Question

Assuming the torsional yield strength of a compression spring is 0.43Sut and the maximum shear stress is equal to 434MPa. What is the factor of safety if the spring is a music wire (A=2170MPa and m=0.146) with a wire diameter of 4mm?

Answer 1

Answer:

factor of safety is 1.756

Explanation:

given data

torsional yield strength = 0.43 sut  

maximum shear stress = 434 MPa

A = 2170 MPa

m = 0.146

diameter = 4 mm

to find out

factor of safety

solution

we know 1 sut is equal to 1772.39 MPa

so 0.43 sut = 0.43 ×1772.39 =  762.128 MPa

so here

factor of safety formula is

factor of safety = yield strength / shear stress   ............1

put here value

factor of safety = 762.128 / 434

factor of safety is 1.756