6. simple stresses and strains (1)

PART II

Mechanics of Deformable Bodies

COURSE CONTENT IN BRIEF

6. Simple stresses and strains

7. Statically indeterminate problems and thermal stresses

8. Stresses on inclined planes

9. Stresses due to fluid pressure in thin cylinders

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.

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.

6. Simple stresses and strains

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.

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

AR

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

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 = σ = R/A = P/A

P

P

P

R

BA

R

P

STRESS

σ = Normal StressSymbol:

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

• The resultant of the internal forces for an axially loaded member is normalto a section cut perpendicular to the member axis.

AP

AF

aveA

0

lim

• The force intensity on that section is defined as the normal stress.

STRESS

Illustrative Problems

A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in figure. Axial loads are applied at the positions indicated. Determine the stress in each section.

BronzeA= 120 mm2

4kN

SteelA= 160 mm2

AluminumA= 180 mm2

7kN2kN13kN

300mm 500mm400mm

Q 6.1

To calculate the stresses, first determine the forces in each section.

For equilibrium condition algebraic sum of forces on LHS of the section must be equal to that of RHS

4kN 7kN2kN13kN

To find the Force in bronze section,

consider a section bb1 as shown in the figure

Bronze

b

b1

Bronze4kN 7kN2kN13kN

13kN 2kN 7kNBronze

4kN 4kN

Force acting on Bronze section is 4kN, tensileStress in Bronze section = Force in Bronze section

Resisting cross sectional area of the Bronze section

=2

22 /33.33120

10004120

4 mmNmm

NmmkN

= 33.33MPa(Tensile stress)

(= )

b1

b

4kN 7kN2kN13kN

2kN 7kN

Aluminum

9kN

Force in Aluminum section

Force acting on Aluminum section is 9kN, (Compressive)

4kN 13kN

Aluminum(= )

4kN 7kN2kN13kN

7kN

steel

7kN

Force in steel section

Force acting on Steel section is 7kN, ( Compressive)

4kN 2kN13kNsteel

Stress in Steel section = Force in Steel section

Resisting cross sectional area of the Steel section

=2

22 /75.43160

10007160

7 mmNmm

NmmkN

= 43.75MPa

Stress in Aluminum section =

Force in Al section

Resisting cross sectional area of the Al section

=2

22 /50180

10009180

9 mmNmm

NmmkN

= 50MPa

(Compressive stress)

Compressive stress

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,

LL

Original lengthChange in length

=

Strain is a dimensionless quantity.

Linear strain,

Linear Strain

strain normal

stress

L

AP

L

AP

AP

22

LL

AP

22

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.


Typical tensile test curve for mild steel

Proportionality limit


Typical tensile test curve for mild steel showing upper yield point and lower yield point and also the elastic range and plastic range

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, that stress is proportional to strain.

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

APP

P Load at proportionality limitOriginal cross sectional area

=

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.

APE

E Load at proportional limitOriginal cross sectional area=


Yield point:

It is the point at which there is an appreciable elongation or yielding of the material without any corresponding increase of load.

APY

Y Load at yield pointOriginal cross sectional area=


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.

APU

U Maximum load taken by the materialOriginal cross sectional area=

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) by the original cross sectional area.

APB

B load at breaking (failure)Original cross sectional area=

True breaking stress = load at breaking (failure)Actual cross sectional area


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



Percentage elongation

A measure of ductility is obtained by measurements of the percentage elongation 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)

Cup and cone fracture for a Ductile Material


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

Stress-strain diagram for a typical brittle material

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 introduced a constant of proportionality that came to be known as Young’s modulus.

stress (σ) α strain (ε)stress (σ) strain (ε) = constant

stress (σ) strain (ε) = E

Modulus of Elasticity

Young’s Modulus= 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 (σ) strain (ε)=E =

LAPL

LL

AP

The value of the Young’s modulus is a definite property of a material

Deformations Under Axial Loading

AEP

EE

• From Hooke’s Law:

• From the definition of strain:

