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 in which there are no rigid solids, there are also no rigid supports. The supports might be much stiffer than the bars, other than they too undergo deformation, to a smaller extent, and in several cases these deformations might be too little to be considered.

Consider the bar display in Figure. Without changing any other data, let us suppose the stiffness of the support, k_s to be 80 kN/mm (i.e. the support will yield through 1 mm for a thrust of 80 kN applied on it).

If P is the force established then Eq. (17) might be rewritten as follows:

δ_t  + δ_e  = d + P/k_s

(Equilibrium requires that P is similar in the support and bar.) Substituting numerical data in Eq. (21).

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

Remember that in the case of steel bolt and copper tube assembly (Figure), the steel bolt is serving as any yielding support to the copper tube.

The support has yielded considerably so in which most of the stress in the bar is relieved.