Deflection at B, C and slope at A, D

Reading of the figure: $AB = 1$ m, $BC = 1$ m, $CD = 3$ m, $DE = 2$ m; hinge at A (x = 0) and roller at D (x = 5 m). Loads: 5 kN at B (x = 1), UDL 1 kN/m over CD (x = 2 to 5), and a 2 kN·m couple at the free end E (taken clockwise). $EI$ is constant. Deflection is taken positive upward.

Reactions

Moments about A (anticlockwise positive):

$$R_D(5) - 5(1) - 3(3.5) - 2 = 0 \Rightarrow R_D = 3.5\ \text{kN}, \qquad R_A = 5 + 3 - 3.5 = 4.5\ \text{kN}$$

Macaulay's method

The UDL ends at D, so it is continued to the end and cancelled by an equal upward load (+1 kN/m beyond D). Bending moment (sagging positive) at a section $x$ from A:

$$EI\frac{d^2y}{dx^2} = 4.5x - 5\langle x-1\rangle - \frac{1}{2}\langle x-2\rangle^2 + 3.5\langle x-5\rangle + \frac{1}{2}\langle x-5\rangle^2 + 2\langle x-7\rangle^0$$

Integrating:

$$EI\,\theta = 2.25x^2 - 2.5\langle x-1\rangle^2 - \frac{1}{6}\langle x-2\rangle^3 + 1.75\langle x-5\rangle^2 + \frac{1}{6}\langle x-5\rangle^3 + 2\langle x-7\rangle + C_1$$ $$EI\,y = 0.75x^3 - \frac{5}{6}\langle x-1\rangle^3 - \frac{1}{24}\langle x-2\rangle^4 + \frac{7}{12}\langle x-5\rangle^3 + \frac{1}{24}\langle x-5\rangle^4 + \langle x-7\rangle^2 + C_1x + C_2$$

Constants

• At A, $y = 0$ at $x = 0$: $C_2 = 0$.
• At D, $y = 0$ at $x = 5$: $0.75(125) - \frac{5}{6}(64) - \frac{1}{24}(81) + 5C_1 = 0 \Rightarrow C_1 = -7.4083$

Slopes at A and D

$$EI\,\theta_A = C_1 = -7.408\ \text{kN·m}^2 \quad\Rightarrow\quad \theta_A = \frac{7.408}{EI}\ \text{rad (clockwise)}$$ $$EI\,\theta_D = 2.25(25) - 2.5(16) - \frac{27}{6} + C_1 = 4.342\ \text{kN·m}^2 \quad\Rightarrow\quad \theta_D = \frac{4.342}{EI}\ \text{rad (anticlockwise)}$$

Deflections at B and C

$$EI\,y_B = 0.75(1) + C_1(1) = -6.658\ \text{kN·m}^3 \Rightarrow y_B = \frac{6.658}{EI}\ \text{downward}$$ $$EI\,y_C = 0.75(8) - \frac{5}{6}(1) + C_1(2) = -9.650\ \text{kN·m}^3 \Rightarrow y_C = \frac{9.650}{EI}\ \text{downward}$$

Answer (with $EI$ in kN·m²): $y_B = 6.66/EI$ and $y_C = 9.65/EI$ (downward); $\theta_A = 7.41/EI$ (clockwise) and $\theta_D = 4.34/EI$ (anticlockwise).