Simple stresses in beams B.TECH 3rd SEM Mechanical by Shashikant

Strength of Material

1

S H A S H I K A N T ' S N O T E S

COURSE CONTENT IN BRIEF

1. Simple stress and strain

2. Statically indeterminate problems and thermal stresses

3. Shearing force and bending moment

4. Stresses due to bending

5. Stresses due to shearing

6. Slope and deflection of beams

7. Stresses due to Torsion in circular shaft

8. Variation of stress at a point

9. Stresses due to fluid pressure in thick and thin cylinder

10. Stability of columns

2


Books for Reference

1. Mechanics of Materials, by E.P.Popov

2. Mechanics of Materials, by E J Hearn

3. Strength of materials, by Beer and Johnston

4. Strength of materials, by F L Singer & Andrew Pytel

5. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa

6. Strength of Materials, by Ramamruthum

7. Strength of Materials, by S S Bhavikatti

3


• Normal stress and strain

• Hooke‟s law

• Modulus of elasticity

• Tension test on ductile and brittle materials

• Factor of safety and allowable stress

• Poisson's ratio

• Shear stress and shear strain

• Modulus of rigidity

• Tapering bar and stepped bar subjected to axial load

CHAPTER – I

Simple stress & strain

4


The subject strength of materials deals with the relations between

externally applied loads and their internal effects on bodies. The

bodies are no longer assumed to be rigid and the deformations,

however small, are of major interest

Alternatively the subject may be called the mechanics of solids.

CHAPTER – I Introduction

The subject, strength of materials or mechanics of materials involves

analytical methods for determining the strength , stiffness

(deformation characteristics), and stability of various load carrying

members.

5


Engineering Mechanics

Mechanics of Solids Mechanics of Fluids

Rigid Bodies Deformable

Bodies

Statics Dynamics Strength of

Materials

Theory of

Elasticity Theory of

Plasticity

Ideal

Fluids Viscous

Fluids

Compressible

Fluids

Branches of Mechanics 6


GENERAL CONCEPTS

STRESS

No engineering material is perfectly rigid and hence, when a

material is subjected to external load, it undergoes

deformation.

While undergoing deformation, the particles of the material

offer a resisting force (internal force). When this resisting

force equals applied load the equilibrium condition exists

and hence the deformation stops.

These internal forces maintain the externally applied forces

in equilibrium.

7


Stress = internal resisting force / resisting cross sectional area

The internal force resisting the deformation per unit area is

called as stress or intensity of stress.

STRESS 8

A

R


gigapascal, 1GPa = 1×109 N/m2

= 1×103 MPa

= 1×103 N/mm2

SI unit for stress

N/m2 also designated as a pascal (Pa)

Pa = N/m2

kilopascal, 1kPa = 1000 N/m2

megapascal, 1 MPa = 1×106 N/m2

= 1×106 N/(106mm2) = 1N/mm2

1 MPa = 1 N/mm2

STRESS 9


AXIAL LOADING – NORMAL STRESS

Consider a uniform bar of cross

sectional area A, subjected to a tensile

force P.

Consider a section AB normal to the

direction of force P

Let R is the total resisting force acting

on the cross section AB.

Then for equilibrium condition,

R = P

Then from the definition of stress,

normal stress = ζ = R/A = P/A

P

P

P

R

B A

R

P

STRESS

10

ζ = Normal Stress Symbol: S H A S H I K A N T ' S N O T E S

Direct or Normal Stress:

AXIAL LOADING – NORMAL STRESS

Intensity of resisting force perpendicular to or normal to the

section is called the normal stress.

Normal stress may be tensile or compressive

Tensile stress: stresses that cause pulling on the surface of the

section, (particles of the materials tend to pull

apart causing extension in the direction of force)

Compressive stress: stresses that cause pushing on the surface of

the section, (particles of the materials tend to

push together causing shortening in the

direction of force)

STRESS

11


• The resultant of the internal forces for an

axially loaded member is normal to a

section cut perpendicular to the member

axis.

