Question

example3 a steel pipe weighting 10 kn is suspended from a couple of steel-wire ropes，the diameter of the rope is 10mm, try to determine the internal stress of the steel-wire rope.

Explanation:

Step1: Analyze Force Equilibrium

The steel pipe's weight \( G = 10\,\text{kN} \) is balanced by the vertical components of the two rope forces. Let each rope's force be \( F \). The vertical component of each force is \( F\cos30^\circ \). So, \( 2F\cos30^\circ = G \).

Step2: Solve for Rope Force \( F \)

From \( 2F\cos30^\circ = 10\,\text{kN} \), we have \( F=\frac{10}{2\cos30^\circ}=\frac{5}{\frac{\sqrt{3}}{2}}=\frac{10}{\sqrt{3}}\approx5.77\,\text{kN} \).

Step3: Calculate Cross - Sectional Area of Rope

The diameter \( d = 10\,\text{mm}=0.01\,\text{m} \), so the area \( A=\pi(\frac{d}{2})^2=\pi(\frac{0.01}{2})^2=\frac{\pi\times10^{- 4}}{4}\approx7.854\times10^{-5}\,\text{m}^2 \).

Step4: Calculate Internal Stress \( \sigma \)

Stress \( \sigma=\frac{F}{A} \). Substituting \( F = \frac{10}{\sqrt{3}}\times10^{3}\,\text{N} \) and \( A = 7.854\times10^{-5}\,\text{m}^2 \), we get \( \sigma=\frac{\frac{10}{\sqrt{3}}\times10^{3}}{7.854\times10^{-5}}\approx\frac{5.77\times10^{3}}{7.854\times10^{-5}}\approx7.35\times10^{7}\,\text{Pa} = 73.5\,\text{MPa} \).

Answer:

The internal stress of the steel - wire rope is approximately \( 73.5\,\text{MPa} \) (or more precisely, using exact calculations: \( \sigma=\frac{10\times10^{3}}{\sqrt{3}\times\pi\times(\frac{0.01}{2})^2}=\frac{10^{4}\times4}{\sqrt{3}\times\pi\times10^{-4}}=\frac{4\times10^{8}}{\sqrt{3}\pi}\approx7.35\times10^{7}\,\text{Pa} = 73.5\,\text{MPa} \))