PHYSICS As LEVEL(FORM FIVE) NOTES – STRENGTH OF MATERIALS

Strength of Materials

Young’s Modulus of Elasticity Y =

A tensile or longitudinal force is a force which stretches or increases the length of a material.

Fig 1: The graph of extension against weight or force.

Key of the Graphs

1. Proportionality limit
2. Elastic limit
3. Yield point
4. The material becomes thin and breaks (breaking point), ε = Permanent extension

Fig 2: The graph of stress against strain.

• S = Permanent strain.
• 0A = Hooke’s law obeyed.
• 0B = Elastic deformation region.
• 0C = Region of plastic deformation.

Features of the Graphs

• The proportionality limit is the maximum stress beyond which Hooke‘s law is not obeyed or the ratio of stress/strain is no longer constant.
• 0A: Hooke‘s law is obeyed i.e. F = kx.
• Stress α Strain i.e. stress/strain = constant.
• Elastic limit is the maximum stress before which the material returns to its original shape and size.
• The yield point is the point beyond which the material changes from elastic behavior to plastic behavior.
• Plastic behavior is a situation where a force may cause a large increase in length unexpectedly and it will not return to its original shape or size.
• Energy stored in a strained wire: Suppose a wire of length is stretched by a force F.
• Work done = Average force × distance.
• This is the elastic energy stored in the wire.
• Energy = Fe (elastic energy stored).
• But, Y =
• F =
• Energy = Fe = ½ × Stress × Strain.
• Energy per unit volume = ½ × Stress × Strain.

Fig 3: Graph of F vs extension and energy if elastic limit is not exceeded.

Energy stored = work done = F × extension.

Total work done = Area under the curve = ½ × 0A × AB = ½ × e × F.

Energy =

If Hooke’s law is obeyed, this agrees with the earlier derived equation.

Plastic behavior is exhibited when the material is permanently strained after the force is removed. The energy is converted to heat.

Example 1: Two vertical wires X and Y suspended at the same horizontal level are connected by a light rod XY at the lower ends as shown. The wires have the same length l and cross-sectional area A. A weight of 30 N is placed at O on the rod, where . Both wires are stretched and the rod XY then remains horizontal.

If wire X has Young’s modulus 1.0 × 10¹⁰ N/m², calculate Young’s modulus for wire Y assuming that the elastic limit is not exceeded for both wires (N&P).

Solution

Since XY is horizontal, the extensions of wires X and Y are equal.

Let F_X and F_Y be the forces in X and Y respectively.

Taking moments about point X:

30 × 1 = F_Y × 3 → F_Y = 10 N

Taking moments about point Y:

30 × 2 = F_X × 3 → F_X = 20 N

For wire X:

Young’s modulus Y_X = (F_X × l) / (A × e_X)

For wire Y:

Young’s modulus Y_Y = (F_Y × l) / (A × e_Y)

Given the extensions are equal and lengths and areas are the same, calculate Y_Y.

Example 2: A rubber cord of a catapult has a cross-sectional area of 2 mm² and an initial length of 0.20 m and is stretched to 0.24 m to fire a small stone of 10 g. Calculate the initial velocity of the object when it just leaves the catapult (Assume Young’s modulus for rubber is b × 10⁶ Pa and the elastic limit is not exceeded).

Solution

Kinetic energy of the fired stone = Energy stored in stretched rubber.

Energy stored = ½ × Force × extension = Fe

Velocity V = √(2 × Energy / mass)

Calculate force F using Young’s modulus and then find velocity V.

Applications of Elasticity

1. In the design of structures, the engineer has to ensure that the stresses applied do not exceed the elastic limit of the structure.
2. Steel has very high strength (high modulus of elasticity) so it is used in construction of girders, springs, etc. It can withstand very high stresses.
3. In construction, I and H cross-section beams are made using steel bars so that the beams can withstand large stresses.

