MecMovies 4.0 : M5.11: Axial Member Subject to Temperature

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$20 - mm diameter steel rod A = 314.16 {mm}^{2} E = 200 GPa α = 12.0 \times 10^{- 6} /^{\circ} C$

Set the force-temperature-deformation relationship equal to zero and solve for $∆ T$ .

$δ = \frac{FL}{AE} + α ∆ TL$

$∴ ∆ T = - \frac{F}{αAE} = - \frac{24, 740 N}{(12.0 \times 10^{- 6} /^{\circ} C) (314.16 {mm}^{2}) (200, 000 N / {mm}^{2})} = - 32.8^{\circ} C$