Answer to Question #118383 in Physics for Ariyo Emmanuel

Question

(3) A 20 cm long steel tube of 15 cm internal diameter and 1.0cm thickness is surrounded by a brass tube of the same length and thickness. The tubes carry an axial load of 150 kN. Estimate the load carried by each.

Es = 21 x 106 N/cm2 Eb = 10 x 106 N/cm2

Answer 1

Given:

Load P: 150 kN

Length L: 20 cm

Steel tube

Internal diameter $d_s$: 15 cm

External diameter $D_s$: 17 cm

E_s = 21x10⁶ N/cm²

Brass tube

Internal diameter $d_b$: 17 cm

External diameter $D_b$: 19 cm

E_b = 10x10⁶ N/cm²

Solution

Areas of steel and brass tubes:

$$A_s=\frac{\pi}{4}(D_s^2-d_s^2)=50.27\text{ cm}^2.\\\space\\ A_b=\frac{\pi}{4}(D_b^2-d_b^2)=56.55\text{ cm}^2.$$

Under the same load strain $\epsilon$ in steel equals the strain in brass:

$$\frac{\sigma_s}{E_s}=\frac{\sigma_b}{E_b},\\\space\\ \sigma_s=\sigma_b\frac{E_s}{E_b}.$$

Load on steel added to the load on brass equals the total load:

$$P_s+P_b=P,\\ \sigma_sA_s+\sigma_bA_b=P,\\ \sigma_b\bigg(\frac{E_s}{E_b}A_s+A_b\bigg)=P.$$

Load carried by brass and steel respectively:

$$P_b=\sigma_bA_b=\frac{PA_b}{\frac{E_s}{E_b}A_s+A_b}=\frac{PE_bA_b}{E_sA_s+E_bA_b}=52.3\text{ kN}.\\ P_s=\sigma_sA_s=\frac{PA_s}{\frac{E_s}{E_b}A_s+A_b}=\frac{PE_sA_s}{E_sA_s+E_bA_b}=97.7\text{ kN}.$$