I have following example (the question is at the bottom):

enter image description here

The load distribution $q$ is constant. I add condition of zero bend in center of the beam, $w_B = 0$, and the assumption $F_{RA} = F_{RC}$. So I have:

$$F_{RA} + F_{RB} + F_{RC} = q\cdot 2 L$$ $$F_{RB} = 2 q L-2 F_{RA}$$

Divide beam to two sections (from left)

I. $x_1\in\{ 0;L\}$

$$M(x_1) = F_{RA} \cdot x_1 - q \frac{x_1^2}{2}$$ $$\frac{\partial M(x_1)}{\partial F_{RB}}=0$$

II. $x_2 \in\{0;L\}$

$$M(x_2) = F_{RA} (L + x_2) + F_{RB}\cdot x_2 - q \frac{(L + x_2)^2}{2}$$ $$\frac{\partial M(x_2)}{\partial F_{RB}}=x_2$$

Use condition $w_B = 0$ $$w_b = 0 = \frac{1}{E\cdot I_z} \left ( \int_0^L M(x1) \cdot \frac{\partial M(x_1)}{\partial F_{RB}} dx_1 + \int_0^L M(x2) \cdot \frac{\partial M(x_2)}{\partial F_{RB}} dx_2 \right )$$

$$0 = \int_0^L M(x2) \cdot \frac{\partial M(x_2)}{\partial F_{RB}} dx_2 = \\ \int_0^L \left ( F_{RA} (l + x_2) + F_{RB}\cdot x_2 - q \frac{(l + x_2)^2}{2} \right ) \cdot x_2 dx_2 $$

Substitute for $F_{RB}$

$$\int_0^L \left ( F_{RA} (l + x_2) + (2 q L-2 F_{RA})\cdot x_2 - q \frac{(l + x_2)^2}{2} \right ) \cdot x_2 dx_2 $$

Solve for $F_{RA}$

$$F_{RA} = \frac{1}{4} q l$$

It's this result correct? It seems suspicious to me. I supposed $F_{RB}\geq F_{RA}.$

  • 1
    $\begingroup$ AISC Steel Construction Manual Tables 3-22c and 3-23 (14th Edition) are extremely useful for gut-checks of this type. slideshow available here $\endgroup$
    – CableStay
    Commented Aug 19, 2015 at 21:18

2 Answers 2


This problem can be simplified by observing the symmetry. This allows us to solve only one side, considering it as a beam of span $L$ which is fixed and pinned:

enter image description here

Now, reactions are really just concentrated loads, so lets replace that pinned support with a concentrated vertical load $P$. To obtain $P$, we first take a look at how the beam deforms without the pinned support (i.e. just the cantilever). A simple cantilever has a total vertical displacement at the end equal to $d_q = \dfrac{q\cdot L^4}{8EI}$. Now, I lied. A reaction isn't just a concentrated load. It's also knowledge of the true displacement. We know that the end of the cantilever actually has a displacement equal to 0. Therefore, the concentrated load $P$ has to create a displacement that cancels out the value we calculated above. Now, a cantilever with a concentrated load at its extremity has a vertical displacement equal to $d_P = \dfrac{P\cdot L^3}{3EI}$.

Now, to cancel out, we need $d_q = d_P$. Therefore, we have

$\dfrac{q\cdot L^4}{8EI} = \dfrac{P\cdot L^3}{3EI} \therefore P = \dfrac{3qL}{8}$

enter image description here

So $F_{RA} = F_{RC} =\dfrac{3qL}{8}$ and $F_{RB} = 2\cdot\dfrac{5qL}{8}$ (it's times two, once for each side of the real beam).

It anyone's interested, the figures were created with Ftool, a free 2D analysis tool.


I can see the partial derivative, $\frac{\partial M(x_1)}{\partial F_{RB}}=0$ on your X1 section is actually not correct, it should be " = x".

An easy hand calculating method of these simple problems is using moment distribution method. Using symmetry we start from left hand span and assume fix ends initially then release them.

RA = RB ql/2 and MA = MB = ql^2/12.
now we release the joint A half of the moment is carried to joint be so now the MB = ql^2/12 + ql^/24 = 3ql^2/24 = ql^2/8.
Now we calculate final RA after moment distribution: RA = ql/2 - ql/8 = 3ql/8.
We set the RC = 3ql/8 too by symmetry. Now RB = 2ql - 6ql/8 = 10ql/8. 10/8 ql = 5/4 ql.


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