Here's where I'm struggling

enter image description here

So I've tried to find the expression for deflection of a simply supported beam acted on by 4 point loads at varying distances.

First I balance out the vertical forces and then I balance out the moment about A. These two equations will get me the reaction forces on edges A and B.

Then I used the double integration method to find the expression for deflection of beam for a section x, 'x' distance away. The two constants can be found using the boundary condition that deflections at both edges (x=0 and x=L) is zero.

I've only used a check mark instead of finding the expression because I just want you guys to check my methodology because I seem to be making a mistake somewhere


Here I've used the same general expression to find the deflection for a single point load 'w' acting on the center of the beam. Found the constant C1 using the condition that deflection is zero (y=0)at x=L.

The expression I get at the end is different from the expression I see mentioned in books or online. It's supposed to be 48 in the denominator and not 32.

If I go through each step it makes sense to me. But the answer is wrong. So I definitely have a fundamental misunderstanding of the method. I would appreciate it if someone could point this out.

All of this has to do with a larger problem I was working on but I noticed something in that problem which made me go back to all these basic derivations and notice that I've been doing something wrong.

Also I'm new to stackechange so I apologise if I messed up the format of the post or didn't conform to the rules of the forum.


I am writing this as an answer to the comment about the superposition method. I started writing it in my original answer, but it was too long and confusing, so I opted to do another answer.

Superposition method.

Let's assume you have this problem.

enter image description here

In that case you only need the following equation.

Formulas for superposition

Essentially what you do is you apply the above equation when only $P_1$ is applied and you apply it again when only $P_2$ is applied.


so starting with the case:

enter image description here

so the application for this problems yields the deflection $y_1$ when only $P_1$ is applied.

Section $0<x<\frac L 2$ $\frac L 2<x<L$
$y_1$ $$\frac{Pbx}{6\cdot l\cdot E\cdot I} \left( L^2-x^2-b^2\right)$$ $$\frac{Pb}{6\cdot l\cdot E\cdot I} \left( \frac{L}{b}(x-a)^3-(L^2-b^2)x -x^3\right)$$

you can further substitute $a=\frac{L}{2}$, and $b=L-\frac{L}{2}$ and simplify

Section $0<x<\frac L 2$ $\frac L 2<x<L$
$y_1$ $$\frac{Pbx}{6\cdot l\cdot E\cdot I} \left( L^2-x^2-b^2\right)$$ $$\frac{Pb}{6\cdot l\cdot E\cdot I} \left( \frac{L}{b}(x-a)^3-(L^2-b^2)x -x^3\right)$$
$y_1$ $$\frac{P\left(L-\frac{L}{2}\right)x}{6\cdot L\cdot E\cdot I} \left( L^2-x^2-\left(L-\frac{L}{2}\right)^2\right)$$ $$\frac{P\left(L-\frac{L}{2}\right)}{6\cdot L\cdot E\cdot I} \left( \frac{L}{\left(L-\frac{L}{2}\right)}\left(x-\frac{L}{2}\right)^3-\left(L^2-\left(L-\frac{L}{2}\right)^2\right)x -x^3\right)$$
$y_1$ $$\frac{P\left(\frac{L}{2}\right)x}{6\cdot L\cdot E\cdot I} \left( L^2-x^2-\left(\frac{L}{2}\right)^2\right)$$ $$\frac{P\left(\frac{L}{2}\right)}{6\cdot L\cdot E\cdot I} \left( \frac{L}{\left(\frac{L}{2}\right)}\left(x-\frac{L}{2}\right)^3-\left(L^2-\frac{L^2}{4}\right)x -x^3\right)$$
$y_1$ $$\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)$$ $$\frac{P}{12\cdot E\cdot I} \left( 2\left(x-\frac{L}{2}\right)^3-\frac{3L^2}{4}x -x^3\right)$$


Similarly you can apply for the second problem when only $P_2$ is applied. Now lets denote $y_2$ as the deflection of problem 2.

enter image description here

Section $0<x<\frac L 4$ $\frac L 4<x<L$
$y_2$ $$\frac{Pbx}{6\cdot l\cdot E\cdot I} \left( L^2-x^2-b^2\right)$$ $$\frac{Pb}{6\cdot l\cdot E\cdot I} \left( \frac{L}{b}(x-a)^3-(L^2-b^2)x -x^3\right)$$

if you check the equation is exactly the same. The only difference is that $a=\frac{L}{4}$, and $b=L-\frac{L}{4}= \frac{3L}{4}$. So substituting a and b and simplify ( I won't do it as thourougly)

Section $0<x<\frac L 4$ $\frac L 4<x<L$
$y_2$ $$\frac{Pbx}{6\cdot l\cdot E\cdot I} \left( L^2-x^2-b^2\right)$$ $$\frac{Pb}{6\cdot L\cdot E\cdot I} \left( \frac{L}{b}(x-a)^3-(L^2-b^2)x -x^3\right)$$
$y_2$ $$\frac{P\left(L-\frac{L}{4}\right)x}{6\cdot L\cdot E\cdot I} \left( L^2-x^2-\left(L-\frac{3L}{4}\right)^2\right)$$ $$\frac{P\left(L-\frac{L}{2}\right)}{6\cdot L\cdot E\cdot I} \left( \frac{L}{L-\frac{L}{4}}\left(x-\frac{L}{4}\right)^3-\left(L^2-\left(L-\frac{L}{4}\right)^2\right)x -x^3\right)$$
$y_2$ (simpified) $$ \frac{P x }{8 \text{EI}}\left(\frac{7 L^2}{16}-x^2\right)$$ $$-\frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right)$$

putting it all together

The total deflection will be equal to $y_1$ and $y_2$. However there is a problem the a and b are different. To be clear I will rewrite the final forms of $y_1$ and $y_2$

Section $0<x<\frac L 2$ $\frac L 2<x<L$
$y_1$ $$\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)$$ $$\frac{P}{12\cdot E\cdot I} \left( 2\left(x-\frac{L}{2}\right)^3-\frac{3L^2}{4}x -x^3\right)$$


Section $0<x<\frac L 4$ $\frac L 4<x<L$
$y_2$ $$ \frac{P x }{8 \text{EI}}\left(\frac{7 L^2}{16}-x^2\right)$$ $$-\frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right)$$

So the solution is to break up the beam into three sections.

  • Leftmost section $0<x<\frac{L}{4}$

in that region the deflection will be equal to:

$$y_{tot}(x) = \color{green}{y_1(x)}+ y_2(x) \qquad\text{ when } 0<x<\frac{L}{4}$$

$$y_{tot}(x) = \color{green}{\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)}+ \frac{P x }{8 \text{EI}}\left(\frac{7 L^2}{16}-x^2\right) \qquad\text{ when } 0<x<\frac{L}{4}$$

  • Middle section $\frac{L}{4}<x\frac{L}{2}$

$$y_{tot}(x) = \color{green}{y_1(x)}+ y_2(x) \qquad\text{ when } \frac{L}{4}<x\frac{L}{2}$$

$$y_{tot}(x) = \color{green}{\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)}+ \frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right) \qquad\text{ when } \frac{L}{4}<x<\frac{L}{2}$$

  • Rightmost section $\frac{L}{2}<x<L$ In the rightmost section the total deflection will be.

$$y_{tot}(x) =\color{green}{y_1(x)}+ y_2(x) \qquad\text{ when } \frac{L}{2}<x<L$$

$$y_{tot}(x) = \color{green}{\frac{P}{12\cdot E\cdot I} \left( 2\left(x-\frac{L}{2}\right)^3-\frac{3L^2}{4}x -x^3\right)} + \frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right) \qquad\text{ when } \frac{L}{2}<x<L$$

Or in another form:

$$y_{tot}(x)= \begin{cases} \color{green}{\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)}+ \frac{P x }{8 \text{EI}}\left(\frac{7 L^2}{16}-x^2\right) \qquad\text{ when } 0<x<\frac{L}{4}\\ \color{green}{\frac{Px}{12\cdot E\cdot I} \left( \frac{3L^2}{4}-x^2\right)}+ \frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right) \qquad\text{ when } \frac{L}{4}<x<\frac{L}{2}\\ \color{green}{\frac{P}{12\cdot E\cdot I} \left( 2\left(x-\frac{L}{2}\right)^3-\frac{3L^2}{4}x -x^3\right)} + \frac{P }{384 \text{EI}}\left(L^3+9 L^2 x+48 L x^2-16 x^3\right) \qquad\text{ when } \frac{L}{2}<x<L \end{cases}$$

final Thoughts

Although there is a lot of computation, here its much easier to repeat and program into a software. You don't need to think too much, you just need to be careful in the calculations.

You can also do the same with the deflection (although a better way IMHO, is to get the deflections and then differentiate.

  • $\begingroup$ @Wren, I found some mistakes in the last equation. Hopefully it will make more sense now. $\endgroup$
    – NMech
    Jul 20 at 12:54
  • $\begingroup$ the mistakes are fine I wasn't really looking for the mathematics of it anyway. This is what I was going to do in the beginning but purely out of intuition. But then disregarded it. I have a doubt though. Since this is a point load and you've defined deflection as a function of x for the two sections on either side of the point load, what is the deflection directly under it? Edit: I'm dumb. I just realised. It's the same. That's how the continuity conditions work. Both expressions would be the same. Sorry. Thanks for this. This reduced the amoutn of work I have to do for the problem by a LOT. $\endgroup$
    – Wren
    Jul 22 at 5:09
  • $\begingroup$ If its done correctly, it shouldn't matter if one use the left or the right. Both functions should yield the same value. Its actually a good way to check for errors. $\endgroup$
    – NMech
    Jul 22 at 5:17

TL;DR (Serious): The problem is that you don't into account that the bending moment equation changes at different section of the beam.

TL;DR: (Tongue in cheek) Don't do this unless you a) really really don't have anything better to do, or b) you are getting paid to do it, c) you really like this sort of problems. :-P

(I wasn't getting paid to do it, so I only did the simple case on a Saturday afternoon (when the rest of the family is asleep). See how long it was. I also feel really lucky to get it right first time without any textbook at hand - just the online tables, for validation of my results. To be honest I wouldn't touch your original problem with a polestick).

As I said I'll just do this case of the simply supported with a load in the middle:

enter image description here

IMHO where you failed to apply the methodology is that you need to partition the beam at different sections and consider the different bending equation in each partition. In the example above, the bending moments on each region are:

Section $0<x<\frac L 2$ $\frac L 2<x< L$
M(x) $\frac{w}{2}\cdot x $ $$\frac{w}{2}\cdot x - w\cdot (x -\frac{L}{2})=\frac{w}{2}(L- x)$$

calculate slope

You can then use those bending equations to (Eventually) find out the slope of the beam. In order to do that you need to integrate each section separately. I.e. integrate the first region $0<x<\frac{L}{2}$. You need to calculate up to x=L/2, because you need to find the slope at the two supports (the beam will have a non zero slope there, and only at x=L/2 is equal to zero).

$$EI\dfrac{d^2y}{dx^2} = M(x)$$ $$\int _0^{\frac{L}{2}}EI\dfrac{d^2y}{dx^2} dx= \int_0^{\frac{L}{2}} \frac{w}{2}\cdot x dx$$ $$\left. EI\dfrac{dy}{dx}\right|_0^{L/2}= \int_0^{\frac{L}{2}} \frac{w}{2}\cdot x dx$$ $$ EI\left(\dfrac{dy(x=L/2)}{dx}-\dfrac{dy(x=0)}{dx}\right)= \int_0^{\frac{L}{2}} \frac{w}{2}\cdot x dx$$

where $\dfrac{dy(x=L/2)}{dx}$ is the slope at x = L/2. However due to the symmetry in this problem $\left.\dfrac{dy}{dx}\right|_{L/2} =0$. Therefore:

$$-EI\dfrac{dy(x=0)}{dx}= \int_0^{\frac{L}{2}} \frac{w}{2}\cdot x dx$$ $$-EI\dfrac{dy(x=0)}{dx}= \frac{w}{2}\left[\frac{x^2}{2}\right]_0^{\frac{L}{2}}$$ $$EI\dfrac{dy(x=0)}{dx}= -\frac{w}{2}\frac{L^2}{8}$$ $$EI\dfrac{dy(x=0)}{dx}= -\frac{wL^2}{16}$$

Therefore the slope at point x=0 (and at x=L) is equal to

$$\dfrac{dy(x=0)}{dx} = \dfrac{dy(x=L)}{dx}= -\frac{wL^2}{16EI}$$

The slope at the end is equivalent to $C_i$ constants that you calculate with the indefinite integral. To me its a bit clearer to use the definite integral (I don't do this everyday, so I'm a bit rusty)

The general equation for the slope for any x from 0 to L/2 will be equal to :

$$\int _0^{x}EI\dfrac{d^2y}{dx^2} dx= \int_0^x \frac{w}{2}\cdot x\;dx$$ $$\int _0^{x}\dfrac{d^2y}{dx^2} dx= \frac{w}{2EI} \int_0^x\cdot x dx$$ $$\left. \dfrac{dy}{dx} \right|_0^x= \frac{w}{2EI} \left[\frac{x^2}{2}\right]_0^x$$ $$\dfrac{dy(x)}{dx} - \dfrac{dy(0)}{dx}= \frac{L=w}{2EI} \left[\frac{x^2}{2}\right]_0^x$$ $$\dfrac{dy(x)}{dx} = \frac{w}{2EI} \left[\frac{x^2}{2}\right]_0^x + \dfrac{dy(0)}{dx}$$ $$\dfrac{dy(x)}{dx} = \frac{w}{4EI} x^2-\frac{wL^2}{16EI}$$ $$\dfrac{dy(x)}{dx} = \frac{w}{16EI} \left(4x^2 - L^2\right)$$

You can check that the above is the correct slope for 0<x<L/2 at mechanicalc or elsewhere.

