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 the force equilibrium

The steel pipe's weight \( G = 10\space kN \) is balanced by the vertical components of the two steel - wire ropes. Let the tension in each rope be \( F \). From the force equilibrium in the vertical direction, we have \( 2F\sin30^{\circ}=G \). Since \( \sin30^{\circ}=\frac{1}{2} \), substituting \( G = 10\space kN \) into the equation: \( 2F\times\frac{1}{2}=10\space kN \), so \( F = 10\space kN \).

Step2: Calculate the cross - sectional area of the rope

The diameter of the rope \( d = 10\space mm=0.01\space m \). The cross - sectional area of a circle is given by the formula \( A=\frac{\pi d^{2}}{4} \). Substituting \( d = 0.01\space m \) into the formula, we get \( A=\frac{\pi\times(0.01)^{2}}{4}=\frac{\pi\times10^{- 4}}{4}\space m^{2}\approx7.854\times10^{-5}\space m^{2} \).

Step3: Calculate the internal stress

The internal stress (normal stress) \( \sigma \) is given by the formula \( \sigma=\frac{F}{A} \). We know that \( F = 10\times10^{3}\space N \) and \( A\approx7.854\times10^{-5}\space m^{2} \). Substituting these values into the formula: \( \sigma=\frac{10\times10^{3}}{7.854\times10^{-5}}\space Pa\approx1.273\times10^{8}\space Pa = 127.3\space MPa \).

Answer:

The internal stress of the steel - wire rope is approximately \( \boldsymbol{127.3\space MPa} \) (or \( 1.273\times10^{8}\space Pa \)).