A

P

A

Favg

A

0lim

• The force intensity on that section is

defined as the normal stress.

STRESS

12


STRAIN

STRAIN :

when a load acts on the material it will undergo deformation.

Strain is a measure of deformation produced by the application of

external forces.

If a bar is subjected to a direct load, and hence a stress, the bar will

changes in length. If the bar has an original length L and change in

length by an amount δL, the linear strain produced is defined as,

L

L

Original length

Change in length =

Strain is a dimension less quantity.

13

Linear strain,


LINEAR STRAIN

strain normal

stress

L

A

P

L

A

P

A

P

2

2

LL

A

P

2

2

14


STRESS-STRAIN DIAGRAM

In order to compare the strength of various materials it is necessary

to carry out some standard form of test to establish their relative

properties.

One such test is the standard tensile test in which a circular bar of

uniform cross section is subjected to a gradually increasing tensile

load until failure occurs.

Measurement of change in length over a selected gauge length of the

bar are recorded throughout the loading operation by means of

extensometers.

A graph of load verses extension or stress against strain is drawn as

shown in figure.

15



Typical tensile test curve for mild steel

16



Typical tensile test curve for mild steel showing upper yield point and

lower yield point and also the elastic range and plastic range

17


Limit of Proportionality :

From the origin O to a point called proportionality limit the stress

strain diagram is a straight line. That is stress is proportional to

strain. Hence proportional limit is the maximum stress up to which

the stress – strain relationship is a straight line and material behaves

elastically.

From this we deduce the well known relation, first postulated by

Robert Hooke in 1678, that stress is proportional to strain.

Beyond this point, the stress is no longer proportional to strain

A

PPP Load at proportionality limit

Original cross sectional area =

18

Stress-strain Diagram


Elastic limit:

It is the stress beyond which the material will not return to its

original shape when unloaded but will retain a permanent

deformation called permanent set. For most practical purposes it can

often be assumed that points corresponding proportional limit and

elastic limit coincide.

Beyond the elastic limit plastic deformation occurs and strains are

not totally recoverable. There will be thus some permanent

deformation when load is removed.

A

PEE Load at proportional limit


19



Yield point:

It is the point at which there is an appreciable elongation or yielding

of the material without any corresponding increase of load.

A

PYY Load at yield point


20



Ultimate strength:

It is the stress corresponding to maximum load recorded during the

test. It is stress corresponding to maximum ordinate in the stress-

strain graph.

A

PUU Maximum load taken by the material


21



Rupture strength (Nominal Breaking stress):

It is the stress at failure.

For most ductile material including structural steel breaking stress is

somewhat lower than ultimate strength because the rupture strength

is computed by dividing the rupture load (Breaking load) ,which is

less than the maximum load which it can bear, by the original cross

sectional area.

A

PBB

load at breaking (failure)


True breaking stress = load at breaking (failure)

Actual cross sectional area

22



The capacity of a material to allow these large plastic deformations is

a measure of ductility of the material

After yield point the graph becomes much more shallow and covers

a much greater portion of the strain axis than the elastic range.

Ductile Materials:

The capacity of a material to allow large extension i.e. the ability to

be drawn out plastically is termed as its ductility. Material with high

ductility are termed ductile material.

Example: Low carbon steel, mild steel, gold, silver, aluminum

23




Percentage elongation

in length

A measure of ductility is obtained by measurements of the

percentage elongation in length or percentage reduction in area,

defined as,

increase in gauge length (up to fracture)

original gauge length ×100

Percentage reduction in

area original area ×100

=

=

Reduction in cross sectional area of necked portion (at fracture)

24



Brittle Materials

A brittle material is one which exhibits relatively small extensions

before fracture so that plastic region of the tensile test graph is

much reduced.