L

• Equating and solving for the deformation,

AEPL

• With variations in loading, cross-section or material properties,

i ii

iiEALP

A specimen of steel 20mm diameter with a gauge length of 200mm was tested to failure. It undergoes an extension of 0.20mm under a load of 60kN. Load at elastic limit is 120kN. The maximum load is 180kN. The breaking load is 160kN. Total extension is 50mm and the diameter at fracture is 16mm. Find:a) Stress at elastic limitb) Young’s modulusc) % elongationd) % reduction in areae) Ultimate strengthf) Nominal breaking stressg) True breaking stress

Q.6.2

Solution:a) Stress at elastic limit,

σE =Load at elastic limit

Original c/s area

MPammN

mmkN

APE 97.38197.381

16.314120

22

b) Young’s Modulus,

GPaMPa

mmN

mmmm

mmkN

LL

AP

E

98.190190980

190980101

98.190

20020.0

16.31460

23

2

(consider a load which is within the elastic limit)

c) % elongation,

% elongation = Final length at fracture – original length

Original length

%2510020050

d) % reduction in area =

%3610016.314

41616.314

2

Original c/s area -Final c/s area at fractureOriginal c/s area

e) Ultimate strength,

Ultimate strength = Maximum load Original c/s area

)(

/96.57216.314

180 22

MPa

mmNmm

kN

f) Nominal breakingStrength = MPakN 29.509

16.314160

Breaking load Original c/s area

g) True breakingStrength =

MPamm

kN 38.79506.201

1602

Breaking load

c/s area at fracture

A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in figure. Axial loads are applied at the positions indicated. Determine the change in each section and the change in total length. Given

Ebr = 100GPa, Eal = 70GPa, Est = 200GPa

BronzeA= 120 mm2

4kN

SteelA= 160 mm2

AluminumA= 180 mm2

7kN2kN13kN

300mm 500mm400mm

Q.6.3

From the Example 1, we know that,

Pbr = +4kN (Tension)

Pal = -9kN (Compression)

Pst = -7kN (Compression)

stress (σ) strain (ε)=E =

LAPL

AEPLL Change in length =

Change in length of bronze = )/(10100120

3004000232 mmNmm

mmNLbr

= 0.1mm

Deformation due to compressive force is shortening in length, and is considered as -ve

stalbr LLL Change in total length =

Change in length of steel section = )/(10200160

5007000232 mmNmm

mmNLst

= -0.109mm

Change in length of aluminum section = )/(1070180

4009000232 mmNmm

mmNLal

= -0.286mm

+0.1 – 0.286 - 0.109

= -0.295mm

An aluminum rod is fastened to a steel rod as shown. Axial loads are applied at the positions shown. The area of cross section of aluminum and steel rods are 600mm2 and 300mm2 respectively. Find maximum value of P that will satisfy the following conditions.a)σst ≤ 140 MPa b)σal ≤ 80 MPac)Total elongation ≤ 1mm,

2P SteelAluminum 2P4P

2.8m0.8m

Q.6.4

Take Eal = 70GPa, Est = 200GPa

To find P, based on the condition, σst ≤ 140 MPa

Stress in steel must be less than or equal to 140MPa. Hence, σst =

= 140MPast

st

AP

2P SteelAluminum 2P4P4P2P 2P

2P2P

2/1402 mmNAP

st

kNNAP st 21210002

140

Tensile

To find P, based on the condition, σal ≤ 80 MPa

Stress in aluminum must be less than or equal to 80MPa. Hence, σal =

= 80MPa al

al

AP

2P SteelAluminum 2P4P4P2P 2P

2P 2P

2/802 mmNAP

al

kNNAP al 24240002

80

Compressive

To find P, based on the condition, total elongation ≤ 1mm

Total elongation = elongation in aluminum + elongation in steel.

stal AEPL

AEPL

stst

st

alal

al

EAPL

EAPL 22

33 1020030028002

10706008002 PP

1mm

1mm

1mm

P = 18.1kN

Ans: P = 18.1kN (minimum of the three values)

Q.6.5

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

WW

A B

LDiameter

d1 Diameter d2

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

B

WW

Axd1 d2dx

Diameter at x, xL

ddd

121 c/s area at x, 21

21

44 kxdd

xkd 1

Change in length over a length dx is

Ekxd

WdxAEPL

dx 214

Change in length over a length L is

L

Ekxd

Wdx0 2

14

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

Put d1+kx = t,

Then k dx = dt


L

Ekxd

Wdx0 2

14

L

Et

kdtW

0 2

4

LLL

kxdEkW

tEkWt

EkW

0100

12

)(1414

14

EddWL

dEdWL

4

42121

Q.6.6

A two meter long steel bar is having uniform diameter of 40mm for a length of 1m, in the next 0.5m its diameter gradually reduces to 20mm and for remaining 0.5m length diameter remains 20mm uniform as shown in the figure. If a load of 150kN is applied at the ends, find the stresses in each section of the bar and total extension of the bar. Take E = 200GPa.

