

only point known

161

PRESSURE

WIND

left-hand

the

shows

a

saw-tooth

loads, the wind

blowing

side of the

right-hand

roof

truss

side

on a roller. being mounted student The structing experience some diflScultyin conmay the funicular The polygon in this case. be begun from construction must the abutment joint

parallelto that

Doubt

EF.

under

oc

is the

oc L

may

should ray

be

between

arise drawn.

to

as

It

the force

where should

he and

the be

line noted

erf, that, so

162

ELEMENTARY

in the

form and

funicular the

CD.

link

GRAPHIC

polygon,

the

connecting

Note

we

get

STATICS

line

lines

the

crossed

a

parallelto of

will

oc

of

action

in this

polygon

BC

case.

Fig. 93.

finding of the reactions and straightforward,and calls for no The

Island

Station

in connection in

dead

loads

with

finding

Roof. this the

alone, and

"

case

The

stresses

further

is

quite

explanation.

only important point method

is the

reactions. assume

the

If these

we

to

to be

consider be

ployed em-

the

uniformly

ROOF

TRUSSES"

equal, and

Consider

now

each

the

has

a

moment

(Rxa:). the

(R) This

has

on

about would

the

tend

right-hand support.

reactions

two

the

total

resultant

the

load. wind

94.

truss

the

the

half

to

which

Fig.

pressure

that

equal

effect

163

PRESSURE

it is obvious

distributed, then will be

WIND

; it will

be

seen

that

right-hand support equal to

This

caus"

a

rotation

tendency

to

R to

about rotation

ELEMENTARY

164

GRAPHIC

is counterbalanced

acting

force distant

support,

Hence

right-hand support. have

downward

a

left-hand

the

at

by

STATICS

for

(say Rg) from

y

equilibrium

the must

we

"

Rxa;=R8Xy. Rxa;

Rs= y

dead-load

The

therefore

reaction

R,

is

load-

dead

i total

=

support

by Rg.

diminished

."

left-hand

the

at

.

.

y

Coming shows

part

readily be

it will

only

of the

dead

g'

where

loads

resultant

wind

found

from

y

We

would

let

that

set

the

reaction

complete

us

suppose

R^,

load-line that

lbs.; it

3600

=

considering

were

will

the be

"

3600x1

1

p

,

^"^'"

=24=12'

therefore

the

(R)

pressure

we

give

of

Now

only).

measurement 2

a:

g'a

now

adf. Fig. 94 (a) enlarged scale, and

an

that, if

seen

we

line

the to

mid-point

is the

(for dead

in

load-line

loads, then

diagram,

stress

loads

external

the

down

the

to

now

mark

^^^,,

R3=^2-=^^^'^"off

g'g equal

to

300

lbs., then

equalsthe reaction R^. The reaction Rg is obtained drawing of the stress diagram by joining fg. The presents no further difficulties. not be forgotten when It must working out stresses

ga

due when

to wind

loads

the wind

on

blows

roofs

fixed

from

the

at

one

end

right,very

only,that different

TRUSSES"

ROOF

will

stresses the

wind

design, obtained and

in

blows

therefore, for

the the

be

maximum

design

the

PRESSURE

WIND

from

obtained from it

the is

wind

calculations.

For

left. that

blowing

possible

obtained

those

essential

loads

165

from

so

when of

purposes the each

obtained

be

stresses

direction, be

used

CHAPTER

AGED

BR

The

of

science

braced

which

large

to

a

to

some

to

deal

extent

here

simple

Fig.

in

this

truss,

with

of

nature

the

and

the

beam

a

a

stanchion, the The

whole

solution

single

the

Design. The

deal

that

nation examibeam

a

changed

kingthat is

members

various

the

is

An

difference,

is

being

inverted

an

to

the now

sion, compres-

tension. loaded

in

load

a

of

the

first

case

166

different

two

load

concentrated

.with

length in

the

to

be

may

or

reality

; tension

compression

with

over

in

stresses

these

to

clearly

and

intended

not

Stanchion."

important

this

bridge,

stanchion.

show in

is

have

single

is

way

It

depend

Structural

on

we

will

95

reversed

completely

Such

with

the

loading,

Single

which

beam

of

braced

either

with

case

braced

works

with

Beam

simplest

of

questions

of

ings load-

The

subjected

are

the

members

work.

bridge

locality.

in

involving

various

the

purposes

the

discussed

Braced

in

the

on

with

in

useful

very

problems

structures

on

extent

adequately

rod

used

such

another

of

stresses

structures

to

finds

solution the

GIRDERS

AND

graphics

of

determination

of

BEAMS

the

in

application

X

distributed

ways

directly

over

uniformly

beam. does

not

call

for

any

:

GRAPHIC

ELEMENTARY

168

introduces

type

pin joints

assume

it will

at

be

once

at

seen

loads

the

If, however,

the to

the

over

it

^w^

NA/

from

FG,

and

this

particular

the

additional

hence,

for case,

is not ing draw-

the

out

in

gonal dia-

necessity.By

a

stress

diagram

for

unequal

loading

it

will

that the

seen

For

a

be

diagonal

becomes

then

96.

no

induced

member

Fig.

stress

that

is

stress

will be

the

diagram,

\o

a

required. are equal

be

steady, it

seen,

is

complete, an

stanchions and

\/v

panel

central make

we

stanchions,

the

(shown dotted) will

member

additional

that

If

interest.

of

head

the

in order

frame, and

deficient

points of

two

or

one

STATICS

sary. neces-

variable

load

both

must

be added.

diagonals

Trapezoidal Truss. "

The

trapezoidal truss,

further used

for

still the

for variable added.

student

of

modification

central

loading

Some in

longer

spans.

in

shown

this

type

The

same

panel applies in double

Fig. 98, is of beam, remark this

counterbracing

case,

must

a as

garding re-

and be

will be experienced by the difficulty will be solving this case, but the difficulty

BRACED

BEAMS

if it is remembered

easily overcome in

stanchion

any

top end.

Truss.

bracing

the

is that

simplest

which

is

have

we

acting

ends

of

with

marked

x,

the

will

the

by the

fixing is

the

point

is

This

AB. the

AB. y, and

the

simple,

can

joint

then

next

this

gives

I,

will

The be

abutment,

the

over

The

point and

Fig. 97.

triangle

efk. We the

equal

polygon eagf,and

load

marked

cd.

time, that

give solve

Taking

in America.

remembeiSng, in FK

forces

(Fig.

must

we

the

load

the

truss

our

obtain

same

stress

Fink

the

of

top

ah, be,

to

modification

and

down

as

begin

now

point

three

the

diagram,

stress

of

its

in

set

order

to

at

in

largelyused

stanchions,

load-line

the

at

the

with

important

the

first

at

stress

acting

load

diagram

adopted

case,

equal loads, AB, BC

In

stress

Another

"

which

99),a type

we

the

to

the

that

x.

Fink

CD,

equal

the

Begin

marked

the

is

169

GIRDERS

AND

point the

on

pass to

the

given

as

joint under

solved

be

is that

polygon ekhl, thereby of

remainder

readily

the

followed

solution from

the

diagram. The

student

is advised

to work

out

a

similar

truss

170

in which

in the

ratio

BoUman in

the

members will

student,

unequal, say AB, BC,

are

2, 3, and Truss.

Fig. 100, the

loads

the

STATICS

GRAPHIC

ELEMENTARY

"

CD,

4.

the

doubtless the

of the

determination

^The of

but

and

truss, shown

BoUman

present

some

diagram

stress

stresses

in

difficulties to can

readily

be

a

I

c.f

Tn".TV.

Fig.

98.

Having set in the line ad, start with the loads down the joint From marked reaction x. ef point, draw e, the and ek parallel to EK. Place the parallelto EF vertical line fk equal to the load AB, with its ends / the lines ef and ek respectively. Pass and k on on of which solution to the joint marked is then y, the obtained

by

the

following construction.

BEAMS

BRACED

given by

be solved

next

the

the

in

stress

now

on

pass

thence

to

the

equals

joint

student in

CD

will find

solving are

Warren

the

under

will the

of

the

joint

the

to

case

the

the load

Fig.

99.

be

readily

completed in

which

Of

the

in

stress

over

valuable

a

joint

171

marked

z

polygon kfgl,remembering

in the EN

GIRDERS

The

triangle eno.

quent procedure examination

AND

We

abutment, AB.

The

followed

force

and

KL.

the

loads

that can

aud

subse-

from

diagram.

instructive

can

an

The

exercise

AB, BC,

and

unequal. Girder.

