Question 7 of 8: Three-hinged frame: reactions and SFD/BMD

Question 7: Three-hinged frame: reactions and SFD/BMD (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 trapezoidal three-hinged frame: pins at feet 1 (0, 0) and 5 (12.5, 0), internal hinge at 3 (8, 6). Flat top at 6 m from joint 2 (2.5, 6) to joint 4 (10, 6). A 30 kN vertical load acts on the top 5 m right of joint 2; a horizontal UDL of $6\tfrac14=6.25$ kN per vertical metre acts on the right leg 4–5 (total $6.25\times6=37.5$ kN, pointing left).

Find. The four support reaction components and the member SFD/BMD extremes.

Approach. Four reaction unknowns (two pins) with three global equations plus one condition of the internal hinge ($\sum M=0$ of one part about the hinge) — a determinate three-hinged frame.

1. Global equilibrium. $\sum F_x$: $H_1+H_5=37.5$. $\sum F_y$: $V_1+V_5=30$. $\sum M_1=0$: $-30(7.5)+37.5(3)+12.5V_5=0\Rightarrow V_5=9.0$ kN, $V_1=21.0$ kN.
2. Hinge condition. Right part about hinge 3: $4.5V_5+6H_5-37.5(3)=0\Rightarrow H_5=12.0$ kN; then $H_1=37.5-12=25.5$ kN. (Both horizontal reactions point inward/right.)
3. Bending moments. Pins: $M_1=M_5=0$. Knee 2: $M_2=2.5V_1-6H_1=... $ ==**$100.5$ kN·m (max)**==. Under the 30 kN load: $M=4.5$ kN·m. Hinge 3: $M=0$. Knee 4: $M_4=18$ kN·m.
4. Right leg 4–5. The horizontal UDL bends the leg parabolically: $M$ rises from 0 at foot 5 to an interior maximum ==**$\approx19.8$ kN·m (2.5 m up)**== and is 18 kN·m (opposite face) at knee 4.