A composite section consists of steel rod of $150 \mathrm{mm}$ diameter and outer brass tube of internal dia. $150 \mathrm{mm}$ and external dia. $170 \mathrm{mm}$ is rigidly fixed over steel rod and having length of $200 \mathrm{mm}$. the composite section carries a load of $600 \mathrm{KN}$. Find the stresses produced in each material and load carried by both the materials. Take $\mathrm{E}_{\text {steel }}=2 \mathrm{x} 10^{5} \mathrm{N} / \mathrm{mm}^2 \text{ and } \mathrm{E}_{brass} =1 \times 10^{5} \mathrm{N} / \mathrm{mm}^{2}$

Area of steel rod = $\frac{\pi}{4}*150^2 = 17671.46 mm^2$

Area of brass tube = $\frac{\pi}{4} * {(outer diameter - inner diameter)}^2$

$=\frac{\pi}{4}*(170-150)^2$

$=314.15mm^2$

Now, let the stresses in steel and brass be $P_s$ and $P_b$ respectively.

Strain in steel = Strain in brass

$\frac{P_s}{E_{steel}}=\frac{P_b}{E_{brass}}$ ---- (since , Strain = $\frac{Stress}{Young's Modulus}$)

$P_s=\frac{E_s}{E_b}8P_b$

$=\frac{2*10^5}{10^5}*P_b$

$P_s=2P_b$ ------------------------------- (1)

Now, load on steel+load on brass = Total load

$P_s*A_s+P_bA_b=600*10^3 N$

Since , Load = stress*area

Hence, $2*P_b*A_s+P_b*A_b=600*10^3 $ ------- from (1)

$2*P_b*17671.46+P_b*A_b=600*10^3$

$P_b=16.82 N/mm^2$ ------------ ( On solving )

Now,$ P_s=2*P_b=2*16.82=33.64 N/mm^2 $

Now, load carried by steel=$P_s*A_s = 33.64*17671.46 = 594.46*10^3 N$

Load carried by brass=$P_b*A_b = 16.82*314.15 = 5284 N$