Question 6 of 9: Two-span beam with a settled central support

Question 6: Two-span beam with a settled central support (24 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The bending moment over the central support, the three reactions, and the shear and bending moment diagrams with the maximum and minimum ordinates in each span.

Approach. Superpose two effects that the slope-deflection equations handle in one pass: the fixed-end moments from the two distributed loads, and the chord rotations produced by the 64 mm settlement. Impose $M_{12}=M_{32}=0$ at the simple ends and $M_{21}+M_{23}=0$ at the central support.

1. Chord rotations and fixed-end moments. With $v_2=-0.064\text{ m}$ and $v_1=v_3=0$, $$\psi_{12}=\frac{v_1-v_2}{8}=+0.008,\qquad \psi_{23}=\frac{v_2-v_3}{8}=-0.008 .$$ The fixed-end moments for a uniformly loaded span are $\mp wL^{2}/12$: $$\text{FEM}_{12}=-\frac{4(8)^{2}}{12}=-21.33,\quad \text{FEM}_{21}=+21.33,\quad \text{FEM}_{23}=-\frac{8(8)^{2}}{12}=-42.67,\quad \text{FEM}_{32}=+42.67\ \text{kN}\cdot\text{m}.$$
2. Assemble the three equations. With $k=2EI/8=5000\ \text{kN}\cdot\text{m}$, $$M_{12}=k\!\left(2\theta_1+\theta_2-0.024\right)-21.33=0,\qquad M_{32}=k\!\left(2\theta_3+\theta_2+0.024\right)+42.67=0,$$ $$M_{21}+M_{23}=k\!\left(2\theta_2+\theta_1-0.024\right)+21.33 +k\!\left(2\theta_2+\theta_3+0.024\right)-42.67=0 .$$
3. Solve for the rotations. The two end conditions give $\theta_1=\tfrac12(0.028267-\theta_2)$ and $\theta_3=\tfrac12(-0.032533-\theta_2)$; substituting into the joint equation leaves $15000\,\theta_2 = 32.0$, so $$\boxed{\theta_2 = +0.002133\ \text{rad},\qquad \theta_1=+0.013067,\qquad \theta_3=-0.017333\ \text{rad}.}$$
4. Find the moment over the central support. $$M_{21}=5000\!\left(0.004267+0.013067-0.024\right)+21.33=-12.0\ \text{kN}\cdot\text{m},$$ and $M_{23}=+12.0\ \text{kN}\cdot\text{m}$, so the two balance as required. In the sagging-positive convention the bending moment over support 2 is $$\boxed{M_2 = -M_{21} = +12.0\ \text{kN}\cdot\text{m}\quad\text{(sagging).}}$$ It is worth separating the two contributions: the loads alone would give $M_2=-(w_1+w_2)L^{2}/16=-48.0\ \text{kN}\cdot\text{m}$ (hogging), while the settlement alone gives $+3EI\delta/L^{2}=+60.0\ \text{kN}\cdot\text{m}$ (sagging). The settlement therefore reverses the sign of the support moment.
5. Reactions from span statics. Taking moments on each span with the end moments known, $$R_1=\frac{w_1L^{2}/2+M_2}{L}=\frac{128+12}{8}=17.50\ \text{kN},\qquad V_{2}^{-}=w_1L-R_1=14.50\ \text{kN},$$ $$V_{2}^{+}=\frac{w_2L^{2}/2-M_2}{L}=\frac{256-12}{8}=30.50\ \text{kN},\qquad R_3=w_2L-V_2^{+}=33.50\ \text{kN},$$ so $R_2 = 14.50+30.50 = 45.00\ \text{kN}$. Check: $17.50+45.00+33.50 = 96.0\ \text{kN} = 4(8)+8(8)$.
6. Locate the zero-shear points and the sagging peaks. In span 1–2, $V=17.50-4x=0$ at $x=4.375\text{ m}$, where $$M_{\max}=17.50(4.375)-2(4.375)^{2}=\boxed{+38.28\ \text{kN}\cdot\text{m}} .$$ In span 2–3, measuring $x'$ from support 2, $V=30.50-8x'=0$ at $x'=3.8125\text{ m}$, where $$M_{\max}=12.0+30.50(3.8125)-4(3.8125)^{2}=\boxed{+70.14\ \text{kN}\cdot\text{m}} .$$
7. Read off the extreme ordinates. Because the support moment has become sagging, the bending moment is positive over the whole 16 m: there is no point of contraflexure anywhere, and the minimum bending moment in each span is the zero at its outer simple support. The shear extremes are $+17.50$ and $-14.50\text{ kN}$ in span 1–2, and $+30.50$ and $-33.50\text{ kN}$ in span 2–3.