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Mechanical Properties of Solids

Elastic Moduli

In the stress-strain curve given below, the region within the elastic limit (region OA) is of importance to structural and manufacturing sectors since it describes the maximum stress a particular material can take before being permanently deformed. The modulus of elasticity is simply the ratio between stress and strain. Elastic Moduli can be of three types, Young’s modulus, Shear modulus, and Bulk modulus. In this article, we will understand elastic moduli in detail.

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Elastic Moduli – Young’s Modulus

elastic moduli

Many experiments show that for a given material, the magnitude of strain produces is the same regardless of the stress being tensile or compressive. Young’s modulus (Y) is the ratio of the tensile/compressive stress (σ) to the longitudinal strain (ε).

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Y = \( \frac {σ}{ε} \) … (1)

We already know that, the magnitude of stress = \( \frac {F}{A} \) and longitudinal strain = \( \frac {ΔL}{L} \). Substituting these values, we get

Y = \( \frac {\frac {F}{A}}{\frac {ΔL}{L}} \)

∴ Y = \( \frac {(F × L)}{(A × ΔL)} \)  … (2)

Now, Strain is a dimensionless quantity. Hence, the unit of Young’s modulus is N/m2 or Pascal (Pa), the same as that of stress. Let’s look at Young’s moduli and yield strengths of some materials now:

Materials Young’s Modulus
Y (109 N/m2)
Elastic Limit
(107 N/m2)
Tensile Strength
(107 N/m2)
Aluminum 70 18 20
Copper 120 20 40
Wrought Iron 190 17 33
Steel 200 30 50
Bone
Tensile 16 12
Compressive 9 12

From the table, you can observe that Young’s moduli for metals are large. This means that metals require a large force to produce a small change in length. Hence, the force required to increase the length of a thin wire of steel is much larger than that required for aluminum or copper. Therefore, steel is more elastic than the other metals in the table.

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Elastic Moduli

Determination of Young’s Modulus of the Material of a Wire

The figure below shows an experiment to determine Young’s modulus of a material of wire under tension.

elastic moduli

As can be seen in the diagram above, the setup consists of two long and straight wires having the same length and equal radius. These wires are suspended side-by-side from a fixed rigid support. The reference wire (wire A) has a millimeter main scale (M) and a pan to place weight.

The experimental wire (wire B) also has a pan in which we can place weights. Further, a vernier scale is attached to a pointer at the bottom of wire B and the scale M is fixed to reference wire A. Now, we place a small weight in both the pans to keep the wires straight and note the vernier scale reading.

Next, the wire B is slowly loaded with more weights, bringing it under tensile stress and the vernier reading is noted. The difference between the two readings gives the elongation produced in the wire. The reference wire A is used to compensate for any change in length due to a change in the temperature of the room.

Let r and L be the initial and final length of the wire B, respectively. Therefore, the area of the cross-section of the wire B is = πr2. Now, let M be the mass that produces an elongation of ΔL in wire B. Therefore, the applied force is = Mg, where ‘g’ is the acceleration due to gravity. Hence, using equations (1) and (2), Young’s modulus of the material of wire B is:

Y = \( \frac {σ}{ε} \) = \( \frac {Mg}{πr^2} \).\( \frac {L}{ΔL} \)

⇒ Y = \( \frac {(Mg × L)}{(πr^2 × ΔL)} \) … (3)

Elastic Moduli – Shear Modulus

Shear Modulus (G) is the ratio of shearing stress to the corresponding shearing strain. Another name for shear stress is the Modulus of Rigidity.

∴ G = \( \frac {shearing stress (σ_s)}{shearing strain} \)

⇒ G = \( \frac {\frac {F}{A}}{\frac {Δx}{L}} \)

= \( \frac {F × L}{A × Δx} \) … (4)

We also know that, Shearing strain = θ

∴ G = \( \frac {\frac {F}{A}}{θ} \)

= \( \frac {F}{A × θ} \) … (5)

Further, the shearing stress σs can also be expressed as

σs = G × θ … (6)

Also, the SI unit of shear modulus is N/m2 or Pa. The shear moduli of a few common materials are given in the table below.

Material Shear Modulus (G)
109 N/m2
Aluminum 25
Brass 36
Copper 42
Glass 23
Iron 70
Lead 5.6
Nickel 77
Steel 84
Tungsten 150
Wood 10

From the table, you can observe that the shear modulus is less than Young’s modulus for the same materials. Usually, G ≈ \( \frac {Y}{3} \).

Elastic Moduli – Bulk Modulus

We have already studied that when we submerge a body in a fluid, it undergoes a hydraulic stress which decreases the volume of the body, leading to a volume strain. Bulk modulus (B) is the ratio of hydraulic stress to the corresponding hydraulic strain.

B = -\( \frac {p}{(\frac {ΔV}{V})} \) … (7)

The negative sign means that as the pressure increases, the volume decreases. Hence, for any system in equilibrium, B is always positive. The SI unit of the bulk modulus is N/m2 or Pa. The bulk moduli of a few common materials are given in the table below.

Material Bulk Modulus (B)
109 N/m2
Aluminum 72
Brass 61
Copper 140
Glass 37
Iron 100
Nickel 260
Steel 160
Liquids
Water 2.2
Ethanol 0.9
Carbon disulfide 1.56
Glycerine 4.76
Mercury 25
Gases
Air (at STP) 1.0 x 10-4

Compressibility (k) is the reciprocal of the bulk modulus. It is the fractional change in volume per unit increase in pressure.

∴ k = \( \frac {1}{B} \) = – \( \frac {1}{Δp} \) × \( \frac {ΔV}{V} \) … (8)

From the table, you can observe that the bulk modulus for solids is much larger than that for liquids and gases. Hence, solids are the least compressible while gases are the most compressible. This is because, in solids, there is a tight coupling between the neighboring atoms.

Solved Examples for You on Elastic Moduli

Q1. A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate:

  1. stress
  2. elongation
  3. the strain on the rod.

Young’s modulus, of structural steel, is 2.0 × 1011 N/m2.

Answer: To solve the problem, let’s assume that the rod is clamped at one end and a force F is applied at the other. This force is parallel to the length of the rod. Therefore, the stress on the rod is:

Stress = \( \frac {F}{A} \) = \( \frac {F}{πr^2} \)

= \( \frac {100 × 10^3 N}{3.14 × (10^{-2} m)} \)

= 3.18 × 108 N/m2

Next, let’s calculate the elongation ΔL.

ΔL = \( \frac {(\frac {F}{A})L}{Y} \)

= \( \frac {(3.18 × 10^8 Nm^{-2})(1 m)}{2 × 10^{11} Nm^{-2}} \) = 1.59 × 10-3 m

∴ ΔL = 1.59 mm

Hence, Strain = \( \frac {ΔL}{L} \)

= \( \frac {1.59 × 10^{-3} m}{1 m} \) = 1.59 × 10-3

This is equivalent to 0.16%.

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