Calculating Factor of Safety

1. Apr 28, 2017

AutumnBeds

1. The problem statement, all variables and given/known data

The 14mm diameter steel bolt with modulus of elasticity of 210GPa and modulus of rigidity of 85 GPa shown in the diagram below holds two components together. The thickness of the bolted interface of the components is 18mm. To 2 decimal places, for the given loading;

Assuming the ultimate tensile stress is 490MN/m2 and the ultimate shear stress is 290MN/m2 determine the factor of safety in operation.

Forces - http://imgur.com/mbTD0Xg

2. Relevant equations

Assuming the ultimate tensile stress is 490MN/m2 and the ultimate shear stress is 290MN/m2 determine the factor of safety in operation.

3. The attempt at a solution

I have resolves the forces and found the following

Direct Stress 35.12 Mpa
Tensile Stress 0.17 Mpa (I think something may be amiss here)
Sheet Stress 26.24 Mpa

But for the factor of safety the notes I have been given say to divide the UTS by the direct stress,

So 490/35.12 = 13.95 (seems rather high). But then do I repeat for the Ultimate sheer stress?

So again 290/35.12 = 8.26.

Then is the lower of the two value?

Any help is greatly appreciated.

2. Apr 29, 2017

tech99

No one answered. I think you have two modes of failure, each having its own FoS. The lowest must be the relevant one.

3. Apr 29, 2017

4. May 1, 2017

PhanthomJay

Bolts subjected to both shear and tension forces have allowable stresses governed by shear-tension interaction equations specified by applicable loads. The safety factor depends on what the results of the interaction equation show. It could be quite less than the safety factor for tension alone or shear alone. As requested, please show your work.

5. May 9, 2017

AutumnBeds

6. May 10, 2017

PhanthomJay

The terms 'direct' stress and 'tensile' stress need definition. Direct stress is tensile stress if no shear stresses exist. Tensile stress is the actual tensile stress on the bolt. In your first image, you are calculating the tensile stress, which you did correctly , except you transposed some numbers it should be 5734, you wrote 5374.

In the 2nd image, you are calculating tensile strain, not stress, and strain is a dimensionless number. I am unsure why you are doing this, though.

In the third image, you have calculated the shear stress correctly.

Now in the 4th image, you calculate a safety factor for tension stress and a safety factor for shear stress, and both are rather large safety factors, but you really can't look at the overall safety factor of the bolt without calculating the effects of both tensile and shear loads applied at the same time, because the shear stress reduces the allowable tensile stress in the bolt, using a combined tension and shear stress formula to determine the allowable tensile stress. Only then can you calculate the safety factor, which will be the lowest of the 3 safety factors so calculated. Are you familiar with combined shear and tension stress formulas? I suspect that the safety factor for shear alone will control, because your actual stresses are rather low, but generally speaking, when shear and tensile stresses are high, the combined stress safety factor will be a lot less than the individual tension and shear alone safety factors.

7. May 10, 2017

AutumnBeds

Thanks,

1. Will redo the calculation and transpose the numbers correctly

2. This most likely just a written error on my behalf

3. Thanks - occasionally I get things correct.

4. No - nothing in the course notes gives guidance on the using the columbines stresses to calculate the fos. If you can offer any guidance on this I would be grateful.

Thanks again.

8. May 11, 2017

PhanthomJay

I
If it's not in your notes, I guess they are not looking for a combined stress formula, which probably isn't necessary on this problem where stresses are rather low and shear alone safety factor controls.
A combined stress formula for bolts might look like ( approxomTely...don't use it at face value $F_t = 1.3 F_u - 5 f_v </= F_u$ where
F_u = max direct tensile syress at failure
F_t = max allowed tensile failure stress
f_v = applied shear load shear stress

If stresses are low then individual shear or tension safety factors control, but when stresses are high, the combined stress governs and must be used.

Last edited: May 12, 2017