What Does a 0 Degree Show in Elongation for Strain Gages?

  • Thread starter Thread starter berdan
  • Start date Start date
  • Tags Tags
    Strain
AI Thread Summary
The discussion centers on understanding the implications of a 0-degree measurement in elongation for strain gauges in a scenario involving pressure acting on a hole. It highlights the confusion regarding the relationship between stress and strain, particularly how strain can occur in directions where no stress is applied, as explained by Hooke's Law and Poisson's ratio. The key point is that while pressure acts perpendicularly, it can still cause strain in the circumferential direction, leading to a greater strain measurement at gauge "a" compared to gauge "b." This counterintuitive result arises because the circumferential strain is influenced by the geometry of the hole and the nature of the applied pressure. Ultimately, the discussion emphasizes the need to revise traditional thinking about stress and strain relationships in such contexts.
berdan
Messages
31
Reaction score
1
Question from one of my exams,I am lgoing crazy becaue this is against everything I understand.
Please please please help me :
http://imageshack.us/f/189/examh.jpg/

Basicaly,I need to found what the "0" degree shows me on elongation.
As I look at the hole in the figure,I see the pressure everywhere is perpendicular to its surfice.
That means,the stress is perpendicular to the surfice,and there is no stress in direction of "a".
All the stress goes in direction of "b".

In my logic,that meansthis is case of simple Hook Law with use of Poisson ratio.Because the stress in the direction of b,then elongation there must be bigger.

But it is not the case.Why why why oh god why!??

Thanks in advance.
 
Physics news on Phys.org
If the pressure causes the hole to enlarge, then (picturing it in 2D for simplicity) the circumference of the hole must increase, meaning 'a' is going to register strain.

A balloon stretches as contained air pressure increases, though pressure acts perpendicularly to the surface. Yes, it's not rigid, but the same idea would apply.
 
NascentOxygen said:
If the pressure causes the hole to enlarge, then (picturing it in 2D for simplicity) the circumference of the hole must increase, meaning 'a' is going to register strain.

A balloon stretches as contained air pressure increases, though pressure acts perpendicularly to the surface. Yes, it's not rigid, but the same idea would apply.


Offcourse,and according to Hook law,there will be strain even in the directions there is no stress.That is,perpendicular directions to stress will also change dimentions,according to Poison ratio.

What I don't understand is:
According to the exam,there is pressure "P" acting inside the hole,meaning it is perpendicular at every point to the hole circumference.

So-the bigger strain will be in the direction of the pressure.So,the strain gage "b",which is in the direction of the pressure (as it is perpendicular to the hole) should move more,than "a".
But infact,the answer is exactly the opposite:

"a" measure strain of 10^(-3)
"b" measures 2.5*10^(-4),which is exactly a*v (v-poison ratio).How can that be?Its supposedto be a=b*v,not the other way around!"b" should bebigger than "a"!

I hate this,nothing makes sense...
 
Imagine the pressure causes the "rigid" body to yield such that the "spherical" hole increases by 1mm in radius all around. (Simplifying to 2 dimensions) This will cause the circumference to increase by 2Pi mm. Hence you see the strain will be 6 times greater circumferentially (gage a) than radially (gage b). This is for 2 dimensions, so is only a rough guide to reality where we are involved with 3 dimensions.

Now, is that a sound basis on which to get you to revise your thinking?
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
Back
Top