Three moment equation: beam with overhang

Reading of the figure: the beam is fixed at 1; span 1-2 = 8 m with $I$ and a 100 kN load at 4 m from 1; span 2-3 = 10 m with $2I$ and 40 kN/m; the overhang 3-4 is 2 m with 20 kN at the free end. Supports 2 and 3 are on rollers. $E$ is constant.

Known moment at 3

The overhang is statically determinate: $M_3 = -20\times2 = -40$ kNm (hogging).

Three moment equation with different $I$

$$M_1\frac{L_1}{I_1} + 2M_2\left(\frac{L_1}{I_1}+\frac{L_2}{I_2}\right) + M_3\frac{L_2}{I_2} = -\frac{6a_1\bar x_1}{I_1L_1} - \frac{6a_2\bar x_2}{I_2L_2}$$

Load terms:

• Span 1-2: free BM peak $= \dfrac{100\times4\times4}{8} = 200$, $a_1 = \tfrac12(8)(200) = 800$, $\bar x_1 = 4$: $\dfrac{6a_1\bar x_1}{L_1} = 2400$.
• Span 2-3: $\dfrac{wL_2^3}{4} = \dfrac{40(10)^3}{4} = 10000$.

Equation at 1 (fixed end): $2M_1\dfrac{8}{I} + M_2\dfrac{8}{I} = -\dfrac{2400}{I}$

$$2M_1 + M_2 = -300$$

Equation at 2: multiply by $I$: $8M_1 + 2M_2\left(8 + \dfrac{10}{2}\right) + M_3\dfrac{10}{2} = -2400 - \dfrac{10000}{2}$

$$8M_1 + 26M_2 + 5(-40) = -7400 \ \Rightarrow\ 8M_1 + 26M_2 = -7200$$

Substituting $M_1 = \dfrac{-300 - M_2}{2}$: $-1200 - 4M_2 + 26M_2 = -7200$, so

$$M_2 = -272.727\ \text{kNm},\qquad M_1 = -13.636\ \text{kNm}$$

($M_1$ is a small hogging moment at the fixed end.)

Reactions

With sagging-positive moments, in span 1-2: $M_2 = M_1 + 8R_1 - 100(4)$, so

$$R_1 = \frac{400 + M_2 - M_1}{8} = \frac{400 + (-272.727) - (-13.636)}{8} = 17.614\ \text{kN}$$

In span 2-3 (taking $x$ from 2): $M_3 = M_2 + 10V_{2R} - 40\frac{10^2}{2}$, so

$$V_{2R} = \frac{2000 + M_3 - M_2}{10} = \frac{2000 - 40 - (-272.727)}{10} = 223.273\ \text{kN}$$ $$R_2 = (100 - 17.614) + 223.273 = 305.659\ \text{kN},\qquad R_3 = 400 - 223.273 + 20 = 196.727\ \text{kN}$$

Check: $R_1 + R_2 + R_3 = 520.00$ kN $= 100 + 400 + 20$.

Bending moment (kNm)

Answer: $M_1 = -13.636$ kNm, $M_2 = -272.727$ kNm, $M_3 = -40$ kNm.