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16-Civ-A1 Elementary Structural Analysis · December 2015

Question 4 of 8: Member forces in two trusses

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 / slope–deflection, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams and three-hinged frames. Sign convention: sagging bending moment positive, upward reactions positive, member tension positive (T) / compression negative (C).

The paper requires Q1–Q4 in full plus two of Q5/Q6/Q7/Q8. For completeness all eight questions are worked here.

Question 4: Member forces in two trusses (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) — cambered roof truss

Given. Bottom chord $L_1(0)\ldots L_5(16)$ at 4 m spacing; $U_1(4,3),U_2(8,5),U_3(12,3)$; pin at $L_1$, roller at $L_5$; $36$ kN ↓ at $L_3$ and $L_4$.

3636 L1U1U2U3L5
4(a): pin at $L_1$, roller at $L_5$; loads 36 kN at $L_3,L_4$.

Approach. Reactions by global statics, then method of sections/joints.

  1. Reactions. $\sum M_{L_1}: R_{L_5}(16)=36(8)+36(12)=720\Rightarrow R_{L_5}=45$ kN ↑; $R_{L_1}=72-45=27$ kN ↑.
  2. $L_1U_1$ (left rafter). Joint $L_1$ vertical: the rafter rises 3 in 5, so $R_{L_1}+\tfrac35 F_{L_1U_1}=0\Rightarrow F_{L_1U_1}=-\tfrac53(27)=\boxed{45\text{ kN C}}$.
  3. $U_1L_3$ (diagonal). A section through $U_1U_2,\,U_1L_3,\,L_2L_3$ and vertical equilibrium of the left part gives $F_{U_1L_3}=\boxed{9\text{ kN T}}$.
  4. $L_3L_4$ (bottom chord). Moments about $U_2(8,5)$ of the left part: $F_{L_3L_4}=\boxed{60\text{ kN T}}$.
MemberForce
$L_1U_1$45 kN C
$U_1L_3$9 kN T
$L_3L_4$60 kN T

4(b) — king-post-style truss

Given. $L_1(0),L_2(4),L_3(12),L_4(16)$; $M_1(4,3),M_2(8,3),M_3(12,3)$; apex $U_1(8,6)$; pin $L_1$, roller $L_4$; $36$ kN ↓ at $U_1$ and at $M_3$.

Determinacy check. With the members shown (no chord between the apex and $M_2$) the truss has $m=13,\ j=8,\ r=3$, so $m+r-2j=0$ — determinate. The rank test confirms that adding a $U_1M_2$ vertical would make it 1° indeterminate, so that line is read as the load arrow, not a member.
36 36 L1M1M2M3U1L4
4(b): loads 36 kN at apex $U_1$ and at $M_3$.
  1. Reactions. $\sum M_{L_1}: R_{L_4}(16)=36(8)+36(12)=720\Rightarrow R_{L_4}=45$ kN ↑; $R_{L_1}=27$ kN ↑.
  2. $L_2M_2$ & $M_2L_3$ (the panel diagonals) and $M_1M_2$. Working the joints from $L_1$ inward gives $F_{M_1M_2}=\boxed{12\text{ kN C}}$.
  3. $L_2M_2$. Joint $L_2$ (only $L_1L_2$, $L_2L_3$, the vertical $M_1L_2$ and diagonal $L_2M_2$) gives $F_{L_2M_2}=\boxed{15\text{ kN C}}$.
  4. $L_2L_3$ (bottom chord). $F_{L_2L_3}=\boxed{48\text{ kN T}}$.
MemberForce
$M_1M_2$12 kN C
$L_2M_2$15 kN C
$L_2L_3$48 kN T

In both trusses the reactions are equal (27 and 45 kN) because the two 36 kN panel loads share the same resultant position; only the internal load paths differ. Choosing a section that cuts the required bar plus two others lets each listed force be found from a single equilibrium equation, which is faster and less error-prone than marching joint-by-joint across the whole truss.