# Factor of safety with an angle

I need help with the following factor of safety.

This is the infomation to apply to the equation.

Tensile Strength = 460MPa

Shear Stress = 280MPa

Bolt Diameter = 8mm

Angle = 50 Degrees

I need to determine the tensile stress and shear stress in the bolt.

I have the equation for Shear stress.

$$\frac{Ultimate Stress}{Allowable Stress}$$

So far i have this.

201.061mm^2 being Area of the 8mm bolt.

$$\frac{9kN}{201.061mm^2} = 0.04476\frac{kN}{mm^2}$$

$$x1000 = 44.76\frac{N}{mm^2}$$

$$= 44.76MPa$$

$$\frac{280MPa}{44.76MPa} = 6.255MPa$$

So Shear Stress = 6.255MPa

And i think i work it backwards to work out tensile strength?

$$\frac{Ultimate Load}{Allowable Load}$$ The problem i have is how to apply the 50 degree angle. I have read to multiply the equation by cos or sin. But i am really stuck on this part.

Thanks.

Attached diagram. Sorry for not attaching this before.

a is 50 degrees. And is what i am stuck on. I do not know how to apply the degrees to the equation. • Drawing a diagram might help. As it is the problem seems to be missing something: 50 degrees between what? On review its likely between the stress plane and tension direction. Sounds like a resolved shear stress problem. Dec 19, 2017 at 19:00
• Possible duplicate of Factor of Safety Dec 19, 2017 at 19:48
• this is all gibberish without the associated diagram or a clear description of the problem. Dec 19, 2017 at 19:50
• Can anyone help please? Dec 20, 2017 at 12:33

You can solve the problem at various levels of complexity. But the first thing to do is to consider the simplest possible situation where all you have is the inclined plane and forces at an angle to the plane. From Euclid, and your figure, that angle is $\theta$ = 50 degrees.
Assume that the bolt is perpendicular to the inclined plane. Then the force can be resolved into components that are normal and tangential to the plane. The two components are $F_n = F \sin\theta$ = 6900 N and $F_t = F \cos\theta$ = 5800 N.
The area of the 8 mm bolt is $A = \pi d^2/4$ = 5e-5 m$^2$.
The normal and tangential stresses are $\sigma_n = F_n/A$ = 137 MPa and $\sigma_t = F_t/A$ = 115 MPa.
The factors of safety are $f_n = \sigma^{\text{allow}}_n/\sigma_n$ = 460/137 = 3.3 and $f_t = \sigma^{\text{allow}}_t/\sigma_t$ = 280/115 = 2.4.