Help with Factor Of Safety (FOS) in Bolt Shear problem

[Jonathan Green]

Homework Statement

A Bolt is in single shear and is tightened so that it exerts a tension force of 60kN, the diameter of the bolt is 25mm and the shear loading is 40kN; Given that the tensile stress should not exceed 350N/mm2 and the maximum shear stress should be taken as 60% of the maximum tensile stress. Determine the maximum factor of safety for the above working condition and discuss the effect of using this value in a safety critical application.

Homework Equations

I know FOS = Failure Load / Allowable Load but I am unsure how to work this out from the data.

The Attempt at a Solution

Got as far as finding out the maximum shear stress is 210N/mm

 

[BvU]

Hello Jonathan,

I'm not an expert in this area, but I'm pretty good at physics exercises.

I suppose your 210 comes from 350 times 0.6 but how you change the dimension from N/mm² to N/mm is a mystery to me.

And: I notice you do not make use of the given 25 mm²; do you have a reason to assume that is not essential for this exercise ?

Make a plan to come to this safety factor and post it. In general: post detailed working so we don't have to suppose anything and make wrong assumptions

 

[Jonathan Green]

Hello Jonathan,

I'm not an expert in this area, but I'm pretty good at physics exercises.

I suppose your 210 comes from 350 times 0.6 but how you change the dimension from N/mm² to N/mm is a mystery to me.

And: I notice you do not make use of the given 25 mm²; do you have a reason to assume that is not essential for this exercise ?

Make a plan to come to this safety factor and post it. In general: post detailed working so we don't have to suppose anything and make wrong assumptions

I am clueless. Yeah it should've been N/mm2. How do you think it should be worked out ?

 

[BvU]

Now all we need is that plan of yours. It doesn't help you if I do the exercise for you. My hint was meant to put you on a track:
determine which of the two has the smallest^(*) safety factor, then for that one:
starting from the far end:

safety factor = max shear / actual shear
max shear = factor x max tensile
for the latter you have to invent a recipe (it's fairly straightforward)
​

(*) I must admit I do not understand the terminology 'determine the max factor of safety' in the problem statement: if it can easily stand the tension with a safety factor of 10 but not the shear with a safety factor of 0.99, then it breaks !

[edit] oh, and 'clueless' doesn't go well in the PF culture

 

[Jonathan Green]

I appreciate this, but it helps the most if I can see it worked out fully. This is not for an assignment I am just having trouble thinking of the calculation. I have been trying at it all week.

 

[BvU]

What's the maximum tensile force for this bolt, given the maximum tensile stress ?

 

[Jonathan Green]

What's the maximum tensile force for this bolt, given the maximum tensile stress ?

60Kn?

 

[BvU]

No, 60 kN (small k, big N) is the actual tensile force. If the maximum would be 60, the safety factor would be one and that's not good at all !

 

[Jonathan Green]

So is it 40kN?

 

[BvU]

Are you guessing or are you joking ? If 40 kN is the maximum tensile force, then it can never exert 60 kN at all !

 

[Jonathan Green]

Im sorry but I don't understand how to work this out. Please would you explain it to me, then I may get it.

 

[BvU]

OK, will try:

You are given that the max tensile stress for the material of the bolt is 350 N/mm². So a bolt with an area of 1 mm² gives up when the tensile force is 350 N.

So: what is the max tensile force for your exercise's bolt ?

 

[Jonathan Green]

So would it be 350N/mm2 dived by 22mm to equal 14N?

 

[BvU]

Wouldn't you think that is a bit small ? Trial and error again ?

Dimensions to the rescue:
What do you get when you divide N/mm² by mm² ? N/mm⁴

If one mm² can pull 350 N, then 10 mm² can pull how much ?

 

[Jonathan Green]

3500N?

 

[Jonathan Green]

So would it be 7700N?

 

[BvU]

Do I hear 8800 ? Think before posting !

a) it was a 25 mm bolt, not a 22

b) 7700 = 350 N/mm² x 22 mm gives you the wrong dimension of N/mm !

 

[Jonathan Green]

Yes sorry. 8800. So is 8800 the max tensile force for your exercise's bolt?

 

[BvU]

Aaargh... remember the first line in #17 !

-- And it's your exercise, not mine

You can NOT get Newtons if you multiply N/mm² with mm .

Your answer in #15 was correct -- I hope you understand that. But your bolt is not 10 mm² (its area). It is 25 mm diameter. What is the area of a bolt with a diameter of 25 mm ?

 

[Jonathan Green]

490?

 

[Jonathan Green]

pie R squared?

 

[BvU]

Do you want to continue until you guess the right answer ? If so, #20 is plain wrong and #21 is a lot better -- except for the question mark , that should have been an exclamation mark.

Now, do we continue this way or can you calculate the radius, area and max tensile force of a bolt with a diameter of 25 mm by yourself ?

[edit] Oops, I was too fast: #20 is only wrong because you left out the dimension (it is not 490 m² but 490 mm²) -- the numerical value is OK.

All in good spirit, of course.

 

[Jonathan Green]

I don't appreciate you patronising me. I came here because I can't work this out. If your going to mock me, please don't reply.

 

[Jonathan Green]

OK. So 490 was correct then ?

 

[Jonathan Green]

490 mm2 to be precise ?

 

[BvU]

I don't appreciate you patronising me. I came here because I can't work this out. If your going to mock me, please don't reply.

My apologies. I am here because I like to help, but sometimes my ironic approach takes over. 490 mm² is the area. 1 mm² can pull 350 N, so how much can the bolt pull (tensile force) ?

 

[Jonathan Green]

is it 171500mm2?

 

[BvU]

490 mm² x 350 N/mm² = 172000 N ( not mm² )

 

[Jonathan Green]

*171500N

 

[BvU]

Good. so the safety factor for the tension is ... ?
And the maximum shear force is ... ?
Which leads to:
The safety factor for the shtension is ... ?