Answer to Question #128735 in Mechanics | Relativity 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

^{Area of steel and brass tubes}

$A_{s}=\frac{\pi}{4}(D^{2}_{s}-d^{2}_{s}=50.27cm^{2}$

$A_{b}=\frac{\pi}{4}(D^{2}_{b}-d^{2}_{b})=56.55cm^{2}$

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

$\frac{\sigma_{s}}{E_{s}}=\frac{\sigma_{b}}{E_{b}}$

$\sigma_{s}=\sigma_{b}\frac{E_{s}}{E_{b}}$

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

$P_{s}+P_{b}=P$

$\sigma_{s}A_{s}+\sigma_{b}A_{b}=P$

$\sigma_{b}(\frac{E_{s}}{E_{b}}A_{s}+A_{b})=P$

Load carried by brass and steel respectively:

$P_{b}=\sigma_{s}A_{b}=\frac{PA_{b}}{\frac{E_{s}}{E_{b}}A_{s}+A_{b}}=\frac{PE_{b}A_{b}}{E_{s}A_{s}+E_{b}A_{b}}=52.3kN$

$P_{s}=\sigma_{s}A_{s}=\frac{PA_{s}}{\frac{E_{s}}{E_{b}}A_{s}+A_{b}}=\frac{PE_{s}A_{s}}{E_{s}A_{s}+E_{b}A_{b}}=97.7kN$