Example 3:

A steel wire of radius 5 mm is fixed at the ceiling. Find the maximum weight which can be hung from it if the allowed strain is ε and Young’s modulus is 2.0 × 10¹¹ Pa.

Solution:

From Y = Stress / Strain, Stress = Y × ε.

Force F = Stress × Area.

Calculate F and hence the maximum weight.

Example 4:

The breaking stress of steel is 8.0 × 10⁸ N/m². Find the greatest length of the steel wire that can hang vertically without breaking if its density is 9.9 × 10³ kg/m³.

Stress = Breaking stress = 8 × 10⁸ N/m²

Force F = Stress × Area

Weight W = F

Calculate length L using W = ρ × g × A × L.

Energy stored in a strained wire is equal to the area under the graph of force F against extension X.

Types of Stress

Stress = Force / Area

SI unit of stress is N/m² or Pascal (Pa).

Dimensions of Stress

There are two main types of stress:

1. Normal stress: force applied perpendicular to the area.

• Normal tensile stress causes an increase in length.
• Normal compressive stress causes a decrease in length.

Types of Strain

Strain = Change in length / Original length (dimensionless)

There are three types of strain:

1. Longitudinal strain =
2. Bulk strain = Change in volume / Original volume
3. Shearing strain = Change in shape caused by tangential force

Definition:

Shearing strain = Angle of shear = angle through which a vertical line is turned by a tangential force.

Types of Moduli of Elasticity

There are three types of modulus of elasticity corresponding to the three types of strain:

• Young’s modulus (Y)
• Bulk modulus (K)
• Shear modulus or modulus of rigidity (η)

I. Young’s Modulus of Elasticity

Y = Stress / Strain

II. Bulk Modulus of Elasticity

K = – (Change in pressure) / (Volume strain)

Examples of moduli of elasticity:

• For steel, K = 160 × 10⁹ Pa
• For water, K = 2.2 × 10⁹ Pa
• For air at normal pressure, K = 1.0 × 10⁵ Pa

High value of K indicates difficulty in changing the volume.

Compressibility

Compressibility is a measure of how hard or easy it is to compress the material. It is the reciprocal of the modulus of elasticity.

Example 1:

A spherical ball is subjected to a normal uniform pressure of 5 × 10⁶ Pa. If the bulk modulus of the material of the ball is 2 × 10⁹ Pa, find the volume strain it suffers.

Solution:

Using K = -ΔP / (ΔV / V)

Calculate volume strain.

Example 2:

The density of ocean water at the surface is 1030 kg/m³. The bulk modulus of water is 2.0 × 10⁹ Pa. If the atmospheric pressure is 1.0 × 10⁵ Pa, what is the density of the ocean water at a depth where the pressure is 500 atmospheres? By what percentage is the water compressed?

Solution:

Calculate density and percentage compression using bulk modulus and pressure difference.

III. Shear Modulus or Modulus of Rigidity

It is defined as the ratio of the tangential stress applied to the shear strain.

η = Tangential stress / Shear strain

or η = τ / γ, where γ is in radians.

The SI unit of η is N/m² or Pa.

Young’s modulus for steel = 20 × 10¹⁰ Pa.

Modulus of rigidity for steel = 8 × 10¹⁰ Pa.

Generally, the modulus of rigidity for solids is much smaller than Young’s modulus of elasticity.

The bulk modulus of elasticity of a gas if the change in the gas is isothermal:

PV = constant

Differentiating with respect to V gives:

P × dV + V × dP = 0

But -V × dP/dV = K, the bulk modulus.

The isothermal bulk modulus of a gas is equal to the pressure of the gas.

The bulk modulus of elasticity of a gas if the change is adiabatic:

PV^γ = constant

Differentiating with respect to V:

γP V^γ-1 dV + V^γ dP = 0

Mathematical analysis shows that K = γP.

The Velocity of Sound

The velocity of sound depends on the density and the modulus of elasticity. By dimensional analysis:

V = √(E / ρ)

where E is the modulus of elasticity and ρ is the density.