# Bending equation and max deflection of UDL and multiple point loads

I have a beam, simply supported with 2 points loads and 2 UDLs. The loads are symmetric about the centre of the beam.

I've tried to derive an expression for the moment along the beam and then via 2 integrations obtained expressions for slope and deflection respectively.

Could someone please let me know whether I'm close with what I have or help me to create an overall moment equation, which i can plot in MathCAD preferably, and also help me with finding the maximum deflection of the beam? Attached is the FBD for the problem.

Below is my attempt at a solution for deflection (In the case where x = L).

Any help is much appreciated.

• Can we assume the loading is symmetrical? At a glance, it seems that the distance the second UDL covers past the right-most nodal force is smaller than the distance B marked on the other side.
– Wasabi
Aug 13, 2018 at 20:08
• Hey Wasabi - we can indeed assume that the loading is perfectly symmetric, my layout was a bit rushed...thanks Aug 13, 2018 at 21:05

Note: I'll be using the following notation:

• $q_1, q_2$ are the distributed loads (where you use $w$, which I reserve for deflections)
• $L = c + 2(a+b)$, the total length of the span
• $\ell = a + b$

First let's get the reactions. Since the structure and the loading is symmetrical, this is trivial, with each support getting half of each load.

$$R = -\dfrac{q_1(L-2a)}{2} - \dfrac{q_2L}{2} - P$$

Let's then use singularity functions to describe the loading on the beam and integrate our way up to the deflection:

\begin{align} \newcommand{\a}[1]{\langle #1 \rangle} q &= R\a{x}^{-1} + q_1\a{x-a}^0 + q_2\a{x}^0 + P\a{x-\ell}^{-1} \\ Q &= R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 + P\a{x-\ell}^0 \\ M &= R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} + P\a{x-\ell}^1 \\ EI\theta &= \dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + \dfrac{P\a{x-\ell}^2}{2} + C_1 \\ EIw &= \dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + \dfrac{P\a{x-\ell}^3}{6} + C_1x + C_2 \\ \end{align}

Now let's use the boundary conditions to find those constants.

\begin{align} \theta\left(\dfrac{L}{2}\right) = EI\theta ={}& \dfrac{R}{2}\left(\dfrac{L}{2}\right)^2 + \dfrac{q_1}{6}\left(\dfrac{L}{2}-a\right)^3 + \dfrac{q_2}{6}\left(\dfrac{L}{2}\right)^3 + \dfrac{P}{2}\left(\dfrac{L}{2}-\ell\right)^2 + C_1 = 0 \\ \therefore C_1 ={}& q_1\left(\dfrac{(L-2a)}{4}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{6}\left(\dfrac{L}{2}-a\right)^3\right) \\ &+ q_2\left(\dfrac{L}{4}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{6}\left(\dfrac{L}{2}\right)^3\right) \\ &+ P\left(\dfrac{1}{2}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{2}\left(\dfrac{L}{2}-\ell\right)^2\right) \\ ={}& q_1\left(\dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\ &+ \dfrac{q_2L^3}{24} \\ &+ \dfrac{P\ell(L-\ell)}{2} \\ w(0) = C_2 ={}& 0 \end{align}

We therefore have the full description of the entire beam. Personally, I prefer splitting the functions up for each beam segment. Therefore, we have:

For $x\in[0,a]$:

\begin{align} Q ={}& R\a{x}^0 + q_2\a{x}^1 \\ ={}& \dfrac{q_1(2a - L)}{2} \\ &+ q_2\left(x - \dfrac{L}{2}\right) \\ &- P \\ M ={}& R\a{x}^1 + \dfrac{q_2\a{x}^2}{2} \\ ={}&\dfrac{q_1(2a - L)}{2}x \\ &+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right) \\ &- Px \\ EI\theta ={}& \dfrac{R\a{x}^2}{2} + \dfrac{q_2\a{x}^3}{6} + C_1 \\ ={}&q_1\left(\dfrac{2a - L}{4}x^2 + \dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\ &+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{L^3}{24}\right) \\ &+ P\left(-\dfrac{1}{2}x^2 + \dfrac{\ell(L-\ell)}{2}\right) \\ EIw ={}& \dfrac{R\a{x}^3}{6} + \dfrac{q_2\a{x}^4}{24} + C_1x \\ ={}&q_1\left(\dfrac{2a - L}{12}x^3 + \left(\dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right)x\right) \\ &+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\ &+ P\left(-\dfrac{1}{6}x^3 + \dfrac{\ell}{2}(L-\ell)x\right) \end{align}

For $x\in(a,\ell]$:

\begin{align} Q ={}& R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 \\ ={}&q_1\left(x - \dfrac{L}{2}\right) \\ &+ q_2\left(x - \dfrac{L}{2}\right) \\ &- P \\ M ={}&R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} \\ ={}&q_1\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x + \dfrac{a^2}{2}\right) \\ &+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right) \\ &- Px \\ EI\theta ={}&\dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + C_1 \\ ={}&q_1\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{a^2}{2}x - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\ &+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{L^3}{24}\right) \\ &+ P\left(-\dfrac{1}{2}x^2 + \dfrac{\ell(L-\ell)}{2}\right) \\ EIw ={}&\dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + C_1x \\ ={}&q_1\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{a^2}{4}x^2 + \left(\dfrac{L^3}{24} - \dfrac{a^2L}{4}\right)x + \dfrac{a^4}{24}\right) \\ &+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\ &+ P\left(-\dfrac{1}{6}x^3 + \dfrac{\ell}{2}(L-\ell)x\right) \end{align}

And finally, for $x\in(\ell,\frac{L}{2}]$:

\begin{align} Q ={}& R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 + P\a{x-\ell}^0 \\ ={}&q_1\left(x - \dfrac{L}{2}\right) \\ &+ q_2\left(x - \dfrac{L}{2}\right) \\ &+ 0 \\ M ={}&R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} + P\a{x-\ell}^1 \\ ={}&q_1\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x + \dfrac{a^2}{2}\right) \\ &+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right) \\ &-P\ell \\ EI\theta ={}&\dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + \dfrac{P\a{x-\ell}^2}{2} + C_1 \\ ={}&q_1\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{a^2}{2}x - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\ &+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{q_2L^3}{24}\right) \\ &+ P\left(-\ell x + \dfrac{L\ell}{2}\right) \\ EIw ={}&\dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + \dfrac{P\a{x-\ell}^3}{6} + C_1x \\ ={}&q_1\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{a^2}{4}x^2 + \left(\dfrac{L^3}{24} - \dfrac{a^2L}{4}\right)x + \dfrac{a^4}{24}\right) \\ &+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\ &+ P\left(-\dfrac{\ell}{2}x^2 + \dfrac{L\ell}{2}x - \dfrac{\ell^3}{6}\right) \end{align}

• Grat for investing so much work into the answer. Aug 14, 2018 at 23:27
• Hi Wasabi - Thanks very much for your effort here - it's much appreciated. I used the exact same approach to get the equation for the beam and integrated using singularity functions. Somehow I came to a different answer! I'll check my algebra...Thanks again! Aug 15, 2018 at 15:05
• Thanks Wasabi - I've checked them and they all work out. You've put a lot of (tedious!) work into this so thanks very much for that :) Aug 15, 2018 at 15:16
• Smart! I’d not considered using a half model-does that mean I have to double any result I get from the moment slope and deflection equations you stated? Aug 15, 2018 at 19:04
• Of course, that makes sense! It’s been a long couple of weeks-thanks for your help and advice Wasabi! Aug 15, 2018 at 19:35