# How to determine the fixed end moment?

I was trying to compute the fixed end moment for the beam (given below) using equilibrium equation for x, y displacement and moment but there are 4 unknown variables reaction and moment at both ends but I have 3 equations only, vertical equilibrium and moment about both ends. could someone help me to get the answer $$-Pab^2/L^2$$? ## It can't be done

Indeed, this problem is unsolvable using only equilibrium equations.

Broadly speaking, structures can be split into three categories:

• hypostatic: when there are too few supports to keep the structure stable. These are commonly called "mechanisms"
• isostatic or statically determinate: when there are exactly as many supports as necessary to stabilize the structure
• hyperstatic or statically indeterminate: when there are more supports than necessary

Here we are looking at a statically indeterminate beam. After all, if you got entirely rid of one of those supports, we'd simply have a cantilever, which is stable.

And the name "statically indeterminate" comes precisely from the fact that the structure can't be solved using only the equations of static equilibrium $$\sum F_x = \sum F_y = \sum M = 0$$.

So you're right, it can't be done.

## Well, not that easily, anyway...

So, how can we solve statically indeterminate beams? Well, that's a huge topic, so I won't even try to walk you through how to get to that specific answer.

But as a general point, one insight is needed to solve such structures, and that is to remember that we adopt the superposition principle. That is, we assume that the internal forces of a beam under loads A and B are equal to the forces due to A plus the forces due to B.

And as far as the beam is concerned, what's the difference between an applied force and a support? Well, none. The support's job is to apply forces on the beam such that the boundary conditions (no deflection or rotation at the support, for example) are satisfied. But as far as the beam is concerned, the support's just another force.

So, if a support is really just an external force and a beam's internal forces can be solved as the sum of its internal forces due to each individual external force, well, then that gives us a way to move forward!

All we have to do is start by pretending the beam only has one of those supports (let's say it's the one to the left). We can then calculate the result for this cantilever beam, which is easy enough.

But a cantilever beam dips and bends all the way to its free end. And we know the real beam actually has a fixed support on that "free end". So, all we have to do now is apply forces on that "free end" such that the cantilever ends up with zero deflection and rotation on that "free end". Once we've found those forces (actually, one vertical force and one concentrated moment), well... we've actually just found the reactions of that support we pretended wasn't there! As for the other support, its reaction is equal to what it got from pretending it was a cantilever at the start plus the result from resisting the "fake support forces".

I'll leave the actual working out as an exercise to the reader.

N.B. what I've described above is known as the Force Method. However, it is more common to use the Displacement Method. That being said, they are fundamentally similar in that they add displacement compatibility equations to the mix to solve statically indeterminate structures, and I find the Force Method to be far easier to grasp at a glance.

And then one day you'll start using analysis software (all of which actually use the Displacement Method under the hood) and forget how to do even basic math. God knows I have...

As Wasabi correctly pointed out, this can't be done using only equilibrium equations. To get the extra equations you need, you must take displacements and rotations into account. Enter Euler-Bernoulli beam theory. To make a long story short, your main equation is now $$\tag{1} \frac{d^2v}{dx^2}=\frac{M(x)}{EI}$$ And recall that a section's rotation is given by $$\theta(x)=\frac{dv}{dx}$$. To calculate the bending moment diagram, we must first "get rid" of the unknown moment at the fixed supports. You can split your problem as follows Now you can calculate the bending moments for each case, which are \begin{aligned} M_1(x)&=\begin{cases} \frac{Pb}{L}x & \text{if }x\leq a \\ \frac{Pb}{L}x-P(x-a) & \text{if } x> a \end{cases} \\ M_2(x)&=\frac{M_1}{L}x-M_1 \\ M_3(x)&=\frac{M_2}{L}x \end{aligned} Notice we just replaced the unknown moment from the fixed support for a positive moment $$M_i$$, which we want to determine.

Now let's proceed to integrate these bending moment expressions as per equation $$(1)$$.

\begin{aligned} v_1(x)&=\frac{1}{EI}\begin{cases} \frac{Pb}{6L}x^3+C_ax+D_a & \text{if }x\leq a \\ \frac{Pb}{6L}x^3-\frac{P}{6}x^3+\frac{Pa}{2}x^2+C_bx+D_b & \text{if } x> a \end{cases} \\ v_2(x)&=\frac{1}{EI}\left(\frac{M_1}{6L}x^3-\frac{M_1}{2}x^2+C_2x+D_2\right) \\ v_3(x)&=\frac{1}{EI}\left(\frac{M_2}{6L}x^3+C_3x+D_3\right) \end{aligned} Now we need boundary conditions to determine the integration constants, these conditions come from the pinned supports and from compatibility, they are $$v_1(0)=0\qquad v_1(a^-)=v_1(a^+) \qquad \theta_1(a^-)=\theta_1(a^+) \qquad v_1(L)=0 \\ v_2(0)=0 \qquad v_2(L)=0 \\ v_3(0)=0 \qquad v_3(L)=0$$ Some algebra to determine these constants and we arrive at \begin{aligned} v_1(x)&=\frac{1}{EI}\begin{cases} \frac{Pb}{6L}x^3+\frac{3PLa^2+PL^3-3PL^2a-PL^2b-Pa^3}{6L}x & \text{if }x\leq a \\ \frac{Pb}{6L}x^3-\frac{P}{6}x^3+\frac{Pa}{2}x^2+\frac{PL^3-3PL^2a-PL^2b-Pa^3}{6L}x+\frac{Pa^3}{6} & \text{if } x> a \end{cases} \\ v_2(x)&=\frac{1}{EI}\left(\frac{M_1}{6L}x^3-\frac{M_1}{2}x^2+\frac{M_1L}{3}x\right) \\ v_3(x)&=\frac{1}{EI}\left(\frac{M_2}{6L}x^3-\frac{M_2L}{6}x\right) \end{aligned}

Now that we have the displacements, we can determine the values for $$M_1$$ and $$M_2$$. For this we need to solve the following system $$\theta_1(0)+M_1\theta_2(0)+M_2\theta_3(0)=0 \\ \theta_1(L)+M_1\theta_2(L)+M_2\theta_3(L)=0$$ Again, doing some algebra leads to \begin{aligned} M_1&=\frac{Pa(L-a)^2}{L^2} \\ M_2&=-\frac{PL^2a+PL^2b+PLa^2-PL^3-Pa^3}{L^2} \end{aligned}

And finally, using that $$b=L-a$$ and $$a=L-b$$ you can simplify this to the values you were originally looking for, which are \begin{aligned} M_1&=\frac{Pab^2}{L^2} \\ M_2&=-\frac{Pa^2b}{L^2} \end{aligned}

And there you have it, the full derivation for the reactions of a fixed-fixed beam with an eccentric concentrated load. It's not overly hard, but the algebra is somewhat cumbersome at times.

Of course, you can extend this method to any loading and boundary condition, even supports at mid-span and such.