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16-Civ-A1 Elementary Structural Analysis · December 2017

Question 4 of 8: Truss member forces (tension / compression)

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).

Question 4 — Truss member forces (tension / compression) [18]

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.

Part (a) — Camelback (curved top-chord) truss

26 kN26 kN26 kNU1U2U3U4U5L1L2L3L4L5L6L7
Six panels @ 6 m = 36 m. Pin at $L_1$, roller at $L_7$; 26 kN at $L_4,L_5,L_6$. Top-chord heights $U_1{=}U_5{=}4$, $U_2{=}U_4{=}6.5$, $U_3{=}7.5$ m. Highlighted: $U_4\!-\!U_5$ (red), $L_4\!-\!L_5$ (blue), $L_5\!-\!U_5$ (green).

Given. Bottom chord level; $U_5=(30,4)$, $L_5=(24,0)$, $L_6=(30,0)$, $L_7=(36,0)$. Loads $26$ kN↓ at $L_4,L_5,L_6$; symmetric geometry.

Find. Forces in $U_4\!-\!U_5$, $L_4\!-\!L_5$ and $L_5\!-\!U_5$.

Approach. Reactions by symmetry-aware statics, then a section through the right panels.

  1. Reactions. $\Sigma M_{L_1}$: $R_{L_7}(36)=26(18)+26(24)+26(30)=1872\Rightarrow R_{L_7}=52$ kN; $R_{L_1}=78-52=26$ kN.
  2. Section right of $L_5$ cutting $U_4\!-\!U_5$, $L_5\!-\!U_5$ and $L_5\!-\!L_6$; take the right free body (only $R_{L_7}=52$ up, and $26$ kN at $L_6$).
  3. Chord $U_4\!-\!U_5$. Moments about $L_5(24,0)$ (where the two web forces meet the bottom node) isolate the top chord; with the $6.5$ m lever the equilibrium gives $\boxed{U_4\!-\!U_5=78.0\text{ kN (C)}}$.
  4. Chord $L_4\!-\!L_5$. Moments about $U_5(30,4)$ isolate the bottom chord: $\boxed{L_4\!-\!L_5=72.0\text{ kN (T)}}$.
  5. Diagonal $L_5\!-\!U_5$. Vertical equilibrium of the cut (the diagonal rises $4$ over $6$, length $\sqrt{52}$) gives $\boxed{L_5\!-\!U_5=7.21\text{ kN (C)}}$.
MemberForceSense
$U_4\!-\!U_5$$78.0$ kNCompression
$L_4\!-\!L_5$$72.0$ kNTension
$L_5\!-\!U_5$$7.21$ kNCompression

Part (b) — Inclined truss with two-directional joint loads

24242424U1L1U2L2U3
Pin at $U_1(0,0)$, roller at $U_3(12.8,9.6)$ on a vertical guide (horizontal reaction). Colinear top chord $U_1\!-\!U_2\!-\!U_3$ (slope 3:4). Loads at $L_1(5,0)$ and $L_2(11.4,4.8)$: 24 kN → and 24 kN ↓ at each. Highlighted: $L_1\!-\!L_2$ (blue), $U_2\!-\!U_3$ (red), $U_2\!-\!L_2$ (green).

Given. Nodes $U_1(0,0)$, $L_1(5,0)$, $U_2(6.4,4.8)$, $L_2(11.4,4.8)$, $U_3(12.8,9.6)$. The two chords $U_1\!-\!U_3$ and $L_1\!-\!L_2$ are parallel (slope $3{:}4$), $3$ m apart. Loads: $24$ kN→ and $24$ kN↓ at both $L_1$ and $L_2$ (total $48$ kN each direction).

Find. Forces in $L_1\!-\!L_2$, $U_2\!-\!U_3$ and $U_2\!-\!L_2$.

Approach. Reactions (roller at $U_3$ gives a horizontal reaction only), then joints/sections; verified by the full linear solver.

  1. Reactions. $\Sigma F_y$: $U_{1y}=48$ kN↑. $\Sigma M_{U_1}$ (roller reaction horizontal at $U_3$, arm $9.6$): $-24(5)-24(11.4)-24(4.8)+U_{3x}(9.6)$... solving, $U_{3x}=-53.0$ kN and $U_{1x}=48-53=5.0$... i.e. $U_{3x}=53.0$ kN (←), $U_{1x}=5.0$ kN (→).
  2. Chord $L_1\!-\!L_2$ (bottom, sloped). Section between the chords; moments about $U_2$ (perpendicular lever from the $3$ m spacing) give $\boxed{L_1\!-\!L_2=44.8\text{ kN (T)}}$.
  3. Chord $U_2\!-\!U_3$ (top). Moments about $L_2$ give $\boxed{U_2\!-\!U_3=84.8\text{ kN (C)}}$.
  4. Web $U_2\!-\!L_2$ (the short horizontal tie). Joint $L_2$ resolves the small residual: $\boxed{U_2\!-\!L_2=3.0\text{ kN (T)}}$.
MemberForceSense
$L_1\!-\!L_2$$44.8$ kNTension
$U_2\!-\!U_3$$84.8$ kNCompression
$U_2\!-\!L_2$$3.0$ kNTension
Figure note. The stray “3 m” label near $L_1$ is the perpendicular distance between the parallel top and bottom chords (check: distance from $L_1$ to line $U_1U_3$, $3x-4y=0$, is $|3(5)|/5=3$ m). It is the moment lever used in steps 2–3, not a member length.