The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5 × 10⁻⁴ is __________ kJ/m³. Assume that material is elastic upto the linear strain of 5 × 10⁻⁴.

Solution

Physics sir · 4 steps

The axes are swapped here: strain is on the y-axis, so Y is stress over strain, the inverse of the slope.

1. Read Young's modulus. Take any marked point, say stress 20 N m⁻² at strain 1 × 10⁻¹⁰. Then Y = stress/strain = 20/(1 × 10⁻¹⁰) = 2 × 10¹¹ N m⁻².
2. Energy density formula. For a linear elastic material, the energy stored per unit volume is the area under the stress-strain line. That area is a triangle.

   u = 1/2×stress×strain = 1/2Yε²

3. Put in numbers. With ε = 5 × 10⁻⁴, ε² = 25 × 10⁻⁸. So u = 1/2 × 2 × 10¹¹ × 25 × 10⁻⁸ = 2.5 × 10⁴ J m⁻³.
4. So. The energy density is 2.5 × 10⁴ J m⁻³ = 25 kJ m⁻³.

The common slip: 50. Dropping the 1/2 and using u = Yε² gives 50, not 25. The stored energy is a triangle's area, not a rectangle's.

The idea to keep. Energy density is the area under the stress-strain line, 1/2Yε². Always check which quantity is on which axis before you read Y from the slope.

Answer from the NTA final answer key. Solution by PrepWiser.