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STRENGTH OF MATERIALS [For Engineering Degree^ Diploma and A. M. I, E, Students]

By S.

RAMAMRUTHAM B.E.,

PrincipaJ,

Modern

{CMh M.l.C.B. New Delhi

Colleise of Engineering,

Auihor of Design of Reinforced Concrete Structures, Design of Steel Structuies, Theory of Structures, Applied Mechanics etc.

DHANPAT 1682, N.\I

RAI

& SONS

SARAK, DELHMiC«006

(H.O.

JULLI^OR)

Design Of Rei^niced Concrete StructnKs

Design of Steel Stmctures

Theory of Straetares Prestressed Concrete

Applied Mechanics Hydraulics. Fluid Mechanics and Fluid Machines Steel Tables

First Edition: 1962

Reprint: 1970

CONTENTS Chapter 1.

Pages

*

‘

Simple Stresses and Strains

Introduction— Definitions, stress, strain, tensile and compresstresses— Sheat stress— Plastic limit— Hooke’s law — sive poisson’s ratio— Modulus of Elasticity -Modulus of Rigidity, Bulk Modulus— Bars of varying section '—Extension of a tapering rod— Composite section -modular ratio- Bar of uniform strength- Equivalent area of composite sections— Temiterature stresses— Hoop stress— Stresses on oblique sections— State of simple shear - Relation between the Elastic constants— Volumetric Strain— Rectangular blodt subject to normal stresses— Diagonal tensile and diagonal compressive stresses— Solved problems 1 to 71 —Problems for exercise. 1

2.

Strain

- 100

Energy— Impact Leading

Strain Energy —Elastic, plastic and rigid members— Stresses due to different types of axial loading —Gradually applied loads— Suddenly applied loadj— Impact loads— Solved problems 72 to 84— Problems for exercise. 101—118 3-

Centre of Gravity and

Moment of Inertia

of Gravity— Definition— Lamina — Moment of an area— Centroid of a uniform lamina— Centroids of laminae of various shapes— Triangle, circle, semicircle, trapezium — Built-in sections— Analytical and graphical methods— Moment of Inertia of a lamina— Definition —Parallel axes theorem— Perpendicular axes theorem— Moment of Inertia of laminae of different shapes - Rectang ilar, ciicular, triangular and composite sections— Solved problems 85 to 104— Problems for Centre

119—157

exercise. 4.

Shear Farces and Bending Moments Definitions -Cantilevers, simply supported beam, fixed

beam,

continuous beams -Conception of Shear Force and Beading Moment— Sign conventions—shear force and Bending Moment diagrams for cantilevers, beams supported at ends. Beams with overhangs— Point of contrafiexure -Member subjected to couples— Members subjected to Oblique loading— Miscellaneous types of members and corresponding S.F.

and B.M. diagrams-;-Iqter-reIation between S F. and B. m! diagrams— To obtain the B.M. diagrams from S.F. diagram- Solved problems 105 to 130 -Problems for exercise.

158-227 5. Stresses in

Beaau

Definition— Pure or 8im{de bending— Theory of simple bending— Netural layer— Neutral axis— Beading Stress distribution-moment of resistance— Assomp^os in the theory of

—

(i7)

Chapter

Pages

simple binding— Practical application of bending equation modulus— Section moduli for different shapesSectioi) Rectangular, triangular, circular, I-section, T-section— Normal force on a partial area of a beam section — Moment of resistance of a partial area of a section -Fliiched beams— Equivalent section— Beams of uniform strength— Shear stress distribution on a beam section— Shear stress distribution on circular, triangular, I and T sections— Shear rectangular, stresses in bolts connecting components in laminated beams. Proportion of B VI and S F. resisted by flange and web of an [ section— Shear centre— Solved problems I3l to 199—

bmm

Problems 6.

228—337

for exercise.

Direct and Bending Stresses Stress

distribution of the section

rectangular column.

