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Properties of Bulk Matter - Elastic Behavior (Hooke's Law, Young's Modulus)

Grade 11ICSEPhysics

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

🔑Concepts

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

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Stress (sigma\\sigma) is defined as the internal restoring force per unit area of the body. Its SI unit is N/m2N/m^2 or PaPa. It is calculated as sigma=fracFA\\sigma = \\frac{F}{A}.

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Strain (epsilon\\epsilon) is the ratio of the change in configuration to the original configuration of the body. It is a dimensionless quantity. Longitudinal strain is given by epsilon=fracDeltaLL\\epsilon = \\frac{\\Delta L}{L}.

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Hooke's Law states that within the elastic limit, the stress developed in a body is directly proportional to the strain produced in it, i.e., textStressproptotextStrain\\text{Stress} \\propto \\text{Strain}.

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Modulus of Elasticity (EE) is the ratio of stress to strain. Young's Modulus (YY) specifically refers to the ratio of longitudinal stress to longitudinal strain.

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

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Yield Point is the point on the stress-strain curve where the material begins to deform plastically. Beyond this point, the material will not return to its original shape.

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Ductile materials have a large plastic deformation range before breaking (e.g., copper), while brittle materials break almost immediately after the elastic limit (e.g., glass).

📐Formulae

textStress(sigma)=fracFA\\text{Stress} (\\sigma) = \\frac{F}{A}

textLongitudinalStrain(epsilon)=fracDeltaLL\\text{Longitudinal Strain} (\\epsilon) = \\frac{\\Delta L}{L}

Y=fractextLongitudinalStresstextLongitudinalStrain=fracF/ADeltaL/L=fracFcdotLAcdotDeltaLY = \\frac{\\text{Longitudinal Stress}}{\\text{Longitudinal Strain}} = \\frac{F/A}{\\Delta L/L} = \\frac{F \\cdot L}{A \\cdot \\Delta L}

Y=fracMgLpir2DeltaLtext(forawireofradiusrtextandmassMtextattached)Y = \\frac{MgL}{\\pi r^2 \\Delta L} \\text{ (for a wire of radius } r \\text{ and mass } M \\text{ attached)}

textElasticPotentialEnergy(U)=frac12timestextStresstimestextStraintimestextVolume\\text{Elastic Potential Energy} (U) = \\frac{1}{2} \\times \\text{Stress} \\times \\text{Strain} \\times \\text{Volume}

U=frac12FDeltaLU = \\frac{1}{2} F \\Delta L

💡Examples

Problem 1:

A structural steel rod has a radius of 10textmm10 \\text{ mm} and a length of 1.0textm1.0 \\text{ m}. A 100textkN100 \\text{ kN} force stretches it along its length. Calculate the stress and the elongation. (Take Y=2.0times1011textPaY = 2.0 \\times 10^{11} \\text{ Pa} for steel).

Solution:

  1. Area of cross-section A=pir2=pitimes(10times10−3textm)2=3.14times10−4textm2A = \\pi r^2 = \\pi \\times (10 \\times 10^{-3} \\text{ m})^2 = 3.14 \\times 10^{-4} \\text{ m}^2.
  2. Stress sigma=fracFA=frac100times103textN3.14times10−4textm2approx3.18times108textPa\\sigma = \\frac{F}{A} = \\frac{100 \\times 10^3 \\text{ N}}{3.14 \\times 10^{-4} \\text{ m}^2} \\approx 3.18 \\times 10^8 \\text{ Pa}.
  3. Elongation DeltaL=fracFLAY=frac(3.18times108textPa)times1.0textm2.0times1011textPa=1.59times10−3textm=1.59textmm\\Delta L = \\frac{F L}{A Y} = \\frac{(3.18 \\times 10^8 \\text{ Pa}) \\times 1.0 \\text{ m}}{2.0 \\times 10^{11} \\text{ Pa}} = 1.59 \\times 10^{-3} \\text{ m} = 1.59 \\text{ mm}.

Explanation:

The stress is found using the force and cross-sectional area. The elongation is then derived from the Young's Modulus formula Y=fracsigmaepsilonY = \\frac{\\sigma}{\\epsilon}.

Problem 2:

Compare the Young's Modulus of two wires AA and BB if wire AA has twice the length and half the radius of wire BB, and both undergo the same extension under the same load.

Solution:

Let LA=2LBL_A = 2L_B and rA=frac12rBr_A = \\frac{1}{2}r_B. Extension DeltaL\\Delta L and Force FF are equal. Using Y=fracFLpir2DeltaLY = \\frac{F L}{\\pi r^2 \\Delta L}, we have: fracYAYB=fracLArA2timesfracrB2LB=frac2LB(rB/2)2timesfracrB2LB=frac2LBrB2/4timesfracrB2LB=2times4=8\\frac{Y_A}{Y_B} = \\frac{L_A}{r_A^2} \\times \\frac{r_B^2}{L_B} = \\frac{2L_B}{(r_B/2)^2} \\times \\frac{r_B^2}{L_B} = \\frac{2L_B}{r_B^2/4} \\times \\frac{r_B^2}{L_B} = 2 \\times 4 = 8. Therefore, YA=8YBY_A = 8 Y_B.

Explanation:

Since FF and DeltaL\\Delta L are constant, YY is proportional to L/r2L/r^2. Substituting the ratios of LL and rr allows us to find the ratio of the moduli.

Elastic Behavior (Hooke's Law, Young's Modulus) Revision - Class 11 Physics ICSE