- #1
Kyle52
- 6
- 0
Homework Statement
A Bolt is in single shear and is tightened so that it exerts a tension force of 39.886kN, the
diameter of the bolt is 20mm and the shear loading is 26.276kN;
Given that the tensile stress should not exceed 290MPa and the maximum shear stress should be
taken as 60% of the maximum tensile stress.
Calculate the factor of safety in operation.
Homework Equations
FOS= ult stress/working stress
A= (2pi d^2) / 4
tensile stress = tensile force / cross secontional area
The Attempt at a Solution
I have taken the cross sectional area as (2pi . 20^2) / 4 = 628.32mm^2
so tensile stress = 39.886kN / 62.32^-3 = 63.48Nm
So for the factor of safety I have taken : (0.6 x 290)/63.48 = 2.74
This is what I have managed to come up with in a combination of notes + textbooks but I have a feeling that I have gone wrong somewhere.
Any help would be great
Thanks