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

minimum required diameter of the steel linkage is 3.57 mm

Step-by-step 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)` =
`(3.142*d^(2) )/(4)` =
`0.7855d^(2)`

`d^(2) = (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