NivaarExam PrepOfficial exam papers ↗

16-Civ-A1 Elementary Structural Analysis · May 2014

Question 4 of 8: Truss member forces (method of sections)

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 and compound structures. Sign convention: sagging bending moment positive; upward reactions positive; tension member forces positive (T), compression negative (C).

The paper directs candidates to answer Q1–Q4, then one of Q5/Q6 and one of Q7/Q8. For completeness all eight questions are worked here.

Question 4: Truss member forces (method of sections) (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)

L₁U₁L₂U₂L₃72 kN60 kN6 m6 m4(a): pin at L₁, roller at L₃; 72 kN horizontal at U₁, 60 kN down at L₂.

Given. Nodes $L_1(0,0),U_1(0,5),L_2(6,0),U_2(6,2.5),L_3(12,0)$ m; pin at $L_1$, roller at $L_3$; 72 kN horizontal (→) at $U_1$, 60 kN down at $L_2$.

Find. Forces in $U_1U_2$, $L_1U_2$, $L_2L_3$.

  1. Reactions. $\sum F_x=0\Rightarrow L_{1x}=72$ kN (←). $\sum M_{L_1}=0$: $12L_{3y} = 72(5)+60(6)\Rightarrow L_{3y}=\boxed{60\ \text{kN}\uparrow}$; $L_{1y}=60-60=\boxed{0}$.
  2. Section & joint equilibrium (solved as a full linear system) give:
  3. $U_1U_2 = \boxed{78\ \text{kN (C)}}$; $\ L_1U_2 = \boxed{78\ \text{kN (C)}}$; $\ L_2L_3 = \boxed{144\ \text{kN (T)}}$.

4(b)

L₁L₂L₃L₄U₁U₂U₃30 kN60 kN30 kN6.25 m3.5 m6.25 m4(b): symmetric gable truss; three 30/60/30 kN loads inclined 3:4 (down-and-right). Pin at L₁, roller at L₄.

Given. Symmetric gable: $L_1(0,0),L_2(6.25,0),L_3(9.75,0),L_4(16,0)$; $U_1(6.25,3),U_2(8,6),U_3(9.75,3)$ m. Pin at L₁, roller at L₄. Three loads inclined 3:4 (down & right): 30 kN at $L_1$, 60 kN at $U_1$, 30 kN at $U_2$.

Find. Forces in $L_1U_1$, $L_1L_2$, $L_2U_2$.

  1. Reactions. Each load resolves to $(0.6F$ →$,\,0.8F\downarrow)$: total $(72\ \rightarrow,\ 96\downarrow)$. $\sum F_x=0\Rightarrow L_{1x}=72$ kN (←). $\sum M_{L_1}=0\Rightarrow L_{4y}=\boxed{44.25\ \text{kN}\uparrow}$; $L_{1y}=96-44.25=\boxed{51.75\ \text{kN}\uparrow}$.
  2. Joint/section equilibrium gives:
  3. $L_1U_1 = \boxed{64.1\ \text{kN (C)}}$; $\ L_1L_2 = \boxed{111.8\ \text{kN (T)}}$; $\ L_2U_2 = \boxed{188.6\ \text{kN (T)}}$.
MemberForce (kN)Sense
4(a) U₁U₂78.0Compression
4(a) L₁U₂78.0Compression
4(a) L₂L₃144.0Tension
4(b) L₁U₁64.1Compression
4(b) L₁L₂111.8Tension
4(b) L₂U₂188.6Tension