"

various

types of girders,

172

ELEMENTARY

which web

bracing,probably

is that

invented

him, the of

of

composed

are

STATICS

GRAPHIC

top the

and

simplest

by Captain Warren,

Warren

parallel top

girder.

The

bottom

and

triangulation system.

These

and and

known

best

called,after consists

simplest form booms,

with

booms

bottom

with

a

triangles are

single usually

Fig. 100.

although this particular shape is not a equilateral, of the girder; but one important special feature girderstruts and point to note is,that in the Warren various all of equal length. There ties are are ways be arranged and this type may in which a girder of loaded, but consider

it will

only

three

be cases.

sufficient

for

our

purpose

to

BRACED

Case shows on

a

the

Set

I.

ah c

It will

Load."

to

from

a

represent W,

and

consideration

of

readily be

produced

to

3*5, hence, by taking moments, of the

will

cut

the

lower

reactions,thus

then the

boom

in the we

fix

101 W

the

positionof

the line of

that

seen

W

value

Fig.

girder of 5 bays, carrying a load joint from the top left-hand end.

second

down

173

GIRDERS

AND

Concentrated

Single

Warren

position of

W.

BEAMS

can

"

RjX5

=

Wx3-5.

Ri=0-7W.

R2X5

=

Wxr5.

R2=0-3W.

action ratio find

of 1*5

the

The

point and

0-7W

therefore

of this

difficulty.Start

problem

at the lower

Fig.

Rj,

and

various

as

then

shown

in

previous case, a

taken

so

should

now

that

ca

=

Fig. 102, every

load, each

load

are

joint over

the

the reaction

bracing, solving

the

met.

Distributed is

present

102.

along

zigzag

points as they II. Uniformly

Case

be

6c=0-3W.

solution

The no

must

c

STATICS

GRAPHIC

ELEMENTARY

174

simply

Load." an

This

expansion

case,

of

the

now joint in the top boom ing carrybeing of equal magnitude. The

176

ELEMENTARY

while its

the

the

as

members

of

of this

furtherance

It will be this

truss

short

as

Linville

idea, the from

given section In

arranging be

subjected to

as

possible.

truss

Fig. 104

that

In

was

duced. intro-

the

struts

giving the

vertical, thus

are

a

the members

made

seen

of

by

it will therefore

structure

that

be

loads

strut

a

influenced

is not

is decreased.

braced

to arrange

compressive

of

length a

STATICS

tie-rod

a

length, the strength

economical

in

of

strength

increases the

GRAPHIC

shortest

possiblelength for any given depth of girder. In the loads in determining the stress, we first set down line af and take the the mid-point gr, since the reactions then the joint start with are equal. We can the left-hand at abutment, and zigzag along the This bracing, solving the joints as they come. method presents no difficulties until the joint at the is reached. foot of the vertical NO The difficulty can, however, be overcome by assuming that the stress in MN is equal to the stress in the member the member fix the point 0 if we of course, We OP. ber rememcan, be equal to CD that the stress in NO must ; but, the load CD first,the by treating the joint under assumptions which point O is fixed without any The mainder remight not be quite clear to the student. solution is quite straightforward, of the although, when to

the

draw

is

loading than

more

symmetrical, one-half

it is

of the

necessary un-

stress

diagram. Pratt

Truss.

truss, known Lattice

"

as

Fig. 105 Pratt

the

Girders.

shows

"

If

a

modification

of this

truss. we

take

two

simple

Warren

BRACED

BEAMS

girders,invert other,

get

we

of them, and

one

arrangement,

an

illustrates

which

fact that

this

type

of

type

of

girder is

and

solve

system

together.

stresses to

each

error,

certainlyhas

majority of cases, by drawing systems.

The

subjected to M

This

a

time one

in

shown lattice

made

up

method girders suggests at once a namely, to splitthe girder up into

Fig.

it

superpose as

one

177

GIRDERS

AND

the

on

Fig. 106,

girder. The of two simple of

its

treatment,

components,

104.

separately,finallyadding method, although

stress

liable

advantages, but, in

many can

more

be

saved

diagram

and to

in

the

the

errors

ated obvi-

serve

both

Fig. 106 is system of symmetrical loads,and hence lattice

girder

shown

ELEMENTARY

178

the

will be

reactions

We

therefore

can

of

If

we

we

at

are

joint in

every

of forces

has

structure

of which

than

more

it would

Hence

with

seem

that

only

economize

the

stress

facts

it

are

a

system

unknown.

is insoluble.

case

gram, dia-

that difficulty

acting on

the

select

diagram

the

two

load.

and

to

draw

to

beset

the

drawn

been

has

total

the

load-line

length. (In

once

the

to half

the

attempt

now

STATICS

equal

down

of its

load-line

the

space.)

each

set

mid-point

fc,the

part

GRAPHIC

If

Fig. 105.

examine

we

the

inverted

girder,it

will

be

the

AB,

CD, EF

loads

forces

can

say

that

seen

this

GH,

and

members

AL

component and

that

from

transmitted,

are

the

through

triangulationsystem

and

the

supports

the

the

Ha.

of

supporting abutments,

Therefore

we

"

Stress in AL

Stress

in Ha

CD

AB

=

GH

=

+

+

-h

EF^^^^

^tJ^. 2

the

At

KA, a

left-hand

which

downward

is

abutment

we

have

a

total

reaction

produced

by

two

distinct

load

to

the

upright system,

due

loads, namely, trans-

BRACED

mitted

direct to the

due

the

to

inverted

calculate

the

system,

the

member

stress

in

through

and

abutment,

179

GIRDERS

AND

BEAMS

transmitted

and

AL,

to

We

AL.

load

downward

a

the

ment abut-

therefore

can

it down

set

as

al

H

K

V.

106.

Fig. in the

load-line

case.

No

further

be

treated

the

point at

vertical

; the

in

the

the

to the

sloping member

point

I coincides

will difficulty

following

left-hand

joint LM,

under

be

order:

abutment, AB.

solving

with

c

in

this

met, if the joints First then

Next first the

go pass

pass

round up

along

the the

joint at^the

180

GRAPHIC

ELEMENTARY

intersection, and inverted the

From

triangle. BC,

joint under

the

then

joint

so

of the

end

of

of

apex

tracing out

on,

one

the

at

the

point rise verticallyto

this

and

N-shaped paths from other. Only one-half

STATICS

the

girder

to

diagram need

stress

of

series

a

the

be

drawn. The

general

more

Fig. 107.

is shown

in

carried

by the

by

upright system.

the

the

only

can

be

calculated

joint under

the

AB

ratio

in the

AE

force AE

The

If any

6

end

X

verticals, the

girder into

AE

4, or

exist

which

AE,

that

two

tions por-

"

set down

should

therefore, is

It will be noticed the

is

is carried

in

stress

4, hence

AB

=

is then

doubt

affect the

divides

2 and

X

the

AB,

AB

load

BO

while

load

The

follows.

as

the

case

system,

affects

loading

unsymmetrical

this

In

inverted

which

one

of

case

$AB.

=

in the to

as

line

ae.

of the

which

readily

be

can difficulty

loads

by tracing out the diagonals forming the sides of the triangles. If the last diagonal finishes then the abutment, the load acting at up directlyon effect the joint,from which the tracing began, has no the end vertical,but if the last diagonal finishes on

overcome

at the

up

be calculated

must

This, of

by

the

end

to

its effect

vertical, then

moments

refers

course,

above

The

shown

members the

of

already explained.

as

loads

on

both

top

and

booms.

bottom

case

end

top

loads.

rule in

have

does

not, however,

Fig. 107 (a),where been

These

introduced verticals

are

a

hold series

good of

immediately intended

to

in the

vertical

below fulfil

a

BRACED

important

very results

attained

made, is doubt. the

If

purpose, are

in

but

refer in

GIRDERS

whether

accordance

question which

a

we

stresses

AND

BEAMS

181

the

with

is open

practical

the

tions assump-

to considerable

that Fig. 106, it will be seen two intersectingdiagonals are

to

any

Fig. 107.

unequal, and

the

vertical

of the these

stresses.

vertical its

top

member end.

object to The

be

attained

members

assumption

carries one-half

Thus, if

is the

we

consider

is

the

duction intro-

equalizationof made

of the the

by

load

load

that

the

acting at CD, it will

ELEMENTARY

182

be

that, without

seen

is carried

of CD

inverted

the

by

STATICS

vertical

the

member

vertical

the

GRAPHIC

system

added,

is

member, ;

when

while,

inverted

the

whole

the

system

CD

vertical, and

the

When

system.

