Thermal Stress Analysis: Determinate vs. Indeterminate Problems

1. Statically Determinate Δ T > 0 Statically Indeterminate Δ T > 0 Lecture # 8 Thermal Stress Objective: To find stress and strain for statically determinate and statically indeterminate problems.

2. Thermal Stress where  = a property of the material, referred to as the linear coefficient of thermal expansion. The units measure strain per degree of temperature. They are 1/oC (Celsius) or 1/oK (Kelvin) in the SI system. ΔT = the algebraic change in temperature of the member L = the original length of the member T = the algebraic change in length of the member

3. Example: The A – 36 steel bar shown in figure is constrained to just fit between two fixed supports when T1 = 30oC. If the temperature is raised to T2 = 60oC, determine the average normal thermal stress developed in the bar.

4. Solution:

6. Example: The rigid bar as shown in figure is fixed to the top of the three posts made of A – 36 steel and 2014 – T6 aluminum. The posts each have a length of 250mm when no load is applied to the bar, and the temperature is T1 = 20oC. Determine the force supported by each post if the bar is subject to a uniform distributed load of 150 kN/m and the temperature is raised to T2 = 80oC.

7. Solution: Equilibrium. The free-body diagram of the bar as shown in the following figure. Moment equilibrium about the bar’s center requires the force in the steel posts to be equal. Summing forces on the free-body diagram, we have, +↑Fy = 0; 2Fst + Fal – 90(103) N = 0 ------(1)

8. Solution: Compatibility. Due to load, geometry, and material symmetry, the top of each post is displaced by an equal amount. Hence, (+↓) st = al ---------------------------- (2) (+↓) st = - (st)T + (st)F (+↓) al = - (al)T + (al)F

9. Solution:

10. Solution: Applying Eq. 2 gives - (st)T + (st)F =- (al)T + (al)F

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