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Mechanical Properties of Solids - Stress and Strain Relationships

Grade 11CBSEPhysics

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

🔑Concepts

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Elasticity: The property of a body by virtue of which it tends to regain its original size and shape when the applied force is removed.

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Stress: The internal restoring force acting per unit area of a cross-section of a deformed body. It is measured in N m−2N\,m^{-2} or PaPa. Types include Tensile, Compressive, and Shearing stress.

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Strain: The ratio of the change in configuration (length, volume, or shape) to the original configuration. It is a dimensionless quantity. Types include Longitudinal strain ΔLL\frac{\Delta L}{L}, Shearing strain θ\theta, and Volume strain ΔVV\frac{\Delta V}{V}.

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

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Young's Modulus (YY): Defined as the ratio of longitudinal stress to longitudinal strain within the elastic limit.

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Bulk Modulus (BB): Defined as the ratio of hydraulic stress to the corresponding volumetric strain. The reciprocal of Bulk Modulus is called Compressibility (k=1Bk = \frac{1}{B}).

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Shear Modulus or Modulus of Rigidity (GG or η\eta): The ratio of shearing stress to the corresponding shearing strain.

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Poisson's Ratio (ν\nu): The ratio of lateral strain to longitudinal strain. For most materials, it lies between 0.20.2 and 0.40.4.

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Stress-Strain Curve: A graph representing the relationship between stress and strain for a material. Key points include the Proportional Limit, Elastic Limit (Yield Point), Permanent Set, and Fracture Point.

📐Formulae

Stress=FA\text{Stress} = \frac{F}{A}

Strain(ϵ)=ΔLL\text{Strain} (\epsilon) = \frac{\Delta L}{L}

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

Bulk Modulus (B)=−PΔV/V\text{Bulk Modulus } (B) = - \frac{P}{\Delta V / V}

Shear Modulus (G)=F/Aθ\text{Shear Modulus } (G) = \frac{F / A}{\theta}

Poisson’s Ratio (ν)=Lateral StrainLongitudinal Strain=−Δd/dΔL/L\text{Poisson's Ratio } (\nu) = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = - \frac{\Delta d / d}{\Delta L / L}

Elastic Potential Energy (U)=12×Stress×Strain×Volume\text{Elastic Potential Energy } (U) = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}

Energy Density (u)=12×Stress×Strain=12Yϵ2\text{Energy Density } (u) = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} Y \epsilon^2

💡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 (a) stress, (b) elongation, and (c) strain on the rod. Given Young's modulus of structural steel is 2.0×1011 N m−22.0 \times 10^{11}\text{ N m}^{-2}.

Solution:

  1. Area of cross-section A=πr2=π(10−2 m)2=3.14×10−4 m2A = \pi r^2 = \pi (10^{-2}\text{ m})^2 = 3.14 \times 10^{-4}\text{ m}^2.
  2. Force F=100 kN=105 NF = 100\text{ kN} = 10^5\text{ N}.
  3. Stress =FA=1053.14×10−4=3.18×108 N m−2= \frac{F}{A} = \frac{10^5}{3.14 \times 10^{-4}} = 3.18 \times 10^8\text{ N m}^{-2}.
  4. Elongation ΔL=(F/A)LY=(3.18×108)(1.0)2.0×1011=1.59×10−3 m=1.59 mm\Delta L = \frac{(F/A)L}{Y} = \frac{(3.18 \times 10^8)(1.0)}{2.0 \times 10^{11}} = 1.59 \times 10^{-3}\text{ m} = 1.59\text{ mm}.
  5. Strain =ΔLL=1.59×10−31.0=1.59×10−3= \frac{\Delta L}{L} = \frac{1.59 \times 10^{-3}}{1.0} = 1.59 \times 10^{-3}.

Explanation:

We first calculate the cross-sectional area in SI units. Then we apply the definition of stress (F/AF/A). Using Hooke's law rearranged as ΔL=FLAY\Delta L = \frac{FL}{AY}, we find the change in length. Finally, strain is calculated as the ratio of change in length to original length.

Problem 2:

Compute the fractional compression of water in the abyss of the ocean, where the pressure is 80.0 MPa80.0\text{ MPa}. Given the bulk modulus of water is 2.2×109 N m−22.2 \times 10^9\text{ N m}^{-2}.

Solution:

  1. Pressure P=80.0 MPa=8.0×107 PaP = 80.0\text{ MPa} = 8.0 \times 10^7\text{ Pa}.
  2. Bulk Modulus B=2.2×109 PaB = 2.2 \times 10^9\text{ Pa}.
  3. Fractional compression is ΔVV\frac{\Delta V}{V}.
  4. From B=PΔV/VB = \frac{P}{\Delta V / V}, we get ΔVV=PB=8.0×1072.2×109≈3.64×10−2\frac{\Delta V}{V} = \frac{P}{B} = \frac{8.0 \times 10^7}{2.2 \times 10^9} \approx 3.64 \times 10^{-2}.

Explanation:

The fractional compression represents the volume strain. By using the Bulk Modulus formula, we relate the increase in external pressure to the relative decrease in volume.

Stress and Strain Relationships - Revision Notes & Key Formulas | CBSE Class 11 Physics