Question 6 of 8: Frame Analysis by Slope Deflection

Question 6: Frame Analysis by Slope Deflection (22 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 shear force and bending moment diagrams for every member, with the maximum and minimum ordinates labelled.

Approach. Slope deflection. Because the members are inextensible and node 4 is fixed, nodes 2 and 3 cannot translate horizontally, and the inextensible columns stop them translating vertically, so there is no sidesway and the only unknowns are the joint rotations $\theta_2$ and $\theta_3$. Write the end moments, impose moment equilibrium at those two joints, solve the two-by-two system, then recover shears from member free bodies.

1. Establish that there is no sidesway and count the unknowns. The beam 1–2–3–4 is inextensible and node 4 is built in, so nodes 2 and 3 are fixed horizontally; the columns are inextensible and their bases are on the ground, so nodes 2 and 3 are fixed vertically. The kinematic unknowns are therefore just $\theta_2$ and $\theta_3$ — a two-equation problem despite five degrees of static indeterminacy.
2. Deal with the overhang first. The 1 m cantilever is statically determinate and applies a known moment at joint 2: $$M_{21} = 16.5\,(1) = 16.5\ \text{kN}\cdot\text{m}$$ It has no rotational stiffness, so it must be entered as a fixed applied moment at the joint and never given a share of the balancing moment.
3. Write the slope-deflection equations. With $\mathrm{FEM}_{23} = -wL^2/12 = -64$ and $\mathrm{FEM}_{32} = +64\ \text{kN}\cdot\text{m}$, and using the modified form $3EI\theta/L$ for the pin-based column 2–5, $$\begin{aligned} M_{23} &= \tfrac{2EI}{8}\left(2\theta_2 + \theta_3\right) - 64, & M_{32} &= \tfrac{2EI}{8}\left(2\theta_3 + \theta_2\right) + 64\\ M_{34} &= \tfrac{2EI}{4}\left(2\theta_3\right), & M_{43} &= \tfrac{2EI}{4}\left(\theta_3\right)\\ M_{25} &= \tfrac{3EI}{4}\,\theta_2, & M_{36} &= \tfrac{2EI}{4}\left(2\theta_3\right) \end{aligned}$$ with $M_{52} = 0$ at the pin and $M_{63} = \tfrac{2EI}{4}\theta_3$.
4. Impose joint equilibrium. Writing $a = EI\theta_2$ and $b = EI\theta_3$, the two joint equations $M_{21} + M_{23} + M_{25} = 0$ and $M_{32} + M_{34} + M_{36} = 0$ become $$\begin{aligned} 1.25\,a + 0.25\,b &= 47.5\\ 0.25\,a + 2.50\,b &= -64 \end{aligned}$$ $$\boxed{EI\theta_2 = 44.0,\qquad EI\theta_3 = -30.0\ \ \text{kN}\cdot\text{m}^2}$$ The clean values are a good sign that the fixed-end moments and the overhang sign are right.
5. Back-substitute for the member end moments. $$\begin{aligned} M_{23} &= -49.5, & M_{32} &= +60.0\\ M_{34} &= -30.0, & M_{43} &= -15.0\\ M_{25} &= +33.0, & M_{52} &= 0\\ M_{36} &= -30.0, & M_{63} &= -15.0 \end{aligned}$$ all in $\text{kN}\cdot\text{m}$. Checking the joints, $16.5 - 49.5 + 33.0 = 0$ at node 2 and $60.0 - 30.0 - 30.0 = 0$ at node 3.
6. Convert to bending moments and find the beam shears. In beam terms the moment at node 2 is $-49.5$ and at node 3 is $-60.0$ $\text{kN}\cdot\text{m}$, both hogging. Free-body equilibrium of member 2–3 gives the end shears $$V_2 = \frac{M_3 - M_2 + wL^2/2}{L} = \frac{-60 + 49.5 + 384}{8} = +46.6875\ \text{kN}$$ $$V_3 = V_2 - wL = 46.6875 - 96 = -49.3125\ \text{kN}$$ The shear passes through zero at $x = 46.6875/12 = 3.8906\ \text{m}$ from node 2, where the sagging peak is $$M_{\max} = -49.5 + 46.6875(3.8906) - 6(3.8906)^2 = +41.322\ \text{kN}\cdot\text{m}$$
7. Finish the remaining members. Beam 3–4 carries no load, so its moment runs linearly from $-30.0$ at node 3 to $+15.0$ at node 4 and its shear is constant at $(15 + 30)/4 = +11.25\ \text{kN}$. Column 2–5 carries $33.0\ \text{kN}\cdot\text{m}$ at the head and zero at the pin, so its shear is $33.0/4 = 8.25\ \text{kN}$; column 3–6 carries $30.0$ at the head and $15.0$ at the base with the ends bending in opposite senses, so its shear is $(30 + 15)/4 = 11.25\ \text{kN}$ and its point of contraflexure sits $2.667\ \text{m}$ below node 3.
8. Extract the reactions and check global equilibrium. The jumps in the beam shear give the vertical reactions $$\begin{aligned} R_5 &= 46.6875 - (-16.5) = 63.1875\ \text{kN}\ \uparrow\\ R_6 &= 11.25 - (-49.3125) = 60.5625\ \text{kN}\ \uparrow\\ R_4 &= -11.25\ \text{kN}\ \ (\text{i.e. }11.25\ \text{kN downward}) \end{aligned}$$ and $63.1875 + 60.5625 - 11.25 = 112.5\ \text{kN}$, exactly the applied total $16.5 + 96$. Horizontally the two column shears, 8.25 kN and 11.25 kN, are balanced by a 3.0 kN reaction at the built-in end 4, and the fixing moment there is the $15.0\ \text{kN}\cdot\text{m}$ already found.