Answer:
minimum required diameter of the steel linkage is 3.57 mm
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 = [tex]2*10^{8} N/m^{2}[/tex]
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/([tex]2*10^{8}[/tex]) = [tex]10^{-5}[/tex] m^2
recall that area = [tex]\pi d^{2} /4[/tex]
[tex]10^{-5}[/tex] = [tex]\frac{3.142*d^{2} }{4}[/tex] = [tex]0.7855d^{2}[/tex]
[tex]d^{2} = \frac{10^{-5} }{0.7855}[/tex] = [tex]1.273*10^{-5}[/tex]
[tex]d = \sqrt{1.273*10^{-5} }[/tex] = [tex]3.57*10^{-3}[/tex] m = 3.57 mm
maximum diameter of the steel linkage d = 3.57 mm
