strength of materials - priodeep.weebly.com · machines, the book also covers the subject of...

STRENGTH OF MATERIALS A COURSE FOR STUDENTS

B Y

P E T E R BLACK

P E R G A M O N P R E S S

OXFORD · LONDON · E D I N B U R G H · NEW YORK

TORONTO · PARIS · F R A N K F U R T

Pergamon Press Ltd. , Headington Hill Hall, Oxford 4 & 5 Fitzroy Square, London W . l

Pergamon Press (Scotland) Ltd. , 2 & 3 Teviot Place, Edinburgh I

Pergamon Press Inc. , 44-01 21st Street, Long Island City, New York 11101

Pergamon of Canada, Ltd. , 6 Adelaide Street East , Toronto, Ontario

Pergamon Press S.A.R.L., 24 rue des Écoles, Paris 5e

Pergamon Press GmbH, Kaiserstrasse 75, Frankfurt-am-Main

Copyright © 1966

Pergamon Press Ltd .

First edition 1966

Library of Congress Catalog Card No. 66-16364

2532/66

PREFACE

T H E contents of this book fulfil the requirements of those studying

for Higher National Certificates and Diplomas in Mechanical

Engineering and will be useful to students of Civil and Structural

Engineering. Together with i ts companion volume Mechanics of

Machines, the book also covers the subject of Applied Mechanics

for those aiming a t Higher National Diploma in Elect r ica l Engineer-

ing.

Only theory essential to simple stress analysis has been given but

this is complete as no steps in the mathemat ics have been omitted.

Discussion of assumptions made and of l imitations of theory as

derived, has been kept to a minimum as i t is felt tha t this can best

and safely be left to the lecturer .

No a t tempt has been made to deal with the test ing of materials,

other than in simple tension. The student will normally gain

experience of the various standard tests in the laboratory, where

also he will undertake pract ical work designed to verify—where

possible— the theoret ical subject mat te r here presented.

The scope of this book is not such as to include a discussion of

the properties of materials, which topic is inseparable from the

subject of Engineering Metallurgy.

A range of worked examples of graduated difficulty has been

included a t each stage to amplify the t ex t , the la ter ones in each

group being intended to give the student some idea of the sort of

question he will meet in his examination paper. The Council of

the Insti tution of Mechanical Engineers and the Senate of the Uni-

versity of London are thanked for their permission to reproduce

questions from past examination papers.

P E T E R B L A C K

l a SM ix

SYMBOLS

The following symbols are used in this book and conform with

B S . 1 9 9 1 :

L Length , span

A Area, cross-section

V Volume

W Weight , load (Note : W\g for mass)

w Weight per unit length, intensi ty of loading

R Radius of curvature

F Force , load, shear force

M Bending moment

Τ Twisting moment, turning moment, torque, periodic t ime

Ε Modulus of elastici ty

G Modulus of rigidity

Κ B u l k modulus

J Polar 2nd Moment of section (torsion constant)

I Moment of Iner t ia , 2nd Moment of area (bending constant)

Ζ Sect ion modulus, Maximum deflection of beam

U S t ra in energy

y Distance from neutral axis, deflection a t any point of beam

r Radius, distance from polar axis , amplitude

d Diameter , depth

k Radius of gyrat ion

ρ P i t ch , pressure

w In tens i ty of loading

χ Displacement, change in length

ν L inear veloci ty

ω Angular veloci ty

/ Linear acceleration

oc Angular accelerat ion

g Gravitat ional acceleration

t T ime

η F requency

l a * XI

xii S Y M B O L S

λ Stiffness, force per unit deflection

π Circumference/diameter rat io

μ Coefficient of friction

η Efficiency

ρ Dens i ty

δ Deflection

σ Poisson's R a t i o

q, fs Shear stress

e Direct strain

φ Shear strain, pressure angle

0 Angle of twist, angle turned through

Σ Sum of

Derived Units

Abbre-

Quant i ty viated Fu l l form

form

Linear velocity ft/s feet per second

Linear acceleration f t / s2

ft/s per second Angular veloci ty rad/s radians per second

Angular acceleration rad/s2

rad/s per second

Stress lbf/in2

pounds force per square inch

Bending moment and) Ibf in pounds force inches

torque J tonf in tons force inches

Densi ty lbf/in3

pounds force per cubic inch

Strain, kinet ic and 1 in Ibf inch pounds force

potential energy J in tonf inch tons force

ft Ibf foot pounds force

Load per unit length lbf/ft pounds force per foot Spring stiffness lbf/in pounds force per inch

Moment of Iner t ia orl Ibf ft s2

pounds force feet seconds squared

2nd Moment of Mass) slug f t2

slug feet squared (slugs = Ihi/g)

2nd Moment of Area in4

inches to the fourth

Section Modulus i n3

inches cubed

Area in2

inches squared

C H A P T E R I

SIMPLE STRESS, STRAIN AND STRAIN ENERGY

Load

Where an engineering component is concerned, this may consist

of forces due to one or more of the following :

1. A stat ionary (dead) load,

2. A change in veloci ty (inertia force),

3 . Ro ta t ion (centrifugal force),

4 . Fr ic t ion ,

5 . Bending,

6. Twisting (Torsion),

7. A change in temperature.

Stress

The application of the load causes a deformation of the compo-

nent which induces an equal and opposite resisting force in the

material, the intensi ty of which is referred to as stress.

LLLLL / / / / / / / /

I ι X

T

F (load)

(Section) A

L (length)

Τ7ΤΤ77Τ7Ύ777777Ύ7

(b) Compressive (fc)

Plan

A (area of 'section)

"I

Li pi

777Π7777Τ777777777777Γ/

(a) Tensile (f+) FIG. 1

1

(c) Shear (fs or q) (L = Arm of shear couple)

2 S T R E N G T H O F M A T E R I A L S

Thus, Stress = Load per unit area (usully lbf/in2)

or / = FjA for simple cases as shown above.

Strain

This is a measure of the deformation produced in the component

b y the load and is denoted b y e. F o r tensile and compressive loads,

the strain is defined as the change in length per unit length, i.e.,

χ

Note tha t this is a rat io and therefore has no units. F o r shear loads

e = x\L = φ radians, since χ is very small.

The effects of shear will be discussed la ter in greater detail .

Elasticity

A material is elastic as long as the strain disappears with the

removal of the load. A limiting value of load exists beyond which

this does not happen and the corresponding stress is called the

Elastic limit

FIG. 2 FIG. 3

Ehstic Limit. After passing this the residual strain is referred to as a Permanent Set.

I n the case of ferrous materials (e.g. steel and iron), the deforma-tion χ is, up to the elastic limit, proportional to the load (Fig. 2) and this is known as Hooke's Law. However, for most non-ferrous

S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 3

metals (e.g. aluminium) the l imit of proportionality occurs before

the elastic l imit.

Now, Stress / = F\A

. ι e

= a constant χ — . χ

B u t F fx is the slope of the graph in Fig . 2 and is constant

within the elastic l imit so t ha t (below the elastic l imit) the ratio

fje is also constant and is clearly the slope of the graph obtained

by plotting stress against strain (Fig. 3 ) . This constant is called the

Modulus of Elasticity (Young's Modulus) and is denoted b y E. I t s

value depends on the material .

Thus , Ε = — e

or Stress = Ε χ Strain.

and Strain e = x/L,

-LI—

L F

A χ

— = / χ — i t follows tha t , if i t were possible

(which, within the elastic limit, i t is not ) to continue loading until

χ = L, then Ε = f. Thus Ε would be the stress required to produce

an elastic extension equal to the original length. F o r steel this

stress would be of the order of 30 χ 1 06 lbf/in

2 and this is the value

of Ε for this material .

/ fL Note: Since change in length χ = eL and e = — , .'. χ = .

Rigidity

When equal and opposite forces are applied as in F ig . 1 (c) so

as to induce a s tate of shear (i.e. forces are parallel and not co-

axial) the resulting couple produces a change in shape in the

material. This is not the case with materials under simple tension

or compression, where the shape remains the same notwith-


standing the deformation due to the load. The resistance offered

by the material to such change in shape is dependent on what is

called the rigidity of the material .

Clearly the material may fail (as shown in F ig . 4 ) a t any one of

an infinite number of sections such as X X of area abed (= A).

As with simple tension and compression, the ratio stress/strain

F

FIG. 4

is constant provided the elastic l imit is not exceeded, and this

constant is called the Modulus of Rigidity and denoted by G. I t s

value is a measure of the abil i ty of a material to retain i ts shape

under load and, as will be shown later, is dependent on the value

of E. Thus, Shear stress = Modulus of rigidity χ Shear strain,

F or fs = Gcp where fs = —,

A

and φ = angle of shear.

Average values (in lbf/ in2) of Ε and G for some common materials

are given in the following table :

Material Ε G

Plain carbon steel (0-1-0-5% C) 29-7 χ 106

11-6 χ 106

Aluminium and its base alloys 10-3 χ 106

3-7 χ 106

Hard drawn copper wire 18-0 χ 106

6-5 χ 106

Cold drawn brass 14-3 χ 106

5-2 χ 106

Cast iron 16-5 χ 106

β·4 χ 10e

Rubber 750 approx. 240 approx.


The Tensile Test

As already stated, extension is proportional to load up to the

limit of proportionality, which, in the case of ferrous metals ,

coincides for pract ical purposes with the elastic limit. When the

extension in a steel tes t piece (see relevant Bri t i sh Standard) is

measured under load and the loading is continued until fracture

occurs, the complete F-x graph has the form shown in F ig . 5

Beyond the elastic l imit ( E . L . , F ig . 5) the rate of extension

increases until a point called the Yield Point ( Y . P. , F ig . 5) is

reached a t which occurs a sudden extension with no increase in load. Fur ther loading results in an extension a t a greatly increased rate until a point of maximum load (Ì, F ig . 5) is reached. Around this point the material begins to extend locally without further load increase, and a " n e c k " is formed a t the midpoint of which fracture finally occurs.

Since / = FjA and e = x/L, the f - e graph is exac t ly similar to the F-x graph up to the point where the neck begins to form, i.e. up to the point beyond which the Sect ion A is no longer constant but diminishing. Thus, although the actual load falls after the point M, due to the reduction in A, the net effect is a continued stress increase as shown in F ig . 6. I f A is assumed constant , i.e. the nominal stress is plotted, the dot ted position of F ig . 6 is obtained.

\ Elastic zone

FIG. 5

ι \ V Uniform Local strain strain

FIG. 6


Proof Stress

Since most metals exhibit no yield point, i.e. no sudden extension

a t any load, some criterion other than yield stress must be used in

the design of non-ferrous components. The quant i ty used is defined

as the stress which will induce in the tes t piece an arbitrary pre-

determined permanent strain, (set) usually 0-001, i.e. the permanent

increase in length after subjection to such stress is OOOIL where L

is the tes t or " g a u g e " length. (See relevant Br i t i sh Standard for

tes t specimens).

This stress is then know as the 0-1 per cent Proof stress and the

corresponding load as the 0-1 per cent Proof load.

FIG. 7

The proof load may be obtained by drawing on the F-x graph (Fig. 7) a straight line parallel to the straight part of the graph from a point χ = 0 0 0 1 L from the origin, on the #-axis, and projecting horizontally (as shown) from its intersection Ρ with the graph.

The 0-1 per cent Proof stress is then the load obtained as above divided b y the test-piece section. I t is usually impossible to fix the position of the limit of proportionality (L.O.P) with any accuracy and the determination of proof stress eliminates the need to t ry .

Alternatively, the proof stress may be obtained directly from the / — e graph by drawing the parallel line from a strain of


•* 6 i n - — * • 12 in — 6 in •

1-25 in dia. lO in dia. 1*5 in dia

Solution

F o r the L H end

F o r the centre portion

F o r the R H end

To ta l reduction

FIG. 8

A1 = 0-785 χ 1-252 = 1-227 in

2.

11

/a

1-227 8-97 tonf/in

2.

0-785 χ 1-02 = 0-785 in

2.

11

0-785 = 14-0 tonf/in

2.

0-785 χ 1-52 = 1-769 in

2.

11

1-769 6-22 tonf/in

2.

X — X^ -\- #2 4~ i

Ε Ε Ε '

1 = — [(8-97 χ 6) + (14 χ 12)

Ε

+ (6-22 χ β ) ] ,

(53-8 + 168 + 37 -3 ) ,

= 0-0207 in.

12500

259

12500

0-1 per cent, i.e. of 0-001. The 0-2 per cent and 0-5 per cent Proof

stresses, also in common use, may be found in the same way.

Note: As a rule, the stress in a component carrying a non-fluctuating tensile load should not exceed two thirds of the specified proof stress.

E X A M P L E . Es t ima te the to ta l reduction in length of the strut

shown when carrying an axial compressive load of 11 tonf. Assume

Ε = 12,500 tonf/in2.


E X A M P L E . A steel tube is secured in a socket by a number of

radial steel pins 0 - 2 5 in. dia. as shown. I f the axial load is such as

to induce a tensile stress of 2 0 tonf/in2 in the tube, find the number

of pins required to limit the shear stress in them to 8 tonf/in2.

FIG. 9

Solution

Tube section = ^ ( 1 - 8 82 - 1 - 7 9

2) = 0 - 2 5 9 in

2.

Load on tube ftAt = 2 0 χ 0 - 2 5 9 - 5 - 1 8 tonf.

Pin section = x 0 - 2 52 = 0 - 0 9 8 in

2.

4

.'. Permissible shear load per pin = fpAp = 8 χ 0 - 0 9 8 = 0 - 7 8 tonf.

.'. Required No. of pins = -77^3-, 0 * 7 8

= 6 - 6 5 , say, 7 .

Note: Such joints usually fail due to excessive bearing stress, i.e. the holes elongate.

E X A M P L E . TWO vert ical steel wires each 0 - 0 5 in. dia. and 4 8 in.

long support a light rod over a span of 1 2 in. A concentrated load

of 6 0 lbf is then applied vertically downwards a t a point 4 in. from

one of the wires.

Calculate: 1 . the load on each wire,

2 . the stress in each wire,

3. the extension in each wire,

4 . the angular displacement of the rod from the

horizontal, assuming i t does not bend.

S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y

Solution

Moments about L H end give

F2 χ 12 = 60 χ 4 .

.·. F2 = 201b f ,

i.e. F1=40lbi.

| * - I 2 i n - * |

111(11 III 11(11111 III III .

0-05 in_ dia.

Fx Stress in L H wire f, = —-, 1 1

A 40

0-785 χ 0 0 52

= 20 ,350 lbf/in2.

Since J1. = - ^ - , Λ / 2 = 10,175 lbf / in

2.

, 4 in

Extens ion in L H wire χ Ε '

20,350 χ 48

30 χ 1 06 3

: 0 - 0 3 2 6 i n .

4 8 in

6 0 l b f

FIG. 10

The extension in the R H wire will be half this amount,

i.e. x2 = 0-0163 in ,

.*. Difference in level of ends == x1 — x2,

= 0 0 1 6 3 in .

Angle required r

_ 0 0 1 6 3

~~ —

Ϊ 2 '

= 0-00136 radian,

180 = 0-00136 χ

π

χ 6 0 , = 4-67 min .

0 - 0 1 6 3 in

12 in

F I G . 1 1

9


Example. Three struts 2 in. dia. and 24 in. long are arranged in

line and equidistant and support a rigid load of 55 tonf. I f Ε for

the material is 12,000 tonf/in2, determine the proportion of the

load carried by each when the centre strut sinks 0-01 in.

Load level

2 4 in 23-99 in

Β A

FIG.12

1

Solution

Area of strut section = 0-785 χ 22,

= 3-14 in2.

Load = Stress χ Section,

= Ε χ Strain χ Sect ion.

.·. Load on A = 12,000 ( ^ 9 ) 3 1 4

>

= 1570 a;.

χ + 0-01 Load on Β and C = 2 [ 12,000

\ 24 = 3140 χ + 31-4 .

x + O-OI

3-14 ,

/ . 55 = 1570α: + 3140a; + 31 -4 .

4170a; = 23 -58 .

23-58

4710 '

= 0-005 in .

.·. Load on A = 1570 χ 0 -005,

- 7-85 t o n f .

S I M P L E S T R E S S , S T R A I N A N D S T R A I N E N E R G Y 1 1

Solution

Steel section As = 0 - 7 8 5 χ 0 - 1 0 52,

= 0 - 0 0 8 6 6 i n2.

Bronze section Ab = 6 ( 0 - 7 8 5 χ 0 - 0 7 82) ,

= 0 - 0 4 5 1 i n2.

τ j -ο ë ç 5 5 - 7 - 8 5 4 7 - 1 5 00 _ , , .*. Load on Â and C = = — - — = 2 3 - 5 7 tont each.

Δ Δ

E X A M P L E . A steel wire 0 - 1 0 5 in. dia. is covered by s ix bronze

wires each 0 - 0 9 8 in. dia. I f the working stress in the bronze is

4 tonf/in2, calculate :

(a) the strength of the combination,

(b) the equivalent tensile modulus.

Es = 2 8 χ 1 06 lbf/ in

2 and Eh = 1 2 χ 1 0

6 lbf/in

2. (L .U.)

Common tensile strain = Λ) /s ç r

ºÃ = -ÅÃ and / b

= 4 tonf/in2.

.*. Stress in steel 9 - 3 3 tonf/ in2.

FIG. 1 3

12 S T R E N G T H O F M A T E R T A L S

Common strain =

Since stress

Tota l load

Tota l section

or Evq,

4 χ 2240

12 χ 1 06 '

0-000747.

Ε χ Strain,

Equivalent Ε χ Common Strain

ZW ι χ — ΣΑ

585 1

(0-00866 + 0-0451)

585

0-0538 χ 0-000747 '

14-58 χ ΙΟ6 lbf/in

2.

0 -000747 '

Strain Energy

This is the work done in the elastic straining of a material , whether under tension, torsion, bending or a combination of one or more of these. Such work (mechanical energy) is stored in the strained material and may be released either under controlled conditions (as in an unwinding clock spring) or suddenly, in which case the result is a vibration. The amount of such strain energy (sometimes referred to as Resilience) is a measure of the abil i ty of a material to resist shock without permanent deformation.

There are three ways of applying a load.

Permissible load = /S^4S + fbAh,

= (9-33 χ 0-00866) + (4 χ 0 -0451) ,

= 0-0808 + 0-1804,

= 0-261 tonf ,

= 585 lbf .


I f / and Ε are in lbf/in2 and V is in i n

3 then U is in in lbf. The

maximum possible value of U is tha t obtaining a t the elastic limit,

i.e. if / E L = elastic l imit stress, then

/ 2

tfmax = -wk- Pe r

unit volume.

2. Suddenly

If , from being held jus t in contac t with the collar, the load is suddenly released, collar and load travel beyond the equilibrium position by an amount equal to the s tat ic deflection x, i.e. the to ta l instantaneous extension is 2x. The maximum instantaneous strain (and hence instantaneous stress) is thus twice the strain corresponding to a gradual loading.

To produce a s tat ic extension of 2x clearly would require a gradually applied load of twice as much, i.e. of 2 I F in which case the graph would be as in Fig . 16, the mean load being W.

1. Gradually

The load starts from zero and increases uniformly up to i ts

final value, e.g. liquid being poured into a tank.

Suppose a load W is held jus t in contac t with the collar Fig . 14

and released gradually. The rod will extend in proportion (like a

spring) and the graph of load against extension will be a straight

line through the origin as in F ig . 15. The final value χ is called the

Static Deflection and when this has been at tained, the load is in the

Equilibrium Position. Then Tensile strain energy U = Average force χ Extens ion

W = —— χ ( = shaded area under

graph), fA fL

= ——- χ —— where 2 Ε

f = max. stress induced,

f2

f2

.*. U = —— χ Volume.


Section A

W

Γ I

-J- ΖΖ1^-Γ, Equili I I posit

-Collar

librium ion

FIG. 1 4

FIG. 1 5 F I G . 1 6

Ι

F I G . 1 7


3. With Shock

I f the load W is permitted to fall freely through a height h

before striking the collar, then the energy absorbed will be tha t due

to a sudden load application plus the

kinet ic energy gained during the fall.

Thus U = Wx + K . E . (where

K . E . = P . E . = Wh)

= Wx + Wh.

This is clearly the to ta l loss in poten-

t ia l energy of the load W. Now, maxi-

mum force in rod = fA where / = max .

stress, induced ;

Mean force in rod

Hence,

Equat ing,

U = fA

JA. 2

χ χ.

ÎA.1L 2 Ε

'///////////////.

t L . Τ -

α; = Wx + Wh and χ

W-^- + Wh,

Fia. 18

IL Ε

i .e.

Then U = shaded area of F ig . 16,

= W χ 2x,

= 2Wx,

= 4 (Area under first graph).

Since the full value of the load is in fact suddenly applied, the

actual graph is the horizontal straight line of F ig . 17 the area under

which is clearly 2Wx. Thus b y applying a load suddenly instead

of gradually, the instantaneous deflection, strain and stress are

doubled and the instantaneous strain energy is multiplied b y four.


Transposing gives - ^ r /2 τ τ " / — Wh = 0 ,

which is a quadratic equation enabling the maximum stress / to be found.

The behaviour of the rod in the foregoing cases of gradual, sudden

and shock application of the load is comparable with tha t of the

spring shown in Fig . 19.

FIG. 1 9

(a) Spring unloaded.

(b) Load released gradually from no-load position X X . The load

remains stat ionary in the stat ical ly deflected (equilibrium)

position.

(c) Load released suddenly from no-load position X X . The load

makes oscillations of which the amplitude (i.e. maximum

displacement on either side of the equilibrium position) is

equal to the s tat ic deflection x.

(d) Load released suddenly from a point above the no-load posi-

t ion X X . Since the spring is initially compressed, the load

already possesses kinet ic energy a t the ins tant of passing

position X X so tha t the behaviour of the spring is similar to

tha t of the rod subjected to shock loading. The amplitude of

the resulting oscillation is thus increased by some amount y

(Fig. 19(d)) .

Note: For a full treatment of vibrations see Mechanics of Machines by the same author.

No load position X-L-

Equilibrium position —


E X A M P L E . A ship of 8 0 0 ton is checked by two 4 in. dia. steel

hawsers when moving a t 0 - 5 f t /s . Assume tha t the kinet ic energy

of the vessel is all converted into strain energy and est imate the

length of hawser required to l imit the tensile stress in the steel to

6 tonf/in2. Take Ε = 1 3 , 4 0 0 tonf/in

2.

Solution

1 W K . E . = — — v 2,

2 g

8 0 0 χ 0 - 5 *

2 χ 3 2 - 2

3 - 1 f t / tonf

/2 Strain energy = - — per unit volume,

IE

χ F (where V = hawser volume in i n3) ,

2 χ 1 3 , 4 0 0

0 - 0 0 1 3 4 V inch tonf.

3 1 χ 1 2 Λ Volume of hawser required =

0 - 0 0 1 3 4 '

2 7 , 7 0 0 in3.

2 7 , 7 0 0

L

= ( 0 - 7 8 5 χ 42) L where

L = length in inches,

2 7 , 7 0 0

2 ( 0 , 7 8 5 χ 42)

1 1 0 5 in,

9 2 ft.

E X A M P L E . A steel rod 1 2 in. long 1 - 2 5 in. dia. has 7 in. of i ts length machined down to 1 in. dia. I f an axial blow induces a stress of 9 - 6 tonf/in

2 in the th icker part, calculate :

(a) the stress induced in the thinner part, (b) the force in the rod, (c) the strain energy of each part in inch tonf,

(d) the to ta l strain energy in inch lbf.

Ε = 1 3 , 5 0 0 tonf/in2.


7 in

in dia.-

1-25 in ο ι α -

Ι 5 in

Blow Solution

Force = Stress χ Sect ion,

. ' . ί = / Λ = / Λ

/« = / i-

- 9-6

l2

1-252

f, = 9-6 -A,

l-O2 '

= 15 tonf / in2,

.·. F χ 15 χ 0-785 χ l2,

= 11-87 t o n f .

'////////////

FIG. 20

Strain energy of th icker part

Q-62

(0-785 χ l - 2 5 2) 5 ,

A . x F

2 £ 1

2 χ 13,500

= 0-0285 inch/ tonf .

Strain energy of thinner part U2 = x F 2 ,

1 52

(0-785 χ l - 02) 7 ,

i n ^ e

1-5 in

ill 4 in

6 in

FIG. 21

2 χ 13,500

= 0-0458 inch tonf .

To ta l strain energy

U=U1 + U%

= (0-0285 + 0-0458) 2 2 4 0 ,

= 166-5 inch lbf .

E X A M P L E . TWO round bars of

the same mater ial have the dimen-

sions shown. I f Β receives an axia l

Ψ blow inducing a stress of 5 tonf/in2,

find the max . stress induced b y a

similar blow on A.


Ε

ü a = A ( 4 x i } + A ( 6 x i , 5 )

(where f1 = stress in reduced part) ,

2ft ±5

Ε +

li

- 1 ( 2 / ? + 4-5/1) and / 2 χ 1-5 = U χ 1-0

(since force is the same, i.e. / 2 = - j ^ - ,

= - 1 ( 2 / ? + 4 · 5 /!

i i Γ 1 ' " 1 - 5 «

= 4 ( 2 / ! + 2 / ? ) '

- A / 2

B u t J 7 A = C 7 n . Since the blow is the same.

. 4

/ l2 1 25

· · ,2 1 25

OT o r

.'. / x = 5-58 tonf/ in2

E X A M P L E . A 1 in. dia. bolt has an effective length of 1 0 in. I f the nut is t ightened until the strain energy is 5 in lbf, calculate :

(a) the stress induced, (b) the extension produced, (c) the force in the bolt . Take Ε = 3 0 χ 1 0 6 lbf/in2.

Solution

Effective volume = 0 -785 χ l 2 χ 1 0 ,

- 7-85 i n 3 .

Solution

U b = 2Ë

X V o l u m e'

52

= - ^ ( i o x i ) ,

* ^ - i n c h tonf .


Strain energy U 2E

χ Volume.

5 = /2

2 χ 30 χ 1 06

5 χ 2 χ 30 χ 1 06

7-85,

Extens ion

7-85

. / - 6-2 χ 1 03 = 6200 lbf/in

2.

jL 6200 χ 10

38-2 χ 106,

Ε 30 χ 1 06

Force in bolt = Stress χ Sec t ion ,

= 6200 χ 0-785 χ l2,

= 48601bf .

0-002 in .

E X A M P L E . I n the arrangement shown, the upper support may

be taken as rigid. Ε = 13,200 tonf/in2.

Es t ima te :

(a) the extension required to

induce a tensile stress of

5 tonf/in2 in the rod,

(b) the height h from which the

annular weight must be re-

leased to induce this stress

on impact .

Assume no energy to be lost in

deforming the stop.

Solution

Required extension χ = ,

5(120)

~ 13,200 '

= 0-0454 in .

///////////////////////

FIG. 22


Loss in Poten t ia l energy = Average force χ Extens ion .

- Μ -

0 · 2 5 ( Α χ 0 - 0 4 5 4 )

h + 0 - 0 4 5 4

— χ 0 - 0 4 5 4

5 0 0 4 5 4 Λ^ Ο Κ

F χ

Ί > 2 5 - Χ 0 , 7 85 Χ 1 ,1

= 0 - 3 5 6 ,

.·. h = 0 - 3 5 6 - 0 - 0 4 5 ,

= 0 - 3 1 in (approx.) .

E X A M P L E . A certain big-end bolt is 1 - 2 5 in. dia. from head to

nut and when the nut is t ightened, the strain energy is 3 in. lbf.

Es t imate the increase in strain energy brought about b y reducing

the dia. to 0 - 1 in. over three quarters of the length, assuming the

tension in the bolt to remain the same.

Solution

i ( 0 - 7 8 5 x l - 2 5z) = F -

r 1-25 in

i

Ι-ΦΟΊΗ

M Υ ι Ο l +

I

FIG. 2 3

Force = Stress χ Sec t ion .

.·. F = ( 0 - 7 8 5 χ 1 - 2 52) / = ( 0 - 7 8 5 χ 1 - 0

2) f±

(where f± = stress in reduced part) ,

• · h h Q2 ,

= 1 - 5 6 / .

Strain energy = — ~ χ Volume .

2 SM


1. For original state

3

Ε

2E (0-785 χ 1 - 2 5

2) L ,

3 x 2

0-785

4-9

1-252 '

2. For modified state

JL 2E

U =IL (0-785 χ 1·252) 0 · 25£ + (0-785 x Ι Ο

2) 0-75 Ζ,,

Ε χ 0 1 5 3 +

2E

2E

(0-785 χ 0-75 L),

= (0-153 + 0-715) , Ε

= 4-9 χ 0-868,

= 4-25 inch lbf .

Increase = 4-25 — 3-0,

= 1-25 inch lbf .

ν//////////////,

E X A M P L E . A load of 0-5 ton is to fall

0-25 in. on to a collar a t the lower end of

a vert ical t ie 72 in. long. Es t imate the

diameter d required to l imit the strain in

the t ie to 0-0005 given tha t Ε = 12,500

tonf /in2.

72 in

Solution

Strain e = = 0 0 0 0 5 ,

72

0 0 0 0 5 χ 7 2 ,

0-036 in .

Stress / = Ε χ St ra in ,

12,500 χ 2240 χ 0-0005,

14,000 lbf/in2.

0-5 ton

I : L 0-25in

FIG. 2 4

or


Strain energy = K . E . of mass + work done during sudden applica-tion.

,\ΙΑ·χ = (0-5 χ 2240) 0-25 + (0-5 χ 2240)a; , where χ = 0-036, Δ

14,000.4 0-036 - 280 + 4 0 - 3 ,

= 320 -3 .

_ 320-3 χ 2

~ 14,000 χ 0 0 3 6 '

= 1-27 i n2.

.·. 0-785 d2 = 1-27,

d2 = 1-62,

.·. d = 1-27 in .

E X A M P L E . A load of 2000 lb is being lowered a t a steady rate of

2 f t / s a t the end of a steel cable of 1-0 i n2 section. I f the pulley

j ams when the load is 30 ft below i t , es t imate the additional stress

induced in the cable by the stoppage taking Ε = 30 x 1 06 lbf/in

2.

Neglect the weight of the cable and assume the pulley axis not to

be deflected.

Solution

Stress before jamming occurs

Load

Section

2000

1-0 ' 3 0 f t

= 2000 lbf/in2.

On jamming

Strain energy of cable = K . E . of load -1- work done during sudden x -loading.

• JL ·· 2E

Cable volume

•v2 + Wx,

-1-0 in2 section

2 0 0 0 Ibf

v = 2 f t / s

2*

FIG. 2 5


where χ

^ χ 30 χ 12 χ 1-0 =

/ 30 χ 12 \

\ 3 0 x 1 06/ ' '

0-000012 / (where / - max. stress

induced).

Ί 9ΑΠΠ Ί

12

1 2000 •

¥X

" 3 2 ¥X 2

"

and /

* * 2(30 χ 106)

+ 2000 (0-000012 / )

(Note tha t the kinet ic energy must be in inch lbf)

i.e. 6 /2 x 1 0 '

6 = 1490 + 0 - 0 2 4 / ,

or /2 = 4 0 0 0 / + (248 χ 10

6)

/ 1 06\

(multiplying b y — I .

.·. /2 - 4 0 0 0 / - (248 χ 1 0

6) = 0 ,

4000 ± V[(16 x 106) + (4 χ 248 χ 10

6) ]

2

= 2000 + 15,880, taking the positive root,

= 17,880 lbf/in2.

B u t stress prior to jamming = 2000 lbf/in2.

.·. Increase due to jamming = 15,880 lbf/in2.

9

E X A M P L E . Two wire ropes each — in. dia. and weighing 16

0-85 lb/ft support a hoist weighing 0-5 ton, the ropes being 80 ft

long when the hoist is a t ground level. Es t ima te the number of

packing cases each weighing 50 lb which m a y safely be onloaded

assuming a permissible stress of 15,000 lbf/in2 and allowing for the

fact t ha t each case may be dropped from a height of 1 ft on to the

platform. T a k e Ε = 30 x 1 0 e lbf/ in 2 and neglect the instantaneous

extension due to impact .

Solution Γ / 9 \

2

R o p e section = 2 0-785 I—J

- 0-496 in2.

S I M P L E S T R E S S . S T R A I N A N D S T R A I N E N E R G Y 25

Weight of ropes = 2(0-85 χ 8 0 ) ,

= 68 χ 2 ,

= 136 lbf .

To ta l dead load

= 136 + (0-5 x 2 2 4 0 ) ,

= 1256 lbf .

Stress due to this

1256 8 0 ft

0-496 = 2525 lbf/in

2.

Energy absorbed in fall =

Wh = —— χ Volume, where lté

• 1-0 f t

/ = max . stress induced due to fall.

Wh χ 2E

FIG. 2 6

/ 2 = Vol

where Volume = 0-496(80 χ 12)

= 477 in3,

(50 χ 12) (2 χ 30 χ 106)

477

= 0-755 χ 108,

/ = 0-867 χ 104,

= 8670 lbf/in2.

To ta l stress due to initial dead load plus impact of one case

= 2525 + 8 6 7 0 ,

- 11,195 lbf/in2.

Stress available for remaining packing cases = 15,000 — 11 ,195 ,

= 3805 lbf/in2.

Load required to induce this = 3805 χ 0-496,

- 1885 lb f .


.*. Number of 50 lb cases = 1885

50 '

= 37-7 say 3 7 .

Hence, to ta l number of cases = 37 + 1,

= 3 8 .

E X A M P L E . The cylinder shown which

weighs 6500 lbf is released from a height

h above the spring, the stiffnes of which

is 3000 lbf/in. I f the maximum stress

induced in the rod is 9000 lbf/in2, find :

(a) the extension in the rod (x),

(b) the reduction in spring length (y),

(c) the value of h.

Assume no losses and tha t the rod

anchorage is immovable.

Solution

Extens ion in rod iL " Ε

9

9000 χ 100

30 χ ΙΟ6 '

= 0-03 in .

Strain energy of rod Ur JL 2E

100 in.

x Volume

9 0 0 02

- 2 in dia.

1— η —

1

• ι

FIG. 27

2(30 χ 106)

418 inch lb f .

92 100

I f W = gradually applied load to produce χ and y,

W

then, = ~2~ x>

W i.e. 418 = — x 0 0 3 ,

W

2

418 0 0 3

27,900 lbf .


Reduct ion in spring length y = Stiffness

27 ,900

3000 '

= 9-3 in .

Strain energy of spring Us

W

27 ,900 χ 9-3

= 129,500 inch lbf .

To ta l work done = Î 7 R + Z7 S,

.·. 6500(A + y + χ) = 418 + 129 ,500 ,

· · · ' ·« · » ^ · .·. A = 19-9 - 9 -33 ,

= 10-57 i n .

Clearly, in comparison with Us the strain energy of the rod is negligible.

Simple Load Shared by Two Materials—Compound Column

Consider the short strut composed of a steel rod and copper sleeve and let the dot ted line (Fig. 28) represent the loaded posi-t ion.

total load

nnnnnin //////////fin/m in π ninr/m//////// Steel . section As

v Copper section Ac

FIG. 2 8

W

2 8 S T R E N G T H O F M A T E R I A L S

I f Es and Ec are the values of Young 's Modulus for s teel and

copper, t hen : χ f f

Common strain = —— = - ~ == . L Es Ec Ε

Stress in copper / c = χ / s , where / s = Stress in steel.

To ta l load W = Ws + Wc,

Ε = fsAs + / s X y X i t .

Es and Et: are known while As and Ac can be calculated from given

dimensions. Hence, if W is known, / s can be found from the above

equation. / L

Then, reduction in length χ = .

Since, Load = Ε χ Strain χ Section,

Load on steel Ws = Es χ — χ As, Ε

and Load on copper Wc = W — Ws.

E X A M P L E . A short steel tube 4 in. inside dia. and 0 - 5 in. th ick is

loosely surrounded by a brass tube of the same length and thick-

ness. Together they carry an axial thrust of 0 - 5 tonf. I f the com-

mon length is 3 * 5 in., es t imate :

(a) The rat io of the compressive stresses in each,

(b) the stress in each,

(c) the load on each, (d) the common reduction in length.

Es = 3 0 x 1 0E, Eb = 1 1 - 8 x 1 0

E lbf/in

2.

Solution

Steel section As = 0 - 7 8 5 ( 5 2 - 4 2 ) = 7 - 0 7 in 2.

Brass section Ah = 0 - 7 8 5 ( 6 2 - 5 2 ) = 8 - 6 4 in 2. x f* = / b

L Es Eh

fh Eh 1 1 - 8

Common strain

= 0 - 3 9 3 .


FIG. 2 9

i.e.

, \ Load on steel

Load on brass

1120

10-47 '

107 lbf/in2.

107 χ 0 -393 ,

42 lbf/in2.

Ws = fsAs = 107 χ 7-07 - 755 lbf.

Wh = fbAh = 42 χ 8-64 = 365 lbf.

Common reduction in length χ = -ψς— , E«

107 χ 3-5

30 χ ΙΟ6 0-0000125 in.

2a SM

Tota l load W = Ws + WH,

Λ 1120 =fsAs +fhAh,

= fsAs + 0-393 fsAh,

= / s (7-07 + 3 - 4 ) ,

29


Tempvraturo Stresses

A change in temperature produces a change in dimension, which,

if prevented sets up a stress. F o r a given material , linear expansion

is proportional to rise in temperature and to length.

Hence, if oc = change in length per degree per unit length,

change in length χ = ocTL,

where Τ temperature change, L = original length,

oc = expansion coefficient.

F o r a rise in temperature, χ is an increase, and if this is prevented

then this is equivalent to compressing a bar of length (L -|- x) back

to a length L, χ

i.e. Strain = — , JU

Stress / = Ε χ s t rain,

χ i.e. / = Ε χ — where χ = aTL.

Ε

E X A M P L E . A steel pipe is 6 0 ft long a t 2 0 ° C . E s t i m a t e :

(a) the elongation due to a rise in temperature to 9 0 ° C , (b) the stress induced by the prevention of this elongation.

T a k e oc = 0 - 0 0 0 0 1 2 / ° C . Ε = 3 0 χ 1 06 lbf/in

2.

Solution

Temperature rise

Elongation

Τ = 9 0 - 2 0 ,

= 7 0 ° C .

ocTL,

0 - 0 0 0 0 1 2 χ 7 0 χ 6 0 ,

0 - 0 5 0 4 f t ,

0 - 6 2 4 in.

Stress / = Ε χ S t ra in ,

oc TL Ε χ L '

- Ε x ocT,

= 3 0 χ 1 0E χ 0 0 0 0 0 1 2 χ 7 0 ,

= 2 5 , 0 0 0 lbf/in2. (approx.)


E X A M P L E . A copper cylinder 6 in. outside dia. has walls 1 in.

th ick. E n d plates are stayed by a 1 in dia. bar of steel which

passes through the cylinder and is jus t t ight a t 12°F . Es t ima te the

stress in cylinder and s tay when steam a t atmospheric pressure is

admitted. Assume the end plates do not distort.

occ = 0 -00001/°F, E, - 15-7 χ 1 06 lbf/in

2,

« s = O0000062 / °F , Es - 30 χ 1 06 lbf/in

2.

Solution

Steel section = 0-785 i n2

Copper section = 0-785 ( 62 - 4

2) ,

= 15-70 in2.

- - * | a cT L I - -

ν///////////////////λ {

1 1

1 J

1 J

__J ^ Common change

I in length

I 12 °F

212 °F

F i g . 3 0

Compression in copper = occTL — χ,

.'. Strain in copper

Extens ion in steel

.*. Strain in steel

- (0-00001 χ 200 L) - a- =

= 0-002 - 4 " · Jj = χ - otsTL,

= χ - (0-0000062 χ 200 L)

= χ - 0 0 0 1 2 4 L,

--=4- - 0 0 0 1 2 4 .

0-002 L - x.


Now, Load = Ε χ Strain χ Sect ion and the loads in steel and copper are equal.

.·. - 0 -00124J 30 χ ΙΟ6 χ 0-785 = 15-7 χ ΙΟ

6

0-002 - j " ) 15-7,

± - 0-00124) - 1 5

·? X

f X 1 5

'7

L J 30 Χ 10° X 0-785

Χ (0-002 - 4" Ε

= 10-5 0-002 - 4-Ε

4- - 0-00124 - 0-021 - 10-5 4" L Ε

11-5 4 · = 0 0 2 2 2 . L

χ

Τ 0 0 0 1 9 3

Stress in steel = 30 χ ΙΟ6 (0-00193 - 0-00124) = 20 ,800 lbf/in

2.

Stress in copper = 15-7 χ 1 06 (0-002 - 0-00193) = 1000 lbf/in

2.

E X A M P L E . The body of a small condenser is made from 8 in.

outside dia. brass tube. The end plates are also brass and are held

in place by four long | in. dia. steel bolts . I f these are t ightened

equally a t 5 2 ° F so tha t each carries 1000 lbf, what will they carry

a t the operating temperature of 212°F? The tube is \ in. th ick .

ocs = 0 -0000062/°F , Es = 30 χ 1 06 lbf/in

2,

ah = 0 -0000095 / °F , Eh= 9 χ 1 06 lbf / in

2.

Solution 9π

Steel Section - 4(0-785 χ 0-3752) = — in

2.

64 15JT

Brass Section = 0-785(8 2 - 7 2) = - j - i n 2 .

Temperature rise = 212 - 52 = 160°F .

Strain in steel es = = -= oc/1 Ε Ε


y 4 - -g- in dia. steel

Common extension

- a hT L -

7 in 8 in I

1

. . . . . .

X j

FIG. 3 1

Hence, 9 χ 1 06

(0-0000095 χ 160) 15π

L

9 χ 15

- (0-0000062 χ 160

00152 - — 1j

L

9π

64 '

30 χ 9 / χ

= 30 χ 1 06

n i . r - 0-000992) ()4 \ /v /

whence, — - 0-00146. Ε

Strain in brass e b - "b T L

* - ocbT - * L L

The loads on steel and brass are equal and Load = Ε χ Strain χ Section.


Stress in steel /s = Εs χ Strain in s teel ,

= 30 χ 1 06

= 30 χ 1 06 [0-00146

= 14,100 lbf / in2.

(0-0000062 + 160) ] ,

Load /Bo l t -Stress χ Section

• 4 . 1 0 0 x | x |

1000 ,

+ 1000 1560 + 1000

= 2 5 6 0 l b f .

E X A M P L E . The two side members of a water cooler are of 7 χ 2 in.

box section aluminium 0-25 in. th ick and 2 4 in. long. Between them

are 256 vert ical copper tubes 0-25 in. outside dia. 0-125 in. bore,

of the same length. Assuming tha t

(a) assembly was carried out at 20°C,

(b) the headers are rigid,

(c) the tubes do not buckle,

(d) the side members remain cold,

calculate, for an operating temperature of 90°C,

(a) the stress in the tubes,

(b) the increase in height of the cooler.

F o r copper ac = 17-1 χ 10"e/ °C and Ec = 15 χ 1 0

6 lbf/in

2.

F o r aluminium Ea = 12 χ 1 06 lbf/in

2.

Solution

Aluminium section Α.. = 2 (3 χ 2) 2

x¥

: 2(6 - 3-75) ,

4-5 in2.

Copper section Ar = 256[0-785(0-25 + 0-125) (0-25 - 0-125) ,

256 χ 0-785 χ 0-375 χ 0 1 2 5 ,

9-43 in2.

Temperature rise Τ = 90 — 2 0 ,

= 70°C.


Compression in copper = occ TL — E.t

0 - 2 5 in -

0 · 2 5 ί η 0 1 2 5 in 1

Tubes-

Side member^

FIG. 3 2

occ TL

.*. Strain in copper = EiX oc'T

.*. Stress in copper / c = Ec χ Strain = Ec

and / { Α = / Λ ,

Ε.Λ

"•'-•it\

Extens ion in copper when free to expand = occTL.

Actual extension in copper = extension in aluminium =

35


i.e.

Hence,

and

f f A

< f 9

'4 3

2 1 /

.·. / c = 15 X 1 06

/ c

17-1

1 0e

15(1195 - 0 175 / c )

χ 70 -2 1 / c

12 x 1 06

15 + 0-175 / c = 1195, whence / c = 4950 lbf/in

2

/ a = 4950 χ 2-1,

= 10,400 lbf/in2.

fsL Extension in aluminium (i.e. in height of cooler) = ,

10400 χ 24

12 x 1 06, '

0-0208 in.

Examples

1. The force required to punch a hole 1-2 in. dia. through a steel plate 0-75 in. thick is found to be 51 tonf. Estimate the shear strength of the ma-terial (18 tonf/in

2).

2. The section of a steel strip is 3-0 χ 0-25 in. and that of a similar copper strip is 3-0 χ 0-125 in. If the two strips are superimposed and rivetted together, estimate the stress in each component when the combination resists a load of 6 tonf applied so that the two sections are in pure tension. Assume Es = 13,000 tonf/in

2 and Ec = 4200 tonf/in

2 (5-17 tonf and 0-83 tonf).

3. Compare the strain energy of a 12 in. length of rod 1-25 in. dia. with that obtainable by reducing 7 in. of the rod to 1-0 in. dia., assuming the maximum tensile stress to be the same in each case (1: 3-45).

4. Two alloy plates are rivetted together at 10°C using copper rivets. Neg-lect the compressive strain in the plates and estimate the stress increase in the rivets resulting from a rise in temperature to 21 °C. Assume Ec — 18 χ 10

6 lbf/in

2, <xc = 0-000005/°C, and a a = 0-000007/°C (1150 lbf/in

2).

5. As a result of a tensile test the following figures were obtained : Load (lbf χ 1000) 3 4 5 6 7 8 9 Extension (inch χ 10

4) 2-4 4-95 7-90 10-70 13-50 16-45 19-10

Load 10 11 12 13 14 15 16 17 18 19 Extension 21-95 25-00 28-00 31-00 34-28 37-10 40-60 44-90 54-00 87-00 Gauge length 2 in. Maximum load 21,000 lbf.


Original dia. 0-5615 in. Load to fracture 14,000 lbf. Fracture dia. 0-375 in. Length after fracture 2-41 in. Plot the graph and find :

(a) Ε (29-5 χ 106 lbf/in

2),

(b) 0-1 per cent proof stress (74,600 lbf/in2),

(c) Percentage elongation (20-5 per cent).

C H A P T E R I I

BEAMS I—BENDING MOMENT

Moment of a Force Producing Bending

The moment of a force about a point is the product of the force

and the perpendicular distance of the point from i ts line of action.

Such a moment, when applied to a component (shaft, beam strut)

in such a way as to result in bending, is referred to as a Bending

Moment and denoted by M. The units are usually lbf/in or tonf/in,

i.e. product of force and length of arm.

Convention: To the left of any section, an anti-clockwise moment is considered

An example of this is a beam built into a wall as shown in

Fig . 33 . This is called a Cantilever.

negative and vice versa.

Beam Rigidly Supported at One End,

With Concentrated Load at the Other

w L

χ X

!x

Ο χ

-moxT x =-WL

M

A

3 8

F i g . 3 3

B E A M S I — B E N D I N G M O M E N T 39

F o r any section X X ,

Bending moment Mx = Moment of W,

= -Wx.

Thus M is a function of .τ, i.e., i t is proportional to x, the distance

from the origin O. I t s graph is therefore the straight line OA.

At the support, where χ = L,

Μιη:ικ = -WL.

The bending moment causes tensile stress (and therefore an

increase in length) in the upper layers of the beam, and compressive

stress in the lower layers. At some intermediate point there will be

a plane of zero stress and i t will be shown later tha t the stress

(tensile or compressive) is proportional to the distance from this

" n e u t r a l " plane.

Cantilever With Several Concentrated Loads

The value of M a t any section will be the sum of the moments

of all forces to the left of the section. Thus, taking an anticlockwise

moment as negative to the left of a section (Fig. 3 4 ) :

I f A = -W^L, ~L2)

MB = -W^L, - £8) - WS(L2 - £,)

Mc = -W.L, - W2L2 - W3L3

and Mc is clearly the maximum value.

The graph of M is thus a series of straight lines as shown.

E X A M P L E . F ind the bending moment a t each of the points A,

Β and C in the system shown in Fig. 35 and draw the graph of M.

Solution

Μ\ = - 0 - 5 ( 1 0 - 5 ) ,

= 2-5 tonf f t .

MB = - 0 - 5 ( 1 0 - 2) - 1(5 - 2 ) ,

= - 4 - 3 ,

= - 7 - 0 tonf f t .

Mc - - ( 0 - 5 χ 10) - (1 χ 5) - (2 χ 2)

^ - 1 4 - 0 tonf f t .

S T R E N G T H O F M A T E R I A L S 40

i

FIG. 34

14 F r o . 35


Simple Span With Central Load

Clearly, load per support = W/2. The bending moment a t any

section is the ne t effect of all forces acting on t ha t par t of the beam

to the left of the section, clockwise direction being taken as

positive.

Thus for the L H half of the beam M is proportional to the

distance from the L H support and i ts graph is the straight line OP.

FIG. 3 6

At any section X X to the left of the load,

W L

Μ„ = —r-x where χ < — . x 2 2


A t the centre where L

w

w L

Μ = ΎΧΎ WL

and this is clearly the maximum value.

At any section Y Y to the right of the load :

- w r L]

Wy

2 Wy + — ,

WL Wy

2 2

W -y) where

L y > τ·

At the R H support when y = L, M — 0

Thus for the R H half of the beam M is proportional to the

distance from the R H support and i ts graph is the straight line

P Q .

Simply Supported Beam With Several Concentrated Loads

The reactions must first be found b y taking moments about point 0 (Fig. 3 7 ) :

Thus,

R2L = WXLX + W2L2 + WsLz

from which R« can be found.

Bt = (W, + W2+ W3) R2. Then

Then,

MA = R ^ .

Mc = R,L3 - WX(LZ - Lx) - W2(LS - Lz).

The graph of M against L will thus be a series of s traight lines as

shown.

B E A M S I — B E N D I N G M O M E N T

W, W,

43

L 3

M

FIG. 37

E X A M P L E . Calculate the reactions and the value of M a t points

A, B , C and D, Fig. 38 . Draw the graph of the lat ter .

Solution Moments about 0 give :

WR2 = (6 χ 2) + (8 X 4) + (10 χ 12 ) ,

= 12 -I- 32 + 120 .

164

16 = 10-25 tonf ,

R1 = (6 + 8 + 10) - 10-25,

- 24 - 10-25,

= 13-75 tonf .


6 8 10 tonf

(13-75) (10-25)

Ö j

FIG. 3 8

Beam With Loaded Ends Overhanging Supports

The diagram represents a wagon axle in which the springs rest

on the ends and carry W each.

I t follows tha t the wheels, which support the axle a t points

distant a from each end, also carry W each.

MA = (13-65 χ 2) = 27-50 tonf ft.

MB = (13-75 χ 4) - (6 χ 2 ) ,

= 55 - 12 ,

= 43 tonf f t .

M(] - (13-75 χ 12) - (6 χ 10) - (8 χ 8 ) ,

- 165 - 60 - 6 4 ,

= 41 tonf f t .

Hence,


FIG. 3 9

FIG. 4 0

W


At any section X X ,

Mx = — Wx, where χ < a.

Hence M is proportional to x, i.e. i t is zero at the ends and has a

value a t the wheels of — Wa.

At any section Y Y , My = -Wy + W(y - a),

= -Wy + Wy - Wa,

= - Wa.

Since a is a constant, the bending moment between the wheels is

also constant and the graph of M is as shown (Fig. 39) :

I f the loads are unequal and/or if the supports are not symmetric-

ally spaced, i.e. c > a then the graph of M is modified as shown,

and the value between the supports is constant only if Wxa = W2c.

The supports themselves may thus share the load unequally and

the load on each must first be found by taking moments about one

of them.

General Case

F o r the system of loads and supports given, (Fig. 41) Moments

about A give

Εφ -ι- Wxa - W2(d - a) -|- Ws{d - a + c)

w, w2 w3 J

•4 d

A c

Β

b * r*— 0 *

R, R 2

FIG. 41

from which R2 and hence Rx can be calculated.


Then,

MA = - W > .

MB = -Wxd + B1(d - a).

Mc = - Wx(a + b) + Rxb - W2(e - c)

( = — W3c al ternatively).

FIG. 4 2

Thus, as shown in F ig . 42 , the introduction of W2 has reduced

the negative bending moment between the supports. I f W2 is

made sufficiently large, the bending moment between the supports

may become positive as shown in F ig . 4 3 .

M

01

FIG. 4 3

I n general, when there is no change in load between two sections

of a beam, the bending moment changes uniformly.


E X A M P L E . Determine the bending moment a t points A , B , C

and D in the system shown and draw the bending moment graph

to scale.

2 tonf

-10 f t -

A

H O f t - - 1 0 f t -

D

-5fH

FIG. 44

Solution

Moments about A give

( i ? 2 x 20) + (2 χ 5) = (4 χ 5) + (6 χ 15) + (2 χ 2 5 ) .

20R2 + 10 = 2 0 + 90 + 5 0 .

20 + 90 + 50 - 10 Ra =

20 7-5 tonf .

.'. R1 = (2 + 4 + 6 - j - 2) — 7-5,

= 14 - 7-5,

- 6.5 tonf .


Hence, Mx = - ( 2 χ 5 ) ,

- - 10 tonf f t .

MB = - ( 2 χ 10) + (6-5 χ 5 ) ,

= - 2 0 + 32 -5 ,

= + 12-5 tonf f t .

Mc = - ( 2 χ 20) + (6-5 χ 15) - (4 χ 10 ) ,

= - 4 0 + 97-5 - 4 0 ,

= + 17-5 tonf f t .

ΜΏ = - ( 2 χ 25) + (6-5 χ 20)

- ( 4 χ 15) - (6 χ 5 ) ,

= - 5 0 + 130 - 60 - 3 0 , = - 1 0 tonf ft [ = al ternatively ( - ( 2 x 5 ) ] .

E X A M P L E . Determine the bending moment a t points A, B , C

and D and draw to scale the graph of bending moment.

4 2 5

8 f t - 9 f t

A Β C D

2 0 f t

R. R 2

M

0^

- 2 0

FIG. 45


Cantilever With Uniform Load

The only force acting to the left of the section X X (Fig. 46 ) is

the weight of the portion x, (i.e. wx) and this acts a t a distance

x\2.

Solution

Moments about A give

(R2 x 20) + (4 χ 5) = (2 χ 3) + (5 χ 12) + (3 χ 1 8 ) ,

20R2 + 2 0 - 6 + 60 + 5 4 ,

_ 6 + 60 -I- 54 - 20 / V

20 '

= 5 tonf.

. \ Ä 1 = ( 4 + 2 + 5 + 3 ) - 5 ,

= 9 t o n f .

Hence, MA = - ( 4 x 5 ) ,

- - 2 0 tonf f t .

Mn = - ( 4 χ 8) + (9 χ 3 ) ,

= - 3 2 + 2 7 ,

= - 5 tonf f t .

Mc - - ( 4 x 17) + (9 χ 12) - (2 χ 9 ) ,

= - 6 8 + 108 - 1 8 ,

= + 22 tonf f t .

MD = - ( 4 χ 23) + (9 χ 18) - (2 χ 15) - (5 χ 6 ) ,

= 92 + 162 - 30 - 3 0 ,

- + 1 0 tonf f t .


Hence, i f x = — wx taking an anticlockwise moment as negative.

W 2

The graph of M against χ is thus a parabola as shown. At the

support when χ = L,

w i i i n ; ix = — and wL = W (the to ta l load)

W L

: . J f . „ : , V = —

I

Για. 46

Simply Supported Beam With Uniform Load

Tota l load = wL,

.'. Load/Support =


At any (Fig. 4 7 ) ,

section X X

wL X X

WX X —-

wL

w/unit length

The graph of M is thus

a parabola. The value of

χ which makes M a max-

imum can be found b y JL differentiating the above

and equating to zero : FIG. 4 7

dx = — wx = 0 for a max. ,

i.e. wx =

χ =

wL

"2

wL

ΊΓ _L

Τ

Hence, Mn

wL L

wL2 wL

2

w (L\2

4

wL2

8

and wL = W (the to ta l load)

FIG. 4 8


w/unit length

Fia. 49

and, from Fig . 51 :

Mb= —w[a 2

b δ2

w ί 0 Λ δ b2\ w Ί w Ίη

= -T(a2 + 2aJ+Tj + Y a b + T b ' w „ w . w w , w ,„

= - ^ r «2 --z-ab --ζ-V + s-eb + — δ

2,

wa

FIG. 50

Thus the maximum value of M between the supports is less w

(numerically) than tha t a t a support b y an amount — b2 and the graph of M is as shown in F ig . 52 . 3 SM

Uniformly Loaded Beam With Simple Supports Not at the Ends

Suppose the beam to be supported a t points A and C (Fig. 49)

such tha t a > 0·5δ and let point Β denote the centre. Then, from

Fig . 5 0 :

M a = — wa ^- = Mc, Δι


U—α =0-5 tM

FIG. 5 3

Clearly, if a < 0-56, Mh becomes positive. B e s t use is made of

the beam when

i.e. when

or when

w w w

Yai + Tb = Ô α ' W 1,2 2

— b2 = wa

2,

8

or when

The graph of M is then as shown in F ig . 54 .