FG

and

of each

half

would

points on load

the

the

on

abutments

by

therefore

equal AR

The is

case

it

r

this is

and

lower

boom.

end

to

down

that

load

we

the

on

while to

load

AL

on

the is

EF^FGX ,

again adopted

diagram,

sponding corre-

abutments,

The

DE

CD

procedure

the

load

is transmitted

verticals.

end

the

the

to

The

the

consist

"

set

noted

by

boom

/BC

1

now

same

being

to half

the

whole

BC, CD, DE,

loads

transferred

direct

the

separated by

of the

loads

the

girder to

be

upper

the

the

now

is carried

boom

lower

all

imagine

and

booms

bottom

One

verticals. EF

and

could

we

removed

to be

bracing

rigidtop

AL,

on

upright

the

by

consider

to

transmitted

being

finallycarried

come

we

effect

their

of the

of

half

other

,

along and

the

carries

only

draw

as

in

followed

was

drawing

the

stress

its

top

at

op

line al.

in the

in the

the

end.

vertical

vertical is

Thus,

and

diagram,

stress

in any

previous

in

mark

the

off op

equal stress

equal

BC ,

to".

Only

a

part

is sufficient

to

of the

stress

show

that

diagonals are The

effect of

the

now

diagram the

is

shown,

stresses

in

the

but

it

secting inter-

equal.

verticals

is therefore

to

equalize

164

ELEMENTARY

this

GRAPHtC

particular case of the

make

back

of the

use

have

we

balance

STaTICS find

first to

tude magni-

the

weight, and, in doing so, we polygon. The student will

funicular

in setting down experience a little difficulty loads, since, at the joint under MN, we have two

doubtless the

forces

EF,

acting, namely, is

which

loads,back

to the

hence

downward

;

draw

of action

of W

polygon

in the

point

to cut

the

must

pass

through

oa'

draw

structure

back

to

The

stress

diagram

Set

magnitude

about

is the

ha'

the

setting

force, EF, which

a6, 6c, cd, and

in the

point y.

produced,

when

From

shown.

as

gives cause

foot

of

The

straightforward.

back

remainder

of

out

is of

de ; the

then

line

first line

the

will

course

the

tude magni-

the the

hence

measure

balance of the

the

whole

member in

the

load-line.

equal

to a' a.

force is EF

next

a'a, acting upwards,

upwards equal to a'a, and ghy leaving ha the downwards.

y,

a

funicular

only point calling for explanation

the

down

and

which

Select

oh, cutting the

polygon,

F,

EG,

becomes

the

MN

xy, and

letter

ah.

Produce

pointsx

balance

balance

MN.

Replace

the

parallelto

of the

joint

member

the last line of the link

polygon,

it, the

construct

x.

the

setting out

load-line

the

Now

o

the

at

and

rays

force,

of the funicular

last link, parallelto

the

In

and, with

in

downward

downward

a

joint.

force

the

polygon, of the

EF

such, it appears

as

pole o

load at the

reaction,

or

all the

included, and

force

the

the

and,

of

sum

for the construction

delete

we

the

to

balance

FG, equal load-line

equal

force

upward

an

of

c/

measure

downwards

/gr

weight acting solution

is

quite

BRACED

GIRDERS

AND

BEAMS

185

Examples. 1. A

trussed

carries

beam

a

of

load

2 tons

at

the

f m

s

K

a

b c

r

\.

^P

d

n

Fig.

centre

of

stanchion

an

18-ft.

2 ft. long

span.

108.

The

directlyunder

beam the

has

load.

a

single Deter-

186

mine

the

material

by

the

if the

2. A

in

stresses

intensity of

section

to the

are

lbs. per

The

the

ft. wide, is

10

supported

and

Find

load

the

the

at

The into

is attached

superstructure

maximum

sq. ft.

beam

6 ins.

X

abutments

at the

trusses

and

the

in

stress

the

ties,and

and

in Fig. 97. type shown 8 ft. long, divide the truss

divisions.

stanchions, and 180

is 9 ins.

the

stanchions, which

equal

stanchion

ft. span

of

trusses

three

the

compressive

bridge,60

two

STATICS

GRAPHIC

ELEMENTARY

of the

head

equivalent

is

in

stresses

to

various

the

members. 3. A divided

The

loads

from

the

the

stresses

4.

four

into

long.

has

trapezoidal truss

in the

5

members

and to the

5 ft.

stanchions

stanchions, reading

respectively. Find

2 tons

loads

in

the

previous

respectively,find

tons

is

ft.,and

40

members.

the

Assuming

by

of the

top

2, 3, and

left,are

2, 3 and

the

of

span

divisions

equal at

a

which

determine central

stresses

diagonal

if the

panel

the

plied sup-

diagonal

in the

stress

in the be

must

be

to

case

is to be tension. 5. A

Fink

truss, 30 ft. long,has four equal divisions,

being

stanchions

the

of each

stanchion

various

members.

6. A 49

the dead

ft. long.

is 2 tons.

load

The

Find

the

at

the

top

stresses

in

the

railway bridge,crossinga river, has

ft.

each

4

The

divided truss

into

is 8 ft.

load is

live load

bridge

is

is

7 It

supported of

bays is

to

two

7 ft. each.

estimated

equivalentto equivalent

on

a

span

Fink The

trusses,

depth

that

the

maximum

35 tons, while

the

maximum

1 J tons

per

of

ft. of

of

length

BRACED

Find

the

maximum

variation

BoUman

a

six

trusses

70

of

span

the

and

members

trains, each

ft.

12

the

the

on

If in

stresses

tons

structure super-

live of

consists

bridge

ft. of

two

termine De-

length.

of the

members

various

the

The

into

maximum

the

per

railway

a

divided

are

deep.

tons, and

weighing

the

to carry

trusses

are

come

may

used

are

The

140

w*eighs which

ft.

and

equal panels

load

in

stress

187

GIRDERS

AND

in stress.

7. Two across

BEAMS

trusses. 8. A

Bollman

ft. and

12

from

is divided

10

and

the

various

4

Warren

4, 6,

5 and

tons

the

each

horizontal.

the

Find

loads

the

girder,in loads

of the

magnitude A

Linville

boom,

a

and

Determine

4 tons

and

12. Determine

2

each

tabulate the

span into

six

tons

at

of

the

the

stresses

of

span

of which from

top joints are members.

in the

would

in

of

then

members

of 60

distributed

to at

in the

a

is divided

equivalent

the

at

joints,what

has

uniformly

a

top joints,a load

stresses

truss

on

Reading

stresses

the

on

lower

of the

ft. deep and

carries

in

stresses

previous question,had,

the

.

is 10

the

members

panels, the

support, the

to the

on

11

60" to

at

the

addition

four

of

4 tons.

If

10.

Find

2, 6, 4, 8,

are

girder,simply supported

left-hand

the

of

depth

a

members.

inclined

are

loads

the

abutment,

has

equal panels. Reading

respectively.

tons

ft.,consists

40

ft. span,

50

into five

left-hand

the

9. A

truss, of

2

be

?

ft.

The

truss

equal panels.

It

load

on

the

upper

each

of

the

ments abut-

intermediate

stresses

in the

joints.

in the

members.

above

truss

if,in

addition

carries

of

three

in

stresses the

addition the

lattice

girder

each

ft. run, of

Determine

Fig. 108,

tons

and

the

ties

from

a

joints

3 of

span

the

at

and

18,

18,

18,

12

tons.

each

if, in in

joint

tons.

ft. and

60

load

of

lower

three

of

consisting a

depth

a

0*6

boom

tons carry

respectively.

tons

stresses.

out

and

end,

the

pier, using

cantilever

a

making all the

respectively. tabulate

girder

carries

diagonals

right-hand

above

of

of

each

the

boom

upper

the

Draw

16.

has

at

the

joints

4

system

while

12,

load

a

the

and

boom,

of

Find

boom,

top

a

consists

at

tons,

has

106,

45^

loads

triangulation

The

boom

top

in

the

on

A

loads

The

loads

Fig.

are

2

the

on

stresses

carries

triangles.

of

the

boom

ft,

the

on

lower

system

the

each

are

lower

16

if

joints,

the

the

in

angles

members

also

to

15.

per

base

abutments

Find

14.

shown

as

triangulation

whose

the

in

joints

load

tona

Each

intermediate

the

the

girder,

ft.

60

triangles,

above

3

lattice

A

span

of

load

a

13.

of

each

STATICS

distributed

uniformly

the

to

boom,

top

of

GRAPHIC

ELEMENTARY

188

slope

longer

are

Determine in

ft.