Example: steel with higher carbon content, cast iron, concrete, brick

25


2

2

1000

)(

)(

in

lbksi

insquareinch

lbpoundspsi

STRESS-STRAIN DIAGRAM: DUCTILE

MATERIALS

26


STRESS-STRAIN DIAGRAM: BRITTLE

MATERIALS

Stress-strain diagram for a typical brittle material

27


STRESS-STRAIN TEST

Machine used to test tensile test specimen test specimen with tensile load

L = gauge length

28


4

29


HOOKE”S LAW

Hooke‟s Law

For all practical purposes, up to certain limit the relationship

between normal stress and linear strain may be said to be linear for

all materials

Thomas Young in 1807 introduced a constant of proportionality that

came to be known as Young‟s modulus.

stress (ζ) α strain (ε)

stress (ζ)

strain (ε) = constant

stress (ζ)

strain (ε) = E

Modulus of Elasticity

Young‟s Modulus =

30

or


HOOKE”S LAW

Young‟s Modulus is defined as the ratio of normal stress to linear

strain within the proportionality limit.

From the experiments, it is known that strain is always a very small

quantity, hence E must be large.

For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2

stress (ζ)

strain (ε) = E =

LA

PL

L

L

A

P

The value of the Young‟s modulus is a definite property of a

material

31


Material Density

(kg/m3)

Young's Modulus

109 N/m2

Ultimate Strength Su

106 N/m2

Yield Strength Sy

106 N/m2

Steel 7860 200 400 250

Aluminum 2710 70 110 95

Glass 2190 65 50 ...

Concrete 2320 30 40 ...

Wood 525 13 50 ...

Bone 1900 9 170 ...

Polystyrene 1050 3 48 ...

32


HOOKE’S LAW: MODULUS OF

ELASTICITY

Below the yield stress

Elasticity of Modulus

or Modulus Youngs

E

E

Strength is affected by alloying, heat

treating, and manufacturing process but

stiffness (Modulus of Elasticity) is not.

Stress-strain diagram for Iron and

different grades of steel

33


STRESS-STRAIN

DIAGRAM

Hard drawn wire materials Various types of nylon and

polycarbonate

34


ELASTIC VS. PLASTIC

BEHAVIOR

• If the strain disappears

when the stress is removed,

the material is said to

behave elastically.

• When the strain does not return to zero after the

stress is removed, the material is said to behave

plastically.

• The largest stress for

which this occurs is called

the elastic limit.

35


ELASTIC VS. PLASTIC

BEHAVIOR

For certain materials, for example, high carbon steel and non-ferrous

metals, it is not possible to detect any difference between the upper

and lower yield points and in some cases no yield point exists at all.

In such cases a proof stress is used to indicate beginning of plastic

strain.

Proof stress is the stress corresponding to a fixed permanent strain

in stress-strain diagram.

For example: 0.1% proof stress indicates that stress which, when

removed, produces a permanent strain or “set” of 0.1% of the

original gauge length.

36


PROOF

STRESS

Determination of 0.1%

proof stress

Permanent deformation or “set”

after straining beyond yield point

37


DEFORMATIONS UNDER AXIAL

LOADING

AE

P

EE

• From Hooke‟s Law:

• From the definition of strain:

L

• Equating and solving for the

deformation,

AE

PL

• With variations in loading, cross-

section or material properties,

i ii

ii

EA

LP

38


Derive an expression for the total extension of the tapered bar of

circular cross section shown in the figure, when subjected to an axial

tensile load W

W W

A B

L Diameter

d1 Diameter

d2

39


Consider an element of length, dx at a distance x from A

B

W W

A

x d1 d2 dx

Diameter at x,

xL

ddd

12

1c/s area at x, 21

2

44kxd

d x

xkd 1

Change in length over a

length dx is

Ekxd

Wdx

AE

PL

dx2

14


length L is

L

Ekxd

Wdx

0 2

14

40


By putting d1+kx = t,

Then k dx = dt

L

Et

k

dtW

0 2

4

LLL

kxdEk

W

tEk

Wt

Ek

W

0100

12

)(

1414

1

4

Edd

WL

dEd

WL

4

4

2121

41


Derive an expression for the total extension of the tapered bar AB of

rectangular cross section and uniform thickness, as shown in the

figure, when subjected to an axial tensile load W.(b is the thickness of

the bar).

