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enter image description here I'm having contradictory results while applying the first theorem of Castigliano for the cantilever beam shown in the figure above. Due to P and the restrictions on point $B$, it will have a bending moment and a shear force on this point as shown in the figure (the cartesian coordinate system considerered is also shown). In terms of these two external loads, we can easily find the distribution of moments and shear forces on any section at a distance $x$ from $B$, i.e $M = M_0 - V_0x$ and $V = V_0$. These two quantities can be then used to calculate the elastic strain energy stored in the beam due the moments and shear forces, i.e

$U = U_1+U_2=\int_{0}^{L} \frac{M^2}{2EI} \,dx+\int_{0}^{L} \frac{V^2}{2AG} \,dx=\frac{LV_0^2}{2AG}+\frac{L(3M0^2-3LM0V0+L^2V0^2)}{6EI}$.

Applying the first theorem of Castigliano, the partial derivative of $U$ in respect to $M_0$ will give the angular rotation on $B$ in the direction of $M_0$. Similarly, the partial derivative of $U$ in respect to $V_0$ will give the linear displacement on $B$ in the direction of $V_0$. But the beam is fixed on $B$, so these derivatives will be zero. The two resulting equations are, respectively, then

$\frac{L(6M_0-3LV_0)}{6EI}=0$,

$\frac{LV_0}{AG}+\frac{L(-3LM_0+2L^2V_0)}{6EI}=0$.

We can use any of these equation for determing $M_0$ in terms of $V_0$. If will do that, we will find, respectively, that

$M_0=\frac{LV_0}{2}$,

$M_0=\frac{2EIV_0}{AGL}+\frac{2LV_0}{3}$.

But these results are clearly different. What did I do wrong here? The math is correct, I checked it many times.

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2 Answers 2

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I think the equations you are using are a little bit disconnected from the cantilever beam case:

  • $\frac{\partial U}{\partial M_0} = 0$ means 0 change of rotation between the ends of the beam, so the end rotations just must be the same
  • $\frac{\partial U}{\partial V_0} = 0$ means 0 change of vertical deflections between the ends, so the end deflections must be the same

This works for cantilever only if $P$ is 0, which would make also $V_0=0$ and $M_0=0$, so there actually may not be a contradiction in your conclusions. For non-trivial cases where $P\neq 0$, this may not be possible to achieve. Even though you are not using $P$ directly, a similar conclusion may be derived from assumption $M(x) = M_0-V_0 x$.

In case of cantilever, you can use Castigliano theorem to find deflection or rotation at arbitrary location along the beam. In general, you should put a zero force or moment at that location and then use these quantities in deriving the elastic energy. Being 0, they will not affect the energy, but you will be able to use them for the derivative to establish the change of deflection or rotation from the known 0 value at B to the value at the location of interest. You should also use force and moment equilibria and the load $P$ in the calculations.

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  • $\begingroup$ I appreciate your answer, but I didn't understand it. The partial derivatives are only valid where the external loads are applied. For example, the partial derivative of $U$ in respect to $M_0$ gives the angular rotation at $B$ (where $M_0$ is applied), not the total angular rotation at $B$ + at the right end of the beam. $\endgroup$ May 20, 2023 at 19:10
  • $\begingroup$ @ThiagoOliveira "The partial derivatives are only valid where the external loads are applied" is essentially true, because zero loads are applied everywhere. If $x=0$, it does not mean $\frac{\partial f(x)}{\partial x}$ cannot be done, does it? $\endgroup$ May 20, 2023 at 20:50
  • $\begingroup$ @ThiagoOliveira Castigliano is really about changes. The strain energy does not know anything about rotations or deflections at any point, it can only encode how these will change along the beam. Just imagine, you would rotate the point B by a small angle; would that change the expression for energy? $\endgroup$ May 20, 2023 at 21:00
  • $\begingroup$ I think I found what went wrong here. For equilibrium, $V_0=-P$ and $M_0=-P*L$. So, among the three external loads $V_0$, $M_0$ and $P$, only $P$ is independent. Thus, the derivative of $U$ in respect to $V_0$ and $M_0$ assuming that they are independent variables as I did above doesn't make any sense. But, if we substitute the expressions for $V_0$ and $M_0$ in $U$ and take the derivative of $U$ in respect to $P$, we will find a value for displacement that agree with the displacement at the right end of the cantilever beam obtained with more elementary methods. $\endgroup$ May 22, 2023 at 22:38
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I think you have made mistakes in the "energy terms", see table below:

enter image description here

https://www.egr.msu.edu/classes/me471/thompson/handout/class09_2005S_Castigliano.pdf

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