Example of Indeterminate Bars:

Let us describe the principle through a numerical example. Let the length and cross-sectional area of the bar be 750 mm and 1200 mm² respectively. Let us also assume, d = 0.12 mm, E = 200 kN/mm², α = 12 × 10^-6 m/m/^oC and ΔT = 32^oC. Using the above data, we can readily calculate the free thermal expansion of the bar is

δ_t  = L × α × ΔT

= 750 × 12 × 10^-6 × 32 = 0.288 mm

But this free expansion cannot take place because of the presence of the rigid restraint after a gap of 0.12 mm.

Assuming the elastic force in the bar as P, the elastic deformation δ_e may be calculated as

δ_e   = PL/AE

Any how the total deformation δ of the bar should be only equal to d. Therefore, the compatibility condition might be expressed as

 δ_t  + δ_e = d

(L × α× ΔT ) + PL/AE = d

From which P =- (L × α × ΔT - d ) AE/ L

or        P = -( α × ΔT - d/L)AE

Substituting the numerical values in Eq. (18),

P =- (0.288 - 0.12) 1200 × 200/750

= - 53.76 kN.

Now, stress induced σ_t = - (53.76 × 1000)/1200 = - 44.8 N/mm²

In many cases, the bar may already be kept between supports and 'd' is zero so that we may write as follows :

P = α× ΔT × AE

or thermal stress σ_t  = E × α × ΔT

Eq. (20) denotes that the thermal stress induced is a function of increase in temperature and elastic and thermal properties and not dependent on the geometric properties of the bar.