The middle

of an eccentrically loaded rule— Core or kernel of

third

—

a section -Circular section—Hollow section Structural section— Walls and pillars— Solved Problems 200 to 223 — 338—370 Problems for exercise. 7.

Masonary Dams Forces acting on a dam— Stress distribution on the base of a dam. Stability of a dam— Minimum bottom width of a dam 371—388 section. Solved Problems 224 to 228,

8.

Deflection of

beams

Member

10.

9.

bending into a circular are— slope, deflection and ladius of curvature — Derivation of formulae for slope and deflection— Cantilever— Propped cantilevers- Beams— Macaulay's Method — Beams subjected to couples— Moment area method -Mohr’s theorems— Relations between maximum bending stress and maximum deflection — Beams of varying section -Strain energy stored due to bending— Law of reciprocal deflections - Bette’s law— The first theorem of Castigliano— Impact loading on beams— Laminated SpringsConjugate beam method— Solved problems 229 to 312 Pro?-, 389—52'’ blcms for exercise.

Fixed and Continoons Beams Fixed beam -Relation between the free B.M. diagram and the fixed B.M, diagram-slope and deflection— Effect of sinking of supports— Fixed beam subjected to couple— Degree of fixing— Advantages and disadvantages of fixing beams— Continuous beam— Clapyron’s thcorm of three moments 528—581 Solved problems 313 to 324— Problems for exercise. Toraioii of Shafts

Pure Torsion— Theory of Pure Torsion —Torsional mement of resistance. Assumptions in the theory of pure Torsion— Polar modulus— H.P. transmitted by a shaft— Torsional Rigidity— Stepped shafts— Composite shafts — Keys— Couplings — Shear and Torsional resiltenoe— Shafts of non^drcular section— Close

coiled helical springs— Torsion of a tapering rod. Problems 325 to 361 —Problems for exercise. 11. Priacipal stresses

Solved

’582-632

and strains

Normal stresses— Tangential or shear streses— Principal

stresses

—Principal planes— Graphical and analytical methods— Ellipse of stress— Determination of principal stresses and strains— Obliquity— Mohr’s circle of stress— Combined bending and Torsion— Strain energy in terms of principal stresses— Equivalent bending moment and equivalent torque— Principal strains— Criterion for failure — Ellipse of strain— Solved prob633—682 lems 362 to 388. Problems for exercise. 12.

Thin Cylinders and Spheres

Thin cylinders— circumferential ond longitudinal Stresses— Riveted cylinderical boilers— Wire bound pipes. Thm spherical shells— Biaxial stresses in doubly curved walls of pressure vessels— Stresses in a conical tank. Solved probkms 342 to 683—705 349. Problems for exercise. 13.

Hiick cylinders and Spheres

Thick cylinders— Derivation of formulae— Lamme’s equations

—Hoop

413— Problems 14.

radial pressure distribution— compound spherical Shells - Solved problems 350 to 706—726 for exercise.

stresses

—Thick

cylinders

and

Colnmns and Struts

—

Introduction Axially loaded compression members— Crushing load— Buckling or critical load or crippling load— Euler’s theory of long columns— Different end conditions— Effective length of colums— Assumptions made in Euler’s theory— Limitations of Euler's formula— Empirical formulae— Rankine’s formula Straight line formula —Johnson's parabolic formula— Formula given by the I S. code— Column •abjected to eccentric loading— Euler's method— Rankinc’s method— Prof Perry’s formula— Columns with initial curvature —Laterally loaded struts— Solved problems 413 to 429. Problems for exercise. 727— 768

—

15.

Rhetcd

Jtrfats

Tjrpes of joints— Lap and butt joints— Failure of a riveted joint— Tearing strength, shearing strength, bearing strengthEfSciency of a joint— Riveted joints in structural steel work— Chain riveting and diamond riveting— Eccentric Riveted conn^ions— Resistance of a rivet against translation and rotation. Solved problems 430 to 442. Problems for -

exercise. 16.

