Two steel rods and an aluminium rod of equal length l0 and equal cross-section are joined rigidly at their ends, as shown in the figure below. All the rods are in a state of zero tension at 0°C. Find the length of the system when the temperature is raised to θ. Coefficient of linear expansion of aluminium and steel are αa and αs, respectively. Young's modulus of aluminium is Ya and of steel is Ys.
Initial length of the system (two identical steel rods + an aluminium rod) at 0 °C = l0
Coefficient of linear expansion of steel and aluminium are αs and αAl, respectively.
Temperature is raised by θ.
So, the change in temperature, Δθ = θ - 0 ° C = θ
Young's modulus of steel and aluminium are γs and γAl, respectively.
If l be the final length of the system at temperature θ,
strain on the system = `(l-l_0)/l_0` ...(1)
Young's modulus = `("Stress") /("Strain")`
Therefore, the total strain on the system = `("Total stress on the system")/("Total Young's modulus of the system")`
Now, total stress = stress due to the two steel rods + stress due to the aluminium rod
`"Stress" = "F"/"A" = αY Δθ`
`"Total stress" = ϒ_sa_stheta + gamma_s a_stheta+ϒ_(Al)a_(Al)theta`
`=2gamma_sa_stheta +gamma_(Al)a_(Al)theta` ...(2)
Young's modulus of the system,
Y = ϒs +ϒs+ϒAl = 2ϒs + ϒAl ...(3)
Using (1), (2) and (3), we get:
Strain on the system = `(2ϒ_sα_sθ + ϒ_(Al)α_Alθ)/(2ϒ_s+γ_Al)`
`=> (l - l_0)/l = (2ϒ_sα_sθ + ϒ_(Al)α_Alθ)/(2ϒ_s+γ_Al)`
`l =l_0[ 1+ (2ϒ_sα_sθ + ϒ_(Al)α_Alθ)/(2ϒ_s+γ_Al)]`
Therefore, the final length of the system will be l0 [1+`(2ϒ_sα_sθ + ϒ_(Al)α_Alθ)/(2ϒ_s+γ_Al)` ] , where l0 is its initial length.