Stress and strain. young modulus

AI Thread Summary
To determine the changes in lengths AB, AD, and AC under stress, the discussion emphasizes the need to apply the equations for plane stress/strain rather than just axial stress/strain. The correct formulas incorporate Poisson's Ratio to account for lateral contraction when a material is loaded in one direction. The user initially calculated changes using a simplified approach, leading to incorrect results, and later recognized the necessity of transforming to the appropriate coordinate system for accurate strain analysis. The conversation also highlights the importance of understanding the geometric relationships between the axes when calculating strains in different directions. Ultimately, the correct methodology involves using the generalized equations for stress and strain in two or three dimensions.
Dell
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in the following question, how do i find the changle in AB, the change in AD and the change in AC?

http://lh4.ggpht.com/_H4Iz7SmBrbk/Sv_hYT83rYI/AAAAAAAAB8s/7l5YRXMxVfQ/s912/Capture.JP what i have done is used the following
\sigmaxx=E*\epsilonxx

\deltaAB=\epsilonAB*AB
since AB is on the x axis,
\epsilonAB=\epsilonxx=\sigmaxx/E=150000000/200000000000=3/400

\deltaAB=\epsilonAB*AB=(3/400)*100*10^-5=75\mum
but, 1st of all this is wrong! the correct answer is meant to be 60\mum, and second of all it seems too simple, why would \nu=0.3 be given if i don't use it?
 
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The equation \sigma=E\epsilon only applies to uniaxial loading of a long, thin object (like a rod or a beam). Do you know of a more general equation for 2-D or 3-D loading configurations?

(This http://john.maloney.org/Papers/Generalized%20Hooke%27s%20Law%20%283-12-07%29.pdf" about stresses, strains, and constitutive equations may be helpful.)
 
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dont think so, what equation can i use?
 
Did you look at the link? Let me put it this way: strain in a certain direction can arise from a load in that direction, but it can also arise from lateral contraction caused by a load in another direction. Know what I mean?
 
Dell said:
in the following question, how do i find the changle in AB, the change in AD and the change in AC?

what i have done is used the following
\sigmaxx=E*\epsilonxx

\deltaAB=\epsilonAB*AB


Maples is right.
You should be using the equations for PLANE stress/strain and not AXIAL stress/strain. For axial strain ONLY, you would solve \epsilonxx=\sigmaxx/E.

For plane strain, use
\epsilonxx=\sigmaxx/E - \upsilon/E*\sigmayy
and comparably
\epsilonyy=\sigmayy/E - \upsilon/E*\sigmaxx.

When you [positively] load a member in direction x, it elongates in that direction. In order to hold the volume of the material constant**, the object will "thin" in the other two directions y and z. The amount that they thin is described by Poisson's Ratio \upsilon.
**Note: The volume of the material can change slightly, but for loading within the normal range of the material's limits, the change will be imperceptible.

Hope this gets you on your way.
 
okay got is, hookes law right??
 
for questions a and b it worked perfectly since AB and AD lie on the X and Z axes and i have all the info i need for them, for c) i need to find the change in the length in AC which is on a differend axis system, let's call it n-y-t. now i know that the angle between them is 53.12 (atan(100/75) ) but i want the other angle,- 90+53.13 since i want the strain which goes along AC and i have the transformations equations for stress/strains
so i looked for the strain\epsilont

\epsilont=\epsilonx*cos2(143.13)+\epsilonz*sin2(143.13)=2.86*10-4

\deltaAB=\epsilont*AB=2.86*10-4*125*10-3=35.75\mum
which is way off, anyone see where I am going wrong?
 
Instead of messing around with different coordinate systems, why not just look at the hypotenuse of the right triangle?
 
i learned that that wouldn't work since these are not vectors, but i figured it out anyways, i was using the wrong angle.

(looking at the answers pythagoras won't work here --> 60^2+20.6^2 is not 60.4^2)
 
  • #10
I'm talking about the hypotenuse of the strained lengths: the hypotenuse of 100.060mm and 75.0206mm is 125.0604mm.
 
  • #11
nice
 
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