Elastic deformation of axially loaded members

[mechengstudent]

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

 

[SteamKing]

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?

 

[mechengstudent]

There is a point A at the end of the 1.5m long bar

 

[paisiello2]

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.

 

[mechengstudent]

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?

 

[mechengstudent]

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?

 

[paisiello2]

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.

 

[paisiello2]

Duplicate.

 

[mechengstudent]

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?

 

[mechengstudent]

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...

 

[mechengstudent]

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...

 

[SteamKing]

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.

 

[mechengstudent]

So what do I type into Wolfram Alpha? What does the formula look like when written out fully, before simplifying etc?

 

[paisiello2]

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.

 

[paisiello2]

Duplicate.

 

[mechengstudent]

And what about the metres and Newtons in the formula?

 

[mechengstudent]

Oh sorry they cancel out, no worries