80

magnitude

members,

long.

from

2, 3, 4, 5, 6, 5 the

the

arm

proportions

Beading

45^

at

loads

stresses

struts.

the

the

the

and of

W

W,

ing distinguish-

CHAPTER

OF

CENTRE

FIGURE

It

force

earth

motion

is be

the

the

body

point,

at

the

possible

were

into

single

a

as

upon

about the

the

earth

original

is

the

cerUre

of

of

condense

the

thep

of A

our

mass,

but

189

the

away

further, will

there the

body

the

fprce

the

magnitude

pulling

application

of

the

C.Q.

term

former

is The

?

is assumed

particle

better

;

of

material

the

that

body.

of

gravity

of gravity

acts

earth

weight

hence

;

it

balance,

of

point

force

position

the

force

direction

"

the

the

scale,

particle,

to

the

apply

its

sense

spring

a

the

centre

to

its

magnitude

the

which

termed

on

the

What

body

a

as

we

know

it towards

the

towards

We

?

know

we

;

on

can

some

"

body

measures

known.

is

the

registered,

which

force

velocity. that

of

spoken

extent

body, drawing

hang

we

is

move

which

by

body

a

will

evident

is

force

earth

what

vertically

it acts

be

the

this

to

it

The

acting.

it

ever-increasing

an

produced,

To

gravity.

specification

if

with

that, if

height,

a

INERTIA

OF

fact

from

^RESISTANCE

"

^MOMENT

"

well-known

a

towards

is drawn

AXIS

SECTION

drop

to

must

from

is

the

towards Since

OF

"

allowed

be

^NEUTRAL

"

MODULUS

"

Gravity.

of

GRAVITY

XI

act,

to

body.

If

of

body

the be

can

in

its

for

it

looked relation

this

point

expression

is

and

common,

more

STATICS

GRAPHIC

ELEMENTARY

190

invariably

is

used

in

this

be

made

up

connection. We

of

can,

however,

assume

number

of

infinite

an

by gravity,and finding

the

parallelforces. The simplest method is to suspend the body,

of

that

it is free

to

any

position

under

of

the

plate

109.

hang

a

through

A,

process.

The

0,

is the

which This

hence we

verticals

two

can

method

is not

the

more

There

0.

point

are

namely,

few

a

two

the

of

line,

plate.

contain

the

and

repeat in

a

the

point

adopted whereby

cases,

simple

bob.

-

always convenient,

be

the

pin,

the

to

Before, however,

complicated

advantage, study figures.

must

means

"

point

a

same

intersect

plate.

other

find the

will

C.G. of the

experimental some

A^

say

"

at

to pend sus-

vertical

will

line

an

plate,

plumb the

point

Thus,

first

the

Transfer

This

the

desire

we

would

on

up

have

we

We

A, and,

take

shaped

-

C.G.

find.

body

a

gravity.

Fig, 109,

whose

another

C.G. of

so

irregular

Select

of like

system

a

on

itself into

finding the

in

C.G.

acted

one

of

action

Fig.

to

resolves

then

resultant

the

body

each particles,

problem

the

of

the

ing consider-

might,

we

with

simpler geometrical cases

triangleand

the

of

extreme

portance im-

parallelogram.

192

ELEMENTARY

"Ach

Btde of

plates,in

GRAPHIC

the

the

EF.

Consider

horizontal

row,

line

same

EF.

Since

from

EF, their moments

these

individual

the

to the

left

The

C.Q.

line EF,

In

similar

a

will be

will

be

obtained.

E

equal

a

all the

plates on

therefore

lies

on

the

C.Q. must

lie

on

the

GH,

be

seen

0

is

4

the

Hence

that

the

It

will

point

the

section inter-

the

also

Rule."

is

O

required C.G.

^

other.

the

intersection

B

and

Every ing correspond-

plates on

the

way

has

line A

equidistant

EF

parallelogram

of the

and

about

all the

from

equidistant

right of EF of EF, hence

side will balance

one

small

two

any

equal weight

balance

plate to

partner the

of

are

opposite, hence

STATICS

gonals. dia-

of

the

yo

find

the

i_ p-

CO.

jji

of

gram, parallelo-

a

the

draw

diagonals

Many

arid

consider steel

solutions

It will

only

plate

required

be

one

of

to find

the the

AF

readily suggest

case.

given

In

position of

BC.

Find

a

of

means

to

is shown

a

it is

and

hole, from

which,

cord, it will remain

BE

the

a

a

the

connection,

Fig, 112 shape, ABCD,

plate be suspended by horizontal position. Draw parallelto

readily be

can

parallelograms,and

sufficient,in this

if the a

intersection.

trianglesand

into

preceding

solution.

their

simple geometrical figures

subdivided two

mark

two

in

parallelto AD, and C.G. of the triangle

BEG, P and

ABCF

and

SR.

The

points S

in the

is the

O

intersection

required hole, i.e.,the C.Q. Funicular Polygon Method.

Fig.

best

solved

The

problem simply

two at

by

the

two

resolves of

systems

right angles.

Let

left-hand the

between N

respectively. position of the R

shape of the

method

shown

leads

soon

into

funicular

itself into like

in

divide a

polygon. finding the

parallelforces, the another, preferably

one

polygon, draw

figure,and them

to

the

consider

us

irregular-shaped plate bounding

C.Q.

the

112.

of

aid

systems being inclined

first the

points

Complicated and irregular figures are

to confusion.

of

also

^Ifthe

"

above

figure be complicated, the

resultant

and

the

in

of the

parallelogram ADEB, Q respectively. Join QP. Find

and

of AFD Join

193

GRAVITY

OF

CENTRE

number

the

case

of

Fig.

113.

two

vertical

the of

the

Taking

horizontal

lines tance dis-

equal parts,as

194

ELEMENTARY

shown

by

GRAPHIC

dotted

the

Mark,

eight parts.

STATICS

vertical

on

a

lines

horizontal

figure, the

mid-points of these

verticals.

These

boundary

of

curve

the

R^ by whole

the

The

intersection

give

the

the

line

the

or

in

process

of the

lengths

reduced

in

the

construction

plane figuresof simpler when funicular the

that

position of inclined

beam

is then lie

C.Q. must

C.G. of the

on

the

in

section, we

of

C.G.

each

Fig.

proportionalto section. now

section

down,

subdivisions

the

which

set

The

the

area

each

of

drawing

the

found

in

the

point

resultant cuts th^ horizontal

centre

figure. beam the

up

into

find

the

points

draw

load-line, ad^

a

and

cd

each

are

respective part funicular

quite straightforward,and is

and

these

a6, be of

of the

determining

as

one

it is known

shown,

as

Only

the section

first divide

line, centre-

a

cast-iron

In

114.

portion. Through

perpendiculars,and

about

centre-line

for

of

cases

becomes

sections.

used

commonly

all

case

required, for

rectangular portions

three

Rg.

found, will

so

The

weight.

in rail and

case

is shown

sections

to

find

so

applicable in

figure is symmetrical

polygon

section

A

is

uniform

the

is the

as

portion pro-

already explained.

resultants,

two

cepts inter-

these

direction

a

the

ak, the

Find

method

by

cut

required C.G.

above

The

is

be

first,preferably at right angles, and

the

the

will

demands.

case

resultant

Repeat

in

in

case

through the raise widths, and

figure. Measure

down

this

line

interceptsbeing increased as

the

the

set them

and

of the

verticals

in

"

the

O,

at

line.

of

polygon

C.G. which

of

the the

OF

CENTRE

When

used

shown

as

of the

Axis.

"

In

exceed

not

alvtayspasses

sectioa The

the

neutral

all beam

the CO.

through

the CO.

axis. so

limit,the

of

ia

arrangement

sections,

the elastic

through

195

line N.A.

the

is called

section

does

beam

a

Fig. 115.

in

Neutral stress

as

GRAVITY

long neutral

the

as

axis

the section.

V

Fig. as

the

centre to

draw

should the

shows

116

metal

is not

line,it the

out to

Compression

east-iron

have

channel

section, and,

symmetrically disposed

becomes

two

work

sectioQ

a

necessary

polygons this the

flange,4

as

example

in

finding

shown.

for

The

about the

a

C.G.

student

himself, assuming

following dimensions, namely in.

x

1'5

in.;tension

flange,6

:

in.

in.; and

x2'5

8

web,

is 1*35

the

CO.

and

1*0 in. from

in.

in. from the

1*5

x

the

the

subjected

with

M

/ 2

of the

line of the with

of

web,

section.

the

strength

bending the

moments, moduliLS

have

we

of

section

to

of

Thus, let

"

=

bending

=

max.