J FIG. 5 2

w w F o r Mh to be zero : 0 = - — a

2 + — b

2,

Ζ o

i.e. a2 = ——

4

or a = 0·5δ ^ =

The graph of M is then as shown in Fig . 53 .

B E A M S 1 — B E N D I N G M O M E N T 55

I f the supports Rx and R2 are not symmetrically spaced, i.e. if

c > a (Fig. 55) then Rx Φ R2 and the graph of M is as shown in

Fig. 56 .

M

- V — F

* I 0 Mo

a*0-35b 1* b J

FIG. 54

Ρ Q

Deflected shape of beam (exaggerated )

FIG. 57

3 *

FIG. 56

FIG. 55


Contraflexure

At points Ρ and Q (Figs. 55 and 57) the bending moment is zero

so tha t a t these points the beam is straight. Also, since M changes

sign a t these points, the curvature also changes sign, i.e. the profile

changes from convex to concave. F o r this reason points such as Ρ

and Q are called Points of Contraflexure.

E X A M P L E . A beam is simply supported and loaded as shown in

Fig . 58 .

F ind the value of W required to make Rx and R2 equal. Calculate,

for this value of W the bending moment a t points A and Β and a t

two points 1-825 ft and 7-50 ft respectively to the right of point A.

Sketch the graph of M using these values.

Solution

Moments about point A give :

(W X 4) + (R2 χ 12) = (6 χ 11) + (12 χ 1 ) - ^ ,

12R2 = 66 + 72 - 4 I F , fu f ρ

138 - m so tha t iio = — . 2 12 Moments about C give :

(W χ 16) + ( 6 x l ) | (12 χ l ) ^ - = R1 χ 1 2 ,

\mx = 6 + 72 + \m, 78 + 161Γ

K l ~ Î 2 * 78 + 16Tf 138 - 4W

Since RX=R2, — = — ,

20 W = 60, i.e. W = 3 tonf.

FIG. 58


Hence, ιΔ

M A — . (3 χ 4 ) = - 1 2 tonf ft.

11 MB = - ( 3 χ 15) + (10-5 χ 11) - (11 χ 1 ) — = + 1 0 tonf ft.

Δ At a point 1-825 ft to the right of point A,

M = - 3(4 + 1-825) + (10-5 χ 1-825) - (1-825 χ 1)

= - 1 7 - 4 7 5 + 19-2 - 1-68,

= 0 (nearly).

At a point 7-50 ft to the right of point A :

7-5 M = - 3 (4 + 7-5) + (10-5 χ 7-5) - (7-5 χ 1) — ,

Δ

= - 3 4 - 5 + 79 - 2 8 ,

= + 1 6 - 5 tonf ft,

1-825

FIG. 59

The graph of M is then as shown.

Examples I I

1. Find the maximum bending moment in a timber joist which is simply supported over a span of 20 ft and loaded with 1 ton at the centre and 2 tons at a point 4 ft from one end. Neglect the weight of the joist itself (9-0 tonf ft).


2. An R.S.J, weighing 133 lb/ft rests on supports 40 ft apart. Calculate the maximum bending moment due to its own weight. (26,400 lbf ft).

3. A beam 22 ft long is simply supported at the LH end and at a point 2 ft from the RH end. At a point 4 ft from the LH end there is a concentrated load of 5 tonf and from this point to the RH end there is a distributed load of 1-0 tonf ft. Draw to scale the bending moment graph and find from it the position and magnitude of the greatest bending moment (55-2 tonf ft at 9-3 ft from the LH end).

4. ABODE is a beam supported at Β and D 10 ft apart. At points A and C and Ε are concentrated loads of 1-5, 2-0 and 1-0 tonf respectively while from Β to D there is a uniform load of 0-5 tonf/ft. The remaining dimensions are : AB = 2 ft, BC = 5 ft, DE = 3 ft. Determine graphically the value of the maximum bending moment (8*25 tonf ft).

5. A joist is simply supported over a span of 40 ft and carries a wall weighing 1-75 tonf/ft from the LH end to the midpoint. In addition there are concentrated loads of 5 tonf at points 6 ft from each support. Calculate the load on each support, draw the graph of bending moment to scale and from it find the maximum value of M and its position (31-25 and 13-75 tonf. 228 tonf ft at 15 ft from the LH end).

C H A P T E R I I I

1st AND 2nd MOMENTS

Centroid

This is the point in a body a t which i ts mass may be assumed

concentrated. I n the case of a lamina i t is the point a t which the

area may be assumed concentrated.

1st Moment of Area

F o r an element of area αλ (F ig . 60) distant xx from an axis Y Y

this is defined as the product axxx.

Total area Α «Σα

FIG. 6 0

The 1st Moment of the whole figure will be the sum of the 1st

Moments of all such elements, v iz : axxx + #2^2 + W + a n

d so

on and this is written Σαχ.

59


Datum

I*—2-0 in—H I

3Oin

Iy

FIG. 6 1

Solution

From symmetry the centroid will be on the axis Y Y . The 1st

Moment of the shaded area is the difference between tha t of the

to ta l area and tha t of the hole.

Thus, 1st Moment of area about Υ Υ Σα χ

= α

ιχι +

α2χ2 + a

sxz

+ and so on.

I f the to ta l area A is assumed concentrated a t the centroid G a t

χ from Y Y , then the to ta l 1st Moment will be the product Ax.

Hence, Ax = Σαχ,

_ . J „ „ , T r Tr J ^ a x 1st Moment of area i.e. Distance of G from 7 7 £ = — - — = .

A Area

Note tha t if the axis Y Y passes through G, then

X = 0 ,

i.e. Ax = 0 ,

. · . 2 > * = o.

Hence the 1st Moment of any figure about an axis through i ts

centroid is zero.

E X A M P L E . F ind the position of the centroid of the box-section

shown.

1 s t A N D 2 n d M O M E N T S 61

10

V

FIG. 6 2

The two rectangular areas comprising the section are (2 χ 10) and

(8 χ 2) with centroids a t 5 and 11 in respectively from Y Y .

3 a SM

Original area as = 3 x 4 -5 ,

= 13-5 in2 a t G s = 2-25 in. from X X .

Area of hole ah = 2-0 χ 2 -0 ,

= 4-0 in2 a t G h - 2-50 in. from X X .

Shaded area A = 13-5 — 4-0 ,

= 9-5 in2 a t £ in. from X X .

.·. 9-5 χ χ = (13-5 χ 2-25) - (4 χ 2 -5 ) .

30-4 - 1 0 0

* = 9-5 -

= 2-145 in from the given datum, i.e. X X .

Example. F ind the position of the centroid for the given Τ

section.

Solution

From symmetry the centroid will be on the axis X X .


(2 χ 1 0 ) 5 + (8 χ 2 ) 1 1

(2 χ 10) + (8 + 2)

100 + 176

36

= 7-68 in. from Y Y .

Area under a Graph —1st Moment

The area enclosed between the graph of y = f(x) and the

x-axis can be found by integration as follows.

FIG. 6 3

Area of element = y dx b

Tota l area A = j y dx n.

Hence,

1st Moment of element about Y Y = y dx χ χ by definition,

= yx dx.

b .'. To ta l 1st Moment about Y Y = J yx dx and this is equal to

a

Ax where A is the to ta l area and X the distance of the centroid from Y Y .


Hence.

J yxdx a

b

J y dx

1st Moment about Y Y

Area

y Again, 1st Moment of element about X X = y dx ~ - (since the

centroid of the element is a t —from X X ) , Δ

y2 * = — A r .

Γ y2

.'. To ta l 1st Moment about X X =^= j ~- dx and this is equal to a

Ay where y is the distance of the centroid G from X X .

Hence, y = • b

j ydx

1st Moment about X X

Area

The above are general expressions and may be applied to particular cases.

Quadrant of a Circle

Since y2 r

2 - x

2, the

equation of the graph is

y = (r2 — x

2)

11" where r is a

constant.

F rom Fig . 64,

Area of element = y dx.

Fi rs t Moment about X X

A y

dx

3 η*

FIG. 6 4

6 4 S T R E N G T H O F M A T E R I A L S

.'. To ta l 1st Moment of quadrant = -^Jy2 dx,

r

= — J (R2 — x2) dx,

1 ~2

r2x — X3

Ύ 1 1 ^3 r

3\

~2 Τ —

Y Area of quadrant A = — π r

2

and Ay = 1st Moment about X X ,

3 nr2

4r

~3π from X X .

The centroid of the quadrant is, from symmetry, also a t a distance

of 4r/37r from Y Y .

Semi Circle

This is composed of two quadrants, the centroid s G,7 of which

lie symmetrically on either side of Y Y a t a distance 4r/3jr from

both axes. Clearly the centroid Gs of the semicircle lies on Y Y a t a

distance of 4r/37r from X X .

Y

•Β' - Ι

\ 4r

\ 3TT

1 I 0

Y Λ 4 r .

3ττ _

FIG. 65


W To ta l mass

g

As before, the 1st Moment of any solid about an axis through G is zero.

Cone : 1st Moment and Position of Centroid

F o r the slice of thickness dx (Fig. 66) : χ

Volume = ny2 dx where, from similar triangles, y = — r,

Tir2 Tir

2

= —rr- χ2 dx and W = — - r - ρ x

2 dx.

h2 h

2

1st Moment about YY = i^f x

2 dx\ χ x = x

zdx,

\ gw I Λ Tota l 1st Moment

= f x3 dx

2r> \ nr

2Q

W gh

2 h

nr*q x*

gh2

nr2q h*

gh2 ' 4 '

nr2h

2q

1st Moment of Mass

F o r an element of mass wjg distant χ from an axis Y Y this is w

defined as the product — x. The 1st Moment of the whole body ^ w

will be the sum of the 1st Moments of all such elements, i.e. Σ — W î w\

g

I f the to ta l mass — = Σ — is assumed concentrated a t the 9 \ gl centroid G a t a; from Y Y , then the to ta l 1st Moment will also be

W _ the product — x.

Wx _ wx Hence, = λ ,

g g

v . W

\ ' ft q 1st Moment of mass

i.e. χ =

2nd Moment of Area

F o r an element of area ax (Fig. 67) distant x1 from an axis Y Y

this is defined as the product axx\.

The 2nd Moment of the whole figure will be the sum of the 2nd

Moments of all such elements, viz. : αλχ\ + a2x\ + and so on

and this is written Σαχ2- I t is denoted b y / . Thus, 2nd Moment of area about Y Y

Iu = Σαχ2 = aiXl + a2

X2 + a3Xl + an(^ SO 0n*

I f Κ is a point in the figure distant ky from Y Y such tha t

Aky = I}J where A = to ta l area,

then **-V-T-

66 S T R E N G T H OF M A T E R I A L S

B u t To ta l 1st Moment =

From similar A's.,(t)r

Equating,

Alternatively, h from base of cone.


Total area A

FIG. 6 7

Parallel Axes Theorem

F o r the irregular plane area of F ig . 68,

Iy = Σαχ2 ° y definition,

= ΣίΦ + r)«],

= ΣίΦ2 + + r2)\,

= Σ{βχ2 + 2xar + a r

2) ,

= ^ ( α ά1) + Σ(^) + Σ(*?2),

= χ2Σα + 2 x i ; ( a r ) + 2/(« r 2) (taking out the

constants) .

B u t the 3rd term Σ(ατ2) = 2nd Moment of area about an

axis through G parallel to Y Y ,

= h-And in the 2nd term Σ(

αν)

= 1 ^ Moment of Area about an

axis through G parallel to Y Y ,

= 0 . This term is therefore zero.

The distance ky is called the Radius of Gyration about Y Y .

Similarly, kx = .


Hence,

Similarly,

ig + Ax2.

l x = Ig+ Ay*

Total area Α=Σα

FIG. 6 8

Thus, if the the value of / about an axis through the centroid G

is known, the value of / about any other parallel axis Y Y can be

found by adding the product Ax2 where χ is the distance between

the two axes .

Rectangle—2nd Moment about an Edge

In Fig . 69, area of element = b ay.

2nd Moment about X X = b ay x y2,

= by2 ay.

b

I

1 dy

Τ

FIG. 6 9

And in the 1st term Σα =

A (the to ta l area).

1 s t A N D 2 n d M O M E N T S

d 69

Tota l 2nd Moment Ix = j by2 dy,

= b

• (in4 if b and d are in inches.)

B u t Ah\ = Ix, i.e. h\ = - j - where ^4 =

ta3 J _

" 3 X

bd 9

d2

, from which kx can be found.

Rectangle—2nd Moment about an Axis Through the Centroid

I n Fig . 70, area of element = b dy

2nd Moment about X X = b dy x y2,

= by2dy.

.'. 2nd Moment of shaded part above X X

= / by2 dy.

0

F I G . 7 0


The total 2nd Moment will be twice this,

d/2

i.e.

B u t Alq. = Ix, i.e.

Ix = 2fby*dy9

- 2b

- — 3

d_

d/2

0

bd*

H T *

" A '

6d3

where A

1

6d,

12 X bd

= —— , from which &r can be found. 12

Note that , from the Parallel axes theorem, considering the axis AA,

/ a A (since ti/2 is the distance between AA and X X ) ,

bdä

12

bd*

bd d*

(as found above).

FIG. 71

1 s t A N D 2 n d M O M E N T S 7 1

E X A M P L E . Find, for the section shown (Fig. 7 1 ) , the values of

IxIyIa and Iz in in4. F ind also the corresponding values of k.

Solution

Area of section A = bd,

= 1 6 χ 6 ,

= 9 6 in2.

_ 16 χ 63

~ 12 '

- 288 in4,

i.e. k, 1-73 in.

_ 6 χ 1 63

~ Ï 2

= 2047 in*.

• / :

2 2 0 47 21 3

i.e. Α.. = 4-62 in.

= Ix + ^ 4 ( 3 )2 (from the parallel axes theorem),

= 2 8 8 -f- ( 9 6 χ 9 ) ,

- 2 8 8 + 8 6 4 ,

= 1 1 5 2 in4,

1 1 52 12 ,

" a

96

i.e. ku = 3-47 in.


Alternatively,

_6*_

3 '

= 12 (as before).

Iz = Ale2 = IX + A{5)

2 (from the parallel

axes theorem),

= 288 + (96 χ 2 5 ) ,

= 288 + 2 4 0 0 ,

= 2688 i n4,

2688 Β =

Je,

96 '

2 8 ,

5-28 in .

E X A M P L E . Show tha t the 2nd Moment of area of a rectangle of

sides b and d about a side of length b is bd2/3. F ind the 2nd Moment

of area of the section shown about

(1) the axis A B

(2) an axis through the

centroid parallel to A B

Solution

Area of element

= bay.

2nd Moment about A B

= b dy χ y*,

= by2 dy.

*. ^ A B = / by2 dy,

- b

ό


F o r the section given.

• 8 χ 63

9 Γ 3 χ 4

31

3 — Δι ί 3 J

y =

= 576 - 1 2 8 ,

= 448 in4.

(8 χ 2) 5 + (2 χ 4) 2

(8 χ 2) + (2 χ 4 )

80 + 16

~ 16 + 8 '

_ 9β

" 2 4 '

= 4 inch, i.e. G lies in the edge of the flange

8 χ 23 \ / 2 χ 4

3

FIG. 73

/ 8 χ 23 \ / 2 χ 4

3 \

_ j>4 128

" Τ " +

" 3 ~ '

192

3 9

= 64 in4.

Alternatively, by the parallel axes theorem,

I G = / A B - (24 x 42),

= 448 - 3 8 4 ,

= 64 in4.

Area Under a Graph—2nd Moment about Y Y

The values of I and k for the area enclosed between the graph of

y = f(x) and the #-axis can be found as follows :


y = f U )

Shaded area A=J ydx

FIG. 74

2nd Moment of element about Y Y = y dx χ x2 by definition,

= yx2 dx.

b

/ . Tota l 2nd Moment about Y Y , Iy = f yx2 dx

and Akff = Iy, where A j y dx a

b jyx

2 dx

h y

J y dx

from which ky can be found.

Arm under a Graph —2nd Moment about XX

In Fig . 74, the whole of the element is a t the same distance χ

from Y Y and the 2nd Moment is simply y dx χ x2. Since this is

not true for the axis X X the 2nd Moment of the strip about X X must first be found.

Considering an element of length dh distant h from X X and forming part of a strip of length y and uniform width dx (Fig. 75) :

Area = dh dx.

2nd Moment about X X = dh dx x h2,

= dxh2dh.

1 s t A N D 2 n d M O M E N T S

y

ο = dx

da;

And -4 A^ftrip — - ^ s t r i p *

dh :

^ s t r i p ^ ^i p

where A = y dx,

dx_ 3 1

3 V X

ydx9

x

- - | - about X X . dx

FIG. 75

75

The area enclosed between the graph of y = f(x) and the a:-axis

(Fig. 74) can be divided into a series of elements of length y for

each of which the radius of gyration is y2/3 (from above) .

Hence, for each strip, 2nd Moment about X X = Area χ k2,

y dx χ 3 '

y dx.

Tota l 2nd Moment

And

-dx.

Ak2 = Ix where A — j ydx,

f i d *

from which kr can be found.

j ydx

Tota l 2nd Moment

Λίπρ = / dxh2dh and da; is constant ,

h3


Common Sections

F o r the channel and I sections shown

jTN A = — (2nd Moment about N.A. of shaded area.)

N.A.

FIG. 76

FIG. 77

E X A M P L E . F ind for the section shown, the values of

(a) Ix,

( b ) / „ , (C) J « , (d)L.

-7 in-|Y

V///////////Ä

I in -

- 5 in-

6 in

V7777/)Ç7777A

4ziin

lOin 12in

Solution

(a) / 7 χ 1 2

3 \ / 6 χ 1 0

3 \

I 12 ) " \ 12 )9

12

= 1008 - 5 0 0 ,

= 508 in4.


Λ / 1 x 73\ / Ι Ο χ 1 » \

( b ) J

- = 2

( - Ï 2 ~ )+

( - ^ 2 — )

= 58 in4.

(c) F rom the parallel axes theorem : Ia = Ix + (Area) 62,

= 508 + [ ( 7 x 1 ) 2 +

+ (10 χ 1 ) ] 3 6 ,

= 508 + 864,

- 1372 in4.

(d) Similarly, Iz = Iy + (Area) 52,

= 58 + (24 χ 25 ) ,

= 58 + 600,

= 658 in4.

Perpendicular Axes Theorem

F o r the element of area a (Fig. 78) forming part of the shaded area in plane Y O X , by definition, 2nd Moment about X X = ay

2

and, for the whole figure

h = Zw2-Similarly, Iy = Σαχ2>

F I G . 7 8


and I. = 2Jaz2> where ZZ is normal to the

plane Y O X ,

= ΣΜν2 + χ2)λ (Pythagoras)

= Σ(αν2 + αχ2),

- ι τ + //7.

iVote: For a thin lamina, the above relation is approximately true for the 2nd Moments of mass about the three mutually perpendicular axes. The relation does not hold good for other three-dimensional solids, for which Iz must be found separately (see under 2nd Moment of mass p. 80 et seq.

Circular Section—2nd Moment of Area

F o r the circular element of radius r and thickness dr (Fig. 79) ,

Area = Length χ Thickness,

= 2nr χ dr.

FIG. 79

Since every part of the element is a t the same distance r from the

axis ZZ, b y definition :

2nd Moment of area about ZZ = 2nr dr χ r2,

= 2nr'6 dr.


2nd Moment of whole Iz = j 2nrz dr,

ο - 2π

ο R

I υ

i.e. / , = £ = D4 (where Z>= 2 Ä ) .

» Δ ΟΔ

This is called the Polar 2nd Moment and denoted (usually) by J.

nD* ... I f D is in inches then J =

32

Since J z = ^4&2, where kz = radius of gyration about ZZ,

/m k\ = -γ-, where ^4 = π ^2,

nRi 1

2 π Α2 '

R2 D

2

" or

τ -

From the perpendicular axes theorem Iz = Iv - f Iy and clearly

r L Hence, l x = γ ,

64

iü2 D

2

I t follows tha t Η = — - or — -4 16

PZcme Surface—Graphical Method for Determination of Position of

Centroid and value of Ig

F o r the irregular plane area A in F ig . 80 , let G be distant y

from any axis X X .

Construction

Draw any axis P P parallel to X X and distant d from it . Draw, on the figure, any line B C parallel to X X . Pro jec t B D and C E . J o i n D and Ε to any point F in X X . Mark the intersections Bx and C 2.

R


Projec t B1D1 and CJ£V J o i n D x and Ej^ to point Γ . Mark the

intersections B 2 and C 2. Repea t for a series of parallel lines such

as B C using the same point F . Then the area A x enclosed b y a

smooth curve connecting points such as B 2 and C 2 is known as the

1st Derived Area and can be measured with a planimeter.

χ F I G . 8 0

The area A 2 enclosed by a similar smooth curve connecting

points such as B 2 and C 2 is known as the 2nd Derived Area and can

be similarly measured.

Then, 1st Moment of area about X X , Ay = Axd,

A,d

(χ may be found similarly).

And 2nd Moment of area about X X , Ix = A2d2.

From the parallel axes theorem, Ig = Ix — Ay2.

2nd Moment of Mass

F o r an element of mass wjg distant xt from an axis Y Y this is defined as the product (wjg) x\. The 2nd Moment of the whole body will be the sum of the 2nd Moments of all such elements


viz (wjg) x\ + (w2/g) x\ + (wjg) %l and so on and this is written

Yj (wig) x2. Although, like the 2nd Moment of area, i t is denoted

by I7J i t is usually called the Moment of Iner t ia . (See also Chapt.

I , Mechanics of Machines by the same author.)

I f Κ is a point in the body distant ky from Y Y such tha t W

— k2 = Iy, where W = to ta l weight,

then,

Similarly,

= Radius of gyration about Y Y .

= Radius of gyration about X X .

F o r the circular element of radius r and thickness dx (Fig. 81) ,

Weight = Volume χ Densi ty,

= nr2 dx χ ρ.

Mass = —dx.

Sphere—2nd Moment of Mass about a Diameter

FIG. 8 1


dx χ 2 '

πρ

πρ

πρ

R2

r4 dx and r

2

(R2 - x

2)

2dx,

(R* - 2R2x

2 + x*) dx.

Since χ varies from zero to R, integration of the above between these

limits will give the 2nd Moment of the R H hemisphere.

Hence, 2nd Moment of whole sphere

η

I_ 2 JR %

^ - ( / T> 4 - 2Rh

2 + A

4) dx,

ο

πρ

(J R*x - 2R

2-

9 \ 3 5 / '

I-

πρ b

_8πρΡ^_

Rb,

W Since / . = — k

2, k

2 I. 4

—f—, where W = —nR3p,

w/g 3

8 π ρ #5

3ί7 log 4:πR

3ρ'

2 W (2R2\

= 4 *a

» Hence 1. 5

Cone—Polar 2nd Moment of Mass

F o r the circular element of radius r and thickness dx (Fig. 82) ,

Weight = Volume χ Density,

= πτ2 dxρ.

Polar 2nd Moment = Mass χ k2,

που* Γ Λ 2nd Moment of whole 7~ (or J) = I x* dx

* v ' 2gh* J

h

0

nqR* χ h

2gh* J 0

9

2gh* X

5 '

~10g~'

1 s t AND 2 n d M O M E N T S 83

Mass = — ax. 9

Polar 2nd Moment = Mass χ k2,

nr2g r

2

9 2

7t0 X = - r ^ - r

4 do: and from similar triangles r = —R,

2g h

πρ ΙχΒγ Ί

= - f — α ;4 dx.

2gh*

FIG. 8 2


W/g ' 3

πΚ4&ρ 3gr

3iü2

I c T '

Hence, I 5 = — ^ — J .

Cylinder—Polar 2nd Moment of Mass

F o r the circular element of radius r, radial thickness dr and length

£ (Fig. 83) ,

FIG. 8 3

Weight = Volume χ Densi ty,

= (2nr dr χ L) ρ.

2πΖ/ρ Mass χ rd r slugs.

Since every part of the element is a t the same distance r from the

axis ZZ, b y definition : 2nLq

2nd Moment of mass about ZZ 9

2πΣρ

r dr χ r2,

rz dr.

W

Since Iz

= k2

z

, where kz

= polar radius of gyration,

. - . i P - i - , where W -


.'. 2nd Moment of whole ΙΎ = %πΣρ Γ 3 ^

ο 2nLQ | r

4|

ß

2ττΧρ i ?4

x —-

2g

w Since = ~ ^ z > where &z = radius of gyration about ZZ,

Λ kl = -^η-, where W = π #2£ ρ (total weight),

nR^Lq g

2g nR2Lq

9

_ R 2

~ ΊΓ' W R

2

Thus, for a disk, / = — k2, where k

2 = ——. Note tha t if the

9 2 Si + R\

cylinder is hollow (e.g. flywheel rim) then k2 — , where

R2 and R1 are the inner and outer radii respectively.

4 SM

R

h-FIG. 8 4


E X A M P L E . Four holes each 4 in. dia. are bored in a steel disk

20 in. dia. 4 in. thick. Their polar axes are 5 in. from and parallel

to tha t of the disk. Take the density of steel as 0-283 lb / in3 and

calculate (in slug f t2) the polar moment of the bored disk, Fig . 84 .

Sohlt ion

Init ial weight of disk - (0-785 χ 2 02 χ 4) 0-283,

- 356 lbf.

k d ~ 2 I 12

356 1 / 1 0 \2

.*. Polar moment before drilling Id = ^ ^ · — I — 1 ,

= 3-84 slug f t2.

Weight removed/hole = (0-785 χ 42 χ 4 ) 0 - 2 8 3 ,

- 14-2 lbf.

« - I i

Polar moment/hole Ih = 14-2 1

32-2 2 12

= 0-0061 slug f t2.

Polar moment of hole about disk axis Ihl = Ih

14-2 / 5 \2

Required

32-2 \ 1 2 /

(from parallel axis theorem),

- 0-0061 Η 0-0764,

- 0-0825 slug f t2.

J = 3-84 - (4 χ 0-0825) = 3-52 slug f t2.

E X A M P L E . Show from first principles that , for a uniform cylinder

of weight W the polar moment of inert ia is given by / = W/Sg

χ (D\ + Dl), where Ώχ and D2 are the outside and inside dia-

meters respectively. Hence, determine the value of / for a single

cylinder diesel engine flywheel consisting of a disk 4-625 in. thick,

20-5 in. outside dia. and 2-625 in. inside dia. Take ρ == 0-28 lb/ in3.

Es t imate the K . E . stored a t 12000 rev/min.

Solution

Mass of elemental ring =


, where ρ = density.

F I G . 8 5

2jrr dr to 2nd Moment about ZZ = χ r

2 2 π ί ρ

χ r3d r .

2 π ί ρ In

/r

3 dr ,

Jit

2ntQ I R \ - R\

^.(Rl-RlURf+Bl)

and n(R\ -R\)tQ=W,

W D — {D\ + Dl), substituting — for R. Sg 2

I n the given case, W = 0-785 (20-52 - 2 -625

2) 4-625 χ 0-28,

= 4 2 0 lbf.

/ = ^ 4 2

? ^ ^ χ 4 f (2° -

2 52 +

2*

6 2 5 2) > 8 χ 32-2 1 2

2

0 0 1 1 3 (420 + 6-9) ,

4-84 slug f t2.

4*

[2nr dr] to

At 1200 rev/min,

S T R E N G T H O F M A T E R I A L S

ω = 60

125-5 rad/s.

K . E . =4· /ω2' Δ

4-84 χ 125-5

2,

= 38200 ft lbf.

Cylinder—2nd Moment about an Axis Through the Centroid

/ / / / / / / / / /

te

.Disk

1

FIG. 86

As already shown, for a circular section

64 and ky le"

Hence, for the infinitely thin disk of thickness da; and diameter D

(Fig. 86) distant χ from an axis through the centroid G,

2nd Moment of mass about Y Y , Iy = Mass χ k*

^ 4 Jo 16

2π χ 1200

88


F rom the parallel axis theorem,

2nd Moment about the axis through G,

(Mass χ . τ2) .

Hence, for the disk

B \ 4 / g lb \ 4

,2

- (πΌ2 9

~ \ 4 ~ 7

D2

— +x*)dx.

.·. F o r the whole cylinder h = 2

j ( χ 2 )2

~ ) (j£ + χ2

) ά χ

'

ο πΌ

2ρ

2g

Ό2 χ

16

L/2

|ο

and πΌ

2

χ 7>ρ

πΌ2ρ ι D

2L L*

2g \ 32 +

"24

π £2 Σρ ID

2 L

2

~ ι ~ χ ι γ 1 ί 6 " + Ί 2 "

IF (the to ta l weight).

W

g

D2 L

2

l ö " +

~Ï2~

W Since / „ = kl, where &σ =

8 g

ë

of suspension i t follows tha t k2

radius of gyration about the axis D

2 L

2

16 + 12 D2

Note tha t if L is small relative to D, then k\ ~ —- ~ k2

b 16

J

E X A M P L E . A rectangular solid of length L, depth d and breadth b

is suspended from a single point with i ts length and breadth hori-zontal. Show from first principles tha t the radius of gyration about the point of suspension is given by

- V ' b

2 + L

2

12


Solution

F o r L and b to be horizontal, the axis of suspension must pass

through the centroid G. As already shown, for a rectangular section

Iy = — and kl = — .

Hence, for the infinitely thin rectangular slice of thickness da;

distant χ from an axis through the centroid G,

A Y

/ S

I 1 1

/

L dx * 2 -

.Slice

Τ - d

ι

FIG. 87

2nd Moment of mass about Y Y , Iy = Mass χ k2,

ρ b2

(bd x d a ; ) ~ χ — g 12

F rom the parallel axis theorem I8 = Iy + (Mass χ χ'

(bdQ · dx^ b2^

l g ! 12

( ^

bdQ lb* dx

g

(where χ varies from zero

L

1 s t A N D 2nd M O M E N T S 91

F o r the whole solid Ig =

2f -y (j£

+ x* )

dx>

ο 2bdq (b2x a?\L/2

\ l 2+

J ) 0 ' 9 \ A Z d/o

2bdQ (b*_ L_ j _ IM

g V12 X 2

+ 3

X 8 J'

to ta l weight,

W lb* + £2

J f /62 + £

2\

Ύ V î 2 ; *

W 9 δ2 4- Ζ

2

Since / „ = — , it follows tha t k2 =

g 89 s

12

' fr2 + L

2

"Î2 '

E X A M P L E . F ind the volume of the solid generated when an equi-

lateral triangle of side a is rota ted once about the #-axis. F ind the

moment of inertia of this solid about the #-axis in terms of a,

π, g and the density ρ. g

Hence show tha t the radius of gyration about this axis, k = ~J^Q~ ·

Solution

Volume of slice = ny2 dx and y

2 = 3x

2, F ig . 88,

= 3πχ2dx.

a/2

7 = 2 / 3πχ2 dx,

υ

Xs

3 0

2π· a

3

Τ* πα

ζ

4


Volume of slice = ny2 dx.

Weight of slice = ny2 dxg .

Mass of slice τι y

2 dxg

FIG. 8 8

πν2 dxo ν

2

2nd Moment = — ^ χ ^ - ,

y* dx and y = y Sx, .*. y1

9πρ

2g • x

4 dx.

α/2

#4 da;, α/2

g J

9πρ α/2

0

9 πρ a5

= "5

X T " X 3 2 '

9πρα5

1st A N D 2nd M O M E N T S 93

w And / = — * » ,

9 7 2 9πρα

5 g πα?ρ

k =WÔg-X-W> W h e re * = F é ? = — :

9πρα5 4

— χ

JE; =

160 " πα3ρ

9α2

3α

"j/4Ö*

Examples I I I

1. A T-section is 1-0 in. thick throughout, 6-0 in. deep overall and has a flange width of 5-0 in. Find the position of the centroid of the section and calculate the approximate value of the 2nd Moment of area about an axis through it and parallel to the flange (3 in. above the tip of the web, 33 in

4).

2. A piece of angle iron 6 χ 6 in overall, is used as a cantilever with one side vertical. If it is 0-5 in. thick throughout, find the 2nd Moment of area of the section about a horizontal axis through the centroid (19-9 in

4).

3. Find the Radius of gyration of the section of a hollow shaft having inter-nal and external radii of 5 and 6 in. respectively. What is the percentage difference between this and the mean radius (5-52 in., 0-36 per cent).

4. Find the 2nd Moment of area of a channel section about an axis through the centroid and parallel to the web given that the section is 8 χ 4 in overall and 1-0 in. thick throughout (18*9 in

4).

5. The section of a beam is an equilateral triangle with its base horizontal. If the side of the section is 5-2 in. find the 2nd Moment of area of the section about a horizontal axis through the centroid (39-6 in

4).

6. Show that, for a triangle of height d base b the 2nd Moment of area about the base is given by bd3/12. The section of an oak beam is rectangular and 8 χ 6 in. Find the 2nd Moment of area about a diagonal (183 in

4).

4a SM

C H A P T E R IV

BEAMS II—SIMPLE BENDING

Simple Bending

A bending moment applied to a beam produces tensile and com-

pressive strains separated by an unstrained or " n e u t r a l " plane as

in Fig . 89.

In Compression

In Tension'

Load producing bendina

Neutral axis of section

FIG. 8 9

Where the beam section and this neutral plane intersect , is

known as the neutral axis of the sect ion; for the symmetrical

section shown this axis is half way down. F o r the unsymmetrical

section of F ig . 90 i ts position must be found.

I t is reasonable to assume t h a t the strain a t any point due to

bending (and hence the stress / ) is proportional to the distance y

FIG. 9 0

9 4

B E A M S I I — S I M P L E B E N D I N G 95

from the neutral plane or axis, so tha t the graph of / against y

is a straight line (shown dot ted in Fig . 90) the origin 0 being any

point on the neutral plane. ( I t will be proved later t ha t this is in

fact the case.)

L e t / be the stress in elemental area a, distant y above the N.A.

Then, compressive force on element is fa and this will increase from

zero a t 0 up to a maximum a t B , where y — h. Similarly, for ele-

ments below the neutral axis, the force will be tensile and will

increase from zero a t 0 down to a negative maximum a t A, where

y = -(d - h).

I f the section A B is to be in horizontal equilibrium, the resultant

compressive force above the neutral axis must be equal and oppo-

site to the resultant tensile force below i t ,

i .e. 27« = °-

Ε As will be shown later, / = — x y where Ε = Elas t ic i ty modulus

R and R = Radius of curvature of beam at the section.

Ey

Hence, Σ x » = 0 ,

Ε

or — Σ W = 0.

Ε .'. Σ

ay

= 0 , since — Φ 0 .

R

Now, b y definition, Σαν i

s the 1st Moment of Area of the section

about the neutral axis, and this can only be zero if this axis passes

through the centroid of the section.

Hence, to find the position of the neutral axis i t is only necessary

to locate the centroid of the section. y ar

As already shown, i ts depth is given by h = — — from the top,

where A is the to ta l sectional area (Fig. 91) .

Having determined the position of the neutral axis, the value of

the 2nd Moment of area of the section about this axis (IN.A . ) m a

y

now be found in a given case b y one of the usual methods. (As

will be seen later, this value, together with those of Ε and Μ, is

required before the stress can be evaluated.)

An expression for the stress a t any point in an elastic beam can

now be derived, the following being the chief assumptions made :

4 a*


FIG. 9 1

L e t the two transverse plane sections A B and CD (Fig. 92) be

a small distance dx apart when the beam is straight. The applica-

t ion of a bending moment M will cause A B and CD produced to

intersect a t some point 0 so tha t B D subtends the small angle do.

Elements such as E F above the neutral axis will be reduced in

length while similar elements below the N.A. will undergo tensile

strain.

ο

f

FIG. 9 2

1.. The material is homogeneous and isotropic.

2. The effects of shear force (which are explained la ter in the

t e x t ) are neglected so tha t a plane section perpendicular to

the axis is assumed to remain plane during bending.

3. I n neither tension nor compression is the limit of proportio-

nali ty exceeded, and in both the value of Ε is the same.


R '

Hence, Compressive stress / = Ε x Strain,

7? *

Similarly, for elements below the neutral axis,

Tensile stress / -— — . Ey

Thus, in general, / = the sign depending on

R the position.

Since R is constant for this small length of beam, it follows tha t

/ oc y as was assumed earlier. Hence / = 0 when y — 0 (i.e. zero

stress a t the neutral plane) and is a maximum when y is a maximum,

FIG. 9 3

i.e. a t the upper and lower surfaces. The stress distribution (graph

of / against y) is shown in Fig . 92. Note tha t the maxima have diffe-

rent signs and are not necessarily the same.

Now the transverse area of the element E F is b dy and this

may, for convenience, be denoted by a.

F o r the element E F in F ig . 92 :

Original length = dx = R d6, where R = radius of neutral plane.

Length after bending = (R — y) d0 = R dd — y dö.

.'. Reduct ion in length = y dO,

ydO i.e. Compressive strain = ,


Then, Compressive force on this element = Stress χ Section,

Ey

Ey Moment of this force about the N.A. = —— χ a X y

Ii

κ ,

I f all such similar moments are added (including those of the

tensile forces below the neutral axis which have the same sign),

the result will be what is called the Moment of resistance of the

beam.

Ε This can be written Σ "5"

α2 /

2

= 2J AV

2 since is constant,

i t R

Now, by definition, the quant i ty 2Jay2 is the 2nd Moment of the

section about the neutral axis . This is denoted b y I. Ε

Hence, Moment of resistance to bending = —I. Since a t any sec-

t ion (within the limit of proportionality) this internal moment of

resistance is equal and opposite to the applied bending moment M WFI p a n WRITE,

B u t

Ε M E M

= - RI °

r - = T

, Ey Ll L f Ε

>=ΊΓ> S O T H AT J = -R

(1)

(2)

Combining (1) and (2) gives the fundamental equation for simple bending, i.e.

L - ϋ - —

B E A M S I I — S I M P L E B E N D I N G 9 9

Since / = My/I i t can be seen that , for a given beam section,

i.e. for a given value of I, the stress varies along the beam in pro-

portion to M and across the section in proportion to y. This is

illustrated in Fig . 9 4 b y the graphs of stress distribution drawn at

equal intervals along a simply loaded cantilever.

Load W

FIG. 94

E X A M P L E . A wooden beam 1 2 χ 6 in. carries a load of 2 tonf

concentrated a t the centre of a 1 2 ft span. F ind the greatest bend-

ing stress in i t when the longer side is

(a) upright and (b) horizontal.

2 tonf 2 tonf

(a) y = 6in

FIG. 95

(b) y=3in

Solution

I n case (a) : bf_ _ 6 χ 1 2

3

~ Ϊ 2 ~ ~ Ϊ 2 8 6 4 in

4,


M„

Λ / »

I n case ( b ) :

WL _ ( 2 χ 1 2 )

~ 4 ~ ~ 4

Mmaxy _ 7 2 χ 6

/ 8 6 4

1 2 χ 63

1 2 = 7 2 tonf in.

0 - 5 tonf in2.

1 2

M is the same as in case (a)

7 2 χ 3

2 1 6

2 1 6 in4.

= 1 -0 tonf in2.

The stress in case (b) is twice tha t in case (a) because, although y

is halved, the value of Ib is only one quarter of tha t of Ia.

E X A M P L E . A horizontal jois t is supported over a span of 2 0 ft.

I t is 1 8 in. deep and the value of I is 1 1 5 0 in4. Determine the value

of the maximum stress if the load per foot (including the weight

of the jo is t ) is 1 - 5 tonf.

Solution

Mma, WL

( 1 - 5 x 2 0 ) ( 2 0 χ 1 2 )

8

9 0 0 tonf in ,

1-5 ton f / f t

- 2 0 f t —

FIG. 9 6

y=9 in

My _ 9 0 0 x 9

' ' m a x ^ " 7 " = 1 1 5 0

- 7 - 0 5 tonf/in2.


E X A M P L E . A jo is t has the section shown. Determine

(a) the value of I N . A. a n

d (b) the bending moment required to induce a maximum stress of 7-5 tonf in

2.

Solution

F o r the solid rectangle 6 χ 12.

6 Χ 1 23 0-875_in

1 N.A. 12

F o r each shaded piece of depth 10-25 and width 2-75,

/Λ .Α. 2-75 χ 10-25

3

12

Net 2nd Moment

6 χ 1 23

^N. A. 12

2-75 χ 10-253

12

= 864 - 4 9 3 ,

= 371 i n4.

- 6 in-

0-5 in-

12 in

FIG. 97

Bending moment M = where y = 6 in. and / = 7-5,

7-5 χ 371

- 6 ;

= 463 tonf in.

E X A M P L E . The 2nd Moment of area about the neutral axis of a beam section 20 in deep is 1670 in

4.

F ind the longest span over which a beam of this section, simply supported, could carry a uniformly distributed load of 1-5 tonf/ft with a maximum resulting bending stress of 7 tonf/in

2.

1 0 2 S T R E N G T H O F M A T E R I A L S

1*5

Load in tonf/in = — - and y = 1 0 in.

wL2 / m ax /

y

1-5 L2 7 χ 1 6 7 0

* 12 8

.'. L2

1 0 '

7 χ 1 6 7 0 χ 9 6

Ï 5

- 7 4 , 8 0 0 .

L = 2 7 3 in.

= 2 2 - 8 ft.

L inches

E X A M P L E . A flat steel strip has a thickness of 0 - 2 in. To what

radius may i t be bent if Ε = 1 3 , 0 0 0 tonf/ in2 and / > 2 0 tonf/in

2?

I f the strip is 1 - 5 in. wide and 7 2 in. long and is supported a t the

centre only, what equal loads suspended from the ends would

produce the same stress?

FIG. 9 8

B E A M S I I — S I M P L E B E N D I N G 1 0 3

Solution

f =

:. β =

Ey_

R '

Ey

t '

1 3 , 0 0 0 χ 0 - 1

y=ο·ι

2 0

6 5 in.

M = W χ 3 6 ,

bd?

1 2

1 - 5 χ 0 - 23

I = -1-5 in-

1 2

0 0 0 1 in4,

My

\ FIG. 99

or

2 0 x 2 2 4 0 =

W =

3 6 χ 0 - 1

0 0 0 1 '

2 0 χ 2 2 4 0 χ 0 0 0 1

3 6 χ 0 - 1

= 1 2 - 4 4 lbf.

FIG. 100

E X A M P L E . The flange of a T-section jois t is 5 in. wide and 0 - 5 in. th ick . T h e web is also 0 - 5 in. t h i ck and has a depth of 4 - 5 in. I f i t is used with the web vert ical and the flange uppermost, find:

(a) the position of the neutral axis (b) the value of /Ν.Λ.·


Solution

The section may be divided into two elements :

ax = 5 χ 0-5 with centroid a t rx = 0*25 in. from the datum,

a9 = 4-5 χ 0-5 with centroid a t r2 = 2-75 in. from the datum.

Datum

FIG. 101

The depth of the centroid G (i.e. of the Neutral Axis N.A.) is then

given b y

Zar (5 χ 0-5) 0-25 + (4-5 χ 0-5) 2-75

~~ A ~~ (5 χ 0-5) + (4-5 χ 0-5)

0-625 + 6-2

~ 2-5 - I - 2-25 '

6-825 4-75

= 1-435 inch from the top.

The 2nd Moment of area of a rectangle about an edge is given by

bd*

~ ~ 3 ~ *

.". ^N.A. 0-5 χ 3 -565

3 5 χ 1-435

3 4-5 χ 0 -935

3

^ + •

45-3 2-96

+ •

3

0-816

6 ' 0-6 0-67

= 7-55 + 4-93 - 1-23,

= 11-25 in4.


E X A M P L E . F ind a suitable spacing for the jois ts supporting a

floor on which the load is to be 336 lbf/ft2 if the max . bending

stress > 1000 lbf/in2. The span is 14 ft and the jo is ts are 12 in

deep χ 4-5 in. wide.

Solution

L e t χ = width of floor per jois t .

J

/ M 12 i

FIG. 102

2nd Moment of section 4-5 χ 1 2

3

12

648 in4.

Permissible Bending moment M = — . y

F o r a Uniform load

il y

1000 χ 648

6 '

108000 lbf/in.

WL

8 8i¥n

L

_ 8 χ 108-000

14 χ 12

= 5150 lbf.

(where L = 14 ft)

Effective area between joists = 14# f t2.

.·. To ta l load W = 14a; χ 3 3 6 .


Equat ing, 14α; χ 336 5 1 5 0 .

5150

14 χ 336

= 1-095 ft,

- 13-1 in.

E X A M P L E . F ind the

depth of the neutral I

axis for the T-sect ion 0-3 i n ^ f

shown. Determine the

maximum tensile and

compressive stresses

when a bending mo-

ment of 4-75 tonf inch

is applied. (Assume a

cantilever.)

Solution

Area of section

A = (3 x 0-3)

_U (4.4 χ 6 -3) ,

= 0-9 + 1-32,

= 2-22 in2.

Depth of N.A.

t 1-25 in

4 - 4 in

3-15 in

h =

- 3in-

0-3 in

FIG. 1 0 3

Ι /0-I5 in

h = 1-55 in t

2-5 in

27 ar

A 9

(3 x 0-3) 0-15 + (4-4 χ 0-3) 2-5

2-22

0 1 3 5 + 3-3

2-22

3-435 2-22 '

= 1-55 in.

3 χ 1-553 3 χ 1-25

3 0-3 χ 1-25

3

( 0-3 χ 3 1 53

r . 1 = o-l in*. 1 Χ.Λ. — ο ο ' Q

1 Q


I f the beam is used as a cantilever, the flange will be in tension.

.'. Max. tensile stress My

, where y =~ 1-55 in.

4-75 χ 1-55

5 1

: 1-45 tonf/in2.

Max. Compressive stress = My

where y = 3-15 in. ,

4-75 χ 3-15

5 1

= 2-94 tonf in2.

E X A M P L E . The shaft of a mechanical hammer is rectangular in

section and pivoted a t a point 60 in. from the head, which weighs

560 lb. I f the head is to be raised from the horizontal with an

init ial acceleration of 5 f t /s2, determine the shaft section required

to l imit the maximum bending stress to 0-75 tonf/ in2 assuming the

depth to be twice the thickness.

6 0 in •

5 f t / s2

5 6 0 lbf

Solution F I G . 1 0 4

W Accelerating force on head F = —/,

9

_ 560 χ 5

32-2

= 87 lbf.

*. Tota l force on end of arm = 87 + 5 6 0 ,

= 647 lbf.

H b

Section on XX

Bending moment M = 647 χ 6 0 ,

= 38 ,820 lbf inch about point O.


From the bending equation -L M

I '

0 - 7 5 χ 2 2 4 0 3 8 , 8 2 0

d/2 ~ M»/12 '

i.e., bd

3

1 6 8 0 - = 3 8 , 8 2 0 4 ·

Δ

.·. bd2

= 1 3 8 - 5 ^and b

:. d3

= 2 χ 1 3 8 - 5 ,

= 2 7 7 ,

Hence, d = 6 - 5 in ,

and b = 3 - 2 5 in .

E X A M P L E . A locomotive coupling rod is 1 0 0 in. long between the

supports, 5 in. deep and 2 in. th ick . The crankpin radius is 1 ft

and the wheel diameter 6 ft. I f the density of the material is

0 - 2 8 3 lb / in3 es t imate the maximum bending stress in the rod

when the locomotive is travelling a t 9 0 ft /s.

FIG. 1 0 5

Solution

The centrifugal force on the rod ac ts radially outwards, and,

like the weight of the rod itself, is uniformly distributed. When the

rod is in i ts lowest position, the centrifugal and gravity forces are

in the same downward direction so tha t the loading is as a t (b)

and the bending moment is a maximum.

Weight of rod W = ( 1 0 0 χ 2 - 5 ) 0 - 2 8 3 = 2 8 3 lbf.

Speed of wheel rim relative to wheel axis = ν = 9 0 ft/s.

Speed of crankpin relative to wheel axis vc = v/3 = 3 0 ft/s.


32-2 χ 1 0 '

7920 lbf.

To ta l distributed load = F + W,

= 7920 + 2 8 3 ,

= 8200 lbf. 2 χ 5 3

2nd Moment of rod section I = — — — = 20-82 in 4. iz

WL Max. bending moment M = ———, where W = 8 2 0 0 ,

8

8200 χ 100

102,500 lbf in.

Corresponding max . stress / = , where y = 2-5 in. , My

I

102,500 χ 2-5

20-82

12,300 lbf/in2.

Beam of Circular Section

A bending moment may be applied to a shaft by a force applied to a crank, pulley or toothed gear in addition to any bending effect due to the weight of the shaft itself. Since the bending stress is given b y / = Myjl i t is required to find the value of / about the neutral axis, i.e. about a diameter. The value of y is clearly the greatest radius.

F rom the theorem of three mutually perpendicular axes :

Iζ =

Ιχ + Iy a n

d clearly Iy = Ix.

/. Iz - 27„

i.e.

W V2

Centrifugal force F = (where r = 1-0 f t ) , gr

283 χ 3 02


Since all points on this element are distant r from the axis ZZ, by

definition :

ι Y 2nd Moment about

Γ ZZ = 2nr dr χ r2,

/ ^T/AC ^ \ \ \ \ V 2 n (

* Moment °^ w n

° le

xf ([ \—τπχ I s = / ^ r 3 d r '

γ or J 3 = . J ( Ä i - Ä i ) FIG. 1 0 7

Hence, I, ~ \ (R{ - />'!) = (D\ - D\).

F o r the circular element in F ig . 107 :

Area = Length χ Width,

= 2πτ χ dr.

I Y

FIG. 1 0 6


I n the case of a solid shaft R2 = 0 and R1 R.

and

J ;

lr

2

nR*

32

ττΧ>4

64

E X A M P L E . A telegraph pole is 9 in. dia. I f the bending stress a t

ground level is not to exceed 500 lbf/in2 find the horizontal force

which may be applied to a s tay wire 36 ft above the ground.

I f the pole were subjected to a wind load of 2 0 lbf/f t2 of projected

area, est imate the maximum bending stress which would be induc-

ed a t ground level if the pole were unstay-

ed, and 4 0 ft high.

Solution

2nd Moment of area about the neutral axis

64 '

_ π χ 94

" 64

= 322 in4.

36 ft

where

M

f

. F

: F (36 χ 12) = ÎL y

500 lbf/in2 and î / = 4 -5 in . / / / / / / / /

500 χ 322

4-5 (36 χ 12)

: 83 lbf.

γτπτττττπτ

Projected area of pole

= Height χ Diameter ,

9

4 0 f t

40 χ 12

9

= 30 f t2.

To ta l load W = 30 χ 2 0 ,

= 600 lbf.

777777777"

FIG. 1 0 8


Bending moment M WL

2

= 6 0 0

at ground level,

4 0 χ 1 2

( ^ )

Max. Bending stress

1 4 4 , 0 0 0 lbf in.

, My

1 4 4 , 0 0 0 χ 4 - 5

3 2 2

= 2 0 0 0 lbf/in2. (approx.)

E X A M P L E . An emergency footbridge consists of two steel pipes

6 i n . bore, 7 in. outside dia., laid parallel and 3 6 in. apart and cover-

ed with planking. The tota l load per foot is 2 2 4 lbf. Es t ima te the

maximum bending stress in the pipes when there is an additional

concentrated load of 4 4 8 0 lbf a t the midpoint of the span which is

of 1 0 ft.

2 cwt / foot 4 0 cwt

7 in 6 in

120 in •

Solution

Tota l load ( 1 0 χ 2 2 4 )

6 7 2 0 lbf ,

R 2

FIG. 1 0 9

4 4 8 0 ,

^ 1 7

^ 2 6 7 2 0

= 3 3 6 0 lbf.

At centre, J f m a x = (R1 χ 6 0 ) - ( 2 χ 1 1 2 ) 5 χ ~ ,

= ( 3 3 6 0 χ 6 0 ) - ( 1 1 2 0 χ 3 0 ) ,

= 2 0 1 , 6 5 0 - 3 3 , 6 0 0 ,

= 1 6 8 , 0 0 0 lbf in.

B E A M S I I — S I M P L E B E N D I N G 1 1 3

/ = where I 6 4

(74 _ 64) , = 5 4 - 2 5 i n

4 per pipe,

1 6 8 , 0 0 0 χ 3 - 5

5 4 - 2 5 χ 2

= 5 3 2 0 lbf / in2.

Alternatively, WL W2L

= L 8 ^ 4

(where W1

W,

Mn

Tota l distributed load,

( 2 χ 1 1 2 ) 1 0 ,

2 2 4 0 lbf ;

Concentrated load,

4 0 χ 1 1 2

: 4 4 8 0 lbf)

1 2 0 ( 2 8 0 + 1 1 2 0 ) ,

1 2 0 χ 1 4 0 0 ,

1 6 8 , 0 0 0 lbf in.

E X A M P L E . A wagon axle is carried in bearings 6 6 in. apart, the

wheel centres being 5 8 in. apart and placed symmetrically between

the bearings. I f the axle carries 1 0 tonf, find a suitable diameter for

the axle assuming a safe stress of 4 tonf/in2.

Corresponding max . bending stress


Solution

d

y = r

ltd*

=

5 t o n f

N - » - A

- 6 6 in -

4 in

5 tonf

58 in -

FIG. 110

Between the wheels, Constant M - 5 χ 4 ,

^ 20 tonf in.

/ = My

20 χ d

4 =

.·. 4 =

64

6 40 · λ

6 40 H I

d = 3-7 in. (say, 3 | i n . ) .


E X A M P L E . A steel pipe 12 in. inside dia., 13 in. outside dia., is

used as a beam. F ind the max . permissible central load on a span

of 8 ft if / m ax > 5 tonf/in2. Neglect the weight of the pipe.

Solution

/ n . a . = ^ ( 1 34 - 1 2

4) ,

= - £ - ( 2 8 , 5 0 0 - 2 0 , 7 0 0 ) , 64

_ π x 7800

64 '

= 383 in4.

4 8 in-

it

FIG. I l l

At centre, W

y

= 24 W tonf in , and / = 5 tonf/in2,

y = 6-5 .

5 χ 383 24Tf =

W =

6-5 '

5 χ 383

6-5 χ 2 4 '

12-28 tonf.

E X A M P L E . F ind the maximum bending stress induced b y i ts own

weight in a 9 in. dia. shaft 30 ft long when i t is simply supported

at the ends, given tha t the density of the material is 0-28 lbf/in3.


Solution

Weight of shaft W = (0-785 χ 92) ( 3 0 χ 12) 0-28,

= 6400 lbf.

- 3 0 f t -

FIG.112

k9in-*H

2nd Moment of section I = π χ 9

4

64

= 322 in4.

-*-"· max WL

8 where L = 360 in ,

_ 6400 χ 360

8

= 288 ,000 lbf in.

/max = MmvaLyjI, where y = 4 - 5 ,

_ 288 ,000 χ 4-5

322 = 4032 lbf/in

2.

E X A M P L E . F ind, in terms of

the diameter D, the dimen-

sions b and d of the strongest

rectangular beam which can

be cut from a cylindrical log

(Fig. 113) .

Solution

. My_ I τ

= Μ [ ^- X — ^

FIG.113

12

ΊηΡ) = M (and<Z

2 = D

2- &

2) ,

bur

6M

~~ b(D* - 62) '


F o r / to be a minimum, bD2 — b

z must be a maximum.

/ . Differentiating, D2 - 3b

2 = 0 ,

D2

Examples IV

1. Calculate the value of the horizontal load which, when applied at the top of a tubular pole, 6 in. outside dia. 0-1 in. thick and 20 ft high, will induce a maximum stress of 12,000 lbf/in

2 (136 lbf).

2. Discuss the assumptions made in the derivation of the equation for simple bending.

3. A diecast bracket is of I-section 0-2 in. thick throughout and 1-4 in. deep, the widths of upper and lower flanges being respectively 1-0 and 0-6 in. Calculate the bending moment which will induce a tensile stress of 1500 lbf/ in

2 in the upper flange and determine the corrseponding stress in the lower

flange (357 lbf in., 2350 lbf/in2).

4. Oil of density 0-0318 lbf/in3 is to be poured into a tank 5 ft square which

is supported symmetrically by the flanges of two parallel cantilevers each of T-section 5 χ 5 χ 0-5 in. If the tank is in contact with the wall and its weight is small in comparison with that of the liquid, determine at what depth of oil the maximum stress in the cantilevers will reach a figure of 10,000 lbf/in

2 (19-5 in.).

5. Estimate the uniformly distributed load which may safely be carried over a simple span of 20 ft by a timber joist 12 in. deep and 6 in. wide if the permissible stress is 800 lbf/in

2 (3850 lbf).

6. A crane is to lift a steel shaft 120 in. long and 2-0 in. dia. by means of two vertical slings. Determine the sling positions which will give rise to minimum stress and find this stress (24-8 in. from each end. 700 lbf/in

2

approx.).

