Correct Answer - Option 3 : 30 N/mm
2
Concept:
For a composite bar, the strain in both the bars will be the same i.e.
\(\epsilon=\frac{P}{AE}=\rm{constant}\)
Calculation:
Given:
ASteel = AAlmunium = 3 cm2 = 300 mm2, Esteel = 3EAl, P = 12000 N.
\(\epsilon=\frac{P}{AE}=\rm{constant}\)
since area is also constant, therefore
\(\epsilon=\frac{P}{E}=\rm{constant}\)
\(⇒\frac{P_{Aluminium}}{E_{Aluminium}}=\frac{P_{Steel}}{E_{Steel}}\)
\(⇒\frac{P_{Steel}}{P_{Aluminium}}=\frac{E_{Steel}}{E_{Aluminium}}\)
\(⇒\frac{P_{Steel}}{P_{Aluminium}}=3\;\;\;[E_{steel}=3E_{Aluminium}]\)
Psteel = 3PAluminium
Total load acting is 12000 N, therefore;
P = Psteel + PAluminium ⇒ 12000 N
4PAluminium ⇒ 12000 N
PAluminum = 3000 N
Psteel = 9000 N
Stress is steel:
\(\sigma_{steel}=\frac{P_{steel}}{Area}\)
\(\sigma_{steel}=\frac{9000}{300}\Rightarrow30\;N/mm^2\)