A bar of mild steel of diameter 75 mm is placed inside a hollow aluminium cylinder of internal diameter 75 mm and external diameter 100 mm; both bar and cylinder are the same length. The resulting composite bar is subjected to an axial compressive load of 10^6 N. If the bar and cylinder contract

Question

A bar of mild steel of diameter 75 mm is placed inside a hollow aluminium cylinder of internal diameter 75 mm and external diameter 100 mm; both bar and cylinder are the same length. The resulting composite bar is subjected to an axial compressive load of [latex] 10^{6} [/latex] N. If the bar and cylinder contract by the same amount, calculate the stress in each.

The temperature of the compressed composite bar is then reduced by 150°C but no change in length is permitted. Calculate the final stress in the bar and in the cylinder. Take [latex] E( ext { steel })=200000 N / mm ^{2}, E ext { (aluminium) }=80000 N / mm ^{2}, \alpha( ext { steel })=0.000012 /{ }^{\circ} C [/latex], [latex] \alpha( ext { aluminium })=0.000005 /{ }^{\circ} C [/latex].

Accepted Answer

The first part of this question is identical in form to P.7.15. Therefore, we can substitute areas of cross section, etc. directly into Eqs (7.38).

[latex]\sigma_{ S }=\frac{E_{ S }}{A_{ C } E_{ C }+A_{ S } E_{ S }} P \quad \sigma_{ C }=\frac{E_{ C }}{A_{ C } E_{ C }+A_{ S } E_{ S }} P[/latex] (7.38)

[latex] A_{ st }=\frac{\pi imes 75^{2}}{4}=4417.9 mm ^{2}, A_{ al }=\frac{\pi\left(100^{2}-75^{2} ight)}{4}=3436.1 mm ^{2} [/latex]

From Eqs (7.38)

[latex] \sigma_{ st }=\frac{10^{6} imes 200000}{(4417.9 imes 200000+3436.1 imes 80000)}[/latex]

i.e.

[latex]\begin{gathered}\sigma_{ ext {steel }}=172.6 N / mm ^{2} \quad( ext { compression }) \\sigma_{ ext {al }}=\frac{10^{6} imes 80000}{(4417.9 imes 200000+3436.1 imes 80000)}\end{gathered}[/latex]

i.e.

[latex] \sigma_{ ext {aluminium }}=69.1 N / mm ^{2} \quad ext { (compression) } [/latex]

Due to the decrease in temperature in which no change in length is allowed the strain in the steel is [latex] \alpha_{ st } T [/latex] and the strain in the aluminium is [latex] \alpha_{ al } T [/latex]. Therefore, due to the decrease in temperature

[latex]\begin{aligned}&\sigma_{ st }=E_{ st } \alpha_{ st } T=200000 imes 0.000012 imes 150=360.0 N / mm ^{2} \quad ext { (tension) } \&\sigma_{ al }=E_{ al } \sigma_{ al } T=80000 imes 0.000005 imes 150=60.0 N / mm ^{2} \quad ext { (tension) }\end{aligned}[/latex]

The final stresses in the steel and aluminium are then

[latex]\begin{gathered}\sigma_{ ext {steel }}( ext { total })=360.0-172.6=187.4 N / mm ^{2} \quad( ext { tension }) \\sigma_{ ext {aluminium }}( ext { total })=60.0-69.1=-9.1 N / mm ^{2} \quad( ext { compression })\end{gathered}[/latex]

Question 7.18: A bar of mild steel of diameter 75 mm is placed inside a hol...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions