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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.

enter image description here

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

enter image description here

Any help is much appreciated.

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  • $\begingroup$ 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. $\endgroup$
    – Wasabi ♦
    Aug 13, 2018 at 20:08
  • $\begingroup$ Hey Wasabi - we can indeed assume that the loading is perfectly symmetric, my layout was a bit rushed...thanks $\endgroup$
    – richyo1000
    Aug 13, 2018 at 21:05

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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
  • $\ell = a + b$

I'll also be making use of the structure and loading's symmetry, and adopting the following model:

half of beam modelled by placing Rz support at midspan

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

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

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  • $\begingroup$ Grat for investing so much work into the answer. $\endgroup$
    – peterh
    Aug 14, 2018 at 23:27
  • $\begingroup$ 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! $\endgroup$
    – richyo1000
    Aug 15, 2018 at 15:05
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    $\begingroup$ 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 :) $\endgroup$
    – richyo1000
    Aug 15, 2018 at 15:16
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    $\begingroup$ 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? $\endgroup$
    – richyo1000
    Aug 15, 2018 at 19:04
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    $\begingroup$ Of course, that makes sense! It’s been a long couple of weeks-thanks for your help and advice Wasabi! $\endgroup$
    – richyo1000
    Aug 15, 2018 at 19:35

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