5 SM

C H A P T E R V

BEAMS III—SIMPLE SHEAR

Shear Force in Beams (F)

Convention

I f the resultant force to the left of a section is upwards as in

Fig . 114 then F is taken as positive.

X X

Positive shear Positive shear

X

J

F I G . 1 1 4

Cantilever with Concentrated Load at the Free End

Considering any section X X of the cantilever in Fig . 115(a) , in

addition to causing a bending movement of — Wx the concentrated

load W also tends to force the piece of length χ downwards relative

to the remainder of the beam, i.e. to shear i t off as in F ig . 115(b ) .

The net transverse force on X X is called the Shear Force and

denoted by F.

In this case F = — W for the whole beam and i ts graph is the

horizontal straight line of Fig . 115(c ) .

Cantilever with Several Concentrated Loads

From 0 to A (Fig. 116(a)) there is no change in load and, there-

fore, F = -W1.

1 1 8

B E A M S I I I — S I M P L E S H E A R

•

119

FIG. 1 1 5

L

F I G . 1 1 6

• (α)

(b)

I F = - W

JL (c)

(α)

(b)


At A there is an increase in load and hence from A to B ,

F = -(W1 + W2)

Similarly from Β to C

F = -{Wx + W2 + W3).

The graph of F is therefore the series of steps shown in Fig . 116 (b ) .

Simply Supported Beam with Single Concentrated Load

(1) Load at Centre

From 0 to A (Fig. 117 (a)), the net transverse force on any sec-

tion X X W

Fa. = and is positive.

W

(a)

Β

Τ

F I G . 1 1 7

(b)

From A to B the net transverse force on any section Y Y ,

W W Fy = W = — , i.e. is negative.

2 2

The graph of F is therefore as in Fig. 117(b).

B E A M S I I I — S I M P L E S H E A R 121

(2) Load not at Centre

Moments about the L H end give :

R2L = Wa, i.e. R2 = Wa

and Wa

L

The graph of F is then as in Fig . 118 (b) and consists of two unequal

steps.

Simply Supported Beam with Several Concentrated Loads

As before Rx and R2 can be found by taking moments . F rom Ο to

A, J P 0 A = Rx (positive). At A the shear force is reduced b y an

amount equal to W1,

i.e. FX]i =B1 - W,.

Similarly,

Then,

F I G . 1 1 8


Similarly, FCd = R 1 - W 1 - W 2 - W ^

= R1 (W1 + W2 + Wz)

Beam with Loaded Ends Overhanging Supports

(./) Supports Symmetrically Spaced

From Ο to A the shear force ( = — W) is downwards to the left

of the section and is therefore negative, F ig . 120.

At A the shear force is reduced b y an amount W and hence be-

tween A and B , F = 0.

A t Β the shear force is reduced by a further amount W and

hence between Β and C,

F = +W.

(2) Supports not Symmetrical

I n th is case Rx and R2 must first be found by taking Moments.

The shear force between the supports will be zero only if the

bending Moments a t A and Β are equal, i .e.

W > = W2c.

Rx will then be equal to W1 and R2 t o W2. F o r all other cases the

graph of F will be as shown in Fig . 121,

F I G . 1 1 9


W W

A Β

R=W R=W

123

-w JL

FIG. 1 2 0

W,

Ο α

W/////JU/f/ff

//ffÜ!ZZL

+w?

FIG. 121

General Case

The value of F a t any section is the ne t transverse force acting on the section. As already stated, i f this force acts downwards to the left of the section considered, i t is taken as negative and vice versa.

I n general, when there is no change in load between two sections of a beam, the shear force remains unchanged,

1 2 4

w,


W 2 W,

ν///////////77777λ

FIG. 1 2 2

E X A M P L E . F ind the shear force a t each of the points A , Β and C

in the system shown in Fig . 1 2 3 and draw the graph of F.

0-5 tonf I 2

_o_

-0-5 •

Li.

- 3 - 5 tonf

FIG. 1 2 3


Solution

FOA = -0-5 tonf (anticlockwise shear) .

(0-5 + 1-0),

1-5 tonf.

F Λ W - (0-5 + 1-0

- 3 - 5 tonf.

2 -0 ) ,

E X A M P L E . Calculate the values of Rt and B2 in the system shown

and draw to scale the graph of F.

6 8 10 tonf

• 2 f H

— 4 f t —

— 1 2 f t -

-16 f t -

5 a S M

FIG. 1 2 4

125


Solution

Moments about Ο give :

1 6 i ? 2 = (6 χ 2) + (8 χ 4 ) + (10 x 1 2 ) ,

= 12 + 32 + 1 2 0 ,

- 1 6 4 .

1U - 10-25 tonf ,

and RX - 6 + 8 + 10 - 10-25,

= 13-75 tonf .

FQK = + 1 3 - 7 5 tonf.

F A N = + 1 3 - 7 5 - 6 ,

- + 7 - 7 5 tonf .

F\ic = +7*75 - 8 ,

= - 0 - 2 5 tonf .

F A ) - - 0 - 2 5 - 1 0 ,

- - 1 0 - 2 5 tonf - - R 2 . The graph of F

E X A M P L E . Draw to scale the graph of shear force for the system

shown in F ig . 125.

is then as shown above.

2 tonf 4 tonf 6 tonf 2 tonf

F

F I G . 1 2 5


Solution

Moments about A give :

(R1 χ 20) + (2 χ 5) = (4 χ 5) + (6 χ 15) + (2 χ 2 5 ) .

20 R2 + 10 .= 20 + 90 + 5 0 ,

20 R2 = 1 5 0 .

E2 = 7-5 tonf .

FH = (2 + 4 + G - I - 2) - 7-5

= 6-5 tonf .

The graph of F is then as shown.

E X A M P L E . Draw to scale the graph of shear force for the system shown in F i g . 126.

Solution


(R2 χ 20) + (4 χ 5 ) ,

= (2 χ 3) + (5 χ 12) + (3 χ 1 8 ) ,

i.e. 20 R2 + 20 = 6 + 60 + 5 4 .

20 R2 = 1 0 0 ,

i.e. R2 = 5 ton f .

R1 = (4 + 2 + 5 + 3) - 5 ,

= 9 tonf .


Cantilever with Uniform Load

Considering any section X X of the cantilever in Fig . 127(a ) , χ

in addition to causing a bending moment of —wx-^-the weight

wx of the piece of length χ induces a shear force of wx on the sec-t ion. This is downwards to the left of X X and is therefore taken as negative,

i.e. Fx = — wx.

5 a*


4 tonf

- 8 f t - - 9 f t - - 6 f t -

· - 5 f t - -20 ft —

F

FIG. 1 2 6

FIG. 127


Since w is constant F oc χ and the graph of F is a straight line as shown in F ig . 127 (b). Clearly Fmax = — wL and occurs a t the support.

Simply Supported Beam with Uniform Load

wL Load per support = ——. Considering any section X X , shear

^ wL force Fx = — wx.

At the L H support, when χ = 0, the second term is zero. _ wL

.". F = —^— and is positive.

FIG. 1 2 8

The value of .F is clearly reduced as χ increases by an amount proportional to x, so tha t the graph is a straight line. At the centre, when χ = L/2,

π wL wL F = = 0

The straight line therefore cuts the #-axis a t χ = L/2. This is also the point of maximum bending moment.


τι wL

Τ

χ = L, F = — wL

wL

The complete graph is therefore as shown in Fig . 128(b) .

Relation Between w, F and M

Consider an element d# of cantilever carrying a load of w per

unit length (Fig. 129(a) ) . I t s weight is wax and the graphs of F

and M are as shown in F igs 129(b) and 129(c ) .

F + dF

FIG. 1 2 9

Since the element is in equilibrium under the forces exerted on i t

by the remaining parts of the beam,

At the R H support when

(α)

Bending movement

(b)

Sheor force

(c)


(2) ACW Moments = C W Moments ,

M + dM (considei

point G ) ,

dx dx i.e. M + F— + (F + dF) — = M + dM (considering

dx ΎΊ dx 1 dx ~Λ ~M M-

J^

+ j P^ ~

+ d j P~ 2 ~

= d J f'

i.e. F dx = dM (neglecting the product of small quanti t ies) ,

dM or F =

dx

Thus the slope of the bending moment graph (i.e. rate of change in M with respect to x) is equal to the shear force. F o r the cantilever shown, dM/dx (and hence F) is negative.

I t follows from the foregoing, t ha t integration of w with respect to χ will give the shear force,

i.e. F = j w dx,

= Area of load intensi ty diagram.

Similarly, integration of F with respect to χ will give the bending moment,

i.e. M = J Fdx,

= Area of shear force diagram.

I t should be noted tha t if w is not constant , i.e. the load per unit length is variable, then w must be expressed as some function of x. I f this is not possible, then the graph of w against χ must be

(1) Upward force = Downward force,

i.e. F + dF = F + wdx,

dF = w dx,

dF or —— = w.

dx

Thus the slope of the shear force graph (Le. rate of change in F

with respect to x) is constant for a uniform load. Since (from Fig . 129(c)) this slope is negative, a downward load must be taken as negative. Then

dF

dx


plotted and the area found b y planimeter, adding squares or mid-

ordinate rule.

Application to Simply Supported Beam with Uniform Load

Referring to Fig . 130 for any section X X . ,

wL χ Mx = ——- χ χ — wx χ — ,

2 2

wL w 2 = ~2~

X ~ YX '

which is the equation of a parabola, is always positive, and is a

continuous function.

wJL 2

ι c d (Mx) ' »"x - d x

ι c d (Mx) ' »"x - d x

0

FIG. 1 3 0


As already shown,

Fx = (Mx) = slope of graph of M, which is positive

when x < L and negative when χ > —}

Δ 2

wL wx (differentiating)

w ( 4 - ) · = w x Distance of X X from centre of beam.

Hence, when

when χ = 0 ,

when χ = L,

F o r max . B .M. ,

i.e.

wL

wL (as previously obtained).

w a; = 0

i.e. — χ = 0, since w Φ 0 ,

o r

The maximum value of M is obtained b y substituting this value of χ in the equation for Μ,

i.e. i f m a x =_ x T _ Y ( _

4 8 w L

2

8 and = IT (the to ta l load) ,

at the centre, where F (i.e. \ is zero. \ ax J


E X A M P L E . Draw to scale the graphs of F and M for the system

shown in Pig. 131 and find the position and value of the maximum

bending moment.

Solution

Moments about L H end give :

12 R2 = (4 χ 3) + (6 χ 6-5) + (3 χ 1 0 ) ,

= 12 + 39 + 3 0 ,

= 6-75 tonf .

Rx = (4 + (3 χ 2) + 3) - 6-75,

= 13-0 - 6-75

= 6-25 tonf .

The graph of F is then as shown and cuts the horizontal a t a point

6-125 ft from the L H end. This gives the position of zero shear and

hence of max . bending moment. F o r any section X X between Β

and C, i.e. a t (x + 5) from the L H end:

Mx = 6-25 (x + 5) - 4(a + 2) - 2a;— , Δ

- 6-25* + 31-25 - 4 τ - 8 - χ2,

= 225χ - χ2 + 23 -25 .

4- (ΜΛ - 2-25 - 2χ = 0 for maximum Β.Μ. , do:

i.e. χ - 1 - 1 2 5 ft,

Hence, confirming the result obtained above, MmuLK occurs a t (x + 5 ) ,

i.e. a t 6-125 ft from 0 .

Substi tuting, l f m ax = (2-25 χ 1-125) - 1-1252 + 23-25 ,

- 2-53 - 1-27 + 23 -25 ,

= 24-5 tonf f t .

Mx = 6-25 x 3 = 18-75 tonf f t .

i f Β = (6-25 χ 5) - (4 χ 2) = 23-25 tonf f t .

Mc = (6 25 χ 8) - (4 χ 5) - (6 χ 3)

- 21-0 tonf f t .

MD = (6-75 χ 2) = 13-5 tonf f t .



4 tonf

- 6 - 7 5

FIG. 131

E X A M P L E . A simply supported jois t is loaded as shown in F ig . 132 F ind the position and magnitude of the maximum bending moment and draw the graphs of F and M to suitable scales.

Solution


(R2 x 40) = (25 χ 1 ) - ^ - + (3 χ 3 0 ) ,

4 0 i ? 2 = 312-5 + 9 0 ,

R2 = 10-06 tonf , whence

and R1 = 17-94 tonf .


From the graph of F

i t can be seen tha t zero

shear (and hence max .

B .M. ) occurs a t point

D. B y measurement

OD = 17-94 ft.

F o r any section X X between Ο and A,

MT = 17-94*; -

( * χ ΐ ) γ

= 17-94a; - — . Δ

From the shear force diagram, this is a maxi-mum when χ = 17*94

3 tonf

I tonf / f t

and

(17-94

χ 17-94)

17-942

- 161 tonf f t .

(17-94 χ 25)

25 - (25 χ 1)

= 136 ton ft.

FIG. 1 3 2

when χ = 10

MB = (10-06 χ 10) (working from R H end),

= 100-6 tonf f t ,

1 02

Mx = (17-94 χ 10) - — ,

= 179-4 - 5 0 ,

= 129-4 tonf f t .



E X A M P L E . A uniform beam 3 0 ft long is simply supported a t

points 9 ft and 24 ft respectively from the L H end. I t carries a

distributed load, the intensi ty of which increases uniformly from

zero a t the L H end to 0 - 8 tonf/ft a t the R H end. Calculate the posi-

tion and magnitude of the maximum bending moment, and the

values of M a t the supports.

Solution. The loading is as shown:

X »d

c

« 9 ft -

wx

* 15 ft—

X

« 6 ft

0-8 tonfAt

F I G . 1 3 3

9 + 1 5

Weight of load A to C = ^—^ j0-8 = 9-6 tonf acting a t Ql9

Weight of load C to D = - | χ 0*8 = 2-4 tonf acting a t G2.

Moments about Β give : R2 χ 15 = (9-6 x 7) + (2-4 χ 1 7 ) ,

= 67-2 + 40 -8 ,

R0 =

1 0 8 .

7-2 tonf .

S1 = (9-6 + 2-4) - 7-2,

= 4-8 tonf .

χ χ B y proportion, loading a t any section X X , wx = — x 0-8 = — .

Art oU

Weight of shaded portion to left of X X ,

Shear force on section Fx = Rx — Wx,

i.e.

giving

= 4*8 — —- and this is zero a t the 60

point of maximum bending moment

,

x = 16-97 f t .

1 3 8 S T R E N G T H 0 E M A T E R I A L S

ΜΓ = - ( 1 0 y 0 - 5 ^ = - 2 5 - 0 tonf f t . 1 0 _

(x + 6 )

Mx = Rlx - (x + 6) 0-5 2 ,

= 8-33x - 0-25(x + 6 )2,

= 8·33.τ - 0-25(a;2 + 12x + 3 6 ) ,

- 8-33^ - 0-25a;2 - Sx - 9 ,

- 5·33.τ - 0·25.τ2 - 9 .

B .M. on section X X between Β and C,

Mx = Βλ(χ - 9) - W^j} 1 6 - 7 7

2 / 1 6 - 9 7

Hence, a t χ = 1 6 - 9 7 , Mmax - 4 - 8 ( 1 6 - 9 7 - 9 ) 6Q I — —

1 6 - 9 73

- 3 8 - 3 - 2 7 - 2 ,

= 1 1 - 1 tonf f t .

When χ = 9 , i.e. a t B , Mv> = - — - - 4 - 0 5 tonf f t . 1 8 0

9 4 3

When χ = 2 4 , i.e. a t C, Mi: ( 4 - 8 χ 1 5 ) - - — = - 5 - 0 tonf f t . 1 8 0

E X A M P L E . F ind, for the system shown, in Fig . 1 3 4

(a) the load carried b y each support, (b) the position of the two points of contraflexure,

(c) the position and value of the maximum bending moment

between the supports.

Solution

Assuming the load (wL) to be concentrated a t the centre of OC :


R2 χ 2 4 = ( 4 0 χ 0 - 5 ) ^ - β ) ,

r2 =

20 *

14

= 1 1 . 6 7 t o n f , 2 2 4

and R1 = 2 0 - 1 1 - 6 7 = 8 - 3 3 t on f .

The graph of F is as shown.

MA - - ( 6 χ 0 - 5 ) - 9 - 0 tonf f t .


At points of contraflexure Mx = 0 ,

0-25x2 - 5·33α; + 9 = 0 (changing signs) ,

x2 - 2h32x + 36 = 0 ,

21-32 ± | / [21-322 - 4(1 χ 36)]

or

χ =

_ 21-32 + ] /3122

2

- 20 f t2 or 1-83 ft to the right of point A.

139

FIG. 1 3 4


Again, dMxjdx = 5-33 - 0·5χ = 0 for max . B .M. ,

0-5x = 5 -33,

χ = 10-66 f t .

I m a x = (5-33 χ 10-66) - 0 .25(10-66)2 - 9 ,

- 57 - 28-6 - 9 ,

= 19-4 tonf f t .

Note tha t this is not the greatest bending moment in the system.

E X A M P L E . Draw to scale the graphs of F and M for the system

shown and find the position and value of the max . B . M .

0-5 tonf 1-5 tonf

F I G . 1 3 5

Solution


(R2 x 6) + (0-5 χ 4) = (4 χ 1)1 + (1-5 χ 4 ) (treating the Δ

distributed load as concentrated at i ts centre) .

6 i ? 2 = 8 + 6 - 2 ,

.·. R2 = 2 tonf ,

R1 = (0-5 + (4 χ 1) + 1-5) - 2-0 ,

= 4 tonf.


MA = - ( 0 - 5 χ 4 )

= - 2 - 0 tonf f t .

MB = - ( 0 - 5 χ 8) + (4-0 χ 4) - (4 χ 1)1 , = - 4 - 0 + 1 6 0 - 8 0 ,

= + 4 - 0 tonf f t .


Mx = - [ 0 - 5 ( 4 + x)] + 4-Oa; - ( a i x l ) - ,

F

= - 2 0 - C )-5x - f 4-0# — ^r- , Δ

• - 2-0 .

Point of zero shear corresponds with point of max. B.M.

0 Π 1 3-5 f t Η

M 4 - l 3 = M m a x.

0

FIG . 136

When χ = 1, i f * = 3-5 - 0-5 - 2-0 = 1-0 tonf f t .

χ = 2 , Mx = 7-0 - 2-0 - 2-0 = 3-0 tonf f t .

x = 3, Mx = 10-5 - 4-5 - 2 0 = 4-0 tonf f t .

Note: that (1) M = 0 at point D so that at this point the beam is straight and (2) M is of different sign on either side of D so that there is a change in curvature here, the deflected shape being as shown dotted.

Since, for the part A B the equation for M is a continuous func-

tion, the value of χ which makes this function a maximum, can be

found by differentiating and equating to zero.

F o r any section X X between A and Β :


Hence,

Thus, 4 - ( Μ τ ) = 3-5 - — = 0 , i.e. χ = 3 - 5 f t . da; 2

J l f I l l ux = (3-5 χ 3-5) - - 2

= 12-25 - 6-12 - 2 ,

- 4-13 tonf f t .

(This is not necessarily the max . B .M. in the system)

FIG. 137

E X A M P L E . The simply supported beam shown, carries a distri-

buted load which increases uniformly from zero a t the L H support

up to a maximum of 1-5 tonf/ft.

Draw the graphs of shear force and bending moment, indicating

the position of the maximum bending moment and i ts value.

Solution

Mean loading Max. loading/2 — 0-75 tonf f t .

To ta l load = (9 χ 0-75) = 6-75 tonf .

Considering this as acting a t G (the centroid of triangle OAC) which

is 6 ft from 0 , Moments about Ο give :

9f t 6 f t

F I G . 1 3 8

(R2 x 15) = 6-75 χ 6 ,

i.e. R2 = 2-70 tonf .

and R% = 4-05 tonf .


X ι η

- x l - 5

jpx = R1 - weight of piece O E D ,

= R1 — area under O D ,

= * < K ~ 2 9 X l

'5

) '

4 0 5 18

F

0

* 6-95 f t— —*\ \ I \

M

-18-76 = M,*».

1 \

1 1

0

FJCI. 1 3 9

F o r any section X X between 0 and A :

Loding wx =


When χ = 3 , Fx = 4-05 - l

'5 * ^ = 3-3 tonf .

When x = 5, Fx = 4-05 - * '5 *

5' = 1-97tonf.

lo

F o r max . B .M. , Fx = 0 , i.e. ^-x2 = 4 -05 ,

χ2 = 48-6 and χ = 6-95 f t .

The weight of the portion O E D may be considered to ac t a t Gx

(the centroid of triangle O E D ) which is \x from X X .

1*5 / 1 Hence, Mx = Rxx Το~

χ2\Τ

χ

18 \ ο

36

dM · χ2

Note: —-r— = 4-05 = — - = 0 for max. B.M., i.e. χ — 6-95 ft as above.) dx 12

33

When χ = 3 , Mx = (4-5 χ 3) - — - 11-4 tonf f t . x 36

x = 5 , = (4-05 χ 5) - — = 16-8 tonf f t .

ou Q3 x = 9, Mx = (4-05 χ 9) - — = 16-2 tonf f t .

6-053 χ - 6 - 9 5 , i V / - (4-05 χ 6-95) 36

= 28-1 - 9-34,

= 18-76 tonf f t .

E X A M P L E . A uniform beam 20 ft long weighing 5 tonf is hinged

at the L H end and is maintained in a horizontal position, by a

vertical prop at a point 12 ft from the hinge. Between hinge and

prop there is a distributed load of 0-75 tonf/ft while a t a point

4 ft to the right of the prop is a concentrated load of 4 tonf. Draw

to scale the graphs of F and M and find (a) the value and position

of the maximum bending moment between hinge and prop and

(b) the position of the point of contraflexure.


Solution

The loading diagram is as shown, the reaction a t the hinge being

vert ical .

0-75 tonf/ft

JX I 0-2 tonf/ft (beam weight) 4 tonf

Hinge

R2(Prop)

FIG. 1 4 0


R2 χ 12 = (0-75 χ 1 2 ) - ^ - + (0-25 χ 2 0 ) - ^ · + (4 χ 1 6 ) , Δ Δ

= 54 + 50 + 6 4 ,

= 1 6 8 . R2 = 14 tonf and R1 = (0-75 χ 12) + (0-25 χ 20) + 4

- 1 4 ,

= 9 + 5 + 4 - 1 4

= 4 tonf.

The graph of F is thus as shown.

FIG. 141

F o r any section X X between hinge and prop : Mx = 4tx — χ

(0-75 + 0-25) j


t o n f f t I

FIG. 142

E X A M P L E . The variation in load intensity over an aircraft wing

is given in the following table :

Distance from wing tip in feet

0 1 3 5 10 15 20 25 30

Load intensity in lbf/ft

0 27 47 60 78 88 95 98 100

Draw the graph of load intensity over the wing using a horizontal scale of 1 in. = 5 ft and a vert ical scale of 1 in. = 50 lbf/ft.

F rom this derive the graphs of F and M.

0 | 5 ÎO 15 20 25 30

Distance from wing tip in feet

FIG. 143

42

At χ = 4 , Mmax - (4 χ 4) - — = 8 tonf f t .

When M = 0 , 4# = ^ - , i.e. a; = 8 ft, giving the position of the Δ

point of contraflexure. The graph of M with the other principal values is as shown.


Section at

Total area to left of section Shear force F = Total load = (A χ 250) lbf.

Section at No. of squares

η Area in in

2

A = rc/100

Shear force F = Total load = (A χ 250) lbf.

A 77 0-77 192 Β 77 + 140 = 217 2-17 542 C 47 + 170 = 387 3-87 967 D 387 + 184 = 571 5-71 1427 Ε 571 + 195 = 766 7-66 1914 F 766 + 199 - 965 9-65 2410

The graph of F against distance from wing t ip is then as drawn

below.

Oj 5 10 15 2 0 25 3 0

Fro. 144

The bending moment M a t any section is the area under the

shear force graph to the left of the section, i.e. f F dx, so tha t having

found the areas under each section of the graph (by counting squa-

res as before) they can be added as required and converted into

bending moments as in the following table :

Solution

The shear force F a t any section is the net load to the left of the

section, i.e. — j w dx so tha t having found the areas under each

section of the graph (by counting squares) t hey can be added as

required and converted into load as in the following table :


Section at

Total area to left of section Bending moment M = Fax lbf ft = {A x 5000) lbf ft

Section at No. of squares

η Area in in

2

A =

Bending moment M = Fax lbf ft = {A x 5000) lbf ft

A 6 0-06 300 Β 6 + 45 = 51 0-51 2550 C 51 + 74 = 125 1-25 6250 D 125 + 120 = 245 2-45 12250 Ε 245 + 170 = 415 4-15 20750 F 415 + 217 = 632 6-32 31600

The graph of M against distance from wing t ip is then as drawn

below.

Wing t ip Wing root

2 1 0 , 0 0 0

2 0 , 0 0 0

3 0 , 0 0 0

FIG. 145

Examples V

1. A beam is simply supported over a span of 20 ft and carries a uniform wall weighing 2 tonf/ft from the left-hand support to the centre of the span. Draw the graph of shear force and determine the position of the point of zero shear. Use this figure to calculate the maximum bending moment (7-5 ft from LH end. 56-25 tonf ft.)

2. A spring is to be made from steel strip 2-0 in. wide and, when deflected, is to be in the form of a circular arc of radius 54 in. Find the strip thickness at which the maximum tensile bending stress attains a value of 17,900 lbf/in

2

and calculate the bending moment exerted by the spring at this stress value (0-064 in., 245 lbf in.).

3. A total load of 5 tonf is distributed over a simply supported span of 20 ft in such a way that the intensity of loading increases uniformly from


zero at the LH end. Determine the position and magnitude of maximum bending moment (12-8 tonf ft at 11-5 ft from the LH end).

4. Over a simply supported span of 30 ft the rate of loading on a beam increases from 1-0 tonf/ft to 3 tonf/ft. Calculate the position and size of the maximum bending moment (228 tonf/ft at 16-2 ft).

5. A joist 18 ft in length carries 2 tonf/ft over the central third from which value the loading rate decreases uniformly to zero at the ends. The supports are at 3 and 6 ft from the ends. Draw to scale the graphs of M and F and state the maximum values (13 tonf ft 10 tonf).

(i S M

C H A P T E R VI

BENDING WITH DIRECT STRESS

Composite Beams

A t imber jois t may be stiffened by a steel plate (or plates) clamp-

ed rigidly to i t so tha t the section is as shown in F ig . 146 :

1 L

(a) ( b )

FIG. 1 4 6

The jois t is then said to be " f l i t ched" .

Since both components bend to the same radius when the load

is applied, then

Common radius R = —ττ^~ =

where Et and Es are the elastic moduli of t imber and steel respec-

verely and 7 t and Is are the corresponding second moments about

the common neutral axis.

Also B . M . carried by t imber M. = LL

and B .M. carried by steel = AZl.

Tota l permissible B . M . for compound beam

M = Mt + Ms.

Note. The section should be symmetrical about the neutral axis and the sections of the component parts should be symmetrical about some vertical axis.

1 5 0

B E N D I N G W I T H D I R E C T S T R E S S 151

E X A M P L E . A steel plate f in. th ick is rigidly sandwiched between

two t imber jois ts as shown. I f the modular ratio (EsjEt) is 20 and

the maximum stress in the t imber is not to exceed 1000 lbf/in2.

calculate ,

(a) the corresponding

stress in the steel and

(b) the permissible uni-

form load (per foot) .

Solution

2nd Moment for t imber

4 χ 1 23

ΝΔ-

12

= 1152 in4.

- 4 in- - 4 in-

ί s:

Permissible Β . M .

fJt

1000 χ 1152

6

= 192,000 lbf in.

FIG. 147

where / t £ 1000 lbf/in2,

. 8 in 12 in

2nd Moment for steel I, = 0-75 χ

12

32 in4.

20 χ 32

1152

= 106,700 lbf in

χ 192 ,000 ,

Stress in steel / s — Msys

106,700 χ 4

32

13,300 lbf/in2.

6*


W L 8 And Ms = — — where 1KS = load carried by steel = — χ Ms

8 W L 8

Mt = — - — where Wt = load carried by t imber = — χ Mt.

Tota l load W Wt I- Ws,

= ^-(Mt + Ms),

8 ( 1 9 2 , 0 0 0 + 1 0 6 , 7 0 0 ) ,

- 9 9 5 5 lbf , i.e. 4 9 7 lbf/ft .

E X A M P L E . A flitched beam 1 0 ft long and simply supported

consists of a steel channel 8 in. deep a t tached to a t imber joist

8 χ 5 in.

F ind the max. permissible central load if (E^Et) = 2 0 and the

stress in the steel > 1 0 , 0 0 0 lbf/in2.

F o r the channel I = 4 6 - 7 2 in4.

Find also the max . bending stress in the t imber a t the load

Solution

8 in

- 5 in 7 S = 4 6 - 7 2 in

4 (given).

5 χ 83

1 2 '

= 2 1 3 - 3 in4.

/ My

R KU M,

Fm. 1 4 8

EtIt

F o r steel, 1 0 0 0 = M s χ 4

4 6 - 7 2 '

4 6 - 7 2 χ 1 0 , 0 0 0

4

1 1 6 , 8 0 0 lbf in .

Mt

= J - χ χ 1 1 6 , 8 0 0 , 2 0 4 6 - 7 2

= 2 6 , 6 5 0 lbf in .


.'. Stress in t imber 26 ,650 χ 4

" 213-3

= 500 lbf/in2.

WL Tota l bending moment = Ms + Mt = ^

W =L(3IS + Jf t),

~ 10 χ 12

(116,800 + 2 6 , 6 5 0 ) ,

_ 4 χ 143,450

~ ~~ Ï2Ô '

= 4780 lb f .

Reinforced Concrete Beams

Concrete is a material which is relat ively weak in tension. Steel

reinforcing rods are therefore embedded in tha t part of a beam

subject to tensile strain e.g., the upper layer in a cantilever and the

lower in a simply supported beam.

General design assumptions are :

1. Perfect adhesion between reinforcement and surrounding concrete.

2. The tensile load is carried by the reinforcement alone.

3. The tensile stress in the reinforcement is uniform over i ts

section.

4 . I n the concrete strain is proportional to stress.

5. Sections which are plane before bending, remain so during bending.

Since such beams are composed of two materials having diffe-rent moduli of elast ici ty, the simple bending formula (/ == My β)

does not hold good i.e., the neutral axis does not pass through the centroid of the section.

I f , as in F ig . 149(a ) , the depth of the neutral axis is h then the shaded area ^ r e p r e s e n t s the section of concrete under compression.


Reinforcing rods I (section on AB) (assumed

uniform)

(a) (b) (c) (d)

FIG. 1 4 9

Assuming tha t the strain a t any point is proportional to i ts

distance from the neutral axis, then

Compressive strain in concrete a t point A _ h

Tensile strain in steel d — h

I f the depth of the reinforcement axes is d then the rods are a t

(dr-h) below the neutral axis.

B u t ec = — and e s = where Ec anidEs are the elastic moduli c Ec Es

for conrete and steel respectively, so tha t

e s ECU °r d - h Ε J,

(1)

The quan t i t y -^ - is referred to as the Steel I Concrete Modular

Ratio.

I n the steel, Tensile force Ft = fsAs shere As = S tee l section.

The average stress in the concrete above the neutral axis is / c / 2

where / c is the maximum stress and this is distributed over the

area Ac = bh. fc

. ' . i n the concrete Thrust Fc = — bh.

Δ Since there is no resultant force normal to the section, the

tota l force (compressive) in the concrete must be equal and


opposite to the to ta l force (tensile) in the steel

i.e. F, = Fc,

or UAs = ffbh (2)

Since, for the concrete, the stress diagram is a triangle, the resul-

t an t force Fc may be considered to ac t a t the centroid of this tri-

angle i.e. a t f h from the neutral axis . Similarly the tensile force

in the steel (P) ac t s a t a distance (d-h) from the neutral axis. Since

these forces are equal they form a couple, the arm of which is given

b y 2 a = —- h + (d — h), ο

or a — d — •— ό Hence, Magnitude of couple (i.e. Moment of resistance of beam),

M = Fta

-/•*(«-!•). M Alternat ively, Fca,

kbh

While the beam remains in equilibrium under the forces acting on

it , the Moment of resistance of the beam is equal and opposite

to the bending moment acting on the beam, i.e. to M.

Note. From Equation (2):

and from Equation ( 1 ) :

A - A . bh

Ύ K = ( Is Fs \-

d-h bh2

2m(d - h) bh2

-d bh Ύ ' where m = Er.

2md — 2mh

.·. bh2 = 2mdAs - 2mAJi

156 S T R E N G T H O E M A T E R I A L S

bh2 + (2mAs) h - (2mdAH) = 0 .

This is a quadratic equation which fixes the value of h and contains

only known quanti t ies.

E X A M P L E . The horizontal flange of a T-beam is 4 in. th ick and

60 in. wide, and the reinforcement in the rib is 15 in. below the upper

surface. I f the stress limits are 16,000 and 600 lbf/in2 8LndEJEc = 15

and the neutral axis is to coincide with the lower edge of the flange,

calculate :

(a) the required section of reinforcement,

(g) the moment of resistance of the beam,

(c) the actual max . stress in each material .

Solution h

d - h ~~ EJS

4 1 5 / c

15

f h = 4 i n Γ d = l5 in f

I

-b = 6 0 in ·

ΦΦΦΦ;

FIG. 1 5 0

/ s ^ s —

_ fc bh 60 χ 4 1

41-25 2-91 in

2.

Moment of resistance M = fsAs (d — — j ,

16,000 χ 2-91 15 -3 / '

= 16,000 χ 2-91 χ 13-67,

= 636,000 lbf in.


Note: If / c is taken as 6 0 0 , then / S = 6 0 0 χ 4 1 - 2 5 =

= 2 4 , 7 5 0 lbf/in2. which is outside the given limit.

E X A M P L E . A concrete slab 7-5 in. th ick has a span of 10 ft and is

reinforced by steel rods 0-5 in. dia., 6 in. apart at 1-5 in. above the

lower surface.

Es t ima te the load which may be carried per square foot, addi-

t ional to i ts own weight, given t h a t : m a x stress in steel >> 18,000 Ε

lbf/in2 and max . stress in concrete > 700 lbf/in2. —^- = 12. Densi ty

of material = 150 lbf/ft 3 (L .U. ) . E*

Solution

i K— b= 12 in—Η

t N-T— 7-5 in Τ ,

J d=6in

1-5 in

0-5 in dia. î

FIG. 1 5 1

Considering a strip of floor 1 ft wide and parallel to the reinforce-ment :

As = 2(0-785 xO-52) = 0-393 in

2.

2 ' U 2 Λ '

12h

2 χ 0-393 '

12h

and

0-785 '

h Esfc fs 12(6 - h)

d - h Ecf/ U * 72 - 12A

h

12h 72 - 12h Equat ing, Q^ ft ,

i.e. 12h2 f (12 χ 0-785) h - (72 χ 0-785) = 0 ,

h2 + 0-785/* - 4-71 = 0 ,

whence h = 1-81 in .

6 a S M


/ 1 2 H e n C e ' t = ¥ 7 8 5

X L-

8 1'

= 2 7 - 7 so tha t if the steel is stressed to its maximum

of 1 8 , 0 0 0 lbf/in2 then,

stress in concrete fc = — ' „ = 6 5 0 lbf/in2

/ c 27.7 ' which is within the permitted figure of 7 0 0 lbf/in

2.

M = fsAs ( d - j } = 1 8 , 0 0 0 χ 0 - 3 9 3 ^ 6 -

WL = 3 8 , 1 0 0 lbf in, and M = ——

.·. To ta l load W{ = 8

* = 2 5 4 0 lbf. 1 0 χ 1 2

B e a m weight Wh = ( 1 0 χ 1 ) - ^ - χ 1 5 0 = 9 3 7 lbf. IΔ

Useful load W = 2 5 4 0 - 9 3 7 = 1 6 0 0 lb over a span 1 - 0 ft wide,

= 1 6 0 lbf/ft2.

E X A M P L E . A concrete beam 1 2 in. wide 1 5 in. deep is reinforced

by 6 steel rods 0 - 7 5 in. dia., the centres of which are a t 2 in. from

the bot tom of the beam which is simply supported over a span of

1 8 0 in. Assume a maximum stress in the concrete of 6 0 0 lbf/in2

and determine

(a) the depth of the neutral axis, (b) the moment of resistance of the beam,

(c) the stress in the steel, (d) the to ta l permissible uniform loading.

Make the usual assumptions and take the steel/concrete modular

rat io as 1 5 .

Solution

Since / S A = l§-bh, where ^ S = 6 ( 0 - 7 8 5 χ 0 - 7 52) = 2 - 5 6 in

2.

Δ

whence / S = 1 3 6 0 / L


Again, — - — = -—- χ - ~ and = Modular ratio = 1 5 , a Lc fs Jtbc

_h / 600 \ _ 6-62

\ — h~ \ 1360A / ~ A 9

' ' 13 -

so tha t A2 = 6-62(13 - A)

or A2 + 6-62A - 86

= 0, giving

A = 6-55 in.

Moment of resistance,

* - τ » ( ' - ϊ ) · 600

χ 12

- « — b= 12 in •

= 13 in - N.A.

/ t \ / t \ A\ /+\ /T\ st\

î h

ι ι V|/ \l) ψ ψ \\} si/ 2 in

/ 6-55 \ x e - β β 1 8 - - J - ) ,

= 3600 χ 6-55 χ 10-8 ,

= 255 ,000 lbf in.

Stress in steel / s = 1360A,

= 1360 χ 6 -55 ,

= 8900 lbf/in2.

FIG. 1 5 2

Bending moment permissible, M = WL

8

.'. Permissible load

Or, load per foot

T Tr 8 χ 255 ,000 B M J_ £ W = —rz τη— = 5-05 tonf.

15 χ 12

5 0 5 w = 180/12

= 0-336 tonf or 750 lbf.

Bending Combined with Direct Stress

The direct compressive stress induced in a short s trut of un-

symmetrical section A b y a compressive load F act ing a t the sec-

t ion centroid G is given b y

/ - F

Αι - ~ Γ · 6 a *


291 ·*>Ή


This may be represented by a horizontal straight line 'mn'

(Fig. 153(a)) normal to the strut axis and distant an amount f(l

(to some convenient scale) from a datum MN.

I n practice, no load acts truly a t the centroid of a section, there

being inevitably some eccentr ic i ty a: (as in F ig . 153 (b)) introducing

a bending moment given b y

M - Fx.

This causes bending about the neutral axis (N.A.) of the section,

so tha t the strut deflects as in F ig . 153(c ) . The bending stress,

which is clearly compressive on the load side of the neutral axis

(i.e. is of the same sign as the direct stress) and tensile on the oppo-

site side, may be represented by the straight line pq intersecting

mn a t the neutral axis . The two maximum values are respectively,

My, ^ , . x / b = —-— at Q (compressive),

and fb = - ^ p - a t Ρ ( tensile) .

The resultant stress is obtained by superimposing these values on

the direct stress,

F My so tha t a t Q : / m ax = — H γ^- (compressive),

-, -p. £ F My2 ί F My2 \ and a t P : / m in = — — ί compressive since — > — — 1.

I f the eccentr ic i ty χ is increased for a given value of F unti l the tensile bending stress (My2\I) is equal to the direct stress, then the resultant stress a t Ρ will be zero as in Fig . 153(d) . Any further increase in χ will result in a net tensile stress a t Ρ (Fig. 153(e)) although the applied load F is compressive. The stress distribution is represented by the shaded area in Figs . 153 (a-e) and shows the effect of increasing eccentr ic i ty for a given load, i.e. of increasing the bending moment. In general, the resultant stress is the alge-braic sum of direct and bending stresses and is given by


I f , instead of increasing the eccentr ic i ty x, the bending moment

is increased by increasing F while keeping χ constant , the direct

stress (F/A) is increased in the same proportion as the bending

IF (increased)

FIG. 1 5 4 FIG. 1 5 5

stress (Fxyjl) so tha t the shape of the stress distribution diagram

is unaltered, although all stress values are increased as shown in

F ig . 155.

Section Modulus

I n the theory jus t given,

Max . compressive bending stress = My1/I = ~JJjj~~

The quant i ty l\yx is referred to as the Compressive Section Modulus

and denoted b y Zv Similarly I\y2 is denoted b y Z2 and called the

Tensile Section Modulus. Thus , where a component is subjected

to a combination of direct and bending stresses, the equation for the

resultant stress m a y be writ ten

, F M

where Ζ is the relevant modulus. The positive sign will give the maximum stress and this will

be of the same kind as the direct stress. The negative sign will give the least stress, the sign of which will depend on the relative magnitudes of the direct and bending stresses.

Clearly, for a section which is symmetrical about the neutral axis, the two values of Ζ are the same.

B E N D I N G W I T H D I R E C T S T R E S S 1 6 3

E X A M P L E . The axis of the load on a tensile test piece is 0 - 0 0 8 in.

from the geometrical axis. I f the tes t piece diameter is 0 - 7 5 in.

and the load a t the first sign of yielding is 9 - 1 6 tonf, find the maxi-

mum corresponding stress. Show diagrammatically how the stress

varies across the section.

FIG. 1 5 6 FIG. 157

Solution

Sect ion

Direct stress

Bending moment

2nd Moment

Distance from N.A

A = — χ 0 - 7 52 = 0 - 4 4 2 in

2.

4

F

1 9 1 6

0 - 4 4 2

M = 9 - 1 6 χ 0 0 0 8 ,

- 0 - 0 7 3 3 tonf in

π

2 0 - 7 8 tonf/in2.

7 = • _ χ 0 - 7 5 4 , 6 4

0 - 0 1 5 5 in4.

0 - 7 5 0 - 3 7 5 in.


.'. Sect ion modulus Ζ = — ,

Ί . . M 0 - 0 7 3 3 -, —_ , . . . 2 Bending stress — = = 1 - 7 7 tonf/m2.

Ζ 0 - 0 4 1 3

l r F M Max. stress / m ax = — + — ,

- 2 0 - 7 8 + 1 - 7 7 ,

= 2 2 - 5 5 tonf /in2.

Min. stress / I N IN = 2 0 - 7 8 - 1 - 7 7 ,

= 1 9 - 0 1 tonf/in2.

The stress distribution is then as shown in F ig . J 5 7 .

E X A M P L E . A short I-section column has a section of 1 0 - 3 in2 the

depth of which is 8 in. Draw to some convenient scale a diagram of

the stress distribution across the section when a vertical load of

2 0 tonf is applied on the web at a distance of 2 - 5 in. from the neutral

axis. Take / as 1 1 0 in4.

Solution

Bending stress *

_ 2 0 χ 2 - 5

1 1 0 / 4 '

= 1 - 8 2 tonf/in2.

F 2 0 Direct stress = — = ——- = 1 - 9 4 tonf/in

2.

A 1 0 - 3

F M Resul tant stress / = — ± —— ,

A Ζ = 1 - 9 4 ± 1 - 8 2 .

/max = 1 - 9 4 + 1 - 8 2 ,

= 3 - 7 6 tonf/in2 (at A , F ig . 1 5 8 ) .

/ M IN = 1 - 9 4 - 1 8 2 ,

- 0 - 1 2 tonf/in2 (at B ) .


^ 2 - 5 i n - » j2 Q

tonf

— 8 i n

FIG. 1 5 8

Datum

E X A M P L E . The maximum stress in a hollow tie 3 in. outside dia. and 0-5 in. th ick is not to exceed the mean by more than 20 per cent. Es t imate the permissible eccentr ic i ty of the load.

Solution

F o r the section, 1-5in, (Fig. 159)

π

y

' = £ - ( 34

- 24

) , 64

Ζ

A

£ ( 8 1 - 1 6 ) ,

3-19 in4

I_ _ 3-19

y ~ 1-5

0-785 ( 32 - 2

2) ,

3-93 in2.

= 2-127 i n3.

166 S T R E N G T H Ο Ε M A T E R I A L S

Max. stress = 1·2 χ Mean stress (where the Mean stress is F/A)

F M

Τ + ΊΓ M

Ζ

Fx

F

l^Z- -4- and M = Fx. A A

F_

T'

- 0 - 2 Ζ A '

0-2 χ 2 1 2 7

3-93 = 0-108 in.

\

— ι

FIG. 1 5 9

E X A M P L E . A wind pressure of 40 lbf/ft2 of projected area is to

cause no tension in the base section of a pillar built of stone having

a density of 140 lbf/ft3. I f the diameter is to be 6 ft, est imate the

height to which i t may safely be built.

Solution I [T]l r~

Weight of column

W = (0-785 x 62

) ä x 140, eftdia.-^ « -

= 3960Λ lbf.

Sect ion of column h

A = 0-785 χ 62,

= 28-25 f t2.

Direct stress

W 3960ft I

A ~ ^ 2 5 " /77mW77ffM7 = 140ft lbf/ft

2 (compressive). y I G \QQ


Projec ted area = 6A f t2.

To ta l wind force = 6h χ 4 0 ,

= 240JUbf .

Bending moment, M = 240h χ - ^ ,

= 1 2 0 A2l b f t .

π χ 64

2nd Moment of section J = — — — = 63-7 f t4 and y = 3 f t .

64

Bending stress My 120Ä

2 χ 3

63-7 '

5-65A2 lbf/ft

2.

W My

Resul tant stress = — ± —j— (and this is to be jus t

zero on the windward side),

W_ My

UOh = 5-65A2,

i.e.

140

24-8 f t .

E X A M P L E . Es t ima te the value of tensile load which will induce

a maximum stress of 10 tonf/ in2 in a 1·5 in. dia. steel rod. The axes

of load and rod are parallel and 0-25 in apart. Determine graphically

the distance from the rod axis a t which the resultant stress is zero.

Solution

R o d section A = 0-785 χ 1-52,

= 1-77 in2.

FIG. 161


2nd Moment π χ 1-5

4

6 4 ~ ~ '

0-248 i n4.

Bending moment M = 0-25F lbf in .

M 025F Bending stress Ζ 0 -248/0-75 '

= 0 - 7 6 ^ lbf/in2.

F F Direct stress -—

A 1-77 '

0 - 5 6 ^ lbf/in2.

Taking tensile stress as positive,

_ F_ M /max - - χ + ^ - '

0 - 5 6 /7 + 0-762<\

1-32^, whence F = 7-56 tonf.

0-56F - 0 - 7 6 ^ ,

- 0 - 2 χ 7-56,

—1-512 tonf/in2 (compressive).

F rom the stress distribution diagram, the resultant stress is zero

a t a distance of 0-55 in. from the axis of the bar, on the side opposite

to the load axis.

FIG. 1 6 2

/min


E X A M P L E . W h a t load may be applied to the cranked strut shown

if the maximum tensile stress is not to exceed 6 tonf/in2? At this

load, what is the maximum stress in the material?

Solution

Sect ion A

2nd Moment

/

χ 42, π

Τ 4π i n

2.

π

64

4π in'

χ 44,

Modulus Ζ = —

4π

ΊΓ' 2π i n

3

M 6F

Ζ ~ ~2π Μ = 6F, .·. Bending stress

and Direct stress

Resul tant stress

Taking compressive stress as positive

- 6

F F

Α

F 6F

4π 2π

F 6F

4π 2π

fd-τ)- ·

F 6π

¥ 7 5 '

- 6 in-

- 4 in d ia .

IF

FIG. 1 6 3

- 6-85 tonf .


• , F 6F Max. compressive stress = - j — - + ,

1

2π

6-85 + (6 χ 6-85)

= - ^ - ( 3 - 4 2 + 4 1 - 1 ) ,

44-5

7-1 tonf/in2.

E X A M P L E . A short column having the section shown supports a

concentrated load of 20 tonf applied a t point O. Calculate the stress

a t points A , B , C, and D taking the web and flanges as rectangular.

0-9 in

0 5 in -

10 in X -

- 6 i n -

|Y ID

| y

FIG. 1 6 4

2t

Solution

Sectional area A = 2(0-9 χ 6) + (8-2 χ 0 -5) ,

= 14-9 in2.


Direct stress 20

14-9 = 1-34 tonf/in

2.

6 χ 1 03 5-5 χ 8 ·2

3

12

500 - 2 5 3 ,

247 in4.

12

/ 0 - 9 x 63\

I i 2 r 8-2 χ 0 -5

3 / 0 - 9 χ

L 2

12 ^

= 0-855 + 32-4 ,

= 33-26 in4.

Bending moment Mx = My = 20 χ 2 ,

= 40 tonf in .

Max. bending stress about X X , fx = Myx

4 0 χ 5

where yx = 5 ,

247 '

= 0-81 tonf/ in2.

4 1 3 From the diagram

f A= - 3 - 0 7 t o n f / i n2

V - 1 - 4 5 t o n f / i n2

f c= + 5 - 7 5 t o n f / i n2

M f D= + 4 - l 3 t o n f / i n

2

FIG. 1 6 5


Solution

The distance of the neutral axis from the edge of the section

nearest to point A is given by :

A

( 1 5 χ 1 ) 0 - 5 + 2 ( 8 χ 1 ) 5 + 2 ( 4 - 5 χ 1 ) 9 - 5 + 2 ( 4 χ 1 ) 1 2

+ 2 ( 2 χ 1 ) 1 4 - 5

- ( 1 5 χ 1 ) + 2 ( 8 χ 1 ) + 2 ( 4 - 5 χ 1 ) + 2 ( 4 χ 1 ) + 2 ( 2 χ 1 )

_ 7 - 5 + 8 0 + 8 5 - 5 + 9 6 + 5 8

1 5 + 1 6 + 9 + 8 + 4 '

3 2 7

~" 5 2 '

= 6 - 2 9 in. ( ~ 6 - 3 in .) .

My j Max. bending stress about Y Y , fy = ——— where yy =-. 3 ,

_ 4 0 χ 4

3 3 - 2 6 '

= 3 - 6 0 tonf/ in2.

The stress distribution across faces A B , B C , CD and D A due to

the direct stress / and the bending stress fx combined is represented

by the dotted lines F G , GH, H E and E F respectively.

Superimposed on this is the bending stress fy represented on

faces D A and B C by the full lines MN and P Q drawn on E F and

GH as datum respectively. The resultant stress distribution over

faces A B and CD is then given b y the full lines N P and QM.

E X A M P L E . The main pillar of an hydraulic press has the section

shown, (Fig. 1 6 6 ) the load being applied a t point A. Es t imate the

values of the max . tensile and compressive stresses in the material

when the operating load is 5 0 tonf and draw to some scale the stress

distribution diagram.


Fo r a rectangle about an edge / = — - . Considering half the section : ό ^N.A. = I [(7-5 χ 6 ·3 3) - (6-5 χ 5 ·3 3) + (3-5 χ 3·7 3) - (2-5 χ 2 ·7 3)

+ (2 χ 8·7 3) - (1 χ 7·7 3) - (1 χ 2 · 7 3) ] ,

= | ( 1875 - 968 + 177 - 48 + 1310 - 456 - 2 0 ) ,

1870 - _

3 ~ '

= 623 in4,

Ι5 ίη 8 in 4 i n

-15 in -- 1 0 in-

- 6 in-

1

FIG. 1 6 6

3-5in7î

yc = 8-7 in

- 4 - 2 9 i n X ^ ^ ^ ,

F I G . 1 6 7

-£^mm^X 3-11 in Datum


Hence for whole section, / n . a . = 1247 in4.

Direct tensile stress = ^— = = 0-962 tonf/in2,

Bending moment , M = 50(6 + 6-3) ,

= 50 χ 12-3 ,

= 615 tonf in .

Mv 615 χ 6·3 Tensile bending stress a t Τ = —p-= — = 3-11 tonf/in

2.

/ 1247 •

Resul tant stress a t Τ = 0-962 + 3 -11 ,

= 4-072 tonf/in2 ( tensile).

. t ν x j. η ~ M y c 615 χ 8-7 Compressive bending stress a t C = — = ————

r ° / 1247 = - 4 - 2 9 tonf/in2.

Resul tant stress a t C = 0-962 - 4 -29 ,

= —3-33 tonf/in2 (compressive).

E X A M P L E . An I-section jois t 3 χ 5 in. overall has a web and

flange 0-3 in. thick. F ind the compressive stress when a short length

is axially loaded with 5 tonf. Determine the stress distribution when

one of the flanges is removed to give a T-section, if the position and

size of the load remain the same.

Solution

Area of section, A = (2 χ 3 χ 0-3) + (4-4 χ 0 -3 ) ,

= 1-80 + 1-32,

= 3 1 2 in2.

5 Compressive stress, / = - Ξ - Τ ^ Γ - ,

= 1-6 tonf/in2.

Reduced area of section A = (3 χ 0-3) + (4-4 χ 0-3)

= 0 - 9 + 1-32 = 2-22 in2.

Depth of neutral axis h = A

(3 χ 0-3) 0-15 + (4-4 χ 0-3) 2-5

1-55 in.


Eccent r ic i ty of load, χ = 2-50 — 1-55,

= 0-95 in .

0-3inj^

5 in 4 - 4 in

Original section

Ν —

FIG. 1 6 8

Bending moment, M = Fx,

= 5 χ 0-95,

- 4-75 t o n f i n .

J N . A . = έ (3 x 1-553) - 1 (3 χ 1-25

3) + è(0-3 χ 1·25

3)

+ J (0-3 χ 3 · 1 53) = 5-1 in

4.

/ m a x , g . + = 4 + i Z ^ i £ = 2 . 2 5 + 2 . 9 4 = g . 1 9 t o n f / i n , .

/ m in = | . 2 - 2 5 - ^ 2 - 2 5 - 1-45 = 0-80tonf/in*,

The stress distribution is therefore as shown above.


E X A M P L E . F ive drums each 4 ft dia. and weighing 6 cwt. are

s tacked as shown. The chassis of the wagon consists of two beams

of uniform and symmetrical section having vertical ends of similar

section rigidly at tached.

FIG. 169

The springs have a span of 2 ft, the ends being 1 ft from the

inside edges of the vert ical members. Determine the loads on the

horizontal and vert ical members, neglecting friction, and draw

the bending moment diagram. F ind also the section modulus for

each beam if the maximum bending stress is not to exceed 6 tonf/in2.

Solution

F o r upper drum A, 2F1 cos 30 = W,

W 1 , Θ

* 1

2cos 30 '

F o r lower drum B , F2 = W + 2F1 cos 3 0 ,

W

2cos 30

= 2W,

--= 12 cwt .


w

F o r lower drum C,

FIG.170

Fs = W + Fx cos 3 0 ,

W = W +

= 1-5JF,

= 9 cwt .

2cos 30 cos 3 0 ,

F± = 2 ^ sin 30

W

2cos 30 sin 3 0 ,

W = — t a n 3 0 ,

Δ

- 3 χ 0-577,

= 1-732 cwt .

Load per support

= 1 ( 9 + 1 2 + 9)

— 7-5 cwt .

Load in cwt

.1-73

Γ Β I D

9 12

7-5 7-5

- 3 8 7

1-73

7-5 7-5

2 3 0 0 = M m

FIG. 171


Hence, Ma - Mh - - ( 1 - 7 3 2 χ 1 1 2 ) 2 - - 3 8 7 lbf ft

Mc = - 3 8 7 + ( 7 - 5 χ 1 1 2 ) 1 = + 4 5 3 lbf ft

Md = - 3 8 7 + ( 7 - 5 + 1 1 2 ) 2 - ( 9 χ 1 1 2 ) 1

= + 2 8 5 lbf ft

Mc = - 3 8 7 + ( 7 - 5 χ 1 1 2 ) 5 - ( 9 χ 1 1 2 ) 4

+ ( 7 - 5 χ 1 1 2 ) 3 = + 2 3 0 0 lbf f t .

The bending moment graph is then as shown in Fig . 1 7 1 , being

symmetrical about the wagon centreline.

E X A M P L E . A short tube 0 - 9 in. outside dia. 0 - 1 in. th ick is to

carry a load the axis of which is parallel to and 0 - 5 in. from the

tube axis. Use the values of working stress given and est imate the

safe load assuming the tube to be made from (a) cast iron, (b) mild

steel. C.I. : 5 0 0 0 lbf/in2 in tension, 2 0 , 0 0 0 lbf/in

2 in compression.

M.S. : 1 5 , 0 0 0 lbf/in2 in both tension and compression.

Required modulus

2 3 0 0 χ 1 2

6 χ 2 2 4 0

= 2 - 0 5 i n3.

A

F F

- f r 1— Datum

Fia. 172

Solution

Inside diameter of tube = 0 - 7 in.,

Sect ion A = ^ ( 0 - 92 - 0 - 7

2) = 0 - 2 5 1 in

2.


2nd Moment of area / = (0-94 - 0-7

4) = 0-0204 in

4,

b4

0-0204 .'. Sect ion modulus Ζ = „ = 0-0454 in

3.

0-45

Bending moment M = 0-5F lbf in where F = Load.

