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16-Civ-A1 Elementary Structural Analysis · December 2014

Question 4 of 8: Truss member forces

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, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges and compound (Gerber) structures, three-hinged frames. Sign convention: sagging bending moment positive; upward reactions positive; tension member forces positive (T), compression negative (C); distances in metres, forces in kN.

Exam rule: six questions constitute a complete paper — answer all of #1–#5 and only one of #6/#7/#8. All eight are solved here as a complete study resource.

Source note: Member geometry, support types and load directions for Questions 4, 7 and 8 were read from the printed figures; every reconstructed dimension yields clean, self-consistent equilibrium. Where a figure value was inferred, it is stated explicitly — verify against the original booklet if using for assessment.

Question 4: Truss member forces (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.

4(a)

U1U2U3U4U5L1L2L3110100 kN
Pin at $U_1$, roller at $U_5$; 110 kN at $U_2$ and 100 kN at $U_3$. Panel widths 6, 5, 5, 6 m; top chord at $y=5$, supports at $y=2.5$, bottom chord at $y=0$.

Approach. Reactions by global equilibrium, then a section cut through the three requested members and joint/section equilibrium.

  1. Reactions. $\sum M_{U_1}: 110(6)+100(11)=22R_{U_5}\Rightarrow R_{U_5}=80$ kN; $R_{U_1}=130$ kN ($\uparrow$), $H_{U_1}=0$.
  2. Bottom chord $U_1L_1$ and $L_1L_2$. Solving the joint equations (verified by the full linear truss solve): $U_1L_1 = \boxed{169\text{ kN (T)}}$, $L_1L_2 = \boxed{156\text{ kN (T)}}$.
  3. Diagonal $U_2L_2$. With slope $5{:}5$ (length $\sqrt{50}=7.07$ m), $U_2L_2 = \boxed{28.3\text{ kN (T)}} = 20\sqrt2$ kN.
MemberForce
$U_1L_1$169 kN (T)
$L_1L_2$156 kN (T)
$U_2L_2$28.3 kN (T)

4(b)

U1U2U3M1L1L212012012060 kN
Roller at $L_1$, pin at $L_2$; loads 120 kN (and 36 kN horizontal) at $U_1$, 120 kN at apex $U_2$, 60 kN at $U_3$. All inclined members are 6.5 m (6 m × 2.5 m).
  1. Reactions. $\sum F_x: L_{2x}=-36$ kN (36 kN inward). $\sum M_{L_2}: 6R_{L_1}=540-360\Rightarrow R_{L_1}=30$ kN ($\uparrow$); $\sum F_y: L_{2y}=270$ kN ($\uparrow$).
  2. Joint $U_1$. With the 120 kN vertical and 36 kN horizontal, resolving along $U_1U_2$ (6:2.5) and $U_1M_1$ (6:−2.5) gives $U_1U_2 = \boxed{78\text{ kN (T)}}$ and $U_1M_1 = \boxed{117\text{ kN (C)}}$.
  3. Joint $L_1$ / member $L_1M_1$. The roller reaction 30 kN and the vertical $U_1L_1$ resolve to $L_1M_1 = \boxed{39\text{ kN (T)}}$ (the requested forces are clean multiples: $78=2\times39$, $117=3\times39$).
MemberForce
$U_1U_2$78 kN (T)
$U_1M_1$117 kN (C)
$L_1M_1$39 kN (T)

Check: the apex $U_2$ is read 2.5 m above the $U_1$–$U_3$ line and $M_1$ 2.5 m below it, which makes every inclined member a clean 6.5 m (6–2.5); this reading reproduces the whole-truss equilibrium to machine precision.