Axial loading problem?

  • Thread starter babalu808
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  • #1
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Axial loading problem??

Im taking a strength and materials class and i cant seem to do this problem. it would be awesome if someone could walk me through how to solve it. i have a link that shows the picture.
Just some information you might need:
Modulus of Elasticity (E) of brass c83400 copper alloy = 101 GPa
Modulus of Elasticity (E) of stainless 304 steel alloy = 193 GPa

Thank you!
http://i54.tinypic.com/2eyvts2.png"
 
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Answers and Replies

  • #2
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Hello babalu and welcome to Physics Forums.

It is usual to offer any thoughts you have on solving the problem, even if you haven't succeeded yet.

Is this coursework or homework? There is a place for that in PF.

Since this is your first post I will offer a suggestion:

The condition to consider is that the strains must be equal. Does this help your thoughts along?
 
  • #3
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This is homework. The only thing I can think of is that the magnitude of the change in length (delta) of each member is the same.
 
  • #4
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Well hasn't that got you started?

What are your expressions for the strains in the three legs?
 
  • #5
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I got Delta=(P*L)/(A*E). But idk how to find the force P. I know it shows two enternal forces. Is that the P I use for the two legs on the left?
 
  • #6
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So what resists the two 40kN forces?

Hint, they do not constitute P directly - These are external loads.

You need to consider the internal forces (P) acting in all three legs.
 
  • #7
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Well there's the force from member EF and the force from the wall. So to find the internal forces do I have to do some global equilibrium kind of thing?
 
  • #8
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The walls and the blue plate are declared rigid.

So
The walls do not move.
The blue plate moves a small amount to the left, it does not bend or rotate.

Since the three legs are connected to the blue plate this causes compression in the two left hand legs and tension in the right hand leg.

Can you see that the contraction for the two left hand legs must therefore equal the extension for the right hand leg? ie [tex]\Delta[/tex] for each leg must be equal.

Furthermore the combination of the Ps for all three legs must add up to the external load to preserve horizontal equilibrium.
 
  • #9
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Yeah I can see that. But idk how to go about solving for it after that.
 
  • #10
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Can you see that by symmetry

PAB = PCD = P

Then the forces in the legs are

In AB P (compressive)

In CD P (compressive)

In EF T (tensile)

Can you draw a free body diagram for the blue plate ( you don't need to post it) and write the equation for horizontal equilibrium?
How many unknowns do you have?
 
  • #11
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Okay i did that. and i have 3 unknowns. force from AB, Force from CD, and force from EF. but i guess its two unknowns because Force AB and Force CD are equal. right?
 
  • #12
5,439
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Yes, two unknowns, P and T.

So what is your equation?
 
  • #13
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What's the problem?

Surely the equation for horizontal equilibrium isn't all that difficult?
 
  • #14
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2p +t = 80
 
  • #15
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Yes, great.

Now we need another equation.

Since nothing becomes detached the extension of EF = the contraction of AB

This gives a second equation connecting P and T. Here is my version. I will leave the resulting arithmetic to you to complete the homework.
 

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  • #16
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Okay, i think i can get it from there. Thank you for you time and patience
 
  • #17
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Hopefully you've learned from your effort, as well as getting your homework done.

:biggrin:
 

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