Taking compressive stress as positive :

F M For cast iron, / m ax = — + ——-, where / m ux = 20,000 lbf/in2,

F 5F Ζ. 20 ,000 = - — - +

F

0-251 0 0 4 5 4

1 5

0-251 0 -0454 ,

= i^(3-99 + 1 1 0 )

= 1 4 - 9 9 ^ whence F = 1330 lbf .

A j , _ F M

And /min — ~^

- 5 0 0 0 = F(3-99 - 11-0) ,

= - 7 - 0 1 ^ whence F = 715 lbf .

The smaller of the two values of F is the safe load for cast iron.

Similarly, for mild steel, we have as before :

either / m ax = 1 4 - 9 9 ^ , but in this case, / m ax = + 1 5 , 0 0 0 lbf/in2,

. 15,000

·' ~ 14-99 '

= 1000 lbf approx.

or / m in = -101F, where / m in = - 1 5 , 0 0 0 lbf/in2,

15,000

7 0 1 '

2140 lbf .

The smaller of the two values of F is the safe load for mild steel.


Examples VI

1. Two rods 0-75 in. in diameter are used to reinforce the section of a concrete cantilever 5 in. wide and are positioned with their centres at a depth of 10 in. Calculate the bending moment required to induce in the con-crete a stress of 560 lbf/in

2 given that Ec = 1-79 χ 10

6 lbf/in

2 and that

Es = 15 Ec (59,500 lbf/in).

2. The horizontal flange of a T-section concrete beam is 18 in. wide and 4 in. thick and is reinforced by 8 rods 3/8 in. dia. with their axes 1-0 in. from the surface furthest from the web. If the web is uppermost and 4 in. thick and the total depth of the section is 18 in., calculate the moment of resistance of the beam assuming stresses of 10,000 and 500 lbf/in

2 and taking

the modular ratio as 12 (Reinforcement 128,900 lbf/in., concrete 105,300 lbf in.).

3. At a certain point the section of a casting is represented by the area enclosed by two circles 10 and 7 in. dia. with centres 1-0 in. apart. Calculate the 2nd Moment of area of the section about an axis normal to the line join-ing these centres and hence find the two values of the section modulus (298 in

4, 73-8 and 50-0 in

3).

4. A square section beam is loaded so that the neutral axis coincides with a diagonal. Show that the section modulus may be increased by removing the upper and lower apices of the section and that the maximum modulus value is obtained when the vertical diagonal has been reduced by one ninth.

5. That part of a G-clamp parallel to the screw axis is of T-section 0-125 in. thick throughout. The flange, which is nearer the screw, is 0-625 in. wide and the total depth is 0-75 in. Calculate the maximum values of tensile and com-pressive stress induced by a clamping load of 75 lbf given that the screw axis is 2-75 in. from the tip of the web (5650 and 9900 lbf/in

2).

C H A P T E R VII

TORSION

Modulus of Rigidity

The application of forces F F to the cube A B C D in Fig. 173 causes

the facte A B to turn through a small angle φ radians to the dotted

FIG. 1 7 3

position A B r This angle is defined as the shear strain.

χ Thus, Shear strain φ = radians (see Chapter I ) .

Within the elastic limit, the ratio Shear stress fs

is a constant . Shear strain φ

This constant is called the Modulus of Rigidity and denoted by G.

FIG. 1 7 4

7 S M 1 8 1


Thus 0 ^- or ψ - i l ψ u

Consider a solid shaft of radius R to be acted on by equal and oppo-

site couples, as shown in Fig . 174, so tha t the radius OA twists

through 0 radians with respect is the other end and adopts the

dotted position O B (Fig. 175). The state is one of pure shear.

FIG. 1 7 5

Clearly, Arc A B == R6 = Lcp,

R6 Λ '

/, _ RO . G6 χ R Λ _ _ _ , !.e. /, - L ,

= kR for a given value of Θ.

Thus for a given twist on a given shaft, the shear stress fs is propor-

tional to the radius, so tha t zero stress obtains a t point 0 .

f GO The equation is normally written — = -γ-, where r has any

value between 0 and R.

I t follows tha t if,

f's = stress a t any radius r, F ig . 176

fs = max. stress a t radius R,

T O R S I O N 183

then ' · ' '· 4 · · . R

Shear force on elemental ring

- f's2nr dr,

— — r 2ir.r j r 2nr dr, R

= -^.2nr2 dr.

Moment of this force about the polar axis

= ~^-2nr2dr x r ,

R

= -4 * 2 j w3d r .

R

Tota l moment of resistance to shear

η

Τ = A x2njr*dr,

= Ι χ 2 π ' Τ '

nR*

Hence, T =

R

nR*fK _ nD3fs

2 ~ 16

R

Radius

FIG. 1 7 6

(lbf in, if D is in inches and fs is in lbf/ in

2) .

This gives the torque which can be t ransmit ted by a given dia-

meter shaft for a given maximum shear stress.

Note: This expression does not apply to hollow shafts, for which see later.

F r o m * above, Τ = L R

nR* . The quant i ty in the brackets ,

nR* nD* i.e. —-— or _ is the polar 2nd moment of area of the shaft 32 section, J.

7*


f±=G6_

R L

Thus, Τ = fs R

χ J ,

i.e. fs R

Τ

T' but

Hence, fs R

Τ

Τ GO

17

This is known as the Torsion Equation and is similar to tha t for

Simple Bending.

Relation between Speed and Shaft Diameter for a Given Max.

Stress and Power

The power t ransmit ted by a shaft carrying a torque Τ lbf ft and

rotating at Ν rev/min is given by

_ 2πΝΤ

'Ρ' "~ 33,000 *

Hence, y = 33 ,000 (h.p.) 1 2π Ν

kx (-^r j if the power is constant .

F o r a solid shaft diameter d :

16 where fs = Max. shear stress,

OR d3 = χ T

7,

= k2T if fs is constant,

/. Nd* = k for a given power and max. stress.

Ï O H S I O N 185

I t follows that , for any given power and permissible shear stress,

the higher the speed, the less the required shaft diameter.

Suppose tha t 100 h.p. is t ransmit ted b y a 3 in. dia. shaft a t

50 rev/min. The diameter required to t ransmit this power a t 50 ,000

rev/min can be found from

E X A M P L E . Determine the shaft diameters required to transmit

15 h.p. a t the following speeds:

50

100 500 The stress in the shaft is not to exceed 3 tonf/in

2 and the

1000 angle of twist is not important . Assume pure torsion. 2000

5000 rev/min.

P lo t D vertically full size, against Ν horizontally and find from

the curve, the diameter of shaft necessary to transmit this power

at 300 rev/min.

Solution

r p nD*fH 33 ,000 h.p. Τ = — Γ 7 Γ - = 0 Λ7 x 12 lbl / in .

16 2πΝ

33,000 χ 15 χ 12 χ 16

2 π2 χ 3 χ 2240

7 1 6 .

Hence, D3 = , and Fig . 177 shows the graph.

i.e.

d - 50 ,000

6 '

~ 1 03 '

d = = 0-3 in.

D3N =


Ν

rev/min

7 1 6 = Iß

Ν D in.

5 0 1 4 - 3 0 2 - 4 3

1 0 0 7 - 1 6 1-93

5 0 0 1-43 1 1 3

1 0 0 0 0 - 7 1 6 0 - 8 9

2 0 0 0 0 - 3 5 8 0-71

5 0 0 0 0 1 4 3 0 - 5 2

2 0 0 0 3 0 0 0 4 0 0 0

Ν r e v / m i n

FIG. 177

E X A M P L E . A 4-0 in. dia. shaft 100 in. long rotates a t 1500 rev/

min.

Calculate (a) torque being transmitted, (b) power being trans-

mitted, (c) to ta l twist of shaft in degrees given tha t G = 12 χ ΙΟ6

lbf/in2 for the shaft material and tha t the measured maximum

shear stress is 10,000 lbf/in2.

Solution nd*

Polar 2nd Moment of shaft section J ----- ,

_ π x 44

= 32 '

25-1 in4.

T O R S I O N 187

Σ = 4* > Torque on shaft Τ = ^ ,

10,000 χ 25-1

2

- 125,000 lbf in. ,

- 10,400 lbf f t .

Power t ransmit ted = 2πΝΤ

33,000 '

2π χ 1500 χ 10,400

33 ,000

= 2970 h.p.

— = Angular twist θ = where R = 2-0 in. ,

10,000 χ 100

~ 12 χ ΙΟ6 χ 2 '

= 0-0416 radian,

= 0-0416 χ - ^ - , π

= 2-38°.

E X A M P L E . A solid circular shaft is to be made of material having

a modulus of rigidity of 12 x 1 0e lbf/in

2. Calculate the required

diameter i f i t is to transmit 100 h.p. a t 3000 rev/min under either

of the following conditions :

( a ) / s m a x> 6000 lbf/in2,

(b) 0 > 0ο8

χ per foot length.

Assume only the Torsion Equat ion .

Solution

Torque 33 ,000 χ h.p.

2^N lbf f t ,

= 2100 lbf in.


/, _ Τ 6000 _ 2100 ( a )

7~ ~ Τ ' ~dJ2~ ~ πάψ2 '

2 χ 6000 χ π # = 32 χ 2100 x d,

32 χ 2100 ri à

12,000 π 9

= 1-785,

.'. =

1-215 in. for safe stress.

— - — · 2 1 00

_ 12 x 1 06

/ 8 π

\ ( ' ~Τ~~~Ί7' - πά*β2 ' 12 \ 60 X 180 / 3

8 since 0 — deg.

32 χ 2100 8π χ ΙΟ6

i.e., d4

π # 60 χ 180

32 χ 2100 χ 60 χ 180

8π2 χ ΙΟ

6

- 9-2.

d2 = 3-03,

i.e. d = 1-74 in. for safe twist.

E X A M P L E . TWO lengths of 3 in. dia. shaft are connected by a

simple flanged coupling having 6 bolts at 8 in. pitch circle diameter.

Determine the bolt diameter so tha t the average stress in the bolts

shall be the same as the max. stress in the shaft when the la t ter

is transmitt ing 4 h.p. at 25 rev/min.

in

I

FIG. 178

T O R S I O N 189

Solution

33,000 x h.p. Τ

2πΝ

33,000 χ 4

2π χ 25

10,100 lbf/in.

12 ,

Τ

Τ 7"

Τ χ 32

where d = shaft diameter,

Τ χ 16

πί23

10,100 χ 16

π χ 33

1900 lbf/in2.

I f (i/, = bolt diameter, Resisting torque of bolts =

- 35 ,800 dl

Equat ing, 35 ,800 ig = 10 ,100 ,

dl = 0-282,

dh = 0-53 in.

6(0-785 df, 1900) χ 4 ,

E X A M P L E . The input to one end of a uniform line shaft is 30 h.p.

a t 150 rev/min. Three identical machines are belt-driven from

identical pulleys a t distances of 12 ft, 20 ft and 30 ft respectively

from the driven end. Take G = 12 χ 1 06 lbf/in

2, assume / i r a ax

= 9000 lbf/in2 and find (a) a suitable shaft diameter and (b) the

tota l angle of twist in degrees.

- 3 0 f t -

- 2 0 f t -

- 1 2 f t - -8ft- - 1 0 f t -

150 rev/min

3 0 h.p.

7 η S M

10 h.p. 10 h.p. 10 h.p.

FIG. 1 7 9


i.e. d3

2 χ 9000π '

7-13, whence d = 1-92 in. (say 2-0 in.).

Since the machines are identical, each pulley transmits 30 /3 , i.e.

10 h.p. The 10 ft portion thus carries ^ of the input torque and

hence the stress in i t is 9000/3 , i.e. 3000 lbf/in2.

Similarly the 8 ft portion carries f of the input torque and has a

maximum shear stress of χ 9000, i.e. 6000 lbf/in2.

0 ~-η£-, .'. Total twist - fl12 + fl8 - I - fl10,

9 θ ο ο ( 1 2 χ 1 2 ) + ^ ο ο ( 8 χ ] 2 )

Gr K

' Gr

+ i g » ( « . « . « . 12

— (10,800 + 48 ,000 + 30,000) rad, Gr

12 χ 186,000 180 , χ —^— deg, 12 χ 106 χ 1 0 π

180 0-186 χ

71

0 = 10-65 deg.

E X A M P L E . A 3*0 in. marine propeller shaft driven a t 100 rev/min is to be replaced by two identical shafts driven a t 720 rev/min and, between them, transmitt ing 80 per cent more power. I f the replacements are to carry a 20 per cent increase in maximum shear stress over the original shaft, est imate a suitable diameter.

Solution

τ m i 33 ,000 χ h.p. \ „ Λ 1 ί. Input torque Τ - — — ± — 12 lbf in ,

^ 3 , 0 0 0 x 3 0 x 12 ^ b f jn

2π χ 150

fs Τ . 9000 _ 12,600

Τ~'Τ' " ~äß~ ~ π ^ / 3 2 '

12,600 χ 32

T O R S I O N

Solution

I f Tx and N1 are the original torque and speed, T2 and N2 are the

torque and speed for replacement then, for the same power :

2πΝ1Τ1

33,000 χ 1-8

I I τ9

/ 2 π 2 \ Γ 2Τ 2

\ 33 ,000

2 Ν2

2 χ 720

1-8 χ 100 '

8-0.

I f d1 and fSl are the original diameter and maximum stress, d2 and

fS2 are corresponding values for the replacements,

nd\fS2 then

i.e.

nd\fs

1 1 6 and T9 =

16

and / e a = 1 .2 / Ä i.

or

1 2 ,

8-0 χ 1-2,

9-6.

Hence, 2-12, where dx = 3-0 in.,

3 0

2-12 '

- 1-42 in.

Hollow Shafts

I n this case the maxi-mum shear stress

fs occurs when r = . FIG. 1 8 0

7;I*

191


Hence, from the torsion equation :

—Γττ- = —r , where J dJ2 J '

U

nd\ nd\

3 2

π

~32~

3 2

(df-di).

Safe torque, Τ - -f^ χ i L (d\ - d\),

i.e. Τ = nfs ( df — d\

1 6 lbf in usually.

E X A M P L E . At what speed must a hollow shaft 6 0 in. long, 3 in.

inside dia. and 6 in. outside dia. rotate in order to t ransmit 5 0 0 h.p.,

assuming fs > 6 0 0 0 lbf/in2 ? Through what angle will the shaft

twist if G = 1 2 χ 1 06 lbf/in

2?

Solution

π J

( 1 2 9 6 - 8 1 ) 3 2

π x 1 2 1 5

3 2

= 1 1 9 in4,

dr τ

h.p. = 5 0 0

Required speed Ν

fsL 6 0 0 0 χ 1 1 9 2 3 8 , 0 0 0 lbf in.

Angle of twist θ =

2πΝΤ

3 3 , 0 0 0

5 0 0 χ 3 3 , 0 0 0

2π χ 2 3 8 , 0 0 0

1 3 2 rev/min.

TL

~GJ '

2 3 8 , 0 0 0 χ 6 0

1 2 χ 1 0 6 χ 1 1 9

0 - 0 1 radian,

0 - 5 7 deg.

χ 1 2 ,

T O R S I O N 193

E X A M P L E . A 4 in. dia. solid shaft is to be replaced by a hollow

shaft 6 in. outside dia. F ind the thickness required to make the

new shaft (a) of equal strength and (b) of equal weight. I n case (a)

determine the ratio of the weights of old and new shafts and in

case (b) the ratio of the strengths.

Solution

Sol id (s) Hol low (h )

FIG. 181

(a) The replacement must carry the same torque,

i.e.

16 16 \ dx

χ d*

d* d \ - d \

Weight hollow

Weight solid

d\ = d \ - dxd*,

= 64 - (6 χ 4

3) ,

= 1296 - 3 8 4 ,

= 9 1 2 .

d2 = 5-5, i.e. thickness

6 0 - 5-5 . = = 0-25 m.

2 0-785(df - d\) LQ

0-785d2 LQ

(? 62 dl - dl ö2 - Γ>·5

2

36 - 30-2 5-8 3

i 6 =i r=

8 a p p r o x -

i.e.


(b) The replacement must have the same cross sect ion:

i.e. 0-785(d2 - d\) = 0-785 d

2.

d \ - d \ = d2.

di = d \ - d2,

= 62 - 4

2,

- 36 - 16 ;

- 2 0 . 6 0 - 4-47

d2 = 4-47 in., i.e. thickness = ---= 0-76 in.

Torque hollow 16 \ d1 ) d\ - rff

Torque solid nfs dß*

_ 64 - 4 -47

4

6 χ 43 '

1296 - 400

6 x 6 4 '

896 7 = " 3 8 4 ' = y W °

x-

E X A M P L E . A torsion spring consists of a steel shaft anchored a t

one end and enclosed b y a steel tube to which i t is rigidly fixed a t

the other end. (Fig. 182.) The arrangement is as shown, the force being applied a t point L .

Determine, for an allowable shear stress of 10,000 lbf/in2,

(a) the rat io of the maximum shear stresses in tube and shaft,

(b) the maximum torque which may be applied,

(c) the angle through which the free end of the tube turns.

The shaft diameter is 1 in. and for the tube the diameters are 1-414 in. outside and 1-300 in. inside, the act ive lengths of each being 24 in. Assume pure torsion and t ake G = 12 χ 1 0

6 lbf/in

2.

Solution TRD

4 7T χ l

4

F o r shaft, J , - ^ L - - — = 0-0982 in4.

F o r tube, j t = JL (Z)4 _. di) = ο.0982(1·4144 - 1-3

4) ,

T O R S I O N 195

FIG. 182

= 0-0982(4 - 2 -85) ,

= 0-113 in4.

Since

hence, for the same torque

Τ

Ύ

fstJ t

fs,

k r

9

J

Ji X rs

0-0982

~ Ö 4 3 ~ X

1-23.

Maximum torque, Τ = —— f s J 10,000 χ 0-113

1-414/2

1-414

1

= 1600 lbf in.

Twist a t free end θ = ds + 6t, and each has the same length,

TL TL

+ GJS

TL

G

1600 χ 24

GJt

V 0-09 12 χ 1 0e

3200

+ 1

0982 0 113

(10-19 + 8-85), 1 0

e

_ 3200 χ 19-04

= 0-06 radian or 3-44 dec.


E X A M P L E . A steel shaft 2 in. dia. is closely surrounded by, and

firmly a t tached to a brass liner 3-5 in. outside diameter. I f the

permissible shear stresses for steel and brass are respectively 3

and 2 tonf/in2, est imate the power which may safely be t ransmit ted

by the compound shaft a t 400 rev/min. ( G B r a ss = 0-45 χ 6r St e ei )

Solution

fs ΟΘ — = —— and tor any one r L

J

section θ and L are constant

A r

f.

i.e., oc G

G oc r.

When, a t the common radius,

or

r = 1,

/%ax /]îmin

f Βτρίη

/ Κ max — χ /

FIG. 1 8 3

1-75

1

= 0-45 χ 3 χ 1-75 (assuming / S m a xt o be the criterion),

= 2-37 tonf/in2,

which exceeds the permitted value, i.e. / L W must be the criterion.

/ = ^ x Cr β

1

1-75

1 χ

x / i w

x 2 ,

For steel, J s = — χ 24 and T{

0-45 1-75

= 2-53 tonf/in2, which is within the limit.

/s«^s where r = 1, r

and fs = 2-53, 2-53 / π χ 2

4 '

1 \ 32

3-97 tonf in .

For brass,

and

T O R S I O N

r

2

3-5

where r = 1*75 and / B

π / 3 ·54 - 2

4 \

1-75 32 \ 3-5

= 4-28 tonf in.

2240 Tota l torque = (3-97 + 4-28) — — ,

\ Δ

= 1520 lbf f t .

2 χ 400 χ 1520 Power at 400 rev/min =

33,000

117 h.p.

Torsional Strain Energy

F rom the torsion equation for a circular shaft,

so tha t the T-θ graph has the linear form shown in Fig . 184.

FIG. 1 8 4 FIG. 1 8 5

197

π / 3 ·54 - 2

4


The shaded area represents the work done in twisting one end

of a shaft (Fig. 185) through an angle θ relative to the other end,

this work being (within the elastic l imit) stored in the material as

strain energy.

Torsional strain energy UT = Average torque χ Angle of twist,

= L x 0, where Τ = ^L χ θ,

Δ L GJO

2

Alternatively,

Again,

2L

Τ

UT

2

T2L

~2GJ '

Τ — χ 0 ,

χ 0 , where 0 TL

~GJ

where Τ = —

and θ

ί£ r

]jL_ rG

1 / , / w isL 2 r rG

fl LJ

~G 2 r2 '

L

where J

G X as" X

"32" a nd f2

32 '

d?_

IT

~G~ X

64

nd2L A.

4G

B u t nd

2L

— Volume of shaft, i.e. UT iL 4 0

4

x Volume

(where fs = max. shear stress).

I f G and fs are in lbf/in2, then the strain energy will be in in. lb.

Coil Springs

A spring may be compressed or extended (as is most common) or twisted axially so as either to wind or unwind it . I n each case

T O R S I O N 199

the work done in deformation (within the elastic l imit) is conserved

in the form of strain energy.

A spring m a y therefore fulfil one of the following functions:

(a) prevention of shock by absorption of kinet ic energy (e.g. vehicle

spring),

(b) restoration of a deflected mechanism to its position of rest

(e.g. indicating instrument control spring),

(c) release of energy under controlled conditions (e.g., clock spring).

Helical Spring—Effects of Axial Load

An axia l load F

(Fig. 186 (a)) can be resolved

into two components acting

a t point Ρ :

1. F cos θ normal to the

axis of the wire,

2. F sin θ parallel to the

axis of the wire.

At point Q on the wire

axis the component F cos 0

tends to produce the effect

shown in F ig . (187(a)) i.e.,

i t induces a direct shear

stress of

F cos θ . . nd%

γ. where A = ——

Λ

A 4

(d being the wire diameter) .

Since F cos Q also acts a t

an instance Ε from point Q

i t applies a torque

Τ = F cos θ x R FIG. 1 8 6

to any section X X (Fig. 186(b)) and hence over all such sections

there will be an additional shear stress increasing from zero a t

the wire axis to a maximum given by

Tr Λ d ——-, where r = —.

J Δ

(This is relatively much greater than the direct shear stress).


F cos0

FIG. 187

F sin Ο

Fcos0

FIG. m

(b)

(α)

(Α)

(b)

T O R S I O N 201

Considering again point Q on the wire axis, the component

i^sinö tends to produce the effect shown in F ig . 188(a) i.e., i t

F sin θ nd2

induces a direct tensile stress of , where A = ——. A 4

This component is also responsible for a bending moment M =

Fsind χ R (Fig. 188(b)) which tends to increase the curvature of

the wire and hence the coil radius R. The maximum bending stress My d nd*

is given by / = — — , where y = —- and I = —-.

As already shown T2L

Torsional strain energy UT = and Τ = FR cos θ,

(FRcose)*L

2GJ '

_ F*R2L I cos20 \

~ 2 \~GJ~I'

Bending strain energy UB = j M2 dx

ο

(see chapter on strain energy of beams). L

M2 r Since M is constant, UB = | do;

2El J ο

M2L and M = FR s inö , 2 # /

(FR sin θ) 2

£

2 7

i ^2. ß

2£ / s i n

20

^ 7

Tota l strain energy U = UT + ? 7 ß

.F272

2L / cos

2<9 \ ^

2. ß

2£ / s in

20

2 \ GJ ) 2 \ EI

F2R

2L î cos

20 s in

20

GJ EI


B u t Work done in deflection = Average force χ Deflection,

„ . F „ F*R*L / c o s2φ s in

20

Lquatmg, 6 = — g — +

Now cos θ = 2 π

^ç

j where ç = number of coils,

Length of wire m spring, L cos Ö

Q w w # r · , 2nFR3n / cos

2Ö sin

2Ö\

SubBtitutmg for £ gives : ί ^ _ _ ß _ + _ j

Although this and previous expressions are based upon the

assumption tha t R and θ are constant , whereas they both vary

slightly with the load F, t hey are in practice sufficiently accurate.

When θ is small ( < 10 deg say) s in2 θ is negligible and cos θ

^ 1-0. The material may thus be considered as being in a state of

pure torsion and the spring described as "c lose-coi led" .

Put t ing cosφ = 1, and ignoring sin2φ gives

2nFRH • T nd* Τ=—GJ—>

W h e r e J s = l T '

_ 2nFR*n 3 2 ~ G

X ~JVF'

64:FR3n

Whence axial deflection

T O R S I O N 203

This expression may be obtained more simply by assuming pure

torsion, as follows :

Work done by F = Torsional strain energy,

F Τ

TO

; J = y , where Τ ^ FR

(Fig. 1 8 9 ) ,

= RO and θ TL

GJ

FRL

GJ 9

= R χ FRL

GJ 9

FR2L

GJ

FR2

, and L = 2nRnm&v\y,

GJ

2nFR*n

χ 2nRn,

η co i l s

GJ ' W h e r e J = l f '

2nFR*n 3 2

G x

Fia. 189

„ , 64, F RH Hence, ο = -r-^— as before.

Gd*

E X A M P L E . Wire 6 ft long and 0 - 1 9 2 in. dia. is formed into a helical spring of 2 in. mean coil dia., a to ta l of 8 in. being allowed for the end fastenings. F ind the axial stiffness in lbf/in.

I f / S > 5 5 , 0 0 0 lbf/in2 and G = 1 2 χ 1 0

6 lbf/in

2, find the maximum

permissible axial load.


Solution

IIIIIIUIIIIIIIIIIIII

w

2 i n -

F I G . 1 9 0

F o r wire,

Τ

Τ

0 - 1 9 2 in

2nRn = 6 ft less 8 in

(2π χ l ) w = (6 χ 12) -

64 .'. η = —— ,

2π

= 10-2 turns.

64:FR*n

k r

or

i.e. F

, Axial stiffness,

F Gd*

~ Ύ ~ 64 RH

12 x 1 0e (0-192)

4

64 χ 13 χ 10-2

= 25 lbf in.

J =

J A =

π ί4

" 3 2 " '

π χ 0-00136

32

0-000133 i n4.

r

U_

r

rR

55,000 x 0-000133

0-192 \

2

76-5 lbf .

x 1

T O R S I O N 2 0 5

E X A M P L E . A helical spring has 1 0 close-coiled turns the outside

dia. of which is 5 t imes tha t of the wire. I f the spring is to stretch

0 - 5 in. under an axial load of 3 0 0 lbf, find (a) the necessary wire

diameter and (b) the max . shear stress.

Take G = 1 2 χ 1 06 lbf/in

2.

Solution

Mean diameter

= M - d,

= 4d,

R = 2d

Deflection

_ 64:FR3n

Gd* 9

6 4 χ 3 0 0 (2df 1 0

* * 1 2 χ ΙΟ6 χ d* '

A

F I G . 1 9 1

_ 6 4 χ 3 0 0 χ 23 χ 1 0

~a¥ ~ 1 2 χ ΙΟ6 χ 0 - 5 '

giving d = 0 - 2 5 6 in. (i.e. r = 0 - 1 2 8 i n . ) ,

d* = 0 - 0 0 4 3 ,

J = =L K

X 0 0 0 4 3'

= 0 - 0 0 0 4 2 2 in4.

Tr

Stress / , = — where Τ = FR,

_ ( 3 0 0 χ 2 χ 0 - 2 5 6 ) 0 - 1 2 8

~ 0 - 0 0 0 4 2 2 '

- 4 6 , 6 0 0 lbf/in2.

E X A M P L E . A length of wire of diameter 0 - 1 4 4 in. is formed into

a spring with a mean coil diameter of 0 - 8 7 5 in. W h a t axial load will

induce a maximum shear stress of 5 0 , 0 0 0 lbf/in2?

Es t imate the number of coils required to give a deflection of

1 - 7 5 in, a t this load. G - 1 2 χ 1 06 lbf/in

2.

206

Deflection


Τ kL } where Τ — FR, r

0-875

2 W - 0-4375i<

7.

/ , - 5 0 , 0 0 0 ,

0 1 4 4

2 = 0-072 in. ,

J „ ( 0 - 1 4 4 )4

,

— x 0-00043

0 0 0 0 0 4 2 2 in4.

0-4375 J

F

50,000 x 0 0 0 0 0 4 2 2

0 0 7 2

50,000 x 0-0000422

0 0 7 2 x 0-4375

67 lbf.

64:FR*n

Gd*ô where R

0-875

64 FR* ' 2

12 χ ΙΟ6 (0-144)

4 x 1-75

64 x 67 ' 0-875 \

3

2 J

12 x 1 0e x 0-00043 x 1-75

64 x 67 x 0-084

— 25-1 coils.

E X A M P L E . A helical compression spring is made of steel rod 0-3

in. dia, and has 10 coils of 1-7 in, outside dia, I t s free (unloaded)

Solution

giving

rL A J " r

T O R S I O N 207

Axial load in 112 224 336 448 560 672 784 1120

Compression in inches

0-26 0-51 0-77 1-04 1-29 1-55 1-81 2-59

Plo t the load vert ically on a base of compression. Selec t a con-

venient pair of values from the graph, find the slope and est imate

(a) the value of G and

(b) the twist per unit length a t maximum load.

Solution

Mean coil radius

Polar 2nd Moment

R = 1-7 - 0-3

2

0-7 in.

I F '

- ^ ( 0 - 3 ) * ,

- 0-000795 in4.

F From the graph slope, ^ = 4 2 6 .

ο

1-7 in

- 0 - 3 in

FIG. 192

4 2 6

J 1 0

δ, i n c h e s

FIG. 1 9 3

length is 10 in. A tes t gave the following figures :


Deflection UFRH

G =

G6

L

Gd* 9

~ d*

64(0-7)3 χ 10 χ 426

(0-3)4

11-6 x 1 06 lbf/in

2 (approx.

T_

~J9

Twist /Unit length — L

Τ

~GJ where Τ

1120 χ 0-7

FR,

11-6 χ ΙΟ6 χ 0-000795

0-085 radian/in.

a t max . load,

E X A M P L E . A helical spring is to have a mean coil diameter of

4 in. and is to deflect 6 in. under an axial load of 1250 lb, the cor-

responding shear stress being then > 25 tonf/in2.

I f wire is available in the following diameters : \, g, JJ,

and f in., find the most suitable, the number of coils required

and the actual maximum shear stress, assuming G = 12 χ ΙΟ6

lbf/in2.

Solution

r J 9

25 χ 2240 1250 χ 2

d\2

d*

π # / 3 2

1250 χ 2

. Since Τ = FR,

χ 32

Deflection

25 χ 2 χ 2240π

= 0-227.

d = 0-61 in. , say 0-625 in . (i.e. § in.)

ô = R6, where à = 6 in.

R = 2 in.

TL

2 TL

GJ

0 = GJ '

and L = 2nRn = 4πη.

Τ = 1250 x 2 = 2500 lbf in.

- 0-015 in4.

π χ 0 -6254

32

6 = 2 χ 2500 χ 4πη

12 χ ΙΟ6 χ 0 0 1 5

T O R S I O N

i.e.

Stress

6 χ 12 x 1 0e χ 0 0 1 5

500 χ 4π

17-2 coils.

Tr

2500 0-625

0 0 1 5 2

52,000 lbf/in2,

23-3 tonf/in2.

E X A M P L E . A helical spring 5 in. mean diameter is required to

absorb 6 inch tonf of energy with a maximum shear stress of

35 tonf/in2. F ind the diameter of steel rod required and the num-

ber of coils if the maximum amount of compression is to be 6 in.

Take G = 5400 tonf/ in2 ( I . Mech. E . )

Solution

Strain energy

4.Γ* 1

3 52

4G per unit volume.

6 = -377(2πΒ* x 4 r < 4(Τ \ 4

2π x 2-5n x - y - j · 4 χ 5400

6 χ 4 χ 5400 χ 4 d

2n

3 52 χ 5 π

2

i.e.,

= 8-56.

8-56

η F

Energy absorbed = — ,

F 6 = — χ 6 giving F = 2 tonf

. Δ

ι κ ζ

Ο Ο Ο Ο Ο

Ό1

Ο Ο Ι ο ο

- 5 in dia-6 in

FIG. 1 9 4

209


~Gd* *

64 χ 2 χ 2-53?a

5400 d*

0 0 6 1 7 ,

i.e. d* = 0-0617 η

Hence, 0-0617^

nz

Required diameter d

E X A M P L E . A petrol engine valve weighing 1 lbf is to be given a

maximum acceleration of 1000 f t / s2 occurring a t a lift of 0-205 in.,

the to ta l lift being 0-475 in. Es t ima te the maximum force in the

spring if i t has a free length of 6-5 in. and is 4-5 in. long when the

valve is shut. Determine the mean diameter of the coils and their

number, taking a wire diameter of 0-128 in. and working a t a

maximum shear stress of 50 ,000 lbf/in2. (G = 12 χ 1 0

6 lbf/ in

2) .

Solution

Ini t ia l compression = 6-5 — 4-5 = 2 in. W

At 0-205 in. lift, i.e. δ = 2-205 in. Acc . force F = /, 9

= w x 1 0 0 0'

= 31-1 lbf.

2-475 Hence a t 0-475 in lift, i.e. δ = 2-475 in. Max. force = — — χ 3 1 - 1 ,

Δ' Δ\)θ = 34-9 lbf.

Deflection

d*

η

_ 8-562

n2

8-562

" 0 0 6 1 7

= 1185 , giving η = 10-6 coils.

1 / 8 - 5 6

- j /0-806 - 0-899 in.

T O R S I O N 211

When d = 0-128 in., J = .

= 0-0000263 in4.

·— , ± —

J r r ο. Λτ >

5 0>

0 0 0 x 0-0000263

34-9 R —

giving R =

0 1 2 8 / 2

50,000 χ 0 0 0 0 0 2 6 3

34-9 χ 0 0 6 4

- 0 - 5 9 i n . , .·. D = 1-18in.

A = , Ζ/ == r

^ , where L = 2nRn,

R

.*. 2 π / ^ = - 7 — x — , fs R

i.e., w = r <

? ^ D 9 , where ό = 2-475 in . lnfsR*

0-064 χ 12 χ ΙΟ6 χ 2-475

2π χ 50 ,000 χ (0-59)2

= 17-35 coils.

E X A M P L E : A compound spring consists of two co-axial close-

coiled helical springs as follows :

Mean coil dia. Ό in. Wire dia. d in. Coils η Free length

Outer 1-75 0-192 10 4 in.

Inner 1-25 0160 8 3-25 in.

I f the combination is subjected to an axial compressive load

of 100 lbf, determine for each spring (a) change in length, (b) load

carried and (c) max . stress.

T a k e G = 12 χ 1 06 lbf/in

2.


Solution

64:FRhi

Od*

F .'. Stiffness λ = — =

Od*

Gd*

64:R3n '

W=IOOIbf Hence for outer spring,

12 χ 106(0 ·192)

4

0 7 5 in

3-25 in

4 in

8 χ (1-75)3 χ 10

38 lbf in .

-1-25 in

-175 in—*j

The outer must be compressed by 0-75 in before the inner begins com-pression. .'. Load to compress outer by this

amount = 0-75 χ 3 8 ,

= 28-5 lbf.

The remaining load of 71*5 lbf now fafo compresses both springs simultane-I Τ

1 ously.

I Again, for inner spring,

12 χ 106(0 -160)

4

λ2 = FIG. 1 9 5

8(1-25) 3 x 8

62-9 lbf in.

.*. Combined stiffness = λ1 + λ2 = 38 + 62-9 ,

= 100-9 lbf/in.

71-5 Common reduction

100-9 0-708 in.

F o r outer δλ = 0-75 + 0-708. F o r inner δ2 = χ = 0-708 in.

= 1-458 in . F9

Fx = Vi - 38 χ 1-458,

= 55-45 lbf .

λ2ο2, 62-9 χ 0-708,

44-55 lbf .

Τ ^ = Ί 7 Γ = 1 1 " 2 " ~7td\ = F l R l Tri*

= 55-45 χ — — χ — χ 2 π

= 34,900 lbf/in2.

— V 0 - 1 9 2 / '

T O R S I O N 2 1 3

And snndarly: = F2B2 — = 4 4 - 5 5 χ _ χ — χ ( _ )

= 3 4 , 9 0 0 lbf/in2.

E X A M P L E . An open-coiled helical spring has 2 0 coils made of 0 - 5 in. dia. steel. The mean coil radius is 5 in. and the coils make an angle of 3 0 ° with the horizontal when the unloaded spring is suspended vertically.

I f Ε = 2 · 5 £ - 3 0 χ 1 06 lbf/in

2 find the axialdeflection for a load

of 4 0 lbf (a) assuming pure torsion and (b) taking bending effects into account .

W h a t approximate percentage error results from making assump-tion (a)?

Solution

(a) Pure Torsion

à = M

g ^n , where Β = 5 i.e., B* = 1 2 5 ,

d - 0 - 5 i.e., d* = 0 - 0 6 2 5 ,

G = 3 0

* 1 06

= 1 2 χ W lbf/in2.

_ 6 4 x 4 0 x 1 2 5 χ 2 0

1 2 χ Ι Ο6 χ 0 - 0 6 2 5 '

= 8 - 5 3 in.

(b) Torsion with bending

2nFB*n / cos20 s in

2ö \ _ _ o _ _

δ= 7Γ-(—γΓΓ- + —ΈΤΓ-)> w h e re

E=25G and cos 0 \ GJ EI j

J = 21 for a circular section,

_ 2nFB*n (cos26 s in

20 \

~~ cos θ [ 2GI +

2-5GI j '

_ 2nFB*n I cos26 s in

20 \

^ GI cos Θ \ ~ ~ 2 ~ ~ +

~ ¥ 5 ~ " j


and cos θ = 0-866 i.e., cos20 = 0-75

sin θ = 0-5 i.e., s in20 - 0-25

64

π χ 0 -54

64

= 0-00307 in4.

2π χ 40 χ 125 χ 20 / 0-75 0-25 \

~ 12 χ ΙΟ6 χ 0-00307 χ 0-866 [ 2

+ 2-5 / '

= 19-66(0-375 + 0-1) ,

- 9-35 in.

Thus assumption (a) underestimates the deflection.

1. Find the strain energy stored in the material of a hollow shaft 48 in. long, 2-0 in. inside dia., 4-0 in. outside dia. when subjected to a torque of 800 lbf ft. Take G = 11-8 χ 10

6 lbf/in

2 (6-3 inch lbf).

2. Determine the torque which will induce in a piece of steel tube 2-0 in. outside dia., 0-125 in. thick, 30 in. long a maximum shear stress of 5 tonf/in

2. Find also the angular displacement of one end relative to the other

when this torque is applied. Take G = 5200 tonf/in2. What power could this

tube transmit under the above conditions? (7-75 tonf in., 1-65 deg, 410h.p.).

3. Design a close-coiled helical spring having a mean coil diameter of ten times the wire diameter, an axial stiffness of 120 lbf/ft and a safe deflection of 1-0 in. Assume a value for G oî 13 χ 10

6 lbf/in

2 and allow a maximum

stress of 15,000 lbf/in2 (21 turns, 1-3 in. dia.).

4. A hollow steel shaft 2-5 in. outside dia. is to be connected via a clutch, to a solid alloy shaft of the same diameter. If the torsional rigidity of the steel shaft is to be 0-8 of that of the alloy shaft, calculate the required inside dia-meter given that Oa = 0-46rs (2-1 in.).

5. The coils of an open-coiled helical spring make an angle Θ with planes normal to the spring axis. If the extension is calculated using the formula for close-coiled springs, find the value of Θ corresponding with an error of 1 per cent (8-1 deg).

Error

82

9-35 '

= 8-8 per cent (low approx).

Examples VI I

C H A P T E R VIII

COMPLEX STRESS I

Shear Stress Resulting from a Tensile Load

A tensile load F produces a direct tensile stress on planes such as

C B (Fig. 196) normal to the load axis, and this stress is given b y :

1 A '

The load m a y be re-

solved into two compo-

nents when considering

i ts effect on any plane

such as CD inclined 0 to

C B :

1. F cos 0 normal to CD

acting over the area

A /cos 0 and intro-

ducing a normal ten-

sile stress given by

i ^ c o s 0

where A = Section.

A /cos 0

F = — c o s

20 .

A FIG. 1 9 6

This is a maximum when cos θ is a maximum, i.e. when 0 = 0 .

Then F

2. F sin θ parallel to CD acting over the same area A /cos 0 and

introducing a shear stress given b y

_ .Fsine '* ~~ ~~A~i

F . = — sm I

/ A J cos θ A

costf == —^7- sin 2 0 . 2A

8* 2 1 5


This is a maximum when sin 26 is a maximum

i.e., when 26 = 90 deg or 6 = 45 deg.

JL = L 2A 2 '

Then,

FIG. 1 9 7

Thus on planes inclined a t 45° to the axis of

the applied stress, the induced shear stress has

half the value of the tensile stress on sections

normal to the load axis.

I t follows that , if the shear strength of a material

is less than half i ts tensile strength, the material

will fail in shear when subjected to a tensile load.

Similarly, if the shear strength is less than half

the compressive strength, failure in shear will

occur during compression. Cast iron fails in this

manner in compression (Fig. 197) .

Complementary Shear

Suppose a shear stress fSi to be induced on opposite faces of a

rectangular element by forces Fx applied as shown in Fig . 1 9 8 :

F[ inducing fS|

F2 -

1 FIG. 1 9 8

Then, clockwise shear couple = Fxx, where Fx = fSl (yz).

I f the equal anticlockwise couple required for equilibrium is

introduced by forces F2 acting (as shown dotted) on the other two

faces.

then. F2y = F,x.

C O M P L E X S T R E S S I 217

I f fSt is the shear stress induced by F2 then F2 = fSl (xz).

Hence, fs2(xz) V = f»I(Vz) x> substituting for Fx and F2.

i.e., Is, /.,· Thus if a body remains in equilibrium under the action of a

system of forces two of which induce a shear stress, an equal shear

stress is automatically in existence on planes a t 90 deg. This is

referred to as a Complementary Shear.

Applied shear

Complementary shear

(a)

From Fig .199 (a) it can be seen that , due to the two shear couples,

the diagonal face A B will be under compression. I f / , v = normal

compressive stress required for equilibrium of portion A D B , then,

resolving forces horizontally :

fx(pS χ S) cos 45 = / , χ S2, where S = length of edge,

i.e. fNpS2 x ~ fsS\

whence, / N = fx.

Clearly there will be a tensile stress (of equal magnitude to the applied shear stress) on the other diagonal section of the cube (CD) as in Fig. 199(c) .

Bulk or Volumetric Strain

As a result of an elastic linear tensile strain on a linear dimen-

sion L,

New length = L + dL9 where dL = increase,

= L + eL where e = strain,

- L(l <•)


Similarly, after compression,

New length — L(l e)

Suppose a rectangular element χ x y χ ζ, to be immersed in liquid

as shown in Fig . 200 . The force acting on each face will bring about a

reduction in the length of each edge

and hence in the volume, i.e., i t will

produce a volumetric strain.

I f the strains in the three mutu-

ally perpendicular directions are

ex, ey and ez then,

New volume

= x(l - ex)y(l - ey)z(l - ez),

= xyz(l + exey + eyez + ezex

FIG. 200

I f the elastic limit is not exceeded

the strains are very small and their

products may be neglected, so tha t

New volume = xyz + xyz ( — ex — ey — e.) ,

= xyz - xyz(ex + ey + β . ) .

B u t , xyz = Original volume.

.'. Volume change = New volume — Original volume

= — xyz(ex -\-ey-\-ez) (the negative sign

indicating a reduction)

-xyz(ex + ey + es) i.e., Volumetr ic strain ev

- [e.

xyz

e,, + ez)

Thus the volumetric strain is the sum of the linear strains, the

negative sign in this case indicating tha t i t is compressive.

F o r an elemental cube, χ = y = ζ.

Hence, ex = ey = ez = e, say.

Hence, ev = — 3e ,

= 3 χ Linear strain.


E X A M P L E . Calculate the reduction in volume of a steel cylinder

2-5 in. dia., 10-0 in. long, when submerged in the sea to a depth

of 36 ,000 ft. Take Κ = 23-3 χ 1 06 lbf/ in

2 and assume tha t the

density of sea water remains constant a t 0·036 lbf/in3.

Solution

The compressive stress is equal to the pressure, i.e., i t is the

product of depth and density

f = hq where h = depth and ρ = densi ty,

= (36,000 χ 1 2 ) 0 - 0 3 6 ,

= 15,500 lbf/in2.

Volumetric strain e v = (see p. 227 ) ,

_ 15,500

23-3 χ 1 06 '

= 0-000667.

Original volume V = — χ 2 ·52 χ 1 0 ,

= 49 in3.

Reduct ion in volume = evV,

= 0-000667 χ 4 9 ,

- 0-0328 in3.

Poisson's Ratio

I n addition to the change in length in the direction of a direct load there is a simultaneous dimensional change in the two direc-tions normal to such a load. Thus a tensile load produces a tensile (longitudinal) strain along i ts own axis and a compressive (lateral) strain along the other two axes resulting in a reduction in the corresponding dimensions.

Within the elastic l imit the ratio of lateral to longitudinal strain is constant for a given material and referred to as Poisson's Ratio.

I t is denoted by σ or 1/m where, for most metals, a lies between 0-25 and 0-33, i.e. m lies between 3 and 4.


Thus Poisson's Ra t io

i.e. Latera l strain

Latera l strain

Longitudinal strain

σ χ Longitudinal, s train,

1

m

Ε f

mE

Since lateral strains are of opposite sign to longitudinal strains

caused by the same load, it follows tha t in the case of three mutually

perpendicular direct loads, the resultant strain in

the direction of any one of them is the algebraic sum

of the relevant longitudinal and lateral strains. fy=6'75

fv

FIG. 201

Solution

B u t ,

E X A M P L E . A tes t piece 2-0 in. wide, 0-5 in. thick

has a gauge length of 4-0 in. Take Ε = 13,500 tonf/in2

and a = 0-25 and estimate, for an axial load of

6-75 tonf :

(a) longitudinal strain,

(b) lateral strain,

(c) volumetric strain,

(d) change in volume.

TU =

6-75

2 χ 0-5

T„ _ 6-75 Ε 13,500

/ , = 0 ,

e. = - ae,j,

= - 0 - 2 5 χ 0 0 0 0 5

= - 0 0 0 0 1 2 5 .

er = eu + ec + ez,

eu + ( - 2 e , )

6-75 tonf/in2,

0 0 0 0 5 ,

0 0 0 0 5 - (2 χ 0 0 0 0 1 2 5 )

0-00025.

C O M P L E X S T R E S S I 2 2 1

Original volume V = 2 χ 0 - 5 χ 4 , =^ 4 - 0 in3,

Volume change = evV = 0 - 0 0 0 2 5 χ 0 - 4 ,

= 0 - 0 0 1 in3, (increase.)

E X A M P L E . I f the test piece in the previous question is also

subjected to an additional horizontal stress of 4 - 0 tonf/in2, the

stress in the remaining direction being zero, calculate the vertical

and horizontal strains when the second stress is (a) compressive

(b) tensile.

Solution

Vertical Strains :

(a) The horizontal compressive stress pro-

duces a reduction in width and hence

an increase in length, i.e., the lateral

strain due to i t is added to the longitu-

dinal strain due to fu.

.'. Resul tant vertical strain

*» = Ε Ε ^- , taking tensile strain as

positive,

1 3 - 5 0 0

7 - 7 5

1 3 , 5 0 0

[ 6 - 7 5 + ( 0 - 2 5 χ 4 - 0 ) ] ,

= 0 - 0 0 0 5 7 3 (tensile).

(b) The effect of fx in a vertical direction is to reduce the strain due to fy.

/// <*ίχ_ Ε '

fy =6-75

f x = 4 - 0

FIG. 2 0 2

f y = 6 -75

'•e- β , = E

1

1 3 , 5 0 0 [ 6 - 7 5 - ( 0 - 2 5 χ 4 - 0 ) ] ,

5 * 7 5

1 3 5 QQ = 0 - 0 0 0 4 2 6 (tensile) FIG. 2 0 3

* fx =4 -0

8;i SM


Horizontal Strains :

I n direction of fx : (a) ex = — -4- -êr (effects of fx and fy are

cumulative),

1 • [4-0 + (0-25 χ 6-75)] ,

13,500

5-69

13,500 = — 0-000421 (compressive).

1 [4-0 - (0-25 χ 6-75)] ,

13,500

2-31

13,500 0-000171 (tensile).

In direction of fz (zero applied stress) :

(a) ez=J±-.?k + ^ , Wh e r e / , = 0 ,

1 [ - ( 0 - 2 5 χ 6-75) + (0-25 χ 4 -0 ) ] ,

13,500

- 0 - 6 9

13,500 -0-000069 (compressive).

1 [ - ( 0 - 2 5 χ 6-75) - (0-25 χ 4-0)]

13,500

- 2 - 6 9 _ 0 · 0 0 0 1 9 8 (compressive).

13,500

E X A M P L E . The three mutually perpendicular stresses acting on

the faces of a rectangular block are :

fy = 5 tonf/in2, vert ically and tensile,

fx = 2 tonf/in2, horizontally and compressive,

f2 = 4 tonf/in2, horizontally and tensile.

I f Ε = 13,000 tonf/in2 and σ = 0-286 calculate the strain in

each of the three directions. I f the dimensions measured along the

axes y, χ and ζ are respectively 4-0 in., 2-0in. , and0-5in . , calculate

the change in length in each direction.

(b)


Solution Taking tensile strain as positive :

S t ra in along y-axis ey = k + - ZJ^~ 2k

Ε 1

13,000

5 - 0-57

1300 '

- 0-00034 ( tensile) .

[5 + 0-286 (2 - 4 ) ] ,

Corresponding change in length

= 0-00034 χ 4 ,

= 0-00136 in. (increase). ^

Strain along a;-axis

FX < <IFZ

Ε Ε Ε

1 = -TTFX + *(fv + /=)L

1

13,000

4-57

[2 + 0-286

( 5 + 4 ) ] ,

13,000

~- - 0 - 0 0 0 3 5 2 (compressive).

Corresponding change in length

= - 0 0 0 0 3 5 2 χ 2 ,

= - 0 - 0 0 0 7 i n . (reduction).

S t ra in along z-axis ez = -{—— Ε Ε

FIG. 2 0 4

*FY

Ε

t =2

¥ [/, + *(/* -/,)], ι

13,000

4 - 0-86

13,000

3 1 4

[4 + 0-286 (2 - 5)]

13,000 = 0-000242 (tensile).

8 a*

2 2 4 S T R E N G T H O P M A T E R I A L S

Corresponding change in length = 0 - 0 0 0 2 4 2 χ 0 - 5 ,

== 0 - 0 0 0 1 2 in. (increase)

E X A M P L E . The shank of a punch is 1 - 0 in. dia. and during the

punching operation carries an axial compressive stress of 1 0 tonf/in2.

I f the socket restricts the lateral (i.e. radial) strain to one third

tha t of the value when unconstrained, find :

(a) the radial stress imposed by the socket ,

(b) the radial strain,

(c) the increase in diameter of the shank.

Assume Ε = 7 5 0 0 tonf/ in2 and a - 0 - 2 5 .

Solution

Clearly the lateral strain is the same in all horizontal directions

i.e. ez = er since f. = fr.

fy=IO

FIG. 2 0 5

This lateral strain is given by :


(taking compressive strain as positive)

1 Τ Ux-<*(fu+fz)]>

= Ύ U.v -o(fy + U)], since f3 = fx.

This is equal to one third of the unconstrained value (fy acting alone),

Ε i.e.

or

ηξϋχ - <r{fy + fx)] =γ

3fx-Mfu+fx)+<rfu = °> 3/, ; -Safu - Safx + fau = 0

3/ j :( l - σ ) = 2 σ / ν ,

i.e., fx = 2 <

3 ( 1 - σ )

2 χ 0-25 χ 10

3(1 - 0-25) = 2-22 tonf/in

2

Hence, radial strain ex

1 [2-22 - 0-25(10 + 2-22) ] ,

Λ Increase in diameter

7500

2-22 - 3 0 5

75ÖÖ 5

0-83

" 7500 '

- 0 - 0 0 0 1 1 , i.e. tensile.

Radia l strain χ Diameter,

+ 0 0 0 0 1 1 χ 1-0

+ 0-00011 in.

I n the absence of the constraint, the increase in diameter would

have been 0-00011 χ 3 or 0-00033 in., i.e. Ε

Relation Between the Elastic Constants

As already shown, a shear stress fs applied to opposite faces of a

cube will, together with i ts complementary shear (shown dot ted in

Fig . 206) produce tensile and compressive stresses / v on the

diagonal faces such tha t fN = fs.


t s — • h H

f s (applied shear) j ι

U ί- H FIG. 2 0 6

Thus, Strain in diagonal = J ~ + ,

- £ < ι -'-»>·

Since the deformation is very small within the elastic range, the χ

increase in length in the diagonal is approximately y, i.e. —j—

nearly. So that , Strain in diagonal

Increase χ ^ ,^

Original length ]/2

1 χ χ f = — — and — = -77, where G = Rigidi ty modulus,

~ 2 Ö '

= γ - ^ - , since fs = fN.

Equat ing expressions for strain we have :

i.e., # = 267(1

Thus, for an elastic, homogeneous and isotropic material , the value

of G depends on tha t of E. Thus if, for steel, Ε is taken as 29-8 χ 1 0

6 lbf/in

2 and a as 0-287 we have ;


Ε Modulus of rigidity G = — ,

2(1 + G)

29-8 χ ΙΟ6

~ 2(1-287) '

= 11-6 χ ΙΟ6 lbf/in

2, approx.

I f the cube is subjected to a fluid stress / on all faces, then, taking

compressive stress as positive :

Linear strain in each edge = ^— — — ,

Ε Ε Ε - i - ( l - & r ) .

B u t , Bu lk or Volumetric strain = 3 χ Linear strain,

- * ( ! - * ) .

And, if Κ = Bu lk modulus,

Volumetric strain = . Ε

f 3 / Equat ing, we have — = — (1 — 2 σ ) , Ε Ε

i.e. Ε = 3K(l - 2σ).

Thus, for steel, using the same figures for Ε and a :

Ε Bulk modulus Κ

3(1 - 2a) 9

29-8 χ 1 06

3[1 - (2 χ 0-287)] '

29-8 χ 1 06

3 χ 0-426 '

23-3 χ 1 06 lbf/in

2. (approx.)

Equa t ing the expressions for Ε :

2 0 ( 1 +σ) = 3K(l - 2 σ ) .

.·. 2G + 2Ga = 3K - 6 Ζ σ ,

or 2Ga + 6 Ζ σ = 3K - 2G}


whence, σ 2G + 6K

9

Ε

but, 1 + a = (from equation for E),

Ε -2G i.e. a = Ö

Equat ing the expressions for a :

Ε -2G 3K -2G

whence, Ε -

G 2G + 6K '

9GK

G + 3K

E X A M P L E . When subjected to an axial load of 4 - 0 tonf the

changes in gauge length and diameter of a standard test piece were

respectively 0 - 0 0 2 8 2 in. and 0 - 0 0 0 1 8 5 in. Es t imate the value of G.

Solution

F o r a standard test piece: Gauge length L = 2 - 0 in.,

Gauge diameter d = 0 - 5 6 4 in.,

Latera l strain = 0'*?°?3J5

= 0 - 0 0 0 3 1 ,

0 - 5 6 4

0 * 0 0 2 8 2 Longitudinal strain e = — = 0 - 0 0 1 4 1 .

.*. Poisson's Ra t io σ = ^ = 0 - 2 2 . 0 0 0 1 4 1

Area of section A = — χ 0 - 5 6 42 = 0 - 2 5 in

2.

4

4 0

.*. Axial stress / = -77^3- = 1 6 tonf/in2.

ί 1 6

Elas t ic modulus Ε = — = 41 = 1 1 , 7 0 0 tonf/in2.

Rigidi ty modulus G = n „ E

x = J1'™® = 4 8 0 0 tonf/in2. 6 J 2 ( 1 + a) 2 χ 1 - 2 2 - —

3K - 2G


Examples V I I I

1. As a result of tests on a certain steel specimen, the values of Ε and a were found to be respectively 29-6 χ 10

6 lbf/in

2 and 0-294. Determine the

values of G and K. Find also the reduction in volume of a cylinder of this material 6 in. long, 2 in. dia. when subjected to a fluid stress of 11,200 lbf/in

2

(11-5 χ 106 and 23-95 χ 10

6 lbf/in

2, 0-009 in

3).

2. A tensile load of 2 tonf resulted in a measured extension of 0-00517 in. on a 2 in. gauge length of a test piece 0-5 in. dia. The measured twist on an 8 in. length of a similar piece of material produced by a torque of 480 lbf in. was 3-2°. Estimate from these figures the value of σ and state why this method of finding it is unsatifactory (0-3 approx.).

3. The strains registered by two gauges mounted at 90° on the surface of a brass pressure vessel 0-125 in. thick were respectively 0-00071 and 0-000167. Take σ = 0-3 and Ε = 12 χ 10

6 lbf/in

2 and estimate the reduction

in the thickness (0-000047 in.).