769 —798

Wdded Couaectioiis The welding process— Advantages of welded connection — Disadvantages of welded connection —Types of weld— Minimum sizes of weld— Effective length- Minimum length— Fillet

weld applied tp the edge of a plate— Angle between fusion faces - Throat thickness -Intermittent

-

fillet

welds— Lap

joints

welds in slots or holes— End returns— Bending about fillet -Permissible stresses in single welds— Combined a 17. in welds— Eccentric welded connections. Solved stresses 799— 830 problems 443 to 462 Fillet

Analysis of Framed Stroctures Perfect frame -Deficient frame— Redundant frame— Reactions supports— Analysis of a truss -Method of joints -Method

at

of section— Graphical method. Solved problems 463 to 489.

831-914 18.

Simple Mechanical Properties of Metals Yield or flow of material— Tensile stress— Stress— Strain diagrams for Mild Steel Specimen— Limit of proportionality— Ultimate stress— Working stress— Factor of safety— Measurement of ductility -Unwin’s Method based on reduction of sectional area— Hardness— Scratch test— Indentation test— testing— Fatigue of metalsBrinnel’s method— Impact Endurance limit. Solved problems 449 to 491. 915—922

19.

Elements of reinforced Conaete General principles of design— Assumptions— Singly reinforced beams-Nctural axis-Lever arm— Moment of resistance— or critical sections— Unbalanced Balanced or economic sections— Under-reinforced and over-reinforced sections—

Doubly reinforced beams— Shear in beams- Shear stressesDiagonal tensile and diagond compressive stresses in concrete-Stirrups-Diagonal reinforcement— Bond stresses— End ancliora.e -Standard hook— Reinforcement— T and L beams- Axially Loaded Columns- Combined bending and direct stresses. Solved problems 492 to 520, Problems for

923— 998

exercises.

Appendix Index

Useful tables.

999-1035 1036—1038

1 Stresses and Strains §1.

btrodactioB

we come across may be

Materials which plastic

and

materials.

rigid

An

elastic

claailQed into elastic,

matoial

undeigoes a deformation when subjected to an external load^ such deformation disappears on the removal oftheloi$ng. plastic material undergoes a continuous deformation Ai^ng the ^riod of loading and the deformation is permanent and materitd does not regain its original dimensions on the removd' df the lAaffiag rigid material does not undergo any deformation when subjected

A

^

A

to

an external loading. In practice no material

We attribute these

is

absolutely elastic nor plastic nor rigid.

when the deformations me within certain Generally we handle a member in its elastic range. Structural limits. members are all generally designed so as to remain in the riastic properties

condition unde r the action of the working loads. §2.

Resistance to Deformation

A material when subjected to an external load system goes a deformation. Against this deformation the ateri al will offer a resistance which tends to prevent the defonnatitin. This material as long as the ntember is resistance is offered by the forced to remmn in the deformed condition.^ This resistance is offered by the material by virtue of its srrengrA. In the elastic stage, the resistance offered by the material is proportional to the deformation brought about on the materia] by the external tna/Ung The material will have the ability to offer the necessary resistance when the deformation is within a certain limit. loaded member remains in equilibrium when the resistance offered by the member against the ddbrmation and the applied load are in equifibriom. When the member is incapable of offering the necessary leristance against the external forces, the deformation wOl continue leading to the failure of the member.

m

A

§3.

Stiem

The is called

the food.

force of resistance offered by a body against the stress. The external force acting on the body

the

The load

is

in tte material of the

b^.

Tj^s of stresses. area

eppUed on the body white the Fig.

1

(a)

shows a rod of uniform tiie aids A and B.

A and subjected to axial leads F «t

is

stress is

naiw

» .

SntSNOTB OF MAtWIAlS 2

,-^„ .i«.io.yyi»niialtottc

longitaillMl

nit »f

”*'*^theiiieii>bwtalakentooiMSiit(>tl«ol)iirBC«iia by the sertioa XX. whicli it is dMded of the part C. consider the equilibrium Let us

See Fig

1

P®"

(W-

is

subjected to the

mlo

.