500mm

Ф = 40mm Ф = 20mm

150kN150kN

500mm1000mm

500mm

Ф = 40mm Ф = 20mm

150kN150kN

500mm1000mm

If we take a section any where along the length of the bar, it is subjected to a load of 150kN.

2

13

MPakN

MPakN

MPakNd

kN

MPakN

46.4774

20150

46.4774

20150

37.1194

40150

4

150

37.1194

40150

23

2min.2,

2.max,222

21

500mm

Ф = 40mm Ф = 20mm

150kN150kN

500mm1000mm

If we take a section any where along the length of the bar, it is subjected to a load of 150kN.

2

13

mmE

kNl

mmE

kNdEd

PLl

mmE

kNl

194.14

20500150

597.02040

50015044

597.04

401000150

23

212

21

mml 388.2 total,

Q.6.7

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.

WW

A B

L

d1d2

bb

W W

A B

x

d1d2

bb

dxConsider an element of length, δx at a distance x from A

depth at x, xL

ddd

121 c/s area at x, bkxd 1

xkd 1

Change in length over a length dx is

EbkxdWdx

AEPL

dx 1


L

EbkxdWdx

01

12 loglog ddkEb

Pee

1212

loglog302.2 ddddEbLP

Q.6.8

Derive an expression for the total extension produced by self weight of a uniform bar, when the bar is suspended vertically.

L

Diameter d

P1x

P1 = weight of the bar belowthe 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,

AEdxxA

AEdxP

AEPL

dx

)(1

The above expression can also be written as

Hence the total extension entire bar

EL

Ex

AEdxxA

LL

22)( 2

0

2

0

AEPL

AELAL

AA

EL

21

2)(

2

2

Where, P = (AL)×= total weight of the bar

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).

PFig. a Fig. b Fig. c

P

P

RR

The resisting forces acting parallel to the surface per unit area is called as shear stress.

Shear stress (τ)

=Shear resistanceArea resisting shear

τ

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

AP

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

τD

C

A

τB'

D

C'

Aτ

B C

Fig. e

The angle of deformation is measured in radians and hence is non-dimensional.

D

τB' C'

Aτ

Fig. e

BC

tanstrain shear ABBB

The angle of deformation is then termed as shear strain

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 (τ) Shear strain (φ) = constant =

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

GShear Modulus or

Modulus of Rigidity=

example: Shearing Stress

• Forces P and P‘ are applied transversely to the member AB.

AP

ave

• 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.

τ

State of simple shear

Force on the face AB = P = τ × AB × t

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

Where, t is the thickness of the element.

τ

A B

CD

Force on the face DC is also equal to P

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 = (τ × AB × t) × AD

P

A B

CD

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

force


Equn.(1) can be written as

If τ1 is the intensity of the shear stress on the faces AD and BC, then P ' can be written as, P ' = τ ' × AD × t

P ' × AB = P × AD= (τ × AB × t)× AD ----- (1)

P

P

A B

CD

P ' P '

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

τ ' = τ

τ

τ

A B

CD

τ ' τ '


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

CD

τ'= τ

τ

τ

τ'= τ

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

CD

τ

τ

τ

τa

AB

CD

τ

τ

τ

τaa

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


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

Resolving the forces in σn direction, i.e., in the X-direction shown

a

Fig.(c).

aa

A

CD

a2

For equilibrium

A σn

CD

τ

τ

a

X

n

n aa

Fx

45cos212

0


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.

