Macaulay's method for an off-centre load

Reactions

$$R_A = \frac{30\times4}{6} = 20\ \text{kN}, \qquad R_B = 10\ \text{kN}$$

Macaulay equations (origin at A, $y$ upward, sagging $M$ positive)

$$EI\frac{d^2y}{dx^2} = 20x - 30\langle x-2\rangle$$ $$EI\frac{dy}{dx} = 10x^2 - 15\langle x-2\rangle^2 + C_1$$ $$EI\,y = \frac{10x^3}{3} - 5\langle x-2\rangle^3 + C_1x + C_2$$

Boundary conditions:

• $x = 0$, $y = 0$ gives $C_2 = 0$.
• $x = 6$, $y = 0$: $\dfrac{10\times216}{3} - 5\times4^3 + 6C_1 = 0$, so $720 - 320 + 6C_1 = 0$, so $C_1 = -66.67$.

(a) Slopes

At A ($x=0$):

$$\theta_A = \frac{C_1}{EI} = \frac{-66.67}{12000} = -5.56\times10^{-3}\ \text{rad (clockwise)}$$

At B ($x=6$):

$$EI\theta_B = 10(36) - 15(16) - 66.67 = 53.33 \;\Rightarrow\; \theta_B = 4.44\times10^{-3}\ \text{rad}$$

(b) Deflection under the load ($x = 2$)

$$EI\,y_C = \frac{10\times8}{3} - 0 - 66.67\times2 = 26.67 - 133.33 = -106.67$$ $$y_C = \frac{-106.67}{12000} = -8.89\times10^{-3}\ \text{m} = 8.89\ \text{mm downward}$$

(c) Maximum deflection

Maximum deflection occurs where the slope is zero. The longer segment is CB, so test $x>2$:

$$10x^2 - 15(x-2)^2 - 66.67 = 0 \;\Rightarrow\; -5x^2 + 60x - 126.67 = 0$$ $$x^2 - 12x + 25.33 = 0 \;\Rightarrow\; x = 6 - \sqrt{36-25.33} = 6 - 3.266 = 2.734\ \text{m}$$

(this lies beyond 2 m, so it is valid.)

$$EI\,y_{max} = \frac{10(2.734)^3}{3} - 5(0.734)^3 - 66.67(2.734) = 68.12 - 1.98 - 182.27 = -116.1$$ $$y_{max} = \frac{-116.1}{12000} = -9.68\times10^{-3}\ \text{m}$$

Answer: $\theta_A = 5.56\times10^{-3}$ rad, $\theta_B = 4.44\times10^{-3}$ rad; $y_{under\ load} = 8.89$ mm; $y_{max} = 9.68$ mm downward at 2.73 m from A.