PH0101 UNIT 1 LECTURE 1

PH0101 UNIT 1 LECTURE 1 1

PH0101 UNIT 1 LECTURE 1

• Elasticity and Plasticity• Stress and Strain• Hooke’s Law• Twisting couple on a cylinder• Worked and Exercise Problems


Elasticity and Plasticity

• The property of the body by virtue of which it tends to regain its original shape or size on the removal of deforming force is called elasticity.

• The property of the body by virtue of which it tends to retain the altered size and shape on removal of deforming forces is called plasticity.


Stress and Strain

• Stress is a quantity that characterizes the strength of the forces causing the deformation, on a “force per unit area” basis.

• The deforming force per unite area of the body is

called stress. The SI unit of stress is the Pascal (abbreviated Pa, and named for the 17th century French scientist and philosopher Blaise Pascal).


• One Pascal equals one Newton per square meter.1 Pascal = 1Pa = 1N/m2.

• Strain is a quantity which describes the resulting deformation.

• Strain is the fractional deformation produced in a body when it is subjected to a set of deforming forces. Strain being ratio has no units


There are following three types of stress and strain

1. Tensile stress and strain 2. Bulk stress and strain3. Shear stress and strain


Tensile Stress and Strain

Initial state

lo

F

F


• Tensile stress at the cross section is defined as the ratio of the force to the cross – sectional area A.

• Tensile stress =

• The tensile strain of the object is equal to the fractional change in length, which is the ratio of the elongation to the original length .

F

F

A

F

o


• Tensile strain =

• Tensile strain is stretch per unit length

oo

o


Bulk Stress and Strain Initial state

F

F

F

F

F

F


• Pressure plays the role of stress in a volume deformation.

• The force per unit area is called the pressure

A

FP


• Bulk stress = , an increase in pressure.• The fractional change in volume that is the

ratio of the volume change to the original volume Vo, is called as bulk strain.

• Bulk (volume) strain =

• Volume strain is the change in volume per unit volume.

P

V

0V

V


Shear Stress and Strainh Initial

state

x

11F

11F


• Shear strain is defined as the ratio of the displacement x to the transverse dimension h

• Shear strain = hx


Hooke’s Law

• Hooke’s law states that within the elastic limit, the stress developed is directly proportional to the strain. The constant of proportionality is the elastic modulus ((or modulus of elasticity).

StrainStress

= elastic modulus (Hooke’s law)


Strain

Plastic range

0Stre

ss

Elastic limit

Elastic range

Permanent set

The plot between stress and strain is called stress - strain diagram. It is clear from the graph that Hooke’s law holds good only for the straight line portion of the curve


Three kinds of elastic moduli

Tensile stress

Bulk stress

Shear stress

Elastic Modulus Definition Nature of strain

Young’s modulus (Y) Tensile strain Change of shape and size

Bulk modulus (B) Bulk strainChange of size but not shape

Shear modulus or Rigidity modulus (S) Shear strain

Change of shape but not size


Approximate elastic moduli Material

Young’s ModulusY(Pa)

Bulk ModulusB(Pa)

Shear ModulusS(Pa)

Aluminium 7.0x1010 7.5x1010 2.5x1010

Brass 9.0 x1010 6.0 x1010 3.5 x1010

Copper 11 x1010 14 x1010 4.4 x1010

Crown Glass 6.0 x1010 5.0 x1010 2.5 x1010

Iron 21 x1010 16 x1010 7.7 x1010

Lead 1.6 x1010 4.1 x1010 0.6 x1010

Nickel 21 x1010 17 x1010 7.8 x1010

Steel 20 x1010 16 x1010 7.5 x1010


Compressibility The reciprocal of bulk modulus is called

compressibility.The unit of compressibility is same as that of reciprocal pressure, Pa-1

Poisson’s ratio Within the elastic limits the ratio of the lateral

strain to the longitudinal strain is constant for the material of the body and is known as Poisson’s ratio and is denoted by .


Worked Example 1

• A steel rod 2.0m long has a cross sectional area of 0.30cm2. The rod is now hung by one end from a support structure and a 550kg milling machine is hung from the rod’s lower end. The Young’s modulus of steel is 20x1010Pa. Determine the stress, the strain and the elongation of the rod.


