Question 4 of 9: Least work — two-hinged portal frame under a uniform load

Question 4: Least work — two-hinged portal frame under a uniform load (18 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 rectangular portal frame with joints numbered 1 (left base), 2 (top left), 3 (top right) and 4 (right base). Both bases are pinned. The beam 2–3 spans 18 m and carries a uniformly distributed load of 22 kN/m over its whole length; both columns are 6 m high. All three members have the same constant $EI$.

Find. The bending moment at each end of beam 2–3 and the bending moment at its mid-span.

Approach. The frame is once statically indeterminate, so choose the horizontal thrust $H$ at the pins as the redundant, write the bending moment in every member as a function of $H$, and impose the theorem of least work $\partial U/\partial H = 0$.

1. Confirm the degree of indeterminacy and pick the redundant. With $m = 3$, $j = 4$ and $r = 4$ the frame is $3(3) + 4 - 3(4) = 1$ times statically indeterminate. The structure and the loading are both symmetric about mid-span, so the vertical reactions follow from symmetry at once: $$V_1 = V_4 = \frac{wL}{2} = \frac{22 \times 18}{2} = \boxed{198\ \text{kN}}$$ and the single unknown left is the inward horizontal thrust $H$ shared by the two pins.
2. Write the moment in each member in terms of $H$. Measuring $y$ upwards from a pin, each column carries only the base thrust, so $$M_{\text{col}} = H\,y, \qquad \frac{\partial M}{\partial H} = y$$ For the beam, measuring $x$ from joint 2, the free bending moment of a simply supported span is reduced everywhere by the constant $H h$ that the thrust supplies: $$M_{\text{beam}} = \frac{wL}{2}x - \frac{w x^{2}}{2} - H h, \qquad \frac{\partial M}{\partial H} = -h$$
3. Impose least work. Because no external force does work through the redundant, the strain energy is stationary with respect to it: $$\frac{\partial U}{\partial H} = \frac{1}{EI}\int M\,\frac{\partial M}{\partial H}\,ds = 0$$ Splitting the integral into the two columns and the beam, $$\frac{2}{EI}\int_{0}^{h}(H y)(y)\,dy + \frac{1}{EI}\int_{0}^{L}\left[\frac{wL x}{2} - \frac{w x^{2}}{2} - H h\right](-h)\,dx = 0$$
4. Carry out the integration. The column term gives $2Hh^{3}/3$. In the beam term, $$\int_{0}^{L}\left(\frac{wLx}{2} - \frac{wx^{2}}{2}\right)dx = \frac{wL^{3}}{4} - \frac{wL^{3}}{6} = \frac{wL^{3}}{12}$$ so the condition becomes $$\frac{2Hh^{3}}{3} - \frac{h\,wL^{3}}{12} + H h^{2}L = 0$$ Solving for the thrust and clearing the fractions, $$\boxed{H = \frac{wL^{3}}{8h^{2} + 12hL}}$$ a result worth memorising for any symmetric two-hinged rectangular portal of uniform $EI$.
5. Evaluate the thrust. Substituting the given data, $$H = \frac{22(18)^{3}}{8(6)^{2} + 12(6)(18)} = \frac{22 \times 5832}{288 + 1296} = \frac{128\,304}{1584} = \boxed{81.0\ \text{kN}}$$ The thrust is inward at both pins, which is the expected sense: a portal loaded on its beam tries to spread, and the supports push back.
6. Read the beam end moments. Each column carries a linear moment running from zero at its pin to $H h$ at the joint, and joint equilibrium transfers that value straight into the beam: $$M_{2} = M_{3} = -H h = -(81.0)(6) = \boxed{486\ \text{kN}\cdot\text{m}\ \text{hogging}}$$ The equality of the two ends is a direct check on the symmetry of the solution.
7. Compute the mid-span moment and locate the contraflexure points. At mid-span the free moment is $wL^{2}/8$, reduced by the same constant $Hh$: $$M_{\text{mid}} = \frac{wL^{2}}{8} - H h = \frac{22(18)^{2}}{8} - 486 = 891 - 486 = \boxed{405\ \text{kN}\cdot\text{m}\ \text{sagging}}$$ Setting $M(x) = -486 + 198x - 11x^{2} = 0$ gives $x = 2.932$ m and $x = 15.068$ m, so the beam hogs over roughly the outer 2.9 m at each end and sags over the middle 12.1 m.
8. Check the answer two ways. First, statics: the sum of the hogging end moment and the sagging mid-span moment must return the free moment, $486 + 405 = 891 = wL^{2}/8$, which it does exactly. Second, the limits: as $h \to 0$ the formula gives $H \to wL^{2}/(12h) \to \infty$ (a very flat frame is nearly a two-hinged arch), while as $h \to \infty$ it gives $H \to 0$ and the beam reverts to a simple span — both are the physically correct trends.