Factor of safety for combined direct and shear loading

[Kyle52]

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

 

[mic*]

so tensile stress = 39.886kN / 62.32^-3 = 63.48Nm

This is incorrect. It should read 62.32*10^-6, as 1m^2 = 1000000mm^2, and the resultant unit is not Nm, it is N/m^2 (or Pa)

 

[mic*]

Actually, I didnt look closely enough at your previous calculations to get to "62.32^-3"...

Where does this figure come from at all? It certainly is not the cross sectional area of a 20mm dia bolt (in m^2).

 

[SteamKing]

Where did you get your cross-sectional area formula from?

 

[Kyle52]

I found it on a website that said to use 2 x pi r^2 if it is in double sheer which this is?

but my tutor prefers us to use (pi d^2)/4 for area so I just put the 2 in front. This is incorrect I am guessing?

 

[mic*]

Your OP says the bolt is in single shear. If that is correct then doubling the area is incorrect.

 

[Kyle52]

Ok I think I might just be going crazy. This is a copy of the diagram that I have wth the question ( sorry about the paint skills I can't copy it from a pdf)

As the shear is acting on both sides I asumed it was double sheer?

Apart from this mis calcultaion are the steps I am going through correct and the numbers are just wrong?

thanks

 

[mic*]

I found this;
http://www.mae.ncsu.edu/zhu/courses/mae314/lecture/Lecture2_Stress-Strain.pdf
Slides 16-18 might be diagramatically helpful.

 

[Kyle52]

Thanks I now understand why it's single and not double. I'm strugling to understand the significance of the tension force and the tension force itt's telling me not to exceed in calculating the working stress on the bolt as the examples I can find online don't seem to have them.

I can calculate the working stress with just : stress=force/area?

 

[mic*]

Engineering is not my speciality. From what I understand, yes, tensile working stress can be calculated as you have stated above, and then factored into the FOS formula you have stated in the OP, using the 290MPa as ult stress.

Shearing working stress - not so sure. When calculating working stress in this instance, does "area" still mean X-sectional area? Is this the area which the force is being applied to?

 

[Kyle52]

I believe that the shear stress area is the cross-sectional area of material with area parallel to the applied force vector. (This is what I have from my class notes)

I think I'm just missing something as there seems to be more information given in the question than I need?

Thanks for trying to help I appreciate it, I wil get my tutor to go though it with me tomorrow if I have no luck on here.

 

[PhanthomJay]

You appear to be mixing up shear and tension loads/stresses and safety factors. You should compare the max shear stress with the actual shear stress to determine the factor of safety against bolt failure in shear. The area of the bolt is still the same, assuming no threads in shear plane.

 

[Kyle52]

So I can find the FOS with just the sheer loading force (sheer force/cross sectional area) and 60% of the 290MPa as the max stress? I don't need to use the tension force that it gives me? Sorry but this is confusing the hell out of me and it's probably so simple once you get it.

 

[PhanthomJay]

So I can find the FOS with just the sheer loading force (sheer force/cross sectional area) and 60% of the 290MPa as the max stress? I don't need to use the tension force that it gives me? Sorry but this is confusing the hell out of me and it's probably so simple once you get it.

Well, not that simple. If the tensile load was externally applied, you would have to use a combined tension and shear stress formula to come up with the safety factor. You probably have not studied that combined stress method yet, but anyway, that does not appear to be the intent of this problem, since it states that the tension force is due to tightening of the bolt (prestressing it by turning the nut with a wrench by a specified amount) and not due to applying an axial tensile load to it. In which case, you can just look at the actual shear stress applied versus maximum allowed shear stress to determine the safety factor.