Castigliano: deflection at 50 kN load, cantilever

Castigliano's second theorem: $\delta=\partial U/\partial P$. Call the 50 kN load $P_2$ at the free end ($x=6$ m from the fixed end) and the 100 kN load $P_1$ at $x=3$ m. Treat both as symbols, differentiate with respect to $P_2$, then substitute the numbers.

Bending moments (x from the fixed end A, hogging taken as negative)

• $0\le x\le3$: $M=-[P_1(3-x)+P_2(6-x)]$, $\;\dfrac{\partial M}{\partial P_2}=-(6-x)$
• $3\le x\le6$: $M=-P_2(6-x)$, $\;\dfrac{\partial M}{\partial P_2}=-(6-x)$

Deflection under the 50 kN load

$$\delta_2=\frac{1}{EI}\left[\int_0^3[100(3-x)+50(6-x)](6-x)\,dx+\int_3^6 50(6-x)^2dx\right]$$ $$\int_0^3[100(3-x)+50(6-x)](6-x)dx=5400\ \text{kN m}^3,\qquad \int_3^650(6-x)^2dx=450\ \text{kN m}^3$$ $$\delta_2=\frac{5850}{EI}\quad(\text{numerator in kN m}^3,\ EI\ \text{in kN m}^2)$$

Flexural rigidity

$$EI=211000\ \text{N/mm}^2\times46875\times10^4\ \text{mm}^4=9.8906\times10^{13}\ \text{N mm}^2=98906.25\ \text{kN m}^2$$ $$\delta_2=\frac{5850}{98906.25}=59.15\ \text{mm}\ \text{(downward)}$$

Check with standard results: tip deflection due to 100 kN at 3 m is $Pa^2(3L-a)/6EI=2250/EI$ and due to 50 kN at the tip is $PL^3/3EI=3600/EI$; total $5850/EI$ ✓.

Answer: vertical deflection under the 50 kN load = 59.15 mm downward.