Mechanical Principals Assignment Question

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Member warned to use the homework template for posts in the homework sections of PF.
I have a long question in my assignment with many part answers and I would just like to know if I have answered them correctly as I have spend a lot of time on it. can anyone help?

My Question is:

The component shown in Fig 1 is made from a material with the following properties and is subjected to a compressive force of 5kN.
upload_2017-1-4_9-51-9.png

Material Properties
Young’s Modulus of -200 GN m^-2
Elasticity Modulus of Rigidity -90 GN m^-2
Poisons ratio - 0.32


Q1a. Calculate the stress in:

(i) the circular section

(ii) the square section

My Solution:
upload_2017-1-4_9-57-33.png

Does This Look Correct?

Q1b. Calculate the strain in:

(i) The circular section

(ii) The square section
upload_2017-1-4_9-59-44.png

Does This Look ok?

Q1c. The change in length of the component
upload_2017-1-4_10-2-22.png


Q1d. The change in diameter of the circular section
upload_2017-1-4_10-7-41.png

Q1e. The change in the 40mm dimension on the square section
upload_2017-1-4_10-8-44.png

Q1f. If the same component were subjected to as shown in FIG 2, calculate the shear strain in:

(i) The Circular section
(ii) The Square section
upload_2017-1-4_10-11-20.png


Any help or advice would be much appreciated. Thanks
 

Answers and Replies

  • #2
mjc123
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1a. Looks OK to me.
1b. Wrong units for strain. Strain is dimensionless (and in this case negative).
1c. Be careful with signs and units. 0.002121 is not equal to -2.121 x 10-3. And -2.121 x 10-3 what? You are mixing units too much, and getting confused. I would convert everything into m (and Pa etc.)
1d. Change in length is not 7.75. What length is being referred to? (Why don't you just do it like 1e?)
1e. OK, but what are the units of the answer?
1f. As there is no Fig 2, this is impossible.
 
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  • #3
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Thanks for your help, I rushed the first part of the assignment because I was working to a deadline. That deadline was removed and now I have more time to think.
I have reworked Q1a although you said it was ok, I couldnt work out how I got the area values I had wrote down and also my answers are now positive? should they be Negative?

upload_2017-1-4_15-23-7.png
 
  • #4
mjc123
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No, you had it right the first time. A uniaxial stress is applied in the z direction. There is no applied stress in the x or y directions. The area to use is ##\pi##r2 or wl. And a compressive stress is negative. (And areas don't have units of mm3. And 1000 mm3 is not 1 m3, nor is 1000 mm2 1 m2. You are causing yourself a lot of unnecessary problems simply by being sloppy.)
 
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  • #5
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upload_2017-1-5_9-48-34.png

I have used the area previously used and obtained a figure the same but I still have a unit measurement difference. I am not sure the first unit of MPa was correct because of the conversion of mm into m. Previously I calculated m from mm by x10^-6 when it should have been x 10^-3. Am I correct in this error?
 
  • #6
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(And areas don't have units of mm3. And 1000 mm3 is not 1 m3, nor is 1000 mm2 1 m2. You are causing yourself a lot of unnecessary problems simply by being sloppy.)
1000mm does equal 1 meter?
 
  • #7
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Ahh I see my problem! 1.6^2 = 2.56 where 1600^2 = 2560000 which is x10^-6. Looks like I have answered my own question there. Sorry about that, I was thinking out loud!
 
  • #8
mjc123
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Yes, but 1000 mm2 does not equal 1 m2. Just picture a circle 30 mm in diameter. Do you think its area is anywhere near 1 m2?
 
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  • #9
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Does this look ok for part b?

upload_2017-1-5_10-25-57.png
 
  • #10
mjc123
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You are mixing units again! That's -7.074 MPa/200 GPa = -0.0354 x 10-3! You got it right (apart from the sign) in part c - 3.3535 x 10-5
 
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  • #11
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upload_2017-1-5_11-2-21.png

Does This look better?
 
