Question

A bimetallic cylindrical bar of cross sectional area 1 $m^2$ is made of bonding Steel (Young's modulus = 210 GPa) and Aluminium (Young's modulus = 70 GPa ) as shown in the figure. To maintain tensile axial strain of magnitude ${10}^{-6}$ in steel bar and compressive axial strain of magnitude ${10}^{-6}$ in Aluminum bar, the magnitude of the required force P (in kN ) along the indicated direction is

• A. 70
• B. 140
• C. 210
• D. 280

Answer 1

The correct option is D 280

$P_{st}+P_{Al}=P$
${\Delta }_{st}+{\Delta }_{Al}=0$

$\frac{P_{st}.\frac{L}{2}}{A{E}_{st}}+\frac{(P_{st}-P)\frac{L}{2}}{A{E}_{AL}}=0$

$\frac{P_{st}}{210}+\frac{P_{st}-P}{70}=0$

$\therefore$ $P_{st}=\frac{3P}{4}$

$Now,{\Delta }_{st}=\frac{P_{st}.\frac{L}{2}}{A{E}_{st}}$

${ϵ}_{st}=\frac{P_{st}}{A{E}_{st}}=\frac{3P}{4A{E}_{st}}$

$\therefore P=\frac{{ϵ}_{st}\times 4A{E}_{st}}{3}$

$=\frac{{10}^{-6}\times 4\times 1\times 210\times {10}^9}{3}$

= 280 kN