Question 6 of 9: Symmetric frame with internal hinges, by slope-deflection (24 marks)

Question 6: Symmetric frame with internal hinges, by slope-deflection (24 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 symmetric frame on two fixed bases: 2 m vertical columns 1–2 and 5–6, inclined members 2–3 and 4–5 rising 2 m over a 1.5 m run, and a horizontal member 3–4 of 8.8 m carrying a 6 kN/m UDL. Joints 3 and 4 are internal hinges.

Find. The end moments, and the shear force and bending moment diagrams with their maximum and minimum ordinates.

Approach. The hinges at joints 3 and 4 make the loaded member a simple span, so the frame reduces to two identical trees connected by an inextensible strut; symmetry then leaves one rotation and one sway parameter on each half.

1. Deal with the loaded member first. Member 3–4 is released at both ends, so it is a simple span: $$R_{3}=R_{4}=\frac{wS}{2}=\frac{6(8.8)}{2}=26.4\ \text{kN},\qquad M_{\text{mid}}=\frac{wS^{2}}{8}=\frac{6(8.8)^{2}}{8}=\boxed{58.08\ \text{kN}\cdot\text{m}}$$ It delivers those 26.4 kN vertically to joints 3 and 4 and no moment at all.
2. Fix the kinematics of one half. Column 1–2 is vertical and inextensible on a fixed base, so $v_{2}=0$. Member 3–4 is inextensible and the response is symmetric, which forces $u_{3}=-u_{4}$ and $u_{3}=u_{4}$ together, hence $u_{3}=u_{4}=0$. Inextensibility of member 2–3 then ties the remaining pair: $$(u_{3}-u_{2})(1.5)+(v_{3}-v_{2})(2)=0\;\Longrightarrow\;v_{3}=\frac{3}{4}u_{2}$$ so one sway parameter $\Delta=u_{2}$ and one rotation $\theta_{2}$ remain.
3. Write the chord rotations. $$\psi_{12}=-\frac{\Delta}{2},\qquad \psi_{23}=+\frac{\Delta}{2}$$ the second following from $\left[(\mathbf{D}_{3}-\mathbf{D}_{2})\cdot\mathbf{n}\right]/L$ with $\mathbf{n}=(-0.8,\,0.6)$ and $\mathbf{D}_{3}-\mathbf{D}_{2}=(-\Delta,\,0.75\Delta)$.
4. Write the two member equations. Member 1–2 is fixed at 1 and member 2–3 is released at the hinge, so $$M_{12}=EI\left(\theta_{2}+1.5\Delta\right),\qquad M_{21}=EI\left(2\theta_{2}+1.5\Delta\right),\qquad M_{23}=1.2EI\left(\theta_{2}-0.5\Delta\right)$$ with $M_{32}=0$ at the hinge.
5. Impose joint equilibrium at 2. $M_{21}+M_{23}=0$ gives $$3.2\,\theta_{2}+0.9\,\Delta=0\;\Longrightarrow\;\theta_{2}=-0.28125\,\Delta$$
6. Write the sway equation by virtual work. For a virtual $\Delta^{*}=1$ the chords rotate by $\psi^{*}_{12}=-0.5$ and $\psi^{*}_{23}=+0.5$ while joint 3 rises $0.75$, so $$-0.5\left(M_{12}+M_{21}\right)+0.5\,M_{23}-26.4(0.75)=0$$ Substituting the member equations gives $-0.9\,EI\theta_{2}-1.8\,EI\Delta=19.8$.
7. Solve and recover the end moments. The pair yields $$EI\Delta=-12.8,\qquad EI\theta_{2}=+3.6$$ and therefore $$M_{12}=3.6+1.5(-12.8)=\boxed{-15.6\ \text{kN}\cdot\text{m}},\qquad M_{21}=\boxed{-12.0\ \text{kN}\cdot\text{m}},\qquad M_{23}=1.2(3.6+6.4)=\boxed{+12.0\ \text{kN}\cdot\text{m}}$$
8. Get the shears and the thrust. The column has no span load, so $$V_{12}=\frac{M_{12}+M_{21}}{L}=\frac{-15.6-12.0}{2}=-13.8\ \text{kN}$$ a horizontal shear of 13.8 kN; the inclined member carries $\left(12.0+0\right)/2.5=4.8$ kN normal to itself, and joint equilibrium puts 13.8 kN of axial compression into member 3–4.

The result is worth reading physically. The frame spreads: joint 2 moves outward by $12.8/EI$ and joint 3 drops by $9.6/EI$, and it is member 3–4, held in compression at 13.8 kN, that stops the two halves from separating further. Each column carries the whole 26.4 kN of its half of the roof load, plus a 13.8 kN horizontal thrust and a 15.6 kN·m fixing moment at the base. Notice too that the column moment changes sign along its length, from 15.6 kN·m at the base to 12.0 kN·m on the opposite face at joint 2, so there is a point of contraflexure $15.6/13.8=1.13$ m above the base — a detail worth marking on the diagram because it is where a lap splice must not be placed.