  • #13
mjc123
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That looks better. But what's with all the ≈ signs? You're asked to calculate a value, not estimate an approximate value. I would calculate it precisely, then round to the appropriate number of significant figures at the end.
 
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  • #14
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I must be mistaken but I thought my use of the algibraic ≈ was suitable for my final answer. I have used it in previous assignments which were marked with distinction but never mentioned. I will stick with the standard = in future.

When you said calculate it precisely, then round to the appropriate number of significant figures at the end. I understand it effects my final answer, but do you think my answer is not accurate enough to be accepted on this occasion?
 
  • #15
mjc123
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Considering the numbers you were given, I think 2 sig figs is OK for an answer, but if you're going on to use those results in other calculations, you should use the more precise values for that purpose.
 
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  • #17
mjc123
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d. Poisson's ratio is positive, so the diameter increases.
e. Wrong - you have used the change in length instead of the strain.
 
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  • #18
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Material Properties
Young’s Modulus of -200 GN m^-2
Elasticity Modulus of Rigidity -90 GN m^-2
Poisons ratio - 0.32
From the Question given, It shows the Poissons ratio to be Negative? I undertand a compression on the top of a cylinder would increase the diameter of the midsection but how would I change the given figure to suit this answer?

If I remove the negative symbol my previous workings for part d, I get the same answer but as a positive. Is this acceptable?

and for part 'e'my reworkings are:
upload_2017-1-6_10-2-36.png
 
  • #19
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1a. Looks OK to me.
1b. Wrong units for strain. Strain is dimensionless (and in this case negative).
1c. Be careful with signs and units. 0.002121 is not equal to -2.121 x 10-3. And -2.121 x 10-3 what? You are mixing units too much, and getting confused. I would convert everything into m (and Pa etc.)
1d. Change in length is not 7.75. What length is being referred to? (Why don't you just do it like 1e?)
1e. OK, but what are the units of the answer?
1f. As there is no Fig 2, this is impossible.
In your first post you said '(Why dont you do it like in 1e?)'

So here is my answer in the same method as 1e and it is positive, and different from my previous calculation.
upload_2017-1-6_10-14-40.png
 
  • #20
mjc123
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I think you are confusing dashes with minus signs. Do you think the Young's modulus is -200 GPa? You haven't used a negative value in your calculations! Poisson's ratio for ordinary materials is positive. When you compress them uniaxially, they expand in the transverse directions. (There are a few exotic materials which have a negative PR, with niche applications.)
What are you doing in part e? You have replaced the change in length with the strain, but you get the same (wrong) answer as before! Didn't you redo the calculation?
 
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  • #21
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I think you are absoloutly correct! I am that concerned with trying to take all of this new information in, I am overlooking an obvious fact. It is poorly set out in the assignment but I should have noticed that anyway!
upload_2017-1-6_10-44-46.png


So as it stands in part d , is my orignial method of working out correct with the correct poissons ratio?
upload_2017-1-6_10-49-55.png
 
  • #22
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What are you doing in part e? You have replaced the change in length with the strain, but you get the same (wrong) answer as before! Didn't you redo the calculation?
I did Re-do the calculation but I noticed I had missed out the multiplication of z in this equation
upload_2017-1-6_10-54-4.png

Also It shows in this equation that v (Poissons Ratio) is a negative figure which makes the answer a positive figure?

Can you recommend another method I should use?
 
  • #23
mjc123
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You have a spurious negative sign on the RHS, but the green answer is correct.
 
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  • #24
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How about this for e?
upload_2017-1-6_11-3-40.png
 
  • #25
mjc123
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Why have you included x60? ##\epsilon x = -\nu\epsilon z## = -0.32 x -1.55 x 10-5 That is your x strain. You multiply that by the width to get the change in width. You got it right the first time!
 
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