_ _

, , F externally

in equilibrium, the part

kli thifoS

J

C oflm a r«i8-

^t

In other words, we may against a possible „ offering the resistance stress. against the deformation is the R resistance This

SnolraUhe^Son

XX is

This part

B

Mie

say,

XX.

R

Son XJf. equal

ai>ove, *hc rwislto^ Obviously for the case mentioned offered by the section to the load. If the resistance

Mdo^ite

defonnatioD be assumed to be uniform across the^ scctioii> in^sity of the resistance per unit area of the section is called At ist&uity of stress or unit stress. In common usage the word sftoss is used to mean the intensity of stress.

iMiml tli9 difc

Intensity of stress

=p=-jR

A

P

=-

.•

A

Let doe to the application of the load the length of the dtange from / to /+45 of Fig. 1 (ii) Compressive stress. to pushing axial loads as shown in Fig. 2, a resistance against a decrease in length. any section such as

be subjec^ set up by

is

XX

Fr..2

This resistance is called a compressive resistance. of the compressive resistance or stress is given by

The

intensity

Let due to the external loading the len^h of the member decrease by dl. The ratio of the decrease in length to the original length is called a compressive strain.

Compressive strain

Decrease in length Original length

SfRENGTH OP MATSRIALS

4

shows a rectangular block of height (no Shear stress. Fig. 3 unity. width and L / and length

Fig. 3

Let the bottom face of the block be fixed to a surface EF. Let a force P be applied tangentially along the top face of the Such a force acting tangentially along a surface is called a block.

hear

force.

For the equilibrium of the block, the surface EF will offer a tangential reaction P equal and opposite to the applied tangential force P. Let the block lx taken to consist of two parts G and Consider the equilibrium to which it is divided by a section XX. of the part G.

H

See Fig. right, the part

ttot

4.

In order the part a resistance

H will offer

G may

not

R along

the section

move from

left

XX

to

such

R=P.

ic)

(a) Fig. 4

Similarly, considering the equilibrium of the part H, that the part G will offer a resistance along the section that R^P.

R

The

resistance

R along

the section

XX is

we

find

XX such

called

a shear

rests-

tmee.

XX

Fig. 5 shows a failure at the section caused by the tangential loads acting on the top and bottom faces of the block. This type of ftflure is called a shear failure. In a shear failure, the two parts into which the block is separated, slide over each other. Hence if udi a shear failure should not occur, the section must be able to offer tangential resistances along the section opposing the force qt file trRAlNS

uz>

5

id

(b>

Hig. 5

brium of tbe system the shear resistance

R shtndd be equal

to the

tangential load P,

R^P. The

intensity of the shear resistance along

the section :

XX

is

called the shear stress.

Sheai stress - ^ ~

R A

Shear resistance Shear area

R _ P

Lx

l

"~L x

1

Shear deformation. Fig. 6 (a) shows a rectangular block subjected to shear forces on its top and bottom faces.

P

Fig. 6

When the block does not fail in shear, a shear deformation occurs as shown in Fig. 6 (6). If the bottom face of the block bo fixed, it can be realized that the block has deformed to the position A^BiCD. Or we can say, that the face ABCD has been distorted to the position A\B\CD through the angle BCBv=^^. Let us now imagine that the block consists of a number of horizontal layers. These horizontal layers have undergone horizontal displacements by dilferenl amounts with respect to the bottom face. can assume that the horizontal displacement of any horizontal layer is proportional to its distance from the lower face of the block.

We

be dh

Let the horizontal displacement of the upper face of the blodc Let the height of the block be /.

STRENGTH OF HATBRiAtS considered any other horizontal layer

shear strain. We could have which is at a distance x from the lower face. say the layer the layer XX. dx be the horizontal displacement of

XX

Let

dx Then shear Since ^

is

strain

very small,

^=tan^= j- = shear

Hence, the angular deformation the shear strain. §4.