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

ll

l

Fig.(a)

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.

dd

bb

lat

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 RATIOLateral strainLongitudinal strain

=l

ld

d

)(

ll

bb

)(OR

Poisson’s Ratio = µ

For most engineering metals the value of µ lies between 0.25 and 0.33

In general

z

y

xPP

Poisson’s Ratio

Lateral strainStrain in the direction of load applied

=x

x

y

y

ll

ll

OR

x

x

z

z

ll

ll

Lx

LyLz

Poisson’s Ratio = µ

In general

Strain in X-direction = εx

z

y

xPxPx

Lx

LyLz

x

x

ll

Strain in Y-direction = εy

Strain in Z-direction = εz

x

x

y

y

ll

ll

x

x

z

z

ll

ll

Load applied in Y-direction

Poisson’s Ratio


=y

y

x

x

ll

ll

OR

y

y

z

z

ll

ll

z

y

x

Py

Lx

LyLz

Py

Strain in X-direction = εxy

y

x

x

ll

ll

Load applied in Z-direction

Poisson’s Ratio


=z

z

x

x

ll

ll

OR

z

z

y

y

ll

ll

y

z

x

Pz

Lx

LyLz

Pz

Strain in X-direction = εxz

z

x

x

ll

ll

Load applied in X & Y direction


z

y

xPxPx

Lx

LyLz

Py

Py


EExy

Strain in Z-direction = εz EExy

EEyx

General case:



Strain in Z-direction = εz

z

y

xPxPx

Py

PyPz

Pz

EEEzyx

x

EEEzxy

y

EEExyz

z

σx

σzσy

σx

σz σy

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.

VdV

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

zyxzyxdV

Consider a cube of side 1unit, subjected to three mutually perpendicular direct stresses as shown in the figure.

Relationship between volumetric strain and linear strain

EEEzyx

EEEzxy

EEExyz

zyxVdV

Volumetric strain

zyxE

21

For element subjected to uniform hydrostatic pressure,

2-13KEor

modulusbulk 213

EK

zyx

321

21

EVdV

EVdV

zyx

VdV

K

Relationship between young’s modulus of elasticity (E) and modulus of rigidity (G) :-

A

D τ

B

τ

aa

45˚A1

φφ

B1

Consider a square element ABCD of side ‘a’ subjected to pure shear ‘τ’. DA'B'C is the deformed shape due to shear τ. Drop a perpendicular AH to diagonal A'C.Strain in the diagonal AC = τ /E – μ (- τ /E) [ σn= τ ]

= τ /E [ 1 + μ ] -----------(1)Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC

C

H

In Δle AA'HCos 45˚ = A'H/AA'A'H= AA' × 1/√2AC = √2 × AD ( AC = √ AD2 +AD2)Strain along the diagonal AC = AA'/ (√2 × √2 × AD)=φ/2 ----(2)Modulus of rigidity = G = τ /φ

φ = τ /GSubstituting in (2) Strain along the diagonal AC = τ /2G -----------(3)Equating (1) & (3)τ /2G = τ /E[1+μ]

E=2G(1+ μ)

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)

(1) A bar of certain material 50 mm square is subjected to an axialpull of 150KN. The extension over a length of 100mm is 0.05mmand decrease in each side is 0.0065mm. Calculate (i) E (ii) μ (iii) G(iv) KSolution:

(i) E = Stress/ Strain = (P/A)/ (dL/L) = (150×103 × 100)/(50 × 50 × 0.05)

E = 1.2 x 105N/mm2

(ii) µ = Lateral strain/ Longitudinal strain = (0.0065/50)/(0.05/100) = 0.26

(iii) E = 2G(1+ μ)

G= E/(2 × (1+ μ)) = (1.2 × 105)/ (2 × (1+ 0.26)) = 0.47 ×105N/mm2

(iv) E = 3K(1-2μ)

K= E/(1-2μ) = (1.2 × 105)/ (3 × (1- 2 × 0.26)) = 8.3 × 104N/mm2

(2) A tension test is subjected on a mild steel tube of externaldiameter 18mm and internal diameter 12mm acted upon byan axial load of 2KN produces an extension of 3.36 x 10-

3mm on a length of 50mm and a lateral contraction of 3.62x 10-4mm of outer diameter. Determine E, μ,G and K.

