Question

A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a stress state greater than 200 N/mm2. If the linkage is to be constructed from solid round stock, what is the minimum required diameter?

Answer 1

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 = $2*10^{8} N/m^{2}$

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/($2*10^{8}$) = $10^{-5}$ m^2

recall that area = $\pi d^{2} /4$

$10^{-5}$ = $\frac{3.142*d^{2} }{4}$ = $0.7855d^{2}$

$d^{2} = \frac{10^{-5} }{0.7855}$ = $1.273*10^{-5}$

$d = \sqrt{1.273*10^{-5} }$ = $3.57*10^{-3}$ m = 3.57 mm

maximum diameter of  the steel linkage d = 3.57 mm