16-Civ-A1 Elementary Structural Analysis · December 2014
Question 7 of 8: Three-hinged frame — reactions and SFD/BMD
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy, method of joints/sections, virtual-work deflections, moment distribution, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges and compound (Gerber) structures, three-hinged frames. Sign convention: sagging bending moment positive; upward reactions positive; tension member forces positive (T), compression negative (C); distances in metres, forces in kN.
Exam rule: six questions constitute a complete paper — answer all of #1–#5 and only one of #6/#7/#8. All eight are solved here as a complete study resource.
Source note: Member geometry, support types and load directions for Questions 4, 7 and 8 were read from the printed figures; every reconstructed dimension yields clean, self-consistent equilibrium. Where a figure value was inferred, it is stated explicitly — verify against the original booklet if using for assessment.
Question 7: Three-hinged frame — reactions and SFD/BMD (20 marks)
Pin at A and pin at E; internal hinge at apex C. Left column 11 m; rafters 13 m (12 m × 5 m); right column 4 m. UDL 9 kN/m over the right rafter's 12 m projection (108 kN); 213 kN horizontal (←) at D.
Given. $A=(0,0)$, $B=(0,11)$, $C=(12,16)$ hinge, $D=(24,11)$, $E=(24,7)$. Two pins (4 reactions) + one hinge = determinate. Find. reactions and member diagram extremes.
Approach. Three global equilibrium equations plus the apex-hinge condition ($\sum M_C=0$ for one half) give the four reactions; internal forces then follow member by member.
Global + hinge equations. $\sum F_x: A_x+E_x=213$; $\sum F_y: A_y+E_y=108$; $\sum M_A=0$ and $\sum M_C^{\,right}=0$. Solving:
$$A_x=\boxed{60},\ A_y=\boxed{80},\quad E_x=\boxed{153},\ E_y=\boxed{28}\ \text{kN}.$$
Check (hinge, left half): $\sum M_C = -12A_y+16A_x = -960+960 = 0$ ✓.
Left column A–B and rafter B–C. Shear $=A_x=60$ kN; the moment grows linearly to $M_B=A_x(11)=\boxed{660\text{ kN}\cdot\text{m}}$, then tapers along the unloaded rafter to $0$ at the hinge C.
Right column E–D. Shear $=E_x=153$ kN; moment $M_D=E_x(4)=\boxed{612\text{ kN}\cdot\text{m}}$, tapering to $0$ at E.
Right rafter C–D. Carries the 108 kN UDL: $M=0$ at the hinge C rising to $612$ kN·m at D (a shallow parabola, peak $\approx612$ near D). The governing moment in the frame is $\boxed{660\text{ kN}\cdot\text{m at B}}$.
Bending-moment ordinates around the frame: 0 (A), 660 (B), 0 (C, hinge), 612 (D), 0 (E), all drawn on the tension face.
