Ritz method for bar with distributed load

Data (units: N, mm)

$AE = 300 \times 70\times10^3 = 2.1\times10^7$ N, $L = 2000$ mm, $q = 6$ N/mm, $P = 12\,000$ N.

The trial function satisfies $u(0) = 0$.

Potential energy

$$\Pi = \frac{AE}{2}\int_0^L\left(\frac{du}{dx}\right)^2dx - q\int_0^L u\,dx - P\,u(L)$$

With $u' = a_1 + 2a_2x$:

$$\begin{aligned} \int_0^L u'^2dx &= a_1^2L + 2a_1a_2L^2 + \tfrac{4}{3}a_2^2L^3 \\ \int_0^L u\,dx &= \tfrac{1}{2}a_1L^2 + \tfrac{1}{3}a_2L^3 \\ u(L) &= a_1L + a_2L^2 \end{aligned}$$

Stationary conditions

$$\begin{aligned} \frac{\partial\Pi}{\partial a_1} &= AE\,(a_1L + a_2L^2) - \tfrac{1}{2}qL^2 - PL = 0 \\ \frac{\partial\Pi}{\partial a_2} &= AE\,(a_1L^2 + \tfrac{4}{3}a_2L^3) - \tfrac{1}{3}qL^3 - PL^2 = 0 \end{aligned}$$

Divide the first by $L$ and the second by $L^2$:

$$\begin{aligned} AE\,(a_1 + a_2L) &= \tfrac{1}{2}qL + P = 6000 + 12\,000 = 18\,000 \\ AE\,(a_1 + \tfrac{4}{3}a_2L) &= \tfrac{1}{3}qL + P = 4000 + 12\,000 = 16\,000 \end{aligned}$$

Subtract: $AE\,\tfrac{1}{3}a_2L = -2000$, so $a_2 = \dfrac{-6000}{AE\,L} = \dfrac{-6000}{4.2\times10^{10}} = -1.4286\times10^{-7}\ \text{mm}^{-1}$.

Then $a_1 = \dfrac{18\,000}{AE} - a_2L = 8.5714\times10^{-4} + 2.8571\times10^{-4} = 1.1429\times10^{-3}$.

End displacement

$$u(L) = a_1L + a_2L^2 = 2.2857 - 0.5714 = 1.714\ \text{mm}$$

Exactness

The exact solution of $AE\,u'' + q = 0$ with $u(0) = 0$ and $AE\,u'(L) = P$ is

$$u = \frac{(P + qL)}{AE}x - \frac{q}{2AE}x^2$$

which is a quadratic. Our coefficients match it ($\frac{24\,000}{2.1\times10^7} = 1.1429\times10^{-3}$ and $\frac{6}{4.2\times10^7} = 1.4286\times10^{-7}$). So the result is exact because the trial function contains the exact solution.

Answer: $a_1 = 1.143\times10^{-3}$, $a_2 = -1.429\times10^{-7}\ \text{mm}^{-1}$; end displacement $= 1.714$ mm; yes, exact.