Question

A steel rod of cross-sectional area $16cm$${}^2$ and two brass rods each of cross-sectional area $10cm$${}^2$ together support a load of $5000kg$ as shown in figure. Find the stress in the brass and steel rod (in kg/cm${}^2$). Take $Y$ for steel $=$ 2.0$\times$10${}^6$ kg / cm${}^2$ and for brass $=$ 1.0$\times$ 10${}^6$ kg / cm${}^2$

• A. $121,161$
• B. $161,120$
• C. $120,140$
• D. $141,120$

Answer 1

The correct option is A $121,161$

Given, $A_s=16c{m}^2$

$A_B=10c{m}^2$

$w=500kg$

Since $\Delta l$ will be same fore steel and brass.

$\frac{F}{A}=Y\frac{\Delta l}{l}$

$F_s+2F_B=\mathrm{5000...}\left(1\right)$

$\frac{F}{A}=$stress (s)

$S_s=\frac{F_s}{A_s}$ , $S_B=\frac{F_B}{A_B}$

$F_s=S_s.A_s$, $F_B=S_B.A_B$

$S_s=Y_s\frac{\Delta l}{l_s}$

$S_s\alpha \frac{Y_s}{l_s}$

$S_B\alpha \frac{Y_B}{l_B}$

$\frac{S_s}{S_B}=\frac{Y_s{l}_B}{Y_B{l}_s}=2\times \frac{2}{3}=\frac{4}{3}$

$S_s=\frac{4}{3}{S}_B$

and, $S_s{A}_s+2\left(S_s{A}_B\right)=5000$..from(1)

$S_s{A}_s+2\times \frac{3}{4}{S}_s.A_B=5000$

$S_s(16+\frac{3}{2}\times 10)=5000$

$S_s\left(31\right)=5000$

$S_s=\frac{5000}{31}\cong 161kg/c{m}^2$

$S_B=\frac{3}{4}\times \frac{5000}{11}$

$S_B\cong 121kg/c{m}^2$