Question 6 of 8: Frame by the slope-deflection method

Question 6: Frame by the slope-deflection method (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. A horizontal beam 20 m long measured from its free left end, on a roller at node 1 ($x = 4$ m), continuing through node 2 ($x = 12$ m) to a built-in end at node 3 ($x = 20$ m). A column 8 m long hangs from node 2 down to a pinned base at node 4. A uniform load of 8 kN/m covers the beam from the free end to node 2, and a 72 kN point load acts at $x = 16$ m, which is the mid-span of member 2–3. All members share the same $EI$ and are inextensible.

Find. The end moments, all reactions, and the shear force and bending moment diagrams with their extreme ordinates.

Approach. Show first that the frame cannot sway, so the only unknowns are the rotations at the roller (node 1) and at the beam–column joint (node 2). Write slope-deflection equations for the three members, using the modified stiffness for the pin-ended column, and solve two simultaneous equations.

1. Rule out sidesway. The beam is horizontal and inextensible, and node 3 is built in, so nodes 1 and 2 cannot translate horizontally. The column is vertical and inextensible and its base is pinned, so node 2 cannot translate vertically, and the roller holds node 1 vertically. Every joint is therefore fixed in position and the chord rotations $\psi$ all vanish. Only $\theta_1$ and $\theta_2$ remain unknown, because $\theta_3 = 0$ at the built-in end and the pinned base is handled by the modified stiffness.
2. Deal with the overhang. The 4 m length to the left of the roller is a cantilever carrying 8 kN/m. It applies a known hogging moment at node 1, $$ M_{\text{over}} = -\tfrac{1}{2}(8)(4)^2 = \boxed{-64\ \text{kN}\!\cdot\!\text{m}} $$ and a known downward force of 32 kN. Because the overhang has no rotational stiffness of its own, it must never be given a share of any balancing moment; it simply sets the boundary condition $M_{12} = -64$ kN·m.
3. Fixed-end moments. For member 1–2 ($L = 8$ m, $w = 8$ kN/m) the standard values are $\mathrm{FEM}_{12} = -wL^2/12 = -42.67$ kN·m and $\mathrm{FEM}_{21} = +42.67$ kN·m. For member 2–3 ($L = 8$ m with a 72 kN load at mid-span) they are $\mathrm{FEM}_{23} = -PL/8 = -72$ kN·m and $\mathrm{FEM}_{32} = +72$ kN·m. The unloaded column has none.
4. Slope-deflection equations. With clockwise end moments positive and $\psi = 0$ throughout, $$ M_{ij} = \frac{2EI}{L}\left(2\theta_i + \theta_j\right) + \mathrm{FEM}_{ij} $$ for the two beam members, while the pin-ended column uses the modified form $M_{24} = (3EI/L)\theta_2$ with $M_{42} = 0$.
5. Impose the two conditions. The boundary condition at node 1 is $M_{12} = -64$, that is $$ 0.25EI\left(2\theta_1 + \theta_2\right) - 42.67 = -64 $$ and joint 2 must balance, $M_{21} + M_{23} + M_{24} = 0$, that is $$ 0.25EI\left(2\theta_2 + \theta_1\right) + 42.67 + 0.5EI\theta_2 - 72 + 0.375EI\theta_2 = 0 . $$ Solving the pair, $$ \boxed{EI\theta_1 = -58.67,\qquad EI\theta_2 = +32.0\ \text{kN}\!\cdot\!\text{m}^2} $$
6. Back-substitute for the end moments. $$ M_{21} = +44.0, \quad M_{23} = -56.0, \quad M_{24} = +12.0, \quad M_{32} = +80.0\ \text{kN}\!\cdot\!\text{m} $$ with $M_{12} = -64.0$ kN·m as imposed. Joint 2 checks exactly: $44 - 56 + 12 = 0$. As an independent check the same frame was re-solved by direct stiffness, treating the members as inextensible bars with rotational and translational freedoms; the end moments agree to three figures.
7. Member end shears. For a member carrying a uniform load, $V_{\text{near}} = (M_{\text{far}} - M_{\text{near}} + wL^2/2)/L$ with the end moments expressed as sagging bending moments. For member 1–2 that gives $$ V_1 = \frac{-44 + 64 + \tfrac{1}{2}(8)(8)^2}{8} = \boxed{34.5\ \text{kN}} $$ falling to $-29.5$ kN at node 2, with zero shear at $34.5/8 = 4.31$ m from node 1. For member 2–3 with the mid-span load, $V_2 = (-80 + 56 + 72\times4)/8 = 33.0$ kN, falling to $-39.0$ kN at node 3. The column carries a constant shear of $12/8 = 1.5$ kN.
8. Reactions. The roller carries the span shear plus the overhang, $34.5 + 32 = \boxed{66.5\ \text{kN}\ (\uparrow)}$. The column takes the two beam shears meeting at node 2, $29.5 + 33.0 = \boxed{62.5\ \text{kN}}$, delivered to the pin at node 4 together with a horizontal thrust of 1.5 kN. The built-in end carries $\boxed{39.0\ \text{kN}\ (\uparrow)}$, a horizontal force of 1.5 kN and a fixing moment of $\boxed{80.0\ \text{kN}\!\cdot\!\text{m}}$. The three vertical reactions total 168 kN, exactly the applied $8(12) + 72 = 168$ kN.
9. Diagram ordinates. The bending moment runs from zero at the free end to $-64$ kN·m at the roller, rises to a sagging peak of $$ M_{\max} = -64 + 34.5(4.3125) - 4(4.3125)^2 = \boxed{+10.4\ \text{kN}\!\cdot\!\text{m}} $$ at 4.31 m past the roller, returns to $-44$ kN·m just left of node 2, steps to $-56$ kN·m just right of it (the column takes the 12 kN·m difference), rises to $+76$ kN·m under the 72 kN load and falls to $-80$ kN·m at the built-in end. The column moment falls linearly from 12 kN·m at node 2 to zero at the pin.