What pressure would a strain of 40 µε indicate?

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The discussion revolves around calculating the maximum allowable pressure for a cylindrical pressure vessel and determining the pressure corresponding to a strain of 40 µε. The maximum allowable pressure, considering a safety factor of 4, is derived to be 0.75 MPa based on a calculated hoop stress of 3 MPa. Participants discuss the application of Young's modulus and the hoop stress equation, emphasizing the need for proper understanding of yield criteria, particularly the von Mises criterion. For the strain calculation, the correct approach involves using the relationship between stress and strain, but participants express confusion over the necessary parameters, such as Poisson's ratio. The conversation highlights the challenges faced by distance learning students in grasping complex engineering concepts.
  • #31
Thanks for that put down! everyone doesn't come from the same level of education. I haven't used the minimal algebra I was taught in school for over 15 years until I started this course. I came on this forum for guidance not answers. I am fully prepared to attempt all of the questions I post first, I was merely expressing my concern at a transposition problem out loud.

Once I get time tomorrow, I will work my way through and post my results.
 
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  • #32
Al_Pa_Cone said:
Thanks for that put down! everyone doesn't come from the same level of education. I haven't used the minimal algebra I was taught in school for over 15 years until I started this course. I came on this forum for guidance not answers. I am fully prepared to attempt all of the questions I post first, I was merely expressing my concern at a transposition problem out loud.

Once I get time tomorrow, I will work my way through and post my results.
I am truly very sorry. No excuse. I won't let that happen again. Here's what you get when you factor the right hand side of the equation (and correct a few arithmetic errors and typos):

$$0.00004=\left[\frac{\frac{1}{0.006}-0.3\frac{1}{(2)(0.006)}}{210000\ MPa}\right]P$$
Is the Young's modulus 210 GPa or 290 GPa (as in post #1)?
 
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  • #33
I haven't taken any offence by your previous comments but thanks for accepting, it wasnt a helpful reply.

I had noticed the 0.0006 typo when it should have been 0.006 but I missed the 210! good spot. I find transposition very interesting and I like to try and solve these problems myself by substituting unknown values with simple numbers to see what works and what is acceptable. The factorising part was a little more advanced to the basic transposition as I couldn't figure out how to get both values of P into 1 P. I will look back over my maths bridging course and try to refresh my memory on how to achieve it.

Thanks for the help!
 
  • #34
upload_2017-1-11_10-28-56.png

Does This look ok?
 
  • #35
upload_2017-1-11_10-36-54.png

Sorry there was a problem with my answer when I didnt use a bracket when typing in 2*0.006 so it seperated.

This is my answer?
 
  • #36
Al_Pa_Cone said:
View attachment 111406
Sorry there was a problem with my answer when I didnt use a bracket when typing in 2*0.006 so it seperated.

This is my answer?
Yes. Of course, you'll need to do the arithmetic on the right hand side to come up with an actual number.
 
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  • #37
Yes, I just wanted to check my method of transposition was correct prior to any further calculations. I have worked out the answer to be 81.88 x 10^-3 MPa

Does this look good?
 
  • #38
Al_Pa_Cone said:
Yes, I just wanted to check my method of transposition was correct prior to any further calculations. I have worked out the answer to be 81.88 x 10^-3 MPa

Does this look good?
Yes, aside from a need to address the significant figures issue. Your answer is the same as 82 KPa.
 
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  • #39
Great! Thank you!
 
  • #40
hello,i am stuck with the last question,how you get the 81.88 Mpa?i tryed the equation you used but it gives different number,didyou used a different equation for the solution?i d appreciate your help,thx in advance
 
  • #41
what have you got so far?
 
  • #42
P = (290000/ 1/0.006 - 0.3 x 1/2 x 0.006) x 0.00004 = 0.0008367398 ,i used the equation that chestermiller suggested you to use,is this wrong?completelly stuck on this problem,didnt plan to combine the two equations,and we miss the poisons ratio that's why i saw your solution but didnt what you got,got any idea?
 
  • #43
I will take a look back through this tonight when i get a chance. I think you may have a rounding error somewhere.
I had to factorise the equation and transpose it to get my answer, but it was a jerk to figure out.

Are you doing your HNC at tees uni as well then?
 
  • #44
thank for your help i m not here to get just the answers because i want to learn these things but i got stuck with this one,i coudnt find a solution, i saw your posts and it was the only way to find a answer but i coudnt get the numbers...,yes i m doing the HNC at tees,very convenient because i work and i can study when i can,,for the first question you divided by 4 to get the right number or the 3 Mpa is enough? I will wait for your reply,thank you again
 
  • #45
I think your problem is the brackets are missing around the denominator part of that equation. It should be:

Also I received some good advice from a guy on here, it was reflected in my mark from the tutor when he commented on my assignment. The advice was to write down the units into the equations as well, because it helps find out where mistakes have been made.

[(290000 G Nm^-2 ) / (1 m / 0.006 m) - 0.3 x (1 m / (2 x 0.006 m)) ]x 0.00004 = P

This input into a scientific calculator would give the answer 81.88 x 10^-3 in Mega Newtons which is equal to 82 kN when rounded
 
  • #46
how many modules have you done then? maybe you could help me with some of my other assignments?
 
  • #47
Hi Al, how did you go with question 5 part c of this assignment? I'm also doing the same tees assignments as you but hit a wall on q5c. I've slipped up by having to work a lot lately and am struggling to get back into it!
 
  • #48
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  • #49
I have made several attempts but keep going wrong, I've followed you last thread on it but can see how you made it work? i can't get the EI side of it out of my head as I've worked that out to be 113.4knm and that don't fit into you way of working. I'm going to work for a couple of hours now and going to have a go when i get back in, if it can't get it I'm going to do question 6 and submit it see if i get a pass!
Out of interest this is my 4th module, how far on are you?
 
  • #50
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  • #52
Fair play to you for sticking with it. I'm into the mechanical side of things so don't really deal with beams but i should be able to understand it.
What side of work do you want to be in? civils? Thanks so much for the start, I'm going to push the EI working out to one side take your start and go back to the lessons.
 
  • #53
by the way I've done analytical methods, Engineering science, Applications of pneumatics and hydraulics so if i can help with those let me know. it was all going well till this bl**dy question!
 
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  • #55
I started as a motorcycle tech and went into heavy rotating plant so we aint that different! I'm happy with my job but wanted to do this as a fall back in case things change.
yes the ones above have all been marked, mainly merits and distinctions with one pass as i sent the wrong sheet in (idiot)
 
  • #56
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  • #57
Thanks, i will look at i when i get back in
 
  • #58
thank you for your reply, just a question,the 290000 is Mn m ^-2 ? and not the Gn ,still i got the same though... P=(290000/166.666666667 - 0.3 x 83.3333333333)x 0.00004 , P= (290000/13863.8888889)x 0.00004, P = 0.0008367398 (i divide the 1/ 0.006 to get the 166... and the 1/2 x 0.006 to get the 83...what am i doing wrong?this is my thrid one i finished analytical methods and Engineering science and after this one i will take the Applications Of Pneumatics And Hydraulics,Engineering Thermodynamics,Fluid Mechanics and Heat Transfer And Combustion
 
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