Calculating the maximum load that can be carried & extension

In summary: Ultimate tensile stress = 420/2.5=168N/mm^2168N/mm^2=168,000,000 Pa168,000,000 x 0.00013125m^2 (131.25mm^2) = 22050N = 22.05 Kn Force = 22.05 KnAt this moment I am assuming i did this correctly. However, the question is asking for the maximum load that can be carried, not the force applied. The maximum load would be the force multiplied by the factor of safety, so the maximum load would be 22.05 Kn x 2.5 = 55.125 Kn. This
  • #1
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Ive been given a question which I'm stuck on and cannot answer. I've only been able to calculate the maximum allowable working stress and other then that I am stuck on how to answer the following question:

A specimen of the same material that was used in the above test (mild steel), had a cross section of 10.5mm x 12.5mm and a length of 0.9m (900mm). On testing this specimen it was found that the ultimate tensile stress was 420 N/mm^2. Calculate:

i) the maximum load that can be carried using a Factor of Safety of 2.5
ii) the extension under this load

So what i currently have worked out is the maximum allowable working stress which is 420/2.5=168 KN/m^2 (is this right?) and the cross section area which is presume is 10.5mm x12.5mm = 131.25mm^2 (is this right too?)

Thank you in advance
 
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  • #2
Mido999 said:
Ive been given a question which I'm stuck on and cannot answer. I've only been able to calculate the maximum allowable working stress and other then that I am stuck on how to answer the following question:

A specimen of the same material that was used in the above test (mild steel), had a cross section of 10.5mm x 12.5mm and a length of 0.9m (900mm). On testing this specimen it was found that the ultimate tensile stress was 420 N/mm^2. Calculate:

i) the maximum load that can be carried using a Factor of Safety of 2.5
ii) the extension under this load

So what i currently have worked out is the maximum allowable working stress which is 420/2.5=168 KN/m^2 (is this right?) and the cross section area which is presume is 10.5mm x12.5mm = 131.25mm^2 (is this right too?)

Thank you in advance
The maximum allowable working stress, with the factor of safety, has units of N/mm2, instead of kN/m2, going by the original text.
The area of the cross section is correct.
 
  • #3
SteamKing said:
The maximum allowable working stress, with the factor of safety, has units of N/mm2, instead of kN/m2, going by the original text.
The area of the cross section is correct.

Thank you but what do i do after is what I am stuck on
 
  • #4
Mido999 said:
Thank you but what do i do after is what I am stuck on
Well, what's your understanding of how to calculate the stress in the sample, if a load P were applied to it?

You have calculated the maximum allowable stress using the factor of safety. Now you need to determine the load P which produces this stress.
 
  • #5
The hint about the stress value for your sample comes from the units of the ultimate tensile stress. Notice that it is N (for Newton) per square mm. Or, to be a bit more obvious: If you had two identical wires, how much force could they hold in combination compared to what one wire could hold alone?

I think that in order to calculate the extension under this load you need more data. The ultimate tensile stress does not tell you what the extension will be at stress much less than that. Do you maybe have something like Young's modulus for the steel?
 
  • #6
The young's modulus from my previous question was 192.31 Kn/mm^2
To find out the load i suppose i would need to maximum tensile force right? is this done by dividing the maximum allowable stress by the cross sectional area so 168/131.25=1.28 Kn? I am not sure whether this is right
 
  • #7
I don't know whether I'm allowed to bump this thread but my deadline is on Friday and I really need to get this done am I on track or far off? Continuing from my previous post
 
  • #8
Mido999 said:
To find out the load i suppose i would need to maximum tensile force right? is this done by dividing the maximum allowable stress by the cross sectional area so 168/131.25=1.28 Kn? I am not sure whether this is right
As has been suggested, consider the units. The max stress is given in units of N/mm2. The area can be calculated in mm2. If you divide the first by the second, what will the units be?
 
  • #9
haruspex said:
As has been suggested, consider the units. The max stress is given in units of N/mm2. The area can be calculated in mm2. If you divide the first by the second, what will the units be?