=

to

position

114.

is termed

what

beam.

depth

"

Fig.

deal

centre

of

centre

The

in.

Figpures. In dealing

Resistance

of beams

STATICS

GRAPHIC

ELEMENTARY

196

(lbs., ins.),

moment

stress

modulus

induced

of section

at sect.

(lbs.p.

sq.

in.),

(ins.^).

ThenM=/x2;. The

value

of

z

varies

with

the

shape

of the

section,

CENTRE

OF

and, in determining its value make

can

of certain

use

Consider

117, and

let

in

sq. in.

per be

At

less and

will

a

equal

be

y from

distance

y' the /x ^

to

tions. construc-

shown

as

maximum

the

distance

any

given section, we

a

simple graphical

that

assume

layer,at

a

for

section, ABCD,

the

us

197

GRAVITY

Fig.

in

duced in-

stress

N.A, is / lbs.

the

will

induced

stress

is the

This

stress

.

y

intensitywhich

is

strip,

cross-sectioned

the

acting on

1 Fig.

whose

area

we

obviously, if the

stress

as

reduce

we

per

take

can

116.

a

this

area

a,

in.,since the total

sq.

Now,

inches.

square

will

we

increase

(/

stress

x

^

x

a

j

if

be

will

now

the

depth

width

lbs. per

until sq.

distributed of

the

the

over

same,

is a' and

area

in.,then

a

the

strip

smaller

have

we

the

area.

let

us

stress

Keeping reduce

its

intensity/

"

/X?^xa=/xa'. y

a'^y'

. .

.

"^~'

"

a

Therefore,

to

"

y

give uniformity of

stress

over

the whole

ELEMENTARY

198

of the

distances

to their

formed

when

have

now

a

from

uniform

stress

by

the

stress

(P) is equal to stress

per .-.

This, of below

the

course,

N.A.

of

/ lbs. the

proportional

In other

bounded the

triangular area,

of

be made

N. A.

the

Fig.

a

STATICS

stripsmust

stripswill be the diagonals

of all the

ends

We

width

the

area,

GRAPHIC

on

by section which

words, the

the

triangles drawn.

are

there

exists

116.

per

in.,so

sq. of the

area

that

the

tptal

trianglemultiplied

sq. in.

/x^. P=i(6x|)x/ =

applies

to

each

triangleabove

and

ELEMENTARY

200

be

N.A.

the

near

from

seen

show

GRAPHIC

examination

an

resistance

the

diagonal vertical It

might of

method section. done

be

and

round

a

well

to

the

a

respectively.

section

general

figure for

resistance

Fig. 120, N.A.

the

This

with

first

thing

Fig.

given to

done

be

may

a

be

by

119.

given instance, but, for unsymmetpolygon must be drawn. sections,the funicular BC at right angles to the and the lines AD

inspection in Draw

119, which

section

square

Fig. 118.

rical

and

Figs.118

will

as

worse,

even

explain briefly the

to

constructing a draw

of

are

figures for

Referring

is to

sections

Many

STATICS

the

N.A., bounding

the

section

on

the

and

left

right

respectively. Complete the rectangle by drawing DC and parallelto the N.A. and equidistant from of the lines, at least, being tangential to one section

at the

should

be

point

noted

most

remote

that, in this

case,

from as

the in

all

AB

it, the

N.A.

It

cases

of

CENTRE

OP

symmetrical sections, both

lines

the section.

as

the

that

of

case

The

sections

N.A.

Draw

verticals

in G

and

the

applicable in

be

in

F.

and K

about

L

N.A.,

E

From

and

in order

F

and

respectively.

the the and raise Join

EF

cut

j^

y\

L

H, which

points

two

are

used

; these

will

lines

given, is

unsymmetrical

are

in E

AB

cut

LO

and

tangential to

line,EF, parallel to the

section

to

will be

also

may

which

any

cutting the KO

wording,

construction

201

GRAVITY

on

ance required resist-

figure. Take a

such so

of

series

lines,

EF,

as

(sg

and

sufficient

obtain

enable points to the sistance complete re-

be

figure

to

in.

In

will

be

drawn

general, it found of

Let

the

transfer

then X

A 2

=

=

=

to obtain

figure by

resistance

the

and

convenient

most

distance

the

C.G.

the

of each

half

suspension method, the

C.Qs.

between

drawing.

in ins.

of resistance

of section

modulus

120.

positionsto

one-half

area,

^^s-

fig.in

sq. ins.

in in.^ units

Then^z=Axx.

Figs. 121 tee

section

It

must,

and and

122 a

show

cast-iron

however,

be

the beam

resistance section

clearly borne

figuresfor

a

respectively. in

mind

that.

although

the resistance

dispositionof

the

it does

necessarily follow

section If

lying outside examine

we

figurefor in the

web

is

metal

section, we

which

highly stressed

more

rsl

than

Pig. 121.

flange,and, tee

use

as

for

if the

section a

the

is

as

strong

of

material

in tension

case,

however,

section

the

be

compression greatly

tension.

For

cast-iron

on

the

metal

metal

a

we

have

compression

is, for

as

in

two stress

all

compression,

economical

an

in

drawn

resistance

the

steel,which

obviously not The

beam.

of

case

strength

one

useless.

Fig. 122

practicalpurposes, the

the

N

A

"

the

that

once

section,

parts of

figureare

shows

at

see

in the

all

that

resistance

Fig. 121,

tee

a

the

the

of the

indication

figuregives an

of efficiency not

STATICS

GRAPHIC

ELEMENTARY

202

is very

one

to

different

cast-iron, whose exceeds

resistance

that

in

figures,

base, as is shown

CENTRE

Fig. 121,

in

shown

base, as draw

and

another in

equidistantfrom that

one

at

point leaat

To

different

is

latter

parallelto this

the

stresB

i^oin

we

and

N.A.

important difference,

tangential to

now

from

remote

the

tenaioo

a

the

section

N.A.

the

portance im-

the

two

resistance

figures,

might

we

this

apply

GH,

203

on

In

it,but with

the

of

123.

and

o" the lines

show

GRAVITY

drawn

Fig.

linee,EF

two

a

OF

section

particular

to

given

a

'^

^

case.

Example. iron

caet-

section,3^

tee

in.

3

in. wide,

^ in. thick

through-

deep ia

A

"

and

What

out.

vxtvld

assuming /,=

section

the web

found

the

at

carry

of the

(a) Gompreaaion Scaling

^iff-^^a.

from

that

between

of a 3-ft.span,

placed uppermost

^=6000

U)a. aq. ina.

drawn

when

above

area

CGs.

full size,it

=

250.

2,= 25x-76=l-875

m_/xg^x

6000x1-875x4

4

" ~

"

L =

1250

?

base.

triangular

aq. in. ; distance .-.

atreaa

Fig. 121,

the

centre

section

Iba. sq. ins. and

2000

*""

load

safe

this

^

12x3

lbs.

the

was

N.A.='75

204

ELEMENTARY

Therefore, concerned,

GRAPHIC

far

as

could

we

tbe

as

STATICS

stress

is

beam

to

compression load

safely

the

lbs.

1250

(6) Tension

Scalingfrom found

that

base.

stress

when

Fig. 123,

the

full size,it

drawn

above

triangular area

N.A.

the

was

l-5

=

sq. ins.

2/=2-5x

/.

/X

w"

. "

1-5

L

4

x

lbs.

830

lbs. at centre, this

is 830

induces

which

3-75

X

123"3

-

Therefore, safe load load

4_2000

g/ X

"

=

the

3-75,

=

tension

maximum

the

being stress

allowable.

It should

will

cast-iron tension

an

at the

will be in

outside

of

layers

than

it would

compression, whose

at that

that

The

shows area

the

to

that

at

point.

limitingvalue the

down

that

appear

stress

seen

of

web, the width

the

stress

therefore

limiting value

Fig. 124

of

section

the

it will be

Fig. 123,

the

keep

to

overstressed

the

and

compression

maximum

acting simultaneously.

mind, however,

than

in

well-designed section

a

examination

is wider

provide, and

the

have

strip necessary limit

that

noted

stresses

From

that,

be

this

is very

given

point

can

the section be borne

It must at

a

point

much

is

a

higher

in tension. resistance

of one-half

of the

figurefor resisttoce

a

rail

tion. sec-

figureis

3'05 sq. ins.,and

C.Qs,

the two

(Note.

of section

Fall-size

"

moment

Inertia

of

in

produced loads, deal

with

the

momevi

138.

