Question 8 of 9: Anti-symmetric portal with loaded columns and free stubs (24 marks)

Question 8: Anti-symmetric portal with loaded columns and free stubs (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 portal carrying an anti-symmetric horizontal load.

Find. Shear force and bending moment diagrams with the maximum and minimum ordinate on each member.

Approach. Anti-symmetry equates the two joint rotations and gives one sway, the stubs are determinate cantilevers, and the storey shear closes the system.

1. Establish the anti-symmetry. Reflecting the frame about its vertical centre line maps one column onto the other and reverses the sense of a horizontal force. Both loads point right, so the reflected load case is the negative of the original: the loading is anti-symmetric. The response is therefore anti-symmetric too, which means $$\theta_{2}=\theta_{3}=\theta,\qquad u_{2}=u_{3}=\Delta,\qquad M_{\text{beam}}(\text{mid-span})=0.$$ Two unknowns, $\theta$ and $\Delta$, instead of three.
2. Replace each stub by its end actions. The 1 m length above the beam is a free-ended cantilever, so it is statically determinate. It delivers to the joint a horizontal force and a couple $$F_{\text{stub}}=wc=12(1)=12.0\ \text{kN},\qquad M_{\text{stub}}=\frac{wc^{2}}{2}=\frac{12(1)^{2}}{2}=6.0\ \text{kN}\cdot\text{m}.$$ Dropping the couple is the classic error here — it still satisfies the storey shear, so it is invisible until the moment diagram refuses to close at the joint.
3. Fixed-end moments of the column below the beam. Treating the 10 m length as a member from the base (end $i$) to the joint (end $j$) with the load acting transversely, $$M^{F}_{AB}=+\frac{wh^{2}}{12}=+\frac{12(10)^{2}}{12}=+100.0,\qquad M^{F}_{BA}=-100.0\ \text{kN}\cdot\text{m}.$$
4. Slope-deflection equations. With $2EI/10=0.2EI$ on the columns and $2EI/10=0.2EI$ on the beam, a fixed base ($\theta=0$) and $\psi=-\Delta/10$ on each column, $$M_{AB}=0.2EI\bigl(\theta+0.3\Delta\bigr)+100,\qquad M_{BA}=0.2EI\bigl(2\theta+0.3\Delta\bigr)-100,$$ $$M_{BE}=M_{EB}=0.2EI\bigl(2\theta+\theta\bigr)=0.6EI\,\theta$$ (the beam has $\psi=0$ because both joints translate horizontally by the same amount).
5. First equation: moment equilibrium at the joint. Summing the column, the beam and the stub couple at joint 2, $$0.2EI\bigl(2\theta+0.3\Delta\bigr)-100+0.6EI\,\theta+6=0\;\Rightarrow\;EI\,\theta+0.06\,EI\,\Delta=94.$$
6. Second equation: storey shear. The total applied horizontal load is $12(11)(2)=264$ kN and anti-symmetry splits it equally, so each base carries $V=132.0$ kN. Since $V=(M_{AB}+M_{BA})/10+\tfrac{1}{2}w(10)$, $$\frac{M_{AB}+M_{BA}}{10}+60=132\;\Rightarrow\;M_{AB}+M_{BA}=720,$$ and substituting the slope-deflection expressions, $$0.6\,EI\,\theta+0.12\,EI\,\Delta=720.$$
7. Solve the pair. Eliminating $\theta$, $$0.084\,EI\,\Delta=663.6\;\Rightarrow\;\boxed{EI\,\Delta=7900\ \text{kN}\cdot\text{m}^{3}},\qquad EI\,\theta=94-0.06(7900)=-380\ \text{kN}\cdot\text{m}^{2}.$$ The frame sways to the right, as it must under a net rightward load.
8. End moments. Substituting, $$M_{\text{base}}=498.0,\qquad M_{\text{column, at the beam}}=222.0,\qquad M_{\text{beam ends}}=\pm 228.0\ \text{kN}\cdot\text{m}.$$ The decisive check is at the joint: the column brings 222.0, the stub brings 6.0, and together they must equal the beam moment, $$222.0+6.0=228.0\ \checkmark$$ Both ordinates on either side of the joint must be shown — the diagram genuinely steps there.
9. Shears and diagram shapes. Each column shear runs linearly from $$V=132.0\ \text{kN at the base to } V=132.0-12(10)=12.0\ \text{kN under the beam},$$ and the stub finishes it off from 12.0 kN to zero at the free top. The column moment is parabolic, $M(y)=-498.0+132.0y-6y^{2}$, with its point of contraflexure at $$y=\frac{132-\sqrt{132^{2}-4(6)(498)}}{2(6)}=4.836\ \text{m}.$$ The beam carries no span load, so its moment is a straight line from $+228.0$ to $-228.0$ kN·m through zero at mid-span, with a constant shear $$V_{\text{beam}}=\frac{2(228.0)}{10}=45.6\ \text{kN},$$ which is also the axial force in each column (down on one, up on the other — the vertical reactions are $\mp45.6$ kN and sum to zero, as they must with no vertical load).