4. Calculate the increase in area of a piece of steel plate 12 in2 when sub-

jected to tensile stresses of 17,900 and 11,200 lbf/in2 at 90°. Take a = 0-3

and Ε = 29-6 χ 106 lbf/in

2 (0-066 in

2).

5. Describe three methods of determining the value of Poisson's Ratio and discuss their relative merits.

C H A P T E R I X

COMPLEX STRESS II

Principal Planes and Stresses

L e t two opposite faces of a block A B C D (Fig. 207) of unit

thickness be subjected to a shear stress fs and let direct stresses

fx and fy both tensile, be applied as shown, the direct stress in the

third direction being zero.

FIG. 2 0 7

F o r equilibrium there must also be a complementary shear

couple of opposite sense to the couple resulting from the applied

shear.

On any inclined section E F there must be acting a normal stress

fN together with a tangential (shear) stress fT in order tha t the

piece E F D (Fig. 208) may be in equilibrium, the direction of fT

depending on the relative magnitudes of fx and fy.

2 3 0

C O M P L E X S T R E S S I I 231

FIG. 2 0 8

Resolving forces || fN (i.e. J_ E F ) :

fN E F - fx F D sin 0 + fs F D cos 0 + / S E D sin 0 + fy E D cos θ

(all terms χ 1-0),

Λ /Ν , F D . F D _ E D . E D

FX s in2 θ + / y cos

2

FR

' 1 - cos20 + /y

1 -{- cos 20 "

L 2 + /y 2

2fs sin 0 cos 0 ,

+ / 5s i n 2 0 ,

FR cos 20 / y /// cos 20

2 ΊΓ 2 '5 '

A + 4 + (/„ /,) c q s 2 Ö + ^ g i n w (1)

Resolving J_ / N (i.e. | [ E F ) :

/ r E F = / S F D s i n 0 - fsED cos 0 - fx¥O cos 0 + / y E D s i n 0 ,

, F D . E D F D E D .

= f s sin2 θ — fs cos

20 — /ajSinO cos0 + / ys i n 0 cos0 ,

= / s( s in2 0 - cos

2 0) + (fy - fx) sin 0 cos 0 ,

= ^ ~ sin 20 - / , cos 20 (2)


F o r maximum or minimum normal stress -^IN = Ö, i.e.,

or

-2 sin 20) + fs(2 cos 20) = 0 differentiating equation 1,

2 / , cos 20 = (fy - / r ) sin 2 0 ,

2fs tan 20 (3)

This gives two values of 20 differing b y 180° and hence there are two values of 0 (0j and 0 2) differing by 90° .

When fT = 0 i.e., when the stress on the oblique plane is wholly direct

^ ~ f * ) sin 20 - / e cos 2 0 , 2

i.e., tan 20 2 / .

h - t.

which is the condition already

obtained for making the direct

stress / i V a maximum. Thus, \

when there is no shear on face

E F i.e., when fT = 0 the stress

diagram becomes as shown in

Fig . 209 .

Resolving \\fx:

fx¥O + / , E D = / N sin 0 E F

F D F D

i.e. fx

or

tan 6

tan 0

sin0

= /N FIG. 2 0 9

/ , = (/.v - / , ) t a n 0

Resolving J_ fx: / S F D + / ^ED - ( / A E F ) cos 0 and E D

E F

i.e.,

F D '

/.v - /// tan 0

(4)

F D

tanfl

F D

sin 0

(5)


U ' ι u

When fy is compressive then, changing i ts sign :

/ A W = * { / , -fy + Wx + / „ ) ' + 4 /2l } >

and is still tensile (positive) even though fy may be greater numeric-

ally than fx.

fs** = * {( /* - /*) - Wx + h? + 4 /2] } '

4 χ o g i v e s : β = (fA - fa){fN - fy),

=~ f% — ÎNÎij — ÎNÎx + fxim -FN-Mtx+TU) + /*/,„

or / , | - (fx + fu)fN + (fjy - / J ) = 0 , which is a quad-rat ic in fN.

Hence , / , = ^ + ^ * V + ^ " ^ ~ ™ , 2

= \IU + fy± M + fv)2 + 2Uu - MÂu + 4/î)]. = I {/* + fu± Wx - /i/)2 + 4/']) ·

Taking the positive sign, the maximum normal stress will be given by

/ A W = \\U + i„ + Wx - /.v)2 +

acting over a plane making an angle θχ with the applied stress fx.

This value is clearly tensile, i.e., of the same sign as fx and fy.

Taking the negative sign, the minimum normal stress will be given by

hmia = h{f, + f,-mf, - / „ ) ' + 4 / * ] } ,

acting over a plane making an angle ö 2 = θ1 + 90° with the stress fx. This is zero when

fx +fu = ]/[(fx-fv)2+*jl], i.e., when β + /» + 2/,./, = /* - 2fJy + /» + 4/?

or when /,./,, = /2.

Thus the least normal stress is tensile (positive) only if frfy > j2

s .

These two values of fN are called the Maximum and Minimum

Principal Stresses, and the planes over which they ac t (and on

which there is no shear) are called the Principal Planes.

The stresses acting on the element are then as shown in Fig . 210 .

Note: F rom E q . 5 : tanO = f* ~

t y , Λ tar i f f -

/ iW ~

fy


and is clearly compressive (negative) since ^[fx + fy)2 + 4/f] >

(fx — fy) always. Thus when fx and fy are of opposite sign, the

principal stresses are also.

FIG. 210

I t should be noted tha t / ^ m ax is not necessarily greater than

fN numerically, i.e. i t is possible for the greatest stress in the

material to be compressive.

Maximum Shear Stress

As already shown in E q . (2), the shear stress on any plane

inclined a t 0 to fx is given b y

fr = ^ ^ ~ *r j sin 20 - / , cos 219.

F o r maximum or minimum shear stress -L- (fr) — 0 , ad

i.e. ^ » ~ (2 cos 20) - / s ( - 2 s i n 2 0 ) - 0 ,

or (fy - fx) cos 20 + 2 / , sin 20 - 0 ,

.·. 2 / , sin 2 0 = - ( / „ - / * ) cos 20

i.e., tan 20 = - ^ ~ *r (6)


This is the negative reciprocal of E q . (3) so tha t the two values

of 0 (90° apart) which i t gives, are 45° in advance of θλ and 0 2

respectively. The planes of maximum shear are therefore midway

between the principal planes as shown in F ig . 2 1 1 .

FIG. 211

E.g. if 0 1 = 2 2 · 5Ο,

20 ! = 45° ,

tan 2 0 x = + 1 - 0 .

Hence, for the planes of maximum shear :

tan 20 = - 1 - 0 ,

20 = 135°,

i.e. 0 = 67-5° = 0, + 45° .

As already shown, fT = ^ L Z À j sin 20 - fs cos 20 (Eq. 2 ) ,

( — fy Han 20 cos 20 - fs cos 2 0 ,


and, from E q . (6), is a maximum when tan 20 ^ /,/ - F,

2/.,

/..• - /.v

' 7 m a x 2

2 / s '

Cv ~ F") cos 20 - /„ cos 20 (sub. for

V "I* I tan 20) .

= - c o s 20

= - c o s 20

4/,· + / V

4/Î

and cos 20 = + ] / - — — ^ r-, (/ t a n

2 20 + 1

±

4/Î + 1

±

4/Î

2 / .

y[(/., - / y )2 + 4 / f

Hence, /7< = ± ' ' ' max - J-

2/., (/ , - /„)» + 4/2

ç

4/«

= ± 1 - tu? + 4/Î.1, = i 2 (Difference of principal stresses)

/ A max / Λ min

Thus each value of / y m ax has the same magnitude and is the complementary shear of the other. Clearly, maximum shear stress occurs when / y m ax and FYMLN are o f opposite sign, since then,

/ y ma x = H / A W - (-/vmin)] (algebraic difference), = \ [/.V'max /-Vmin 1 '

Since all solids are three-dimensional there may be three

principal stresses acting over three mutually perpendicular

principal planes. I f the third principal stress is zero, as is usually

the case, then, if the other two principal stresses are of the same


sign, the maximum shear stress in the material will be given by

7'raax ~~ 2 (AVmax ^ ) - 2 either, /

/ V m a x =

2 lt\ or 0] = /Λ

2

(i f

^ m a x > / A m i nn u m e r i c a l l y ) ,

Principal Strains

I f a material is assumed isotropic, i.e. the elast ici ty is the same

in all directions, then, under complex stress, the greatest and least

direct strains will occur in the directions of the principal stresses

and will be linear functions of such stresses. Such strains are

referred to as Principal Strains.

Suppose principal stresses fx / 2 and / 3 to ac t upon the element

shown in Fig . 212. E a c h will produce the same strain as i t would

if acting independently. Thus

Principal strain in direction of ft :

fx <*h - ah 1 e Ε Ε

h Ε ~¥^2 ~'~

Similarly,

and 6 0

~Ê ~ ~E~ ^3 + / 1 ) '

FIG. 2 1 2

I f / j is the greatest principal stress and / 2 the least principal stress, then greatest principal strain difference :

, i \ fz

Hence.

h σ Ε ~Ë h - ft

Ε fl -

Ε /ι —

+ /ι>.

1 + a) and 1 + a = —-E_

2G (see p. 226)

2G


i.e. /i - /. = 2 / 3

/

Thus the maximum shear strain is

equal to the greatest difference in

the principal strains.

E X A M P L E . The stresses applied

a t a point in a material are as

shown. Determine the principal

stresses and the angles made with

the horizontal b y the planes over

which they ac t .

G '

= Maximum shear strain

5 t o n f / i n2

Solution

f,

fy

ÎS

AVmax =

2

- 5 ,

3 FIG. 2 1 3

\% + fv + l/[(f*- fyf + 4 / ? ) ] }

i [ 2 + ( - 5 ) + V(2- ( - 5 ) )2 + ( 4 χ 3

2) |

* [ - 3 + V(72 + 3 6 ) ]

9 - 2 2 = - 1 - 5

and

F rom E q . ( 3 ) :

- 1 - 5 + 4 - 6 1 = + 3 - 1 1 tonf/ in2 (tensile)

- 1 - 5 — 4 - 6 1 = — 6 - 1 1 tonf/in2 (compressive).

2 / , 6 tan 20 =

tu - U '

6

- 5 - ( + 2 ) '

- 0 - 8 5 7 . " - 7 '

2 0 ! = 1 8 0 - 4 0 - 6 ,

= 1 3 9 - 4 , θ1 = 6 9 - 7 deg.

Since Maximum shear stress / r m ax = — / 2 ) ,

C O M P L E X S T R E S S Π

f N i m j x= +3ΊΙ tonf/ in-

fN m i n

= ~

6' " +onf/ in

2 (Comp.)

FIG. 2 1 4

3 (Applied shear couple anticlockwise)

Complementary shear

15

F r o . 2 1 5

239


Alternatively, tan θ1 = /.V max

_ 3 - 1 1 - ( - 5 )

3

- 2 - 7 .

Λ 9 0 - 0 X = 2 0 - 3 deg, i.e. θχ = 6 9 - 7 deg as before.

The effect of reversing the applied shear couple i.e. of making i t

anticlockwise as in Fig . 2 1 5 is to t i l t the principal planes the other

way to get a mirror image of F ig . 2 1 4 as in F ig . 2 1 6 .

This is equivalent to measuring 0λ downward from the direction

of the horizontal stress.

E X A M P L E . A beam having the section shown in Fig . 2 1 7 sustains

a bending moment of 1 5 0 tonf in. and a shear force of 8 tonf.

Assume the shear stress to be uniform over the web section and

find the value of the maximum principal stress a t a point 5 in.

above the neutral axis.

f N = + 3 · Ι I t o n f / i n2

"max.

f N = - 6 · Ι I + o n f / i n2 (comp.)

FIG. 2 1 6


Solution

2nd Moment about N.A.

3-5 χ 1 23 3 125 χ 1 1

3

I = 12 12

- 503-5 - 346-5 = 157 in4.

Direct bending stress

4 M

y ν

j — — — , where y = 5 in . 150 χ 5

157 '

= 4-78 tonf/in2.

3 Web section — 11 χ —-,

8

= 4-125 in2.

ft Mean shear stress /,

I in

FIG. 2 1 7

4-125

*. Maximum principal stress :

1-95 tonf/in2.

/ A W = i [ / + l / ( /2+ 4 /

2) ] ,

= -ΐ-{4·78 + ]/[4-782 + (4 χ 1-94

2)]},

= l [4 -78 + V(22-9 χ 15-1)] ,

= £[4·78 + 6·1β] ,

10-94

12 in

- 5-47 tonf/in2.

Möhr's Circle:

From this can be determined graphically :

(a) Principal stresses / iv m ax

a n (i / ; v mi n>

(b) Max. shear stress / r m a x> (c) Normal and tangential stress on a given section.


Construction: Se t off 0 A X = fy (Fig. 2 1 8 ) ,

0 A 2 = fx,

A ^ j = fs = A 2 E 2 , normal to Ο Α 2Α χ .

FIG. 218

Bisec t A 2A j a t C and with C as centre draw a circle radius C E X .

J o i n Έ2ΟΕ1 and mark D 2 and D 2 . L e t D 1 C E 1 = 2Θ1 (i.e. twice the

angle between fx and plane of principal stress).

S e t off 2Θ ACW from CEX. Obtain point Ρ and project to A.

Inser t point Q. Then,

ODi = / A W = * & + / „ + UU - fy? + 4/f]} a t φί + 90) to fx, OA = fN = Normal Stress on Sect ion a t θ to fx.

O D 2 = / A W = Η/χ + /. - VtO« - fy? + 4/f]} a t (β, + 90) to /,. A P = fT = Tangent ia l stress on section a t θ to fx.

C Q = /tw = ± i V[(/x - / y )8 + 4/ s

2] a t 45° to

0 X = -| (angle DjCEj^ as measured).

When using the aforementioned construction, the following rules

must be observed :

1. S e t off tensile (i.e. positive) stresses to the right of point 0

and compressive to the left,

C O M P L E X S T R E S S I I

2 . Shear stress to be set off down

from Av and up from A 2 ,

3 . Se t off 2 0 A C W from 0ΈΧ,

4 . Measure 2Θ1 A C W from C E 2 .

E X A M P L E . The stresses applied a t

a point in a material are as shown in

Fig . 2 1 9 . F ind graphically the principal

stresses and the angles made with the

horizontal by the planes over which

they act . Ske tch the stress diagram.

15

FIG. 2 1 9

FIG. 2 2 0

243


Solution

f N M I N= - 6 - l l t o n f / i n2 (Comp.)

F i d . 2 2 1

(a)

fr =33

FIG. 2 2 2

The principal planes and stresses arc as shown in Fig. 221 :

E X A M P L E . F ind, using the figures of the previous example, the

effect of : • f N m j n= + 3 - 1 1 t o n f / i n

2

(a) reversing the di-

rection of fy,

(b) reversing the di-

rection of fu and fr,

(c) interchanging the

values of fy and fx in

(a) above.

Theories of Elastic Failure

I n simple tension a material is said to have failed when a per-

manent strain occurs, i.e. when the elastic l imit is exceeded. (For

steels, which obey Hooke's Law, the elastic l imit corresponds

approximately with the limit of proportionality).


(b)

I

FIG. 2 2 3

(c)

FIG. 2 2 4

The solution to (c) is left as an excercise.


The elastic limit in simple tension is usually thought of as a

definite value of tensile stress and, for simple tensile loading, the

working (operating) stress is usually limited to a value leaving a

margin of safety suitable to the conditions.

In a complex stress system there exist other quantities (see

below) one of which, according to the theory adopted, may be the

criterion of failure, i.e. lead to permanent strain. The value of this

quanti ty corresponding with the simple tensile elastic l imit is then

taken as the limiting value, with which the calculated actual value

in a given case is then compared.

The main hypotheses of failure are considered below in relation

to two-dimensional stress systems only.

1. Maximum Principal Stress Theory (Rankine)

This theory appears to hold good for brit t le materials generally

and, according to it , failure occurs when, irrespective of the values

of the other principal stress(es), the maximum principal stress

reaches the simple elastic limit stress,

i.e. when / iv m ax = ± / e (where fe = simple elastic l imit)

so that , for failure not to occur :

L{(fX-fy) + N(fX-fy)2 + ±M>fe

Note: If, numerically, / N m in > /jvm ax then the former value must be used.

If, in a given case, the principal stresses / j v m as

a n (l ÎN^

A RE

expressed as fractions of fe (i.e. as dimensionless numbers) and

plotted against one another, then the boundary within which the

plotted points must lie for safety (i.e. for failure not to occur) will

be the square E A B C D (Fig. 225) in which O E = O F = 1-0. In

other words, according to the Rankine Theory, failure will occur

when such points fall outside this perimeter, i.e. when the ratio

of either principal stress to the elastic l imit exceeds unity.

F o r combined bending and torsion (see later) i t should be noted

tha t fj\'msix is positive and fNmfn negative so tha t in this case the

fourth quadrant of Fig . 225 is the relevant one.


FN M I,

f.

Β 1-0 F i

Ε - 1 - 0 0 1-0

c -1-0 I

FIG. 2 2 5

Maximum Shear Stress Theory (Coulomb, Tresca and Guest)

This theory gives a good approximation where ductile materials are concerned and, according to i t , failure occurs when the maxi-mum shear stress is equal to the shear stress value ( / e/2) correspond-ing with the simple tensile elastic limit i.e. when

FR = — ·

' 1

MAX 2

Now, in general, / r m ax — ±2" (difference between greatest and least principal stresses) so that , when the two principal stresses are alike (of the same sign), the third principal stress being zero,

E i the r :

/ T W = ±i(AvMAX - ° ) IF

/ A W > /vmm numerically, 9*


i.e. ΊΫΨ. = ±FTM&X = ± A a t elastic limit,

•'• Aw > ±fe for ^ f e t y ( o r : - ^ - > ± 1 - 0 j .

Or: / T W = ±WN^ - 0 )

i.e. » ±FE for safety (or: > ± 1 - 0 ) .

I f the points representing the rat io of the principal stresses to

FE are plotted as before, then, for like principal stresses (1st and 3rd

quadrants), they must not be outside the boundaries E A F and

GCH (Fig. 226) .

FIG. 2 2 6

Again, when the two principal stresses are unlike, i.e. of opposite

sign (2nd and 3rd quadrants) then

FTAXNX = i 2 (Ανmax ~~ / min ) '

Since the limiting value of / r m ax is -—then, for safety: Δ

ÎNU


> ± 1 ,

i.e. Ji^- = > ± JL 9 thus fixing points U and V .

fe fe *

Clearly other points obtained from the above equation lie on the straight lines E H and F G so tha t the complete perimeter is E A F V G C H U E .

3. Maximum Strain Energy Theory (Haigh)

This theory also gives a good approximation where ductile materials are concerned and, according to i t , failure occurs when the strain energy per unit volume is equal to the value (β/2Ε)

corresponding with the simple elastic limit.

Suppose the three principal stresses to be fx / 2 and / 3 . Then, as already shown :

Principal strain in direction of / x ex = ~- —~ (/2 + /3) Ε Ε

Work done in lateral straining = χ La te ra l strain,

= - Γ Χ Ψ (k + h) Per unit volume,

Net work done by / 3 per unit volume = —\- — ——- (/ 2 + / 3 ) , ZE ZE

= w[/î-/iff(/. + /s)L

and since the minor principal stress is negative,

/jVmax + /iVmin > ± FE >

i.e. I ^ + h ^ u _ > ± 1

JE Je

Put t ing the first term equal to zero gives points F and H (Fig. 226) while points Ε and G result from putting the second term equal to zero.

When the two principal stresses are equal numerically then

2/Vnun 2

/ i V „


and work done by / 2 per unit volume = — [ / f — / 2 σ ( / 3 + / J ] , 2z^

and work done by / 3 per unit volume = —3— [/3 — /^( /χ + / 2)] . JILL/

Hence, to ta l strain energy per unit volume,

Υ = M - FMH + H) + 11- H°IH + H) + /i - HO (H + /,)],

= 2 j - [ / i2 + /! + /I - σ(/χ/2 + y, + HH + HK + HK + KM,

= [/? + / ! + / ! + 2 σ ( / 1 / 2 + HF3 + KH)L ·

Equat ing this t o — g i v e s

/? + /I + /§ - 2ff(/j/2 + / 2/ 3 + /,/,) = /J, and, if the third principal stress is zero, then

/! + /I - 2σΑ/2 = /;, or, using the previous notation,

/ Anax ~ /iVmin / mai AVmin =

/ c '

This is the equation of an ellipse the major and minor axes of

which intersect a t the origin, the major axis being inclined a t some

angle φ (Fig. 227) to the horizontal axis.

When FNMIN = 0 , then FNMIA = ±FE,

when FNMAI = 0 , then FNMIN = ±FE. Thus the ellipse passes through the points E , F , G and H (Fig. 226) .

To find the inclination φ and the magnitude of the major and

minor semi-axes, put

/ iv m ax = x c os

Ψ - Y s in

Ψ

and FNMIN = X sin φ - Y cos φ (Fig. 227)

Then, X2 cos φ + Y

2 s in

2 φ — 2 Χ Υ sin φ cos<p - f X

2 s i n

2^

+ Y2 cos

2ç> + 2 X Y sin φ cosç? — 2 σ ( Χ

2 sinç? cosç? + X Y cos

2«??

— X Y sin2ç9 — Y

2 sin φ coscp) = F

2

E,

i.e. X2 + Y

2 - 2 σ ( Χ Υ cos2<p + ( Χ

2 - Y

2) sin<p cos<p) = /

2.


When cos2<p = 0 then 2φ = 90 and φ = 45° . At this value of φ the

term in X Y disappears and the equation becomes

/ X2 — Y

2 \

X 2 _i_ Y2 _ 2σ I sin 2φ\ = /2 and sin 2φ = 1 0

Λ Χ 2 + Y2 _ σ{χ2 _ Υ2 ) = f2

or Χ2( 1 - α) + Υ

2( 1 + σ ) = /

2

;

Χ2 Υ

2

i.e. _ + 7 2 —= 1

·

Je Je

This is the equation of an ellipse of major semi-axis fel]'(l — σ)

passing through the origin and inclined at 45° to the horizontal axis . The minor semi-axis is / e / j / ( l + σ) at 90° to the major axis and also passing through the origin.

when a = 0-25, Major semi-axis = J * ^ ~ 1-15L, 0-75

Minor semi-axis = ^ ^ 0 - 8 9 / e .

The ellipse is shown in Fig . 228 and, for safety according to this theory, the points defined by the principal stresses must be within i t s perimeter.


The equation of the ellipse can be written :

I te I \ U I max / A min

fl FNM LN

*.

1-0 F

6 0-63 ! , Ε

\

i 0 8 2

/ ι / 1 /

•0

H

FIG. 2 2 8

Hence, when /A min /·>

2 σ

γ^- and is of the same sign,

te

i.e. [k^L^2(\ -σ) = 1,

whence, /ΑΓΠ 1

— ± 0 - 8 2 , when or = 0-25 . /E ^ 1/2(1 - σ

When the principal stresses are of different sign, then

7 A W \2 , »ff*

x2

whence, /Λ

= ± 1

]/2(1 +σ) ± 0 - 6 3 , when σ = 0-25 .


4. Shear Strain Energy Theory ( Von Mises and Hencky)

A good approximation in the case of ductile materials is also

given by this theory, which is to be preferred when the mean

principal stress (see below) is compressive. According to this theory

failure will occur when the to ta l shear strain energy is equal to the

value /?/66r a t the simple elastic limit. This energy is tha t part of

the to ta l strain energy which produces distortion, i.e. i t is the

to ta l strain energy less the strain energy due to volumetric stress.

Suppose the three principal stresses to be denoted by / j / 2

and / 3 .

Now, U = K/i + H + H) + - U) + WX - FS),

and

and

(Multiplying out will show tha t this is so.)

H = i(/i + H + H) + OFT - H) + - H) H = WX + H + H) + - H) + UH - /«)·

F I G . 2 2 9

The first term \(FX + / 2 + / 3 ) is the mean principal stress and this acting volumetrically on all three faces (Fig. 230) produces a volumetric strain unaccompanied b y distortion since the shear stress is everywhere zero.

f|+f2+f3 3

The second and third terms ac t as shown in F igs . 231 and 232 and are

3 3

FIG. 2 3 0

9 a SM


FIG. 2 3 2 FIG. 2 3 1

proportional to the three corresponding maximum shear stress

values since these values are themselves proportional to the

differences in principal stress. The strains due to these result in

distortion and the energy of distortion, i.e. the shear strain energy,

is therefore given by

Us = To ta l strain energy—Volumetric strain energy.

Now, Volumetric strain

ev = βι + e2 + e 3,

fi < r ( / a + h) , U o(U + fi) Ε E E Ε fi+U + h 2σ

Ε fi + /, + fs

Ε

Ε (1 - 2a)

(fi +U +fz)>

And Volumetric strain energy

Mean stress χ Strain,

Ε Ε

(1)

1 /ι + U + h χ e„

fi+f* + U v fi+U + fz 7^ X ~ Ε (1 -2a),

6E (/i + / , + hf (1 - 2σ)

1

(2)

To ta l strain energy = _ [ / « + / » + / » - 2 σ ( / 1/ 2 + / 2 / 3 + / 3 Λ ) ] (3) (from p. 250) J±

C O M P L E X S T E E S S I I

[2(/? + / I -Ι- / I ) - 2 ( / j / 2 + / 2 / 3 + y j ] ,

and 1 + σ = ,

= ~

/ a )2 + ( /a ~

/ a )a + ( /s ~

/ i ) 2 ]-

I n Simple tension a t the elastic limit f1 = fe and / 2 = f3

1

"· U s =

W( 2 / ΐ) =

~ 6 # "

Equat ing, A - [ f t - / , ) « + ( / , - / , ) · + ( / , - hf\ = A

i.e. ( / a - / 2 )2 + (/, - k)2 + (k - hf = 2/?·

I f the third principal stress is zero, then,

(A - kf + / ! + / ! = 2 / f ,

or, / Î - 2A/ 2 + / ! + / ! + /? = 2/? ,

i-e., /? + / ! - / i / , = / | .

Or, using the previous notat ion,

/ iVmax /iVmin / ^ m a x / ^ m i n =

/e

Hence, Shear strain energy

u. = u- u„

= 4 WM + {L 1 IL

~ M HK + HK + / S / L )]

- A f t + / , + / , ) « ( l - ä r ) ,

= - ^ - { 3 ( / ? + fl+ il) - 6 σ ( / χ / 2 + kh + Μι)

- (1 - 2er) [/} + / I + / f + 2 ( Λ / 2 + /2/3 + / a / x ) ] } ,

= -^W+fl+il) ( 3 - H - 2σ)

- ( / ι / 2 + / 2 / 3 + / 3 / ι ) ( 6 ( Τ + 2 - 4 σ ) ] ,

= A { ( 2 + 2σ[(/? + / f + / i ) - ( / x/ 2 + kh + kh)l},

1 + Σ

Γ Λ / / 2 ι ;2 ι 42

255

ι (6)

9 a*


i.e.

or

X2 - I - Y

2 -

X2 + Y

2 -

X2 ^

9

2

X2 Y

2

~~2 2"

3 Y2

whence, X

2 Y*_

9 / 2 2 /2

= /I-

= 1. (8)

The major semi-axis is thus ] /2/ e so tha t the ellipse passes through

points A and C Fig . 233 while the minor axis is | / f fe or 0 ·82/ β

approx.

f.

F

/

y/ /

/ \

g /

/

/E 0

\

Ao

H - 1 - 0

FIG. 2 3 3

Dividing through by fe, gives

/ /iVmax \2 , (jj^mmY _ /^max /A

7min _ τ (H\

\ fe ) \ fe ) fe ' ' This is the equation of an ellipse similar to the one obtained

under the previous (Haigh) theory. Put t ing / iv m ax = X cos9? — Y

sin<p and fNmm — X s m <

P + Y COSÇJ as before and substituting in

E q . (6) gives, for φ — 45 deg.

X2 — Y

2

- sin 2φ ^ /2, where sin 2ç? = 1,


Again, when / Λ Ί η Ιη = 0 , !ψ± = ± 1 from E q . (7) le

and when fNmBki = 0 , = ± 1 ,

le so tha t the ellipse, within which the points must lie for safety

(according to this theory) passes also through E , F , G and H which

are common to the perimeters derived from theories 1, 2 and 3.

5. Maximum Principal Strain Theory (St. Venant)

According to this theory, which should not be generally used since i t overestimates the strength of ductile materials, failure will occur when the greatest principal strain reaches the value fejE

corresponding with the simple elastic limit, i.e. when, assuming principal stresses fx / 2 and / 3 , either,

4 " - ! - ( / • + /») = ± j - . i-e. h - * ( / , + / , ) = ±U,

or A _ I L ( / , + / L ) = ±1±, i.e. / , -σ( / ,+/ 1 )= ± fe,

or A _ £ . ( / 1 + / i ) = ± A . , i.e. h-a{h + ft) = ±fc.

F o r a two-dimensional stress system, / 3 = 0. Then, either

/i - tf/2 = ±/<<> or /2 - ff/i = ±/e-Using the previous notation, the above will be written

F o r like principal stresses, when / i Y m In = 0 , fNmax = ±fe,

and when fNmax = 0 , fNlûin = ±fe.

Thus the perimeter passes through the points E F G and H as in the previous four cases.

When fNmln = fNmax then, (taking a = 0-25) ,

ι _ /A7

ma x ι ι

/Armax ^ — = t / c ;

. / iVmax _ , _ fNmln

"h 3 / „ '

FIG. 2 3 5


I FIG. 2 3 4


This fixes points J and K . F o r unlike principal stresses,when fNmin = — / A W

/ ^ m a x , l ™ / iVmax , ^ /^Vmin tnen fNm + = ± / e , whence ^ = ± • β

This fixes points R and S.

Composite Graphical Representation

B y superimposing the perimeters corresponding with each of the

five theories, a composite figure is obtained which is usually referred

to as a B e c k e r Diagram.

1. Rankine Square A B C D 2. Coulomb, Tresca, Guest Figure A F G C H E 3. Haigh Ellipse P F N G Q H M E 4 . Von Mises, Hencky Ellipse A F G C H E 5. S t . Venant Rhombus J F R G K H S E

Combined Bending and Torsion

The shear stress induced in a shaft by torsion is given by

Tr , τΦ = ——, where J =

J ' 32 '

T h e tensile and compressive bending stresses are given by

Mr _ T i d4

/ = — — , where I 64

The criteria for failure are usually :

(a) Max. principal stress, i.e. tha t on a plane where the shear stress is zero, or where the stress is entirely normal,

(b) Max. shear stress, i.e. failure is due to shear. Br i t t l e materials fail in tension—hence (a) is here used and ductile materials fail in shear—hence (b) is here used.

Consider the greatly magnified element E G F D (Fig. 236) : The shear stresses on E G and F D are accompanied by equal shears on F G and E D . The bending stress (tensile or compressive) acts normally to E G and F D .

L e t fN = normal stress, 1 required for equilibrium of

fT == tangential (shear) stress j portion E F D (Fig. 237) .


W

FIG. 236

f

FIG. 237

Comparing the system from which the principal stress formulae

were derived with the shaft stress system, i t will be seen tha t

/* = /

/ir = 0

Hence, substituting in \{FX + FY ± ]/L(FX ~ + 4 / ? ] } , Mr Tr

M a x . p r i n . s t r e s s , / A W = i [ / + l / ( / 2 + 4 / 1 ) ] , w h e r e / — , / , = —

Min. prin. stress, FNMIN = \[F - | / ( / 2 + 4 / 2) ] .

Note: 1. Since ^(/2 + 4/

2) is always > / it follows that fNm&K is tensile and

fNmin is compressive, i.e. principal stresses are of opposite sign. 2. For the other side of the neutral axis, the bending stress is compres-

sive. This alters the values of fNmax and fNmla but not the signs. 2fs 2fs

Again, substituting in - — ^ - 7 - , tan 2 0 = 7 ^ . Iy ~ fx /


This gives the two angles θ1 and 0 2 made by the principal planes

with the bending stress / .

Final ly, substituting in ± | - ftJf + 4 /2] ,

Max. shear stress /rm ax = ±j V(/2 + 4/Î)

Summary—Combined bending and torsion :

FIG. 2 3 8

Normal stress, / γ = / sin 20 - f /> sin 2 6 ,

Max./Min. prin. stresses / A W = \ ]/[f(±) V(/2 + 4 /

2) ]

( m i n )

2f on planes given by tan 26 = γ-, 90 deg apart over which

fT = 0 /

Shear stress, / r = \f sin 26 - f fs cos 2 6 ,

Max. shear stress / 7 m ax = ± \ | / ( /2 + 4 /

2) .

FIG. 2 3 9


On planes given by tan 2Θ = / / 2 / s , 90 deg apart and halfway between principal planes.

Equivalent Torque

As already shown,

1 Mr / a w = g" [/ + V(/2 + 4 /

2) ] , where / = — and /,

4Mr

nr* + lQM

2r

2 \6T

2r

2

(nr* ( t i t *

2M 1 4r

T r r3 2 π

4Γ

r

2 M 2 :±5- + - £ - y ( J | f i

Ufr

7 r r4/ 4

4 J f r

~πτΓ

Tr

~7~' Tr

2Tr

τη. ÎN m a x • [M + | / ( M

2 + T

2) ] ,

i.e., / λ · , M + ]j(M2 + T

2) and similarly,

Now, —— fs = —— / s = Torque for a solid shaft under shear stress

only. Since the e x p r e s s i o n — — / A m ax has the same form, it is re-Δ

ferred to as the Equivalent Twisting Moment and denoted by ΤE.

Thus, Τ Ε; = M + ]/(ilf2 + Τ

2) (Rankine Formula for brit t le

shafts.)

Again, as already shown,

1 fi - V ( / 2 + 4 / f ) , where / = _ and fs = —

(from above) ,

4 K 1 6 1 f

2r

2 1 6 T

2r

2

+ (nr* (nr*f

Λ ii

1 4r

2 Ivf-

2 nr

3 y (71/2 +y 2 ) >


i.e. —— / r m ax = ]/(M2 + T

2) = Equivalent torque as before.

Δ

Thus TE = ]/(M2 + T

2) (Coulomb Formula for ductile shafts.)

The above expressions apply only to solid shafts.

F o r a hollow shaft J = n(R* — r4) /2 and must be separately

calculated. I t can then be substi tuted in the equation

TE / Λ ' ΜΧ

From the foregoing an equivalent torque may be described as

tha t combined bending and twisting moment (]/(M2 + T2) or

M + ] / ( l f2 - f T

2), depending on the criterion of failure) which will

induce a given maximum principal/shear stress a t radius r when

used in the torsion equation,

V(^2 + r 3 ) /rm a. i.e., J

where ]j(M2 -f- Τ

2) — ΤE when the criterion of failure is the

maximum shear stress.

J r '

where M - f ]I(M2 + T

2) = T

7^ when the criterion of failure is the

maximum principal (tensile) stress.

Note tha t an expression for equivalent torque can be derived from each of the other theories of elastic failure.

E X A M P L E . F ind a suitable diameter for a mild steel shaft required

to t ransmit 9 h.p. a t 1 3 0 0 rev/min. given tha t the permitted

maximum shear stress is 6 0 0 0 lbf/in2. I t may be assumed tha t the

bending moment act ing is of equal magnitude to the torque.

Solution ... n m 3 3 , 0 0 0 χ h.p. „ r f I orque transmitted, Τ = —— lbf f t ,

2τιΝ

3 3 , 0 0 0 χ 9 >

2π χ 1 3 0 0

4 3 6 lbf in.

Hence this is also the value of M.


Since the criterion is the maximum shear stress, the Coulomb

Formula must be employed.

= ]/(4362 + 4 3 6

2) ,

= V(2 x 4 ·362 χ 1 0 0

2) ,

= 100 1/38,

= 617 and / w = 6000 lbf/in2.

3 617 χ 2

"T " π x 6000 '

= 0 0 6 5 5 . ,

whence, r = 0-403 in.,

13 d = 0-806 in , say, — - i n . dia. J

16

E X A M P L E . The maximum torque carried b y a crankshaft is

40 tonf ft and the corresponding bending moment is 28 tonf ft. I f

the safe tensile and shear stresses are respectively 6 tonf/in3. and

3-5 tonf/in2 find the required shaft diameter b y (a) Rankine

Formula , (b) Coulomb Formula .

Solution

1. Rank ine :

— / j V m ax - M + ]/(M2 + T

2) (equivalent torque)

= 28 + V (28 2 + 4 0 2 )

= 28 + 48-83 ,

= 76-83 tonf f t ,

= 922 tonf in. , and fNmax = 6 tonf/in2.

922 χ 2

π x 6

= 97-8 .

• r = 4 - 6 in,, i.e. d = 9-2 in.


2. Coulomb:

max |/(if

2 + T

2) ,

= 48-83 (from above) ,

= 585 tonf in. and fTmaK = 3-5 tonf/in2.

585 χ 2 * γ* —

π χ 3-5

- 106-3.

r = 4-735 in. i.e., d = 9-47 in.

A suitable diameter is therefore 9 J in.

E X A M P L E . F ind the principal stresses in a propeller shaft and the inclination to the axis of the planes over which they ac t given tha t (a) the thrust results in an axial compressive stress of 1 tonf/in

2

and (b) the engine torque results in a shear stress a t the shaft surface of 4 tonf/in

2.

f N m in = - 4 - 5 3

= 3 -53

f = -

f s= 4

FIG. 2 4 0

Solution

/ A w = i~[/ + V ( / 2 + 4 / f ) ] , = + + (4 χ 4^ ) ] ,

= * ( - ! + 1/65),

= ^ ( - 1 + 8-06) = = 3-53 tonf/ in2.


tan 20 = - - y - >

_ 2 x 4

- 1 ' =r. 8 .

/. 201 = 82°54 ' ,

i.e. 01 = 41°27A.

.·. 0 a = 41°27/ + 90 - 131°27'

The planes and stresses are then as shown in Fig . 240 .

E X A M P L E . I f M = 10,000 lbf ft, Τ = 12,000 lbf ft and

/ * m a X ^ 1^,000 lbf/in2 est imate a suitable diameter for the shaft

and calculate the value of the greatest tensile stress in the material .

( I . Mech. E . )

Solution

Using the Guest Formula :

TE = —— fSmax = ]/(M2 + Τ

2), where M = 10,000 lbf f t .

Δ

Τ = 12,000 lbf f t .

= ]/[(100 χ 106) + (144 x 1 0

e) ] ,

= 1 03 1/(100 + 144 ) ,

- 15,650 lbf f t . ,

= 188,000 lbf in. and fSmax = 10,000 lbf/in2.

3 _ 188,000 χ 2

' ' T

10,000 π '

= 11-94,

r = 2-29 in., i.e., d = 4-58 in.

Using the Rankine Formula to find the max . principal stress :

TE = ^fNm&x = M + y(M2 + T

2),

= 10,000 + 15 ,650 ,

= 25 ,620 lbf f t ,

= 308,000 lbf in. , (and r = 2-29 in.)

_ 308,000 χ 2 • · m a x - π χ 2. 2 93

= 16,375 lbf/in2.


E X A M P L E . A meta l tube 3 in. inside

dia. 0-125 in. th ick is held rigidly a t one

end in a vert ical position and a flexible

cable 0-125 in. dia. is wound on the upper-

most 2 in., the ends being brought out as

shown. Es t imate the maximum tensile

stress in the tube when the pull on each

end of the cable is 240 lbf.

Solution

Effective cable radius

FIG. 2 4 1

810 lbf in .

2 4 0

2 4 0 lbf 2 4 0

L - 2 L

FIG. 2 4 2 FIG. 2 4 3

At any point in the tube distant L from the top :

Bending moment M = 240 L - 240 (L-2),

= 240 L - 240 L + 4 8 0 ,

= 480 lbf in .


Equivalent torque TE = M + ]/(M2 + T

2),

= 480 + ]/(4802 + 8102),

= 480 + 1/(230,400 + 656,100),

- 480 + 1/886,500,

- 480 + 940,

- 1420 lbf in.

Outside diameter = 3 + (2 χ 0-125) = 3-25 in.

Polar moment J = (3-254 - 3 4),

= -g- (10-55 + 9) (10-55 - 9),

_ π χ 19-55 χ 1-55 32 '

= 2-98 in4.

Σ± = Ί ^ - / 1 4 20 w 3 2 5

j r > . . / A m : ix 2. 9 8 2

= 773 lbf/in2.

E X A M P L E . The shaft in the system shown is 6 in dia. and trans-

mits 250 h.p. a t 80 rev/min the maximum torque being 1-25 χ Mean.

Neglect the weight of the shaft and find the max. tensile and shear

stresses induced in i t .

12 in 9 6 i n -

2 -5 t o n f

FIG. 2 4 4


/ 1-25 χ 33 ,000 χ 2 5 0 \

\ 2π χ 80 j 12 = 246,000 lbf in .

Moments about the L H end give: i? 2(12 + 96) = 2-5 χ 12 ,

R2= 2

'5

* 12

= 0-278 tonf .

108

M a x . B .M. , M = 0-278 χ 2240 χ 96 ,

= 5 9 , 7 0 0 lbf. in. T T f

3

F o r failure in shear, TE = — / S m ax = ]/(M2 + T

2),

= ] /(59,7002 + 2 4 6 , 0 0 0

2) ,

= 253,000 lbf in .

.'. /.nax = 253,000 χ π χ 3

3 '

= 6000 lbf/in2 ( = 2-68 tonf / in

2) .

Fo r failure other than in shear, TE = - y Avmax/min

= M ± ]/(M2 + T

2) ,

= 59 ,700 ± 253 ,000 ,

= + 3 1 2 , 7 0 0 lbf in. ,

or = - 1 9 3 , 3 0 0 lbf in.

.·. / , V m ax = 312,700 χ = 7350 lbf/in2 (tensile),

TZ Χ ο

( = 3-28 tonf / in2) ,

2 a n d Avmin = - 1 9 3 , 3 0 0 χ — = 4550 lbf/in

2 (compressive).

71 X od

E X A M P L E . A shaft 4-0 in dia. carries a flywheel weighing 2-0 tonf

a t the centre of a simply supported span of 8-0 ft. The centroid of

the flywheel is 0 -1 in. from the axis of rotat ion. I f the shaft t ransmits

1000 h.p. a t 360 rev/min., es t imate the maximum stress a t the

centroid, (a) when this is vert ically below the axis, (b) when this is vert ically above the axis.

Solution

Max. torque, Τ = 1-25 lbf f t ,

1-25 χ 33 ,000 χ 250


Solution

Angular velocity

Centrifugal force

W Fe = —œ

2r,

2π χ 360 C0=—60—

= 12jz;rad/s.

ω = 1420.

1 —4-0 f t •

12 χ 2240 \ Έ, Λ Λ

( - 3 2 5 - ) 1 4 20

1640 lbf .

~Ϊ2

| 4 · 0 in

Flywheel weight

W = 2 χ 2 2 4 0 ,

= 4480 lbf . 2 ± F C

FIG. 2 4 5

The centrifugal force must be added to or subtracted from the flywheel weight when the centroid is below or above the axis respectively.

(a) Load per support = £(4480 + 1640) = 3060 lbf .

Maximum B . M . = 3060(4 χ 12) = 147,000 lbf in .

' 3 3 , 0 0 0 χ 1000 \ Torque t ransmit ted = χ 12 = 175,000 lbf in.

2π χ 360 I

Equivalent torque TE = M + ]/(M2 + T

2),

= 147,000 + ]/(147,0002 + 1 7 5 , 0 0 0

2) ,

= 147,000 + 2 2 8 , 0 0 0 ,

= 375,000 lbf in .

/ i V ma Now, ^-y-, where r = 0-1 and J = , J Δ

π2 χ 2*

C O M P L E X S T R E S S I I 2 7 1

.'. Max. principal stress / Λ o. l (

3 7 5 , 0 0 0

8π

1 4 9 0 lbf/in2 (tensile).

(b) Load per support = ± ( 4 4 8 0 - 1 6 4 0 ) - 1 4 2 0 lbf .

Maximum B . M . = 1 4 2 0 ( 4 χ 1 2 ) = 6 8 , 0 0 0 lbf in. and Τ is

unchanged.

/ . TE = 6 8 , 0 0 0 + } / 6 8 , 0 0 02 + 1 7 5 , 0 0 0

2)

= 6 8 , 0 0 0 + 1 8 8 , 0 0 0 ,

= 2 5 6 , 0 0 0 lbf in.

Hence, / 2 5 6 , 0 0 0

\ 8ττ

1 0 2 0 lbf/in2 (tensile).

E X A M P L E . A hollow shaft is to t ransmit 5 0 h.p. a t 2 5 0 0 rev/min

and must withstand a simultaneous bending moment of 2 9 0 lbf in.

I f the internal diameter is to be 0 - 6 5 of the external diameter and

the maximum principal stress is not to exceed 5 tonf/in2 calculate

suitable shaft diameters.

1 2 ,

Solution

3 3 , 0 0 0 χ h.p. Torque Τ = ^

_ f 3 3 , 0 0 0 χ 5 0 \

~ \ 2π χ 2 5 0 0 J

= 1 2 6 0 lbf in.

Equivalent twisting moment,

= 2 9 0 + ] / ( 2 9 02 + 1 2 6 0

2) ,

= 1 9 0 + 1 2 9 0 ,

= 1 5 8 0 lbf in.


F o r a hollow shaft, J = ( £4

- r4

) and r = 0-65Ä,

[R* - (0 -65Ä)4] ,

£4[ 1 - (0 -65 )

4] ,

(1 - 0 - 1 7 8 ) ,

nR* = l - 2 9 i ?4i n

4.

2

π

π

Τ nR*

2

0-822

Again, r£_E_

J

1580

1-29I?4 :

R3

/Λ'»

R

5 χ 2240

(where / Λ ι Μχ > 5 tonf / in2) .

R

1580

5 χ 2240 χ 1-29 '

= 0-109.

R = 0-478 in., i.e. D = 0-956 in. (say 1 in.) .

.*. Internal diameter = 0-65 in.

E X A M P L E . A 6 tonf propeller on a 9 in. dia. shaft overhangs the

bearing by 18 in. and exerts a thrust of 15 tonf when the input to

i t is 4000 h.p. a t 300 rev/min. Calculate the principal stresses a t

the point of support :

(a) a t the ends of a vert ical diameter and

(b) a t the ends of a horizontal diameter.

6 t o n f

FIG. 2 4 6


Solution

Bending moment,

Torque,

M = (6 χ 2 2 4 0 ) 1 8 ,

= 242 ,000 18 lbf in. ,

Τ = / 33 ,000 χ 4000 \

V 2π χ 300 / '

840,000 lbf in.

y = 4-5 in

I = 322 in4.

J χ 94,

32

644 in4.

Bending stress,

Direct stress,

My _ 242,000 χ 4-5

~~Γ ~ 322

F_

~A

± 3 3 8 0 lbf/in2.

15 x 2240 , = - 5 3 0 lbf/in2 (compres-

0-785 χ 92 sive).

ι j j . Tr 840,000 x 4-5 Shearstressduetotorsion = — - = —— = 5870 lbl/nr*.

J 644

Λ Τ Ι , . ^ . Ι ^ · 4 IF 4 6 χ 2240 Max. shear stress due to bending = — χ — - = — χ -γτ-^ζ τττ

= 282 lbf/in2. (see Chapter X I I )

F o r point A :

and fNmS]

max 2

1

F o r point C :

- 5 3 0 ) + 3380 / = 2 8 5 0 , = 5870 J

{2850 + ]/[28502 + 4(5870

2) ]} = 2 = 7475 lbf/in

2,

{2850 - ] /[28502 + 4(5870

2) ]} = - 4 6 2 5 lbf/in

2.

- 5 3 0 - 3 3 8 0 J / = - 3 9 1 0 , = 5870 1

··· fit** = γ { - 3 9 1 0 + V[3910* + 4(58702) ]} = 4245 Ibf/in

2,


a n d /jvmin = I T { - 3 9 1 0 - ]/[3910

2 + 4(5870

2) ]} = - 8 1 5 5 lbf/in

2.

Δ

For point Β

• f r

- 5 3 0 = / - 5870 + 282 = 6152

.'. /.v«,« = γ { - 5 3 0 + ]/[5302 + 4 ( 6 1 5 2

2) ] } , - + 5 8 8 5 lbf/in

2,

and / i V m in = - - { - 5 3 0 - }/[5302 + 4 ( 6 1 5 2

2) ] } , = - 6 4 1 5 lbf/in

2.

The maximum compressive stress is therefore 8155 lbf/in2 a t

point C and the maximum tensile stress is 7475 lbf/in2 a t point A.

E X A M P L E . Calculate for points X and Y on the shaft shown, the

principal stresses and illustrate their directions on a suitable

diagram. Assume clockwise rotation when viewed from the input

end.

100 rev/min 1«·—3ft—·+«—3 f t -

m C_ Input

•γ 150 h.p.

3 tonf

FIG. 2 4 7

- 3 f t » h 3ft—»j

I

6 in dia.

3 tonf

F o r bending only :

Compressive stress a t X = Tensile stress a t Y ,

My and / where M = 3(3 χ 12) tonf in.

= 108 tonf in.,

π

64

~2 ~ ~ 2 :

108 χ 3

y

x 64 = 62 in4,

3 in.

62

± 5 - 2 tonf/in2.


F o r torsion only :

where J = 21 = 124 in4,

/ 33 ,000 χ 150

~ \ 2π χ 100 12 χ 1-5,

= 142,000 lbf in.

= 63-3 tonf in.

63-3 χ 3

~ Ϊ 2 4 '

= 1-53 tonf/in2.

= I- { - 5-2 + ]/[5·22 + (4 χ 1-53

2)]}

= H - 5 - 2 + 6-0)

= - 2 - 6 + 3 0

= + 0 - 4 tonf/in2.

fantn = ~2

*6 - 3-0

= 5*6 tonf/in2.

X : h + V ( / » + 4/î)]

tan 2θα1 = r-

t Λ 2 ö n = 30°30

Γ,

2 χ 1-53

- 5 - 2 = + 0 - 5 9 .

fln = 15°15 '

At Υ = Κ + 5 - 2 + 6-0) ,

= + 5 - 6 tonf/in2.

/ Λ η , η = έ ( + 5 · 2 - 6 · 0 ) ,

= - 0 - 4 tonf/in2.

= - 0 - 5 9 .

θν1 = 74°46 ' . (see Fig . 248)


Examples I X

1 . Calculate the values of the principal and maximum shear stresses and the inclinations of the planes over which they act given that

fx = 8-0 tonf/in2 (horizontal),

fy = 3 0 tonf/in2 (vertical),

fz = 0, fs = 3-0 tonf/in

2 clockwise in the plane of fx and fy.

Check the values obtained by means of the Möhr Circle and insert them on a suitable sketch (9-4, 1-6 and ±3-9 tonf/in

2, θ1 = 25-1 deg).

2. State the five main theories which have been advanced with the object of setting up a criterion of elastic failure under complex stress. Discuss briefly their relative merits.

Fio. 248


3. The bending moment in a 2 0 in. dia. solid shaft rotating at 300 rev/min is numerically equal to 40 per cent of the torque. If the maximum shear stress in the material is 5 tonf/in

2 calculate the power being transmitted

(778 h.p.).

4. The flywheel of a single cylinder gas engine weighs 2-0 tonf, has a radius of gyration of 2-5 ft and is mounted on a 6-0 in. dia. shaft which overhangs the bearing by 1-25 ft. If, as a result of fluctuation in torque, the flywheel has a maximum instantaneous angular acceleration of 4-8 rad/s

2 determine

the maximum shear stress induced in the shaft (0-89 tonf/in2).

5. Find the ratio of twisting and bending moments in a solid circular shaft given that the minimum principal stress is numerically equal to one fifth of the maximum principal stress (TjM = ^(20)/4).

10 SM

C H A P T E R Χ

BEAMS IV—DEFLECTION

Flexure and Radius of Curvature

I n accordance with the sign convention, a clockwise bending

moment to the left of a section is taken as positive. Such a bending

moment induces a " sagg ing" curvature i.e. concave when viewed

from above as in Fig . 249 .

Positive bending moment M

Negative Q| (sagging) detlection y

Graph of slope | f

FEG. 2 4 9

P a r t of a sagging beam of varying radius of curvature, i.e.,

deflected b y a varying positive bending moment is shown in

Fig . 250 .

2 7 8

B E A M S I V D E F L E C T I O N 279

Slope of beam over element ds,

$ - = UnO, dx

·' dx* = sec 0

2

dx

differentiating,

sec20 4 - - -τ—·

ds dx

sec20 x — x sec0

R

(and sec20 = 1 + t a n

20 ) .

. 1 &y

" R dx2 (1 + tan

20)

3'

and tan 0 = dy_

dx

dx2 1 + V.

FIG. 2 5 0

Since deflection is small within the elastic limit, dy/dx is of the

order of small quantities and i ts powers may by neglected.

1 d2y

Thus, in this case — = - — - . R dx

2

Similarly, for a negative bending moment resulting in a "hogging "

curvature, i.e., convex when viewed from above:

1 d2y

— = — · Also, from the bending equation,

l M xl x . , 1 d2y M

so tha t m general : _ = ± _ = —

Hence, when / is constant,

This is known as the Differential Equation of Flexure.

ί M d2y\

Note tha t when / is not constant —— = ± Ε —— . V I dx

2 J

10*


I f both sides of this equation are integrated with respect to χ

then, ±EI^- = f M dx, where M is expressed in d x

J terms of x.

i.e., slope 4 - = ± —!- f]\ldx. ax EI J

Again, deflection y = ± -^j- j*JM dx.

The above general expressions may be applied to specific cases of

loading, care being taken to obtain the correct sign for d2yfdx

2.

Referring to F ig . 249 i t can be seen tha t the slope dy/dx in-

creases from negative through zero to positive so tha t d2y/dx

2 is

+ v e . in this case. Conversely, for the same axes, a hogging (nega-

t ive) bending moment makes d2yldx

2 negative.

Horizontal Cantilever with Concentrated Load at Free End

FIG. 2 5 1

F o r any section X X distant χ from the origin 0 :

MX = EI—^ = - W(L-x),

:. EI^r- = -W(LX -^L) + A, and when x = Q,^L = 0 , dx \ 2 / da;

Hence, EI^- = - W (LX -dx \ 2 /

Λ - 4 = 0 .

B E A M S I V — D E F L E C T I O N 281

dy from which the s l o p e - — a t any point can be found.

ax

f x2 1 x

z \

And, Ely = -WlL— - — — I + B, and when χ = 0 ,

y = 0, .'.B = 0.

Hence, Ely = - W ( γ χ2 -

from which the deflection y a t any point can be found.

At the free end, y = ymAX and χ = L,

Ely* -'(τ»-τ)-

(Τ ~ ~ 6 ~ ) ' = — WL*

WL3

Thus, denoting the maximum deflection ymsix by Z, we have

3 EI '

The negative sign shows tha t the deflection is downwards from the origin considered.

E X A M P L E . A horizontal cable carrying a tension of 300 lbf is

a t tached to the upper end of a wooden pole 6 in. dia. 12 ft high,

the lower end being embedded in the ground.

I f Ε = 1·0 x 1 06 lbf/in

2, find (a) the maximum bending stress

in the pole and (b) the deflection a t the top.

Solution

2nd Moment of area of section

64 ' π

144 in

X

y = ¥ = 3 i n .

64

= 63-65 in4.

and M = 300 χ 144

= 43 ,200 lbf in.

77777^

-300 Ibl

-6 in dia.

\777,

Ρίο . 252


.*. Ground level bending stress a t y = 3 / My_

I 9

43,200 χ 3

= 2035 lbf/in2.

Deflection a t upper end,

1

WL3

EI

300 χ 1 4 43

1-0 χ ΙΟ6 χ 63-65

= 4-7 in (approx.)

E X A M P L E . A short horizontal canti lever 12 in. long, 3 in. wide

and 5 in. deep of material having Ε = 13,500 tonf/in2 carries a

concentrated load a t the free end. I f a spirit level half-way along

registers an incline of 1 in 2000 , est imate the value of the load.

Solution

« L = 12 in - >>

1 1

1 u _

dx; 1

J Υ

i I

3 in

- 5 in

FIG. 2 5 3

2nd Moment I 12 '

_ 3 x 53

12 9

= 31-25 in4.

Mx = EIiâ== - w(L~x)> EI

ày

dx •W Lx — A,

B E A M S I V — D E F L E C T I O N 2 8 3

and when χ = 0, ^L = 0, t\ A = 0 . dx

.·. BIEL da;

dy

•W

and when χ = 6 , -τ- = — ——— , ' dx 2 0 0 0 '

1 3 , 5 0 0 χ 3 1 - 2 5

2 0 0 0 = -WÇ72 - 1 8 ) ,

whence, W = 3 - 9 tonf.

E X A M P L E . A cantilever of circular section is 4 8 in. long. Estimate the diameter necessary to prevent the maximum stress from exceeding 9 0 0 0 lbf/in

2 when there is a concentrated load of 5 0 0 0 lbf

at the free end. Calculate also the corresponding deflection at the load point, taking Ε = 2 9 x 1 0

E lbf/in

2.