W W

A B

L

d1 d2

b

b

42


W W

A B

x

d1 d2

b

b dx

Consider an element of length, dx at a distance x from A

depth at x,

xL

ddd

12

1c/s area at x, bkxd 1

xkd 1


length dx is

Ebkxd

Wdx

AE

PL

dx 1

43



length L is

L

Ebkxd

Wdx

01

12 loglog ddkEb

Pee

12

12

loglog303.2

ddddEb

LP

44


Derive an expression for the total extension produced by self weight

of a uniform bar, when the bar is suspended vertically.

L

Diameter

d

45


P1 x

P1 = weight of the bar below

the section,

= volume × specific weight

= (π d2/4)× x ×

= A× x ×

Diameter

d

dx dx

element

Extension of the element due to weight of the bar below that,

AE

dxxA

AE

dxP

AE

PL

dx

)(1

46


The above expression can also be written as

Hence the total extension entire bar

E

L

E

x

AE

dxxAL

L

22

)( 2

0

2

0

AE

PL

AE

LAL

A

A

E

L

2

1

2

)(

2

2

Where, P = (AL)×

= total weight of the bar

47


SHEAR STRESS

Consider a block or portion of a material shown in Fig.(a) subjected to

a set of equal and opposite forces P. then there is a tendency for one

layer of the material to slide over another to produce the form failure

as shown in Fig.(b)

P

The resisting force developed by any plane ( or section) of the block

will be parallel to the surface as shown in Fig.(c).

P Fig. a Fig. b Fig. c

P

P

R R

The resisting forces acting parallel to the surface per unit area is called

as shear stress.

48


Shear stress (η) = Shear resistance

Area resisting shear

η

If block ABCD subjected to shearing stress as shown in Fig.(d),

then it undergoes deformation. The shape will not remain

rectangular, it changes into the form shown in Fig.(e), as AB'C'D.

B

Fig. d

Shear strain

A

P

This shear stress will always be tangential to the area on which it acts

η D

C

A

η B'

D

C'

A

η

B C

Fig. e

49


The angle of deformation is measured in radians and hence is

non-dimensional.

D

η B' C'

A η

Fig. e

B C

tanstrain shear AB

BB

The angle of deformation is then termed as shear strain

50

Shear strain is defined as the

change in angle between two

line element which are

originally right angles to one

another.


SHEAR MODULUS

For materials within the proportionality limit the shear strain is

proportional to the shear stress. Hence the ratio of shear stress to shear

strain is a constant within the proportionality limit.

For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2

Shear stress (η)

Shear strain (θ) = constant =

The value of the modulus of rigidity is a definite property of a

material

G Shear Modulus

or

Modulus of Rigidity

=

51


EXAMPLE: SHEARING STRESS

• Forces P and P‘ are applied

transversely to the member AB.

A

Pave

• The corresponding average shear stress is,

• The resultant of the internal shear force

distribution is defined as the shear of the

section and is equal to the load P.

• Corresponding internal forces act in the

plane of section C and are called

shearing forces.

• The shear stress distribution cannot be

assumed to be uniform.

52


Double shear

Consider the simple riveted lap joint shown in the Fig.(a). When load is

applied to the plates as shown in the figure the rivet is subjected to shear

forces tending to shear it on one plane as indicated.

