How to calculate deflection of a cantilever beam subject to point loading

Question

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.

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 $EI\dfrac{dy}{dx} = -\dfrac{Px^2}{2} = -\dfrac{10\cdot2^2}{2} = -\dfrac{40}{EI}$.

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):

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