Solution

FIG. 2 5 4

Bending moment, M = 5 0 0 0 χ 4 8 ,

= 2 4 0 , 0 0 0 lbf in.

/ = 9 0 0 0 lbf/in2,

y = - ~ , where d is the required diameter,


Since,

9000 -

My

240,000 χ 4"

nd1

64

120,000 χ 64

Hence,

9000π

= 2 7 1 .

d = 6-5 in.

r π χ 6 ·44

64

= 86 in4.

1 WL3

Deflection a t free end, Ζ = —- χ 3 EI '

1 5000 χ 4 83

χ 3 " 29 χ ΙΟ6 χ 86 '

= 0-074 in.

E X A M P L E . An I-section jois t 10 χ 6 in. overall has a web-J in.

thick and flanges ^ in. thick. I t is built into a wall with the web

vertical for use with lifting tackle , the overhang being 7 ft 6 in.

Additional support is given to the free end by a steel t ie bar 1 in.

wide and f in. thick, the upper end of which is rigidly fixed 24 ft

above. Take Ε for cantilever and tie as 13,400 tonf/in2 and est imate

the deflection produced by a concentrated load of 5 tonf at the free

end. F ind also the ratio of the tensile stress in the tie to the maxi-

mum bending stress in the jois t .

Solution F

Tensile stress in tie bar /

Extension in tie bar χ

1 χ 0-375

Ε '

F 24 χ 12 χ 0-375 Ε

—, F .


1

"Î2

19QQ

~ Ϊ 2 ~ '

158 in4.

(6000 - 4 1 0 0 ) ,

7 f t 6 in

5 tonf

24 ft

FIG. 255

Deflection a t free end of cantilever,

Ζ = WL

3

3EI

(5 - F) 9 03

3E χ 158 '

and this is equal to the tie bar extension.

10 a SM

where W = 5 - F, L = 90 in,

F o r cantilever section,

/N.A. = - ^ - [ ( 6 x 103) - (5-625 χ 9

3) ] ,

286 S T R E N G T H O F M A T E E I A L S

Equat ing, 770 4r= ! 5 4 0 Ε

110F = 7700 - 1540.F,

7700 F =

2310 '

3 - 33 tonf.

Hence,

.'. Stress in tie

Bending stress

Z = 1 ^ 4 ^ = 0-192 in. 13,400

3-33

0-375 = 8-9 tonf/in

2.

My

.*. Stress ratio

_ (1-67 x 90) 5

158

= 4-75 tonf/in2,

8-9

4-75 = 1-87.

E x A M P L E . A cantilever is made from steel for which Ε = 13,000

tonf/in2 and has a uniform section for which / = 200 in

4. I f a

concentrated load of 5 tonf is applied a t the mid-point, calculate,

assuming only the differential equation of flexure :

(a) the slope of the beam a t the load point, (b) the deflection a t the load point, (c) the deflection a t the free end, if L = 240 in.

Solution

FIG. 2 5 6


At any section X X :

ΜΎ = EI

dx

dx2

I20x

-5(120 - x),

r2 1 faj

+ A, and when χ = 0 , = 0 , dx

Λ ^ = 0 .

A t the load point, i.e. when χ = 1 2 0 :

EI dy_

dx - 5 (120 χ 120)

1 2 02

500 χ 144

/ . Slope *L dis-

in tegrat ing again :

- 3 6 , 0 0 0 ,

36 ,000

13,000 χ 200 '

- 0 0 1 3 8 5 .

Ely = - 5 ^

2 1 #

3

1 2 0 T - I T + B} and, when a; = 0 , y = 0 ,

.'. J5 = 0

= - 3 0 0 . T2 + —X

s

Ό

At the load point, i.e. when χ = 1 2 0 :

Ely = - ( 3 0 0 χ 120 2 ) + ( - | - x 120 3J,

- 4 , 3 3 0 , 0 0 0 + 1,450,000,

- 2 , 8 8 0 , 0 0 0

13,000 χ 200 '

— 1-107 in., i.e. downwards from the origin 0,

= Z2 (Fig. 256)

In the diagram, Z1 = Slope χ 120 ,

= - 0 - 0 1 3 8 5 χ 120 ,

- - 1 - 6 6 0 in.

10a*


Deflection a t free end, Z% = Zx + Z2,

= - 1 - 6 6 0 - 1-107,

= - 2 - 7 6 7 in.

E X A M P L E . I f the deflection a t the top of the 2 in. dia. pole

illustrated in Fig . 258 is to be zero and the bending stress is not

to exceed 2000 lbf/in2, find

(a) the rat io P i / P 2 and

(b) the values of Ρλ and

P 2 ( I . Mech. E . ) .

Solution

2nd Moment,

0-785 in4.

Bending moment, M = i l y

where y = 1-0 in ,

2000 χ 0-785

1 0

= 1570 lbf in.

F o r Ρ2 alone,

Mx = E I ^ = P2(l20-x)) dx*

When χ = 0 ,

7T = °> dx

0 .

When χ

dy

120 ,

EI dx

dx

1 2 02 -

7 2 0 0 P 2

1 2 02\

EI and

dy

dx

Straight part

g l

120 in

//////////{///

FIG. 2 5 7

I oft

F I G . 2 5 8

2 0 f t

'//////A

;. Ζ = 864,000 — f - · EL

120

B E A M S I V — D E F L E C T I O N

y»2 1 >

289

Integrat ing again, J J J ! , = P 1 ( 1 2 0 1 -

and y = 0 when χ = 0 , .'. Β = 0 .

When χ = 1 2 0 , „ / 1 2 0

3 1 2 0

3\

6 ) '

= 5 7 6 , 0 0 0 P 2 and y

: . z 2 = 5 7 6 , 0 0 0 A .

Tota l deflection due to P 2 , Z 3 = Z 3

= (864,000 + 576 ,000) - = f , ILL

:. Zz = 1 , 4 4 0 , 0 0 0 - ^ .

F o r Px alone, deflection a t end = Τ

P1 χ 2 4 03

X EI '

i.e., Ζ = 4 ,608,000 Pi EI '

Since this is to be equal and opposite to Z3,

4,608,000 = 1 , 4 4 0 , 0 0 0 ,

Ρτ 1-440 5

" P2~ 4-608 ~~ 16 '

Mmax = 1570 = 1 2 0 P 2 - 2 4 0 P 1 / •

[—*~p\

g where P 1 = — P 2 , /

= P* 3 / 24C ) in

= P 2( 1 2 0 - 7 5 ) , /

15 //////////////

Hence, P1 = 10-9 lbf. FIG . 2 5 9


- [L -x ] y =z 'mox.

Τ PIG. 2 6 0

At any section X X distant χ from the origin Ο :

L - x .τ) (-

——{L2 — 2Lx + χη,

E Id y

dx

dy

A - 0 .

EI^-= - — L2x - Lx

2 + — - ) from which

dir 2 \ 3

and when χ = 0 ,

Hence,

dy the slope ——at any point can be found.

ÇX.X

w / _ _ χ2 _ X '

3 l a :

4. _

• · ί ί » = - τ Κ - £ τ + 8 Τ ι + ί ' and when χ = 0 , y = 0,

τ τ T IT w / £2

Hence, Ai?/ w I L

2

5 = 0 .

L x

z + — a;

4 ] from which the

3 12

deflection y a t any point can be found.

Horizontal Cantilever with Uniformly Distributed Load


A t the free end, y

:• Ely™, =

and χ = L,

2 \ 2 3 1 2 ) '

\2 3 +

1 2 / '

and wL = W (the to ta l load) ,

wL* / l

" 2 ~

wL*

8 and ym Ζ

i.e., 1 WL3

The negative sign shows tha t the deflection is downwards from the origin considered.

E X A M P L E . A uniform cantilever has a depth of 1 8 in. and a

length of 1 4 ft. A load of 4 tonf is to be concentrated a t the free end

and a second load of 6 tonf is to be uniformly distributed throughout

the length. I f the maximum bending stress is to be limited to

5 tonf/in2 find the least necessary value of / .

I f Ε = 1 3 , 0 0 0 tonf/in2, find the deflection a t the free end.

W,=4 tonf

Solution

At the fixed end, M„ ( 4 χ 1 4 ) +

= ( 5 6 + 4 2 ) 1 2 ,

= 9 8 χ 1 2 ,

= 1 1 7 6 tonf in.

( · * τ ) 1 2


j — tJL } where y = 9 in ,

_ 1176 χ 9

" 5 '

= 2117 in4.

The tota l deflection is the sum of the deflections due to each load

acting alone, i.e. 1 WXU 1 W2L*

3 EI 8 EI '

L3 / 4 6_\

~ " ^ T i 3~ + 8"/ '

_ (14 χ 1 2 )3 25

~ 13,000 χ 2117 X

"Î2 '

= 0-36 in.

0-25 in

E X A M P L E . A cantilever consists of a triangular steel plate of

density 0-28 lbf/in3 and having the dimensions shown. Assume only

the differential equation

of flexure and calculate

the deflection a t the free

end when loaded uniform-

ly with 4 lbf/in2 of plan

area.

The 2nd Moment of the

section varies linearly

from zero a t the free end,

so that , for any section

X X

χ Ix = —1 , where

Ε

I — value at support.

I f w = load/in2, then load -p IG

causing bending on sec-tion X X = w xbx/2 and this acts a t χβ from X X .

Hence, EL dx

2

bx ••= —w X

X x 3-


i.e., Ε xl d

2y

L dx*

wBx2

EI

When χ

dy_

dx

6

dy wB xz

~dx =

6~~ 3~

wB χ*

wbx2

6

(1)

where b = —B, Ε

+ Ä (2)

18 4 Ax + B (3)

FIG. 2 6 3

When χ = L ]

y = 0 J " Hence, in E q . (3)

in E q . (2)

0 =

i n E q . (3)

V 18

whence A = wBL*

18

χ L* + ^ — — x L j + B, 72

whence J5 = wBIA

~24T

wB wBLz wBL*

When χ = 0 , A/ i / W

^ and = jr/ the to ta l load,

W ^3

1 2 ^ 7

To ta l load/in2 = (0-28 X 0-25) + 4-0 = 4-07 lbf.

' 6 χ 10\ W = 4-1

= 122-1 lbf.

and I = 6 χ 0 -25

3

12 128


Λ Deflection, 1 2 2 - 1 χ 1 0

3 1 2 8

y = 1 2 χ 3 0 χ 1 0

6 χ 1

= 0 0 4 3 5 in.

E X A M P L E . A tota l load of 2 0 tonf is distributed over a uniform

section cantilever 1 0 ft long, the intensi ty of loading varying

uniformly from zero a t the free end to a maximum a t the fixed end.

Obtain an expression for the bending moment in terms of distance

from the free end and plot this a t intervals of 2 ft over the length,

using the following scales : 1 cm = 2 ft, 1 cm = 1 0 tonf ft. Es t ima te

the deflection a t the free end, given tha t 7 = 1 6 0 0 i n4 and

Ε = 1 3 , 4 0 0 tonf/in2.

Solution

Tota l load W = Mean loading χ L,

/ M a x . rate of loading\

.'. Max. rate of loading = 2W

L

Max. rate of loading

70

tonf f t

FIG. 2 6 4


d% R a t e of loading w

x = EI (i.e. proportional t

o x)

2W χ

EI d^r

da;4

2W

L2 χ x.

Shear force

When χ = 0

Fx = 0

. " . - 4 = 0

d3y 2W tx*

EI d

3y

dx3

W

L2 χ x*

Bending moment, Mx = EI d

2y

When χ = 0

i f , = 0

. \ 5 = 0

= - -

dx2

W

W a;3

ΊΪΧΤ + Β·

where i f = 20 tonf

and L = 1 0 f t ,

20

3 χ 1 02

= - 0·0667;ζ3.

χ χ3,

The table of bending moments is then as given in Fig . 264 .

F rom above, „Td

2y W

EI dy

dx

W fx*

and when χ = L, dx

= 0

3 L2 \ 4

L4

+ C

Hence dy

El-rr- = dx

;. Ely =

Λ + ( 7 = 0 i.e., C=—. 4 4

I 4 4 / '

(a;4 - £

4) .

3 £2

1 2 £2

IT /a ;5

1 2 L2

At any section XX:


when χ = L1

y = o|

.·. Ely

Hence, deflection y

and is a m a x . when χ

Simply Supported Beam ivith Central Concentrated Load

ο* * Για. 265

At any section X X distant χ from a central origin Q (Fig. 265) ,

„T&y W I L \

5 5 '

TF / a ;5 „ 4 Z

5\

= - w ( - 5 - - ^+

— ) -

1 2 Ä J £ » I 5 5 j = 0 .

12EIIß[ 5 / '

_ WL3

15EI '

2(10 χ 1 2 )3

~ _

15 χ 13,400 χ 1600 '

= — 0 4 7 0 6 in, i.e. downwards.


and 4 " = 0 , when χ = 0 , A = 0 . ax

EI — W

<L

dx

_W(L * \

~ 2 [Τ* 2 J d?/

from which the slope - j — a t any point can be found.

W / L χ* 1 ζ3 \ _ _

Again, Ä J „ = —(— — - γ—j + 5 and y = 0

when χ = 0 , .*. -B

I f / L ^2 z

3\

= τ ( — - τ ) from which the deflection y a t any point can be found.

At a support, when L

% = γ , y = 2/max = " Ζ ,

where Ζ is the maximum deflection (at the centre) .

W/L L* 1 L*\

·'· Α / ^ = τ ( τ χ Ί Γ - β X" T J ' _ J f / Iß_ _ L*_\ ~ ~ 2 ~ \ ~ Î 6 ~ ~~ 48 ) '

if J L) V 32 96 / '

P F £3

48

1 TfX3

2/max = ~ # ] p ' upwards relative to Q.

Hence, Ζ 1 WL*

48 A /

The negative sign shows tha t the deflection is downwards relative to the origin 0 .

The fractions which precede the quant i ty WL3/EI are known as

Deflection Coefficients.

E X A M P L E . T W O self-aligning bearings 10 ft apart support a 2-5 in. dia. shaft. A t the centre of the span is a pulley to which a transverse load of 600 lbf is applied. Find, from first principles, in degrees, the inclination of the shaft a t the bearings.


Find also, by any method, the deflection of the shaft a t a point

3 ft from a bearing. Ε = 30 x 1 0e lbf/in

2.

Solution

F o r shaft section,

= 1-92 in4.

π / = — χ 2-5*,

L = I0 f t =120 in

FIG. 2 6 6

At any section X X ,

and

EI

d2?/ W (L \

dy M

dy_

dx

dx 2 \ 2 " 2

0 when χ = 0 , . ' . - 4 = 0 .

1 T dy W iL2 1 L

2

Ί ΊΤ dy W (L* 1 L2\

* ^ = ^ ( τ - 2 Ί γ ) ' w h e n *

. dy

L_

2

W (— —

" dx _ 2ËF\Ï 8~

WL2

WEI'

600 χ 1 2 02

16 χ 30 x 1 0e χ 1-92

= 0-00938.


Hence, angle of inclination when χ = —

θ = t a n1 0 -00938 , = 0°36 ' .

W l L χ2 1 χ

3 \

Again, Ely = — ^— — — — j + Β and y = 0 ,

when # = 0 , Β = 0 ,

_ TT / £ a2

2 \ 4

_ IT / 120

χό

— [—— χ 2 42

2 \ 4 when χ = 24 in

Hence. 600 χ 2 4

2

2 43\

" 6 ~ ; '

i.e., a t 3ft from one end.

( 3 0 - 4 ) , 2 χ 30 x 1 0e χ 1-92

0-078 in., upwards from the lowest point, i.e.

from the centre.

Max . deflection, WL

3

EI 1

600 χ 1 2 03

48 χ 30 χ ΙΟ6 χ 1-92

= 0-375 in .

.·. Required deflection = 0-375 - 0-078,

= 0-297 in., downwards from the original

position.

E X A M P L E . Three identical joists are built into a wall a t the same level and equidistant, the distance between adjacent joists being equal t o the protruding length. A fourth jois t of the same section but twice the length is mounted across the free ends of the other three, and a vertical load concentrated a t the centre. Show tha t the central cantilever will deflect twice as much as the outers.

Solution

I f ρ = upward force exerted b y end cantilever on cross member, then force producing deflection in cross member is : W — P, when Ρ is the upward force due to centre cantilever.


From the diagram, Z 3 = Z2 — Zx,

1 (W - P)L* _ 1 PP 1 plz

48 # 7 3 EI 3 EI

(W -P)L* ( P - y)

48 3

2pZ,3 (W - 3p)l*

" ' 48 3

8 ( J F - 3 p ) = p ( y )3 and j = 2 ,

8 I f - 24p = 8p

3 2 ^ = SW

W i.e., ρ = — .

[sub-for (W - P) and (P - p)]

W W Hence, p=W-2p = W - -^ = - - .

Δ Δ

The central cantilever thus carries twice the load of an outer and

hence deflects twice as much.

31, = wL IL

wL (L

~ 2

(j~X)-W(-2 -X)

4

wL2

8

wxz

~ 2 ~

l/L X ,

2 2

W iL* ο r \

2{-T + x 2 - L x ) wL

2 wLx wL

2

wx* 8 + •

wLx

w/unit length

wL

2

FIG. 2 6 8

• • ^ d e »_

2 U r

when x = 0 , - ^ = 0, A = 0.

Simply Supported Beam with Uniformly Distributed Load

At any section X X distant χ from a central origin Ο :



„ T dy w IL2 x

z\ , . . . t. . dv

*. EI -p— = — I -£-χ ^-1 from which the slope a t any

point can be found.

w iL2 χ

2 1 x*\

• ' • m y = Y [ - T x Y - J x T ) + B w h en *YZ0\ :. Β = ο!

_ R M> / L2x

2 X

s- \ , . , , . , .

.'. Ely = — — — I , from which y a t any point can be 2 \

8 1 21 found.

A t the support y = y ^ and χ = γ ,

w / i2 i

2 1 L

4 1

w / i2 1 £

4\

• • • Ä i y - - T l T x T - l 2 x l 6 J ' w ί L* \

~ ΊΓ\32" ~~ 102/

• ( - - — ) \ 3 2 1 9 2 /

w £4 / 1

~ 2 ~

w L4 5

2 1 9 2 and w £ = ΡΓ the to ta l load.

5 w L3

' y - = 3 8 4 X ΎΓ ( u p W a r d 8)

Λ Ζ == ~3§ïië~ ( d o w n w a r d s)-

E X A M P L E . A uniform beam of length L and depth d is to carry

a to ta l load of IT tonf uniformly distributed. I f the maximum

deflection is not to exceed L / 4 0 0 when the maximum bending stress

is 8 tonf/in2, show tha t L > 20d. The beam is simply supported

and Ε = 1 3 , 5 0 0 tonf/in2.

Solution

5 WL* „ L Maximum deflection Ζ = — - rrr— and Ζ > ——

3 8 4 EI 4 0 0

. _ L _ _ _ 5 _ W X3

' " 4ÖÖ ~ 3 8 4 ^ 7 '


i.e., 3 8 4 EI

2Ô50" W 9 where I = My

f WL d 1

8 XT * T

L2

or

3 8 4 χ 1 3 , 5 0 0

2 0 0 0 χ 1 2 8

L = 20-25<Z.

χ Ld

E X A M P L E . A brass strip 3 in. χ 0 - 7 5 in. χ 1 2 0 in. weighing

7 6 lbf is slightly curved. The centre deflection is 1 - 6 4 in. when

supported a t the ends with the 3 in. width horizontal. This becomes

1 * 1 3 in. when the strip is turned completely over. F ind the sag due

to the original curvature and determine the value of E.

Case©

3 8 lbf

FIG. 2 6 9

Solution

I f χ — sag due to curvature,

then deflection due to weight == 1 - 6 4 -

and

Hence,

B u t deflection Ζ

where I

1 1 3 + 0 - 2 5 5

/. Ε •-

χ, in case ( 1 )

= 1 - 1 3 + x, in case ( 2 ) .

1 - 6 4 - χ = 1 - 1 3 + x,

2x = 1 - 6 4 - 1 - 1 3 ,

χ = 0 - 2 5 5 in.

5 χ

WL3

3 8 4 " EI

3 χ 0 - 7 53

1 2 0 - 1 0 5 5 in

4.

χ 7 6 χ 1 2 0

3

3 8 4 " 0-1055.Ë7 '

5 χ 7 6 χ 1 2 03

1 - 3 8 5 χ 3 8 4 χ 0 - 1 0 5 5 '

1 1 - 7 x 1 0E lbf/in

2.


Uniformly Loaded Cantilever, Supported at Free End

As already shown, without the support Ρ

1 WL3

8 EI

^ ι h

0 • L

k z,

/ W = total load = wL

FIG. 2 7 0

I f the free end is propped up to the same level as the wall sup-

port then

Z 2 - Z 1 = 0 9

1 PL* ~t · . . .

and is positive, i.e. upwards. where Z0

3 EI

Thus, λ ™ 3 EI

1 WL3

8 EI whence Ρ = —W.

Uniformly, Loaded Beam, Supported at Ends and Centre

As already shown, without the support Ρ

= 5

WL3

1 - 384 # J '

An upward force Ρ will produce a centre deflection

1 PL3

Z2 = + 48 ,E7


whence,

I t follows tha t each end support carries —-^ W.

FIG. 2 7 1

Assuming the prop Ρ not to sink, i t will, if elastic, shorten by an il Ρ

amount χ = — , where / = —— and I = length of prop. Sect ion

E X A M P L E . A short vert ical elastic strut of height h and section A supports the free end of a horizontal canti lever of length L and 2nd Moment of area I.

I f the lower end of the strut does not sink when a to ta l load of W is distributed uniformly along the cantilever, show tha t the load i t carries is given b y

3 AWL* F =·

24ΖΛ + SAL*

(E has the same values for strut and cantilever.)

Hence, for the centre to be raised to the level of the ends,

Z 2 - Zx = 0 ,

i.e. 2 2 = ^ι·

1 PL* 5 WL*

48 EI "~ 384 EI


Solution 1 WL

3

Deflection a t free end due to W = ——=y—, downwards. 8 LI

1 FL3

Deflection at free end due to F = ————, upwards. 3 LI

FIG. 2 7 2

The difference between these deflections is the reduction (x) in the

height of the strut.

B u t , χ = , where / = direct stress in strut, L

_ . F A

" Τ x ¥ "

_ Fh 1 i f L3 1 FL

3 . t

H e n c e ' Ü F = 8" " I T ~ ¥ ·anrl £ c a n c e l s o u t -F_

• • 1 WL

3 FL

3

8Ih BIh

„ , 1 L3

01 Fl^r + m i.e.

WL3

SIh

I 3Ih + . 4 £3 \ _ 1 T £

3

\ / 8 J A '

so that ,

whence,

F(SIh + ^ L3) =

F =

3 AWL3

8 '

_ 3 £ ] f L3

_

2 4 / f c + 8 J L3

E X A M P L E . A B e a m of I section 1 4 in. deep and having I = 4 3 9 i n4

carries a load of 1 5 tonf uniformly distributed over a simple span

of 3 0 ft. I t is then propped a t the centre so tha t the deflection there


FIG. 2 7 3

Before Rz is introduced, the centre deflection is given by

5 WL*

3 8 4 X ~ËT '

5 1 5 χ 2 2 4 0 ( 3 0 χ 1 2 )3

3 8 4 3 0 χ ΙΟ6 χ 4 3 9

1-55 in.

The upward deflection produced by R3 is given by χ

Ί KK Bs ( 3 0 χ 1 2 )3

i.e. 1-55 = - 4 - χ ν

'

4 8 ΕΙ

4 8 3 0 χ ΙΟ6 χ 4 8 9 *

1-55 χ 4 8 χ 1 0Β χ 4 8 9

Β = 3

~ 3 02 χ 1 7 2 8 '

= 2 1 , 0 0 0 lbf,

= 9 - 3 6 tonf.

D D 1 5 - 9 - 3 6 R

i = R

2 = Ö = 2 - 8 2 tonf.

is zero. F ind

(a) the load on each of the three supports,

(b) the positions of the points of contraflexure,

(c) the magnitude and position of the max . positive bending

moment,

(d) the bending moment and stress a t the centre.

Ske tch the Bending moment diagram. Ε = 3 0 x 1 0E lbf/in

2.

Solution


F o r any section X X , Mx = 2-S2x - (o-5x χ -jj + 9-36 (x - 1 5 ) ,

working from the L H end.

= 12Λ8χ - 0-25z2 - 140 ,

= 0 a t points of contraflexure.

12-18 ± V[12-182 - (4 χ 0-25 χ 140)]

·"' X

" 0^5 '

= 18-72 f t , or 30 ft which can be ignored.

Thus the points of contraflexure are a t (18-72 — 15) or 3-72 ft from

the centre or 11-28 ft from the ends.

Alternatively, working from the R H end,

= 2·82χ - (θ·5χ x | = 0 ,

x2 = 4 χ 2-82o:

i.e., # = 11-28 ft.

= 2-82 - 0·5.τ = 0 for max. positive B .M. , i.e. χ = 5-64 ft. da:

.·. M = (2-82 χ 5-64)

- (0-25 χ 5 -642) ,

= 15-90 - 7 - 9 5

= 7-95 tonf ft.

(posit ive).

At the centre, F l G

* 2 74

M = (2-82 χ 15) - (0-5 χ 1 5 ) 7 - 5 ,

= 13-8 tonf ft.

(13-8 χ 2240 χ 12) 7 .·. Max. stress in beam =

439

5900 lbf/in2.

T h e B . M . diagram is therefore as shown.

Beam Subjected to Uniform Bending Moment

As already shown, if the supports are symmetrically placed as

in Fig . 275 then the bending moment between the supports is


uniform and equal to — Wl. Since M = EIjB, i t follows tha t R is

constant between the supports, i.e., this part of the beam bends into

an arc of a circle.

F o r any section X X between A and Β

MX = E I ^ = - Wl

EI dy

dx

and, when

•Wlx + A

L dy

0 = -Wl^- + A, Δ

w giving A = — x IL.

: . E I ^ - == - wix dx

WIL

from which the slope a t

any point between the

supports can be found.

4 L

»

1 x

X Z=

ymox

FIG. 2 7 5

Again, Ely = - WI χ χ

2 WILx

i.e.

Wl

wi

+ Β and, when χ = 0 , y = 0 ,

.*. 5 = 0 .

2 # /

( L * - χ2),

( 2 λ Ε — α*2),

from which the deflection a t any point between the supports can be found.

11 SM


The maximum deflection Ζ occurs a t the centre, when χ = —

\ L x j -Ζ wi r 2EI

WIL2

ML2

(since Wl = M) SEI ' SEI

which is positive i.e., upwards from 0 and relative to point A .

E X A M P L E . A copper rod 48 in. long 0·75 in. dia. was supported

horizontally a t points 6 in. from the ends. When loads of 25 lbf

were hung from the ends the measured deflection a t the centre was

0-1 in.

A tes t length of 6 ft was then subjected to a torque of 600 lbf in

and the twist found to be 1-078 deg. Es t imate the value of Pois-

son's Ra t io .

Solution

- 4 8 in-

25 lbf 25

— 36 in —

FIG. 2 7 6

Π oL 25 lbf

Between the supports the bending moment is uniform and is

given b y :

i f = - (25 χ 24) + (25 χ 1 8 ) ,

- - 25(24 - 18 ) ,

= - 25 χ 6

= - 150 lbf in.

Deflection a t centre, Ζ

o-i

.*. Ε

ML2 ι γ π /(Λ and 7 = _ ( 0 · 7 5 )

4,

SEI

150 χ 3 62

8 # χ 0 0 1 5 5 '

150 χ 3 62

0 1 χ 8 χ 0-0155

15-7 χ 1 0β lbf/in

2.

0-0155 in4.


where J = -Ξ- (0-75)4 - 0-031 in

4.

Hence G 6 00 x 6

0 0 3 1 χ 0-0188 '

= 6-18 χ 1 06 lbf/in

2.

1^-7 γ 1 0e

Since Ε =2G{1 + a), l + , = _ _ = 1 4 7 .

a = 0-27.

Simply Supported Beam with Ν on-central Concentrated Load

Moments about Ο give :

R2L = Wa, i.e. R2

Wa

Similarly,

FIG. 2 7 7

Wb R, =

Wb F o r any section X X to the left of W, Mx = — - x x.

L

Τ GO τι I n torsion, — = —— where θ = 1-078 χ ——

J L 180

= 0-0188 radian.

1 1 *


dx L Jx dx — kW J (x - a) d(x — a)

Thus the 1st term of E q . (1) must be integrated with respect to χ

while the 2nd term must be integrated with respect to the bracket

(χ - a),

riTdy I Wb χ2 \ JTI_/ x - a \

2

1Χ>·' A / d H — ΧΎ + Λ)- kW(—2— ) • ( 2)

wrîere A is a constant .

Similarly, Ely = χ £ + A* + B) - (^J, (3)

integrating the last term with respect to (x — a) as before.

Wb F o r any section X X to the right of W, Mx = — χ χ

Ε — W (χ — a).

Thus, in general, to avoid using two equations, d2y Wb

Mx = EI = -j- x x - kW(x - α), (1)

where k = 0 for χ < a, i.e. where the bracket in the 2nd term is

negative, and k = 1 for χ > α, i.e. where the bracket is positive.

Note: If there are two loads there will be three equations, the last of which can be employed in the same manner, terms including negative brackets being omitted as required. The method applies for any number of loads and is due to Macaulay.

X

dy Wb Γ Γ Integrating, EI - j — = -j— \ xdx — kW J (x — a) dx.

ô χ α χ

B u t , kW J (x - a) dx = kW j (x - a) dx + kW j (x - a) dx, o o o

χ

= 0 + kW j (x - a) dx a

(because k = 0 , when χ < a), x - a

= kW I (x - a)d(x - a).


At the L H end when χ = 0, y = 0 and k = 0, i.e. the term in

(x — a) is omitted. Hence, in E q . ( 3 ) :

0 = B.

At the R H end when χ = L, y = 0 and k = 1, i.e. the term in

(x — a) is included. Hence, in E q . (3)

W

i.e.,

and (L

so t ha t

0

A

a) =

Wb L*

6 (L-a)\

W Wb

~6L χ L

2

b, L = α I δ, L2 = a

2 + 2a& + 6

2.

Hence, =

6 £

6 L

Wb

W

6

6

6L

Wba2

a2 + 2a6 + &

2)

21Ta&2 J F 6

3

6 L

χ x°

foe3

W_

6

6 L

(a - a )3 +

6 L *

2Wab2

6 L 6 L

(substituting for A in E q . (3 ) ,

a2bx 2ab

2x

bx

W_

6

iL 6

- (ζ2 — a

2

- ( a2

2ab) -

(x — a )3

ix — a)

bx

2ab - x2) + (x - a)

and a2 + 2a& + b

2 = L

2,

i.e. a2 + 2a6 = L

2— b

2,

(L2 - b

2 - x

2) + (x - a)

3

Thus the downward deflection y a t any point distant χ from the L H end is given by


the term (x — a)z being omit ted when χ < a i.e. to the left of the

load. A t the load point, when χ = α,

y W

W

6EI

Wa2b

2

3EIL

ή - (L2 - b

2- a

2)

E and L = a + δ,

ab [a

2 I- 2ab + 62) - b

2 - a

2

The negative sign indicates tha t the deflection is downwards from

the origin considered. The deflected shape is as shown dotted in

Fig. 278 .

I 1

FIG. 278

The maximum deflection occurs when the slope dy/dx is zero and

in Fig . 278 this is clearly to the right of W, i.e. χ > a.

If , for convenience, a is made greater than b as in Fig. 279, then

zero slope occurs when χ < a.

Fie. 279


A 6L 6 L

Wb 9 Wba2 2Wab

2

0 = — — χ χ2

or, 0 = bx2

i.e., 0 = x2

2L 6L 6L

ba2 2ab

2

3 3

a2 2ab

3 3 '

a2 + 2ab

3

L2 - b

2

3

b2

(and a2 + 2ab = L

2 - b

2),

(or 3x2 - L

2 - b

2),

P i L L

2 3L

2

Note tha t since b < —, .'. b2 < —-, i.e. 3x

2 > ——from above,

2 4 4

or χ2 >

L2

Τ L

and χ >

T o obtain the maximum deflection, the above expression for χ

may be substituted in E q . (3), the 2nd term being ignored.

The slope a t the ends may be found by calculating the value of the constant A and substituting in E q . (2), care being taken to omit terms as required. The actual inclinations of the beam (or shaft) in degrees are given by θ = t a n

- 1 dy/dx for each end value

of dy/dx, i.e. when χ = 0 and χ = L.

F o r more than one load the deflection a t any point is the sum of the deflections due to each load acting alone.

E X A M P L E . T W O self-aligning bearings 120 in. apart carry a steel countershaft 2-5 in. dia. At 4 0 in. from one end of the shaft is a pulley on which the resultant transverse load due to a bel t is

Thus, in E q . (2) when dyjdx = 0, the 2nd term is omitted and

the constant A has the value already found i.e.,

Wba2 2Wab

2


1 6 0 0 lbf

- 4 0 in- 8 0 in -

Fig . 2 8 0

Moments about R H end give :

1 2 0 ^ = 600 χ 8 0 , whence R1 = 400 lbf and R2 = 200 lbf.

Working from the L H end :

= (8000 χ 14,400) - (8000 χ 6 4 0 0 ) .

1 2 0 4 - - 8 0 0 02,

Λ Α = - 5 3 3 , 0 0 0 .

600 lbf. Determine in degrees, the inclinations of the shaft at the

bearings. Ε = 30 χ 1 06 lbf/in

2.

Solution

'hi X

Μτ = El^r- = 40Os - 600 (x - 40 dx

2 (1

r3 fx — 4 0 \

3

and Ely = 200— - 300 ί j +Ax+B.

For L H end, .τ < 40 , .'. 2nd term omitted. Also when

(2)

(3)

χ = 0, y = 0, ,\B = 0.

F o r R H end, χ > 40 , 2nd term included. Also when

χ = 120, y = 0 .

Hence in E q (3) above : 0 = ^2. χ 1 2 03 - 100(120 - 4 0 )

3 + 120 ,4 ,


F o r R H end, χ = 120 and 2nd term is included since χ > 4 0 ,

dy

dx E I ^ - = (200 χ 120

2) - 300(120 - 4 0 )

2 - 533 ,000 ,

= 2 ,880,000 - 1,920,000 - 5 3 3 , 0 0 0 ,

= 4 2 7 , 0 0 0 .

dy 427 ,000

' ' dx ~ 30 χ ΙΟ6 χ 1 92

.·. ΘΗ = t a n1 0 -00742,

= 0°25 ' .

= + 0 0 0 7 4 2 ,

E X A M P L E . A beam 20 ft long and simply supported a t i ts ends

carries a load of 0-5 tonf a t a point 12 ft from the L H end. The

beam is of I-section 1\ in. χ 5 in. overall, flanges and web being

each 0-3 in. thick. Take Ε = 13,500 tonf/in2 and find

(a) the deflection under the load and (b) the maximum bending stress.

Solution 2-5 χ 5

3 2-2 χ 4 - 4

3

' N . A . - ja ja ,

= ^ ( 3 1 2 - 5 - 1 8 7 ) ,

_ 125-5

Ϊ 2 ~ '

- 10-46 in4.

I I a S M

Hence in E q . (2) above: EI^- = 200χ·2 - 300(χ· - 4 0 )

2- 5 3 3 , 0 0 0 .

dx

F o r L H end χ = 0 and 2nd term is omitted since χ < 4 0 .

E I ^ = - 5 3 3 , 0 0 0 and / = -^r(^X = 1-92 in4,

dx 64 \ 2 /

. dy 533 ,000

" d ^ - 30 χ 10* χ 1-92 ^Q

'Q Q 9 2 5

>

.·. ΘΣ = t a n1 0 -00925,

= 0°32 ' .


Ζ = ZEIL '

α = 12 χ 12 = 144 in.

b = 8 χ 12 = 96 in.

L - 20 χ 12 - 240 in.

-2-5 in-

0-3 in

V/////,

0-3 in 0-5 tonf

5 in

12 f t -

- 20 f t -

RL

F i g . 2 8 1

F i g . 2 8 2

.'. Ζ = 0-5 χ 1 4 4

2 χ 9 62

3 χ 13,500 χ 10-46 χ 240 '

= 0-942 in.


RL χ 20 = 0-5 χ 8 ,

Max. B . M . occurs a t Load point.

0-5 χ 8

20

0-2 tonf.

BL x 12 ,

0-2 χ 12 χ 12 ,

28-8 tonf in.

y 28-8 χ 2-5 / m a x

i Um a x ^ J 10*46

= 6-9 tonf/in2.

Wa*b* At the load point,

where


E X A M P L E . A brass beam 3 6 in. long of l i n . square section is

simply supported a t i ts ends and carries a concentrated load of

2 4 lbf a t a point 6 in. from the R H end.

Take Ε = 1 4 x 1 06 lbf/in

2 and calculate

(a) the deflection a t mid span,

(b) the deflection a t the load point and

(c) the position and magnitude of the maximum deflection.

Solution


3 6 2 ^ = 2 4 χ 6 ,

2 4 χ 6 R1

R2 = 2 0 lbf.

0 - 5 χ 0 - 53

Ϊ 2 ~

3 6 = 4 lbf.

0 - 0 0 5 2 in4.

- 3 0 in-

R, = 4

2 4 lbf

- 6 in-

R2=20

FIG. 2 8 3

0-5 in

Έ2Ά Ό·5ίη

Mr = EI d

2y

dx2

dy E I ^ = dx

Ely =

4;r - 24(.τ - 3 0 ) .

4x2

Λ Λ (χ - 3 0 )2

2 4 - + A.

2x3

~ 3 ~

( 1 )

(2)

(3)

F o r L H end χ < 3 0 ,

y = 0, Β = 0 .

.'. 2nd term omit ted. Also when χ = 0 ,

l i a *


F o r R H end χ > 30 , .*. 2nd term included. Also when

x = 36 , y = 0 .

3 63 12

.·. 0 = 2 χ — — ( 3 6 - 3 0 )3 + 36-4 (sub. in E q . (3 ) ) , ό ό

= 31,104 - 864 h 36.4 , A - - 3° f * ° .

=-- - 8 4 0 .

At mid-span, χ = 18 and the 2nd term is therefore omitted in

E q . (3) above.

1 83

Hence, Ely = 2 χ — (840 χ 18 ) , ο

= 3888 - 15,120

= - 1 1 , 2 3 2 .

Π>

2 3 2 fti^r-

• •y

= - 14 χ IQ« χ 0-0052 = " « j i5

^

At the load point :

I f aa6

2 24 χ 3 0

2 χ 6

2

J 3EIL 3 χ 14 χ ΙΟ

6 χ 0-0052 χ 36 :

F o r max . deflection :

i/L*-b* Ί/ ( 3 6 + 6 ) (36 - 6) i / 4 2 x 3 0 x = ν—γ—= r — s = I/—*—

= | /420 = 20-5 in.

20· 53

Λ EIym = 2 x — (840 χ 20 -5 ) ,

= 5850 - 172,000,

= - 1 1 , 3 5 0 . 11,350

·· * ~ = - 14 χ 10« χ 0-0052 =

E X A M P L E . I f Ε = 13,600 tonf/in2, and I = 1200 in 4,


find, for the beam shown

(a) the position and value of the maximum deflection and

(b) the deflection under each load.

4 10 t o n f

8 f t » | « » ~ 4 f f » |

R L( = 6 - 4 )

L = 2 0 f t -

FIG. 2 8 4

Solution

Moments about L H end give: RR χ 20 = (4 χ 8) + (10 χ 1 2 ) , whence RR = 77-6 tonf. and RL = 6-4 tonf.

F o r any section X X , Mx = EI = 6·4£ - i(x - 8)

- 10(χ - 12) (1)

(χ - 1 2 )2

(2)

A t L H end, when χ = Ο, y = Ο in and the 2nd and 3rd terms in E q . (1) are omit ted; .'. Β = 0 . A t R H end, when χ = 2 0 , y = 0. Hence in E q . (3),

0 = ^ χ 2 03) - ( | - χ 1 2

3) - (4- x δ 3) + 2 M >

whence,

3 " " / \ 3

A = - 3 2 6 in ft units.

The deflection is a maximum at the point of zero slope. I f this is assumed to be between the two loads, then in E q . (3),

0 = 3 ·2^2 - 2(x - 8 )

2 - 326 (neglecting 3rd term),

= 3-2x2 - 2(x

2 - I6x + 64) - 3 2 6 ,

= l-2x + 32.τ - 4 5 4 .

- 4 3 2 ± V{322 - 4 [ 1 · 2 ( - 4 5 4 ) ] } _ 32 -j- 56-6

2 χ 1-2

= 10-25 ft (ignoring negative value).

2-4

, ^ = 6 ^ - 4 i ^ - 1 0 i ^ ^ + ^ dx 2 2 2

n n a:3

ft (α - 8 )3 , (.τ - 1 2 )

3

Λ Ely = 3-2 — - 2 v ;

- 5 — — + Ax + Β Ο Ο Ο

(3)


Putt ing this value in E q . (3) and changing all units to inches,

1 3-2(10-25 χ 1 2 )3

2(2-25 χ 1 2 )3

(326 χ 122)(10·25 χ 12)

4!r (1980 - 1 31

- 5 8 0

° ) ' EL

ΙΟ3 χ 3830

13,600 χ 1200 = - 0 - 2 3 5 in.

Under the 4 tonf load, i.e. when χ = 96 in. the 2nd and subsequent

terms are ignored.

3-2 χ 9 63

EI

- 0-226 in.

(326 χ 122 χ 96)

Under the 10 tonf load, i.e. when χ = 144 in. the 3rd term only is

ignored.

EI

3-2 χ 1442 2 χ 4 8

3

3 (326 χ 12

2 χ 144)

= - 0-230 in.

E X A M P L E . Pa r t of the mechanism of a recording machine

consists of a steel beam 12 in. χ 0-25 in. χ 0-125 in. supported a t

A and Β as shown. Determine the load required a t point C halfway

between the supports to give a movement of 0-1 in. at the free end.

W h a t will be the ratio between this movement and tha t a t the load

point? Take Ε = 30 χ 1 06 lbf/in

2 and assume only the differential

equation of flexure.


Solution

0-1 in

FIG. 2 8 6

EI

T7r dhj WIE \

dy _ W (Lx _ Χ2

\ Λ _ Δ

d ^ " ~ 2 " U Ύ) + Α

'

ay when x = 0, —2- == 0 , ;.A=0.

dx

L L1T dy W IL L 1 L » '

• è l - w (El Ελ -

"~dx~ JET \~4 8 /

a t B).

WL*

10ËT'

The slope a t Β is tha t of the straight part, i.e.

and

Equa t ing ,

_ 0-25 χ 0 -1253

~ Ϊ2

= 0-0000407 in4.

0-1

7

0-25 in •

t 0-125 in N-

1

FIG. 2 8 7

W χ 52

.'. W =

16 χ 30 χ ΙΟ6 χ 0 0 0 0 0 4 0 7

0-1 χ 480 χ 44-2 7 χ 25

11-1 lbf.


W ί Lx2 χ3 \

Integrating again, Ely = ~2~l~~g jg~J + B> w n e n

% = 0, y = 0, ,'.B=0.

At the ends where y is a max. and equal to the centre deflection,

L x = - .

W iL L2 1 L

3

2ËI \T X ~ 4 ~ " 6"

X ~ 8 ~

WL3

4SEI

1 2 1 χ 53

4 8 χ 3 0 χ Ι Ο6 χ 0 - 0 0 0 0 4 0 7

= 0 - 0 2 3 8 in.

Λ Deflection ratio = 0 - 1

0 0 2 3 8 = 4 - 2 .

E X A M P L E . A beam freely supported over a span of 2 5 ft has

concentrated loads of 2 tonf and 8 tonf a t points 5 ft and 2 1 ft

respectively from the L H end. Find, in terms of Ε and / , the

deflection a t a point 8 ft from the L H end. I f the beam is now

additional supported a t this point, so tha t all three supports are a t

same level, find the magnitude and direction of the load on each

support.

Solution

In general,

Ely =

Ely,

6

8

ΊΓ

8

1Γ

bx (L2

+ {χ - a )3

4 χ 8 ( 2 52 - 4

2 - 8

2)

2 5

3 2 ( 6 2 5 - 1 6 - 6 4 )

2 5

8 χ 2 2 χ 5 4 5

6 χ 2 5

( 8 - 2 1 )3

(neglecting negative term),

- 9 3 0 .


W2=2 W,=8

- o 2 = 5 - - b 2= 2 0

- a , =21 Hb ,=4 -H

x=8

L = 2 5

FIG. 2 8 8

20 χ 8 ( 2 52 - 2 0

2 - 8

2)

25

160(625 - 400 - 64)

(8 - 5)*

25 + 27

_ / 160 χ 161

~~ \ 25

= 344 + 9 ,

= 3 5 3 .

. • . ^ + ^ 2 = 930 + 353 = 1 2 8 3 , giving y1 + y2

1283

Upward deflection due to R3 = ZEIL

J R O / 82 χ 17

2

, where a = 8

and b = 17 ,

R^/i EI \

2 4 7 # 3

~ ^ 7 ~

3 χ 25

^. 247 _ 1283 Equat ing, — - R3 - jß , = = 512 tonf.

247 EI * EI 7

Moments about L H end give,

(R2 χ 25) + (5-2 χ 8) = (8 χ 21) + ( 2 x 5 )

168 + 10 - 41-6 R9

25

136-4

~ 2 5 ~ '

5-46 tonf.

.·. R1 = 10 - (5-2 + 5-46)

= —0*66 tonf. i.e. downwards.


73 WL3

χ Z a 648

N EI,

w w

FIG. 2 8 9

i 1

W

FIG. 2 9 0

Solution

E a c h cantilever carries W and has a length of L/2.

.*. Deflection a t free end = -ί- W

EIr

WL3

2iEIv

F o r the arrangement shown,

W

6EI

bx (L

2 - b

2 - χ

2)

+ (χ - α )3

W

6EI

bx (£2 _ δ

2 _ a;2

since .τ < a.

I n the given case, χ =

κ .1 — ι

FIG. 2 9 1

.*. Centre deflection due to each load = -— W

6 t f ( / , / 2 )

X 1

Τ x Τ

Example . I f J c a n l i l c v cr = 2 / b c a m, in Fig. 289,

show tha t


Tota l beam deflection a t

centre

2 χ W

3EIr 6 \ 9

9EIC \ 9

WL* I _1_ _

L2

Τ

L2

~4~

9EIC

WL* I 23

EIC 9 3G '

ZA = J F L

3 / 23

23 PTL3

324 X

1

EIC \ 324 ^ 2 4 / '

73 WL* χ 648

Encastré Beam with Central Concentrated Load

This type of beam has the ends built in, but without longitudinal constraint. The shape under load is as shown dotted, and, since the slope a t the ends is zero, there must be a fixing couple a t each end in addition to the reaction. This is CW a t R H end and ACW at L H end.

F rom Symmetry , R = R = W/2 and ML = MR.

F o r any section X X to the left of W, F ig . 292 ,

π τ cry vvχ

αχ 2 2

W r3 r

2

(1)

(2)

(3)


At the L H end, when χ = 0 , 4^ = 0 , Λ A = 0 .

dx Also when χ = 0 , ?/ = 0 , δ = 0

Λ'οίβ: All constants of integration are zero for built-in beams.

-φΞ^Ι0·75ΐη

At the centre, when

Hence in E q . (2), ο

or

W 1? L — χ Ύ - ML χ γ

WL2

16 '

WL

8

Hence E q . (2) becomes :

dx

Wx2

WLx

From this the slope a t any point to the left of W can be obtained. And E q . (3) becomes:

Ely Wx'

~I2~ - WLx2


Prom this the deflection a t any point to the left of W can be

obtained. The deflection is a maximum a t the centre where χ = L/2.

W EIymax = - j ^ " x

L*

~ 8 ~ ~~

WL

16

* - y max EI y "96" ~ ±) 6 4 /

or Z= 192

WL3

or Z= 192 EI

4 '

i.e. downwards from the origin considered.

I n E q . ( 1 ) :

Wx

2

Wx

ML,

WL

2 8

The B . M . diagram is

thus the normal triangle

A B C less the rectangle

A D E C of height WL/S

and the ne t B . M . is re-

presented by the shaded

area.

When χ = L / 4 there

are two points of con-

traflexure, Ρ and Q.

Since

WL

8 '

= 0 .

w L

The Shear force diagram is as shown. The beam is thus twice as strong and four t imes as stiff as when simply supported.

M

y? Datunrr

t i

FIG. 2 9 3


Β.ΛΙ. Diagram jar Fixing Moments alone

I f a fixing moment is applied as shown a t point E , then the

reaction a t the L H end must be downwards for equilibrium. Hence

the reaction a t Ε (the moment of which is zero about E ) must be

upwards.

The directions of these reactions R,{ and R{ will be unchanged by

the introduction of a fixing moment Mr provided tha t this is less

than Mu.

Μ κ

FIG. 2 9 4

For point Ε , clearly, Clockwise Moment = Anticlockwise Moment,

i.e. MR - ML = RL χ L.

For any section X X , Fig . 295 ,

Mx = — RLx — ML ^ where RL = ——j-—-fcom abovej .

Since this is a straight line law, the B . M . diagram for unequal

fixing moments alone is as shown in Fig. 296 .

1 X

* — X

1 ^

Ι 1

FIG. 2 9 5

Encastré Beam with Non-Central Concentrated Load

The reactions will be unequal and so will the fixing moments required a t the two ends.

η . - )

FIG. 2 9 7

X X ,

M* =

ΕΙ^ΖΞ =

Ri

x -

ψ(

χ - a) - ML

EI

F o r any section X X ,

dx2

dx

(x — a)2

MLx

BL X>_ W {x - af 2 3

(1)

(2)

(3)


When χ = L,

F I G . 2 9 6


And in E q . (3), 0 = ^ L3 - (L - a)

3 -

.'. —- L3 = (L — a )

3 + ( a n d £ - o = 6 ) ,

Ό Ό 2

Wa2b

Similarly, MR = ^2

Wb2 2 Wab

2

Hence from E q . (4), n L = —JJ- + — x

Wb2

• (L + 2a)

Wa2

Similarly, RR =~jjr(L +

2b)

Now, E I

^ = R

L Τ ~ Ύ{ Χ

-= ^

(Eq*

( 2 ) )'

At the R H end, when χ = L, y = 0 and —— = 0 . da;

Hence in E q . (3), 0 = - ^ " ^2 - - « )

2 ~

Λ J | l / v2 = -51 (£ _ « ) 2 + j f A£ ( a nd L - a = b)

Wb3 M L

W&3 I f , i l f f

Equat ing (5) and (4), + 3 = + 2

(which eliminates RL),

ML ML Wb* Wb3

L L I ? 1? '

Wb2

whence, ML = —jj- [L — b) and L — b = a.

Wab2

.'. Mr = —— -£ 2

(5)

• 7? W h" . 2

M' < (4)


The maximum deflection occurs a t the point to the left of W where

dy —— = 0 and χ < a. dx

Wb2

Wab2

Hence in E q . (2), 0 = —JJ- {L + 2a) ^ „

(ignoring 2nd term and substituting for BL and ML).

Wab2 Wb

2

e. — - — χ = L 3 (L + 2a)Y

L + 2a x

or

L 2 9

2aL

L 4- 2a for max . deflection.

B y substituting the above for χ in E q . (3), the actual maximum

deflection can be found.

F rom E q . (1) , B

Mx = RL χ - ML

when χ < a.

The B . M . diagram is thus

the normal triangle A B C

less the trapezoid A D E C

due to the fixing moments.

T h e ne t B . M . is then

represented by the shaded

area, D E beingthe datum.

The Shear force dia-

gram is as shown.

E X A M P L E . The ends of

the built-in beam shown

are a t the same level, / for

the section is 387 i n4 and

Ε for the material is

13,500 tonf/in2.

Assume only the differential equation of flexure and find the

position and magnitude of the maximum deflection.

κ\//////////////////77Λ

FIG. 2 9 8


Solution

L e t the fixing moments and vertical reactions be as shown.

Then E I ^ = Mx = RLx - ML - \0{x - 12)

„Tdy ΒΣ 9 „ i . f - 1 2 )2

(1)

(2)

10 tonf

- 1 2 f t -

- 2 0 f t -

FIG. 2 9 9

R r r3 x

2

ο

(3)

When χ — 0, both y and dy/dx are zero so tha t both A and Β are zero.

At the R H end when χ = 2 0 , = 0 and « = 0 . dx

I n E q . (2) 0 - ~ x 2 02 - 20ML - 5(20 - 12 )

2,

Δ

:. 200RL = 2 0 i l f i + 3 2 0 ,

10 -H 1 6 .

And in E q . (3) 0 = χ 203 - χ 20

2 - - | (20 - 12)

3,

υ Δ ό 8000 π „ Λ Λ 1, 2560

.·. — - £ Λ = 200JÎZ. + — — ,

Λ n n = — M L + 0-64. (5)

(4)


Equat ing (5) and (4), 10

+ 1-6 = ^ M L + 0 - 6 4 ,

- 20

20

1-6 - 0-64,

Hence,

ML = 19-2 tonf ft. = 230 tonf in.

10

3-52 tonf.

1-6,

When dy/dx = 0 , y is a maximum and this point clearly is between RL and W.

In E q . (2), 0 = — 19*2.1' (ignoring the term (χ — 12) ) .

Max. deflection ymilx

3 ·52

2 χ 19-2

3-52 '

10-9 ft ( = 131 in.).

1

EI 1 3 1

2

1

13,500 χ 387

- ( 2 · * . , , .

x KM2 3-52

- - 0 - 1 2 6 in .

Encastré Beam with uniformly Distributed Load

The fixing moments ML and MR are again equal while each reaction is half the to ta l load wL.

F o r any section X X to the left of the centre,

MX = M ^ = ^ - M L - W X , * - (1)

^Tày wL χ2 w x

3

ΎΊΤ wL χ3 ^r x

2 w x*

(2)

(3)


At the L H end, when χ = 0 ,

ax

\ A = 0 .

( - -

X w/unit length V///

WX

ο - *L

L" T

Straight

FIG. 3 0 0

and when

At the centre, when

χ = 0 ,

y = 0 ,

5 = 0 .

L

4 =o. da;

. Λ L2

Hence in E q . (2), 0 = —— χ —— 4 4

_ M r Χ —

L3

τχ—' L

M L x Y

wL3 wL

3

16

wL3

wL3

24

W J y2

12

48

— --M 16 48 / '

(= WL/12. As will be seen later from the graph, this is the max. value.)


From this the deflection a t any point to the left of the centre can

be obtained.

The deflection is a maximum a t the centre when χ = L\2 wL L* _ wL

2 L

2 L*

% m a x - 72" x

8 - ~2Γ X Τ " 24

X 16

wL χ L* ( 1 1 1 \ . . £ = ί / , η η ν = 77T 1 — - — - ΤΓ777" I and wL = W,

1 WLZ

i.e. downwards from the origin considered. F rom E q . (1) ,

wL2

wLx w _

~ 2 2 12

The graph of M is thus the normal parabola A B C less the

rectangle of height wL2\\2 and the net B . M . is represented by the

shaded area.

The height of the shaded portion above the line D E is

w/S (L — 2xp)2, where L — 2xp = base of small parabola and

xp = distance of point of contrariexure Ρ from L H end.

w wL2 wL

2

From the graph, — (L - 2xp)2 = — — .

SL2

(L - 2xp)2 = L2 - 12

L2^

Hence E q . (2) becomes

^T dy wL 0 wL2 w 0

dx 4 12 6

F rom this the slope a t any point to the left of the centre can be

obtained. And E q . (3) becomes

^T wL _ wL2 . w .

Ely = —— χ3 - —τ—χ

2 - —x

4.

* 12 24 24


i.e. L - -E = 2xp, whence xp - ~ ^1 - - ^ - j ,

= 0-21L approx.

E X A M P L E . A uniform built in beam carries a distributed load of w/it over two thirds of its length, beginning a t the left hand end. F ind the values of the fixing moments given tha t for a point load the fixing moments a t left and right hand ends are respectively Wab

2jL

2 and Wa

2bjL

2 where a f b — L and a is the distance from

the load to the left hand end.

Solution

Fixing couple required a t L H end due to load element w άχ

FIG. 3 0 1

.'. To ta l fixing couple required a t L H end will be the sum of such

couples between χ = 0 and χ = 2L/3,


ç 0 b

^

WOB2 ' / / ^ '; ; - WOB

* - L - - - - -I

— * A h L -x v

^ ^ w/unit length >^

FIG . 3 0 2

2 A/3

i.e., M l

^ l ß J ( Ι Λν

* 2 L : r2 :i

'3)

clr>

ο

w TJ2 16 4 \ 2 r_

Fixing couple required a t R H end due to load element w àx

w ax χ x2(L — x) w

= - ρ = -jj (Lx2 - -τ

3) dx.