Shear stress η (in double shear) = P/2A

But the joint with two cover plates, shown in Fig.(b), the rivet is

subjected to possible shearing on two faces, which is called as double

shear. In such cases twice the area of the rivet is resisting the applied

forces so that the shear stress set up is given by

Fig. a Fig. b

P

P

P

P

53


A

F

A

Pave

Single Shear

A

F

A

P

2ave

Double Shear

Examples

54


PIN SHEARING

STRESSES

• The cross-sectional area for

pins at D

26

2

2 m104912

mm25

rA

MPa102m10491

N105026

3

,

A

PaveC

• The force on the pin at C is equal

to the force exerted by the rod BC,

example: Shearing Stress

To find the shearing stress in

pin.

55

Rod BC


• The cross-sectional area for pins at D,

& E 26

2

2 m104912

mm25

rA

• The pin at A is in double shear with a

total force equal to the force exerted

by the boom AB,

MPa7.40)m10491(2

kN40

2 26,

A

PaveA

example: Double shear 56

Rod AB


η

State of simple shear

Force on the face AB = P = η × AB × t

Consider an element ABCD in a strained material subjected to shear stress, η as shown in the figure

Where, t is the thickness of the element.

η

A B

C D

Force on the face DC is also equal to P

57


P


The element is subjected to a clockwise moment

Now consider the equilibrium of the element.

(i.e., ΣFx = 0, ΣFy = 0, ΣM = 0.)

P × AD = (η × AB × t) × AD

P

A B

C D

But, as the element is actually in equilibrium, there must be another pair of forces say P' acting on faces AD and BC, such that they produce a anticlockwise moment equal to ( P × AD )

For the force diagram shown,

ΣFx = 0, & ΣFy = 0,

But ΣM = 0

58

force



Equn.(1) can be written as

If η1 is the intensity of the shear stress on the faces AD and BC, then P ' can be written as,

P ' = η ' × AD × t

P ' × AB = P × AD

= (η × AB × t)× AD ----- (1)

P

P

A B

C D

P ' P '

(η ' × AD× t ) × AB = (η × AB × t) × AD ----- (1)

η ' = η

59

η

η

A B

C D

η ' η '



Thus in a strained material a shear stress is always

accompanied by a balancing shear of same intensity at right

angles to itself. This balancing shear is called

“complementary shear”.

The shear and the complementary

shear together constitute a state of

simple shear

A B

C D

η'= η

η

η

η'= η

60


Direct stress due to pure shear

Consider a square element of side „a‟ subjected to shear stress

as shown in the Fig.(a). Let the thickness of the square be

unity.

Fig.(b) shows the deformed shape of the element. The length of

diagonal DB increases, indicating that it is subjected to tensile stress.

Similarly the length of diagonal AC decreases indicating that

compressive stress.

a

A B

C D

η

η

η

η

a

A B

C D

η

η

η

η a

a

Fig.(a). Fig.(b).

61



Now consider the section, ADC of the element, Fig.(c).

Resolving the forces in the X-direction shown

a

Fig.(c).

a

a

A

C D

a2

For equilibrium

A ζn

C D

η

η

a

X

n

n aa

Fx

45cos212

0

62



Therefore the intensity of normal tensile stress developed on

plane BD is numerically equal to the intensity of shear stress.

Similarly it can be proved that the intensity of compressive stress

developed on plane AC is numerically equal to the intensity of shear

stress.

63


Poisson‟s Ratio:

Consider the rectangular bar shown in Fig.(a) subjected to a tensile

load. Under the action of this load the bar will increase in length by

an amount δL giving a longitudinal strain in the bar of

POISSON‟S RATIO

l

ll

Fig.(a)

64


The associated lateral strains will be equal and are of opposite sense

to the longitudinal strain.

