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16-Civ-A1 Elementary Structural Analysis · December 2016

Question 5 of 8: Frame by moment distribution / slope-deflection

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 (Ch. 2), method of joints/sections (Ch. 3), shear & moment diagrams (Ch. 4), virtual-work deflections (Ch. 8–9), influence lines (Ch. 6), moment distribution / slope–deflection (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged and two-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. Because every structure is defined entirely by its figure, each drawing has been read from the printed figure and redrawn to scale below.

Question 5: Frame by moment distribution / slope-deflection (18 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.

Geometryhorizontal beam $\text{(1)}(0)\!-\!\text{(2)}(8)\!-\!\text{(3)}(14)\!-\!\text{(4)}(15)$ m; column $\text{(2)}\!-\!\text{(5)}$ drops $4$ m to a fixed base
Supportsfixed at (1) and (5); roller at (3); free overhang tip (4)
Loading$12\text{ kN/m}$ on (1)–(2); $8\text{ kN/m}$ on (2)–(4)
Memberssame $EI$, axially rigid

Find. All reactions and the member shear/moment diagrams.

12 kN/m8 kN/m8 m6 m1(1)(2)(3)(4)(5)
Q5 — two-fixed-end beam propped by a roller and a fixed-base column (indeterminate to 4°).

Approach. The frame is indeterminate to the 4th degree with no sidesway (the fixed-base column and the beam to fixed (1) hold joint (2) in place). Slope-deflection with the two unknown joint rotations $\theta_2,\theta_3$ ($\theta_1=\theta_5=0$; the $1\text{ m}$ overhang applies a known $-8\tfrac{1^2}{2}=-4\text{ kN}\cdot\text{m}$ at (3)) yields the member-end moments; a stiffness-matrix solution confirms them.

  1. Fixed-end moments. Beam (1)–(2): $\text{FEM}=\pm\tfrac{12(8)^2}{12}=\pm64$; (2)–(3): $\pm\tfrac{8(6)^2}{12}=\pm24$ (adjusted for the overhang). Solving the two slope-deflection equations gives $\theta_2,\theta_3$.
  2. Joint moments. At (1): $M=\boxed{67.75\text{ kN}\cdot\text{m}}$ (hogging). At (2) the beam delivers $56.5$, the column $15.0$ and the right beam $41.5$ — they balance ($-56.5+41.5+15.0=0$). Column base (5): $M=\boxed{7.5\text{ kN}\cdot\text{m}}$. Overhang moment at (3): $-4\text{ kN}\cdot\text{m}$.
  3. Reactions. $R_{(1)}=49.41\text{ kN}\uparrow$, $R_{(5)}=76.84\text{ kN}\uparrow$, $R_{(3)}=\boxed{25.75\text{ kN}\uparrow}$ (sum $=152=12(8)+8(7)$ ✓). The bent column develops a shear $H=\tfrac{15+7.5}{4}=5.63\text{ kN}$, balanced by equal horizontal reactions at (1) and (5).
  4. Span peaks. Sagging maxima are $+33.97\text{ kN}\cdot\text{m}$ (at $x\approx4.12$ m in (1)–(2)) and $+15.7\text{ kN}\cdot\text{m}$ (in (2)–(3)). Shear steps: $+49.4\to-46.6$ at (2)$^-$, up to $+30.25$, down to $-17.75$ at (3)$^-$, up to $+8.0$, to $0$ at the free tip.
Bending Moment — top beam (kN·m)−67.8+34−56.5+15.7−4col: M_top=15, M_base=7.5
Q5 — bending-moment diagram of the beam line; the jump at (2) is the column moment (15 kN·m).