Elasticity
Original Scanned PDF – View Notes
Elasticity and Plasticity
The source gives quartz, phosphor bronze, etc. as examples of bodies described as elastic in nature.
The source gives paraffin wax, wet clay, etc. as examples of bodies having plasticity.
Deforming Force
Restoring Force
Stress
It is a scalar quantity. Its SI unit is N m−2 and its dimensional formula is [M L−1 T−2].
Types of Stress
1. Normal Stress
The deforming force per unit area acting perpendicular to the area is called normal stress.
2. Bulk Stress / Volume Stress
When a body is immersed in liquid, liquid molecules strike the surface from all directions. The force applied by liquid molecules per unit area is called bulk stress.
3. Shear Stress / Tangential Stress
The stress in which force is applied tangentially on two opposite faces and changes the shape but not the volume is called shear stress.
Strain
Types of Strain
1. Longitudinal Strain
The ratio of change in length to the original length is called longitudinal strain.
2. Volumetric Strain
The ratio of change in volume to the original volume is called volumetric strain.
3. Lateral Strain
The ratio of change in diameter to the original diameter is called lateral strain.
4. Shear Strain
When a body is subjected to shear stress, the angle between the displaced position and original position is taken as shear strain.
From the figure:
tan θ = Δx/x
For small θ, tan θ ≈ θ.
θ = Δx/x
Elastic Limit and Hooke’s Law
Elastic Limit
Hooke’s Law
If F is deforming force and ℓ is elongation:
ℓ ∝ F
ℓ/A ∝ F/A
Since F/A = stress and ℓ/L = strain, for a given body L/A is constant.
Within elastic limit, stress is directly proportional to strain.
Experimental Verification of Hooke’s Law
A spring is suspended vertically. A scale pan and a pointer are attached to the lower end of the spring. The pointer is used to check the scale reading.
First, note the pointer reading when the pan is empty. Add a known weight to the pan and again note the reading. The difference between the two readings gives the elongation of the spring for that load.
Continue adding weights step by step and note the corresponding elongation. Plot load applied against elongation.
From the straight-line graph through the origin:
Types of Modulus of Elasticity
1. Young’s Modulus of Elasticity (Y)
Let L = original length of wire, ℓ = elongation, A = cross-sectional area and F = applied deforming force.
Y = (F/A)/(ℓ/L)
Y = FL/(Aℓ)
Its SI unit is N m−2 or pascal (Pa).
2. Bulk Modulus of Elasticity (B or K)
Let F = applied deforming force, A = cross-sectional area, V = original volume and ΔV = change in volume.
K = (F/A)/(−ΔV/V)
K = −FV/(AΔV)
3. Shear Modulus of Elasticity (η)
η = (F/A)/θ
η = F/(Aθ)
Energy Stored in a Stretched Wire
Consider a uniform metallic wire of original length L and cross-sectional area A. When a force F elongates it by ℓ, Young’s modulus is:
Y = (F/A)/(ℓ/L) = FL/(Aℓ)
Therefore, F = YAℓ/L … (i)
If the wire is stretched through a small length dℓ:
dW = F dℓ = (YAℓ/L)dℓ … (ii)
Total work done in stretching from 0 to ℓ:
W = ∫0ℓ (YAℓ/L)dℓ
W = (YA/L)[ℓ2/2]0ℓ
W = YAℓ2/(2L) … (iii)
Using equation (i):
W = ½Fℓ
This work done is stored in the wire in the form of potential energy.
Energy Density
P.E. per unit volume = [YAℓ2/(2L)] / (AL)
= ½Y(ℓ/L)2
Since stress/strain = Y:
P.E. per unit volume = ½ × Stress × Strain
Solved Numericals from the Scanned Notes
Numerical 1 – Copper and Steel Wires Connected End to End
Question: A copper wire and a steel wire of the same cross-sectional area and lengths 1 m and 2 m respectively are connected end to end. A force stretches their combined length by 1 cm. Find the elongation of each wire.
Lc = 1 m, Ls = 2 m
Yc = 1.2 × 1011 N m−2
Ys = 2 × 1011 N m−2
ℓs + ℓc = 0.01 m … (i)
Using Y = FL/(Aℓ) for both wires:
Ys/Yc = (Lsℓc)/(Lcℓs)
2 × 1011 / 1.2 × 1011 = 2ℓc/ℓs
ℓs = 1.2ℓc … (ii)
Using (i): 1.2ℓc + ℓc = 0.01
ℓc ≈ 4.5 × 10−3 m
ℓs ≈ 5.5 × 10−3 m
Numerical 2 – Rubber Cord of a Catapult
Question: The rubber cord of a catapult has cross-sectional area 1 mm2 and unstretched length 10 cm. It is stretched to 12 cm and released to project a missile of mass 5 g. Calculate the velocity of projection.
A = 1 mm2 = 10−6 m2
L = 0.1 m, ℓ = 0.02 m
m = 5 × 10−3 kg
Y = 5 × 108 N m−2
Y = FL/(Aℓ)
5 × 108 = F(0.1)/(10−6 × 0.02)
F = 102 N
½mv2 = Fℓ/2
v = 20 m s−1
Numerical 3 – Wire Carrying a 15 kg Mass
Question: A wire of length 2.5 m and cross-sectional area 1 × 10−6 m2 has a mass of 15 kg hanging from it. Find the extension and energy stored if Young’s modulus is 2 × 1011 N m−2.
