Elastic deformation of axially loaded members

AI Thread Summary
The discussion revolves around calculating the displacement of a bar under a distributed axial load using the correct formula. The original poster mistakenly applied the formula for constant axial loads instead of integrating for a distributed load. The correct approach involves using the integral of P(x)/A(x)E(x) over the length of the bar, where P(x) is defined as the distributed load. Clarifications were provided that A(x) and E(x) remain constant for a prismatic beam, and the integration should not include additional variables for these constants. The conversation emphasizes the importance of correctly interpreting the problem and applying the right mathematical methods for accurate results.
mechengstudent
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Hi all,

I have a textbook question which I am stuck on and should be quite basic.

There is a bar that has a cross-sectional area of 1750mm2, and E = 220GPa.

http://imgur.com/gU29J1q
Edit: okay link is not working, it is http://imgur.com/gU29J1q

I am asked to find the displacement with the loading w=500x^1/3 N/m.

In class we have been using the formula displacement = PL/AE, where

P=load
L=length of member
A=cross sectional area tangent to load
E=Young's or elastic modulus

So far I have this

500(1.5m)^1/3 N/m x 1.5m
0.00175m^2 x 2.2x10^11 N/m^2

So it would seem all the units would cancel out apart from the m^1/3

However to me this didn't make much sense as the question is only asking for the answer to three significant figures and I got 1.467x10^-6 m^1/3

Can someone tell me if I am doing this correctly?

Thanks
 
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You have a bar where the axial load is distributed longitudinally. Exactly at what location are you supposed to calculate the axial deflection of the bar?
 
There is a point A at the end of the 1.5m long bar
 
I think you made a good attempt with trying to cancel out the units. The fact you came out with a strange unit for length should tell you did something wrong some where.

The mistake you made was using the wrong formula. The formula used is for a constant axial load. You have a distributed axial load here.

To solve this problem you need to develop your own formula by looking at how PL/AE is derived. Once you set this up and do the integral calculus you should arrive at a more sensible answer.
 
Yes thank you, I just found the right formula to use. It states the displacement is = integral between 0 and L of P(x)/A(x)E(x)

However I still don't know how to do it unfortunately. They give me w in the diagram, is this supposed to be P in the equation? And how would you substitute x in? Do just make it 500*1.5m^1/3 N/m? Or do I actually leave x in the equation?
 
Yes thank you, I just found the right formula to use. It states the displacement is = integral between 0 and L of P(x)/A(x)E(x)

However I still don't know how to do it unfortunately. They give me w in the diagram, is this supposed to be P in the equation? And how would you substitute x in? Do just make it 500*1.5m^1/3 N/m? Or do I actually leave x in the equation?
 
You're trying to substitute in x = L = 1.5m too soon. x does not equal 1.5m as it is variable that ranges from 0 to 1.5m.

But otherwise you guessed right, P(x)= w = 500x^1/3. Stick that in there and do the integration.
 
Duplicate.
 
Last edited:
Okay thanks. And for the A(x) and E(x) am I just using the area multiplied by x, and the elastic modulus multiplied by x?
 
  • #10
This is what I am putting into Wolfram Alpha - 'definite integral of (500x^(1/3))/((0.00175x)(2.2x10^(11)x)) between 0 and 1.5'

However I don't know how to interpret the answer...
 
  • #11
This is what I am putting into Wolfram Alpha - 'definite integral of (500x^(1/3))/((0.00175x)(2.2x10^(11)x)) between 0 and 1.5'

However I don't know how to interpret the answer...
 
  • #12
mechengstudent said:
Okay thanks. And for the A(x) and E(x) am I just using the area multiplied by x, and the elastic modulus multiplied by x?

No. A(x) and E(x) represent the cross-sectional area and the modulus of elasticity of the beam as functions of the length variable, x. For a prismatic beam made of a single material, A(x) and E(x) will both be constants.
 
  • #13
So what do I type into Wolfram Alpha? What does the formula look like when written out fully, before simplifying etc?
 
  • #14
Put in the same formula but drop the extra x's that you put in for A(x) and E(x).

Generally, when someone defines a function F(x) this means the value of F depends on the variable x without having to state exactly what the function is. In this specific case P(x) = 500x^1/3 but A(x) = A and E(x) = E since these are constant along the length of the beam.
 
  • #15
Duplicate.
 
  • #16
And what about the metres and Newtons in the formula?
 
  • #17
Oh sorry they cancel out, no worries
 
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