Determine working stress and greatest allowable load

In summary, the conversation discussed determining the working stress and greatest allowable load for a mild steel specimen with a diameter of 12.5 mm and 25 mm, respectively, using the ultimate tensile stress and a factor of safety of 3.5. The calculations involved finding the ultimate tensile stress, applying the factor of safety, and using the resulting value to determine the maximum allowable load for the larger diameter specimen. The final answer was 108.6 kN.
  • #1
ziziu
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"Determine working stress and greatest allowable load"

Homework Statement



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.

Homework Equations


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.

Greatest allowable load, is this the same as maximum load?

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.
 
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  • #2
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
Sorry yes that wasn't meant to have 2 there.
 
  • #4
ziziu said:

Homework Statement



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.

Homework Equations





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.

Greatest allowable load, is this the same as maximum load?

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.
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
 
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  • #5
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?

where the max allowable load is the load above which you start compromising your safety factor.

Could you explain this further please. I can't really understand what your saying.
 
  • #6
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. Let's 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?
 
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  • #7
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.
 
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  • #8
Last edited:
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  • #9
Oh nice one. Thank you guys. Very helpful site.

Seems pretty simple once you know how..

Thanks again.
 

1. What is the definition of working stress?

The working stress is the maximum stress that a material can withstand while still retaining its structural integrity and functionality. It is important to determine the working stress in order to ensure the safety and longevity of a structure or component.

2. How is working stress calculated?

The working stress is calculated by dividing the maximum load a material can withstand by its cross-sectional area. This gives the stress in units of force per unit area, such as pounds per square inch (psi) or newtons per square meter (Pa).

3. What factors affect the determination of working stress?

The determination of working stress is affected by several factors, including the material's tensile strength, yield strength, and elastic modulus. The type of loading (tension, compression, bending) and the geometry of the structure or component also play a role in determining the working stress.

4. What is the greatest allowable load?

The greatest allowable load is the maximum amount of force that a structure or component can withstand without experiencing failure or permanent deformation. It is determined by considering the working stress, safety factors, and any applicable building codes or regulations.

5. Why is it important to determine the working stress and greatest allowable load?

Determining the working stress and greatest allowable load is crucial for ensuring the safety and reliability of a structure or component. It helps engineers and designers select the appropriate material and design for a given application, and also helps to prevent overloading and potential catastrophic failures.

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