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Mechanical Properties of Solids - Young's Modulus, Bulk Modulus, and Shear Modulus

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Stress: It is the restoring force per unit area developed inside a body when it is deformed. It is expressed as σ=FA\sigma = \frac{F}{A} and measured in N/m2N/m^2 or PaPa.

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Strain: The ratio of the change in dimension to the original dimension of the body. It is a dimensionless quantity.

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Hooke's Law: Within the elastic limit, the stress is directly proportional to the strain produced. The constant of proportionality is called the Modulus of Elasticity.

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Young's Modulus (YY): It is the ratio of longitudinal stress to longitudinal strain. It describes the resistance of a material to change in length. Y=Longitudinal StressLongitudinal StrainY = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}}.

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Bulk Modulus (BB): It is the ratio of hydraulic stress (pressure) to the volumetric strain. It describes the resistance of a material to change in volume. B=−ΔPΔV/VB = -\frac{\Delta P}{\Delta V / V}. The negative sign indicates that volume decreases with increasing pressure.

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Shear Modulus (GG or η\eta): Also known as the Modulus of Rigidity, it is the ratio of shearing stress to shearing strain. It describes the resistance to change in the shape of a body. G=Shearing StressShearing StrainG = \frac{\text{Shearing Stress}}{\text{Shearing Strain}}.

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Compressibility: It is the reciprocal of the Bulk Modulus (k=1Bk = \frac{1}{B}). Gases have higher compressibility compared to solids and liquids.

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Poisson's Ratio (σ\sigma): The ratio of lateral strain to longitudinal strain. For most materials, its value lies between 00 and 0.50.5.

📐Formulae

Young’s Modulus (Y)=F⋅LA⋅ΔL\text{Young's Modulus } (Y) = \frac{F \cdot L}{A \cdot \Delta L}

Bulk Modulus (B)=−P⋅VΔV\text{Bulk Modulus } (B) = -\frac{P \cdot V}{\Delta V}

Shear Modulus (G)=FA⋅θ where θ is the shear strain\text{Shear Modulus } (G) = \frac{F}{A \cdot \theta} \text{ where } \theta \text{ is the shear strain}

Strain Energy Density (u)=12×Stress×Strain\text{Strain Energy Density } (u) = \frac{1}{2} \times \text{Stress} \times \text{Strain}

Poisson’s Ratio (σ)=−ΔR/RΔL/L\text{Poisson's Ratio } (\sigma) = -\frac{\Delta R / R}{\Delta L / L}

💡Examples

Problem 1:

A structural steel rod has a radius of 10 mm10 \text{ mm} and a length of 1.0 m1.0 \text{ m}. A 100 kN100 \text{ kN} force stretches it along its length. Calculate the stress and the elongation. (Take Young's Modulus for steel Y=2.0×1011 PaY = 2.0 \times 10^{11} \text{ Pa})

Solution:

Given: r=10 mm=10−2 mr = 10 \text{ mm} = 10^{-2} \text{ m}, L=1.0 mL = 1.0 \text{ m}, F=100 kN=105 NF = 100 \text{ kN} = 10^5 \text{ N}. Area A=πr2=π(10−2)2=3.14×10−4 m2A = \pi r^2 = \pi (10^{-2})^2 = 3.14 \times 10^{-4} \text{ m}^2. Stress=FA=1053.14×10−4=3.18×108 Pa\text{Stress} = \frac{F}{A} = \frac{10^5}{3.14 \times 10^{-4}} = 3.18 \times 10^8 \text{ Pa}. Elongation ΔL=F⋅LA⋅Y=(3.18×108)×1.02.0×1011=1.59×10−3 m=1.59 mm\Delta L = \frac{F \cdot L}{A \cdot Y} = \frac{(3.18 \times 10^8) \times 1.0}{2.0 \times 10^{11}} = 1.59 \times 10^{-3} \text{ m} = 1.59 \text{ mm}.

Explanation:

We first calculate the cross-sectional area of the rod. Stress is found by dividing the force by this area. Finally, the elongation is derived using the Young's Modulus formula rearranged as ΔL=Stress⋅LY\Delta L = \frac{\text{Stress} \cdot L}{Y}.

Problem 2:

The average depth of the Indian Ocean is about 3000 m3000 \text{ m}. Calculate the fractional compression, ΔVV\frac{\Delta V}{V}, of water at the bottom of the ocean, given that the Bulk Modulus of water is 2.2×109 Pa2.2 \times 10^9 \text{ Pa} and density ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3.

Solution:

Pressure at depth hh is P=ρgh=1000×9.8×3000=2.94×107 PaP = \rho g h = 1000 \times 9.8 \times 3000 = 2.94 \times 10^7 \text{ Pa}. Using Bulk Modulus formula: B=PΔV/V  ⟹  ΔVV=PBB = \frac{P}{\Delta V / V} \implies \frac{\Delta V}{V} = \frac{P}{B}. ΔVV=2.94×1072.2×109=1.336×10−2≈1.34%\frac{\Delta V}{V} = \frac{2.94 \times 10^7}{2.2 \times 10^9} = 1.336 \times 10^{-2} \approx 1.34\%.

Explanation:

First, we calculate the hydrostatic pressure at the given depth. Then, using the definition of Bulk Modulus, we find the ratio of change in volume to original volume, which represents the fractional compression.

Young's Modulus, Bulk Modulus, and Shear Modulus Revision - Class 11 Physics CBSE