Determining yield stress via a bar subjected to tensile force

In summary, the student is trying to solve a homework problem that involves using the wrong area for the cylinder. They should use the area of a circle as the force acting on the circular end. They also need to remember that quantity equals number plus unit. Until they get the hang of this, they should write their units in their calculations.
  • #1
Al_Pa_Cone
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Homework Statement



upload_2017-1-6_13-50-58.png

Homework Equations


All Equations Below

The Attempt at a Solution


upload_2017-1-6_13-52-20.png
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  • #2
You are using the wrong area for the cylinder. What should it be?
 
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  • #3
Do I use the area of a circle as the force is acting on the circular end? pi r^2
 
  • #4
Al_Pa_Cone said:
Do I use the area of a circle as the force is acting on the circular end? pi r^2
Of course. The force is equal to the stress times the cross sectional area.
 
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  • #5
upload_2017-1-9_11-14-43.png

Does This look better?
 
  • #6
Once again, 1 m2 is NOT 1000 mm2! And GN m is not a unit of force.
 
  • #7
I should have noticed that given the fact I only did it last week! As for the units, I think GPa is the correct unit to use in my answer? I am getting mixed up by force measured in Newtons and Newton meters as a measurment. I have reworked my answer again?
upload_2017-1-9_12-55-36.png
 
  • #8
Al_Pa_Cone said:
I should have noticed that given the fact I only did it last week! As for the units, I think GPa is the correct unit to use in my answer? I am getting mixed up by force measured in Newtons and Newton meters as a measurment. I have reworked my answer again?
View attachment 111305
Looks OK, except for the units. GPa is a unit of stress, not force. It would be better to just give the answer in N.
 
Last edited:
  • #9
That would be 117.6 Pascals then.

Thanks!
 
  • #10
No it wouldn't. You have a stress in GPa and you multiply it by an area in m2. What are the units of the answer?
 
  • #11
Sorry I did not read the last comment properly,
Chestermiller said:
Looks OK, except for the units. GPa is a unit of stress, not force. It would be better to just give the answer in N.
I only caught half of it in my emails (Looks OK, except for the units. GPa) and assumed it was ok.

As my undertanding goes with these units:

A Pascal is a unit which can be used for internal pressure acting in all directions equally within a material
A Newton is a unit which can represent force which acts upon a structure or an object
meters and millimeters is obviously a measurement of distance
meters^2 is a measurement of 2 dimensional area
meters^3 is a measurement of 3 dimensional area

The part where I get mixed up is when I have a GN m^-2

a Giga Newton is type of unit and a m^-2 is another but when do you choose to drop the distance/ area measurement out of the final answer giving a Force answer in simply Newtons or Giga Newtons rather than Newton meters ect

So... I am unsure whether my answer should be in Newtons or Newton meters
As the calculations go I presume they are ok so my attempted answer is:
upload_2017-1-9_13-57-21.png
 
  • #12
Much better.
 
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  • #13
You don't "choose to drop the distance/area measurement", as if this was optional. It disappears as a result of being multiplied by a distance/area measurement. N m-2 * m2 = N.
Remember, quantity equals number plus unit. Until you get the hang of this I would strongly advise you to write your units in your calculations, e.g.
Force = 6 x 10-3 GN m-2 * 1.96 x 10-5 m2 = 1.176 x 10-7 GN
And you should just learn the basic units for common quantities, e.g:
Force: N = kg m s-2
Energy: J = N m = kg m2 s-2
Pressure and Stress: Pa = N m-2
Strain: m/m = dimensionless (sometimes we use "units" like με = 10-6 m/m)
etc.
 
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  • #14
I think I need to read up some more on common units and measurements, It was introduced into the earlier assignments without covering in the maths bridging course I completed before starting this HNC,. Once again I thank you!
 

1. What is yield stress?

Yield stress is the amount of stress that a material can withstand before it permanently deforms or breaks. It is an important measure of a material's strength and is typically determined through tensile testing.

2. How is yield stress determined through a bar subjected to tensile force?

In order to determine yield stress, a material specimen is placed in a tensile testing machine and gradually subjected to a pulling force. As the force increases, the material will stretch until it reaches its yield point, at which it will begin to permanently deform. The maximum force at which this occurs is the yield stress.

3. What factors can affect the determination of yield stress?

The determination of yield stress can be affected by factors such as the composition and structure of the material, the testing conditions (such as temperature and rate of loading), and any imperfections or defects in the material.

4. What is the significance of yield stress in material testing?

Yield stress is a critical parameter in material testing as it provides important information about a material's strength and its ability to withstand applied forces. It is also used in the design and manufacturing of materials and structures to ensure they can withstand the expected loads and stresses.

5. Can yield stress change over time?

Yes, yield stress can change over time due to factors such as material degradation, changes in environmental conditions, or applied loads. This is why it is important to regularly retest materials to ensure their properties are still within the desired range.

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