An elastic specimen gives (strain, tensile stress in MPa) readings (0, 0), (0.0005, 95), (0.001, 190), (0.0015, 285), (0.002, 380). Its unstretched volume is 0.000011 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 190000000000 Pa; energy density 380000 J/m³; stored energy 4.18 J; unloading returns along the loading line. Young modulus is stress divided by strain in the linear region. The endpoint stress is 380 MPa = 380000000 Pa, so E = 380000000/0.0020 = 190000000000 Pa. Strain has no unit. The triangular area under this stress–strain line is ½σε = ½ × 380000000 × 0.0020 = 380000 Pa, equivalent to 380000 J/m³ of stored elastic energy. Multiply this energy density by the original volume: U = 380000 × 0.000011 = 4.18 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 10

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Tensile test measurements

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Tensile test measurements — supplied values
Strain	Stress / MPa
0	0
0.0005	95
0.001	190
0.0015	285
0.002	380