Solve Castigiliano Method for Continuous Beam Problem

[Zouatine]

Homework Statement

Problem:
a continuous beam ABC simply presses, loaded with a distributed load q, the length of the beam is 2L.
EI=constant

find the reactions in the beam ,we use Castigiliano method.

Homework Equations

- U=(1/2EI)∫M^2 dx (U: Energy)
- dU/dP=d (d: displacement)

The Attempt at a Solution

first we have 3 reactions (Ra,Rb,Rc) in the supports.
∑F=0→ Ra+Rb+Rc=ql
∑M/A=0 → Rb+2Rc=3ql/2
for the moment flexing : 0≤x<L
- M(x)= -Ra*x .
L≤x<2L
M(x)=-Ra*x+Rb*(x-L)-(qL/2)*(x-L)^2
my problem is how to find the energy equation??
Thanx.

 

[haruspex]

You seem to have some sign inconsistency.
In your force balance you have Ra and Rb with the same sign, but in the torque balance they have opposite signs.
What is your sign convention?

To get the energy, you have to integrate (M²/(2EI)).dx no?

 

[Zouatine]

First thanks for your answer Sir ,
I made a mistake ∑F=0→ -Ra+Rb+Rc=ql
- the positive sense for the moment is the clockwise.
yes I know to get energy I have to integrate (M^2/(2EI)).dx , but it's complicated to integrate from 0 to L M^2/(2EI);dx with (M(x)=-Ra*x and from L to 2L M^2/(2EI) with M(x)=-Ra*x+Rb*(x-L)-(qL/2)*(x-L)^2 .
I saw in some document this formula :du/df=(1/EI)*∫M*(dM/df) dx (f: force) but I do not know how they are found that.
thanx

 

[haruspex]

this formula :du/df=(1/EI)*∫M*(dM/df) dx

That is obtained from the first formula simply by differentiating wrt f. Because the integration bounds are fixed, you can simply differentiate through the integral sign as though it wasn't there. You then use the chain rule to get from d(M²)/df to 2M dM/df.

Can you find dM/df?

 

[Zouatine]

thanx, so to find dM/df , I choose Ra so,
1- dM/dRa =-x
2- dM/dRa =-x
and after that
du/dRa=(1/EI)*∫(-x*Ra)*(-x) dx+(1/EI)*∫(-x*Ra+Rb*(x-L)-(qL/2)*(x-L)^2)*(-x) dx =0
with (0<x<L) and in second (L<x<2L).
and I find Ra in this equation, just one more question about this du/df=(1/EI)*∫M*(dM/df) dx
Mf (x) it is according to f , when we drift : we find this 2Mf(x) * dMf(x)/df
thanx sir

 

[haruspex]

Mf (x) it is according to f , when we drift : we find this 2Mf(x) * dMf(x)/df

Sorry, I don't understand your question. What is "Mf(x)"? If you mean literally M multiplied by f(x), I do not see any f that is a function of x. If you mean M as a function if f and x, that would be written M=M(f,x).

 

[Zouatine]

Sorry. My english is not good my question was how you found this :
d(M^2)/df =2M* dM/df

 

[haruspex]

Sorry. My english is not good my question was how you found this :
d(M^2)/df =2M* dM/df

It's just the chain rule of differentiation.
https://en.m.wikipedia.org/wiki/Chain_rule

 

[Zouatine]

thanx , Sir