I have a continuous beam across a center column (not sure whether I've draw it correctly or not)

enter image description here

The Displacement and force diagram is as shown:

enter image description here

The first diagram is the bending moment, second the axial force, the third is the displacement.

Now, what is the boundary condition at $R_A$, $R_B$ and $R_C$?

From what I can infer, it seems to be

$w(0)=w(L)=w(2L)=0$ (corresponds to the deflection at the three support)

$M(0)=M(2L)=0$, or $\frac{d^2w(0)}{dx^2}=\frac{d^2w(2L)}{dx^2}=0$ (corresponds to the moment).

But I suspect that I am still missing some boundary conditions in order to derive the complete displacement/force diagram for the continuous beam. Are there any boundary conditions that I've missed?


Well you are missing the compatibility of slopes at the middle support:

$$\frac{dw(L^-)}{dx} = \frac{dw(L^+)}{dx}$$

In the case of symmetric geometry & loading, the slope of the beam at the middle support will be zero.

Since the bending moment has no derivative at x=L, you will need to derive the deflections of the two halves separately and 'join' them with compatibility.

Update: derivation of beam deflection formula:

Starting from the Euler-Bernoulli beam equation (assuming constant EI), and taking x from the outer supports towards the center: $$ q = EI \frac{dw^4}{dx^4} $$

Integrating four times:

$$ EI \frac{d^3 w}{dx^3} = q x + A $$ $$ EI \frac{d^2 w}{dx^2} = q \frac{x^2}{2} + Ax +B $$ $$ EI \frac{d w}{dx} = q \frac{x^3}{6} + A\frac{x^2}{2} + Bx + C $$ $$ EI w = q \frac{x^4}{24} + A\frac{x^3}{6} + B\frac{x^2}{2} +Cx + D$$

Noticing the problem is symmetric the boundary conditions are: $$ w(0)=w(L) = 0$$ $$ \frac{d w(L)}{dx} = 0 $$ $$ \frac{d^2 w(0)}{dx^2} = 0 $$

Therefore we can immediately see that: $ B = D = 0$

We now have two equations with two unknowns (A,C). Solving we find: $$ A = -\frac{3 L q}{8}$$ $$ C = \frac{L^3 q}{48}$$

We can now substitue all the constants back into the equation for w. Simplifying results in:

$$ w = \frac{q x}{48EI} (L-x)^2 (L+2 x) $$

Which is identical to the result referenced here (note their coordinate system has x=0 at the centre). Also notice how this is exactly the same result as a propped cantilever. This is due to symmetry, meaning that the beam slope at the center is zero (which is the same boundary condition as a cantilever support).

You can also substitute into the bending moment equation:

$$ M = EI \frac{d^2 w}{dx^2} = \frac{1}{8} q x (4 x-3 L) $$

| improve this answer | |
  • $\begingroup$ a) the compatibility of slope condition is only helpful if one can model the continuous beam as two span beams. How is it helpful in this case? b) Why do you say that the bending moment is discontinuous at $L$? the moment diagram in my question clearly shows that it is continuous. $\endgroup$ – Graviton Aug 5 '16 at 9:08
  • $\begingroup$ Would be grateful if you can elaborate a bit and if you can show how your boundary conditions can lead to the displacement/bending moment diagram for continuous beam. $\endgroup$ – Graviton Aug 5 '16 at 9:09
  • 1
    $\begingroup$ @Graviton, a) I will update the question with the derivation. b) you are right, I was being a bit loose with my mathematical terminology. What I mean is the bending moment has no derivative at x=L. $\endgroup$ – atom44 Aug 5 '16 at 10:53

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