Yielding of Supports:

In our previous analysis of thermal stresses we have supposed the bars to be placed among rigid supports. As we have already realized that there are no rigid solids, there are also no rigid supports. The supports might be much stiffer than the bars, but they too undergo deformation, to a smaller extent, and in several cases these deformations might be too small to be considered.

Consider the bar illustrated in figure. Without modifying any other data, let us suppose the stiffness of the support, k_s to be 80 kN/mm (that means the support shall yield by 1 mm for a thrust of 80 kN applied on it).

If P is the force developed then Eq. might be rewritten as follows :

δ_t + δ_e  = d +(P/k_s)

 (Equilibrium needs that P is similar in the support and bar.) By substituting numerical data in Eq.

0.288 - (P × 750)/ 1200 × 200 = 0.12 + (P/ 80)

P (0.0125 + 0.003125) = 0.288 - 0.12

P = 0.168/0.015625

= 10.752 kN

 Note down that in the case of steel bolt & copper tube assembly, the steel bolt is serving as any yielding support to the copper tube.

The support has considerably yielded so that mostly stress in the bar is relieved.