Answer:
The minimum diameter of steel is 7.32 mm
Explanation:
Given;
load raised by the steel, F = 4000 N
steel rating (tensile stress), σ = 95 MN/m² = 95 N/mm²
Area of the steel is given by;
[tex]A = \frac{F}{\sigma} \\\\\pi r^{2} = \frac{F}{\sigma} \\\\r^{2} = \frac{F}{\pi \sigma} \\\\r = \sqrt{\frac{F}{\pi \sigma}} \\\\r = \sqrt{\frac{4000}{\pi (95)}}\\\\r = 3.66 \ mm[/tex]
Thus, the minimum diameter of the steel = 2r
D = 2(3.66 mm)
D = 7.32 mm
Therefore, the minimum diameter of steel is 7.32 mm
