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16-Civ-A1 Elementary Structural Analysis · December 2016

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

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 4: Truss member forces (tension / compression) (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.

4(a) — symmetric-ish roof truss

Given. Bottom chord $L_1(0,0)\,L_2(3,0)\,L_3(9,0)\,L_4(15,0)\,L_5(18,0)$; top nodes $U_1(3,4),U_2(9,6.5),U_3(15,4)$. Pin at $L_1$, roller at $L_5$. Loads: $36\text{ kN}$ down at $U_1$ and $U_2$, $24\text{ kN}$ down at $U_3$.

Find. Forces in $L_2\!-\!L_3$, $U_1\!-\!U_2$, $L_3\!-\!U_3$.

L1L2L3L4L5U1U2U336 kN36 kN24 kN36 m6 m3
4(a) — the truss (determinate: $m{+}r{=}2n=16$).

Approach. Support reactions from global equilibrium, then a vertical section through the three unknown members combined with joint resolution; every value is confirmed by solving the full $16\times16$ joint-equilibrium system (residual $<10^{-12}$).

  1. Reactions. $\sum M_{L_1}=0:\;R_{L_5}(18)=36(3)+36(9)+24(15)=792\Rightarrow R_{L_5}=44\text{ kN}$; $R_{L_1}=96-44=52\text{ kN}$.
  2. $L_2\!-\!L_3$. Section just right of $U_1$ and take moments about $U_1(3,4)$: the bottom chord tie carries $\boxed{39.0\text{ kN (T)}}$.
  3. $U_1\!-\!U_2$. The upper rafter is in $\boxed{42.0\text{ kN (C)}}$.
  4. $L_3\!-\!U_3$. The interior diagonal from the centre node to $U_3$ carries $\boxed{6.93\text{ kN (T)}}$.

4(b) — cantilever (stepped) truss

Given. $L_1(0,0),U_1(0,4.5),B_1(6,0),L_2(6,4.5),U_2(6,9),L_3(12,9),U_3(12,13.5),L_4(18,9)$; pins at $L_1$ and $B_1$. Loads $30\text{ kN}$ at $U_1$, $60\text{ kN}$ at $U_2$ and $U_3$, $30\text{ kN}$ at $L_4$ (all down).

Find. Forces in $U_1\!-\!U_2$, $L_2\!-\!L_3$, $U_2\!-\!L_2$.

L1U1B1L2U2L3U3L4306060306 m6 m6 m4.54.5
4(b) — the truss cantilevers to the right off the two left supports.

Approach. Only two supports ($L_1,B_1$), both on the left, so the structure cantilevers the far loads back — the members carry large forces. Solve the complete joint system (determinate, $m{+}r=2n=16$).

  1. $U_1\!-\!U_2$ (rising top chord): $\boxed{200\text{ kN (T)}}$.
  2. $L_2\!-\!L_3$ (rising diagonal): $\boxed{200\text{ kN (C)}}$.
  3. $U_2\!-\!L_2$ (vertical): $\boxed{150\text{ kN (C)}}$.
4(a)4(b)
$L_2\!-\!L_3$39.0 kN T$U_1\!-\!U_2$200 kN T
$U_1\!-\!U_2$42.0 kN C$L_2\!-\!L_3$200 kN C
$L_3\!-\!U_3$6.93 kN T$U_2\!-\!L_2$150 kN C