How Do You Calculate Strain Energy in Beams Using Integral Equations?

[smr101]

I can't work out the strain energy using the integral equation here, question (b)(i).

I've got the bending moment for the first beam as 400kN.

E and I are given, I'm guessing you can take '2EI' out of the integral and just integrate the M^2, is that correct?

 

[SteamKing]

I can't work out the strain energy using the integral equation here, question (b)(i).

I've got the bending moment for the first beam as 400kN.

E and I are given, I'm guessing you can take '2EI' out of the integral and just integrate the M^2, is that correct?

To calculate the strain energy of the beams, you need to know the function of M w.r.t. the length coordinate of each beam, which is s here.

So the strain energy is more accurately represented by the following integral:

$U=\int_0^L \frac {M^2(s)}{2EI} \, ds$

If EI is a constant value over the length of the beam, it can be moved before the integral sign, like any other constant.

 

[smr101]

To calculate the strain energy of the beams, you need to know the function of M w.r.t. the length coordinate of each beam, which is s here.

So the strain energy is more accurately represented by the following integral:

$U=\int_0^L \frac {M^2(s)}{2EI} \, ds$

If EI is a constant value over the length of the beam, it can be moved before the integral sign, like any other constant.

This is what I thought would be the case, I did the following:

2EIU = M^2*s

U = 400,000^2 * 4 / (2 * 200x10^9 * 60.8x10^-6)

= 26.315 kJ

The solution states the answer is 8.772 kJ.

Which strangely enough is exactly a third of the answer I got it...

 

[SteamKing]

This is what I thought would be the case, I did the following:

2EIU = M^2*s

U = 400,000^2 * 4 / (2 * 200x10^9 * 60.8x10^-6)

= 26.315 kJ

The solution states the answer is 8.772 kJ.

Which strangely enough is exactly a third of the answer I got it...

In each beam case, M(s) ≠ constant over the length of the beam.

Remember drawing shear and bending moment diagrams for beams with various loadings? The integral of the strain energy in bending is the area under those bending moment curves squared term-by-term, then divided by 2EI. That's the calculation you are supposed to do here.

 

[smr101]

In each beam case, M(s) ≠ constant over the length of the beam.

Remember drawing shear and bending moment diagrams for beams with various loadings? The integral of the strain energy in bending is the area under those bending moment curves squared term-by-term, then divided by 2EI. That's the calculation you are supposed to do here.

I'm not understand what you mean by 'squared term by term'. What terms?

Again, I've noticed for the second beam doing what I did in my last post and dividing by 3 gives the correct solution.

 

[SteamKing]

I'm not understand what you mean by 'squared term by term'. What terms?

I meant point-by-point.
The bending moment curve for a cantilever beam loaded at the end is this:

​

In order to evaluate the integral
$U=\int_0^L \frac {M^2(s)}{2EI} \, ds$
you must square the ordinates of the bending moment diagram.

Since the ordinates of this bending moment diagram vary linearly with the coordinate s, which runs along the length of the beam, you can find a symbolic expression for M(s) as a function of s and then square that expression and evaluate the integral for U.

Again, I've noticed for the second beam doing what I did in my last post and dividing by 3 gives the correct solution.

The bending moment diagram for the second beam is similar to the one below:

​

Just as in the case of the cantilever beam, you can find a symbolic expression for M(s) for the simply-supported beam and use the square that expression for M(s) to evaluate the integral which gives U. In this case, it might be easier to split the integration interval 0 ≤ s ≤ L into two parts.