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Mechanical Properties of Solids - Hooke's Law

Grade 11CBSEPhysics

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

🔑Concepts

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Hooke's Law states that within the elastic limit, the stress developed in a body is directly proportional to the corresponding strain produced in it.

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The law is mathematically expressed as Stress∝StrainStress \propto Strain, or Stress=E×StrainStress = E \times Strain, where EE is the Modulus of Elasticity.

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The Proportionality Limit is the point on the stress-strain curve up to which Hooke's law is strictly valid.

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Elastic Limit is the maximum stress that a material can endure without undergoing permanent deformation.

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Modulus of Elasticity (EE) is a characteristic property of the material and its value depends on the nature of the material and the type of deformation (longitudinal, shearing, or volumetric).

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Hooke's Law is only applicable in the linear region of the stress-strain curve.

📐Formulae

σ=E⋅ϵ\sigma = E \cdot \epsilon

Y=Longitudinal StressLongitudinal Strain=F/AΔL/L0Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F/A}{\Delta L/L_0}

G=Shearing StressShearing Strain=F/AθG = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{F/A}{\theta}

B=−Hydraulic StressVolumetric Strain=−ΔPΔV/V0B = - \frac{\text{Hydraulic Stress}}{\text{Volumetric Strain}} = -\frac{\Delta P}{\Delta V/V_0}

Modulus of Elasticity (E)=StressStrain\text{Modulus of Elasticity } (E) = \frac{\text{Stress}}{\text{Strain}}

💡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. 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. Stress =FA=100×103 N3.14×10−4 m2=3.18×108 N m−2= \frac{F}{A} = \frac{100 \times 10^3\text{ N}}{3.14 \times 10^{-4}\text{ m}^2} = 3.18 \times 10^8\text{ N m}^{-2}.
  3. Elongation ΔL=(F/A)LY=(3.18×108 N m−2)(1.0 m)2.0×1011 N m−2=1.59×10−3 m=1.59 mm\Delta L = \frac{(F/A)L}{Y} = \frac{(3.18 \times 10^8\text{ N m}^{-2})(1.0\text{ m})}{2.0 \times 10^{11}\text{ N m}^{-2}} = 1.59 \times 10^{-3}\text{ m} = 1.59\text{ mm}.

Explanation:

We first calculate the cross-sectional area in SI units. Then, applying the definition of stress (F/AF/A), we find the internal restoring force per unit area. Finally, we rearrange the Young's modulus formula Y=σΔL/LY = \frac{\sigma}{\Delta L/L} to solve for the extension ΔL\Delta L.

Problem 2:

A wire of length LL and radius rr is clamped at one end and a force FF is applied to the other end, producing an extension ll. If another wire of the same material but length 2L2L and radius 2r2r is stretched by the same force FF, what will be the extension?

Solution:

From Hooke's Law, Y=FLAl  ⟹  l=FLπr2YY = \frac{FL}{A l} \implies l = \frac{FL}{\pi r^2 Y}. For the second wire: l′=F(2L)π(2r)2Y=F(2L)π(4r2)Y=12(FLπr2Y)=l2l' = \frac{F(2L)}{\pi (2r)^2 Y} = \frac{F(2L)}{\pi (4r^2) Y} = \frac{1}{2} \left( \frac{FL}{\pi r^2 Y} \right) = \frac{l}{2}.

Explanation:

Since the material is the same, Young's modulus YY remains constant. By substituting the new dimensions into the extension formula, we see that doubling the length increases extension, but doubling the radius (which squares in the area term) decreases extension by a factor of four, resulting in a net extension of half the original value.

Hooke's Law - Revision Notes & Key Formulas | CBSE Class 11 Physics