^

in radian

strain.

measure represents

Elastic limit

A material is said to be elastic when it undergoes a deformation on the application of a loading such that the deformation disappears on the removal of the loading. When a member is subjected to an

When the its section svill olfer a resistance or stress. removed, obviously the stress will vanish and the deformaBut this is true when the deformation caused tion will also vanish. by the loading is within a certain limit. For every maieriaJ the property of assuming or regaining its previous shape and size is exhibited on the removal of the loading, when the intensity of stress If the loading is so is within a certain limit called the elastic limit. large that the intensity of stress exceeds the elastic limit, the member If after exceeding the loses to some extent its pioperty of elasticity. elastic limit the loading is removed, the member will not regain its original shape and a residual strain or permanent set remains. axial

loading,

loading

is

Hooke's law. It is observed that when a material is loaded such that the intensity of stress is within a certain limit, the ratio of the intensity of stress to the corresponding strain is a constant which is characteristic of that material. Intensi ty of stress

^constant.

strain

In the case of axial loading, the ratio of the intensity of tensile or cotnpressive stress to the corresponding strain is constant This ratio

is

called

denoted by

E

Young’s Modulus or Modulus of Elasticity* and

is

e In the case of shear loading also, the ratio of the shear stress to the corresponding shear strain is found to be a constant when the shear deformation is within a certain limit. This ratio is called Shear Modulus or Modulus of Rigidity and is denoted by C, or G.

N

fS.

Units

In this book SI andMKS units are adopted to express quantities of various magoitudes. The following nomenclature is

SiSdT

tlllPLK STRESSES

AND STRAMS

KILO

7

MILUIQT^

I0«

MEGA GIGA

MICRO 10-« NANA 10-*

10«

10»

TERRA

PICA

1012

I0-»*

In the SI units force is generally expressed in newtats. newton ikN) means 1000 newtons. In the

MKS units force is expressed in kg

was to express force

in

Stress intensity

is

kg

The

kilo-

(the earlier practice

wt).

expressed in various forms

newtonlmn^, newtonim*, kglcm\ kglnfi 1

J\r/meife2= 10-« iV/mm*

1

Njmm^ =10® N/metr^-

Problem

A rod of

!.

l

mega Newtonlmei^^

25 mm in ^tnketer and 200 cm an axial pull of 450(1 kg. Find {i) the

steel is

The rod is subjected to long. intensity of stress, (it) the strain,

and (Ui) ehngation. TUdce E—Tl X lOA

kglcm*. SoIutiOD.

Diameter of

Length of rod

PoU

rod=d=25 mm=2‘5 cm —1—200 cm =P=4500 kg

.

(i) Intensity

of stress

—

kx2’5*

=A— ntP

Area of the section

=/

P

^

'

.

«=4 909 cnr

4500

,

7 kglotfi

Strain

916'7 f “ .£’“21x10®“^

(Hi) Elongation

=d/=StramXorigina! length

(ii)

» 00004365 X 200 cm ‘0-0873 cm.

mm

in diameter tmd 2 metre long Problem 2. (SI) A steel rod 25 subjected to an axial pull of 45 kN. Find (/) the intensity of stress, (ii) the strain, tmd (Hi) elongation. Take E—2(X) GNfmetr^. is

Solution. Diameter of rod =25

Length of rod Puii

Area of rod

mm

**2 metn

P=45

=-J

jt)^=45000

N

(25)*-490-9 mnfl

=490-9 X10“« metr^ (i) Intensity

of jtresr=/—

P

STRENGTH

8

45000

MATERIALS

..

.

“490'9X 10r« ‘=9167 X /O® Nlmetr^~=9r67 Njmm^

MN/metr^

m>9J'67 (h) Strain

—d/^Strain X original length =0’0004583 x 2 metre

(h7) Elongation

=^0 0009166 metre •sO'9166 Probicin

metre long. the the

member tie

A wooden

3.

