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16-Civ-A1 Elementary Structural Analysis · December 2017

Question 6 of 8: Slope-deflection analysis of a frame

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, 16-Civ-A1 Elementary Structural Analysis, December 2017, 3 hours, closed book (approved Sharp or Casio calculator only). A complete paper is six questions: candidates answer all of Q1–Q5 and exactly one of Q6, Q7 or Q8. All eight questions are worked below — Q6, Q7 and Q8 each in full — so the set is a complete study resource. Marks: Q1 [6], Q2–Q5 [18 each], Q6–Q8 [22 each].

Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy & stability (Ch. 2), method of joints/sections (Ch. 3), shear & moment diagrams (Ch. 4), deflections by virtual work (Ch. 8–9), influence lines (Ch. 6), slope–deflection and moment distribution (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — compound (Gerber) beams, internal hinges and condition equations, indeterminate frames. Sign convention: sagging bending moment positive (tension on the underside); truss forces tension (T) positive, compression (C) negative.

Reading the exam figures. Every support, member and load below was read directly from the printed figures. The Q4(b) truss carries a stray “3 m” label that is the perpendicular spacing between the two parallel chords (confirmed below), and the Q8 frame carries a stray “4.8 m” dimension that is not a load — it is the offset from the beam down to the lower dimension line (the columns are 4 m, stated on both outer columns).
Choose one of Q6, Q7, Q8. The exam requires exactly one of these; all three are worked in full below so the study set is complete.

Question 6 — Slope-deflection analysis of a frame [22]

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.

12 kN/m30 kN6 m8 m1 m3 m3EI4EIEI23451
Fixed at 2; beam 2–3 ($3EI$, 6 m); a column 3–1 ($EI$, 3 m) to a pin at 1; beam 3–4 ($4EI$, 8 m) under UDL 12 kN/m; roller at 4; 1 m overhang 4–5 with 30 kN at the tip.

Given. Fixed support at 2; interior column 3–1 (pin at 1); roller at 4; free tip 5. Relative rigidities $3EI$ (2–3), $4EI$ (3–4 and 4–5), $EI$ (column). Loads: UDL $12$ kN/m on 3–4, and $30$ kN↓ at the tip 5.

Find. All member end moments and the $V$, $M$ diagrams.

Approach. Because both the fixed end and the inextensible beam 2–3 lock node 3 against translation, and the column against vertical movement, there is no sidesway. The unknowns are the joint rotations $\theta_3,\theta_4$; the overhang applies a known moment at 4. Write slope-deflection equations and enforce joint moment balance.

  1. Fixed-end moments (member 3–4). $\text{FEM}_{34}=-\dfrac{wL^2}{12}=-\dfrac{12(8)^2}{12}=-64$, $\text{FEM}_{43}=+64$ kN·m; other members unloaded.
  2. Overhang moment at 4. The $30$ kN at $1$ m gives $M_{45}=-30$ kN·m, so joint 4 balance requires $M_{43}=+30$.
  3. Slope-deflection relations (stiffness $2E I_{\text{mem}}/L$; column pinned at 1 uses $3EI/L$): $M_{32}=2EI\theta_3$, $M_{34}=EI(2\theta_3+\theta_4)-64$, $M_{31}=EI\theta_3$, $M_{43}=EI(2\theta_4+\theta_3)+64$.
  4. Solve joint equations. Joint 4: $EI(\theta_3+2\theta_4)=-34$; joint 3 ($M_{32}+M_{34}+M_{31}=0$): $EI(5\theta_3+\theta_4)=64$. Hence $EI\theta_3=18$, $EI\theta_4=-26$.
  5. End moments. $M_{23}=18,\ M_{32}=36,\ M_{34}=-54,\ M_{43}=30,\ M_{31}=18,\ M_{45}=-30$ kN·m (joint 3 checks: $36-54+18=0$).
  6. Reactions & span values. Column shear $6$ kN (so $H_2=6$); beam 3–4 shear runs $+51\to-45$ kN, zero at $4.25$ m from 3 where $M_{\max}=+54.4$ kN·m (sag); vertical reactions $R_1=60$, $R_4=75$ kN, $R_2=9$ kN↓.
-9+51-45+30Shear force (kN) [beam line]x
Shear (beam line 2–3–4–5): $-9$ on 2–3; $+51\to-45$ on 3–4; $+30$ on the overhang.
+18-36 || -54+54.4-30Bending moment (kN·m) [beam line 2–3–4–5]x
Bending moment (beam line). Note the step at joint 3 ($-36$ in 2–3 to $-54$ in 3–4) equal to the $18$ kN·m column moment; sag peak $+54.4$ near mid-span 3–4; $-30$ at the roller 4.
LocationMoment (kN·m)
Fixed end 2 ($M_{23}$)$+18$
Joint 3 — beam 2–3 / beam 3–4 / column$-36$ / $-54$ / $-18$
Beam 3–4 max sag$+54.4$ (at $4.25$ m from 3)
Roller 4 ($M_{43}$)$-30$
Reactions $R_1,R_4,R_2$$60\uparrow,\ 75\uparrow,\ 9\downarrow$ kN
Check. An independent 2-D frame finite-element model (large $EA$ for inextensibility, column base pinned, roller at 4) reproduces every end moment and reaction above to the stated precision.