Question 8 of 8: Three-hinged frame — reactions and internal-force diagrams

Question 8: Three-hinged frame — reactions and internal-force diagrams (22 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 three-hinged frame: pinned feet $1(1,1)$ and $5(14.5,0)$, crown hinge $3(12,6)$; top beam $2(0,6)\!-\!3\!-\!4(13.5,6)$; inclined legs $1\!-\!2$ and $4\!-\!5$. Loading on the top beam: $32.5\text{ kN}$ down at $x=6\text{ m}$ and $2\text{ kN/m}$ over $0\!-\!12\text{ m}$ (resultant $24\text{ kN}$ at $x=6$).

Find. The four reaction components and the SFD/BMD (max/min per member).

Approach. Three hinges make the frame determinate: three global equilibrium equations plus the condition that the bending moment at the crown hinge is zero (taken on the right half).

1. Total load. $P=32.5+2(12)=56.5\text{ kN}$ down, resultant at $x=6\text{ m}$.
2. Crown-hinge condition (right of 3). Only the foot 5 acts on the right part: $2.5\,V_5+6\,H_5=0$.
3. Global equilibrium. $\sum M_1:\;13.5\,V_5+H_5=56.5(5)=282.5$; combined with the hinge condition, $V_5=\boxed{21.6\text{ kN}}$, $H_5=\boxed{9.0\text{ kN}\ (\leftarrow)}$. Then $V_1=56.5-21.6=\boxed{34.9\text{ kN}}$, $H_1=\boxed{9.0\text{ kN}\ (\rightarrow)}$.
4. Bending moments. Zero at both pins and at the crown. Knee $2$: $+79.9\text{ kN}\cdot\text{m}$; under the $32.5\text{ kN}$ load ($x=6$): $\boxed{-93.6\text{ kN}\cdot\text{m}}$ (largest magnitude); knee $4$: $\mp32.4\text{ kN}\cdot\text{m}$.