To calculate the slope for $\frac{L}{2}<x<L$, you need to do the same as above starting from L/2. So:

$$\int _\frac{L}{2}^{x}EI\dfrac{d^2y}{dx^2} dx= \int_\frac{L}{2}^{x} \frac{w}{2}\cdot (L-x)dx$$ $$\int_\frac{L}{2}^{x}\dfrac{d^2y}{dx^2} dx= \frac{w}{2EI} \int_\frac{L}{2}^{x} (L-x) dx$$ $$\left. \dfrac{dy}{dx} \right|_\frac{L}{2}^{x}= \frac{w}{2EI} \left[Lx-\frac{x^2}{2}\right]_\frac{L}{2}^{x}$$ $$\dfrac{dy(x)}{dx} - \dfrac{dy(\frac{L}{2})}{dx}= \frac{w}{2EI} \left(Lx-\frac{x^2}{2} -\left(\frac{L^2}{2}-\frac{L^2}{8}\right)\right)$$ $$\dfrac{dy(x)}{dx} -0 = \frac{w}{2EI} \left(Lx-\frac{x^2}{2} -L^2\frac{3}{8}\right)$$ $$\dfrac{dy(x)}{dx} = \frac{w}{16EI} \left(8Lx-4x^2 -3L^2\right)$$

Again if you check the slope for x =L in the above equation you'll get $$\dfrac{dy(x=L)}{dx} = \frac{w}{16EI} \left(8L^2-4L^2 -3L^2\right)= \frac{w}{16EI} L^2$$

just as the tables predict.


Now that you know the slope at the two ends you are able to proceed to the deflection. To recap you know that

Section $0<x<\frac L 2$ $\frac L 2<x<L$
$\dfrac{dy(x)}{dx} $ $\frac{w}{16EI} \left(4x^2 - L^2\right)$ $$\frac{w}{16EI} \left(8Lx-4x^2 -3L^2\right)$$

So, integrating from 0 to L/2:

$$\int_0\frac{L}{2} \dfrac{dy(x)}{dx} dx = \int_0\frac{L}{2} \frac{w}{16EI} \left(4x^2 - L^2\right)dx$$ $$\left[y(x)\right]_0^\frac{L}{2} = \frac{w}{16EI} \int_0\frac{L}{2} \left(4x^2 - L^2\right) dx$$

$$\big[y(x)\big]_0^\frac{L}{2} = \frac{w}{16EI} \int_0^\frac{L}{2} \left(4x^2 - L^2\right) dx$$ $$y\left(\frac{L}{2}\right)-y(0) = \frac{w}{16EI} \left[4\frac{x^3}{3} - L^2x\right]_0^\frac{L}{2} $$

However we know that $y(0)=0$, therefore:

$$y\left(\frac{L}{2}\right) -0= \frac{w}{16EI} \left(\frac{4}{3}\frac{L^3}{8} - \frac{L^3}{2}\right) $$ $$y\left(\frac{L}{2}\right) -0= \frac{wL^3}{16EI} \left(\frac{4}{24}- \frac{1}{2}\right) =\frac{wL^3}{16EI} \left(\frac{4}{24}- \frac{12}{24}\right) $$ $$y\left(\frac{L}{2}\right) =\frac{wL^3}{16EI} \left(- \frac{8}{24}\right) =\frac{wL^3}{16EI} \left(- \frac{1}{3}\right) $$ $$y\left(\frac{L}{2}\right) =-\frac{wL^3}{48EI}$$

You can also obtain the same integrating from L/2 to L. You can do it for exercise.

Final thoughts.

Hopefully, I have explained the procedure, and how mindblowingly tedious and boring it is. Its very useful to do it once (or twice) in your life to get the points, but I advice strongly to use the tables and the superposition method.