POISSON‟S RATIO

The bar will also exhibit, reduction in dimension laterally, i.e. its

breadth and depth will both reduce. These change in lateral

dimension is measured as strains in the lateral direction as given

below.

d

d

b

blat

Provided the load on the material is retained within the elastic range

the ratio of the lateral and longitudinal strains will always be

constant. This ratio is termed Poisson’s ratio (µ)

POISSON‟S RATIO Lateral strain

Longitudinal strain =

ll

dd

)(

ll

bb

)(

OR

65


Poisson‟s Ratio = µ

For most engineering metals the value of µ lies between 0.25 and

0.33

In general

z

y

x P P

Poisson‟s Ratio Lateral strain

Strain in the direction of

load applied

=

x

x

y

y

ll

l

l

OR

x

x

z

z

ll

ll

Lx

Ly Lz

66


Poisson‟s Ratio = µ

In general

Strain in X-direction = εx

z

y

x Px Px

Lx

Ly Lz

x

x

l

l

Strain in Y-direction = εy

Strain in Z-direction = εz

x

x

y

y

l

l

l

l

x

x

z

z

l

l

l

l

67


Poisson‟s Ratio Load applied in Y-direction



load applied

=

y

y

x

x

l

l

ll

OR

y

y

z

z

l

l

ll

z

y

x

Py

Lx

Ly Lz

Py


y

y

x

x

l

l

l

l

68


Poisson‟s Ratio Load applied in Z-direction



load applied

=

z

z

x

x

ll

ll

OR

z

z

y

y

ll

l

l

y

z

x

Pz

Lx

Ly Lz

Pz


z

z

x

x

l

l

l

l

69


Load applied in X & Y direction


z

y

x Px Px

Lx

Ly Lz

Py

Py


EE

xy


EE

xy

EE

yx

70


General case:




z

y

x Px Px

Py

Py Pz

Pz

71

EEE

zyxx

EEE

zxy

y

EEE

xyzz

ζx

ζz

ζy

ζx

ζz ζy


Bulk Modulus

Bulk Modulus

A body subjected to three mutually perpendicular equal direct stresses

undergoes volumetric change without distortion of shape.

If V is the original volume and dV is the change in volume, then

dV/V is called volumetric strain.

Bulk modulus, K

A body subjected to three mutually perpendicular equal direct stresses

then the ratio of stress to volumetric strain is called Bulk Modulus.

V

dV

72


RELATIONSHIP BETWEEN VOLUMETRIC STRAIN AND

LINEAR STRAIN

Relative to the unstressed state, the change in

volume per unit volume is

eunit volumper in volume change

1111111

zyx

zyxzyx

dV

Consider a cube of side 1unit, subjected to three

mutually perpendicular direct stresses as shown

in the figure.

73


RELATIONSHIP BETWEEN VOLUMETRIC STRAIN

AND LINEAR STRAIN

EEE

zyx

EEE

zxy

EEE

xyz

zyx

E

21

zyxV

dV

Volumetric strain

74

For element subjected to uniform hydrostatic pressure,

2-13KE

or

modulusbulk 213

E

K

zyx

321

21

EV

dV

EV

dVzyx

V

dVK

75

Relationship between young’s modulus of elasticity (E) and

modulus of rigidity (G) :-

A

D η

B

η

a

a 45˚

A1

θ θ

B1

Consider a square element ABCD of side „a‟ subjected to pure shear

„η‟. DA'B'C is the deformed shape due to shear η. Drop a perpendicular

AH to diagonal A'C.

Strain in the diagonal AC = η /E – μ (- η /E) [ ζn= η ]

= η /E [ 1 + μ ] -----------(1)

Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC

C

H

76


In Δle AA'H

Cos 45˚ = A'H/AA'

A'H= AA' × 1/√2

AC = √2 × AD ( AC = √ AD2 +AD2)

Strain along the diagonal AC = AA'/ (√2 × √2 × AD)=θ/2 ----(2)

Modulus of rigidity = G = η /θ

θ = η /G

Substituting in (2)

Strain along the diagonal AC = η /2G -----------(3)

Equating (1) & (3)

η /2G = η /E[1+μ]