=

determining

ins. the

) tions deflec-

ternal ex-

to

inertia

section.

This on

the

shape

of

the

must

be

cal-

section,and

4'5

In

of

and

X

beam

a

qoantity depends size

of

centres

is termed

what

the

of

305

of 8ection=6

have

we

=

depth "

application of

the

on

the

ins.

45

Modulus

205

between

distance

the

IB

GRAVITY

OP

CENTRE

. _

culated

axis

neutral

sections, the is the

to

axis ; in beam

fixed

some

reference

with

reference.

of

axis

methods

Graphical

can

in employed again be determining this important

If we quantity. a body situated,

have with as

respect

shown

in

to

The

mrK

respect be

a

to

Fig. 125, then

moment

at

a

about

r

of inertia

which

from of

the

Referring we

of inertia of any

the moment

distance

xy=l=^r^.

particleof

inertia

axis xy,

an

little particlem

'"'K-124.

wish

P, P being the end

xy

will be

whole to

to find

equal to body with

Fig, 126, the

elevation

moment

let

m

of

of the axis

206

GRAPHIC

ELEMENTARY

of reference.

Set

down

ah

Join and

6b

IV.,

oh

and

oa

and

draw

have

moment

of

Now

set

an

and

down

in is

(abxr)

a

of ins.

parallelto

md

explained

as

wi, select

number

even

a6xr=cdxH;

oa

Chapter the

first

P.

about

m

represent

mc

respectively. Then,

we

to

H

pole o, preferably making

STATICS

cd

select

and

Fig.

pole o', making,

a

if

126.

From o"' draw possible,H^ equal to H. parallelto o'c and o^d respectively. Then again we have efxW^cdxr.

o^'e and

o7

abxr But

cd= H

,,efxW=^xr. id.

e/ X

.-.

H^

X

H

=a6 =

The of

X

abx

about

If the

above

and

that

so

in

P,

or

the

is full

finding the

r

termed

are

drawn

size, then value

the

second

of inertia

moment

diagrams r

X

r2=mr2=Ip.

quantity {ab x r^) is m

r

of

so

there

of

m

moment

about

P.

that

1 in. =

l

will

be

culty diffi-

no

lb.,

Ip,but, in general,both

from

Now

STATICS

GRAPHIC

ELEMENTARY

208

diagrram it will be

the

"5=

V

"

ef

(ms.)

r

that

seen

"

"

=

^

r^

r

/.

mr^

ef x

=

xg^xpxHxr,

"

T

(e/X g2 X p) X Hj

=

Thus

let

scales

in

used

lbs.,and

10

.*. Ip

=

215

=

2144

82

X

X

shown

1-25

X

of

can

a

ins.

2'15

=

l-25(Hi

X

=

Fig. 127,

of inertia,so

of inertia which

these

of the

of

System

readily

of inertia in

moment

from

10

of Inertia

the moment

moment

ef

equals

H

=

125

in.),

lbs. f t.2 units.

construction

are

ft.,and

8

The

lbs. ft.* units,

2160

Moment

the

equals

in.

1

are

ft.

12

15x12x12, =

as

equal

r

original diagram

the

1 in.

lbs. and

15

lbs. ft.^ units.

e/xg2xpxHiXH,

=

mr2=

above

equal

m

H

X

of

a

but

be

applied of

system it will

found, does beam

individual

Forces." to

The mine deter-

forces, such

be

seen

that

equal the (shown dotted) not

section

forces have

been

derived.

approximation can, however, be obtained, by of strips, dividing the section into a large number whose The section in are axes parallelto the N.A. A

close

this 8

case

ins.

ins.=12

X

is of

cast-iron, the compression flange being

4 ins. =32

sq.

sq.

ins.,and

ins.,the tension

the

web

12

ins.

x

flange 6 2 ins.

=

ins.

x

2

24 sq. ins.

CENTRE

We

first set down

12

respectivelyto Select

a

inches, and the

resultant

the

CO.

GRAVITY

a", be and

pole

o,

equal

H

making

an

funicular

the

R,

cd

209

to

32, 24

and

scale.

some

draw

of the

OF

will

which

Let

section.

even

number

of

Draw

in

polygon.

of

course

us

assume

pass

through

that

the

line

Fig. 127. of action about

of

which

R

is,in this wish

we

polygon, so

reference.

find

If necessary,

of the system. funicular

to

Then

we

the

case,

that have

the

axis moment

produce each

one

the

cuts

of

reference, of inertia

links the

"

Moment

of AB

about

O

Moment

of BC

about

O

Moment

of CP

about

0

=

=

=

+(aV x H), (6V x H), {c'd'x H). -

-

of the axis

of

GRAPHIC

ELEMENTARY

210

These

STATICS

of the

moments

respective forces

about

first

the

intercepts,multiplied by H, give

axis

the

of

reference. The

intercepts

drawn

made

possible,be the

cut

Then

axis

the

equal

H.

to

reference

of

required

being

xy

polar

measured

in

and

to the

H^

should,

Produce

pg

and

and

x

following dimensions original drawing:

if to

is

respectively.

y

a:i/xHxH^,

=

scale.

proper

The

polygon

distance

of inertia

moment

fresh

a

funicular

second

a

The

"pqrBt

as

down

set

now

selected, and

o'

pole

are

from

taken

were

the

"

Point

ins. from

0-7*5 H=Hi=3 scale

Force Linear

under

of section

side

ins.

scale

lbs.

in. =20

1

1 in. =2

ins.

iri/=3'l ins. .".

of

Moment

about

0

inertia

of

of

given system

forces

"

=31x22x20x3x3 =

I for the

calculated

The

2232

lbs. ins.2 units.

given

section

beam

is 2532

ins.* units.* It will

applied to Much

the

on *

In

the

areas,

a

same

Note

the

discrepancy

is

considerable,

particularsolution

accuracy

largernumber

of inertia can,

of

beam

a

be

however,

of sections, and

be

cannot

section. obtained

proceeding

lines. difference

previous t.e,

this

find the moment

greater

by taking

the

that

seen

obviously

that

so

be

case

to inches

\

we

between have

lbs, inches^ taken

forces

and as

inches^

units.

proportional

to

CENTRE

The

from

funicular

constructed to Professor

of

section

a

of the

measurement

a

211

GRAVITY

of inertia

moment

The

OP

specially

a

method

the

polygon,

of

area

mined deter-

be

can

due

being

Mohr.

construction

is first divided

is shown into

up

Fig. 128.

in

of

number

a

section

The

parts as shown,

perpendiculars dropped from the C.G. of each portion. A load-line is set down, as shown, in the Let the total length of this line equal x ins., line ay. and

then

the

polar

Construct

Let

of

means

of the

area

funicular

the

by

area

H

distance

is made

equal

ins.

-

its

measure

Measure

planimeter.

a

and

polygon

to

also

the

section.

Ai=area

of funicular

polygon

A2=area

of section

sq. ins.,

in

of inertia

I=moment

in sq. ins.,

in in.* units.

ThenI=AiXA2. It

should

further

be

noted

that

where

2=",

y=

y

the

of

distance

the

greatest

strained

fibres from

the

N.A. The the

size

section

undernoted

shown

is the

dimensions

same were

Fig. 124,

in

as

taken

from

the

and full-

drawing. Area

of funicular

Area

of rail section

Distance .'.

""

polygon (Ai)=4-6

of extreme

In.a.=4-6

3-25"

(A2)=10'0 fibres from

X

10=46

sq. ins.,

sq. ins..

N.A.=3*25

ins.* units.

ins.

212

ELEMENTARY

It will same

be

shown

case

obtained

this

STATICS

figure

from

is

the

resiatance

Beam

Section-

"

interesting adaptation of Mohr's of

practically the figure

Fig. 124.

in

Feppo-concpete an

that

aeen

that

as

GRAPHIC

a

ferro-concrete

beam

Fig.

129

method

section.

In

shows to

such

the a

beam

the

stresses, while the

tension

arranged

is

concrete

the

Elastic In

determining

f

"

^

b

c

Fig.

section,

into

beam

of each each be

first divide

we a

series of

strip a

stripand

noted

divided

that into

horizontal the

only

a

strips;the

"

__-"^

and

z

for such

a

o"

compressive draw

line.

areas

the

side of the

through

Measure

in the

portion of reason

I

129.

the

up

of

and strips,

set out

that

of concrete

values

^

all

takes

shown

has

of steel

modulus the

sive compres-

steel reinforcement

modulus

Elastic

the

take

to

Experiment

stresses.