(i) E = Stress/Strain = (2 ×103 × 50)/ (π/4(182 – 122)× 3.36× 10-3)

= 2.11× 105N/mm2

ii) μ=lateral strain/longitudinal strain = [(3.62 ×10-4)/18]/[(3.36 × 10-3)/50]

= 0.3 iii) E = 2G (1 + μ)

G = E / 2(1+ μ) = (2.11 × 105)/(2 × 1.3) = 81.15 × 103N/mm2

iv) E = 3K(1 -2 μ)

K = E/ [3×(1-2 μ)] = (2.11×105)/{3×[1-(2 × 0.3)]} = 175.42 ×103N/mm2

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

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

Isotropic: If a material exhibits uniform properties throughout in all directions ,it is said to be isotropic.

Anisotropic: If a material does not exhibit uniform properties throughout in all directions ,it is said to be anisotropic or nonisotropic.

Q.6.9

A metallic bar 250mm×100mm×50mm is loaded as shown in the figure. Find the change in each dimension and total volume. Take E = 200GPa, Poisson's ratio, µ = 0.25

250

400kN50100

2000kN

4000kN

4000kN

400kN

2000kN

Stresses in different directions

100250

400kN502000kN

4000kN

100250

50 MPamm

Nx 80

501001000400

2

MPamm

Ny 160

10025010004000

2

MPamm

Nz 160

5025010002000

2

Stresses in different direction

MPa80

MPa160

MPa160 EEEzyx

x

410416016080

EEEx

mml

lll

x

x

x

x

1.0

104250

4

EEEzxy

y

3101.116080160

EEEy

mml

lll

y

y

y

y

005.0

101.150

3

MPa80

MPa160

MPa160

mml

lll

z

z

z

z

09.0

109250

4

EEExyz

z

410980160160

EEEz

MPa80

MPa160

MPa160

3

44

44

25050100250102102

102109114

mmdVVdV

VdV

zyxVdV

To find change in volume

410280E2-1

1601608021

21

EVdV

EVdV

zyx

Alternatively,

MPa80

MPa160

MPa160

Q.6.10

A metallic bar 250mm×100mm×50mm is loaded as shown in the Fig. shown below. Find the change in value that should be made in 4000kN load, in order that there should be no change in the volume of the bar. Take E = 200GPa, Poisson's ratio, µ = 0.25

250

400kN50

100

2000kN

4000kN

We know that

zyxEVdV

21

In order that change in volume to be zero

0

210

zyx

zyxE

kNP

PMPa

y

y

y

y

6000100250

240

240

016080

MPa80

MPa160

MPa160

The change in value should be an addition of 2000kN compressive force in Y-direction

Exercise Problems

Q1. 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 700mm600mm

steelaluminum

brass20kN 15kN 15kN 10kN

Ab=700mm2Aa=1000mm2

As=800mm2

Ans: σb=28.57MPa, σa=5MPa, σs=12.5MPa, δl = - 0.142mm

Q2. 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

200kN200kN

500mm500mm1000mm

Q3. 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

Q4. 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: 27.6, 19.5mm

Q5. 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

WA

CB300 450

Q6. 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 CB

2000 N

3m

D

Q7. 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 270mm320mm

300mm2 450mm2

250mm2

10kN40kN20kN

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

Q8. 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

Q9. 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

Q10. 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

Q11. 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

90kN3030

Ans: 0.125mm

Q12. 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,

Q13. Draw a typical stress-strain diagram for mild steel rod under tension and mark the salient points.

Q14 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

Q15. Derive the relationship between Young’s modulus and modulus of rigidity.

Q17 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.

Q18 Find the extension of a conical rod due to its own weight when suspended vertically with its base at the top.

Q19. 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.

Q16 Derive the relationship between Young’s modulus and Bulk modulus.

Q20. For a given material E=1.1×105N/mm2& G=0.43×105N/mm2 .Find bulk modulus & lateral contraction of round bar of 40mm diameter & 2.5m length when stretched by 2.5mm. ANS: K=83.33Gpa, Lateral contraction=0.011mm

Q21. 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.

6. simple stresses and strains (1)

Documents

direction of force stress

resisting force internal

internal resisting force

section ab normal

intensity of stress

tensile force

force intensity

compressivetensile stress