2L/3

Λ M«=Jjf ( L a

2- *

3) dx,

0

_ W

\L — - — *LIZ

~ U \ X

~ 3 ~ ~ " T 0 '

E X A M P L E . F ind the reactions and fixing moments for the system

of loading shown in Fig . 3 0 3 , and sketch the graph of M.


12 tonf

FIG. 3 0 3

F o r any section X X as shown :

Mx=EI^jL= RLx - M j - \2{x - 5)-w(x - 10) { X

^ (1)

dy EI-?- = RL— - MLx - 12

dx L

2 L

2

(x - 5 )2 w (x - 1 0 )

3

+ A (2)

^ T R t XZ , r X

2 (χ - 5

3) w (a - 1 0 )

4

6 + Ax + £

(3)

When dy/dx = 0, χ = 0 and functions in brackets are ignored, .*. A = 0. Also when dy/dx — 0, y — 0 and 5 is similarly zero.

When χ = L

dx

y = ο

In E q . (2) 0 = ^ L2 - i f 7 i - 6 ( i - 5 )

2

Δ

-—(L - 1 0 )3 (and Ζ = 2 0 ) ,

2 02

0-75 - 6 ( 2 0 - 5 )

2 _ ( 2 0 - 10 )

3,

or

= 2 0 0 £ L - 2 0 i f L - 1350 - 125 ,

1475 = 200Ä,, - 2 0 J f L (4)

Solution


·. I n E q . (3) 0 = - ^ - Z3 - ^ L* - 2(L - 5 )

3 - ^ (L - 10)*,

_ 8000 400 = -Rz, —w ML— 6750

0-75

or

6 ^ 2 24

= 1 3 3 3 J ? L - 200ML - 6750 - 3 1 3 ,

7063 = 1 3 3 3 i ? L - 2 0 0 J f L

(4) χ 10 gives 14,750 = 2 0 0 0 i ? L - 2 0 0 i f L

(6) - (5) gives 7687 = 666RL

(10 ,000) ,

(5)

(6)

i.e.

F rom (4)

RL = 11-53 tonf ,

.·. RR = (0-75 χ 10) + 12 - 11-53

- 7 -97 tonf .

20ML = (200 χ 11-53) - 1475 ,

2306 - 1475 M L = 20 '

= 41-56 tonf f t .

Moments about L H end give :

MR = ML + (12 χ 15) + (0-75 χ 1 0 )5 - (RL χ 2 0 ) ,

= 41-56 + 180 + 37-5 - 230-6 ,

= 28-46 tonf f t .


Μ I

FIG. 3 0 4

12 SM


E X A M P L E . A beam A B has a clear span of 3 0 ft and is built in at

A and B . I t carries a load which varies from zero a t A to 1 tonf/ft

a t Β plus a concentrated load of 1 0 tonf a t a point 1 0 ft from A . Determine the reactions and fixing moments a t A and Β and the

deflection a t the centre of the span, given tha t Ε = 1 3 , 2 0 0 tonf/in2

and I = 7 0 5 in4.

Solution


Intensi ty of loading χ 1

F k j . 3 0 5

Tota l load on part of length

χ = χ χ Mean load,

Η τ χ ! » ) ·

and this ac ts a t # / 3 from X X .

1 χ

2 X 30 τ -

l 0 (x

R \X x°

EI dx

χ·

180

.2 ^4

= R x ~2

1 0 ) - MA,

10{x - 1 0 ) - Mx,

1 0 )2 - MAx - f A,

x* 10_.

EIy = Rx-3 6 0 0

J _ 0 (χ - 1 0 )

3 - Μ χ - •Ax + Β.

At the L H end when χ = 0 , y = 0 and dy/dx = 0 while the 3rd term is omitted since χ < 1 0 . Hence both A and Β are zero.

At χ = 30

= 450ÁΛ - 1125 - 2000 - 307lf A .

. '. 3 0 3 / A - 4 5 0 Α Λ - 3125

- 4500ÄA - 6750 - 13,330 - 4 5 0 i f A

4 5 0 i l f A = 4500ÄA - 20 ,080

(1)

(2) 450itf A = 4 5 0 0 ^ A - 20 ,080

F r o m ( l ) : 3 0 0 J f A - 4 5 0 0 f l A - 3 1 , 2 5 0 ,

i .e. , (2) - (1) gives 150MA = 11 ,170 ,

/ . i f A - 74-4 tonf f t .

Substi tut ing in E q . (1 ) : 4 5 0 i ? A = (30 χ 74-4) + 3125

whence RA — 11-87 tonf ,

/. R = (15 + 10) - 11-87

= 13-13 tonf .

Moments about Β give: MB =» MA + (10 χ 20) + 15

- (11-87 χ 3 0 ) ,

= 74-4 + 200 + 150 - 356 -1 ,

= 68-3 tonf f t .

( 1 5 3 \ ΊÊ5 i n

n ' 8 7 > ! T r ) - i « J τ - - r < 1 5 ^ 1 0 » '

= 6670 - 211 - 208 - 8 3 7 0 ,

= - 2 1 1 9

and Ε = 13,200 χ 1 22 tonf/ft

2,

705 I = " j 2 T

ft

4. (Note foot units.)

2119 χ 12*

' 'y " 13,200 χ 1 2

2 χ 705 '

= - 0 - 0 3 2 7 ft = - 0 - 3 9 2 in .



Moment-Area Method

L e t F ig . 306 be part of a bending moment diagram and Fig . 307

the deflected shape.

FIG. 3 0 0

In tercept cut off by) tangents at a and b j |

( + v e when tangent at a ~ i ~ strikes the vertical above ' the tangent at b )

dz

'b Deflectei

shape

FIG. 3 0 7

The area under the graph of M between points a and b on the deflected shape is given by

b

A - j'Max. a

The first moment of this area about the J f - a x i s is given by b

Ax = j Mx dx, where χ = distance of centroid of A from M-axis .

S i n c e " pL^-l^M Λ [*>L = _ L fMdx dx

2 EI J dx2 EI J

i.e., dy

b _ A

~dx a

= ~ËÎ


Thus Slope a t b — Slope a t a = I/ΕΙ χ Net Area under graph of M between a and b .

Now dz = χ άθ

dx = ΧΊΓ'

where — = M

M x dx

Αχ ; . j dz=-^jj Mxdx i.e. Ζ

a

Thus, for a given vertical, In tercept Ζ = —^-r χ 1st Moment of Area. EI

I f a p o i n t o f zero slope is taken as point b [i.e. (dy/dx)h = 0 ] , and the vert ical is drawn through point a, then the tangent from b will

FIG. 3 0 8

be horizontal as in Fig . 308 and the intercept Ζ will be the upward deflection of point a relative to point b .

Since ^ χ ( d r - a n

d ( <^τ ) b ~ ° '

Slope a t a = -φ-El

Ax and, relative to b, Upward deflection at a - Ζ — -ρ—.

El


I t is convenient to split the B . M . diagram into simple shapes and

find the first moment of area of each. Account must be taken of the

sign of I f , i.e. of positive and negative areas.

Simply Supported Beam with Central Concentrated Load

For the piece of beam ab :

Shaded area

WL

4

WL2

16

Λ Slope a t a

dy

L\ 1

A

EI

WL2

WEI FIG. 3 0 9

dy Intercept Ζ = Deflection a t centre where —— = 0,

dx Ax

EI '

WL2

16

WL*

4:$EI 9

WL*

where χ = ~

ο

1 L X EI

X T'

L_

Ύ L^

48EI

upwards from b,

, downwards from a.

Simply Supported Beam with Uniform Load

F o r the piece of beam ab :

2 Shaded area A = (Enclosing rectangle),

ό _2(WL L

WL2

24 '


Uniformly Loaded Cantilever

F o r the piece of beam ab.

Shaded area A = — (Enclosing rectangle) , «5

3 \ 2

WI?