213

GRAVITY

OP

CENTRE

the

the C.G. area

of

line ag ; it should section

will be obvious

need

be

from

the

ELEMENTARY

214

Through

diagram. oa

GRAPHIC

the

Measure

of the

area

Let

a'

total

off aA

(15

=

set

ah off to the

the

various

r

I

Then

N.A.

scale

of the

will

only

sr

has

be

one

to

the

when

be

to

(for steel) (15 x =

==ahx

the

scale

to

is measured

been

set

out.

half

size, the

funicular

polygon

scale, then same

the

H

its full-size

in the

scale,

some

measuring to

and

x2B,.

pqr

drawn

of the

horizontal

section.

drawn

drawn,

quarter of

I

construct

The

is measured

also

will be reduced Now

mind

been

reduced

Join

N.A.)=area

section and

a

and

ab, be, etc.,have

beam

section

drawn

length

pqr

to which

if the

area

be

area

section

to the

in

borne

of the

will be

section

be

The

the

ah

about

(for section

o,

to

for ag.

used

shown.

the

give

ins.

ins.,being careful

was

pole

will

which

Thus

the

to

of ins.

in sq. ins.

a')sq.

sq.

as

as

pqr.

area

scale

pqr

this must

and

x

a')

number

off

bars,

area=(m

polygon

general,the

In

of steel

X

mark

reinforcingbars. bar

m

same

points

funicular

through

X

and

even

of each

area

=

Then Set

steel

number

m=

vertical

a

preferably an

this

H, making

=

raise

a

STATICS

value, and

obviously

if the

proportion. steel

sq^.

area) x sq^

GRAPHIC

ELEMENTARY

216

STATICS

/c= crushing strength of

Let

of

distance

n=

N.A. Then

2=

greatest strained

particlesfrom

in ins.

L_2-?Li, and

"

in lbs. p. sq. in.

concrete

the

of resist. =^-^

mom.

n

n

Examples. 1. Draw

Fig. the

irregular figure,such

any

of the

CO.

funicular

the

by

Find,

113.

is shown

as

in

polygon method,

area.

Carefullycut out the figure and check your result by the suspension method. circular discs, 2. A steel plate is composed of two centre joined together by a rectangular piece, whose line passes circle. One through the centre of each is 3

circle

4*5 ft. and the

from

distance

The

C.G.

3. A

Draw

the

4.

Check

the

load

4 ins.

square,

(6)

A

circular

deep

and

5. A

tension

tee

ft. diam. circles

1*2 ft. wide.

1

wide,

on

beam

ins.

the

is

Find

modulus

of

and

find

4-ft. span

if the

stress

sq. in. "

diagonal vertical,

ins. diam.

in. metal both section

X

3 ins. wide.

calculation

(solid),

section,8 ins.O.D.and

section,

flange is 8

find

figuresfor

section, 4*5

cast-iron

a

side, with

circular

4 ins.

on

by

lbs. per

resistance

A

{d) A

the

deep by

is 6 ins.

result

2500

(a)

hollow

of

centre

figure and

your

to exceed

(c) A

beam

resistance

Draw

is 2*25

piece.

the safe distributed is not

to

centre

rectangular

section.

other

the

rectangular piece is

of the

the

and

ft. diam.

4

ins.I.D.,

throughout,

6

ins.

bases. is 18

3 ins. ; the

ins.

deep.

The

compression flange

CENTRE

ins,

6

C.G.

in.

If

X

OP

and

;

web

the

the resistance

Draw

217

GRAVITY

Find

in. thick.

1^

figureand

the

find the modulus

of section.

Calculate if the

distributed

safe

the

is not

in tension

stress

load

for

exceed

to

8-f t. span

an

2000

lbs. per

sq. in. 6. Draw

horizontal

a

forces,P, Q

S,

and

about

system

following

the

values

the

point

a

scale

its

the resistance

method,

1bs. + 6

check

the the

+8,

-10-7,

S + 10

+11-10.

depth

is 8

section

of

value

in

shown

of Question

z

to such

its C.G.

Find

ins.

Fig.124

Draw

of section.

find its modulus

of

moment

your

of

inertia

forces have

the

Q+6

figure and

the

A

"

the

Using

8.

when

the rail section

out

that

A

Apply

ins. from

8

of

moment

long.

:

P+8

7. Draw

8 ins.

AB,

ins.,and

ins.,5

3

Find

respectively.

line

7, find, by Mohr's

inertia

of

found

from

section, and

the

the

previous

question. 9. A 24

ins.

consists

reinforced

deep

to the

ends,

find

load

exceed

600

Find

(a)

Position

(6)

Value

of I

(c)

Value

of

beam

a

of

of

1

lbs. per

maximum ton

ins. wide

per

sq. in.

and

which

"

N.A.,

(total),

z.

of this section the

is 18

of the reinforcement,

centre

of 6 bars, 1 in. diam.

Assuming the

beam

concrete

simply supported safe

span

foot-run, if jc

for must

a

at

tributed disnot

CHAPTER

XII

WALLS

RETAINING

A

wall,

RETAINING

given

name

earth

sustain

to

is

daTa

The

opinion.

complicated

sustain

water

exert

a

wall, and that

P

must

will

tend

to

magnitude P

X

H.

of

Now,

such

as

the the

which

dam

or

Obviously P,

on

to

the

wall

wall,

that about

overturning of 218

moment

the

to

is

shown,

the

water

back

the

we

used

as

hydrostatics

seen

weight

very

tary elemen-

very

of

normally

the

overturn

and

arrangement

130.

be

a

are

difference

with

wall

Fig,

knowledge

readily

time,

wide

a

duced in-

stresses

and

of

case

The

pressure,

act

is

on

here.

pressure.

our

It

face.

subject

retaining

in

the

term

acting

present

only problems

form,

the

discussion

treated

is the

crudest

must

much

connection

in

forces

the

at

the

although

of

straightforward

deal a

its

in

be

most to

which,

and

can

The

the

employed

are

waterworks

nature

whole

one,

nature

have

of

subject

The

the

quantities

are

the

in

used.

and

walls

as

pressure,

while

is

masonry,

particularly

commonly

more

retaining

still

water more

earthworks,

with

structures

or

used

is

expression

of

such

to

of

generally

of

the

teaches the

on

us

inner

the

force, P,

will

the

point

O,

the

being

wall, W,

equal

acting

at

to a

distance

from

H^

O, tends

is obtained

and

wall

the

In

conditions becomes

we

the

combine

the

two

W

find

their

resultant, then

the

and

of

base

the

of

line

the

cut

middle

Fig. 131,

to

ABCD

let

represent of

cross-section should

noted

be

of the

wall.

In

this

A=Cross-sectional

t(;=weight =

depth

that, in

As

lbs.

shown

of material

the

deal

130.

problems

out

with

1-ft.

length

"

of wall

of wall

in reservoir

acting at by

let

case

Then =Aw

working

always

area

of water

Fig.

a

dam.

the

retaining walls, we

A

joints, it

the

at

other

action

must

within

Referring

on

certain

of its width.

third

It

never

such

be

must

resultant

the

must

turning. over-

P

between

relationship the

point

of

restrictions.

and

that

point

pose im-

P

W

this

stresses

forces

and

the

on

stabilityand

to

necessary

just equals Wxff,

just be

ensure

regarding

certain

If

to

of

condition

The

PxH

practice,however,

reached, and,

be

then

steady by

wall

the

moment.

when

would

keep

to

resistingthe overturning balance

219

WALLS

RETAINING

in sq.

ft.,

in lbs. per

cubic

ft.,

in ft.

W=Axlxw;. the

C.G. of the

triangle GHC,

section. the

pressure

exerted from

by

the

water

on

surface

the

at

zero

the

wall

to

maximum

C.G.

X

acts

pressure

at

parallelogramof forces,

B

A

CK=^

where

point K,

a

the aid of the

With

at the bottom.

A2 lbs.

=31-2 total

a

gradually increases,

P=Axlx|x62-4.

.-.

This

STATICS

GRAPHIC

ELEMENTARY

220

find

the

resultant

of P

and

W.

Trisect DC

all the

F, and

and

ditions con-

necessary

of

of

the

line

the

resultant

the

middle

of the

stability if

fulfilled

be

will

base

the

E

in

R

of

action

(R) third

cuts

(EF)

base.

