Thermal ShockDOE-HDBK-1017/2-93THERMAL STRESSIn the simple case where two ends of a material are strictly constrained, the thermal stress canbe calculated using Hooke's Law by equating values of from Equations (3-1), (3-2), and Dll(3-3).E= =(3-3)stressstrainF/ADllor=(3-4)DllF/AEaDT =(3-5)F/AEF/A =EaDTwhere:F/A =thermal stress (psi)E=modulus of elasticity (psi)a=linear thermal expansion coefficient (°F^{-1})DT=change in temperature (°F)Example: Given a carbon steel bar constrained at both ends, what is the thermal stress whenheated from 60°F to 540°F?Solution:a=5.8 x 10^{-6}/°F (from Table 1)E=3.0 x 10^{7} lb/in.^{2} (from Table 1, Module 2)DT=540°F - 60°F = 480°FStress = F/A = EaDT = (3.0 x 10^{7} lb/in.^{2}) x (5.8 x 10^{-6}/°F) x 480°FThermal stress = 8.4 x 10^{4} lb/in.^{2} (which is higher than the yield point)Rev. 0Page 3MS-03