It is

is

tie is 7'

mm.

wide, 15 cm deep and I'SO of 4500 kg. The stretch of Find the Young's Modulus for

5 cm

subjected to an axial pull

found

to

be 0 0638 cm.

material.

Area of tie=j4*=7‘5x 15=112’5 cm^

SoIntioB.

=/>-4500 kg

Pull .'.

hg/cm~

Stress

Change Strain

in length

Original length

Young's Modulus

_ 0 0638 =00004253 ISO

= E— f e

40

“cF0W4253 ProUem

A

load of 400

kg/cm-.

kg has

to he raised at the end of a must not exceed 800 kgjcm^ what is the minimum diameter required ? If^ai will be the extension of 3' 50 metre length of wire ? Take E—2x /(/> kgfem^. steel wire.

4.

If the unit stress in the wire

SoIntioB.

Load on the wire= 1F=400 kg

=A=*y

Area required

Let the diameter of the wire be

ueP •••

^=

d cm

0-5

^

d-yj ^^0-7979 cm 7'979

mm

£xt0irjiMi=«d7’=Strainx original length

cm^

MMPU STRESSES AMO STRAINS

9

800

X3S0 an 2X10* =m0’l4 cm =ar4

mm

ProUen 5. (SI) A wooden tie is 60 mm wide, 120 mm deep and 1‘50 metres long. It is stdtjected to an axial pull of 30 kN. The stretch of the member is found to be 0‘625 mm. Find the Young's Modulus for the tie material. SoiathM.

Area ofthetie=X==60x 120= 7200 mm®=-7200xl0’* metr^

Puli=P=30

ifciV=30,000 JV

Stre88-/=

^=

_ Mrain-c=

72^^-6 =

Change

.

^

0‘625

in length

Qjiginal

10* Nlmetre^

len^ “ rsx 1000

= 4 167X 10-« Young's Modulus—

E=— e

4167X10*

-

,

,

W=«-

=10^ Nfmetr^ =10xl(F N/metre^

=10 GNJmetr^. ProUem 6 tensile

divisions in the O' 00 1

(SI).

load of 40 kN.

200

mm, find the

A 20 mm diameter

brass rod was subjected to a The extension of the rod was found to be 254

mm

extensometre.

If each division

is

modulus of brass.

electric

Stdathm. 7C

Area of the rod=y4=-^ (20)*

mm*

=3 14’ 16 mm* =314*16 xl0-« metre*

PuIl=P=40 r

Stress- /=•-

^ ^

itiV= 40,000

_

N

40,000 3j4*i6’x i0-«

^

= l•2732Xl08^/mefre* Lmgth of specimen=/=200 mm «d7= 254 XO’OOl =0*254 mm

Extension

Strain=e=

=0*00127

,

etpial

to

STKENGTIi OF tIAlEUAXS

10

r2732x

t

Modulus=E=

Young's

‘e

“

108

Njmetr^

0 00127

=^1002-5xmNlmetr^ =‘l00-25xl(fl Nlmetr^

= 100-25

GNImetre^.

Problem 7. A hollow steel column has to carry an axial load of 200,000 kg- If the external diameter of the column is 25 cm, nd the Take the ultimate stress for the steel column to be internal diameter. 4800 kglcm^ and allow a load factor of 4. Soiotion.

=D =25 cm

External diameter

-d

Internal diameter

Load on

the column

=

200,000 kg

Ultimate stress

=4800 kgjcm^

Factor of safety

=4

Safe stress =/=

Ultimate stress Factor of safety

4800

4

= 1200 kgjcrn^

Sectional area required

W

200,000

“l200~ ( 252

= V6-61

(7w2

-

=480

kg.lem*.

-

( 3 25)2

Total extension

dl‘=dh+dh-\-dlz

= 4'^i+=|-/2+^.fa ( fth +A&+/3iyi

2 x lO®

[565-8x18+1274x26

+480x16]

cm.