It should, however, be

noted

cases

Fig. 131.

above

not

of universal

Example. the

base

the outer

conditions Determine

1

has

the

dam^

masonry

and

face

many

practice

where

the

resultant

falls

outside

middle

third,

the so

that

conditions,

is

application. A

"

in

occur

rule, although giving ideal

the

at

that

ft. 8

a

water

ins,

wide

batter of 1 rises

graphically if the

ft highy is

8

at

in

to

wall

the

is

jt

top

of

safe

wide

top, while

Under

8.

the

5

certain the

under

tvall. these

Take

conditions.

221

WALLS

RETAINING

weight of

the

lbs.

150

as

masonry

cvbic ft

per

P=31-2xA2=31-2x8x8, lbs.

=1995 Note face

this

that

force

normally

acts

the

to

inner

AB.

W==(^-+^)xSxm Find

of

equal

ac

It will be

ad.

third,

middle

lbs. wall

the

R

that

06

scale, and

ab

the

cuts

wall

the

intersect

to

parallelogram

the

that

P

point O, in

and

probably

a,

equal join the

outside

base

may

W.

represent

to

lbs.,to

1995

to

seen

so

of

action

the

in

section

vertical

a

Complete

lbs.

4000

to

4000

line of

the off

=

1200,

X

O, draw

and, through

Mark

3-33

C.G.

the

Produce

=

not

be

safe. solution of

The

for earth

retaining walls than

of the

earth

the shows

a

the

on

like

section

top and

other

to

assume

a

level

shown

of

flush

plane, such by

the

by the

line the

with of

a

the

as

CE, and

"p is

the

being

level

Earth,

nature, will be found when

The

tipped

slope

inclination

p, is termed

termed

earth

which

Fig. 133

of the wall.

slope

C.G.

of

account

wall.

the

top

similar

definite

angle

angle

of

back

culties diffi-

more

of the thrust

retainingwall,the

a

certain

chieflyon

magnitude

the

on

materials a

measured while

exerts

case,

of stability

the

presents

pressure

preceding

the

uncertainty

the

problems dealing with

the

the

assumed of the

natural

angle of

on

to

is

line, slope,

repose.

222

GRAPHIC

ELEMENTARY

Obviously,none CE

have

can

retain than

of the

any

its natural the

earth

eflfect on

to

the

slope without

friction

of

its

own

ig the angle of friction for

STATICS

the

right

the line

of

it is able

wall, since

assistance

any

to

other

particles.The angle p the earth composing the and

mass,

ledge know-

our

of

mechanics

teaches

that

us

a

earth, particle of lying on this plane, will just be on the point of slipping It is only down. the triangular portion of

earth, BCE, the

which

wall

is

required to support, and the problem is, mutKiuPiu^uiuM B

find

to

force

horizontal

of earth

mass on

the

of this

132.

the many that

years due

to

Rankine's an

has

found

most

Professor

of

problem,

theory

and

which

country

for is

Raukine.

Theory."

examination

advanced

the solution

in this

favour

exerts

theories

been

towards Fig.

this

wall.

Various have

what

out

Rankine's

the

theory

conditions

is based

on

influencingthe

224

ELEMENTARY

fact that the

GRAPHIC

earth

the

lying

slide

down

wall.

It

the

in

due

the

to

fact, that

obtained which

is

This show

consider

A

"

the wall

and

is 45".

Damp

ft.and

masonry

170

134

being equal The

shows

prism,

whose

14

120)

wall

the

plane

this

method,

a

the

plane OF.

a

angle BCE.

the

To

of rupture. will

we

now

This

ft.

lbs.

=

weight, W, of

if fulfilled

=

to

calling

W^

in the

the

the

(?"^)

14

the have

we

Now,

of the X

angle of cubic

of

170

=

point 0^

weight W^

=

of the

(Jx5*75

C.G. of the

calculate

wall, thus 8330

BDK,

mately approxi-

find the and

BDK

angle

measure

point O,

length X

whether

bisector

the

found

lbs.

required

lbs. per

120

of

cubic ft.

is BDE,

4825

1 ft.

Determine

vxill.

triangle BDE,inthe

Hence,

ft.

5

clay which

of damp

DE,

be

section

vxill is

arrangement,

Draw

is

the top, the hack

at

hank

U)s. per

will

BE

ft high,

14

weighs

day

the

section, ABDC,

W

of

C.G. of the

length

X

of

45".

to

the

5*75

X

bisects

of stabilityare

repose

find

down

OF

ft.wide

2

the top

the conditions

and

to slide

as

the pressure

flush with

Fig.

than

is

being perpendicular.

to sustain

is

less

moment

retaining wall,

the base

at

ia much

turning over-

simple example.

a

Example. wide

the

the

that

however,

to

overturn

overturning

application of

the

to

tend

and,

that

is known

plane

slope,i.e.,

triangular portion BCF,

tends

situated

so

tend

so

maximum

BCF

when

natural

portion BCE

smaller

the

the

(Fig.133),will

shown,

effect of the

that

and

plane

been

has

above

BCE

triangular portion

STATICS

lbs.

"

'

the

Now, Coulomb's

theory

between the

first to

have

we

the

earth

force, P, will

takes

no

and

the

act

at

action

Now,

point

a

of W^ if

normally

P intersect

friction had

no

B,

and

the

would

to

=

on

existed

plane have

of

any

friction

wall, hence

the

back

P.

of the

wall,

134.

DM

where

and

earth, BDE, reaction,

M

of

account

back

of

magnitude

the

find

Fig.

and

225

WALLS

RETAINING

of been

iDB. the

The

plane

between

of

the

'rupture,DE, normal

to

lines

of

rupture. mass

of

then

the

DE,

but,

226

ELEMENTARY

since the

of action

45"

with

we

0

by drawing

the

normal

R, and

determine

their

represent

the

the

P,

line of action

line

and

ac

W, and

de=

will

resultant

allowance student on

Rebhann's

to

this

The

shown.

by drawing vertical through W, and

of

the

produce

base

conditions

dg

within of

methods,

exact at

off

=

parallelogram. The the

cut

the

friction

Mark

it in d,

to

to

of

135.

more

is advised

angle

equilibrium,hence

a

action

cut

hence

an

values

Draw

of

found

for earth

in

are

complete

For

fulfilled.

the

be

third, FG,

middle are

of P

plane, as

B's

obtain

can

making

individual

Fig.

the

45", we

line

triangle of forces abc. to

STATICS

is a

to

forces,W^

can

the

friction

of

angle

Hue

three

GRAPHIC

the

consult

back the

the

stability

making

of the

various

wall, books text-

subject. Method.

"

It

might

be

advantageous

to

RETAINING

indicate used

for

many

construction

and

seems

to

and

in

of

than

the

top

of

by

the

line

BEG.

semicircle,

and

and

radius

BE,

to

AD.

Make

FK

draw

FG

parallel

join

GK.

Then

centre

FGK

resultant

=

of

the P

and

with be

of

line W the

split

point

a

describe

right

at

is

to

to

F, and

in

equal

FGK

BAD

angles

BC

cut

slope

FG

and

termed

the

ft.,

sq. in

earth

lbs. per

cubic

wall

on

in

ft., lbs.

ft.

per

of

action

of

up

into shown

to

P

\

the

BA,

parallel

be

now

aid

lbs.

P=(Axlxti;)

equal

may

as

to

length.

BL

off

in

pressure

Then Mark

DE

triangle

of

weight

te;=

P

of

area

=

the

triangle.

pressure A

the

being

angle

an

B

With

135

line

construct

draw

Fig.

natural

which

on

D

BC.

Let

The

from

to

earth

A

rises

case,

wall.

BC,

At

equal

2^,

the

AC

wall,

this

in

theory,

practice.

retaining

a

engineers.

quite satisfactory

are

actual

which,

earth,

indicated

a

of

been

has

Coulomb's

on

which

results with

section the

higher is

is based

agreement a

level

give

which

Continental

by

years

Rebhann's

shows

construction

graphical

one

227

WALLS

and to

combined

in

horizontal the

force

the in

parallelogram its

through

of and

triangle

L

natural the

draw

slope.

usual

'forces. vertical above.

way P

can

ponents, com-

EDINBURiiH

LIMITED

COLSTONS

PRINTERS

LIST

A

OF

BOOKS

PUBLISHED

WHITTAKER 2, White

A

Catalogua

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books

"

St., Paternoster

Hart

be

BY

sent

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London,

Square,

full details

free

post

CO.,

of

the

E.C.

following

application.

on

s.

Practical

H.

Adams,

Survey

Engineers, Alexander, F.

Allsop,

J. O., and Analysis J. R.

into Work Steel

F.

net

Work net

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Light and

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Practical

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Light

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