I think it's either N/mm^2 or Kn/mm^2? - I'm sorry for being so slow by the way guys, I'm pretty new to this field so I hope you guys can bear with me
 
  • #10
Mido999 said:
I think it's either N/mm^2 or Kn/mm^2? - I'm sorry for being so slow by the way guys, I'm pretty new to this field so I hope you guys can bear with me
Treat N and mm as unknowns in an algebraic expression. You have a = N/mm2 and b = mm2. You want a result of N. Will you get that by dividing a by b? If not, what operation will give the desired result?
 
  • #11
haruspex said:
Treat N and mm as unknowns in an algebraic expression. You have a = N/mm2 and b = mm2. You want a result of N. Will you get that by dividing a by b? If not, what operation will give the desired result?

I gave it another try and got the following:

Cross section = 10.5mm x 12.5mm = 131.25mm^2
0.9m length =900mm
Ultimate tensile stress = 420/2.5=168N/mm^2

168N/mm^2=168,000,000 Pa
168,000,000 x 0.00013125m^2 (131.25mm^2) = 22050N = 22.05 Kn
Force = 22.05 Kn
At this moment I am assuming i did this correctly i acquired the force, but is the force "the maximum load that can be carried" as the question is asking? or do i do something else from here?

radius= 22.05/pie x 168= 0.4177817256
0.4177817256 square rooted = 0.2044m

Diameter = 2x0.2044=0.4088m=408.8mm
408.8m/2=204.4
pie x (204.4^2) = 131253.7304

extension = E = FxL / A x change in Length
so Change in length = 22.05 x 900/ 131253,7304 x 192.31 (this is the youngs modulus i got from my previous question)
= 0.000786mm extension

Now i feel as I've gone wrong somewhere here but I am not sure
 
  • #12
Mido999 said:
I gave it another try and got the following:

Cross section = 10.5mm x 12.5mm = 131.25mm^2
0.9m length =900mm
Ultimate tensile stress = 420/2.5=168N/mm^2

168N/mm^2=168,000,000 Pa
168,000,000 x 0.00013125m^2 (131.25mm^2) = 22050N = 22.05 Kn
Force = 22.05 Kn
At this moment I am assuming i did this correctly i acquired the force, but is the force "the maximum load that can be carried" as the question is asking? or do i do something else from here?

The SI symbol for the Newton is 'N'. The force should be written as 22.05 kN.

radius= 22.05/pie x 168= 0.4177817256
0.4177817256 square rooted = 0.2044m

Wait a minute here! When did the specimen cross section become circular? According to the OP, the cross section was 10.5 mm x 12.5 mm, which suggests a rectangular cross section. Did you omit some information from your problem statement?
Diameter = 2x0.2044=0.4088m=408.8mm
408.8m/2=204.4
pie x (204.4^2) = 131253.7304

extension = E = FxL / A x change in Length
so Change in length = 22.05 x 900/ 131253,7304 x 192.31 (this is the youngs modulus i got from my previous question)
= 0.000786mm extension

Now i feel as I've gone wrong somewhere here but I am not sure

It's not clear what units were used for your value for Young's modulus.

Always try to include units in your calculations so that the numbers are intelligible to someone else trying to follow your work.
 
  • #13
SteamKing said:
The SI symbol for the Newton is 'N'. The force should be written as 22.05 kN.
Wait a minute here! When did the specimen cross section become circular? According to the OP, the cross section was 10.5 mm x 12.5 mm, which suggests a rectangular cross section. Did you omit some information from your problem statement?It's not clear what units were used for your value for Young's modulus.

Always try to include units in your calculations so that the numbers are intelligible to someone else trying to follow your work.

Thank you for that, so is 22.05 kN (the force) the maximum load that can be applied? I am not too familiar with this subject
The young's modulus calculation was stress/strain so 0.2/1.04 x10 to the power of -3 = 192.307692 = 192.31 kN/mm^2 (this was the previous question which I've managed to complete)

And how did you determine that the specimen was of rectangular shape from the cross section? My lecturer told me that to find the diameter of a mild steel bar is to find the radius and then times the radius by 2 to get the diameter?
 
