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.

Find. All member end moments, hence the shear force and bending moment diagrams with the maximum and minimum ordinate of each member.

Approach. Show first that the frame cannot sway, so the only unknowns are the rotations of joints 2 and 3. Write the four slope-deflection equations that contain them, add the two joint-moment equations, solve, and recover shears from member statics. The same numbers follow from moment distribution and the distribution table is given as a check.

1. Show that there is no sidesway. Members are inextensible, so joint 2 cannot move horizontally (member 1–2 ties it to the built-in joint 1) and cannot move vertically (member 2–5 ties it to the pin at joint 5). Joint 3 cannot move horizontally for the same reason through member 2–3, and its roller stops it moving vertically. The kinematic unknowns are therefore only the rotations $\theta_{2}$ and $\theta_{3}$, and no sway correction is needed — which is what makes the problem tractable by hand.
2. Replace the cantilever by its end action. Member 3–4 carries 4 kN at 1 m from joint 3, so it delivers a known 4 kN·m to the joint together with a 4 kN vertical force. In the clockwise-positive convention used throughout, $$M_{34} = -4\ \text{kN}\cdot\text{m}$$ and joint 3 must satisfy $M_{32} + M_{34} = 0$, that is $M_{32} = +4$ kN·m.
3. Write the fixed-end moments. For the uniformly loaded member 1–2 and the centrally loaded member 2–3, $$\begin{aligned} \text{FEM}_{12} &= -\frac{wL^{2}}{12} = -\frac{6(8)^{2}}{12} = -32\ \text{kN}\cdot\text{m} \\ \text{FEM}_{21} &= +32\ \text{kN}\cdot\text{m} \end{aligned}$$ $$\begin{aligned} \text{FEM}_{23} &= -\frac{PL}{8} = -\frac{16(6)}{8} = -12\ \text{kN}\cdot\text{m} \\ \text{FEM}_{32} &= +12\ \text{kN}\cdot\text{m} \end{aligned}$$ Member 2–5 carries no load, so both of its fixed-end moments are zero.
4. Set down the slope-deflection equations. With $M_{AB} = (2EI/L)(2\theta_{A} + \theta_{B}) + \text{FEM}_{AB}$ and $\theta_{1} = 0$, and using the modified form for member 2–5 because its far end is a pin free to rotate, $$\begin{aligned} M_{12} &= 0.25\,EI\theta_{2} - 32 \\ M_{21} &= 0.5\,EI\theta_{2} + 32 \\ M_{25} &= EI\theta_{2} \end{aligned}$$ $$\begin{aligned} M_{23} &= \tfrac{1}{3}EI\,(2\theta_{2} + \theta_{3}) - 12 \\ M_{32} &= \tfrac{1}{3}EI\,(2\theta_{3} + \theta_{2}) + 12 \end{aligned}$$
5. Impose joint equilibrium and solve. Joint 2 carries three member ends and joint 3 carries two, so $$\begin{aligned} M_{21} + M_{25} + M_{23} &= 0 \\ M_{32} + M_{34} &= 0 \end{aligned}$$ which reduce to $$\begin{aligned} 2.1667\,EI\theta_{2} + 0.3333\,EI\theta_{3} &= -20 \\ 0.3333\,EI\theta_{2} + 0.6667\,EI\theta_{3} &= -8 \end{aligned}$$ and give the pleasingly simple result $$\boxed{EI\theta_{2} = EI\theta_{3} = -8\ \text{kN}\cdot\text{m}^{2}}$$ Both joints rotate anticlockwise, which is what the sagging of the two loaded spans demands.
6. Back-substitute for the end moments. Feeding the two rotations back through the five equations, $$M_{12} = -34, \quad M_{21} = +28, \quad M_{25} = -8, \quad M_{23} = -20, \quad M_{32} = +4, \quad M_{34} = -4$$ all in kN·m. Joint 2 checks exactly, $28 - 8 - 20 = 0$, and joint 3 checks as $4 - 4 = 0$. Moment distribution reproduces these figures: the distribution factors at joint 2 are $0.5/2.1667 = 0.231$ to member 2–1, $1.0/2.1667 = 0.461$ to the column and $0.667/2.1667 = 0.308$ to member 2–3, and joint 3 has a single distributing member so its factor is unity against the known cantilever moment.
7. Recover the member shears. Take each member as a free body loaded by its span load and its two end moments. For member 1–2, $$\begin{aligned} V_{1} &= \frac{6(8)}{2} + \frac{M_{12} + M_{21}}{-8} = 24 + \frac{-34 + 28}{-8} = 24.75\ \text{kN}\uparrow \\ V_{2} &= 48 - 24.75 = 23.25\ \text{kN} \end{aligned}$$ For member 2–3 the same treatment gives 10.67 kN at joint 2 and 5.33 kN at joint 3, the column carries a constant shear of $8/3 = 2.67$ kN, and the cantilever carries 4 kN throughout.
8. Assemble the reactions. The built-in end takes $$\begin{aligned} V_{1} &= 24.75\ \text{kN}\uparrow \\ H_{1} &= 2.67\ \text{kN}\rightarrow \\ M_{1} &= 34\ \text{kN}\cdot\text{m} \end{aligned}$$ the roller at joint 3 takes $5.33 + 4 = 9.33$ kN upward, and the pin at joint 5 takes $23.25 + 10.67 = 33.92$ kN upward together with 2.67 kN to the left. Vertical equilibrium closes on the total applied load, $24.75 + 9.33 + 33.92 = 68.0$ kN, which is $6(8) + 16 + 4$.
9. Locate the extreme ordinates. On member 1–2 the shear passes through zero at $x = 24.75/6 = 4.125$ m, where $$M = -34 + 24.75(4.125) - 3(4.125)^{2} = \boxed{+17.05\ \text{kN}\cdot\text{m}}$$ and the moment is zero at $x = 1.741$ m and $x = 6.509$ m. The most negative ordinate on the whole frame is the built-in moment, $-34.0$ kN·m. On member 2–3 the moment runs from $-20$ kN·m at joint 2 up to $-20 + 10.67(3) = +12.0$ kN·m under the 16 kN load and back to $-4.0$ kN·m at joint 3, which matches the cantilever exactly. The column varies linearly from $-8.0$ kN·m at joint 2 to zero at the pin.