Question 7 of 9: Flexibility (force) method — portal with combined gravity and horizontal load

Question 7: Flexibility (force) method — portal with combined gravity and horizontal load (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 portal with pinned bases at joints 1 and 4, an 18 m beam 2–3 and 6 m columns, all of the same constant $EI$. The beam carries 13.2 kN/m over its full length and a horizontal point load of 52.8 kN acts at joint 3, directed towards the right.

Find. The redundant horizontal reaction, all support reactions, and the shear force and bending moment diagrams with their extreme ordinates.

Approach. Release the horizontal restraint at base 4 to obtain a determinate primary structure, compute the horizontal opening $\Delta_{10}$ produced by the applied loads and the flexibility $f_{11}$ produced by a unit inward force, then close the gap with $\Delta_{10} + X f_{11} = 0$.

1. Choose the release and analyse the primary structure. Replacing the pin at 4 by a horizontal roller leaves a determinate frame. Horizontal equilibrium gives $H_1 = -P$ (the base pushes 52.8 kN to the left), and taking moments about joint 1, $$V_4 = \frac{w L^{2}/2 + P h}{L} = \frac{13.2(18)(9) + 52.8(6)}{18} = \frac{2138.4 + 316.8}{18} = \boxed{136.4\ \text{kN}}$$ $$V_1 = wL - V_4 = 237.6 - 136.4 = \boxed{101.2\ \text{kN}}$$ These vertical reactions are unaffected by the redundant, because the redundant is horizontal and both bases lie at the same level.
2. Set up the unit-load system. Apply a unit inward horizontal force at base 4. Because both bases are at the same level the unit force has no lever arm about joint 1, so it produces no vertical reactions at all: the entire system is $H_1 = +1$ balancing the unit force. The resulting moment diagram is $$m = y \ \text{in each column (}y\text{ from the base)}, \qquad m = h \ \text{in the beam}$$ constant along the beam because the only force acting is horizontal at base level.
3. Evaluate the flexibility. Both columns contribute the same $y^{2}$ integral and the beam contributes a constant $h^{2}$: $$f_{11} = \frac{1}{EI}\left[2\int_{0}^{h}y^{2}\,dy + \int_{0}^{L}h^{2}\,dx\right] = \frac{1}{EI}\left(\frac{2h^{3}}{3} + h^{2}L\right) = \frac{1}{EI}\left(144 + 648\right) = \frac{792}{EI}$$
4. Evaluate the opening under the UDL. With no horizontal base force in the primary structure the columns are moment-free under the UDL alone, so only the beam contributes: $$\Delta_{10}^{(w)} = -\frac{h}{EI}\int_{0}^{L}\left(\frac{wLx}{2}-\frac{wx^{2}}{2}\right)dx = -\frac{h}{EI}\cdot\frac{wL^{3}}{12} = -\frac{6(13.2)(5832)}{12\,EI} = -\frac{38\,491.2}{EI}$$ The negative sign records that the release opens outwards, against the unit inward load — the frame is spreading, exactly as in Question 4.
5. Evaluate the opening under the horizontal load. Now the left column does carry moment, running from zero at the pin to $Ph$ at joint 2, and the beam moment falls linearly from $Ph$ at joint 2 to zero at joint 3, while column 4–3 is moment-free: $$\Delta_{10}^{(P)} = -\frac{1}{EI}\left[\frac{Ph^{3}}{3} + \frac{Ph^{2}L}{2}\right] = -\frac{P h^{2}}{EI}\left(\frac{h}{3}+\frac{L}{2}\right) = -\frac{52.8(36)(2+9)}{EI} = -\frac{20\,908.8}{EI}$$ Both effects push the release the same way, so they add rather than cancel.
6. Solve the compatibility equation. Superposing, $$\Delta_{10} = -\frac{38\,491.2 + 20\,908.8}{EI} = -\frac{59\,400}{EI}$$ and enforcing zero horizontal movement at the real pin, $$\Delta_{10} + X f_{11} = 0 \qquad\Longrightarrow\qquad X = \frac{59\,400}{792} = \boxed{75.0\ \text{kN}}$$ directed inwards at base 4. Note that $EI$ cancels, which is why the question can be set without giving a numerical value.
7. Recover the base reactions. Superposing the primary and redundant systems at base 1, $$H_1 = -P + X = -52.8 + 75.0 = \boxed{22.2\ \text{kN inwards}}$$ so both bases push inwards, but by very different amounts: 22.2 kN on the left and 75.0 kN on the right. Horizontal equilibrium checks out, $22.2 + 52.8 = 75.0$.
8. Draw the column diagrams. Neither column carries a transverse load, so each has constant shear equal to its base thrust and a moment increasing linearly from zero at the pin: $$M_{2} = H_1 h = 22.2(6) = \boxed{133.2\ \text{kN}\cdot\text{m}}, \qquad M_{3} = X h = 75.0(6) = \boxed{450.0\ \text{kN}\cdot\text{m}}$$ Both are hogging on the beam. The asymmetry is entirely the work of the horizontal load: without it both corners would carry 245 kN·m.
9. Draw the beam diagrams. The beam shear falls linearly from $+101.2$ kN at joint 2 to $-136.4$ kN at joint 3 and vanishes at $x = 101.2/13.2 = 7.667$ m, where $$M_{\max} = -133.2 + \frac{101.2^{2}}{2(13.2)} = -133.2 + 387.93 = \boxed{+254.73\ \text{kN}\cdot\text{m}}$$ The moment passes through zero at $x = 1.454$ m and $x = 13.879$ m, so the sagging region is displaced towards the lightly loaded left-hand corner.