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Easy stress/strain Q

  • #1
Hullo. Hoping someone can help me with this Q off a statics past paper.
It's one of those things that I'm sure is ridiculously easy but I seem to be missing something (possibly very obvious.)
It's in the attatchment by the way.

my attempt:

strain = extension / o.g.length = 12 / 5000 = 0.003

E = stress / strain

therefore stress = E * strain = 210000 * 0.003 = 630

Which isn't a given answer so must be wrong. Can anyone please point out where I'm going wrong?
 
Last edited:

Answers and Replies

  • #2
Ooer managed to neglect actually attatch the problem. Here it is (I hope)
 

Attachments

  • #3
783
9
You don't have any units but by the values you have listed I am assuming that they are metric, 12mm, 5000mm ... However, you should check your double check your math and your units.
 
  • #4
Oh yeah, I was working in mm just because all of the potential answers were in mm so i figured it was easier to not have to convert at the end. If it was a unit error it would be out by a factor of 10 though wouldn't it?
 
  • #5
783
9
If I divide 12/5000 = 0.0024. Check your math again.
 
  • #6
Oops, that was a mis-type. the question says 15 mm, not 12 mm.
so it should say 15 / 5000 = 0.003
 
  • #7
783
9
What is the given answer?
 
  • #8
It's in the attatchment.
I dont know which one is correct but it has to be one of the given ones (400, -400, 14, -14 or 414 [all in N/mm^2])
 
  • #9
1,041
4
Somethin's strange about those answers. Can't see the attachment, but are you doing shear stress by any chance?
 
  • #10
I dont think the attatchment was working before, but it seems to be working now.
 
  • #11
1,041
4
OK, I see the attachment, but still don't understand the answers. I think it must be a language thing, i.e., the question states the problem in a way that I'm not used to. I would understand it if we just hadn't jacked against the concrete. Please post the explanation when you get it; I'd love to see how I misunderstand this.
 
  • #12
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,136
476
By jacking against the concrete to put tension in the rod, (and compression in the concrete), then both the concrete and the steel see the same magnitude of load prior to release of the jacking force. Thus, it appears that initially the compressive stress in the concrete is 30 times less than the tensile stress, since its area is 30 times more, and the applied forces on the steel and concrete are the same. The answer choices do not support this, so as you noted earlier, something is amiss in the problem statement or problem's choices.
 
  • #13
1,041
4
By jacking against the concrete to put tension in the rod, (and compression in the concrete), then both the concrete and the steel see the same magnitude of load prior to release of the jacking force. Thus, it appears that initially the compressive stress in the concrete is 30 times less than the tensile stress, since its area is 30 times more, and the applied forces on the steel and concrete are the same. The answer choices do not support this, so as you noted earlier, something is amiss in the problem statement or problem's choices.
If you assume that you tensioned the steel without involving the concrete, snugged the nuts, then released the tensioning force, you can get one of the answers in b, but not in a (unless I've made an arithmetic error). I'm thinking there must be assumptions in this particular class that we don't know about???
 

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