Pam

smkgA

FStress 8

25

2

108.1100.3

)/8.9()550(

410

8

100.91020108.1

PaPa

YStressStrain

o

Elongation = = (strain) x o = (9.0x10-4) (2.0m)

= 0.0018m = 1.8mm


Worked Example 2

• A box – shaped piece of gelatin dessert has a top area of 15cm2 and a height of 3cm. When a shearing force of 0.50N is applied to the upper surface, the upper surface displaces 4mm relative to the bottom surface.

What are the shearing stress, the shearing strain, and the shear modulus for the gelatin?


Pam

NfaceofareaforcegentialstressShear 333

101550.0tan

24

1333.034.0

cmcm

heightntdisplaceme

strainShear

stress 333PaShearModulus

strain 0.133 2.5kPa


Twisting Couple on a Cylinder(or Wire)

• The twisting of a structural member about its longitudinal axis by two equal and opposite torques is expressed through a certain angle.

• The stress seen in this situation is not tensile or compressive, it is said to be shearing or shear stress.

• The strain in this case is measured by an angle in unit of radians



• Let us consider a cylindrical rod of length l and radius r with its upper end fixed.

• Let a twisting couple-be applied to the lower end of the rod in a plane perpendicular to its length and let the rod twist through an angle θ (radians).

• While the rod is twisted restoring couple acts in the opposite direction and in the position of equilibrium, the twisting couple is equal and opposite to the restoring couple.


• To calculate this couple, let us consider the solid cylinder to be made up of a larger number of concentric thin walled cylinders.

• Let us consider one such hollow cylinder of radius x, and radial thickness dx.

• When the rod is twisted through an angle θ, the angle through which the rim of the cylinder is sheared is ф.


• From this it is clear that with x, ф varies. ф has the maximum value when x is the greatest. i.e., the strain is maximum on the outermost part of the cylinder and minimum on the innermost.

• In other words, the shearing stress is not uniform through out the material.

i.e. BB’=l ф Also BB’=x θ

:. ф = xl


• If N is the rigidity modulus,

F

shearofanglestressshearing

N

NxHence

F =N ф=


Now the face area of the hollow cylinder = 2πxdx

:. Total shearing force on this area = 2πxdx. Nx θ / l

= 2πN . θ. x2.dx l


Therefore, moment of this force about the axis 00’ = 2πN θ x2dx . x l

= 2πN θ x3dx l

Total twisting couple of the cylinder can be obtained by integrating this expression between limits x = 0 and x = r.


:. Total twisting couple on the cylinder =

dxxl

Nr

3

0

2 rr xNdxx

lN

0

43

0 422

lrNr

lN

242 44


If θ = l radian, we have Twisting couple per unit twist C= πNr4/2l

• This twisting couple per unit twist of the wire is called the torsional rigidity or modulus of torsion of the cylinder of wire.

• It is evident form this relation that the couple required is proportional to the fourth power of the radius.


Note: Hollow Cylinder For a hollow cylinder of the same length l and of

inner radius r1 and outer radius r2,

Twisting couple of the cylinder

dxxlNC

r

r

322

1

dxxlN

r

r

32

1

2


If θ = 1 radian, twisting couple per unit twist

C= (r24 – r1

4) lN

2

2

14

2 4 r

r

xlN

)(2

4

14

2rr

lN


Worked Example 3 A wire of length 1 meter and diameter 1 mm is fixed

at one end and a couple is applied at the other end so that the wire twists by π/2 radians. Calculate the moment of the couple required if rigidity modulus of the material = 2.8 ×1010 N/m2.

Rigidity modulus of the material N = 2.8x1010 N/m2

Angle twisted by wire θ = radians2


Required couple C

= 4.3 x10-3 N-m.

lrN

2

4

12)105.0()14.35.0()108.2(14.3 4310


Worked Example 4 • A wire of length l m and diameter 1mm is

clamped at one of its ends. Calculate the couple required to twist the other end by 90o. Given N = 2.8 × 1010 N/m2.

• The torque required to twist the free end of a clamped wire of length through θ radian will be

lrN

2

4


For θ = 90o = π/2 radian , C =

N= 2.8x10-10 N/m2, l =1m

r = 5mm = 0.0005 m

:.C

2 10 4 42.8 10 (5 10 )4

-34.32×10 Nm

4

42 Nr


Exercise Problem • A wire of length 1 metre and diameter 1mm is

clamped at one of its ends. Calculate the couple required to twist the other end by 90o. Given the modulus of rigidity is 298GPa.

radianso

290 Hint :

Twisting couple = 4N r

2l

-34.313×10 Nm


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PH0101 UNIT 1 LECTURE 1