Question 8 of 8: Three-hinged gable frame (answer one of Q6/Q7/Q8)

Question 8: Three-hinged gable frame (22 marks) (answer one of Q6/Q7/Q8)

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 gable: pins at feet $1(0,0)$ and $5(24,0)$, eaves $2(0,7)$ and $4(24,7)$, crown hinge $3(12,12)$; $span 24\text{ m}$, eave $7\text{ m}$, apex rise $5\text{ m}$. UDL $8.45\text{ kN/m}$ (vertical, on the horizontal projection) over the left rafter and UDL $4.26\text{ kN/m}$ (horizontal, $\rightarrow$) over the left column. Find. Reactions and the SFD/BMD (max/min ordinates).

1. Loads. The rafter UDL over its $12\text{ m}$ horizontal projection totals $8.45(12)=101.4\text{ kN}\downarrow$ at $x{=}6$; the column UDL totals $4.26(7)=29.82\text{ kN}\rightarrow$ at $y{=}3.5$.
2. Global equilibrium. $\sum M_{1}=0:\;V_5(24)=101.4(6)+29.82(3.5)\Rightarrow \boxed{V_5=29.7\text{ kN}}$, so $\boxed{V_1=71.7\text{ kN}}$.
3. Crown-hinge condition. $M=0$ at the crown; taking the right half (crown to support 5, no applied load), $\sum M_{3}=0$ gives $\boxed{H_5=-29.7\text{ kN}}$ (i.e.\ $29.7\text{ kN}\leftarrow$), and horizontal equilibrium leaves $\boxed{H_1\approx0}$ — the wind thrust is reacted almost entirely at support 5.
4. Member moments. The left column, pinned at $1$ and loaded by its UDL, hogs to $\boxed{-103.5\text{kN}\cdot\text{m}}$ at the eave $2$; the crown is a hinge ($M=0$); and the largest moment is at the right eave $4$, $\boxed{|M_4|=207.9\text{kN}\cdot\text{m}}$ ($=H_5\times7$).