I am trying to solve the following beam deflection problem:

The cantilever beam is subjected to the point load at C.

  1. Generate the equation for the elastic curve by using the double integration method.
  2. Find the maximum deflection and slope if L = 3 m and P = 10 kN acted at 2 m from A.

A cantilever beam is fixed on the left (point A). Force P is shown acting in the downward direction at the rightmost end of the beam (point C). The length of the beam is L.

Ans: $y_{max} = -46.67/{EI}$

For part 1, my answer is $EI\dfrac{d^2y}{dx^2} = - Px$.

For part 2, my answer is: $$\begin{gather} EI\dfrac{dy}{dx} = -\dfrac{Px^2}{2} +c_1 \\ EIy = -\dfrac{Px^3}{6} +c_1x +c2 \end{gather}$$

  • at $x= 0$, $y = 0$, so $c_2 = 0$
  • at $x = 0$ , $\dfrac{dy}{dx} = 0$, so $c_1 = 0$

So, $EIy = -\dfrac{Px^3}{6}$.

$EIy$ max occurs at $L=3$, so $EIy_{max} = -\dfrac{10\cdot3^3}{6} = -45$, but the answer is $EIy_{max} =-46.67$. What have I done wrong?

For the slope at x = 2, my answer is $EI\dfrac{dy}{dx} = -\dfrac{Px^2}{2} = -\dfrac{10\cdot2^2}{2} = -\dfrac{40}{EI}$.


1 Answer 1


Both answers are incorrect.

Your answer to (a) would imply that the bending moment is equal to zero at $x=0$ and increases linearly until $x=L$. This is only true if you put $x=0$ at the free end of the beam and $x=L$ at the fixed support, in which case you'd have to change your constraints. What you actually need is:

$$EI\dfrac{d^2y}{dx^2} = -PL + Px$$

For (b), you didn't notice that the behavior is different in such a beam. You can't simply use the equation above and be done with it. After all, this equation describes a beam under load, but the span from $x \in [2, 3]$ is not under load and therefore its beam equation is different (it follows a linear path).

The easy way to solve it is to start by finding the slope and deflection at $x=2$. For this, we can use the equation above:

$$\begin{gather} EI\dfrac{dy}{dx} = -PLx + \dfrac{Px^2}{2} + c_1 \\ \dfrac{dy}{dx}(x=0) = 0 = c_1 \\ EIy = -\dfrac{PLx}{2} + \dfrac{Px^3}{6} + c_2 \\ y(x=0) = 0 = c_2 \\ \therefore EI\dfrac{dy}{dx} = -PLx + \dfrac{Px^2}{2} \\ \therefore EIy = -\dfrac{PLx^2}{2} + \dfrac{Px^3}{6} \end{gather}$$

From which we gather that

$$\begin{align} \dfrac{dy}{dx}(2) &= -\dfrac{20}{EI} \\ y(2) &= -\dfrac{26.667}{EI} \end{align}$$

From $x = 2$ to $x = 3$, the deflection is linear, meaning that the slope is constant, so the result above for $\dfrac{dy}{dx}$ is the same at $x=3$. Knowing this, we can find the deflection at $x=3$ by finding the linear component and then adding it to $y(2)$ (where $\ell$ is the linear span):

$$\begin{align} \delta &= \dfrac{dy}{dx}(2)\cdot\ell = -\dfrac{26.667}{EI}\cdot1\\ \therefore y(3) &= y(2) + \delta = -\dfrac{46.667}{EI} \end{align}$$

  • $\begingroup$ are you sure that y(4 ) = y(2) + $$\delta$$ ? in my working ? i found that y(2) = -46.67/EI directly without adding the $$\delta$$ $\endgroup$ Commented Nov 28, 2016 at 11:16
  • 1
    $\begingroup$ @kelvinmacks: Why wouldn't it be? And in your original post you found 45, not 46.67, so I have no idea what you did. Another (equivalent) way of finding it would be to find the beam equation at $x \in [2,3]$ considering the initial conditions at $x=2$. But yes, since the deflection is linear at $x \in [2,3]$, one can just calculate the results at $x=2$ and then add the linear variation with the constant slope. And please read the MathJax tutorial for how to write math here. $\endgroup$
    – Wasabi
    Commented Nov 28, 2016 at 11:34
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    $\begingroup$ @kelvinmacks: $EI\frac{d^2y}{dx^2}$ is equal to the bending moment $M$. $M(x=0) = -PL$ and $M(x=L) = 0$ and it follows a linear diagram. That behavior is described by $EI\frac{d^2y}{dx^2} = M = -PL+Px$. As I wrote in my answer, your answer of $M = -Px$ means $M(x=0) = 0$ and $M(x=L) = -PL$, which only makes sense if you put your origin ($x=0$) at the free end. But then your constraints are incorrect since they are at $x=0$, while they'd have to be at $x=L$ for your answer to work. $\endgroup$
    – Wasabi
    Commented Nov 28, 2016 at 12:29
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    $\begingroup$ @kelvinmacks: I read your answer, I know what you did, no need to repeat it all here. The problem there is that you set $x=0$ at the load, but then you put the constraints at $x=0$ as well ("when x = 0 , y = 0", "x = 0 , dy/dx = 0"), when they would then have to be at $x=L$ instead (so $y(x=L)=0$ and $\frac{dy}{dx}(x=L) = 0$). $\endgroup$
    – Wasabi
    Commented Nov 28, 2016 at 13:15
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    $\begingroup$ @kelvinmacks: I've already told you what's wrong with your answer. Twice. Explicitly. I suggest you take some time to carefully read my answer and study it. If you can't see what is wrong after my explicit descriptions, then I honestly don't what else I can do for you. $\endgroup$
    – Wasabi
    Commented Nov 28, 2016 at 13:25

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