An elastic specimen gives (strain, tensile stress in MPa) readings (0, 0), (0.0005, 70), (0.001, 140), (0.0015, 210), (0.002, 280). Its unstretched volume is 0.000006 m³. Assume linear elastic behaviour throughout this loading interval. (a) Find Young modulus. (b) Use the area under the stress–strain graph to find energy stored per unit volume and the total stored energy at strain 0.0020. (c) Explain what unloading should show.

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Young modulus 140000000000 Pa; energy density 280000 J/m³; stored energy 1.68 J; unloading returns along the loading line. Young modulus is stress divided by strain in the linear region. The endpoint stress is 280 MPa = 280000000 Pa, so E = 280000000/0.0020 = 140000000000 Pa. Strain has no unit. The triangular area under this stress–strain line is ½σε = ½ × 280000000 × 0.0020 = 280000 Pa, equivalent to 280000 J/m³ of stored elastic energy. Multiply this energy density by the original volume: U = 280000 × 0.000006 = 1.68 J. The area is not itself the total energy unless the volume is included. Under the stated linear elastic model, unloading retraces the straight line and returns to zero strain with the stored energy recoverable. A residual strain or a loading–unloading loop would contradict that idealisation and indicate permanent deformation or energy dissipation.

Elastic energy from stress–strain measurements 5

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Tensile test measurements

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Tensile test measurements — supplied values
Strain	Stress / MPa
0	0
0.0005	70
0.001	140
0.0015	210
0.002	280