# Determine working stress and greatest allowable load

1. Dec 29, 2013

### ziziu

"Determine working stress and greatest allowable load"

1. The problem statement, all variables and given/known data

The maximum load in a tensile test on a mild steel specimen of diameter 12.5 mm is 95 kN, calculate the ultimate tensile stress. Also, determine the working stress and greatest allowable load on a rod of the same material 25 mm in diameter, given that the factor of safety is to be 3.5.

2. Relevant equations

3. The attempt at a solution

I'm only in the first term of my course and still new to this. I believe i've worked out the ultimate tensile strength here;

Pi x 6.25^2 = 122.718mm^2
95000/122.718 = 774.1326

now, where i'm struggling is the second two parts. Am i using the same force for this, the 95 kN or do i need to work out a new force using factor of safety and the new diameter?

I know factor of safety = Stress at Failure/Max working stress.

What i have tried and found out to be wrong is;

Pi x 12.5^2 = 490.87
95000/490.87 = 193.534

193.534/3.5 = 55.3

I'm not asking for the answer. Just a point in the right direction would be great.

Thanks.

Last edited: Dec 29, 2013
2. Dec 29, 2013

### SteamKing

Staff Emeritus
Well, the radius of a 12.5 mm dia. shaft is 6.25 mm, not 6.252 mm.

Your calculations lack units. You should get into the habit now of always including units in any calculation.

3. Dec 29, 2013

### ziziu

Sorry yes that wasn't meant to have 2 there.

4. Dec 29, 2013

### PhanthomJay

you solved for the ultimate tensile stress in part a ( units?) so per your equation the max allowable working stress is the ult stress divided by the safety factor. So now that you know the max allowable working stress, you can now calculate the max allowable load , where the max allowable load is the load above which you start compromising your safety factor. The max allowable load is a design safe working load and should not be confused with the ultimate failure load for that larger diameter rod.

both
confused with the max ultimate load

5. Dec 30, 2013

### ziziu

Thanks for your reply although i still don't fully understand. I understand what you are saying about the ult stress divided by the safety factor to give me working stress. But is the question not asking me to do this for the new diameter?

774.1326 mm^2 is for the diameter 12.5 in part A. It's asking me for the working stress and max allowable load for the new diameter 25 or no? So i don't really have the ult stress for the new diameter? Perhaps i'm totally missing something obvious.

From my understanding of what your saying these would be my next calculations;

774.1326/3.5 = 221.181 This would be the working stress?

Could you explain this further please. I can't really understand what your saying.

6. Dec 30, 2013

### SteamKing

Staff Emeritus
You are still resisting the suggestion to include units with your calculations. It makes it hard for third parties to review your results without working out the calculations from scratch.

Sample 1 provides you with a method to determine the ultimate tensile stress for a certain mild steel material. Lets call this ult. stress 'x'.

Sample 2 is made out of the same mild steel material as Sample 1, but its diameter is larger. It is reasonable to assume that the ult. stress of Sample 2 is the same as Sample 1, or 'x'.

Knowing that the ult. stress is the same value for both Samples, what tensile load produces the ult. stress 'x' in Sample 2? If a Factor of Safety of 3.5 is applied to this load, what is the max. tensile load which can now be applied to Sample 2?

7. Dec 31, 2013

### ziziu

I'm still a bit lost.

Max allowable load = Working stress x Area ?

Pi x 12.5^2 = 490.87mm^2 Is the area of the 25mm diameter.

221.181 MPa x 490.87mm^2 = 108.56 kN

Max allowable load = 108.56 kN?

Pretty much a guess.

Last edited: Dec 31, 2013
8. Dec 31, 2013

### nvn

Last edited: Dec 31, 2013
9. Dec 31, 2013

### ziziu

Oh nice one. Thank you guys. Very helpful site.

Seems pretty simple once you know how..

Thanks again.