L = 2.5 m, A = 1 × 10−6 m2, Y = 2 × 1011 N m−2
F = mg = 150 N
Y = FL/(Aℓ)
2 × 1011 = (150 × 2.5)/(1 × 10−6 × ℓ)
ℓ = 1.875 × 10−3 m
E = Fℓ/2
E ≈ 0.14 J
Numerical 4 – Steel Wire from Density Data
Question: A uniform steel wire of density 8000 kg m−3 weighs 20 g and is 2.5 m long. It lengthens by 1 mm when stretched by a force of 80 N. Calculate Young’s modulus and the energy stored.
ρ = 8000 kg m−3, m = 2 × 10−2 kg, L = 2.5 m
ℓ = 1 × 10−3 m, F = 80 N
ρ = m/V = m/(AL)
A = 1 × 10−6 m2
Y = FL/(Aℓ)
Y = 2 × 1011 N m−2
E = Fℓ/2
E = 40 × 10−3 J
Numerical 5 – Force Required to Double the Length
Question: What force is required to stretch a steel wire of cross-sectional area 1 cm2 to double its length?
A = 1 cm2 = 10−4 m2
For doubling the length, ℓ = L.
Y = 2 × 1011 N m−2
Y = (F/A)(L/ℓ)
2 × 1011 = F/10−4
F = 2 × 107 N
Numerical 6 – Work Done in Stretching a Steel Wire
Question: Calculate the work done in stretching a steel wire 100 cm in length and cross-sectional area 0.030 cm2 when a load of 100 N is slowly applied before the elastic limit is reached.
L = 1 m
A = 0.030 cm2 = 3 × 10−6 m2
F = 100 N, Y = 2 × 1011 N m−2
Y = FL/(Aℓ)
ℓ = 1.67 × 10−4 m
W = Fℓ/2
W = 8.33 × 10−3 J
Numerical 7 – Steel Wire of Density 7800 kg m−3
Question: A uniform steel wire of density 7800 kg m−3 weighs 16 g and is 250 cm long. It lengthens by 1.2 mm when stretched by 80 N. Calculate Young’s modulus and the energy stored.
ρ = 7800 kg m−3, m = 0.016 kg, L = 2.5 m
ℓ = 1.2 × 10−3 m, F = 80 N
ρ = m/(AL)
A = 8.2 × 10−7 m2
Y = FL/(Aℓ)
Y = 2 × 1011 N m−2
E = Fℓ/2
E = 4.8 × 10−2 J
Numerical 8 – Steel Cable and Elevator
Question: A steel cable of cross-sectional area 3 cm2 has an elastic limit written in the source as 2.4 × 108 Pa. Find the maximum upward acceleration that can be given to a 1200 kg elevator if the stress does not exceed the stated fraction of the elastic limit.
A = 3 × 10−4 m2
The handwritten working then takes the allowable stress as 3 × 107 Pa.
Using the source relation:
Stress = F/A = ma/A
3 × 107 = (1200a)/(3 × 10−4)
a = 20 m s−2
The calculation above is preserved as written in the supplied notes rather than silently replacing the source method.
Numerical 9 – Vertical Brass Rod
Question: A vertical brass rod of circular section is loaded by placing a 5 kg weight on top of it. If its length is 50 cm and radius of cross-section is 1 cm, find the contraction and energy stored.
m = 5 kg, L = 0.5 m, r = 0.01 m
A = πr2 = π × 10−4 m2
Y = 0.91 × 1011 N m−2, F = mg = 50 N
Y = FL/(Aℓ)
ℓ ≈ 8.74 × 10−7 m
E = Fℓ/2
E ≈ 2.18 × 10−5 J
Numerical 10 – Repeated Steel-Wire Work Problem
Page 14 repeats the steel-wire work calculation using L = 100 cm, A = 0.03 cm2, F = 100 N and Y = 2 × 1011 N m−2.
ℓ = 1.67 × 10−4 m
E ≈ 8.3 × 10−3 J
Numerical 11 – Punching a Hole in a Steel Plate
Question: The source asks for the force required to punch a hole of diameter 1 cm through a steel plate 4 mm thick, using a shearing strength of 3.5 × 107 N m−2.
Diameter d = 1 cm = 0.01 m
Thickness t = 4 mm = 0.004 m
Sheared area = circumference × thickness = πdt
F = Shearing strength × πdt
F = 3.5 × 107 × π × 0.01 × 0.004
F ≈ 4.398 × 103 N
Page 15 – Second Numerical Partly Unreadable
The second numerical on page 15 is too faint in the supplied scan to identify its complete question, given values and intermediate steps with confidence. The visible part shows an elasticity/energy calculation ending with a small energy value, but the full data are not readable enough to reproduce faithfully. It has therefore not been guessed or reconstructed.
Numerical 12 – Steel Wire, 24 g and 250 cm
Question: A steel wire of density 8000 kg m−3 weighs 24 g and is 250 cm long. It lengthens by 1.2 mm when stretched by a force of 80 N. Calculate Young’s modulus and energy stored in the wire.
ρ = 8000 kg m−3, m = 24 × 10−3 kg
L = 2.5 m, ℓ = 1.2 × 10−3 m, F = 80 N
ρ = m/(AL)
A = 1.2 × 10−6 m2
Y = FL/(Aℓ)
Y ≈ 1.4 × 1011 N m−2
E = Fℓ/2
E = 0.048 J
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