Question 5 of 9: Plane frame jacked at mid-span

Question 5: Plane frame jacked at mid-span (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 symmetric portal, columns 4.5 m with pinned bases, beam 4 m + 4 m, $EI=8.0\times10^{4}$ kN·m$^{2}$ in every member, all members inextensible, and a jack that raises joint 3 by 0.020 m. There is no applied load of any other kind.

Find. The bending moment and shear force diagrams, the force the jack must deliver, and the reactions the jacking induces.

Approach. Exploit symmetry to reduce the frame to a single unknown rotation, write slope-deflection for the half frame with the chord rotation produced by the jack, then recover shears and reactions.

1. Reduce the problem by symmetry. Geometry and imposed displacement are both symmetric about joint 3, so the response is symmetric: the sway vanishes ($u_{2}=u_{4}$ from the inextensible beam and $u_{2}=-u_{4}$ from symmetry give $u_{2}=u_{4}=0$), joint 3 does not rotate, and $\theta_{4}=-\theta_{2}$. The columns are inextensible with pinned bases, so $v_{2}=v_{4}=0$ and only $\theta_{2}$ remains unknown.
2. Compute the chord rotation the jack imposes. Joint 3 rises 0.020 m while joint 2 stays put, so for the horizontal member 2–3$$\psi_{23}=\frac{v_{3}-v_{2}}{L}=\frac{0.020}{4}=0.0050\ \text{rad}.$$This is the entire loading: the frame carries no applied force, only a prescribed geometry change.
3. Write the two member equations. The column has a pinned base, so its far-end moment is zero and its modified stiffness applies, while the beam is a full slope-deflection member with no span load:$$M_{21}=\frac{3EI}{h}\theta_{2},\qquad M_{23}=\frac{2EI}{L}\left(2\theta_{2}-3\psi_{23}\right),\qquad M_{32}=\frac{2EI}{L}\left(\theta_{2}-3\psi_{23}\right).$$No chord rotation appears in the column term because the frame does not sway.
4. Enforce moment equilibrium at joint 2. Setting $M_{21}+M_{23}=0$,$$\left(\frac{3EI}{4.5}+\frac{4EI}{4}\right)\theta_{2}=\frac{6EI}{4}\,\psi_{23}\;\Longrightarrow\;1.6667\,\theta_{2}=1.5(0.0050),$$so$$\theta_{2}=\boxed{0.00450\ \text{rad}}.$$The rotation is a shade less than the chord rotation itself, which is the physical signature of a fairly flexible column restraining a stiffer beam.
5. Recover the end moments. Substituting $EI=8.0\times10^{4}$ kN·m$^{2}$,$$M_{21}=+240\ \text{kN}\cdot\text{m},\qquad M_{23}=-240\ \text{kN}\cdot\text{m},\qquad M_{32}=-420\ \text{kN}\cdot\text{m},$$and by symmetry $M_{34}=+420$, $M_{43}=+240$ and $M_{45}=-240$ kN·m. Joint 2 balances ($240-240=0$) and so does joint 3 ($-420+420=0$).
6. Translate into diagram ordinates and reactions. In sagging terms the beam carries $+240$ kN·m at joints 2 and 4 and $-420$ kN·m over the jack, varying linearly because there is no span load, with a point of contraflexure$$x=\frac{240}{240+420}(4)=1.455\ \text{m}$$from each outer joint. The beam shear is constant at $(M_{23}+M_{32})/L=-165$ kN in each half, so the jack must push$$F_{\text{jack}}=2(165)=\boxed{330\ \text{kN upwards}},$$each pinned base must be held down with 165 kN, and each column carries a constant shear of $240/4.5=53.3$ kN, the two acting in opposite directions so that the horizontal reactions self-equilibrate.

The hold-down is the practical point of the question. Jacking a frame that was built with pinned bases generates uplift at those bases, and 165 kN of tension has to be delivered by anchor rods or by the dead load of the footing. A designer who reads only the moment diagram will miss it.