E=2G(1+ μ)

77


Substituting in (1)

E = 2G[ 1+(3K – 2G)/ (2G+6K)]

E = 18GK/( 2G+6K) E = 9GK/(G+3K)

Relationship between E, G, and K:-

We have

E = 2G( 1+ μ) -----------(1)

E = 3K( 1-2μ) -----------(2)

Equating (1) & (2)

2G( 1+ μ) =3K( 1- 2μ)

2G + 2Gμ=3K- 6Kμ

μ= (3K- 2G) /(2G +6K)

78


Working stress: It is obvious that one cannot take risk of loading a

member to its ultimate strength, in practice. The maximum stress to

which the material of a member is subjected to in practice is called

working stress.

This value should be well within the elastic limit in elastic design

method.

Factor of safety: Because of uncertainty of loading conditions,

design procedure, production methods, etc., designers generally

introduce a factor of safety into their design, defined as follows

Factor of safety = Allowable working stress

Maximum stress

Allowable working stress

Yield stress (or proof stress) or

79


Homogeneous: A material which has a uniform structure throughout

without any flaws or discontinuities.

Malleability: A property closely related to ductility, which defines a

material‟s ability to be hammered out in to thin sheets

80

Isotropic: If a material exhibits uniform properties throughout in all

directions it is said to be isotropic.

Anisotropic: If a material does not exhibits uniform properties

throughout in all directions it is said to be anisotropic or nonisotropic.


SAINT-VENANT’S

PRINCIPLE

• Loads transmitted through

rigid plates result in uniform

distribution of stress and

strain.

• Concentrated loads result in

large stresses in the vicinity of

the load application point.

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SAINT-VENANT’S

PRINCIPLE

• Saint-Venant’s Principle:

Stress distribution may be

assumed independent of the mode

of load application except in the

immediate vicinity of load

application points.

• Stress and strain distributions

become uniform at a relatively

short distance from the load

application points.

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b


The normal stress at a particular point may

not be equal to the average stress but the

resultant of the stress distribution must

satisfy

A

ave dAdFAP

83


Exercise Problems

1 An aluminum tube is rigidly fastened between a brass rod and

steel rod. Axial loads are applied as indicated in the figure.

Determine the stresses in each material and total deformation.

Take Ea=70GPa, Eb=100GPa, Es=200GPa

500mm 700mm 600mm

steel aluminum

brass 20kN 15kN 15kN 10kN

Ab=700mm2

Aa=1000mm2

As=800mm2

Ans: ζb=28.57MPa, ζa=5MPa, ζs=12.5MPa, δl = - 0.142mm

84


Example 7

2. A 2.4m long steel bar has uniform diameter of 40mm for a

length of 1.2m and in the next 0.6m of its length its diameter

gradually reduces to „D‟ mm and for remaining 0.6m of its

length diameter remains the same as shown in the figure. When

a load of 200kN is applied to this bar extension observed is

equal to 2.59mm. Determine the diameter „D‟ of the bar. Take E

=200GPa

Ф = 40mm

Ф = D mm

200kN 200kN

500mm 500mm 1000mm

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Exercise Problems

3 The diameter of a specimen is found to reduce by 0.004mm when

it is subjected to a tensile force of 19kN. The initial diameter of

the specimen was 20mm. Taking modulus of rigidity as 40GPa

determine the value of E and µ

Ans: E=110GPa, µ=0.36

4 A circular bar of brass is to be loaded by a shear load of 30kN.

Determine the necessary diameter of the bars (a) in single shear

(b) in double shear, if the shear stress in material must not exceed

50MPa.

Ans: 27.63mm, 19.54mm

86


Exercise Problems

5 Determine the largest weight W that can be supported by the two

wires shown. Stresses in wires AB and AC are not to exceed

100MPa and 150MPa respectively. The cross sectional areas of

the two wires are 400mm2 for AB and 200mm2 for AC.