WL χ L

Υ FIG. 310

Intercept Ζ = Deflection a t centre where = 0 ,

Αχ Λ 5 L 5L = -τ—-, where χ = χ — = ——, since curve is

EI 8 2 16 u v parabolic,

WL2 1 5L

~~~24"~ x

ΎΓ x

le"' = "384"

X ~W'

uP

w a r ds f r om b-


Slope a t a = + - τ — , since tangent a t b strikes the vertical above a,

. (dy\ WL2

Intercept Ζ = Deflection of a relative to b,

Ax

where χ = 3L

(Since

curve is parabolic.)

WL2 3L

1 WLZ

FIG. 311

Leaf, Laminated or Plate Springs

The simplest type has one leaf and is an inverted, simply sup-

ported centrally loaded beam as shown.

Maximum stress / = My

WL t 12

A W L

2 bt*


Central deflection

1 WL* Ζ = — - X 48 EI

9

WL* 12

4 8 # bt*

1 WL*

1~EbF'

τ

FIG. 3 1 2

Curvature

F o r a spring to be jus t flat under the operating load, i t must be given an initial curvature. The value of this may be obtained as follows :

B y geometry :

R2 = (R- Z)

2 + (AJ

R2 = R

2 — 2RZ + Z

2 + —

4 L* É

.·. 2RZ-Z* = —, — r r 4 /

Z(2R -Z) = ¥-, /

2Â=ºæ+æ> / . „ L* Ζ „_ i.

/ n L2 \ FIG. 3 1 3

\o r

~8Z A

P PR O X

- j

Thus if L is known and Ζ is specified, R may be calculated.

12 a S M


Multi-leaf Type

L e t a rhombus of thickness t be supported along i ts diagonal A B and loaded with W\2 a t the ends of the other diagonal of length L

as shown in Fig . 314 .

At any distance χ from the L H end :

X 2i71/l)X Width of section =

x nb =

2nd Moment /

Bending Moment M

Depth of section y

L/2 L

I 2nbx \

\ L ) X

__ nbt*x

12 ~ 6L

Wx

2 t

~2'

, My Wx t 6L 3 WL .'. Stress on section / = — — = — — χ χ —

2 2 nbt*x 2 nbt2

This is I jn of t ha t with a single rectangular plate. I t is also indepen-

dent of χ and hence is uniform throughout the span.


I f the rhombus were cut along the dot ted lines as in F ig . 3 1 5 ( a ) .

and the pieces of width b/2 assembled as shown in F igs . 315 (b) and

315 (c ) , i t would become a mult ideaf spring having η leaves of

width b, except for the pointed ends.

To form the eye, the ends of the centre (main) leaf are made

rectangular .

( 0

xn leaves of width b

F I G . 3 1 5

Since the beam is inverted, EI à

2y

άχ2 -M.

. à2y

άχ2

M

Wx _1_ 6L

Ί Γ X Ύ + nbFx~ 9

3WL

nEbt*

Integrat ing,

12 a*

ày

dx

3WL

nEbfi χ x + A


and when

Hence,

Integrat ing,

and when y

dx U

' X

" 2 ' 3 W L L

• 0 = - « X T + I ' A 3

da;

2 wJ^W3

3 J F L 3 J F Z2

χ a; + • nEbt*

3WL

2 nEbt* '

nEbt* X

2 +

2

3 WL2

= 0 , x = 0, ,\B = 0.

The maximum deflection occurs when χ L / 2 .

χ .τ + B ,

W Z2

_ WL 1_ Iß_ £

- J

~ ~ m_B^s x

J X

Τ +

Ί nEbt3

3WL3

nEbt3

3 WL3

( ϊ - ϊ ) ·

8 wJEW1

Quarter Elliptic Type

I n the expressions already ob-tained, ΡΓ must be written for W/2 (Le.2Wfor i f ) a n d Z f o r X / 2 (i.e. 2 £ for L ) .

This gives :

2W χ 2L 6WL

' 4

and

nbt2

2W(2L)*

nbt2

6WL*

8 nEbt* nEbt*

E X A M P L E . A vehicle spring has plates 3 in. χ 0-375 in. and has a FIG. 3 1 6

span of 36 in. I f the stress is not to exceed 15 tonf/in

2 when the load a t mid-

span is 0-5 tonf, est imate the required number of leaves and


calculate the deflection. Ε = 30 x 1 06 lbf/in

2. W h a t is the init ial

radius, if the spring is jus t flat under load?

Solution

0-25 font

3 WL

2 nbt2

0-25 tonf

FIG. 3 1 7

15 χ 2240 3 0-5 χ 2240 χ 36

~~2 X

3 (0 ·375)2 η '

3 χ 1120 χ 36 : 2 χ 3(0 ·375)

2 15 χ 2240

4-27.

A suitable number of plates is therefore 5 since a lesser number will

give a stress in excess of tha t specified.

0 - 3 7 5 in

3 in-

Deflection

Radius

FIG . 318

3 WL3

8 nEbt*5

3 x 1120 χ 3 63

8 χ 5 χ 30 χ ΙΟ6 χ 3 (0 -375)

3 '

0-825 in .

R = ~8Z approx

3 62

8 χ 0-825 '

: 197 in .


E X A M P L E . Calculate the leaf thickness for a semi-elliptic spring

if i t is to deflect 5 in. when centrally loaded with 1500 lbf and the

corresponding stress is not to exceed 75,000 lbf/in2. The effective

length is to be 40 in. F ind a suitable number of leaves and state

their width. Take Ε = 30 χ 1 06 lbf/in

2.

Solution

W= 1 5 0 0 lbf

Fro. 3 1 9

4 3 WL

nr A A A 3 1 5 00 X 40

Stress / - — x —rpr , Λ 75,000 = τΓΤΓζ 2 nbt- 2nbt

2

3 x 1500 χ 40 nbt

2 =

Deflection Ζ

75,000 χ 2 '

- 1-2.

3 WL* _ 3 χ 1500 χ 4 03

8 nEbt* ' ' * 8^(30 χ 106) bt*

3 χ 1500 χ 4 03

nbt* 5 χ 8 χ 30 χ 1 0

6

- 0-24.

nbt* _ 0-24

* ~ÜW = 1

~ 1-2

= 0-2 in.

1-2 B u t nbt

2 - 1-2, nb = — ,

tz

1-2

" (0-2)2

- 3 0 .

A suitable number of leaves would be 10 and this makes b = 3 in.


8 x 5

4 0 in .

/ _E_ . l_-UL Ί β ~ Ύ ' " 2 " Ε

/ 7 5 , 0 0 0 χ 4 0 \

I 3 0 χ 1 06 j '

0 - 2 in.

1 / Ε T MR WL I = — — , where M =

I R' ·· Ε 5

4

Hence / 1 5 0 0 x 4 0 \ 4 0

\ 4 / 3 0 χ 1 06

B u t ,

0 - 0 2 in4.

(nb) t*

nb =

1 2

1 2 /

1 2 χ 0 0 2

( 0 - 2 )3 '

_ 1 2 χ 0 0 2

0 - 0 0 8 '

= 3 0 , as before.

E X A M P L E : A cantilever leaf spring 2 4 in. long has 6 leaves 3 in.

wide and 0 - 3 7 5 in. thick.

W h a t load will produce a max . stress of 4 0 , 0 0 0 lbf/in2 and what

will be the corresponding deflection? Ε = 3 0 χ 1 06 lbf/in

2.

Alternative Solution to Previous Example

L2

In i t ia l radius R = , where Ζ = Max. deflection. oZ

4 02


Solution

f = GWL

nbt2

0 - 3 7 5 i n :

3 in

4 0 , 0 0 0 =

W =

6W χ 2 4

6 χ 3 ( 0 · 3 7 5 )2 '

4 0 , 0 0 0 χ 6 χ 3 ( 0 - 3 7 5 )2

6 χ 2 4

= 7 0 0 lbf.

- 2 4 i n -

\ n = 6

FIG. 3 2 0

Deflection Ζ = 6WL*

nEbt*

6 χ 7 0 0 χ 2 43

6 χ 3 0 χ Ι Ο6 χ 3 ( 0 - 3 7 5 )

3

2 in.

E X A M P L E . A locomotive plate spring has 1 6 plates 5 in. wide and

0 - 5 in. thick, and the span is 4 0 in. Take Ε = 3 0 χ 1 06 lbf/in

2 and

calculate the stress and deflection per ton of load.

Solution

Deflection

3_ WL*

" 8 X

nEbt* '

Stress

/ 3 WL

3 χ 2 2 4 0 χ 4 P3

: 2 χ 1 6 ( 3 0 χ 1 0

6) 5 ( 0 - 5 )

3

0 - 1 7 9 in.

nbt2 '

_ 3 χ 2 2 4 0 χ 4 0

~ 2 χ 1 6 χ 5 ( 0 - 5 ) 2

= 6 7 2 0 lbf/in2 ( = 3 tonf/in

2)

16 leoves(n)

I 2 2 4 0 lbf

FIG. 3 2 1


Summary of Conventions used in the Theory of Simple Bending

To achieve correct mathemat ical correlation between quantities

the following conventions must be observed.

The origin Ο should be taken

whenever possible a t the left of the beam as in Fig . 322 (a) which

illustrates (for the sake of argument) a simply supported beam

uniformly loaded.

Rela t ive to this natural origin the load

produces a negative deflection a t all points as in Fig . 322 (b ) .

A t all points such as Ρ to the r ight of

mid-span, àyjàx is clearly positive and the complete slope curve

(graph of àyfàx against x) is as in F ig . 322(c) .

Since the slope of the àyjàx curve (i.e. d2v/da;

2)

increases to a maximum a t mid-span and then decreases, the graph of d

2.y/dx

2 against χ is as shown in Fig . 322 (d) and is

wholly positive. F rom the differential equation of flexure: à

2y

M = EI -Tj^g- so tha t if the value of J is constant , the graph of

M (Fig. 322 (e)) will be identical in form and sign t o F ig . 322 (d),

the vert ical scale being multiplied by EI.

I f the curve of F ig . 322 (d) is differentiated

again to give à3y/àx

3 the derived graph has the form shown in

Fig . 322 (f) since the slope of the curve is initially positive and

decreases uniformly, changing sign a t mid-span.

Again, the shear force is given b y :

dM _ _r d2y

F = — — where Μ - EI —4 da; dx

2

^ da;3

The graph of F against χ (Fig. 322 (g)) is therefore identical in form and sign to tha t of F ig . 322( f ) , the vert ical scale being multiplied b y EL


FIG. 3 2 2

(α)

( b )

( c )

(d)

(e)

( f >

(g)

( h )


Final ly , the load intensi ty is given by :

dF

dx

= EI

where F = EI

d'y

à3y

dx*

M x = + Rx

4- ve. m o m e n t

(»)

+ ve. shear

( b )

and since dF/dx is everywhere negative, the graph of w against

χ is the straight line of F ig . 3 2 2 ( h ) .

F rom the foregoing i t is clear that , to obtain the correct signs

for M and F the following conventions must be used :

1. To the left of any

section X X (Fig.

323(a) ) a clockwise

moment is considered

positive and vice

versa.

2. I f the resultant trans-

verse force to the left

of a section X X

(Fig. 323(b ) ) is up-

wards, then the shear

force is considered

positive and vice

versa. FIG. 323

Examples X

1. The flexural rigidity (EI) of a uniform section beam is 120,000 tonf in2.

If the beam is simply supported over a horizontal span of 96 in. and loaded vertically with 5 tonf at a point 36 in. from the left hand end, determine the deflection at mid-span (0-7 in. approx.).

2. Equal end couples of 30,000 lbf in are applied to a 72 in. length of aluminium shaft 2 0 in. dia.. If Ε =10 χ 10« lbf/in

2 find

(a) the slope at the ends (7-9 deg) (b) the deflection at the midpoint (248 in.). 3. A beam ABCD carries a uniform load from A to Β and is simply

supported at Β and C so that AB = CD. Find the ratio of AB to AD so that the slope at A and D shall be zero. (AB = 0-21 AD approx.).

4. The wind load on an unstayed vertical mast 32 ft high may be assu-med to increase from zero at a point 7 ft from the top to a maximum of 25 lbf/ft at the base. Find the value of the horizontal force which, when applied at the top, would bring about zero resultant deflection at this point. (42.5 lbf).


5. The span of an Encastré beam is 36 ft and the section has a 2nd Moment of Area of 400 in

4. Find the position and magnitude of the maximum

deflection due to a vertical load of 16 tonf at a point 12 ft from the left hand end. Assume Ε = 13,000 tonf/in

2. (0-87 in at 20-5 ft from LH end,).

6. A single leaf return spring is Od in. thick and has an initial radius of curvature of 60 in. The span is 30 in. while the width is 4-0 in at the centre and decreases uniformly to zero at the ends. Determine the stress in the spring when it is just flat, (25,000 lbf/in

2).

C H A P T E E X I

STRAIN E N E R G Y OF BENDING

I T HAS been shown that , in simple tension or compression the work done in straining is given by

f2

U = x Volume,

where / is the uniform tensile or compressive stress across the

section under load and is within the elastic limit.

The application of a simple bending moment to a beam induces

a varying stress across the section.

FIG. 3 2 4

F o r a small length of beam dx between two plane transverse sections, consider the transverse area dA (distant y from the neutral axis) over which the direct stress due to the application of a simple bending moment (M) is / .

Volume of element = dx χ dA.

361


2EI2

M2y

2

Strain energy between sections = ν ρ ρ x

^1 c

^ '

- 1 1 ± . Σ ΆΛ x y2

2EI2

(taking out the constants).

Clearly Σ^Α x y2 is the 2nd Moment of Area of the section

i.e., / .

.'. Strain energy between sections i.e. for the length da;

1 χ M

2dx.

2EI

Thus for a length of beam L :

Tota l strain energy U = JM2 dx

where M is expressed in terms of x.

Note: Tf / is not constant it must be expressed in terms of χ and included in the integral.

The foregoing is a general expression and may be applied to

part icular cases of loading, as follows :

(a) Simply Supported Beam with Central Concentrated Load

From a support to the centre

of the span : X w

X

- L -t \ 2 5

FIG. 3 2 5

w

M2 W

2x

2

f2 (My\

2

Strain energy of element - - — - χ dx χ dA, where /2 = ——

2E \ I j

χ da; χ d i .

S T R A I N E N E R G Y OF B E N D I N G 363

L/2

F o r naif the beam, £7 = / 2i£i J 4

Fo r whole beam, Î7

L / 2

\2EIJ 4 ~ ~

L W 2

U X

4

^ 2 Z3

~ 3 ~ o

12J0J 8 '

96EI '

(b) Cantilever with Uniform Load


, , χ ΜΎ = wx χ —,

wx*

i f2 W W

4 ' FIG. 3 2 6

·. u 2Elj

w2x* dx

w* 2EI

w x Z , s , and i.e. w2 = —.

1 jf2

4 0 ^ 7 X

X2

W2L*

40EI '

L\


(c) Simply Supported Beam with Uniform Load


l i r wLx χ Mx = — wx χ — ,

wLx wx2

,_0 w2L

2x

2 w

2x*

Ml = — r - + - r -

/ wLx wx2

w2L

2

X X2 + -

^ ( = ^x totol load)

FIG. 3 2 7

w2L

χ χό.

1 Γ / w2£

2

w* χ

2 # /

w2L

2 χ

3 w

2 x

5

— χΎ + —χΎ

w2L

2

w2£

X a:3 da;,

x*

2 ~ X T 1 /M )

2Z

2

T, W2

2 U l 2 " x i + ¥ x i

w2L

χ £4 ,

m W / 1

I P X3

" 2 ^ F

TF2Z

3

2 4 0 ^ 7

120

(d) Beam Subjected to Uniform Bending Moment

An example of this is tha t part of a wagon axle between the

wheels.

Since M\I = Ε/Β, Λ Β = EIjM (all of which are constant .)

Hence R is constant and the part A B bends into a circular arc .

S T R A I N E N E R G Y O F B E N D I N G 3 6 5

Thus, U = 2Elj

ο

m CA

2ËïJ dX

>

M* dx,

M2L

2EI '

E X A M P L E . A 2 in. dia. shaft is simply supported over a span of 3 0 in. and loaded a t the centre unti l the bending stress is 1 5 , 0 0 0 lbf/in

2. Calculate the value of the load and est imate the strain

energy stored. Ε = 3 0 x 1 0E lbf/in

2.

Solution

π 7

= 6 4Χ 2 4

'

y = 1 in.

/ = 1 5 , 0 0 0 lbf/in2.

|W

- 3 0 in -

0 - 7 8 5 in4.

-ψ. ^

FIG. 3 2 9


Max. B .M. -WL

Strain energy

and M

WL

W

U

1± y

9

i l

y '

ML Ly '

4 χ 1 5 , 0 0 0 χ 0 - 7 8 5

3 0 χ 1

1 5 7 0 lbf.

W2L

3

9QEI '

1 5 7 02 χ 3 0

3

9 6 χ 3 0 χ Ι Ο6 χ 0 - 7 8 5

2 9 - 4 in lbf

E X A M P L E . A spring steel r ibbon 0 - 1 2 5 in. wide, 0 - 0 2 in. th ick is

wound on a 2 0 in. dia. cylinder. F ind the max . stress in the steel

and the strain energy stored per foot. Take Ε = 3 0 χ 1 06 lbf/in

2.

Solution

I =

U

bd3

Ί 2 ~ '

0 1 2 5 χ 0 0 23

ΙΟ in r

12

1

/ = 0 - 1 2 5 in

FIG. 3 3 0

1 2 χ 1 0

Ey

~R~

3 0 χ Ι Ο6 χ 0 - 0 1

" ÏÔ 9

= 3 0 , 0 0 0 lbf/in2.

M2L 1 EI c%/\

, where M = —— and L = 20π, 2E1 M

EIL

2R2 9

30 χ Ι Ο6 χ 20π

0 - 0 2 ίη

S T R A I N E N E R G Y O F B E N D I N G 367

U per ft length = 0-785 χ 12

20π

0-15 in lbf.

(e) Simply Supported Beam with Non-central Concentrated Load

Moments about the L H end give

Wa,

Wa

R2L

R9

L

Λ X I » X, x 2 *2 X, x 2 *2

Χι x2

FIG. 3 3 1

η - wo

Similarly, Wb

F o r any section X X X ! in part a : il/,. = / ijA'j L

\ r ψ2^2

\ F o r part a of beam: Un = / ——— χ dx, ILL J LJ

WW

2EIL*

W2b

2a

s

QEIL2


Similarly, for part b of beam :

W2a

2

L2

0

0

W2a

2 I

2EIL2 I

W2a

2b*

6EIL2

F o r whole beam : U = Ua + Ub,

W2b

2a* + W

2a

2b*

.r| da:,

6EIL2

W2a

2b

2(a + 6)

£7 =

6EIL2

W2a

2b

2

and α

6 # L L

IF Since work done in straining = — y where y = deflection a t load point,

W -y

W2a

2b

2

whence y — Wa

2b

2

ZEIL 2 ü

6EIL '

(a result already obtained.)

E X A M P L E . A steel jo is t having I = 8 0 i n4 carries a uniform load

of 4 tonf/ft over a span of 1 0 ft. Es t imate the elastic strain energy,

assuming Ε = 3 0 χ 1 06 lbf/in

2.

Solution

Tota l load

Span

where

W = 4 χ 1 0 χ 2 2 4 0 ,

= 8 9 , 6 0 0 lbf.

L = 1 0 χ 1 2 ,

= 1 2 0 in.

W2L*

" 2 4 0 ^ 7 '

# = 3 0 χ 1 06 lbf/in

2,

7 - 8 0 in4.

4 t o n f / f t

- 10 f t -

FIG. 3 3 2


He nee U = (89-6 x 1Q

3)

2 x 1 2 0

3

240 χ 30 x 1 0e χ 80 '

89 ·62 χ ΙΟ

6 χ 1728 x 1Q

3

24 χ 3 χ 8 χ ΙΟ6 χ 1 0

3

89·62 χ 1728

2 42

- 24 ,000 in lbf

- 2000 ft lbf ,

= 0-89 ft tonf.

E X A M P L E . A shaft 60 in. long 2 in. dia. carries a pulley weighing

100 lbf a t a point 20 in. from a bearing. Calculate the strain energy

and hence find the deflection a t the pulley. Ε = 30 χ 1 06 lbf/in

2.

Solution

64 I =

17 =

J T X 24

64 0-785 in

4

6EIL '

1 0 02 χ 2 0

2 χ 4P

2

6 χ 30 χ ΙΟ6 χ 0-785 χ 60

64 χ 1 08

108 χ 0-785 χ 1 08 '

0-75 in lbf.

6 0 m -

2 0 in

1 0 0 lbf

I FIG. 333

Deflection under load

2 0-75 χ 2 0 0 1 5 1 in.

E X A M P L E . A steel wire is to be of such diameter tha t , when wound on a 72 in. dia. drum, there shall jus t be no permanent strain on removal. Take Ε = 13,200 tonf/in

2, assume an elastic l imit of

14 tonf/ in2 and calculate

(a) the permissible maximum wire diameter,

(b) the 2nd Moment of area of the wire section, (c) the bending moment on the wire,

(d) the strain energy stored per turn.


Solution

From the bending equation :

. Ey d / = — ' where y = —,

14

. d :

R

13,200 d

~ ~ 8 6 ~ ~ X

2"'

14 χ 36 χ 2

13,200

0-076 in.

Fo r the wire section

nd*

"64~ '

π χ (0-076)4

FIG. 334

Bending moment

64

0-00000164 in4.

M -EI

13,200 χ 0-00000164

36

- 0-000603 ton in.

= 1-35 lbf in. and is uniform.

F o r uniform bending moment,

M*L Strain energy U

2ΕΙ '

1·352(72π)

2 χ 13,200 χ 2240 χ 0 0 0 0 0 0 1 6 4 '

where 72π = length per turn,

= 4-25 in lbf .

E X A M P L E . A shaft of length Lin., dia. d, carries a concentrated

load W lbf,

(a) a t the centre of a simply supported span

(b) in direct tension.

S T R A I N E N E R G Y OF B E N D I N G 371

Determine from first principles the ratio of L to d for the strain

energies in the two states to be equal.

Solution

FIG.335

(a) At any section X X

2

/ , / 2

u- = -arf M2dxx2> 0

L / 2

EI S

ψ2χ2 —: dx,

0

W2 Ι . τ

31

7/

2

4 £ 7 I 3 |o

W2L* Λ r nd*

where I = 96EI

9 " " " "

M - 64

9

2W2L*

3End* *

(b) Area of section = .

71 Volume of section = ~-d

2L.

4


Tensile stress / =

iL 2E

nd2

χ Volume, 0 .

Hence, if

then

Un

2W2L

3

~ ( nd2 J 2E

2W2L

End2 '

2W2L

V///J, nd

2L

Τ

3End* End2

L2 = 3d

2

or L = dp.

E X A M P L E . The 2nd Moment of area of the

section of a canti lever increases linearly from FIG. 3 3 0 zero a t the free end to I a t the support.

Deduce, from a consideration of strain energy, an expression for

the deflection a t the free end when a transverse load W is concen-

t ra ted there.

A cantilever spring consists of a triangular steel plate having

the dimensions shown:

F I G . 3 3 7


Calculate the deflection a t the free end when a load of 60 lbf

is applied there. Ε = 30 χ 1 06 lbf/in

2.

Solution


2nd Moment Ir

Bending moment

Strain energy U

W*L L* i.e. U =

W2L*

4EI

B u t U = ^WZ, where Ζ = deflection a t load point.

Equat ing,

WL3

1 W2L*

±WZ = ——, 2 4 # / '

whence,

I n the given case,

Ζ =

/ =

2JW '

bd3

w 6 χ 0 -25

3

where I = value a t support.

12

1

"Ϊ28 '

Deflection, Ζ 60 χ 1 0

3

2 χ 30 χ 1 06

128

128

1 03 '

0-128 in .

13 S M


Examples X I

1. A uniform beam the section of which has a 2nd Moment of area of 35 in

4 supports a dead load of 3 tonf at the centre of a simple span of 120 in.

Take Ε = 13,000 tonf/in2 and calculate

(a) the strain energy of bending (0-37 in tonf), (b) the deflection at the load point (0-24 in.).

2. Deduce an expression for the strain energy per unit length of a beam at a point where the section has a 2nd Moment of area of I and the bending

3. A steel bar of square section is subjected to a uniform bending moment (a) in a plane parallel to one of the sides and (b) in the plane of a diagonal. If the maximum stress (within the elastic limit) is / tonf/in

2 deduce in each

case (in terms of / and E) an expression for the recoverable strain energy

4. A cantilever of length L has the form of a right conical frustum, the axis being horizontal and the diameter at the free end being d. Derive an expression for the strain energy of bending due to a vertical load W at the free and given that the diameter at the built-in end is 2d. Hence deduce an expression for the deflection at the load point.

(Note: Change limits and integrate the expression for U with respect to diameter)

5. Derive, in terms of stress and Modulus of Elasticity, a formula for the strain energy of a centrally loaded leaf spring having η leaves of width b

moment is M

(a) -fijjj x Volume, (b) χ Volume.

and

and thickness t, the length of the main leaf being L

C H A P T E R X I I

SHEAR STRESS DUE TO BENDING

Shear in Beams

The shear stress produced b y a vert ical shear force is accompan-

ied by an equal complementary shear in a horizontal plane. I n a

laminated spring, this results in relative motion between the leaves.

The two shears produce tensile and compressive stresses in the

web of an I- joist which m a y cause failure by buckling.

A reo Δ

C D

M + d M

dy y

km TT A L A

E F

FIG. 3 3 9

My — b.dy

= stress x section =force due to M

( M + d M ) y j b .dy =force due t o ( M + d M )

FIG. 3 4 0

13* 3 7 5


a x i s : Resul tant force || N.A. =

( M +^

M ) y χ b dy — χ bdy,

dM bydy.

y max

dM .'. To ta l pull on piece of section A and length dx = j^-j-bydy.

y

This is resisted by the complementary shear stress q acting on the area b dx. The shear face produced = qb dx.

Equat ing, dM

qb dx = J — by dy ,

y

I'/max

' " • '/ 7Λ./

Λ·"

, 1· '

/

and

"UNIX

J bydy 1st Moment of area A,

where 2/ is the distance from the N.A. of the centroid of this area.

Hence q = ^ ^ , since = T7 (the shear force.) 76 dx

Clearly this also gives the value of shear stress on the vertical

section a t y from N.A.

7. Rectangular Section

It w/4y////

It -A d

FIG. 3 4 1

Considering the forces on the element distant y from the neutral

S H E A R S T R E S S D U E T O B E N D I N G 377

Shaded area A = b ^— ?/j ,

F rom the neutral axis the distance of the centroid of this area

bd3

2nd Moment / = — . I Ζ

.*. Shear stress on a vert ical section a t any distance y from the N.A.

FAy

d \( d/2 - y

2

12F (d τ - y

12

d - 2y\ y + *

yx

bd3 \ 2

12F ld-2y\( 4y + d - 2y \

bd3

6F (d - 2y) (2y -f- d)

~bd? 4 '

6F (d2 - 4y

2 ' _ 6F (d

2 - ±y

2 \

~ ~h¥\ 4 Γ bd3 \ 4

6F / d2 - y

2

I f q is plotted against y, the result is a parabola. Clearly q is a

max . when y = 0 and a min when y = d/2.

3 F 3 Put t ing j , = 0 , 2 L N A X= - — = -

E X A M P L E . A t imber jois t 3 in. wide by 6 in. deep is simply sup-ported over a span of 20 ft and carries a central concentrated load of 200 lbf. I f Ε = 1 χ 1 0

6 lbf/in

2, ca lcula te :

(a) Max. bending stress, (b) Slope a t supports in degrees,

d


(c) Deflection a t load point,

(d) Max. shear stress,

(e) Bending and shear stresses a t a point 2 in. above the N.A.

and 5 ft from a support.

Solution

3 in, 2 0 0 lb*

I- 5.η---•I

'31 0 m

2 0 f t 4100 lbf

FIG. 3 4 2

/ =

At the centre,

Slope a t ends,

Centre deflection,

3 χ 63

12 = 54 in

4.

M = 100 χ 10 χ 12

= 12,000 lbf in .

When y - 3 in., / m ax -12,000

54

667-0 lbf/in2.

dy WL2 200(20 χ 1 2 )

2

"cbT " 16EI ~ 16 χ 1 χ ΙΟ6 χ 54

= 0-01335 radian = 0-76 deg.

WL* Ζ = 48EI

= 0 0 1 3 3 5 χ

3

dy L

dx 3

Max. shear stress, qnVAK = — χ -r-r = - x F_

bd

20 χ 12

3 ^

3 100 3 x 6

1 065 in.

8-33 lbf/in2.

At 5 ft from support, M =- 100 χ 5 χ 12 ,

= 6000 lbf in.

When y = 2 in

A = 3 in2

y = 2-5 in.

F = 100 l b f

δ = 3in

S H E A R S T R E S S D U E TO B E N D I N G

6000 X 2

379

54

= 222-0 lbf/in2.

FAy _ 100 χ 3 χ 2-5

Λ 54 χ 3 '

= 4-63 lbf/in2.

2. / — S e c t i o n

(a) Web

The complementary shear

in the web is on longitudinal

planes parallel to the N.A.

(Fig. 343) and a t 90° to the

shear applied to the section

which is parallel t o the sur-

face of the web.

F o r any section X X dis-t an t y from the neutral axis (Fig. 344)

FIG. 343

A

- H h — b I FIG. 3 4 4

Applied shear


where A = shaded area above X X (including tha t of the flanges),

y = distance of centroid of this area from the N.A. (found from y = J £ a r / A ) ,

b = web thickness.

The graph of q against y is again parabolic in form, though q

is not zero where the web joins the flange, since A is not zero a t

this point. The shear stress is again a max . when y = 0, i.e. on

the neutral axis .

F I G . 3 4 5

"Jl Q m o x .

(b) Flange

Since the shear stress is parallel to the surface of the material , i t must ac t inwards from the extremities of the flange towards the centre as in Fig . 345 .

F o r any section Y Y distant χ from the centre (Fig. 346) ,

FIG. 3 4 6 9x =

FAy

lb

S H E A R S T R E S S D U E TO B E N D I N G 381

where A = shaded area to the left of Y Y , y = distance of centroid of this area from the N.A., b =r- flange thickness.

Since A is the only variable, the graph of q against a; is a straight

line, the max . shear stress being a t the centre.

In comparison with qy, qx is usually negligible.

E X A M P L E . I n I-section jois t 8 in. χ 4 in. overall may be taken

as 0-5 in. thick throughout. Es t imate the maximum shear force

which m a y be applied given t ha t the longitudinal shear stress is

not to exceed 5 tonf/in2.

Solution

4 χ 83 3-5 χ 7

3

12 12

2048 - 1200

Ϊ 2 '

- 4 in -

ΥΖΖΖΖΖ7ΖΖΖΖΖΆ

0 5 i n -

_ 848 _ H T '

= 70-67 in4.

Area of section above neutral axis

(shaded),

A = (4 χ 0-5) + (3-5 χ 0 -5 ) ,

= 2 + 1-75

= 3-75 in2.

Ν 8 in

0 - 5 in

FIG. 347

The distance of the centroid of this area from the neutral axis is

given b y :

y

(4 χ 0-5) 3-75 + (3-5 χ 0-5) 1-75

3-75

= 2 1·75

2

3-75

= 2 + 0-817,

= 2-817 in.

13 a SM


At the neutral axis, qn

Permissible shear force F =

F Ay

F χ 3-75 χ 2-817

70-67 χ 0-5

70-67 χ 0-5 χ 5

3-75 χ 2-817

= 16-75 tonf.

E X A M P L E . A beam has the

section shown, and carries

a shear force of 20 tonf.

Determine the maximum

shear stress and the ratio

(Ζιη;ιχ/(/ ιη<';»η·

- 5 in

Solution

5 χ 53 3 χ 3

3

12 12

= i ( 6 2 5 - 8 1 ) ,

544

- 3 in-

Fio. 348

12 '

= 45-3 in4.

Shaded area above N.A. A = | ( 52 - 3

2) ,

= 8 in2.

Λ

Γιο. 349

4 — 3 in 5 in

S H E A R S T R E S S D U E TO B E N D I N G 3 8 3

Distance of centroid of A from neutral axis,

_ 2 > » = — '

_ ( 5 χ 1 ) 2 + 2 ( 1 χ 1 - 5 ) 0 - 7 5

" 8

_ 1 0 + 2 - 2 5

" 8 '

_ 1 2 - 2 5

~ ~ ~ 8 ~ ~ '

= 1 - 5 3 1 in.

_ 2 0 x 8 x 1 - 5 3 1

4 5 - 3 χ 2 '

- 2 - 7 tonf/in2

Çmeixn

. gmax = 2 - 7

imoan 1 * 2 5

E X A M P L E . An I-section jo is t is 1 4 in. χ 8 in. overall and is

0 - 5 in. th ick throughout. Determine the rat io of max . to mean

shear stress a t a section where the upper flange thickness has been

reduced b y planing off 0 - 2 5 in. of metal . F o r the mean value, as-

sume tha t the shear is taken b y the web alone.

Solution

Ini t ia l section = 2 ( 8 χ 0 - 5 ) -f ( 1 3 χ 0 - 5 )

= 8 + 6 - 5 ,

= 1 4 - 5 in2.

Sect ion removed = 8 χ 0 - 2 5 ,

= 2 in2.

Sect ion remaining Aj — 1 4 - 5 — 2 ,

= 1 2 - 5 in2.

Web section Aw = 1 3 χ J ,

= 6 - 5 in2.

= 2 J = ^ = 1 - 2 6 tonf/in».

= 2 - 1 7 .

13a*


h = 7 - 8 5 in

y = 5-15 in

7-6 in

Ό 2 5 in

I 3 in

h =

« ι in

F E G . 3 5 0

Σαν (8 χ 0-25)0-125 + (13 χ 0-5)6-75 + (8 χ 0-5)13-5

12-5 0-25 - f 43-85 + 54

98-1

12-5

12-5

= 7-85 in.

Sect ion above N.A., A = (7-85 - 0-25] 0-5 + (8 χ 0-25) ,

= 3-8 + 2 0 ,

= 5-8 in2 (shaded).

F o r this area y

^ Ν . Λ . =

(7-6 χ 0-5) 7-6/2 + (8 χ 0-25) (7-6 + 0-125)

14-4 + 15-45

5-8

0-5 χ 1 33

5-8

29-85

5-8 = 5-15 in.

12 + (0-5 χ 13) 1-1

2 +

8 χ 0 -253

Ϊ 2

8 χ 0 -52

12 + (8 χ 0-25) (7-85 - 0 -125)

2 +

+ (8 χ 0-5) (5-9 - 0 -25 )2

91-5 + 7-87 + 0 0 1 + 119-4 + 0 0 8 3 3 + 1 2 8 ,

347 in4.

j-* 8 m - - H


= 1 1 2 .

lb I Aw Ib F 347 χ 0-5 (taking least value of b)

E X A M P L E . Show that , for a square section beam simply supported

with a diagonal horizontal, the shear stress a t a point y from the

neutral axis is given b y

F q = ( a V ^ — %y) (a)'2 + fy) where a is the length of the side.

Hence show tha t q is a max . when y = α/]/32

Solution

From similar triangles :

b ap (a/1/2) - y a/p

= 91 a

F I G , 3 5 1

FAy Ι F FAy Aw 5-8 χ 5-15 _ κ

- X " T T - = -7ΓΤΖ 7Z~Z X 6-5

386

F o r this area

S T R E N G T H OF M A T E R I A L S

b ( a = ~2\~P Ρ

1

2

^ I a χ 2 — a

ρ a

Ρ - y

1 / a

ïy , « 3 31/2 '

F o r a triangle about one edge, I =

y + τ 3]/2 1 _ 3

base χ (height)3

Λ / Ν . Λ . = 2

12

α]/2 / α ^3

12 \ p

Shear stress

F Ay F

α4/12

6F

ni

a

Ρ

12

2y

y]\ir 31/2

a

ρ

^ I a 2

( p -y

'2y a

α4 V 3 + 6

6F (ayy2 a2

a4 V 6 6

3

4y2

~ 6 ~

LL (ayp + a2 - 4y

2)

6F (ayp a2 2y

2 _ ay \

3 +

~6 3 " 3 ψ 2 / '

6i^ / ai/1/2 ( a2 2#

2 ayp \

— (2ay 1/2 + 2 a2 - 8?/*)

F2

-.—(ap-2y)(ap + ±y).

Shaded area b


For max . q :

= (4a V(2) - 8 t / - 2a \'(2) - 8y),

^ r < M ( 2 > - l < * ) = 0 (and 0 = ^ - 1 6 )

which is negative,

16y = 2a p a

1/32

E X A M P L E . An overhead rail is made from 2 in. square section

with the diagonal horizontal and supports a load of 0*75 tonf

Find, for the load point

(a) the max . shear stress,

(b) the mean shear stress,

(c) the shear stress a t the neutral axis.

Y o u may assume the expressions derived in the previous example.

Solution _ a

(a) q is a max. when y = 1/32 '

2

1/32 '

2

1/(2 χ 16) '

2 p ~ 4 p ~~ 4 '

0-75 = 1 Γ ( 1 ' 5 ^ 2 X 3 ] / 2 )

0-75 χ 9

32

0-212 tonf/in2.

3. Solid Circular Sectio?i

The width of the shaded area A is not constant so tha t an ex-pression for Ay in terms of constants must first be derived.

Width of strip = b1 dz and -j^j r

— cosö, i.e. bx = 2r co s0 ,

= 2r cos θ dz.

1st moment of strip = 2r cos θ dz χ ζ and ζ = r sin θ,

dz i.e. - ττΓ = r cos 0,

dd

= 2 r2 cos θ sinÖ dz and dz = r cos θ do,

- 2 r3 cos

20 sine d0 .


F 0*75 (b) AVERAGE = ~ ^ =

= 0 1 87 Ί Θ Ϊ ΐ Ί^ '

F c) when y = 0 , g N. A. = — (α J/2 χ a | / 2 ) ,

= — 2 a2

2 a4

_ ^

~ ä2

= (ZUVCRAGC

= 0-187 tonf/in2.

FIG. 3 5 2


and cos-^- = 0 .

( Ύ \ = 2 r

3 I — J u

2 du J , where u = cos0,

- 2 r3

2 r3

d^ = — sin0 d0 ,

u3

3 Jçj

= — I cosJ — — COS (fj

2r3

.'. Ay = — - c o s3 φ.

ό

FIG. 353

Since 0 varies between φ and JZ /2,

π/2

Tota l 1st Moment Ay = f 2 r3 co s

20 sin0 dö,

ψ π/2


Shear stress q — ———, where 1 = —— and b = 2r cos φ, lb 4

T

2 r3 „ \ 4 1

χ ( 2 r * \ 4

- — C O S30 9 — r „ _

3 7

/ π Γ4 2 r cos <P

4i^ cos2 99

3nr* and cos

2 99 = 1 — sin

2 φ,

1 - —

4 f Λ ν2

3nr2 \ r

2

which is the equation of a parabola.

4:F 4 When y = 0 , g m ux = ^ ^2 = — χ Mean shear stress.

E X A M P L E . Determine the ratio Ljd which willmake the maximum

shear stress in a short length L of solid circular shaft dia. d equal

to one quarter of the maximum bending stress when the shaft is

(a) simply supported with W a t the centre and (b) built in a t one

end and loaded with W a t the other. F o r a semicircular area, the

centroid is a t 4Λ/6π from the diameter.

FIG. 354

Solution

The shear stress is given by q = a n

d is a maximum when b —d.


Then

y

I

nd2

: " ΊΓ '

M

nd*

~6T

(given),

where F = Shear force.

The bending stress is given by /

d

My

when y

B u t ,

i.e.

M =

F =

9 m a x

M

/ n u

d_ 64

and is a max.

32M

nd* "

1 32M 16F

3nd2 ~JX"^cW

M 2d

F ~Ύ

WL

4

iL 2

WL 2

4 Χ

ΐ Γ

_2d

ΊΓ

whence, FIG. 355

d 3 "

M=WL\ . ΤΓ£ 2d

F=W \ " W " " 3 " '

£ 2

w

whence, FIG. 35(5

Ζ

/ y ^ - - - L •

/


Examples X I I

1. A T-section is used as a beam with the flange uppermost. Sketch the distribution of shear stress across the section.

2. The average shear stress over the section of a hollow beam 5 in square outside and 1-0 in. thick when supported with two sides horizontal is 2-22 tonf/in

2. Calculate the corresponding value of the maximum shear stress

and sketch the distribution graph (2-7 tonf/in2).

3. Particulars of two dissimilar beam sections are given below :

Web Flange Total Distance of N.A. from Section Thickness Thickness Depth outer edge of flange

in. in. in. in.

/ 0-20 0-35 2-5 5-0 Τ 0-50 0-50 5-0 4 0

The beams are simply supported over the same span and each is loaded at the centre. If the maximum bending stress is the same in each case, find the ratio of the maximum shear stresses. (/ = 3-4Τ approx.).

4. The vertical section of a horizontal beam is an isosceles triangle of base 4-0 in. and height 6-0 in. Draw a diagram showing the distribution of shear stress at a point where the shear load is 6 tonf. (L. U.)-

5. A beam which is initially square in section, the side of the square being 4-0 in., has a longitudinal hole 3-0 in. dia. bored symmetrically through it. Calculate the value of the maximum shear stress induced by a vertical shear load of 5 tonf, the neutral axis being horizontal and parallel to a side (1-66 tonf/in

2).

C H A P T E R X I I I

STRUTS

Struts Subject to Axial Load

When the length of a s trut is great in relation to i ts sectional

dimensions failure will not occur due to the compressive stress but

due to bending since no strut can be truly straight, no load truly

axial, and no material t ruly homogeneous.

Such bending under axial load is called buckling and the load

which produces i t is referred to as the Buckling, Crippling or

Critical Load.

When such failure occurs, the strut remains in equilibrium in the bent position, as shown in Fig . 357 :

Euler Formulae

Assuming

(a) tha t direct compressive stress is

negligible and

(b) the ends are pin-jointed (i.e. free to

change their s lope) :

d2y

M = EI dx

2 = —Py (where I = least valuefor secion).

d2y P_

da:2 + EI

0 ,

y = ο

i.e.

Rewriting,

0 , where k2 =

EI

Ρ = k2EI.

(D2 + k

2)y = 0 .

393

Γ FIG. 357


Hence the Auxiliary Equat ion is m2 + k

2 — 0 ,

Hence, General solution is

When χ = 0 ]

V = 0 J

m2 - — k

2,

m = 0 ± jk.

y — A cos kx + ß sin

0 - .4 cosO -I . B s i n O .

When χ = L

y = 0

- A -h 0 .

Λ = 0

.'. y = Β sin for.

0 = 5 sin kL.

The constant Β cannot be zero since this would make y zero for all

values of x.

Hence sin kL = 0 , i.e. kL — π (taking least value),

4

k2 =

whence, Ρ =

71

u — χ i£7 from above. E"

The safe load will be this value divided by a suitable safety factor.

I f one end is built in (i.e. fixed) so tha t a change in slope a t this end is prevented, then, if there is no lateral restraint a t the other end (Fig. 358) a bending moment M must be introduced to maintain equilibrium.

The strut is now equivalent to half a strut of length 2L loaded as in Fig . 357.

Hence, Critical load = •

i.e. Ρ =

π" (2L)

2

1 π2ΕΙ

1 L2

χ EI,

S T R U T S 395

I f both ends are built in as shown in Fig . 359 there will be two

points of contraflexure C, C, distant. L / 4 from each end.

The piece of strut between them, of length Lj2 is exac t ly similar

in shape to the pin-jointed strut of F ig . 357 since the bending mo-

ment a t the points C is zero. Hence for this case,

Critical load = — = — τ - χ EI,

τ ) -

4 n

2EI

I f the free end of the strut in F ig . 358 is prevented from moving

laterally by a horizontal force F (Fig. 360) then, for any point dis-

t an t χ from the fixed end :

Mx=EI$?jL= -Py + F(L-x),

I I ι

F I G . 3 5 8 F I G . 3 5 9 F I G . 3 6 0


R

RL Equat ing constants, k2A = RL, i.e. A = - 7 0 -

Hence, RL Rx

'Ύ2 W

R

F El — χ — [L-x).

F Hence the particular integral ν = — (L — x).

T o find the complementary function, the R H side of the equation

must be made zero.

Thus, g . + iV = 0,

or D2y + k

2y = 0 (using operator D).

d2y P F /r

d2y P F /τ

d^ +

-ÊIy = W

{ L~

x h

d2y F P

or —— -f k2y = R(L — χ), where R = ·— and i

2 = —— .

d.r2 j \ η EI EI

Assuming the particular integral ν = A -f Bx (i.e. equal in de-gree to R H side),

then, = Β dx

d2v

and - ^ = 0 .

Substi tution of these values in the equation gives

0 + k2{A + Bx) = i ? (L - x)9

i.e. 4 M + k2Bx = RL - Rx.

Equat ing coefficients of x, k2B = — R, i.e. B =

S T R U T S 397

i.e.

Also

y = u + ν

F '. y = A cos kx + Β sin 4# -f — (X — #)

d#

Ρ — 4 4 sin kx -f 5 4 cos kx —

( 1 )

(2)

when χ = 0

y = 0 0 = A cos 0 + Β sin 0

P L

P L (Sub. in (1)) ,

=4 + •

i.e.

when χ

ày

άχ

i.e.,

A = -P L

Ρ - 4 4 sin 0 + 5 4 cos 0 - — (Sub. in (2))

= 5 4 - P_

P'

Β = F

kP

0 = A cos 4 L + Β sin 4 L (Sub. in (1)) When χ = L

y = 0

since the third term becomes zero.

.'. Β sin 4 L = —A cos 4 L

or tan 4 L A_

Β *

Then the auxiliary equation is :

m 2 _|_ jfc2 = ο

m2 = — 4

2

m = 0 ± ?4

Hence the complementary function w = 4 cos kx + 5 sin 4#.

Since the general solution is the sum of particular integral ν and

complementary function u

398 S T R E N G T H ΟΓ M A T E R I A L S

kL — t a n- 1 ( —

Β

tan FL kP\

P F /

.'. kL = t an"1 kL (or tan kL = kL)

B y tr ial and error, kL = 4 4 9 radians, = 257°24 '

a n d t a n 2 5 7 ° 2 4 ' = tan 77°24 ' = 4 4 9 .

4 4 9 . k =

Ä:2 -

L

20-2

""Σ2"

Hence for this case, Critical load = 20-2

Ρ

W '

EI L

2 '

Since 2 π2 = 19-75 i t may be assumed tha t the cri t ical load is given

approximately by n

2EI

Ρ = 2 -L

2

Thus the axial crippling load is a constant χ π

2ΕΙ

IF • and this is

summarized in Fig . 361 in which the crippling loads are expressed

as a multiple of tha t for the simplest (first) case :

0 25 Ρ

" '//At///,

®

2 Ρ opprox.

FIG. 3 0 1

S T R U T S 399

In pract ice, some change in slope always occurs a t the ends, even

when 4 4

f ixed" . The constants must therefore be considered care-

fully before use.

F o r the first case, F — ——— , LP

since I = ^4&2, where A = section,

& = least radius of gyration.

I f the compressive stress P/A is plot ted vertically against the slen-

derness rat io Ljk for a given mater ial and end conditions, a curve

will be obtained from which the limiting value of L\k (correspond-

ing to the yield stress) can be obtained.

Assuming Mild Steel and pin-jointed ends, π2Ε = 9-87 χ 13 ,400 ,

= 132 ,000 .

P/Ä 132,000

Ρ/Α

ι / / 132,000 \

η Pia ) - L>k

2 66000 235 5 26500 163

10 13200 115 15 8800 94 20 6600 81 25 5300 73 30 4410 65

From the curve F ig 362 a limiting value of Ljk 81 corre-

sponds with the Mild Stee l yield stress of 20 tonf/in2.

I n other words, if L < 81 k the strut material will yield before

buckling occurs.

P_ π2Ε _

1 , e" -4 ~^ ( Z # )

2 : compressive stress a t failure,

or — - ] /π Ε -

k ~~V F/A ~ slenderness ratio.


100 200

L/k

FIG. 3 6 2

Consider the following sections :

R2 (a) Solid circular, about a diameter,

(b) Thin tube dia. d,

4

^ 1 6 '

.·. k = L i.e., L = 81 L = 20 d approx.

i.e., j /8 L - 81 - p - = 28-7 d approx.

Other sections may be similarly t reated.

F o r the Euler Theory, which neglects the direct stress, a slender-

ness ratio of about 80 corresponds with the Mild Steel yield point

and gives L = 20 d as the approximate l imit.

Thus in practice the Euler Formulae are used only for 4

' long

columns", i.e., having a minimum slenderness ratio of about 120.

This keeps the direct stress a t the buckling point down to about

S T R U T S 401

one third of the yield stress (see curve) and gives

L = 1204 , where h =

= 120 χ

- 30rf.

Rankine-Gordon Formula

This is an empirical formula i.e. based on experimental results, and is used where both direct and bending stresses are important, i.e. where L < 30d.

The crippling load is given by

Ρ A4

' 1 + "(ΤΊ where fc = yield stress, A section ( in

2) , L — length (in), Tc — least

radius of gyration, a — a constant which depends on material and end conditions.

F o r mild steel, the values of a for the four cases considered are

(a)

(b)

(o)

(d)

7500

1

7500

4 / 1

) Ο θ ) '

9 \ 7500

1 / 1

4 \ 7500

Note tha t case (d) which is the stiffest, has the least value of a i.e., Ρ has the greatest value.

The safe load is obtained by dividing the crippling load Ρ by a suitable safety factor .

x E X A M P L E . Compare the strengths of two steel struts 2 in 2 in. χ 100 in. and 2 in. χ 2 in. χ 30 in. given tha t Ε = 30 χ ΙΟ

6

lbf/ in2 and fc = 21 tonf/in

2.

The columns have one fixed and one free end.


Solution

I = 9 3

12 '

- 1-334 in4.

Euler buckling load,

Ρ = 71

fif for the long strut, 4 L

2

π2 x 30 x 1 0

s x 1-334

4 χ 1 0 02

= 9880 lbf ,

= 4-41 tonf.

FIG. 3 6 3

Rankine-Gordon

[ where k2 = -4-

\ A

Hence

1-334

22 = 0-334 , and a =

7500

21 χ 2240 χ 22

1 + / 3 0

2 \ '

7500

84 χ 2240

1 4 χ 900

7500 χ 0-334

84 χ 2240 1 + 1-44

84 χ 2240

2 4 4

= 77,150 lbf ,

34-4 tonf .

E X A M P L E . A strut having rounded ends is 12ft long and of circular section. When freely supported a t the ends a load of 9 lbf a t the centre produces a deflection of 0-375 in. Determine the Euler crit ical load. I f Ε = 30 χ 1 0

6 lbf/in

2, find the diameter of the bar.

S T R U T S 4 0 3

Solution

Ζ

EI =

Ρ =

cl*

. d

J _ WL*

4 8 El

WL*

4 8 Z "

n2EI

Ζ = 0·375 in

W=9lbf

FIG. 3 6 4

L2

2 WL* 1 71 X~4SZXT2

n2LW

4SZ 9

π2 x 1 4 4 x 9

4 8 x 0 - 3 7 5 9

> Ί2π2,

7 1 0 lbf.

P = n

2EI

L2

PL2

6 4 n2E

9

64PL2 6 4 x 7 1 0 χ 1 4 4

2

π*Ε π2 x 3 0 x 1 0

E

^ 1 - 0 2 = 1 - 0 0 5 in, say, l i n

= 1 - 0 2 .

E X A M P L E . A 1 4 s.w.g. ( 0 - 0 8 in. th ick) round tube of 1 in. out-side dia. and length 4 0 in. is to be used to t ransmit thrust to an aircraft control. I t may be assumed to be pin-jointed a t each end. Use Rankine ' s Formula to determine i ts failing load taking a - 1 / 3 0 0 0 and fc = 1 9 tonf/in

2.

Obtain the Euler buckling load for the same strut under the

same loading conditions.

404 S T R E N G T H Ο Γ M A T E R I A L S

Solution

π I = — [ I4 - (0 -84)

4] ,

^ ( 1 - 0-498) ,

0-025 in4.

Ο Ό 8 in

A = - ^ - [ l2 - (0 -84)

2] ,

4

= j - ( l - 0-706) ,

- 0-2325 in2.

* 2 = T>

0 0 2 5

0-232 '

= 0 1 0 6 .

Pn = /c4

4 0 in

-ffl 19 χ 0-232

1 1600

3000 \ 0-106 1 + •

π2ΕΙ

L2 9

_ π2 χ 13400 χ 0-025

Î6ÔO

= 2-03 tonf.

19 χ 0 2 3 2

1 + 5 0 1

I in

FIG. 3 6 7

= 0-73 tonf.

taking Ε = 13,400 tonf/in2

E X A M P L E . A s trut is composed of two T-sections riveted back to

back to form a cruciform section 6 in. χ 8 | in. overall. E a c h T-sec-

t ion is 6 in. χ gin. χ 4 | in. The effective length is 20 ft and the

ends are rigidly secured. F ind the maximum safe load using a

safety factor of 5 given tha t fc = 21 tonf / in2 and tha t a = 1/30,000

in the Rankine-Gordon Formula .

S T R U T S

Solution

0-625 χ 8-753

+ g /2·687 χ 1 ·253^

1

~ " Ϊ 2

_ 428-5

12 '

= 35-8 in4.

12

(418-0 + 10-5) ,

Z2-687 χ l - 2 53\

A 12 I

0 - 6 2 5 i n -

I 2 5 in

- 2 - 6 8 7 i n -

8 - 7 5 in

— 6 in

FIG. 366

1-25 χ

12

= 22-5 + 0-15

= 22-6 in4.

+ • 7-5 χ 0 -625

3

12

Area of section, A = (6 χ 1-25) + (7-5 χ 0 -625) ,

= 7-5 + 4 -68 ,

= 12-18 in2.

22-6 Leas t value of i,2 _ Iy\i

12-18 = 1-86.

14 S M

405


Rankine buckling load Ρ fcA

where L = 20 ft,

21 χ 1 2 1 8

30 ,000 \ 1

21 χ 1 2 1 8

1 + 1 033 '

2 02 χ 1 2

2

b 8 6

21 χ 12-18

^ 0 3 3 125-5 tonf.

.'. Safe load = 125-5

5

= 25-1 tonf.

E X A M P L E . S ta te the fundamental assumptions made in deriving

the Euler Formula for a strut hinged a t the ends. A straight length

of steel bar 6 ft long and 1 in. χ | in. in section is loaded axial ly

until i t buckles. Assume Euler 's formula to apply and est imate

the max . central deflection possible before the material passes

the yield point, if the la t ter is 21 tonf/ in2 and Ε = 30 χ 1 0

6 lbf/in

2.

Solution

Assumptions :

1. Uniform elast ici ty throughout,

2. No eccentr ic i ty of axial load a t ends, 3. Buckl ing load independent of degree of deformation, 4 . E las t ic limit not exceeded.

1 χ 0 -253

Leas t value of / = —τ , Λ = 0-25 in2. 12

= 0-0013 in4,

0 0 0 1 3

0 1 2 5 = 0-0104 in

3.

Buckling load, Ρ = π*ΕΙ

Ζ2

π2 χ 30 χ ΙΟ

6 χ 0-0013

7 22

= 74-5 lbf.

S T R U T S 4 0 7

Direct stress, — Ρ

A

7 4 - 5

0 - 2 5 *

2 9 8 lbf/in2.

Bending stress,

Μ __Ρδ (where Ô = ~2 -2, required deflection),

7 4 - 5 0

" 0 - 0 1 0 4 '

= 7 1 5 0 < 5 .

Permissible stress,

Ρ = 2 1 χ 2 2 4 0

Λ 4 7 , 0 4 0

δ

M

2 9 8 + 7 1 5 0 ( 5

4 7 , 0 4 0 - 2 9 8

7 Ï 5 0

4 6 , 7 4 0

7 1 5 0 *

6 - 5 3 in.

|«-K)in-»| j

0-25 in

L = 72 in

FIG. 3 6 7

E X A M P L E . A stanchion with fixed ends and 3 6 ft high, consists of an R . S . J , having a web 0 - 5 in. thick, flanges 0 - 8 8 in. th ick and measuring 1 5 in. χ 6 in. overall. E a c h flange is reinforced by a plate 1 2 in. wide and § in. th ick, b y means of rivets. Es t ima te the safe axial load, assuming a factor of safety of 5 and taking Ε = 3 0 χ 1 0

6 lbf/in

2.

Solution

The least value of / must first be found.

(12 χ 1 6 - 2 53

1 2

- 4 2 9 0 - 1 6 8 7

= 1 5 3 9 in4.

j _ ^ 6 χ 1 5 3 j _ ^ 5 - 5 χ 1 3 - 2 4 3 j

1 0 6 4

14*


- 0 6 2 5

"0-88 in — 15 i n —

-13-24 in-

0-5 in

6in

12 in

3 in

FIG. 3 6 8

1-25 χ 1 23 1-76 χ 6

3

12 1

12

= 180 + 31-7 + 0 1 3 8

= 211-8 in4.

+ • 13-24 χ 0 -5

3

12

This is the value required. Since both ends are fixed and L > 30 d the Euler crippling load is

given b y n

2EI

Ρ = 4 L

2

4 π2 χ 30 χ 1 0

β χ 211-8

(36 χ 1 2 )2

1,345,000 lbf,

600 tonf.

Hence safe load = 600

= 120 tonf .

S T R U T S 409

E X A M P L E . F ind the Euler critical load for a hollow C.I. column 6 in. outside dia. J in. th ick 2 0 ft long and hinged a t bo th ends. Ε = 5000 tonf/in

2.

Compare this with the Rankine-Gordon critical load taking 1

a = "Ï6ÔO"

a n d fc = 3 6 t 0 nf / i n 2'

F o r what length of column would the two formulae give the same critical load?

Solution

π

4-5*),

= — (1296 - 4 1 0 ) ,

886π

64 43-5 in

4.

Ρ = π

2ΕΙ π

2 x 5000 χ 43-5

L2 ~ 2Â& *

= 37-3 tonf. (Euler)

A = 0-785 [D2 - d

2),

= 0-785 ( 62 - 4 - 5

2) ,

= 12-4 in2.

~ A

P =

43-5

12-4

LA

3-51 ,

1 + a (L/k)2

36 x 12-4

L=240 in

4 4 6

1 + • / 2 4 0

2\ 1 + 10-25

1600 \ 3 - 5 1 /

39-8 tonf. (Rankine)


F o r Same Load : π

2ΕΙ

L2 1 + a (Ljk)

2 9

fcAL2 = π

2ΕΙ + π

2ΕΙα

IF9

L2

fcA-π

2ΕΙα

L2(fcA -

L2 =

k2

π2ΕαΑ)

π2ΕΙ

= π2ΕΙ

= π2ΕΙ.

and k2 1_

~Α

fcA - π2ΕαΑ

9-87 χ 5000 χ 43-5

L

(36 χ 12-4)

2 ,150,000

4 4 - 382

183 in. = 15-3 ft.

^9-87 χ 5000

= 3 3 , 6 0 0 .

1600 χ 12-4

0 - 1 2 5 m

0 · I in

y//

-1 in -

FIG. 3 7 0

" "Γ 0-5 in

Solution

Sectional area, A

E X A M P L E . A symmetrical steel

connecting rod has the section

shown and a length between

bearing centres of 10 in. Take the

crushing stress as 21 tonf/in2 and,

using appropriate values of the

constant a calculate the R a n k i n e -

Gordon crippling load (a) in the

plane of oscillation and (b) in the

plane a t 90 deg to the web.

(2 χ 0-5 χ 0-125) + (0-65 χ 0 -1) ,

0 125 + 0 0 5 7 ,

0-2 in2.

0-5 χ Ρ 0-4 χ 0 -753

12

0-5 - 0 1 6 9

12

0-331

12 9

0-0276 in4.

12

S T R U T S 4 1 1

L^Ak* Λ * -0

"0 2 76

0 - 2

= 0 1 3 8

ΙΟ2

7 2 5

ίήτγ (AND A = W ) ' ι +

2 1 χ 0 - 2 4 - 2 3 - 8 3 tonf.

lyy

7 2 5 1 - 0 9 6 6 + 7 5 0 0

/ 0 - 1 2 5 χ 0 - 53\ / 0 - 7 5 x 0 - l

3\

= 2 l 1 2 J + l 1 2 ] '

0 - 0 3 1 2 5 + 0 - 0 0 0 7 5

1 2

0 0 3 2

1 2 '

= 0 - 0 0 2 6 7 in4.

Im. = Akf. . . k1t

0 - 0 0 2 6 7

yy — ^Ltoy . . toy — ^

= 0 - 1 3 3 ,

1 02

ky 0 0 1 3 3 7 5 2 0 .

SINCE a = — x 2 1 x 0 - 2 / . 1 1

s

1 / 7520 \ + T \ 7 5 0 0 /

' * ~" 1 + 1

( 1 5 20

) \ " 4

" 7 5 00

4 - 2

1 - 2 5 1 3 - 3 6 tonf.

E X A M P L E . A tubular strut 6 0 in long 2 in. outside dia. is to carry

an axial load of 2 tonf with a factor of safety of 6 based on the

Euler crit ical load for pin-jointed ends. Es t ima te the thickness

required, given tha t Ε = 1 3 , 4 0 0 tonf/in2. Es t ima te also the factor

of safety according to the Rankine-Gordon formula, assuming

the constants to be 2 1 tonf/in2 and 1 / 7 5 0 0 .


Solution

Safe load = 2 tonf ,

Λ Euler crippling load Ρ

= 2 x 6

= 12 tonf.

π2ΕΙ

Ρ = — w h e r e L = 6 0 . L2

12 = π

2 χ 13,400

6 02

π

or

x EI" ( 2 * - ^ ) ,

16 - =

Em. 371

12 χ 6 02 x 64

i.e., t

13 ,400π3

= 6-654.

. & = 16 - 6-654,

= 9-345, giving d = 1

-2

-1

·7 8

- 0 - 1 2 5 in.

Area of section A = 0-785 ( 22 - 1-75)

2,

= 0-736 in2.

/ = " 6 Τ ( 2 4 _ 1 ' 7 5 4 ) ·

. — (16 - 9 -346) ,

6·65π

k2

64

I 0-362

A " 0-736

0-326 in4.

= 0-443

S T R U T S 4 1 3

Rankine crippling load = fcA

1 + a(L/k)2

2 1 χ 0 - 7 3 6

1 + 6 0

2

7 5 0 0 χ 0 4 4 3 ,

1 5 4 5

1 + 1 - 0 8 '

1 5 4 5

2 - 0 8 '

= 7 - 4 3 tonf.

.'. Corresponding safety factor Crippling load

Actual load

7 4 3

2

3 - 7

E X A M P L E . A stanchion is to be constructed from two channel sections as shown, each having Ixx — 4 - 3 i n

4 and Iyy = 5 5 i n

4 and

a section of 5 - 6 7 5 in2. The ends are to be fixed and the effective

1 length is to be 2 5 ft. Take a = — — - and L = 2 1 tonf/in2 in

° 7 5 0 0 x 4 ' the Rankine Formula and est imate the distance apart to give equal strength in both directions. F ind the safe load for a factor of 7 .

i

0 - 8 in

FIG. 3 7 2

Solution

B y the Paral le l axis theorem, from Fig . 3 7 3 ,

Ixx = 4 - 3 + 5 - 6 7 5 h2 for each section (where h = (d /2) + 0 - 8 ) . 14 a SM

4 1 4 S T R E N G T H OF M A T E R I A L S

Y

0-8 in

IY

FIG. 373

d

and

4-3 + 5 - 6 7 5 ^ + 0-8

/ d \2

= 8-6 + 1 1 - 3 5 ( — + 0 - 8 J

= Iyy = 2 x 5 5 = 1 1 0 in4.

'd \2

1 1 0 = 8-6 + 1 1 - 3 5 ^ - + 0-8 j

id \2 1 1 0 - 8 - 6

d d - + 0-8 = 2 9 9 , . ' . — = 2 1 9 ,

I = Ak2,

Buckl ing load, Ρ = -

h2 — ΊΑ

n χ —

ny

icA

8-93

d = 4 - 3 8 in.

= 9 - 6 9 .

1 + .(. 2 1 χ 1 1 - 3 5

1 1 0

1 1 - 3 5

and the ends are fixed,

1 + • 1

7 5 0 0 χ 4

2 1 χ 1 1 3 5

~~ 1 + 0 - 3 0 9 6 '

= 1 8 2 tonf.

( 2 5 χ 1 2 )2

9 - 6 9

S T R U T S 4 1 5

Hence, for a factor of safety of 7 : Safe load = 1 8 2

7

= 2 6 tonf.

E X A M P L E . A short length of mild steel tube 0 - 1 in. th ick 3 - 0 in.

outside dia. yielded a t 3 5 tonf/ in2 when tes ted in compression,

the t es t figures also giving a value of 1 3 , 5 0 0 tonf/in2 for E.

The crippling load on a length of 7 ft when tes ted as a strut

with pin-jointed ends, was 1 7 tonf. Compare this with the Euler

crippling load. F ind also the constant a in the Rankine—Gordon

formula, assuming fc in this formula to be the yield stress.

Solution

I

π χ 1 9 - 6

6 4 = 0 - 9 6 1 in

4.

6 1 - 4 ) ,

2 - 84) ,

0-1 in

L 7 χ 1 2 = 8 4 in. FIG. 374

Euler crippling load = n

2EI

L2

π2 χ 1 3 , 5 0 0 χ 0 - 9 6 1

8 Ϊ2

= 1 8 - 1 tonf.

Area of section, A = 0 - 7 8 5 ( 3 + 2 - 8 ) ( 3 - 2 - 8 ) ,

= 0 - 7 8 5 χ 5 - 8 χ 0 - 2 ,

= 0 - 9 1 in2.

14a*

/.a 1

1 Q 56

416 S T R E N G T H 0 E M A T E R I A L S

fcA

1 +

35 χ 0-91

1 + 84

2α

1 0 5 6

31-85

1 + 6685α

31-85

17 '

= 0-875,

0-875

6685 '

= 1-875,

7630

Rankine cripppling load, Ρ

Λ 17

.·. 1 + 6685 α

Λ 6685 α

α

FIG. 375

Solution

E X A M P L E . A b lock and tackle is

suspended from the apex of a tripod

as shown, and used to raise a load

W. E a c h leg makes an angle of

60 deg with the ground and consists

of a pipe 2-5 in. outside dia., 2-0in.

inside dia., 10 ft long.

Take the Rankine constants as

1/7500 and 21 tonf/ in2 and est imate

the maximum load which may be

lifted, assuming a safety factor of 5.

/ = - ^ ( 2 - 5 * - 2 - 0 * ) ,

= - £ - ( 6 - 2 5 + 4) (6-25 - 4 ) ,

A = 0-785 (2-52 - 2

2) ,

= 0-785 χ 2-25 = 1-77 in2.

/ = Ak2, Λ k

2 =

1-77

= 0-64.

S T R U T S 417

Rankine crippling load,

ι +

where £ = 120 in. ,

21 χ 1-77

1 + χ 1 2 0

2 '

7500 0-64

3 7 2

1 + 3 '

= 9-3 tonf.

.*. Safe load per strut,

9-3

5 '

= 1-86 tonf.

F FIG. 376

F cos 3 0 ,

Since there are three struts,

• iL ·' 3

Λ W = 3 χ 1-86 χ 0 -866,

= 4-82 tonf.

w Τ

τ \ /////////////////////m FlG-377

Examples X I I I

1. The force required at the end of a clutch operating lever is 56 lbf and is transmitted by means of a pin-ended steel rod 10 in. long. Take Ε = 29-8 lbf/in

2 and determine a suitable rod diameter given that the factor

of safety based on the Euler Formula is to be 10 (0-25 in.).


2. Draw a curve showing the theoretical relation between crippling stress and slenderness ratio up to Ljk = 150 for a material for which crushing stress and elastic modulus are respectively 10 and 4000 tonf/in

2. Show how

and explain why an experimentally obtained curve would differ from the above.

3. The section of a pin-ended strut is 0-375 χ 0-125 in. If the strut is to fail at an axial load of 356 lbf, find the required length given that Ε = 28-9 χ 10

6 lbf/in

2 (10 in.).

4. Each corner of a column 30 ft high 18 in. square consists of an angle 4 χ 4 χ 0*5 in. nominally, the angles being braced together. Calculate the Rankine-Gordon safe load using appropriate coefficients for steel and pin-jointed ends and assuming a factor of safety of 6. The centroid of an angle section is 1-17 in. from an outer side and the value of / about an axis through the centroid and parallel to a side is 5-46 in

4 (I. Mech. E.) (41 tonf).

5. Calculate, using the Rankine-Gordon Formula, the safe load for a pin-jointed strut 120 in. long composed of two steel T-sections 5 χ 2-5 χ 0-5 in.

with the 5 in. sides riveted back to back. Take a = a n

d / = 30 tonf/in2

and assume a factor of safety of 6 (10 tonf).

C H A P T E R X I V

THICK AND THIN CYLINDERS

Stresses in a Thick Tube

L e t Fig . 378 represent the section of a th ick tube under internal

and external pressure, i.e. both surfaces subject to compressive

stress. Due to the internal pressure on the tube ends there will be

an axial tensile stress fz. I n addition to this, any element as shown

in F ig . 379 will be subject to a radial stress fy and a hoop stress fx.

Since the cylindrical element in F ig . 379 is in equilibrium,

Downward force due to stress on section A B

= Upward force tending to fracture cylinder across A B

i.e., 2fx dr = fy2r - (fy + dfy) 2(r + dr)

= 2rfy-(fy + dfy){2r + 2är)

= 2rfy - (2rfy + 2r dfy + 2fy dr + 2dfy dr)

= 2rfy - 2rfy - 2rdfy - 2fydr - 2dfydr.

.*. 2fx dr = —2r dfy — 2fydr (neglecting the last t e rm) , J 1

Λ fx = - ^ " ^ 7 - fy (dividing through by d r ) ,

I f fz = axial tensile stress, then, assuming transverse sections to remain plane :

or y (1)

Axial strain, ez

fz Φ afy Ε Ε ^ Ε

i.e. 4 - - f y ) ]

4 1 9

(2)

S T R E N G T H OF M A T E R I A L S

FIG. 3 7 8

Unit length

fy + dfv

F =f, dr

FIG. 3 7 9

4 2 0

T H I C K A N D T H I N C Y L I N D E R S 421

Now the quant i ty (jx — fy) in E q . (2) must be a constant since

Ε, σ and fz are constant .

L e t fx — fy = 2a so tha t fx = ftJ + 2a . Substi tut ion of this value

in equation (1) gives

Transposing, — 2 x — dr,

i.e. J fy + a

J J r

log e (/// + α) = — 21og er -f constant and 2 log er = l og er2

\°ge(jy + ») + l o g e r2 = constant ,

log e

f 2 (/// +

a) = constant ,

^2(/// + = constant .

L e t r2( / y + a) = 6 , then — a

And = fy + 2 a , whence / r = — + a

(3)

(4)

Equat ions (3) and (4) are known as Lame's Formulai and give the

radial and hoop stresses respectively a t any radius r between R2

and Rv

At the inner surface where r = R 2 ,

fy = / 2 and .·. / a = — _ α

(and / 2 = applied internal

pressure).

At the outer surface where

r = Bl9

fy = fx and ; . f 1 = — - a

(and fx is usually zero).

Thus the constants α and b

can be found and their values substi tuted in the equation for hoop stress. FIG. 3 8 0


F o r zero external pressure the stress distribution is as shown

in Fig . 3 8 0 , the difference in height of the curves being 2a.

E X A M P L E . A cylinder 8 in. outside dia. and 4 in. inside dia. has

an internal fluid pressure of 3 0 0 0 lbf/in2. Calculate the longitudinal

stress (fz) assuming this to be uniform.

Solution

4 i n •—• I 3 0 0 0 = tx

FIG. 3 8 1

Area over which pressure ac ts , Ax = 0 - 7 8 5 χ 42,

Tensile force in pipe, F = 3 0 0 0 χ 0 - 7 8 5 χ 42,

= 3 7 , 7 0 0 lbf.

Area over which force acts , A2 = 0 - 7 8 5 ( 82 — 4

2) ,

= 0 - 7 8 5 ( 8 + 4 ) ( 8 - 4 ) ,

= 0 - 7 8 5 χ 1 2 χ 4 ,

= 3 7 - 7 in2.

Tensile stress =3 7

'7 0

° 3 7 - 7 '

= 1 0 0 0 lbf/in2.

E X A M P L E . The delivery pipe to a diesel engine injector is 6 - 0 mm

outside dia. and has a bore of 1 - 4 mm. Determine the maximum

hoop stress in the material if the injector orifice opens a t 2 0 0 0 lbf/ in

2.

Solution ΕΎ = 3 - 0 mm = 0 - 1 1 8 in.

R2 = 0 - 7 mm - 0 - 0 2 7 5 in.

T H I C K A N D T H I N C Y L I N D E R S

Hoop stress,

Radia l stress,

b

R=3O mm

FIG. 3 8 2

At R2:

At JV

2000 =

0 =

0 -02752

b

0 0 0 0 7 5 5

b

- a,

0-1182

b

0-0139

Substi tut ing for a in equation (1) above:

b b

— a, .*. a = 0 0 1 3 9

2000 0 0 0 0 7 5 5 0-0139 '

1 1

\ 0 0 0 0 7 5 5 0-0139

= 6(1320 - 72)

= 1248&,

.*. b 2000

1248 — = 1-6, :.a =

1-6

0 0 1 3 9 = 1 1 5 .

423

(1)

(2)


The hoop stress is a maximum a t R2.

I, + 1 1 5 , *

m ax ~ 0-000755

= 2120 + 1 1 5 ,

= 2235 lbf/in2.

E X A M P L E . A cylinder 8 in. outside dia. and 4 in. inside dia. has

an internal fluid pressure of 3000 lbf/in2.

F ind the max . and min. hoop stresses.

Solution

Hoop stress

Radia l stress

fx =Τ + °" r2

5 0 0 0 i

2 0 0 0

FIG. 3 8 3

At inner surface r = R2 = 2,

fy = 3 0 0 0 ,

3 0 0 0 = — - a .

T H I C K A N D T H I N C Y L I N D E R S 4 2 5

At outer surface

fy = 0,

• • • 0 = 4 - *

a = 1 6

• Κ 3 0 00

Ι Α

. . b = —— χ 1 6 ,

Min. hoop stress =

Max. hoop stress =

= 1 6 , 0 0 0 Hence a = 1 0 0 0 .

1 6 , 0 0 0

— T T " + 1 0 00

> 42

2 0 0 0 lbf/in2

1 6 , 0 0 0 + 1 0 0 0

22

5 0 0 0 lbf/in2.

E X A M P L E . A hollow forged boiler drum 7 6 - 4 in. outside dia. and

5 - 2 in. th ick was tes ted before entering service to 2 8 5 5 lbf/in2.

FIG. 3 8 4


Calculate the maximum and minimum hoop stresses under test .

Solution

Rt = 38-2 in.

R2 = 33 0 in.

At inner surface, r = R2 = 3 3 ,

33:

At outer surface, r = Rx — 38-2 ,

tu = 2855 = 4r - α (1)

b_

b a = 38·2

2

Substi tut ing for α in (1) above:

b 2855

3 32 38 -2

2

= b( b [ ~ W ~ ISS 2") 9

= δ(0·000915 - 0 -000685) ,

= 0-00023 &

b = 12-4 χ 1 06, a = 8 5 0 0 .

b ( 12-4 x 1 06 \

/ « Ü N - ^ + a , -( 38,22 J + 8 5 0 0 ,

= 8500 + 8 5 0 0 , = 17,000 lbf/in2.

4 b 1 1 2 4 x 1 0

6 \ rt ΛΛ

fcmax = - j j * +a>

= [ 332 j +

8 5 00 >

- 11,400 + 8 5 0 0 , = 19,900 lbf/in2.

E X A M P L E . F ind the longitudinal and hoop stresses a t the outer

surface of a hydraulic pipe 2 in. bore, 4 in. outside dia. a t the

operating pressure of 9 tonf/in2. Es t ima te also the increase in

outer dia. when under load. The pipe is steel and σ = 0-26,

Ε = 13,400 tonf/ in2.

Solution

Force on end = 9 χ 0-785 χ 22.

Area of section = 0-785(42 - 2

2) .

(2)

T H I C K AND T H I N C Y L I N D E R S

.*. Longitudinal stress,

427

U 9 χ 0-785 χ 2

2

~ 0-785 χ 12

= 3 tonf/ in2.

Radia l stress,

fy = 4- - α· A t R,

9 = - p - - a ,

a = b - 9 .

At i ^ :

0 = ^ - « ,

6 α = τ·

FIG. 3 8 5

b_

Τ 6 = 1 2 ,

Hoop stress,

At Rx:

= 9 ,

Λ a = 3 .

+ α,

~~ "22~ 3 5

= 6 tonf/in2. (and fy = 0)

Circumferential strain = 4τ — - τ τ and σ = 0-26 ,

Ε Ε 1

[6 - (0-26 χ 3)] 13,400

= 0-00039 .

'. Increase in diameter on 4 in. = 0-00039 χ 4 ,

= 0 0 0 1 5 6 in.

R. = 2 in


E X A M P L E . Draw the curves of radial and hoop stress on a base

of radius for a tube 6 in. outside dia. and 3 in. inside dia. when sub-

jec ted to

(a) an internal pressure of 9 0 0 0 lbf/in2 only,

(b) an external pressure of 6 0 0 0 lbf/in2 only,

(e) pressures as a t (a) and (b) together.

Solution

(a) Radia l stress fy = — a ( tensile),

Λ 9 0 0 0 ==T^~a

b b and 0= — - a , : . a = — .

Λ 9 0 0 0 = B

2 - 2 5 9 '

= 6 ( 0 - 4 4 5 - 0 1 1 1 ) ,

= 0 - 3 3 4 6 ,

.·. b = 2 6 , 9 0 0 and a = 3 0 0 0

r r2 26,900

r2

26,900 / y = — ^ - - 3 0 0 0

26,900 fx = +

3 0 00

1-5 2-25 11,900 8900 14,900 2 0 4Ό0 6720 3720 9720 2-5 6-25 4300 1300 7300 3-0 9-00 3000 0 6000

(b) Radia l stress, /„ = ^ — a (compressive),

b b


and 6000 = — - a ,

ό

_b_ b__

~ 9 _

2-25 ' = 6(0-111 - 0-445) ,

= - 0 - 3 3 4 6 ,

- - 1 8 , 0 0 0 and a = - 8 0 0 0 .

r r2 18,000

r2

18,000 fy- r2 + 8 0 0 0

18,000 fx= ~2 8000

1-5 2-0 2-5 3 0

2-25 4-00 6-25 9-00

- 8 0 0 0 - 4 5 0 0 - 2 8 8 0 - 2 0 0 0

0 +3500 +5120 +6000

-16 ,000 -12 ,500 -10 ,880 -10 ,000

(c) F o r inner surface 9000 = 1-5

2

b F o r outer surface 6000 == — — a

ό

. • . ^ i - 9 0 0 0=A_ e 0 0 0

Ύ ( Ά - ΐ ) = 3 0 0 0' whence b = 9000

and a = - 5 0 0 0

r r2 9000 9000

fy- r2 +5< > U U

9000 fx = — - 5000

1-5 2-25 4000 +9000 - 1 0 0 0 2 0 4-00 2250 + 7250 - 2 2 7 5 2-5 6-25 1440 + 6440 - 3560 3-0 9-00 1000 + 6000 - 4 0 0 0

The diagrams of stress distribution are shown in Fig . 386 .

E X A M P L E . A hydraulic press has a ram dia. of 8 in. and operates

on an internal pressure of 1500 lbf/in2. Es t ima te the wall thickness

needed to prevent the hoop stress from exceeding 3600 lbf/in2.

F ind the increase in inside diameter under load, taking a = 0-25

and Ε = 16-8 χ 1 06 lbf/in

2.


Ι 5·

0 0 0Γ Hoop f, w

10,000 - / ^ S . Rodial fy 7 7 % .

5000 - ^ 7 / / 7>

2 O- (a) internal pressure only

-5° o ° - , b

9, ° / ° W I

-ιο,οοοί 7 /

10.000 ρ \ /

5000 - τ ·. ^ r ^ 7 % lf v

A Tensile ^<77/////

0 - - - ! {////// ( b) E x + e r n al Pressure only

~ - 5000 - ^ Compressive/ / / / / / / A

- 1 0 . 0 0 0 - / ' / / / / j J A * *

- 1 5 . 0 0 0 - y/>^ >Ä>

\ / ^ s " 6 0 0 0 l b f / i n2

- 2 a o o oL χ /

iQOOOr-

5 0 0 0 - ////^ZffiTTtf*

/ / / / / / / / / / / \ ^0 +h

Internal and ^ 0 *LZ / / / / / ex+ernal pressure

§ 9 0 0 0 -^7 ( /J* - 5 0 0 0 - l b f / i n

2 < ^ /

- 1 0 . 0 0 0 - \ v £ ^

\ 6 0 0 0 lb f / in2

- I 5 , 0 0 0 L \ .

FIG. 3 8 6

T H I C K AND T H I N C Y L I N D E R S

Solution

Radia l stress

Hoop stress

At R2,

b

1500 = — — a 42

R, = (4 + t ) in

f , = 0

FIG. 3 8 7

At Rl9

AtR2,

Adding (1) and ( 3 ) :

3600 = A + « 42

1βΟΟ + 3600 = Α + ^ = |

b = 5100 χ 8 ,

= 40 ,800

40 ,800 /0 Έ . _ α = — ^ 1500 (Subst . in E q .

= 2550 - 1500

- 1050

431

(1)

(2)

(3)


Substi tuting these values in E q . ( 2 ) :

4 0 , 8 0 0 0 = — 1 0 5 0 ,

R\

R\ 4 0 , 8 0 0

1 0 5 0

= 3 8 - 9

\ Rx = 6 - 2 5 i.e. t = 6 - 2 5 - 4 = 2 - 2 5 in.

. ι. , ± 1 5 0 0 χ 0 - 7 8 5 χ 82 , Λ , Λ 11 0

Longitudinal stress, / , = 0 - 7 8 5 ( 1 2 - 52 - 8

2)

= 1 0 42 L B F/

M 2-

.'. Circumferential strain a t inner surface

3 6 0 0

Ε ( τ x I 5 0 0 ) - (τ x i o 4 3 7 1 4

1 6 - 8 χ 1 06 9

0 - 0 0 0 2 2 1

,\ Increase in circ. = 0 - 0 0 0 2 2 1 χ 8π

·. Increase in dia. = 0 - 0 0 0 2 2 1 x 8 = 0 - 0 0 1 7 7 in .

FIG. 3 8 8

E X A M P L E . Es t imate the ratio of thickness to inside dia. required

to limit the max . hoop stress in a thick pipe to 1 -6 t imes the internal

pressure, the external pressure being negligible. I f such a tube

T H I C K AND T H I N C Y L I N D E R S 433

has an inside diameter of 4 in., est imate the increase in this diam-

eter when the internal pressure is 6 tonf/in2 given tha t Ε = 13,40

tonf/in2 and Poisson's R a t i o is 1/3*5.

Solution

Radia l stress,

b_

r2 fy

Hoop stress fx

AtRl9 0

a and

b + a.

AtR9

At Bl9

and

i.e.

giving

a =

_b_

b

Rl

_ A A Rl

D9

2-6 _ 0-6

_ Ri — R% 2R,

fx (max) : ' * 6 fp

FIG. 3 8 9

/a'max

b_

R\

\r% + R*)~~

~Rj + ~Rj

R\ + R\~ \R\ + R\J

1·6/ 2 (given).

' • e 4 ( ^ - 3 f ) '

or

1 6

~Rj

Ik R9

1-6

Rl

2 ·08 R2 — R2

2&>

1-08 0-54

I n the given case, D2 = 4 in, i.e. Dr = 8-32 in,


Area of section = 0·785(8·322 - 4

2)

Force on end = 6(0-785 χ 42)

6 χ 0-785 χ 42

0-785(8 · 3 22

1-81 tonf/in2.

42)

Max. hoop stress

Circumferential strain

1-6 χ Internal pressure = 1 - 6 χ 6

= 9-6 tonf /in2.

9^6

Ε

1

-α) -M (6 - 1-81)

9-6 + 3-5

Increase in diameter =

13,400

9-6 + 1-2

13,400

= 0-000806.

Increase in circumference

π 4π χ 0-000806

0-00322 in.

E X A M P L E . A steel tube 4 in. inside dia. and 6 in. outside dia. is

to have a second tube of the same material 7 in outside dia. shrunk on

i t , the shrinkage allowance being such tha t the radial pressure be-

tween the two tubes is to be 2 tonf/in2. Take Ε = 13,000 tonf/in

2

and a = 0-25 and calculate :

(a) hoop stress a t inner surface of outer tube,

(b) increase in internal diameter of outer tube,

(c) hoop stress a t outer surface of inner tube, (d) reduction in external diameter of inner tube.

Solution

F o r outer tube,

FIG. 3 9 0 giving


And fy=2 a t r = 3 ,

' · 32 12-25

6 (0 -1111 - 0-0816) = 0 -02956 .

6 = 67-8 1 67-8 Λ at r = 3 : / r = — - + 5-53

and ο, = 5-53 32

= 7-53 + 5-53 ,

= 13-06 tonf/ in2 ( tensile) .

Increase in internal dia. = [13-06 + (2-5 χ 2 ) ] - ^ -

1 3 , 0 0 0 η

= 0-00624 in.

F o r inner tube, = 0 a t r = 2 ,

6 0 = — - a giving α = — ,

And / y = 2 a t r = 3 ,

and a = - 3 - 5 8 J 32

= - 1 - 5 9 - 3-58 ,

= - 5 - 1 7 tonf/in2

(compressive).

Reduct ion in external dia. = , ^ \„„ (δ·17 + •^-\ 13,000 \ 4 / π

= 0-00261 in.

Shrinkage allowance on diameter = 0-00624 + 0 -00261 ,

= 0-00885 in.

E X A M P L E . Determine the interference fit per in. of dia. necessary

to produce a radial pressure of 12 tonf/ in2 a t the common surface

of a compound tube. The inner and outer diameters are 4 in. and

10 in. and the inner tube has an outer diameter initially of 8 in.

W h a t must be the initial inner diameter of the outer tube? Take

Ε = 13,500 tonf/in2 and assume Lame 's Formulae.


FIG. 3 9 1

F o r outer tube fy = 0 a t r = 5 in.

Als

25

/„ - 12 a t r = 4 in.

12 = — - a = 6(0-0625 - 0 -04) , 4

2

= 0-02256.

6 = 533

a = 21-3 : . a t r = 4 : fx =

533 + 21 -3 ,

Hoop strain f. 54-6 12σ

Ε ^ Ε ^ 13,500 +

' 1 3 , 5 0 0

= 33-3 + 21 -3 ,

= 54-6 tonf/in2.

(1]

Solution

Hoop stress

Radia l stress


F o r inner tube , fv = 0 a t r •-= 2 in.

:. ο b b

: α = Γ Also, fy = 12 a t r = 4 in.

.'. 12 b

δ(0·0625 -

- 0 - 1 8 7 5 6

.·. b = - 6 4 1

a = — 16 I .'. a t r — 4 :

64

= - 4 - 16

= —20 tonf/ in 2 (i.e. compressive)

Hoop strain = _ + _ / „ = - _ _ + _ _ (2)

Strain difference a t common diameter = (1) — ( 2 ) ,

54-6 12σ

+ • 13,500 13,500

20 12<7

+ 13,500 13,500

74-6 = 0 0 0 5 5 3 . 13,500

Change in inner diameter of outer tube = 0-00553 χ 8 = 0-044 in

Required initial diameter = 8 — 0-044 = 7-956 in.

Thin Cylinders

When the thickness is small in comparison with the diameter,

the hoop stress m a y be considered uniform and the radial stress

negligible. Referring to K g . 392 :

Area of element = r do χ L.

Radia l force

15 SM

= Lrad χ p.


FIG. 3 9 2

pd2 pd

(2)

Thus,

Hence the vessel will fail under hoop stress, i.e. split along its

fx - fz length. N . B . Max. shear stress = — .

Horizontal component = Lrάθ χ ρ χ sind

= pLr sin θ dö

Tota l force on half cylinder to the right of Y Y :

π

F = pLr / s i n O d o ,

ο

= I - c o s e i S ,

= pLr [ — cos η — ( — cos Θ)]

= pLr[-(-l) - ( - 1 ) ] ,

= pLr( + l + 1)

= 2pLr.

This is resisted by the hoop

stress acting over the section

tL on each side.

Λ 2(fjL) = 2pLr,

Tota l force on an end

nd2

= p χ — •

This is resisted by the longitu-dinal stress /- acting over the section π dt.

.. /- x ndt = ρ x —^—,


E X A M P L E . A torpedo shell 20 in. dia. is 0-375 in. thick. Es t ima te

the hoop stress induced b y an internal pressure of 2000 lbf/in2

assuming the stress to be uniform.

Solution

Hoop stress fx = ,

2000 χ 20

~~ 2 χ 0-375 '

neglecting any external pressure,

= 53 ,300 lbf/in2.

E X A M P L E . A s team boiler 6 f t dia. is to operate a t a pressure of

300 lbf/in2 above atmospheric pressure. Es t ima te the plate thick-

ness required to limit the hoop stress to 5 tonf/in2.

Solution

Hoop stress fx

i.e. t

pd

"~2t'

pd

Ύ 300(6 χ 12)

2(5 χ 2 2 4 0 ) '

: 0-964 in. (say 1-0 in.)

Tensile Stress in a Thin Rim Due to Rotation

I f the thickness of the rim or hollow cylinder is small in relation to the diameter, then the stress m a y be assumed uniform.

Sect ion of element = r άθ χ t referring to Fig . 393 ,

Volume of element = r άθ t χ b

Weight of element = r άθ tb χ ρ, where ρ = density.

^άθώρ Mass of element =

g 15 a SM


.*. Centrifugal force on element = T

Ü>Q χ v_ g r

where ν = peripheral speed, ( = cor)

btqv2

άθ.

Vert ical component of this btqv

2

dö χ sin 0.

YZZA

FIG. 393

F = centrifugal F sin θ ι >*force on

[s(Q element

f ( b t ) = force per side resisting rupture

.*. Tota l vertical force tending to rupture rim across a horizontal diameter,

btqv2

btqv2

f g

btqv2

g

Μρν2

g

2btqv2

g

s inödf l ,

π

— cos θ , ο

[ — cos τι — ( —cosO)] ,

[ _ ( _ ! ) _ ( _ ι ] ) ?


I f ν is in ft/s and ρ is in lbf /ft3, then / will be in lbf/ft

2.

E X A M P L E . Determine the maximum safe speed of a cast iron

flywheel 5 ft dia. if the density is 0-27 lbf i n3 and the safe stress

is 5000 lbf/in2.

Es t imate the bursting speed given tha t cast iron fails in tension

a t a stress of 7 tonf/in2. t ake Ε = 16 χ 1 0

6 lbf/in

2.

, ρν2 pœ

2r

2 , Λ

/ = — = - , where r = 2-5 f t . 9 9 0*27 χ 1 2 3

5000 χ 12 2 = — — χ 2 · 5 2ω 2 (working in ft units.),

i.e.,

32-2 5000 χ 32-2

w " 0-27 χ 12 χ 6-25 '

= 7 9 5 0 .

ω = 89-1 rad/s.

.". ^safe = 852 rev/min.

F o r the flywheel to " b u r s t " :

/ = 7 χ 2240 χ 1 22 lbf/ft

2.

2 7 χ 2240 χ 1 22 χ 32-2

* ' ω

0-27 χ 12 χ 625

- 2 4 , 9 0 0 .

.'. ω = 158 rad/s.

.*. Ν = 1510 rev/min.

E X A M P L E . Show tha t , for a flywheel r im made of C.I. having a

density of 0-272 lbf/ in3 the safe speed in rev/min is given by

Ν = 30 ? where / = permissible stress in lbf/in2.

outside diameter r = in ft.

and this must be equal to (stress in rim) χ (section resisting rup-ture) .

whence, / =


Solution

As already shown,

QV2

Stress / = - — , where υ = ων, 9

ρω2/·

2 - 2πΝ

— - and ω =-- ,„ν , 17 υΟ

" ? 3600 Χ

·

π2ρτ

2

where / is in lbf/ft2, ρ is in lbf/ft

3, r is in ft, g is in f t / s

2.

900gr(/ χ 122)

π2Γ

2(ρ χ 12

3)

putting / into lbf/in2 and ρ into lbf/in

3.

900a / χ 12π

2ρ " r

2 '

_ -ι / 900 χ 32-2 ι /_/_

|/ 12π2 x 0-272

X y r

2 '

= y 9 o o x | / l ,

= 30 j/A rev/min.

E X A M P L E . The stress in a petrol engine flywheel is not to exceed 5000 lbf/in

2 a t 4170 rev/min. Es t ima te a suitable outside dia.

taking ρ = 0-28 lbf/in3.

I f the axial width of this flywheel is to be 2 in. and the polar moment of inertia is to be 0-167 slug f t

2, determine from first prin-

ciples a suitable radial thickness.

Solution

Α ι ι 2π x 4170 . . . Angular veloci ty ω = — = 437 rad/s.


Peripheral rim speed ν = œRx, where Rx = outer radius in ft,

= 437 Rx.

Centrifugal stress, / = - — , where ρ = density in lbf/ft3

and / = stress in lbf/ft2.

.·. 5000 χ 1 22 =

R\

0-28 χ 1 23

(437ÄJ)2

32-2

5000 χ 1 22 χ 32-2

0-28 χ 1 23 χ 4 3 7

2 '

= 0 -25 .

·. R1 = 0-5 ft ( = 6 i n . ) , i.e. A = 1-0 ft.

E lementa l mass

2nd Moment

2nrdr χ bρ

9

2nrdr χ ϋρ

where b — width,

9 χ r

2,

I -

2πδρ

2π bρ

χ r3dr.

r3 dr,

2nbQ I R{ - R\

Axis

FIG. 3 9 4


12 32-2 \ 4 0-167 χ 12 χ 32-2

485π 2

0-0425 = 0 0 6 2 5 - Rl

Λ Rl = 0-020,

i.e. R2 = 0-375 ft,

= 4-5 in.

Hence, Radia l thickness / = Rx — i£ 2 ^ 6 — 4 -5 ,

= 1-5 in.

E X A M P L E . A flywheel is to have / = 200 slug f t2 when using

material having a density of 0-261 lbf/in3. Determine from first

principles the outside dia. diameter and radial thickness of the

rim so tha t the induced stress shall not exceed 600 lbf/in2 a t

300 rev/min. Neglect the effect of the spokes and assume an axial

width of 6 in.

Solution

Λ ι ι ·. * u ι 2π x 300 Angular velocity οι wheel, ω = 60

= 31-42 rad/s.

Peripheral speed of rim, ν = œRx

(where Rx = outer radius infeet),

= 31-42 i ^ f t / s .

QV2

Centrifugal (or hoop) stress, / = - — , where ρ is in lb/ft3,

/ is in lb/f t2.

.'. 600 χ 122 = ° '

2^

1 2 3( 3 1 - 4 2 ^ )

2

where ρ = 0-28 χ 1 23 lbf/ft

3 = 485 lbf/ft

3, δ = - ^ - f t .

_ 2 485 / 0 - 54 — Ri

Hence, 0-167 = 2π χ — - χ

i.e.


R\ 12 χ 0-261 x 3 1 - 4 2

2'

= 6-25.

Rx = 2-5 ft, i.e. outside dia. = 5 ft.

Mass of element

2nr dr bq

g /

2nd Moment

2nr dr bq

g

2nbq

g

'.I =

x r*

χ r3d r .

2nbq ί FIG. 395

r3 dr ,

n2

2nbq ( R\ - R\

g \ 4

I f / is to be in slug f t2,

6 then b = = 0-5 ft,

ρ = 0-261 χ 1 23 = 450 lb/ft

3,

R± = 2 - 5 ft (already found).

where R2 = required min. radius.

200 χ 2 χ 32-2 _

π x 0-5 χ 4 5 0

18-2 = 39 - ß4.

22f = 20 -8 .

R2 = 2-14 f t .

600 χ 32-2


Change in Volume Under Load

As already shown, the hoop stress fx in a thin cylinder under in-

1

ternal pressure is given by — and is twice the longitudinal stress /-. Referring to Fig. 3 9 6 :

2t

Axial strain, = ^- yy^^ I

Ε Ε /Y \ \

and fx = 2fS9 f i Λ

Hence e. = 4τ — I

\\ J

= ^ ( 1 - 2 σ ) \^===^ Circumferential or Hoop strain, a

t a

e = L - 2 k where / = Α ε ε

9 h 2 ' L^rz^ — il _ — ~ 2 $ ' FIG. 3 9 6

^ - é ( ' - ï ) -The diametral strain is the same as this, since diameter and circum-

ference increase in the same ratio. Fur ther , the diameter increases

in both directions so tha t :

.*. Rad ia l thickness, t = 2-5 — 2-14 ,

= 0-36 f t ,

= 4-32 in .


2tE \ 2

i .e. e0=^(l-2i>-a).

UE 4tE 2tE UE '

* d ' 1 a + 2-a\

Change in Volume = Volumetric strain χ Original volume,

= evV,

Vpd

= (1*25 — σ) and is clearly an increase in

this case.

E X A M P L E . Es t imate the increase in capaci ty in a 7 ft diameter shell 16 ft in length when subjected to an internal pressure of 120 lbf/ in

2 (gauge) given tha t Ε = 30 χ 1 0

6 lbf/in

2, σ = 0-287

and tha t the skin is nominally 0-875 in. thick.

Solution

Original volume V = j (7 χ 12)2(16 χ 12) = 1-125 χ 1 0

6 in

3.

Volumetric strain, ev = - 7 ^ - (1-25 — σ ) , tE

- 1 2 0 ( 7 X 1 2 ) ( 1. 2 5 _ 0. 2 8 7 );

0-875(30 χ 106)

== 370 χ 10~6.

Change in volume = (370 χ 10"6)(1-125 χ 1 0

6) ,

= 416 in3 (increase).

E X A M P L E . An internal pressure of 750 lbf/in2 is applied to a steel

tube 6-0 in. dia. 0-125 in. thick, 24 in. long, the ends of which are plugged.

Volumetric strain, ev = ez + 2ex,

-Α<>-*>+·£Κ). pd 2apd 2pd 2apd


Axia l s t ra in , e. = 4 1

Circumferential strain, ev

Ε Ε 29-8 x 1 0e

χ [9000 - (0-28 χ 18,000)J ,

0 0 0 0 1 3 3 .

U «h 1 Ε Ε 29-8 χ ΙΟ

6

χ [18,000 - (0-28 χ 9000) ] ,

= 0 0 0 0 5 2 .

Volumetric strain ev = ez + 2ex,

= 0 0 0 0 1 3 3 + (2 χ 0 0 0 0 5 2 ) ,

= 0 0 0 1 1 7 .

.'. Increase in capacity of tube = 0-00117 χ 62 χ 2 4 j ,

- 0-79 in3.

Radia l strain due to hoop and axial stresses,

e - -2L^?L " Ε Ε '

29-8 x 1 0e

= - 0 - 0 0 0 2 7 .

0*28 (18,000 -f 9 0 0 0 ) ,

Volumetric strain of tube, er = ex -f e,f + e_,

= 0-00052 - 0 0 0 0 2 7 + 0 0 0 0 1 3 3 ,

= 0 0 0 0 3 8 3 .

I f Ε = 29-8 χ 1 06 lbf/in

2 and σ = 0-28, find:

(a) hoop stress,

(b) axial stress,

(c) increase in capacity,

(d) increase in volume of tube material .

Solution

Hoop stress fx = % = - 18,000 lbf/in*.

Axial stress, / . = ^- = 9000 lbf/in2.


.·. Change in material volume = 0-000383 (π x 6 χ 24 χ 0-125)

= 0-0216 in3.

E X A M P L E . A storage t ank is made of material 1-0 in. th ick and

consists of a cylinder 8 ft long, 4 ft inside dia. the ends of which are

closed by hemispheres. I f , initially, i t is full of water a t atmospheric

pressure, determine how much additional water must be pumped

in to bring about a tes t pressure of 500 lbf/in2. Take Ε and σ for

the tank material as 29-8 χ 1 06 lbf/ in

2 and 0-287 respectively and

assume the circumferential strains a t the junct ion of cylinder and

hemisphere to be the same. F o r water assume Κ = 0-32 x 1 0e lbf/in

2.

Considering the cylindrical part first and assuming the hoop stress uniform :

Longitudinal stress, / . = = 6000 lbf/in2.

.*. Hoop strain, ev = — ^—, assuming the radial stress f%

Solution

Hoop stress, /., = — = 500 χ 48

2 χ 1 = 12,000 lbf/in

2.

pd

to be zero,

^ [12,000

10,278

(0-287 χ 6000)]

and Longitudinal strain, Ε

= — [6000 - (0-287 χ 12,000)] Ε 2556

~ Ί Γ ~

but, Volumetric strain e. = ez + 2e ;


Jt 7t and Original volume, V = ~^d2L = — χ 4 8 2 χ 9 6 ,

= 173,000 in3.

Hence, Increase in volume = Volumetric strain χ Original volume,

= evV,

2 3'

1 0 2 χ 173 ,000 ,

29-8 χ 1 06

- 134 in3.

Considering next the hemispherical ends :

Volumetric strain ev = 3eL., where ec has the value found above,

_ 30,834

Ε ' 4

Original volume V = -— nr3, where r = 24 in, ό

= φ π χ 2 43,

= 57 ,800 in3.

Hence, Increase in volume = evV,

30 ,834

29-8 χ 1 06

60 in3.

χ 5 7 , 8 0 0 ,

Reduct ion in volume of original water

5 0 0 (173,000 + 57,800)

0-32 χ 1 06

360 in3.

.*. Additional water necessary = 134 -\- 60 -f 360 = 554 in3.


Examples X I V

1. A cylindrical vessel 8-0 in outside dia. is made of sheet 0-2 in. thick and subjected to an internal pressure of 800 lbf/in

2. Calculate the hoop and longi-

tudinal strains when the cylinder is subjected to an axial load of 10,000 lbf

(a) in tension (ex = 0-000522., e s = 0-0000865) and (b) in compression (ex = 0-000566,. ez = -0-000221).

2. Show that the maximum shear stress occurs at the inner surface of a thick tube when inder internal pressure and that its value is \ (fx + fy) where fx and fy are respectively the hoop and radial stresses.

3. Find the ratio of thickness to inside diameter for a tube subjected to internal pressure when the ratio of internal pressure to maximum hoop stress is 0-5. Find the alteration in the thickness of such a tube when the internal pressure is 5 tonf/in

2 given that ο = 0-304 and Ε = 13,000 tonf/in

2

(L.U.) (0-73., 0-0031 in.).

4. One end of a steel tube 1-0 in. outside dia. and 0-125 in. thick is closed and the other screwed into a pressure vessel. Find what pressure will induce a maximum stress of 20,000 lbf/in

2. If the effective length is 10 in. estimate,

neglecting end effects, the increase in internal volume under this pressure. (5600 lbf/in

2, 0-0062 in

3).

5. An iron pipe 8-0 in. outside dia., 1-0 in. thick failed under test at an internal pressure of 3-1 tonf/in

2. Determine, using a factor of safety of 4,

the safe pressure for a pipe of the same material and inside diameter but 1-5 in. thick. Assume failure in both cases to be due to the maximum hoop stress. (1-07 tonf/in

2).

I N D E X

Beams composite 150 direct stress in 97 encastré 327 reinforced concrete 153 shear stress in 118

Bending moment 38 simple theory of 94, 357 with direct stress 159 with torsion 259

Flexure differential equation of 279

Lamé theory of thick pipes 421

Load critical or buckling 393 gradual 13 shared by two materials 27 shock 15 sudden 13

Columns long 393 short 27

Contraflexure 56 Cylinders

compound 434 thick 419 thin 437

Deflection coefficient 297 of beams 278 of coil springs 199 of leaf springs 348 Macaulay's method 312 moment-area method 344

Maximum principal strain theory 257 principal stress theory 246 shear strain energy theory 249 shear stress theory 246 strain energy theory 249

Modulus of Bulk 227 of Elasticity 3 of Rigidity 4, 181 of a section 162

Moment of Inertia cone 82 cylinder 84 disk 85 sphere 81

Elastic constants 225 failure, theories of 245 limit 2 modulus 3

Euler theory of struts 393

Neutral axis 94

Parallel axes theorem 67 Perpendicular axes theorem 77 Poisson's Ratio 219 Power transmitted by shafts 184

First Radius Moment of Area 59 of curvature 95, 278, 349 Moment of Mass 65 of gyration 67, 399

453

454 I N D E X

Rankine-Gordon theory of struts 401

Relation between elastic constants 225 between w, F and M 130, 358

Second Moment of Area 66 Moment of Mass 80

Shear complementary 216, 379 force 118 strain 4, 181 stress 4, 182, 234 stress due to bending 118, 375 stress due to tension 215

Simple bending assumptions 96 equation 98 theory 94

Slenderness ratio 399 Springs

close coil helical 202 leaf or laminated 348 open coil helical 199

Strain lateral 219 longitudinal 2, 219 principal 237 shear 181 volumetric 217, 227, 447

Strain energy in bending 361, 201 in tension 12 in torsion 197, 201

Stress complex 230 distribution in beams 96 due to bending 97 due to rotation 439 due to temperature change 30 due to twisting 182 in beams 94 in cylinders and pipes 419 in struts 399 Möhr circle of 241 principal 230 proof 6 shear 234 simple 1

Struts long 393 short 27

Theorem of parallel axes 67 of perpendicular axes 77

Torque equivalent 262 in hollow shaft 191 in solid shaft 182

Torsion simple theory of 181 with bending 259

Volumetric strain 217, 227, 447

strength of materials - priodeep.weebly.com · machines, the book also covers the subject of...

Documents