=0‘0255 cm. 10. A bras.i bar having a cross-sectimud area of 10 sq. subjected to axial forces shown in fig. 9. Find the total change in length of the bar. Take E> ’^^ J OSx Kfi kg.Icnfi.

ProUem

cm.

is

B 5,000kg.

\

C

8,ooaJsshai 'i1 by the two should bc\nC\ th it thc\ estend by the same Let the stress ifiionsiry on the sections of the lopes 4B and

ih) In

ropes

AB

ftmount.

CB be pi

the

and

ard Pi

respccci\cl>\

Biquatmg the extensions of the two

m P‘i

h'0^_

J3

U

^

4‘5n ^

'

4

.

in

BqI tennon

^

r >pe

- jpi

in i4B4-tcnsion in

kg

CB

-*

W

i

wc

]ia\e.

^X SIMPLE STRESSES

AND

=

19

STRAINS

500

/»i+P2=—j,

4

500

3P8+;»2=— 7 3

^~

500 8

p2’^26 8 kgicnfl. Pi =

and

35‘7 kglcm^.

26 8

Downward movement of the

pulley

^Extension of the rope

AB or CB

^V662 cm. Problem 15. A steel subjected to a pull of St,

tie

rod 4 cm.

in diameter

add 2

long is

To what

length the bar should be bored centrally so that the total extension will increase by 20% under the same pull, the bore being 2 cm. diameter.

Take E^2000

tjern^.

et

at Fig. 14.

ilutioo.

*

4

«

— (4)2«47c

cm^.

«

4n

/ 'E

‘

're

X 2000

x200=

Extension after the bore

is

l_

5«

made=r2x

J'5J5

6 Lei the bar be bored to a length cl reduced section A' "=4ji

—" 4

I

metres.

Area at the

(2)®

4nr-ic=>3n

enfi.

Extension of the rod

imttrn

I

pi. i


and pb respectively,

AND STR MNS

SIMPLE STRESSES

Load on

steel

31

+ load on brass = Total load on the compound

tube

psAs-{-pbAit=P

i.e.,

E»

pbA»+pbAi>’=‘F

Eb

]=/> rEiiA$'\~Ei^Eo

,

"]

'•'‘[fiAfBx)’ Extension of the

compound tube =df ==ExteRSion of steel or brass tuoc tb

dh

P-

4\E,A,+EbAb

\i }

E be

Young’s Modulus of a tube of area the same load and undergoing the same extension. Let

" *-( {Aa+At,)E jP/

carrying

)

PI

(E^Ab^ Eo Ah)

(Aa-\-Ab)E

F.«(«- 0 01

)

4m0

2-2xl0*ij/cii# ) kg./em?

Stress in

/»,— 14000 3 kg.laifl.

Load OB steel+Ioad on alumimam =totaI load on the poflopoate ptAf\-p»A.^P (>-0015) 44000 x 38*4H-14000»X 3^36=6000011: i

SIMNCTH OP IMIBUAU 16938+48I 04*=60+25-4 2174048-85-4

/.

85-4

«

*- 21^045=0 03928 an. /».=(S-0015) 44000 kg.Icm.*

-(0 03928-0 015) 44000 kg./an.*

~ 1068-32 kg.lem.^ 14000 5

and

= 14000 X 0 03928 kgjcn^. =549 92 kg.fcnt,^ §9.

EquWaleot Area of a Compomid Sectioa

compound Suppose a column consists of a concrete oolumn reinforced with steel

•

•

bars.

Let the gross area of the

oolumn be

^

>4.

Let the area of

steel

^ ^ ^O^RALL / AR£Ag^A

ai ARfACf

be

Si sna^A^

A$.

Actual area of con-

mte-^AcMA—As).

^

^ '

Let the stresses in conCrete and steel be pc and p,

~ '

p.

it^)ectively.

