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
Force to be supported
Corresponding area
=>a=(15xx10^3)/(75xx10^6) = 2xx10^-4\ m^2
We know that for a circular cable of diameter
a=(pid^2)/4
=>d=sqrt((4a)/pi)
Inserting calculated value we get
d=sqrt((4xx2xx10^-4)/pi)
d=0.016\ m