Question 6 of 8: Frame by slope deflection — shear and moment diagrams

Question 6: Frame by slope deflection — shear and moment diagrams (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. Take node 1 at the origin. The frame consists of a vertical column 1–2 of height 5 m and stiffness 1.5EI, fixed at node 1; an inclined member 2–3 running from the top of the column down to a pin support at node 3, 12 m horizontally and 5 m vertically away, hence 13 m long, of stiffness 1.3EI; a 2 m horizontal overhang 5–2 at the top level, of stiffness EI, free at node 5; and a 2 m horizontal overhang 3–4 at the base level, also of stiffness EI, free at node 4. A uniform 5 kN/m acts downward over the entire 16 m horizontal projection of the sketch, from node 5 to node 4.

Find. The member end moments, the reactions, and the shear force and bending moment diagrams with the maximum and minimum ordinate of each member labelled.

Approach. Argue that there is no sidesway, replace each overhang by the known moment it delivers to its joint, write slope deflection for the column and the inclined member, solve the two joint equations, and integrate for the diagrams. The 5-12-13 geometry makes every fixed-end quantity exact.

1. Establish the degrees of freedom, and rule out sidesway. The unknown displacements are the rotations of joints 2 and 3. Node 1 is fixed and node 3 is a pin, so neither translates; the column 1–2 is inextensible and vertical, which fixes the vertical position of node 2, and the member 2–3 is inextensible, which fixes its distance from node 3. Two independent constraints on a point in a plane leave it nowhere to go, so $\Delta = 0$ and no sway correction is required. That argument is worth writing out, because it is what the words "all members are inextensible" are in the question to license.
2. Convert the projected load on the inclined member into a fixed-end moment. The 5 kN/m acts per metre of horizontal projection. Per metre of member it is $w(L_h/L)$, and only the component perpendicular to the member bends it, a further factor $L_h/L$, so $$w_{\perp} = w\left(\frac{L_h}{L}\right)^{2} = 5\left(\frac{12}{13}\right)^{2} = 4.260\ \text{kN/m}.$$ The fixed-end moment then collapses to a projection-only formula, which is worth remembering: $$\text{FEM} = \frac{w_{\perp}L^{2}}{12} = \frac{w L_h^{2}}{12} = \frac{5(12)^{2}}{12} = \boxed{60.0\ \text{kN}\cdot\text{m}}$$ so $M^{F}_{23} = -60$ and $M^{F}_{32} = +60$ kN·m in the clockwise-positive convention used below.
3. Replace each overhang by the moment it delivers. Both overhangs are determinate cantilevers carrying $5(2) = 10$ kN, so each applies a moment of $5(2)^{2}/2 = 10$ kN·m to its joint and contributes no stiffness — it must never be given a distribution factor. In end-moment terms, $M_{25} = +10$ and $M_{34} = -10$ kN·m; the signs differ because the two overhangs point in opposite directions along the global $x$ axis, so the same physical hogging enters the two joints with opposite rotational sense.
4. Write slope deflection and form the joint equations. With $M_{ij} = (2EI/L)(2\theta_i + \theta_j) + M^{F}_{ij}$ and $\theta_1 = 0$, $$M_{12} = 0.6EI\,\theta_2,\qquad M_{21} = 1.2EI\,\theta_2,$$ $$M_{23} = 0.2EI(2\theta_2 + \theta_3) - 60,\qquad M_{32} = 0.2EI(2\theta_3 + \theta_2) + 60.$$ Joint 2 requires $M_{21} + M_{23} + M_{25} = 0$ and joint 3 requires $M_{32} + M_{34} = 0$, giving $$1.6\,EI\theta_2 + 0.2\,EI\theta_3 = 50, \qquad 0.2\,EI\theta_2 + 0.4\,EI\theta_3 = -50 .$$
5. Solve, and use the integers as the sign check. Eliminating, $$EI\theta_2 = 50,\qquad EI\theta_3 = -150$$ and back-substitution gives $$M_{12} = +30,\quad M_{21} = +60,\quad M_{23} = -70,\quad M_{32} = +10 \ \text{kN}\cdot\text{m} .$$ Every value is an integer and both joint equations close exactly ($60 - 70 + 10 = 0$ and $10 - 10 = 0$), which is the confirmation that the overhang moments were entered with the right signs — the commonest failure on this question family produces four-decimal end moments instead. An independent direct-stiffness solve of the same frame reproduces all four values.
6. Recover the reactions. The column carries no transverse load, so its shear is constant, $$H_1 = \frac{M_{12} + M_{21}}{5} = \frac{90}{5} = \boxed{18.0\ \text{kN}}$$ acting to the right at the base and, by global horizontal equilibrium, to the left at the pin. The total applied load is $5(16) = 80$ kN with its centroid 6 m to the right of the column line, so taking moments about node 1, $$V_3 = \frac{80(6) + 30}{12} = \boxed{42.5\ \text{kN}},\qquad V_1 = 80 - 42.5 = \boxed{37.5\ \text{kN}}$$ both upward, with a 30 kN·m clockwise fixing moment at node 1.
7. Diagrams for the inclined member. Its end shears follow from the end moments and the perpendicular load, $$V_2 = \frac{w_{\perp}L}{2} - \frac{M_{23} + M_{32}}{L} = 27.69 + 4.62 = 32.31\ \text{kN},\qquad V_3 = 55.38 - 32.31 = 23.08\ \text{kN},$$ so the shear falls linearly from $+32.3$ kN at node 2 to $-23.1$ kN at node 3, crossing zero $32.31/4.260 = 7.58$ m along the member. The peak sagging moment there is $$M_{\max} = M_{23} + \frac{V_2^{2}}{2w_{\perp}} = -70 + 122.50 = \boxed{+52.5\ \text{kN}\cdot\text{m}}$$ with the member hogging $-70$ kN·m at node 2 and $-10$ kN·m at node 3, and points of contraflexure 2.62 m and 12.55 m along it.
8. Diagrams for the column and the overhangs. The column carries a constant shear of 18.0 kN and a moment that runs linearly from $+30$ kN·m at the fixed base to $-60$ kN·m at node 2, changing sign 1.67 m above the base. Each overhang carries a shear rising linearly from zero at the free tip to 10 kN at its joint and a moment falling parabolically from zero to $-10$ kN·m. Plotting the roof line 5–2–3–4 as one developed diagram makes the structure of the answer visible: the moment steps by 60 kN·m at node 2, which is precisely what the column removes from the joint, and it is continuous at node 3, where nothing else frames in.