I have the following cantilever beam problem where I need to find the deflection of the tip at P. Sketch of the situation

I tried to compute the deflection of the beam using Castignliano' s theorem as described in this similar thread. Where my moment is defined as $$ M(x) = w_1 \cdot x \cdot (L_1 - \frac{x}{2}) + w_2 \cdot x \cdot (L_1 + L_2 -\frac{x}{2}) + w_3\cdot x \cdot (L_1+L_2+L_3 - \frac{x}{2}) + P \cdot (L_1+L_2+L_3-x) $$ Then I defined:

$$ EI(x) = EI_1 \cdot H(L_1-x) + EI_2 \cdot H(x-L_1) \cdot H(L_1-L_2 - x) + EI_3 \cdot H(x-L_1 -L_2) $$ Next I calculate the deflection using Castignliano's theorem: $$ \delta_j = \int_0^L \frac{M(x)}{EI(x)} \frac{\partial M(x)}{\partial P_j}dx $$ However, when I apply this method the integral I obtain increases and then it starts to decrease, which it should not. If my formula is indeed correct (which I am ot sure of) I think the probem has to do with the integral bounds I use. Should I use $$ \delta_{P} = \int_0^{L_1+L_2+L_3} \frac{M(x)}{EI(x)} \frac{\partial M(x)}{\partial P}dx $$ Or should I split the integral up?


2 Answers 2


The equation for the moments is not correct. you need to create three different equations for moment for each length. So assuming x starts from the support at the left hand side, then M(x) should be given by a piecewise equation:

$$M(x) = \begin{cases} ... & 0<x<L_1 \\ -w_2 \cdot \frac{(L_1+L_2- x)^2}{2} - w_3\cdot L_3(L1+L2-x+\frac{L_3}{2}) & L_1<x<L_1+L_2 \\ -w_3 \cdot \frac{(L_1+L_2+L_3 - x)^2}{2} & L_1+L_2<x<L_1+L_2+L_3 \\ \end{cases}$$

I've written the last two. Verify their validity (if the sign is opposite then it's just a matter of convention). I'll leave to you the first one, as an exercise.

(The way you had your equation set up, $w_1$ affected the moment on the last segment, which is not correct. )

As you can imagive, since you have a piecewise equation for moments, it is only natural that the integral has to be split in the same regions.

$$\int_{0}^{L_1}... dx +\int_{L_1}^{L_1+L_2}\ldots dx+\int_{L_1+L_2}^{L_1+L_2+L_3}\ldots dx$$

  • $\begingroup$ Thank you for your reply. If I use your logic then the missing line should be: $$-w_1 \cdot \frac{(L_1-x)^2}{2}- w_2\cdot L_2 \cdot (L_1 -x + \frac{L_2}{2}) - w_3 \cdot L_3 \cdot (L_1 + L_2 - x +\frac{L_3}{2})$$ correct? Furthermore, if I apply this method how would we find the deflection of the tip at P then since it is not inlcuded in the moment equation. Is the statement for P correct and can it simply be added to the last line? $\endgroup$ Commented Jan 30, 2021 at 13:32
  • $\begingroup$ You were right about the P (I completely missed it). However the third term does not look correct. $\endgroup$
    – NMech
    Commented Jan 31, 2021 at 19:36
  • $\begingroup$ So the term I provide for the missing line is incorrect... Did you derive the second and third line using a free body diagram? I only followed what seemed like the logic you used. But apparently, I do not comprehend the pattern. $\endgroup$ Commented Feb 1, 2021 at 20:10

As others have pointed out, you need to consider the deflection in each segment due to the force over it, and the effect of forces carried over from the other segment(s). I think it is easier to draw the free body diagram, and work from the free end towards the fixed end. You can write the equation of deflection for each interval, then sum all the deflections, which represents the total deflection of the free end.

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