Ritz solution of beam under uniform load

Choice of trial function

$w = a\sin(\pi x/L)$ gives $w(0) = w(L) = 0$, which satisfies the essential boundary conditions of a simply supported beam.

Potential energy

$$\Pi = \frac{EI}{2}\int_0^L \left(\frac{d^2w}{dx^2}\right)^2dx - q\int_0^L w\,dx$$

With $w'' = -a\left(\dfrac{\pi}{L}\right)^2\sin\dfrac{\pi x}{L}$, $\displaystyle\int_0^L\sin^2\dfrac{\pi x}{L}\,dx = \dfrac{L}{2}$ and $\displaystyle\int_0^L\sin\dfrac{\pi x}{L}\,dx = \dfrac{2L}{\pi}$:

$$\Pi = \frac{EI}{2}\,a^2\left(\frac{\pi}{L}\right)^4\frac{L}{2} - qa\,\frac{2L}{\pi}$$

Stationary condition

$$\frac{d\Pi}{da} = EI\,a\left(\frac{\pi}{L}\right)^4\frac{L}{2} - \frac{2qL}{\pi} = 0 \Rightarrow a = \frac{4qL^4}{\pi^5EI}$$

Substituting values:

$$a = \frac{4(10\,000)(4)^4}{\pi^5(1.6\times10^6)} = \frac{1.024\times10^7}{306.02\times1.6\times10^6} = 0.02091\ \text{m}$$

Mid-span values

• Deflection: $w(L/2) = a = 20.91$ mm.
• Bending moment: at $x = L/2$, the magnitude of $M = EI\,w''$ is $M = EI\,a\left(\dfrac{\pi}{L}\right)^2 = 1.6\times10^6(0.02091)(0.6169) = 20\,641$ N m.

Comparison

The deflection (the primary variable) is accurate; the moment (a second derivative) is less accurate because differentiation reduces accuracy.

Answer: $a = 20.9$ mm; $w_{max} = 20.9$ mm; $M_{max} = 20.6$ kN m.