Ans: 33.4kN

W A

C B

300 450

87


Exercise Problems

6 A homogeneous rigid bar of weight 1500N carries a 2000N load

as shown. The bar is supported by a pin at B and a 10mm

diameter cable CD. Determine the stress in the cable

Ans: 87.53MPa

3m

A C B

2000 N

3m

D

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7. A stepped bar with three different cross-sectional areas, is

fixed at one end and loaded as shown in the figure. Determine

the stress and deformation in each portions. Also find the net

change in the length of the bar. Take E = 200GPa

250mm 270mm 320mm

300mm2 450mm2

250mm2

10kN 40kN

20kN

Ans: -33.33, -120, 22.2MPa, -0.042, -0.192, 0.03mm, -0.204mm

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8 The coupling shown in figure is constructed from steel of

rectangular cross-section and is designed to transmit a tensile

force of 50kN. If the bolt is of 15mm diameter calculate:

a) The shear stress in the bolt;

b) The direct stress in the plate;

c) The direct stress in the forked end of the coupling.

Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa

90


Exercise Problems

9 The maximum safe compressive stress in a hardened steel punch

is limited to 1000MPa, and the punch is used to pierce circular

holes in mild steel plate 20mm thick. If the ultimate shearing

stress is 312.5MPa, calculate the smallest diameter of hole that

can be pierced.

Ans: 25mm

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10. A rectangular bar of 250mm long is 75mm wide and 25mm

thick. It is loaded with an axial tensile load of 200kN, together

with a normal compressive force of 2000kN on face

75mm×250mm and a tensile force 400kN on face

25mm×250mm. Calculate the change in length, breadth,

thickness and volume. Take E = 200GPa & µ=0.3

Ans: 0.15,0.024,0.0197mm, 60mm3 S H A S H I K A N T ' S N O T E S

Exercise Problems

11 A piece of 180mm long by 30mm square is in compression under

a load of 90kN as shown in the figure. If the modulus of elasticity

of the material is 120GPa and Poisson‟s ratio is 0.25, find the

change in the length if all lateral strain is prevented by the

application of uniform lateral external pressure of suitable

intensity.

180

90kN 30

30

Ans: 0.125mm

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12. Define the terms: stress, strain, elastic limit, proportionality

limit, yield stress, ultimate stress, proof stress, true stress, factor

of safety, Young‟s modulus, modulus of rigidity, bulk modulus,

Poisson's ratio,

13. Draw a typical stress-strain diagram for mild steel rod under

tension and mark the salient points.

14. Diameter of a bar of length „L‟ varies from D1 at one end to D2

at the other end. Find the extension of the bar under the axial load

P

15. Derive the relationship between Young‟s modulus and modulus

of rigidity.

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17. A flat plate of thickness „t‟ tapers uniformly from a width b1at

one end to b2 at the other end, in a length of L units. Determine

the extension of the plate due to a pull P.

18. Find the extension of a conical rod due to its own weight when

suspended vertically with its base at the top.

19. Prove that a material subjected to pure shear in two perpendicular

planes has a diagonal tension and compression of same

magnitude at 45o to the planes of shear.

16. Derive the relationship between Young‟s modulus and Bulk

modulus.

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95

20 For a given materials E=1.1×105N/mm2& G=0.43×105N/mm2

.Find bulk modulus & lateral contraction of round bar 40mm

diameter & 2.5m length when stretched by 2.5mm.

ANS: K=83.33Gpa, Lateral contraction=0.011mm

21. The modulus of rigidity of a material is 0.8×105N/mm2 , when

6mm×6mm bar of this material subjected to an axial pull of

3600N.It was found that the lateral dimension of the bar is

changed to 5.9991mm×5.9991mm. Find µ & E. ANS: µ=0.31,

E= 210Gpa.


Simple stresses in beams B.TECH 3rd SEM Mechanical by Shashikant

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