Strain in

Concrete— Strain

^

in Steel

P*^mpc

...(#)

where

E, ‘s

called the

modular ratio between

OQilcrete.

Load on

steel -{-load

on concrete =load on column

P*At-\-pcAt^P

pd.A»~i~mAt)

=P

P.^

iL-_

steel

and

SIMPLE STRESSES

AND

STRAINS

A,"‘(A—A$)

But

_

P A — At-f-mAt

P ^‘’°A+{m-l)A. Suppose in place of the composite section, a plain concrete column of area A-\-{m— l)A, had been provided, the stress in concrete

P A-\-(m—ljAt

Hence for determining the stress in concrete, we may consider that the given reinforced concrete column is equivalent to a plain concrete column whose sectional area

=At^A-{-(m — This area >(»/+/»c/*=Ph ...(/«)

Let

Sa>

adO

CCi and DDi.' These pressures

on the

be the depressions of the pillars AAj, BBi, depressions are proportional to the respective

pillars.

Where AT is constant

8a— AlPa.

SaoiF,

h=KPi, dc=--KPe

h=KPi — Average of depressions of A O

Depression of

and

C

—Average of depressions of B and

D

KiPa+Po) "

__8.+8.

-

2

2

“

“

2

K(P^+P4 2

P.+Pt^Po+Pi Rewriting the above equations.

P.+P.+/>.+Pi-P

...(0

Pc+Pd=^

...(iV)

Pi+Pc=~

...(f/f)

P«+Po=P»4-P1 «

eq.

(ff/),

1

we

get

..(v)

SaiFU SntESSES and strains

SI

Substituting the value of Pd in equation (v/)

y)

J*.-P.+Pd-P.= -|—

2(a+b)~/|

/-2(a+b)| Thus the pressures on the

pillars are.

P.= -J|3f-2

(u+b)|

P*= £-{/+2 (*-^l2xfO-AfC and »e=18 5 x W'^jC. {.A.M.LE. May, 1965) Solution. Area of each pillar— .4— 5 cm.* ProUem

pt^

pillars,

Initial stresses

At 15®C each

pillar

carries-^ tonnes.

Stress in each pillar—

toimeslem*.

-I333’33kg.lcm.*

I

STRENGTH OF MATERIALS

66

of temperature alone. Let the stresses due of temperature aloue be pc kg Icm.^ (compressive) in copper

Stresses due to rise

to rise

and

steel. p* kg-tern.^ (tensile) in

Lit the change in length of each

member be

I^oicTl—^-

«.r/+-§'

,,T+^^c,cT-Pf PL Ec P*

•

•

.

.£i«r(«.-«.)=(ll5-15)(18-5-12)10-« Er

P'

I

looxesxio'^

‘lx 10*"*'0'8xl0« I’Ll jPl ==650 8 2

^0

2/>.+5/>.=2600 pc At

Also

...(i)

= pc At

p.X5 -prXlO p.=2p. Substituting in Eqn«='480 1920

^+3ip.-19» Pk’^S76 kg./cm,* (jeompressive)

x

SIMMOTH »' Total load

V44fa+Anf,^7m r-M/.+4 32x3/.-2000 14'4(>/.-2000 f

.8

2000.

/•“144 **•/lcai 69. A steel bar 40 mm. x 40 mm. iOO cm. long in a tonnes. Taking x 104 section is subjected to an axial pull of 12 ratio O' as calculate poisson’s the alterations in the 3. kg-lcm^ ond length and sides of the bar during the extension. (AMIE, Winter 1974)

E =2

Solutioii.

Tensile stress

P 12S00 =800 -/=-^ = 4

^

kg.jcm^

800

f Linear or longitudinal strain=e«=“-^=2j^-j-Q6*=4x 10“Lateral

strain=e»=e*= ntJb

= -0'3x4x10”« --r2xl0"* /.

Increase in length “‘Cx

x original length

=(4xl0~«) 300=0-72