  • #14
Mido999 said:
Thank you for that, so is 22.05 kN (the force) the maximum load that can be applied? I am not too familiar with this subject
The young's modulus calculation was stress/strain so 0.2/1.04 x10 to the power of -3 = 192.307692 = 192.31 kN/mm^2 (this was the previous question which I've managed to complete)

Instead of writing E = 192.31 kN/mm2, write E = 192.31 MPa instead. You might want to check your calculation of the extension of the specimen.
And how did you determine that the specimen was of rectangular shape from the cross section? My lecturer told me that to find the diameter of a mild steel bar is to find the radius and then times the radius by 2 to get the diameter?

I think you are taking what your instructor said out of context. Not all cross sections are circular unless specified as such.

This is the text from your OP:
"A specimen of the same material that was used in the above test (mild steel), had a cross section of 10.5mm x 12.5mm and a length of 0.9m (900mm)." [emphasis added]

This suggests to me that the cross section is rectangular (two different dimensions given). If the cross section were indeed circular, all you would need to say is that the specimen had a diameter (or radius) of so many mm.
 
  • #15
SteamKing said:
Instead of writing E = 192.31 kN/mm2, write E = 192.31 MPa instead. You might want to check your calculation of the extension of the specimen.I think you are taking what your instructor said out of context. Not all cross sections are circular unless specified as such.

This is the text from your OP:
"A specimen of the same material that was used in the above test (mild steel), had a cross section of 10.5mm x 12.5mm and a length of 0.9m (900mm)." [emphasis added]

This suggests to me that the cross section is rectangular (two different dimensions given). If the cross section were indeed circular, all you would need to say is that the specimen had a diameter (or radius) of so many mm.

What is the formula to find the diameter of a rectangular specimen?
 
  • #16
Mido999 said:
What is the formula to find the diameter of a rectangular specimen?

Disregard my previous question, what about this for the extension:

22.05 x 900/131,25 x 192.31 (young's modulus from before) = 0.786mm extension?
 
  • #17
Mido999 said:
Disregard my previous question, what about this for the extension:

22.05 x 900/131,25 x 192.31 (young's modulus from before) = 0.786mm extension?
Yes, this appears correct.

I should correct my previous post. Young's modulus of 192.31 kN/mm2 should be equivalent to 192.31 GPa, instead of 192.31 MPa.
 
  • #18
SteamKing said:
Yes, this appears correct.

I should correct my previous post. Young's modulus of 192.31 kN/mm2 should be equivalent to 192.31 GPa, instead of 192.31 MPa.

Thank you very much for your and everybody that posed on this thread's help!
Would you say that I've finished this question with the answers I've provided?
 
  • #19
Mido999 said:
Thank you very much for your and everybody that posed on this thread's help!
Would you say that I've finished this question with the answers I've provided?

Have I completed this question or do I still need to work out the maximum load?
 
  • #20
Mido999 said:
Have I completed this question or do I still need to work out the maximum load?
You worked out the maximum load to answer part i) of the question.
 

What is the maximum load that can be carried?

The maximum load that can be carried refers to the maximum weight that a structure or object can support without collapsing or failing. It is an important factor to consider in engineering, construction, and transportation industries.

How is the maximum load calculated?

The maximum load is calculated using a combination of factors such as the strength and stability of the material, the design of the structure, and the external forces acting on it. Engineers use mathematical equations and computer simulations to determine the maximum load that a structure can bear.

What is the role of extension in calculating the maximum load?

Extension is the elongation or stretching of a material under stress. It plays a crucial role in calculating the maximum load as it determines the amount of deformation a structure can undergo before reaching its breaking point. Engineers must consider the extension properties of materials to accurately determine the maximum load that can be carried.

How does the type of material affect the maximum load that can be carried?

The type of material used in a structure directly impacts the maximum load that it can carry. For example, steel is much stronger and can bear a higher load than wood. The properties and characteristics of the material, such as density, elasticity, and tensile strength, must be taken into account when calculating the maximum load.

What safety measures should be taken when determining the maximum load that can be carried?

Safety is of utmost importance when determining the maximum load that can be carried. Engineers must consider a significant margin of safety to account for unforeseen circumstances and to ensure that the structure can withstand the expected load without failing. Regular inspections and maintenance are also crucial in ensuring the safety and longevity of structures that carry heavy loads.

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