
#1
May2206, 04:13 AM

P: 5

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? 



#2
May2206, 06:58 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

And the work you have done so far towards solving the problem is... ?
<see my signature> 



#3
May2306, 04:39 AM

P: 5

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 10E05 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????????????????????????? 



#4
May2306, 07:14 AM

Emeritus
Sci Advisor
PF Gold
P: 11,154

Stress and strain questionBefore 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. 



#5
May2306, 05:58 PM

P: 5

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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