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16-Civ-A1 Elementary Structural Analysis · December 2013

Question 8 of 8: Reactions and SFD/BMD of a hinged frame

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Notes on this paper

National Exams, December 2013 — 98-Civ-A1 Elementary Structural Analysis. Three-hour, closed-book examination (approved Sharp/Casio calculator only). Format: six questions constitute a complete paper — answer all of #1–#5 and any one of #6, #7 or #8. All eight questions are solved below for completeness.

Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy, method of joints/sections, virtual-work deflections, moment distribution, slope–deflection, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges and compound structures. Sign convention: sagging bending moment positive; upward reactions positive; tension member forces positive (T), compression negative (C).

Question 8: Reactions and SFD/BMD of a hinged frame (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.

50 kN 20 kN/m 1 2 (hinge) 3 4 4.5 m 3.5 m 4.5 m 1.5 m
Frame 8: pins at feet 1 and 4, an internal hinge at node 2; 50 kN down on the top member (1.5 m right of node 2); a 20 kN/m horizontal (leftward) load over the 6 m rise of member 3–4.

Given. Trapezoidal frame 1(0,0)–2(4.5,6)–3(8,6)–4(12.5,0) m; pins at feet 1 and 4; an internal hinge at node 2; rigid corner at node 3. Loads: 50 kN downward on top member 2–3, 1.5 m right of node 2 (at x = 6 m); a horizontal UDL of 20 kN/m acting leftward on the inclined member 3–4, distributed over its 6 m rise (resultant 120 kN ← at elevation y = 3 m). Determinate (two support pins + one internal hinge).

Find. The four reaction components and the SFD/BMD with extreme ordinates.

  1. Global moment about foot 1. $\sum M_1 = 0$: $-50(6) + 120(3) + V_4(12.5) = 0 \Rightarrow V_4 = -4.8$ kN, i.e. $\boxed{4.8\ \text{kN}\downarrow}$ at foot 4.
  2. Vertical equilibrium. $V_1 = 50 - (-4.8) = \boxed{54.8\ \text{kN}\uparrow}$.
  3. Hinge condition on member 1–2. With no load on member 1–2, moment zero at hinge 2 requires $6H_1 = 4.5V_1$, so $H_1 = 0.75(54.8) = \boxed{41.1\ \text{kN}\rightarrow}$; member 1–2 is therefore a two-force strut carrying $\sqrt{41.1^2+54.8^2} = 68.5$ kN in compression.
  4. Horizontal equilibrium. $H_4 = 120 - 41.1 = \boxed{78.9\ \text{kN}\rightarrow}$.
  5. Moment diagram. Member 1–2: zero throughout (two-force member). Member 2–3: 0 at the hinge, $-82.2$ kN·m under the 50 kN load, $-91.8$ kN·m at node 3. Member 3–4: from $-91.8$ kN·m at node 3, the leftward UDL bows the leg to a peak $\boxed{-141.8\ \text{kN}\cdot\text{m}}$ (about 4.7 m up from foot 4) before returning to zero at pin 4 — this is the critical moment in the frame.
−82 −92 −142 (max) hinge (0) 1 4
BMD 8 (developed along members 1–2–3–4): zero on the two-force strut 1–2, rising to −82 under the 50 kN load and −92 at node 3, peaking at −142 kN·m on leg 3–4.
QuantityValue
Foot 1 (pin)$H_1=41.1$ kN →, $V_1=54.8$ kN ↑
Foot 4 (pin)$H_4=78.9$ kN →, $V_4=4.8$ kN ↓
Member 1–2two-force strut, 68.5 kN compression, M = 0
Under 50 kN load (member 2–3)M = −82.2 kN·m
Node 3M = −91.8 kN·m
Member 3–4 (≈4.7 m up)M = −141.8 kN·m (maximum)
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