Question 5 of 8: Frame analysis by slope-deflection, with shear and bending moment diagrams

Question 5: Frame analysis by slope-deflection, with shear and bending moment diagrams (20 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. A continuous member 1–2–3–4 propped at joint 2 by a short column 2–5.

Find. All member end moments, then complete shear force and bending moment diagrams for every member, labelled with the maximum and minimum ordinates.

Approach. No joint can translate, so the only unknowns are the rotations at joints 2 and 3; write the slope-deflection equations for the four members, impose moment equilibrium at those two joints, and recover the shears and diagrams from the resulting end moments.

1. Fixed-end moments and the overhang moment. Only span 2–3 carries a distributed load, so with the sign convention that a clockwise moment applied by the joint to the member end is positive, $$\begin{aligned} M^{F}_{23} &= -\frac{wL^2}{12} = -\frac{9(16)^2}{12} = -192\ \text{kN}\cdot\text{m}, \\ M^{F}_{32} &= +192\ \text{kN}\cdot\text{m} \end{aligned}$$ The 4 m overhang is statically determinate, so it applies a known moment at joint 3 rather than an unknown one: $$M_{34} = -11(4) = -44\ \text{kN}\cdot\text{m}$$
2. Write the slope-deflection equations. With $M_{AB} = \dfrac{2EI}{L}\left(2\theta_A + \theta_B\right) + M^{F}_{AB}$ and $\theta_1 = 0$, and with the far end of the column released at the pin so that $M_{52} = 0$ can be used to eliminate $\theta_5$, $$\begin{aligned} M_{12} &= 0.25E\theta_2, \\ M_{21} &= 0.50E\theta_2 \end{aligned}$$ $$\begin{aligned} M_{23} &= 0.50E\theta_2 + 0.25E\theta_3 - 192, \\ M_{32} &= 0.50E\theta_3 + 0.25E\theta_2 + 192 \end{aligned}$$ $$M_{25} = \frac{3EI}{L}\theta_2 = 1.50E\theta_2$$ where the column has been given its reduced pinned-end stiffness $3EI/L$ because joint 5 carries no moment.
3. Impose equilibrium at the two free joints. At joint 2 three members meet, and at joint 3 the span moment must balance the known overhang moment: $$M_{21} + M_{23} + M_{25} = 0 \;\Rightarrow\; 2.50E\theta_2 + 0.25E\theta_3 = 192$$ $$M_{32} + M_{34} = 0 \;\Rightarrow\; 0.25E\theta_2 + 0.50E\theta_3 = -148$$ Solving the pair, $$\begin{aligned} E\theta_2 &= +112.0\ \text{kN}\cdot\text{m}^2, \\ E\theta_3 &= -352.0\ \text{kN}\cdot\text{m}^2 \end{aligned}$$
4. Back-substitute for the end moments. $$M_{12} = \boxed{+28.0}, \quad M_{21} = +56.0, \quad M_{23} = -224.0, \quad M_{32} = +44.0, \quad M_{25} = +168.0\ \text{kN}\cdot\text{m}$$ Three independent checks confirm the set: joint 2 balances, $56.0 - 224.0 + 168.0 = 0$; joint 3 balances, $44.0 - 44.0 = 0$; and the carry-over relation for a member whose far end is fixed holds exactly, $M_{12} = \tfrac{1}{2}M_{21} = 28.0$ kN.m. Moment distribution started from the same fixed-end moments converges on the identical figures.
5. Member end shears. For a member of length $L$ carrying a uniform load $w$, the end shears follow from $V_A = \dfrac{wL}{2} - \dfrac{M_{AB} + M_{BA}}{L}$ and $V_B = wL - V_A$. Hence $$\text{1--2:}\ V = \frac{28.0 + 56.0}{8} = 10.5\ \text{kN (constant)}$$ $$\text{2--3:}\ V_2 = \frac{9(16)}{2} - \frac{-224.0 + 44.0}{16} = 72 + 11.25 = 83.25\ \text{kN}, \quad V_3 = 144 - 83.25 = 60.75\ \text{kN}$$ $$\text{2--5:}\ V = \frac{168.0 + 0}{2} = 84.0\ \text{kN (horizontal)}$$ and the overhang carries a constant 11.0 kN.
6. Support reactions, as a check on everything above. Assembling the member end shears at each support, $$R_1 = 10.5\ \text{kN downward}, \quad R_3 = 60.75 + 11.0 = 71.75\ \text{kN upward}, \quad R_5 = 10.5 + 83.25 = 93.75\ \text{kN upward}$$ $$\sum F_y:\ -10.5 + 71.75 + 93.75 = 155.0 = 9(16) + 11 \;\checkmark$$ The downward reaction at the fixed end is real, not an error: the stiff column at joint 2 lifts that end of the frame, and member 1–2 has to be held down.
7. Bending moment ordinates. On member 1–2 the moment runs linearly from $+28.0$ kN.m at the fixed end to $-56.0$ kN.m just left of joint 2, crossing zero at 2.67 m from joint 1. At joint 2 the diagram steps by 168.0 kN.m, the moment carried away down the column, so it restarts at $-224.0$ kN.m. Along the 16 m span, $$M(x) = -224.0 + 83.25x - 4.5x^2$$ whose peak is where the shear vanishes, at $x = 83.25/9 = 9.25$ m: $$M_{max}^{+} = \boxed{+161.03\ \text{kN}\cdot\text{m at } 9.25\ \text{m from joint 2}}$$ The span ends at $-44.0$ kN.m over the roller, which the overhang reproduces exactly as $-11(4)$. The column carries a linear diagram from zero at the pin to 168.0 kN.m at joint 2.