How Much Weight Can an 8mm Diameter Steel Bar Safely Support?

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Discussion Overview

The discussion revolves around determining the maximum working load that an 8mm diameter steel bar can safely support, considering the ultimate tensile strength of steel and a factor of safety. The scope includes mathematical reasoning and technical calculations related to material strength and safety factors.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant states that the ultimate tensile strength of steel is 470 MPa and seeks to calculate the maximum working load for an 8mm diameter steel bar.
  • Another participant asks for clarification on the work done so far towards solving the problem.
  • A participant presents calculations using the formula Stress = Force / Area, arriving at a force of 23624.78 N, and questions the application of the factor of safety.
  • Another reply confirms the initial calculation of maximum load but questions the application of the factor of safety, suggesting that the allowable load should not be five times the maximum load.
  • A different participant argues that the allowable load should be calculated by dividing the maximum load by the factor of safety, resulting in a maximum working load of approximately 472.5 kg or 4724.96 N.

Areas of Agreement / Disagreement

Participants express differing views on how to apply the factor of safety in calculating the maximum working load, leading to conflicting conclusions about the allowable load. The discussion remains unresolved regarding the correct interpretation of the factor of safety in this context.

Contextual Notes

There are unresolved aspects regarding the definitions of maximum load and allowable load, as well as the implications of applying the factor of safety in calculations. Participants have not reached a consensus on these points.

dblanche
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The Ultimate tensile strength of steel is 470MPa.

Assuming a factor of safety of 5 the maximum working load an 8.00mm diameter round steel bar should support against gravity is?
 
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And the work you have done so far towards solving the problem is... ?

<see my signature>
 
I have used every formula that I can think of, I've used the following equations:



Stress= Force / Area
470 x 10E06 = force / 5.03 x 10E-05
Force = 23624.78 N

then multiply 23624.78 by the factor of safety being 5
= 118123.9 Newtons?


Don't really know how to do it, could you please give us some hints?
 
dblanche said:
I have used every formula that I can think of, I've used the following equations:
Stress= Force / Area
470 x 10E06 = force / 5.03 x 10E-05
Force = 23624.78 N
This is correct. You've calculated the maximum load.

then multiply 23624.78 by the factor of safety being 5
= 118123.9 Newtons?
There's a small error here. Should the allowable load be 5 times the maximum load ? What happens to the rod if you apply a load greater than the maximum load ?

Before you do any calculation involving numbers, first write down all the necessary equations using symbols to represent the various quantities involved. Plug in the numbers only after you've arrived at a final equation with the required unknown on one side, and all the knowns on the other.
 
No the allowable load would not be 5 times the maximum load, you divide this maximum load by 5 to give the maximum working load, which is 472.5 kg or 4724.96 Newtons........

Thanks for your help
 

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