Find the minimum diameter of an alloy cable, tensile stress75 MPa (75,000,000 Pa), needed to support a Force of 15 kN (15,000 N)?

1 Answer
May 22, 2018

Given tensile strength =75 xx10^6\ MPa=>75 xx10^6\ Nm^-2
Force to be supported F=15xx10^3\ N

Corresponding area a=F/"tensile strength"

=>a=(15xx10^3)/(75xx10^6) = 2xx10^-4\ m^2

We know that for a circular cable of diameter d

a=(pid^2)/4
=>d=sqrt((4a)/pi)

Inserting calculated value we get

d=sqrt((4xx2xx10^-4)/pi)
d=0.016\ m