Question 5 of 8: Frame with an internal hinge, analysed by slope deflection

Question 5: Frame with an internal hinge, analysed by slope deflection (20 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 continuous beam on a pin, two propped columns and a roller, with an internal hinge in the first span.

Find. The member-end moments, the reactions, and shear and bending moment diagrams with every maximum and minimum ordinate labelled.

Approach. Use the hinge to detach the first span as a simple beam, treat what remains as a two-unknown slope-deflection problem in the rotations of joints ③ and ④ (no sidesway, because the beam is inextensible and pinned at ①), then recover shears by span equilibrium.

1. Establish that there is no sidesway. The members are inextensible, so the beam cannot change length; joint ① is pinned, so it cannot move horizontally; therefore joints ③, ④ and ⑤ cannot translate horizontally either. The columns are inextensible, so joints ③ and ④ cannot move vertically. Only rotations are unknown, and there are two of them.
2. Detach span ①–②. Both of its ends are moment-free (a pin at ①, a hinge at ②), so it is simply supported: $$R_{1} = \frac{24(8)}{2} = \boxed{96.0\ \text{kN}}$$ and an equal 96.0 kN is delivered through the hinge to the rest of the structure. Its own maximum sagging moment is $wL^{2}/8 = 192$ kN·m at $x = 4$ m.
3. Turn ②–③ into an overhang. The 2 m of beam between the hinge and joint ③ now carries the 96.0 kN point load at its tip plus its own 48 kN of distributed load, so it applies to joint ③ a known moment $$M_{32} = 96.0(2) + 24(2)(1) = \boxed{240\ \text{kN}\cdot\text{m}}$$ and a shear of $96 + 48 = 144$ kN.
4. Fixed-end moments. Writing end moments positive clockwise on the member, for the 16 m span $$FEM_{34} = -\frac{wL^{2}}{12} = -\frac{24(16)^{2}}{12} = -512\ \text{kN}\cdot\text{m}, \qquad FEM_{43} = +512\ \text{kN}\cdot\text{m}$$ and for the 9 m span, whose far end is a roller, the modified value $$FEM_{45} = -\frac{wL^{2}}{8} = -\frac{24(9)^{2}}{8} = -243\ \text{kN}\cdot\text{m}, \qquad FEM_{54} = 0 .$$
5. Slope-deflection equations. With $EI\theta$ as the unknowns and no chord rotations, and using $4I$ for beams and $I$ for columns, $$M_{36} = \theta_{3}, \qquad M_{47} = \theta_{4}$$ $$M_{34} = \theta_{3} + 0.5\theta_{4} - 512, \qquad M_{43} = 0.5\theta_{3} + \theta_{4} + 512$$ $$M_{45} = \tfrac{4}{3}\theta_{4} - 243$$ where the last line already carries the reduced $3EI/L$ stiffness of the pin-ended span.
6. Joint equilibrium. Setting the sum of end moments to zero at each rigid joint, $$\text{joint 3:}\quad 240 + (\theta_{3} + 0.5\theta_{4} - 512) + \theta_{3} = 0 \;\Rightarrow\; 2\theta_{3} + 0.5\theta_{4} = 272$$ $$\text{joint 4:}\quad (0.5\theta_{3} + \theta_{4} + 512) + (\tfrac{4}{3}\theta_{4} - 243) + \theta_{4} = 0 \;\Rightarrow\; 0.5\theta_{3} + \tfrac{10}{3}\theta_{4} = -269$$ Solving the pair, $$\boxed{EI\theta_{3} = 162.26\ \text{kN}\cdot\text{m}^{2},\qquad EI\theta_{4} = -105.04\ \text{kN}\cdot\text{m}^{2}}$$ The opposite signs are physically right: the long 16 m span drags joint ③ one way and joint ④ the other.
7. Back-substitute for the end moments. $$M_{34} = -402.3, \quad M_{43} = +488.1, \quad M_{45} = -383.0\ \text{kN}\cdot\text{m}$$ $$M_{36} = +162.3, \quad M_{63} = +81.1, \quad M_{47} = -105.0, \quad M_{74} = -52.5\ \text{kN}\cdot\text{m}$$ Each rigid joint checks: $240 - 402.3 + 162.3 = 0$ and $488.1 - 383.0 - 105.0 = 0$. In sagging-positive terms the beam moments are $-240$ kN·m just left of joint ③, $-402.3$ kN·m just right of it, $-488.1$ kN·m at joint ④ from the left and $-383.0$ kN·m from the right.
8. Span shears. For the 16 m span, equating the moment at its right end to the value found, $$V_{3} = \frac{488.1 - 402.3 + 24(16)^{2}/2}{16} = 186.6\ \text{kN}, \qquad V_{4}^{L} = 186.6 - 384 = -197.4\ \text{kN}$$ The shear vanishes 7.777 m into the span, where $$M_{\max} = -402.3 + 186.6(7.777) - 12(7.777)^{2} = \boxed{+323.4\ \text{kN}\cdot\text{m}}$$ For the 9 m span, $V_{4}^{R} = (383.0 + 972)/9 = 150.6$ kN, the roller reaction is $216 - 150.6 = \boxed{65.4\ \text{kN}}$, and the sagging peak 6.273 m along is $+89.2$ kN·m.
9. Reactions and the overall check. Each column carries the jump in beam shear across its joint: $$R_{6} = 186.6 - (-144) = 330.6\ \text{kN}, \qquad R_{7} = 150.6 - (-197.4) = 347.9\ \text{kN}$$ Adding the four vertical reactions, $96.0 + 330.6 + 347.9 + 65.4 = 840$ kN, which is exactly $24 \times 35 = 840$ kN. The two column shears, $(162.3 + 81.1)/4 = 60.9$ kN and $(-105.0 - 52.5)/4 = -39.4$ kN, do not cancel; their resultant of 21.5 kN is taken in axial thrust along the beam by the pin at ①, which is why the frame does not sway.