Mathematical Representation:
Thus, since the two materials are rigidly joined as a compound bar and subjected to the similar temperature rise, every material will attempt to expand to its free length position but each will be affected through the movement of the other. The higher coefficient of expansion material will try to pull the lower expansion material to its free length, other will be held back through the latter to its own free length position. In practice, a compromise will be reached along with both extending to a general position in among the individual free length positions. This, in effect, is equivalent to a contraction within bar 2 from its free length position and an expansion of bar 1 from its free position. Therefore, the higher coefficient of expansion material develops tensile stresses, while the temperature of the compound bar increases. It will be vice-versa while the temperature decreases. From Figures (c) to (f), it is clear that
Extension of bar 1 + Contraction of bar 2 = Difference in free lengths
Let the stresses in bars 1 and 2 are σ1 and σ2 because of the temperature change. Then the above rule could be written as
σ1L/E1 + σ2 L/E2 = (α2- α1 ) L ΔT
Because there are no external forces acting on the compound bar, for equilibrium, the compressive force in bar 2 should be equivalent to the tensile force in bar 1. This means that
σ1 A1 = σ2 A2
From the above two expressions, σ1 and σ2 can be written as
σ1 = A2 E1 E2 (α2 - α1 ) ΔT/ A1 E1 + A2 E2 ,
and σ2 = A1 E1 E2 (α2 -α1) ΔT/ A1 E1 + A2 E2
The extension of the compound bar, i.e. ΔL is given as
ΔL = L α1 ΔT + (α1/E1) L
ΔL = L α1 ΔT + A2 E2 (α2 - α1) LΔT/ A1 E1 + A2 E2
ΔL = (A1 E1 α1 + A2 E2 α2) L ΔT/ A1 E1 + A2 E2