What is the Factor of Safety for Overlapping Bolted Metal Plates with 12 Bolts?

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In summary, the problem involves determining the factor of safety for two overlapping metal plates bolted together with 12 bolts, each with a diameter of 8mm. The ultimate shear stress of the bolt material is 300MPa and the total force to be carried by the joint is 20kN. The equation for stress is stress=force/area and the equation for factor of safety is ultimate stress/actual stress. After calculating the stress and factor of safety, the final answer is a factor of safety of 9 or 9.05 (rounded to one significant figure).
  • #1
sl8rbaby
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Homework Statement


Two overlapping metal plates are bolted together with 12 bolts. The diameter of each bolt is 8mm. If the ultimate shear stress of the bolt material is 300MPa determine what factor of safety applies if the total force to be carries by the joint is 20kN.


Homework Equations


Stress = force/area
factor of safety = ultimate stress/actual stress (I think)


The Attempt at a Solution


stress=force/area
stress=20kn/∏r^2 x 12
stress=20,000N / (∏ X 0.004)^2 X 12
= 10.55 MN/m^2

factor of safety = ultimate stress/actual stress
factor of safety = 300MPa/10.55 MN/m^2
= 300 x 10^6 / 10.55 x 10^6
= 28.42

Is this correct, cause I am not that confident? also what is the unit? is there one?
 
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  • #2
Your equations are good but you squared Pi in error. Would you expect the safety factor to have units? The stress unit N/ m^2 is called a Pascal (Pa).
 
  • #3
PhanthomJay said:
Your equations are good but you squared Pi in error. Would you expect the safety factor to have units? The stress unit N/ m^2 is called a Pascal (Pa).

Thanks for the reply, I thought that to work out the area of a circle which in this case is the bolt, the equation is ∏r^2? or are you saying I only need to square the radius first then times it by pi?

and no I assume there's no units as its a ratio
 
  • #4
ok think I've sussed it:

300,000,000 x ∏ x 0.004^2 x 12 = 180,955 = max force

max force / actual force = factor of safety

180,955 / 20000 = 9.05

I sure that's the final answer!
 
  • #5
sl8rbaby said:
Thanks for the reply, I thought that to work out the area of a circle which in this case is the bolt, the equation is ∏r^2? or are you saying I only need to square the radius first then times it by pi?
Yes, you have to be careful with use (or non-use) of parentheses when writing equations. The area of a circle is ∏r^2 or ∏(r^2) or (∏)r^2 or (∏)(r^2) or ∏*r^2, but it is not (∏r)^2. Also when you wrote "stress=20kn/∏r^2 x 12" , that should have read ""stress=20kN/12∏r^2" or some correct variation thereof. The way you wrote it implies you are multiplying by 12 instead of dividing by 12.
And no I assume there's no units as its a ratio
correct
 
  • #6
sl8rbaby said:
ok think I've sussed it:

300,000,000 x ∏ x 0.004^2 x 12 = 180,955 = max force

max force / actual force = factor of safety

180,955 / 20000 = 9.05

I sure that's the final answer!
Well, OK, but you should round it off to 9 (one significant figure).
 
  • #7
Did you ever find the correct solution to this question? I am currently stuck on the same one
 
  • #8
sl8rbaby never came back again so you are unlikely to get a direct answer .

You can have some help on this problem though - ask the question afresh in a new thread here on homework section .
Best to give us your own statement of the question and your own attempts at solution rather than just copy what was originally posted .
 

1. What is a factor of safety?

A factor of safety is a numerical value that represents the ratio of a structure's or material's maximum safe load to its actual or expected load. It is used to ensure that the structure or material can withstand unexpected or excessive loads without failure.

2. How is a factor of safety calculated?

A factor of safety is calculated by dividing the maximum safe load by the expected or actual load. It is typically expressed as a decimal or a percentage. For example, if a structure can safely hold 10,000 pounds and is expected to be under a load of 5,000 pounds, the factor of safety would be 2 (10,000/5,000).

3. Why is a factor of safety important?

A factor of safety is important because it ensures the safety and reliability of a structure or material. It takes into account any potential errors or uncertainties in design, material properties, or external forces. A higher factor of safety indicates a greater margin of safety against failure.

4. What is the recommended factor of safety?

The recommended factor of safety varies depending on the application and industry. Generally, a factor of safety of 1.5 to 2 is considered safe for most structures and materials. However, in high-risk industries such as aerospace or nuclear engineering, a higher factor of safety may be required.

5. Can a factor of safety be too high?

Yes, a factor of safety can be too high. While a higher factor of safety provides a greater margin of safety, it also means that the structure or material may be overdesigned and potentially more expensive. It is important to strike a balance between safety and cost efficiency when determining the appropriate factor of safety for a specific application.

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