1. PH0101 UNIT 1 LECTURE 1 Elasticity and Plasticity Stress and Strain
Hooke’s Law
Twisting couple on a cylinder
Worked and Exercise Problems
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2. 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.
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3. 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).
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4. 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
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5. There are following three types of stress and strain
Tensile stress and strain
Bulk stress and strain
Shear stress and strain
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6. Tensile Stress and Strain
Initial
state lo
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7. 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 .
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8. Tensile strain is stretch per unit length
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9. Bulk Stress and Strain
Initial
state
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10. Pressure plays the role of stress in a volume deformation.
The force per unit area is called the pressure
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11. 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.
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12. Shear Stress and Strain
Initial
state
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13. Shear strain is defined as the ratio of the displacement x to the transverse dimension h
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14. 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).
= elastic modulus (Hooke’s law)
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15. Strain
Plastic range
Stress
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
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16. Three kinds of elastic moduli
Elastic Modulus
Definition
Nature of strain
Young’s modulus (Y)
Tensile strain
Change of shape and size
Bulk modulus (B)
Bulk strain
Change of size but not shape
Shear modulus or Rigidity modulus (S)
Shear strain
Change of shape but not size
Bulk stress
Tensile stress
Shear stress
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17. Approximate elastic moduli
Material
Young’s Modulus
Y(Pa)
Bulk Modulus
B(Pa)
Shear Modulus
S(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
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
17 x1010
7.8 x1010
Steel
20 x1010
7.5 x1010
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18. Compressibility Poisson’s ratio
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 .
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19. 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.
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20. Elongation = = (strain) x o = (9.0x10-4) (2.0m) = 0.0018m = 1.8mm
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21. 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?
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22. PH0101 UNIT 1 LECTURE 1

23. 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
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24. PH0101 UNIT 1 LECTURE 1

25. 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.
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26. 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.
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27. i.e. BB’=l ф Also BB’=x θ
:. 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.
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28. If N is the rigidity modulus,
Hence
F =N ф=
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29. Now the face area of the hollow cylinder = 2πxdx
:. Total shearing force on this area = 2πxdx. Nx θ / l
= 2πN . θ. x2.dx l
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30. Therefore, moment of this force about the axis 00’ = 2πN θ x2dx . x l
Total twisting couple of the cylinder can be obtained by integrating this expression between limits x = 0 and x = r.
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31. :. Total twisting couple on the cylinder =
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32. Twisting couple per unit twist C= πNr4/2l
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.
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33. Twisting couple of the cylinder
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
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34. If θ = 1 radian, twisting couple per unit twist C= (r24 – r14)
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35. Worked Example 3 Angle twisted by wire θ = radians
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 = radians
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36. Required couple C
= 4.3 x10-3 N-m.
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37. 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
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38. For θ = 90o = π/2 radian , C = N= 2.8x10-10 N/m2, l =1m
r = 5mm = m
:.C
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39. 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.
Hint :
Twisting couple =
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40. THANK YOU
THANK YOU
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