Two steel rods and an aluminium rod of equal length l_{0} 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 α_{a} and α_{s}_{,} respectively. Young's modulus of aluminium is Y_{a} and of steel is Y_{s}.

Steel |

Aluminium |

Steel |

#### Solution

Given:-

Initial length of the system (two identical steel rods + an aluminium rod) at 0 °C = l_{0}

Coefficient of linear expansion of steel and aluminium are α_{s}_{ }and α_{Al}, respectively.

Temperature is raised by θ.

_{}So, the change in temperature, Δθ = θ - 0 ° C = θ

Young's modulus of steel and aluminium are γ_{s} and γ_{Al}_{, }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 = ϒ_{s} +ϒ_{s}+ϒ_{Al} = 2ϒ_{s} + ϒ_{Al} ...(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 l_{0} [1+`(2ϒ_sα_sθ + ϒ_(Al)α_Alθ)/(2ϒ_s+γ_Al)` ] , where l_{0} is its initial length.