Factor of safety in operation: angles, bolts, tensile strength

[Oliver]

Hello all, Probably a very simple question when you know how to do it. I am just struggling to find information that is correct.



So far:

Bolt @ 47°
Ultimate tensile strength = 509MPa
Shear Strength = 357MPa
Bolt Diameter = 17mm
F = 20KN

Divide the force of the bolt between the axis of the bolt and the perpendicular to the axis

i) along the axis = 20Cos47 =14.640KN
ii) along perpendicular to the axis = 20Sin47 = 14.627 KN

Area of cross section of the bolt = PiD²/4 = Pi17²/4 = 226.980mm²

Direct tensile stress = DF/A = 13.640/226.980
= 0.06009KN/mm²
= 60.09N/mm²
= 60.09MPa (as 1Nmm² = 1MPa)

Shear/tangential stress = SF/A = 14.627/226.980
= 0.06444KN/mm²
= 64.44N/mm²
= 64.44MPa

I know it is correct up to this point, the next bit I think I've done wrong and information on the equations seems to vary depending on the source.

Shear FoS = Ultimate tensile strength / shear stress = 509/64.44 = 7.90
Tensile FoS = Shear Strength / Direct tensile stress = 357/60.09 = 5.94

So is the factor of safety in operation = the lowest number? so 5.94??

Could anyone help me with this please, Thank you for looking.

 

[Oliver]

Jut had this confirmed as being correct. This was for BTec lvl 3, hope it helps someone in the future.

 

[Beasty729]

I beleive you've mixed up your cos and sin when finding the components of the force:

along the axis should be 20sin47 and perpendicular should be 20cos47