For a simply supported beam with a UDL if the deflection is assumed to be:


What is the formula for calculating the slope?

I thought it would be:


However, working through examples where I calculate the values for the constants this does not appear to be the case. At the midpoint of the beam when I should be getting zero deflection, the value I get using dV/dx is completely wrong.

How do I obtain θ?

  • $\begingroup$ Why do you expect an approximation method to produce exact results? Please show how you set up the problem, and what boundary conditions you chose to use to solve for the coefficients. Since the problem is overly constrained, there are a lot of possible solutions you could have come up with. $\endgroup$ – Phil Sweet Apr 18 at 0:49

The Euler-Bernoulli beam equation is as follows:

$$q = \dfrac{\partial^2}{\partial x^2}\left(EI\dfrac{\partial^2 \delta}{\partial x^2}\right)$$

where $q$ is the distributed load along the beam, $\delta$ is the deflection, and $EI$ is the beam's stiffness (assumed constant). So, the deflection is the fourth integral of the applied load. From this we can derive that, for the case of a uniform load along the entire beam of uniform stiffness:

$$\begin{align} Q &= \int q\text{d}x \\ &= qx + Q_0 \\ M &= \int Q\text{d}x \\ &= \dfrac{1}{2}qx^2 + Q_0x + M_0 \\ \theta &= \dfrac{1}{EI}\int M\text{d}x \\ &= \dfrac{1}{6}qx^3 + \dfrac{1}{2}Q_0x^2 + M_0x + \theta_0 \\ \delta &= \int \theta\text{d}x \\ &= \dfrac{1}{24}qx^4 + \dfrac{1}{6}Q_0x^3 + \dfrac{1}{2}M_0x^2 + \theta_0x + \delta_0 \\ \end{align}$$

where $q$ is the constant uniform load and $Q_0$, $M_0$, $\theta_0$ and $\delta_0$ are the shear force, bending moment, rotation and deflection at the beam's origin, respectively.

So, the problem you're having is that your initial equation is incorrect: a beam under a UDL's deflection is a quartic function. And for small deflections and rotations, you are correct: we can approximate the rotations (in radians) as equal to the tangent of the deflection, i.e. its derivative.

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  • $\begingroup$ +1, but I don't think this is what he is asking. The Rayleigh-Ritz method is an approximation scheme and you can choose the number of terms in the basis set. So you could, for example, Find a cubic that has work and energy balanced, and has some boundary conditions satisfied - x0 = 0, X1 = 0, and 1 more of your choosing, maybe equal slope at the endpoints. Or you could choose to use midpoint slope = 0 for your third one if you wanted. Until we know how he constructed the approximation, we can't answer his question. $\endgroup$ – Phil Sweet Apr 18 at 0:30
  • $\begingroup$ As an approximation method, there is no reason to expect the slope to be zero at the midpoint unless that was part of the construct. $\endgroup$ – Phil Sweet Apr 18 at 0:30
  • $\begingroup$ If you use a sensible method to find the constants, the OP's cubic function should give quadratic shape (d = 0) which is symmetrical about the mid point. But it you don't use a sensible method, you might get anything. $\endgroup$ – alephzero Apr 18 at 1:09

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