Question 6 of 8: Two-pinned portal frame — reactions and diagrams

Question 6: Two-pinned portal frame — reactions and diagrams (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. Symmetric portal: pin at (1)$=(0,0)$ and (4)$=(13,0)$; corners (2)$=(3,4)$, (3)$=(10,4)$; $13\text{ kN/m}$ UDL on the $7\text{ m}$ top beam (2)–(3); inclined legs rise $4$ m over $3$ m. Uniform $EI$.

Find. Reactions and the $V,M$ diagrams for members (2)–(3) and (3)–(4).

Approach. With two pinned bases the frame is indeterminate to the 1st degree; the horizontal thrust $H$ is the redundant. By symmetry the vertical reactions are equal. For a uniform-$EI$ two-pinned frame $H$ follows from $\displaystyle H=\frac{\int M_0\,m_1/EI\,ds}{\int m_1^2/EI\,ds}$ — $EI$ cancels, which is why the problem needs no numerical $EI$.

1. Vertical reactions. Total load $=13(7)=91\text{ kN}$; by symmetry $R_{(1)}=R_{(4)}=\boxed{45.5\text{ kN}\uparrow}$.
2. Horizontal thrust. Releasing one horizontal reaction and enforcing zero relative horizontal displacement gives $H=\boxed{43.1\text{ kN}}$ (inward at each base).
3. Corner moments. $M_{(2)}=M_{(3)}=\boxed{-35.96\text{ kN}\cdot\text{m}}$ (hogging).
4. Member (2)–(3). With hogging ends $-35.96$ and free sagging $\tfrac{13(7)^2}{8}=79.6$, the mid-span moment is $-35.96+79.6=\boxed{+43.66\text{ kN}\cdot\text{m}}$ (sagging). Shear runs $+45.5$ at (2) to $-45.5$ at (3), zero at mid-span.
5. Member (3)–(4). The unloaded leg carries a straight-line moment from $\boxed{-35.96\text{ kN}\cdot\text{m}}$ at (3) to $0$ at the pin (4); axial compression $\approx62.3\text{ kN}$, constant shear $\approx7.2\text{ kN}$.