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Mechanical Properties of Solids - Stress-Strain Curve

Grade 11CBSEPhysics

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

🔑Concepts

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The Stress-Strain curve is divided into several regions. The first is the Proportional Limit (OA), where stress is directly proportional to strain, following Hooke's Law. Point B is the Elastic Limit or Yield Point, beyond which the material will not return to its original shape. Between B and D, the material exhibits plastic behavior. D is the Ultimate Tensile Strength point, and E is the Fracture point.

Stress-strain curve showing proportional limit A, yield point B, ultimate strength D, and fracture point E.
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Ductile materials have a large plastic deformation region between the yield point and the fracture point (distance between B and E is large). Brittle materials, however, fracture almost immediately after the elastic limit (distance between B and E is very small).

Comparison of ductile and brittle stress-strain curves.
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Elastomers are substances that can be stretched to cause large strains but do not obey Hooke's law. Even though they return to their original shape, the stress-strain curve is non-linear and does not have a well-defined plastic region, such as in the case of aortic tissue.

Stress-strain curve for an elastomer showing non-linear elastic behavior.
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The slope of the linear portion (OA) of the stress-strain curve represents the Young's Modulus (YY) of the material. A steeper slope indicates a higher Young's Modulus, meaning the material is more rigid and harder to deform.

Linear part of stress-strain curve showing slope as Young's Modulus.

📐Formulae

Stress (σ)=FA\text{Stress } (\sigma) = \frac{F}{A}

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

Hooke’s Law: σ∝ϵ⇒σ=Yϵ\text{Hooke's Law: } \sigma \propto \epsilon \Rightarrow \sigma = Y \epsilon

Y=StressStrain=F/AΔL/L=F⋅LA⋅ΔLY = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta 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. Given Young's modulus for steel is 2.0×1011 Pa2.0 \times 10^{11} \text{ Pa}, calculate the stress and the elongation.

Solution:

First, calculate the cross-sectional area: A=πr2=3.14×(10−2 m)2=3.14×10−4 m2A = \pi r^2 = 3.14 \times (10^{-2} \text{ m})^2 = 3.14 \times 10^{-4} \text{ m}^2 Stress σ\sigma calculation: σ=FA=100×1033.14×10−4≈3.18×108 Pa\sigma = \frac{F}{A} = \frac{100 \times 10^3}{3.14 \times 10^{-4}} \approx 3.18 \times 10^8 \text{ Pa} Elongation ΔL\Delta L calculation: ΔL=σ⋅LY=(3.18×108)×1.02.0×1011=1.59×10−3 m=1.59 mm\Delta L = \frac{\sigma \cdot L}{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:

Stress is found by dividing the applied force by the area of the rod. Then, using Hooke's Law within the proportional limit, the elongation is derived from the definition of Young's Modulus.

Problem 2:

During a tensile test, a specimen reaches a maximum load of 50000 N50000 \text{ N} before necking. If the original diameter was 12 mm12 \text{ mm}, find the Ultimate Tensile Strength.

Solution:

Radius r=6 mm=6×10−3 mr = 6 \text{ mm} = 6 \times 10^{-3} \text{ m}. Area A=π(6×10−3)2=1.13×10−4 m2A = \pi (6 \times 10^{-3})^2 = 1.13 \times 10^{-4} \text{ m}^2. Ultimate Tensile Strength=FmaxA=500001.13×10−4≈4.42×108 Pa\text{Ultimate Tensile Strength} = \frac{F_{max}}{A} = \frac{50000}{1.13 \times 10^{-4}} \approx 4.42 \times 10^8 \text{ Pa}

Explanation:

The Ultimate Tensile Strength corresponds to the stress value at the highest point (D) of the stress-strain curve.

Problem 3:

A wire of length L=2 mL = 2 \text{ m} and cross-sectional area A=10−6 m2A = 10^{-6} \text{ m}^2 is stretched by a load. If the stress-strain graph for the wire is a straight line passing through the origin and the point where stress is 108 N/m210^8 \text{ N/m}^2 and strain is 5×10−45 \times 10^{-4}, calculate the Young's Modulus of the material and the force applied.

Linear stress-strain graph for the calculation of Young's Modulus.

Solution:

  1. Young's Modulus is the slope of the stress-strain graph: Y=StressStrainY = \frac{\text{Stress}}{\text{Strain}} Y=1085×10−4=2×1011 N/m2Y = \frac{10^8}{5 \times 10^{-4}} = 2 \times 10^{11} \text{ N/m}^2
  2. Force can be found from stress: Stress=FA\text{Stress} = \frac{F}{A} F=Stress×A=108×10−6=100 NF = \text{Stress} \times A = 10^8 \times 10^{-6} = 100 \text{ N} The Young's Modulus is 2×1011 Pa2 \times 10^{11} \text{ Pa} and the force is 100 N100 \text{ N}.

Explanation:

In the elastic region, the ratio of stress to strain is constant and equal to Young's Modulus. Force is the product of stress and the cross-sectional area.

Problem 4:

Consider two wires W1W_1 and W2W_2 made of different materials. In a test, W1W_1 fails at a strain of 0.020.02 and W2W_2 fails at a strain of 0.250.25. If both have the same elastic limit, which wire is more likely to be used for making springs and which for making sheets?

Comparison of two wires W1 (brittle) and W2 (ductile) showing different failure strains.

Solution:

W1W_1 has a very small plastic region (fails at 2%2\% strain) and is therefore brittle. W2W_2 has a large plastic region (fails at 25%25\% strain) and is ductile.

  • W1W_1 (Brittle/High Elasticity): Better suited for applications where deformation must be recovered, like springs (assuming high yield strength).
  • W2W_2 (Ductile): Better suited for making sheets (malleability) or wires (ductility) because it can undergo permanent deformation without breaking.

Explanation:

Ductility is the ability of a material to undergo significant plastic deformation before rupture. Materials with high failure strain are ductile, while those with low failure strain are brittle.