Question 5 of 9: Slope-deflection for a frame jacked at joint 3 (16 marks)

Question 5: Slope-deflection for a frame jacked at joint 3 (16 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 stepped frame of three 5 m members: a horizontal member 1–2 from a pin at joint 1, a vertical member 2–3 dropping to joint 3, and a horizontal member 3–4 running to a pin at joint 4. No applied loads; a jack raises joints 2 and 3 by 0.02 m.

Find. The end moments, the shear and bending moment diagrams with their extreme ordinates, and the force the jack must deliver.

Approach. The imposed lift is a prescribed displacement, so the chord rotations are known at the outset and the only unknowns are the two joint rotations, which anti-symmetry reduces to one.

1. Fix the kinematics. Member 1–2 is horizontal and inextensible from a pin, so $u_{2}=0$; member 3–4 likewise gives $u_{3}=0$; member 2–3 is vertical and inextensible, so $v_{2}=v_{3}$ — which is why lifting joint 3 necessarily lifts joint 2 by the same 0.020 m. There is no sway degree of freedom at all.
2. Write the chord rotations. With $\Delta=0.020$ m upward, $$\psi_{12}=-\frac{\Delta}{L}=-0.004,\qquad\psi_{23}=0,\qquad \psi_{34}=+\frac{\Delta}{L}=+0.004$$ the vertical member has no chord rotation because both its ends rise together.
3. Write the slope-deflection equations. Joints 1 and 4 are pins, so those members take the modified form: $$M_{21}=\frac{3EI}{L}\left(\theta_{2}-\psi_{12}\right),\qquad M_{34}=\frac{3EI}{L}\left(\theta_{3}-\psi_{34}\right)$$ $$M_{23}=\frac{2EI}{L}\left(2\theta_{2}+\theta_{3}\right),\qquad M_{32}=\frac{2EI}{L}\left(2\theta_{3}+\theta_{2}\right)$$
4. Use anti-symmetry and solve. Rotating the frame through $180^\circ$ about the mid-point of member 2–3 maps it onto itself and reverses the imposed lift, so $\theta_{3}=-\theta_{2}$. Joint 2 equilibrium $M_{21}+M_{23}=0$ then gives $$\left(\frac{3EI}{L}+\frac{2EI}{L}\right)\theta_{2}=-\frac{3EI\Delta}{L^{2}}\;\Longrightarrow\;\theta_{2}=-\frac{3\Delta}{5L}=-0.0024\ \text{rad}$$
5. Recover the end moments. With $EI/L=5000$ kN·m, $$M_{21}=3(5000)\left(-0.0024+0.004\right)=\boxed{24.0\ \text{kN}\cdot\text{m}}$$ and by anti-symmetry $M_{23}=-24.0$, $M_{32}=+24.0$, $M_{34}=-24.0$, with $M_{12}=M_{43}=0$ at the pins.
6. Get the shears and the jack force. Each horizontal member carries $$V=\frac{M_{21}}{L}=\frac{24}{5}=\boxed{4.8\ \text{kN}}$$ while the vertical member has $M_{23}+M_{32}=0$ and therefore zero shear and a constant moment of 24 kN·m. Both pins hold down with 4.8 kN, so the jack must push up with $\boxed{9.6\ \text{kN}}$.

The shape that falls out is unusually clean: the bending moment rises linearly from zero at each pin to 24 kN·m at the inner joint, and the column simply carries that 24 kN·m unchanged from top to bottom. Because the column takes no shear, it also takes no horizontal force, so both horizontal reactions are exactly zero — the entire jacking effort is resisted by a pair of 4.8 kN hold-down forces at the pins. That is the practical lesson of the question: jacking a member of a rigid frame does not just deflect it, it demands anchorage at supports that were never intended to carry uplift.