Loaded Components Two Factors of Safety

[Tracey3]

Homework Statement

Ok so I have been presented with a bolt which has shear and tension acting upon it. I have done my best to 'draw' and recreate the problem with the information provided.
I am trying to figure out how, one would go about choosing which of the two safety factors should be used and justify why.

2/3. Equations and attempt at calculations
I personally have attempted to subject each Ultimate stress to opposite working stress (i.e. Tau U/ Sigma W) and see how they would change and pick the one that would drop more (Not part of image). This would have turned the shear FOS to 1.88 and tensile FOS to 3.97.

I think it looks better illustrated in the picture but here is the calculations in text:
Bolt Diameter=10mm
σ_u=390MN/m^2
τ_u=220MN/m^2

Shear= A/H=cos⁡50 A/12=12cos50=7.71KN
Tensile= O/H=sin⁡50 O/12=12cos50=9.19KN
A_O=(πD^2)/4=78.54mm^2
τ=F/A=7710/78.54=98.17MPa

σ=F/A=9190/78.54=117 MPa

FOS= (Ultimate stress)/(Working stress)

Shear FOS= τ_U/τ_W =220/98.17=2.24

Tensile FOS= σ_U/σ_W =390/117=3.3

 

[PhanthomJay]

Homework Statement

Ok so I have been presented with a bolt which has shear and tension acting upon it. I have done my best to 'draw' and recreate the problem with the information provided.
I am trying to figure out how, one would go about choosing which of the two safety factors should be used and justify why.

2/3. Equations and attempt at calculations
I personally have attempted to subject each Ultimate stress to opposite working stress (i.e. Tau U/ Sigma W) and see how they would change and pick the one that would drop more (Not part of image). This would have turned the shear FOS to 1.88 and tensile FOS to 3.97.

View attachment 218173

I think it looks better illustrated in the picture but here is the calculations in text:
Bolt Diameter=10mm
σ_u=390MN/m^2
τ_u=220MN/m^2

Shear= A/H=cos⁡50 A/12=12cos50=7.71KN
Tensile= O/H=sin⁡50 O/12=12cos50=9.19KN
A_O=(πD^2)/4=78.54mm^2
τ=F/A=7710/78.54=98.17MPa

σ=F/A=9190/78.54=117 MPa

FOS= (Ultimate stress)/(Working stress)

Shear FOS= τ_U/τ_W =220/98.17=2.24

Tensile FOS= σ_U/σ_W =390/117=3.3

Hi Tracey, welcome back!
Your question is a good one. Actually, when a bolt is simultaneously subjected to shear and tension, the individual tensile and shear safety factors lack meaning, because of combined tension/shear interaction curves that generally show a reduced or non existent overall safety factor. Such interaction curves are obtained by tests, Code equations, Von Mise combined stress equation, etc.
For example, suppose your loads produced bolt shear and tension stresses with a 1.1 safety factor against tension failure and a 1.1 safety factor against shear failure, with tension and shear loads applied individually. You might tend to then assume that the overall bolt safety factor against failure is1.1. If the loads you used were high atDand would be seldom in occurrence, maybe once every 50 years or so, and be temporary in nature, you might assume the bolts to be OK with But this would be a huge mistake, because the bolt will fail under that loading.

 

[Tracey3]

Hi Tracey, welcome back!
Your question is a good one. Actually, when a bolt is simultaneously subjected to shear and tension, the individual tensile and shear safety factors lack meaning, because of combined tension/shear interaction curves that generally show a reduced or non existent overall safety factor. Such interaction curves are obtained by tests, Code equations, Von Mise combined stress equation, etc.
For example, suppose your loads produced bolt shear and tension stresses with a 1.1 safety factor against tension failure and a 1.1 safety factor against shear failure, with tension and shear loads applied individually. You might tend to then assume that the overall bolt safety factor against failure is1.1. If the loads you used were high atDand would be seldom in occurrence, maybe once every 50 years or so, and be temporary in nature, you might assume the bolts to be OK with But this would be a huge mistake, because the bolt will fail under that loading.

@PhanthomJay thank you for the response. I was looking into this and it makes the most sense to me what you said. I have talked with my lecturer today and she refused to give me a direct answer but rather said that I should switch the two FOS values around and see how the Working stress changes. (i.e change shear FOS to 3.3)
Based on that we should be able to pick and justify which FOS we should use. I still don't see how that would be an appropriate way of going about this or what difference does it make.

Am I missing something obvious here?

 

[PhanthomJay]

@PhanthomJay thank you for the response. I was looking into this and it makes the most sense to me what you said. I have talked with my lecturer today and she refused to give me a direct answer but rather said that I should switch the two FOS values around and see how the Working stress changes. (i.e change shear FOS to 3.3)
Based on that we should be able to pick and justify which FOS we should use. I still don't see how that would be an appropriate way of going about this or what difference does it make.

Am I missing something obvious here?

That makes no sense to me. I would use the combined stress formula of the governing steel code, or use a Von Mise combined stress formula, which is a sort of sq root of the sum of the squares approach that compares the calculated Von Mise stress with the ultimate tensile strength to get a safety factor against failure. Using this latter approach, I get a safety factor of 1.9.
This is probably too advanced for your course, so take it for what it's worth.