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16-Civ-A1 Elementary Structural Analysis · May 2015

Question 8 of 8: Three-hinged gable frame

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 / slope–deflection, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams and three-hinged frames. Sign convention: sagging bending moment positive; upward reactions positive; member tension positive (T), compression negative (C).

The paper directs candidates to answer Q1–Q5 and then one of Q6/Q7/Q8. For completeness all eight questions are worked here.

Question 8: Three-hinged gable frame (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.

2.5 kN/m30 kN1234512 m9 m
Three-hinged gable frame: pins at 1 and 5, crown hinge at 3; 2.5 kN/m on the left rafter (horizontal projection) and 30 kN horizontal at point 4 level.

Given. Pins at 1 $(0,0)$ and 5 $(21,-3)$, crown hinge at 3 $(12,9)$; eaves 2 $(0,4)$ and 4 $(21,5.25)$; both rafters slope $12{:}5$. Loads: $2.5$ kN/m over the left rafter (horizontal projection, $0\!-\!12$ m) and a $30$ kN horizontal load (leftward) at $(21,0)$. Find. Reactions and SFD/BMD.

Approach. A three-hinged frame is determinate: three global equations plus $M=0$ at the crown give the four reaction components. Then walk each member with a free body to build the moment diagram.

  1. Resultant of the UDL. $2.5\times12=30$ kN down at $x=6$ m.
  2. Global + crown equations. $\sum M_1=0:\;21A_{5y}+3A_{5x}=180$; crown release (right part) $9A_{5y}+12A_{5x}=270$. Solving: $A_{5y}=6,\ A_{5x}=18$ kN.
  3. Remaining reactions. $\sum F_y:\ A_{1y}=30-6=\boxed{24\ \text{kN}\uparrow}$; $\sum F_x:\ A_{1x}=30-18=\boxed{12\ \text{kN}\rightarrow}$; and $A_5=(\boxed{18\ \text{kN}\rightarrow},\ \boxed{6\ \text{kN}\uparrow})$. (The $30$ kN load passes through node 1’s level, so it drops out of $\sum M_1$.)
  4. Bending moments. $M=0$ at both pins and the crown. The left eaves (joint 2) carries $A_{1x}(4)=48$ kN·m; the UDL bows the left rafter to a reversed $-24.2$ kN·m at $x=7.6$ m; the right eaves (joint 4) carries $9$ kN·m; and the right column peaks at $\boxed{54\ \text{kN}\cdot ext{m}}$ where the $30$ kN load acts — the largest moment in the frame.
48 @eaves−24.20 (hinge)54 @load−9Q8 BMD (kN·m) along frame axis — |M|max = 54 at the 30 kN load
Bending-moment diagram traced along the frame axis (kN·m); $M=0$ at both pins and the crown hinge.
QuantityValue
Reaction at 112 kN →, 24 kN ↑
Reaction at 518 kN →, 6 kN ↑
$M$ at left eaves (2)48 kN·m
$M$ on left rafter (reversed)−24.2 kN·m
$M$ at right eaves (4)9 kN·m
$M$ max (at 30 kN load)54 kN·m
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