  • $\begingroup$ Thank you for the thorough response. You're right about where my misunderstanding lies. After going through yours and r13's response I think I have a full understanding of what to do for my solution. But I don't think I recognise the tables you mentioned or the superposition method for deflection. Can you share links for the method if you don't mind? $\endgroup$
    – Wren
    Jul 20 at 5:57
  • $\begingroup$ @Wren I started writing it in my original answer, but it was too long and confusing, so I opted to do another answer. Check my other answer in the same post. $\endgroup$
    – NMech
    Jul 20 at 8:54
  • $\begingroup$ Hello again! Not really sure how to message someone directly in this forum hence commenting here. Can you let me know where you got the section of the table from? I'd like to know the deflection formulae for other support types. I found some online but they didn't seem very exhaustive. $\endgroup$
    – Wren
    Aug 4 at 3:53
  • $\begingroup$ you just need to google for "beam deflection equations", and they sift thought the results to find your particular loading (uniform, concentrated), and the supports (simply supported, fixed, cantilever etc) $\endgroup$
    – NMech
    Aug 4 at 6:54

enter image description here

$EI\dfrac {d^2_y}{dx^2} = M$


  1. Write equations for the moment in segment AB and BC as a function of x.

  2. Integrate the moment of each segment to get the slope, $EI\dfrac{dy}{dx} = EI\theta_x$

  3. Integrate the slope of each segment to get the deflection, $EIy = EI\Delta_x$

  4. Apply boundary conditions to set up equations for the constants.

  5. Solve the equations, and plug the resulting constants back to the respective equations.

Between $x = 0$ and $x = L/2$ (just before the concentrated load),

$M_{AB} = \dfrac {W}{2}x$

$EI\theta_{AB} = \dfrac {W}{4}x^2 + C_1$

$EI\Delta_{AB} = \dfrac {W}{12}x^3 + C_{1}x + C_2$

Between $x = L/2$ and $x = L$,

$M_{BC} = \dfrac {WL}{2} - \dfrac {W}{2}x$

$EI\theta_{BC} = \dfrac {WL}{2}x - \dfrac {W}{4}x^2 + C_3$

$EI\Delta_{BC} = \dfrac {WL}{4}x^2 - \dfrac {W}{12}x^3 + C_{3}x + C_4$

Apply boundary conditions:

At $x = 0$, $\Delta_{AB} = 0$

$EI\Delta_{AB,x=0} = \dfrac {W}{12}x^3 + C_{1}x + C_2 = 0$

---> $C_2 = 0$

At $x = L$, $\Delta_{BC} = 0$

$EI\Delta_{BC,x=L} = \dfrac {WL}{4}x^2 - \dfrac {W}{12}x^3 + C_{3}x + C_4 = 0$

---> $\dfrac {WL^3}{6} + C_3L + C_4 = 0$ -----(1)

At $x = L/2$, $\theta_{AB} =\theta_{BC}$ & $\Delta_{AB} = \Delta_{BC}$

$EI\theta_{AB,x=L/2} = EI\theta_{BC,x=L/2}$,

$\dfrac {W}{4}x^2 + C_1 = \dfrac {WL}{2}x - \dfrac {W}{4}x^2 + C_3$

---> $C_1 - C_3 = \dfrac{WL^2}{16}$ ----- (2)

$EI\Delta_{AB,x=L/2} = EI\Delta_{BC,x=L/2}$

$\dfrac {W}{12}x^3 + C_{1}x + C_2 = \dfrac {WL}{4}x^2 - \dfrac {W}{12}x^3 + C_{3}x + C_4$

$\dfrac{C_1L}{2} + C_2 - \dfrac{C_3L}{2} - C_4 = \dfrac {WL^3}{24}$

From previous condition, $C_2 = 0$,

---> $\dfrac{C_1L}{2} - \dfrac{C_3L}{2} - C_4 = \dfrac {WL^3}{24}$ ----- (3)

Now, there are 3 equations for the 3 unknowns, $C_1, C_3$ & $C_4$, you can solve them simultaneously.

Please check my math for mistakes.

Here is a good example for your further study. https://wethestudy.com/engineering/explaining-the-double-integration-method/

  • $\begingroup$ Thank you for the link. Comparing your solution to mien and reading through the example in the link you shared gave me exactly what I was looking for. Basically my understanding of segments, and how to use continuity/boundary conditions was non existent. $\endgroup$
    – Wren
    Jul 20 at 5:54
  • $\begingroup$ You are welcome. That's been the reason I decided to provide an answer. The important thing to learn here is its concept in utilizing calculus to solve engineering problems. Later, you will learn many more methods, then you would appreciate the efforts of the people before us in finding much simpler solutions that benefiting the engineering community immensely. $\endgroup$
    – r13
    Jul 20 at 12:00

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