16-Civ-A1 Elementary Structural Analysis · December 2013
Question 4 of 8: Truss member forces
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, December 2013 — 98-Civ-A1 Elementary Structural Analysis. Three-hour, closed-book examination (approved Sharp/Casio calculator only). Format: six questions constitute a complete paper — answer all of #1–#5 and any one of #6, #7 or #8. All eight questions are solved below for completeness.
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).
Truss 4(a): pin at L₁, roller at L₅; 62.4 kN down at U₁ and U₂, 15.6 kN horizontal at U₃. Members solved are shown in red.
Given. Bottom chord $L_1(0,4),L_2(9.6,0),L_3(15.6,0),L_4(21.6,0),L_5(31.2,4)$; top chord $U_1(9.6,8),U_2(15.6,8),U_3(21.6,8)$ (metres). Pin at $L_1$, roller at $L_5$. Loads: 62.4 kN down at $U_1$ and $U_2$; 15.6 kN horizontal (→) at $U_3$.
Find. Forces in $L_1U_1$, $U_1L_2$, $U_1L_3$ (T/C).
Reactions. $\sum M_{L_1}=0$ and $\sum F=0$ give $L_{1x}=15.6$ kN (←, balancing the horizontal load), $L_{1y}=72.4$ kN ↑, $L_{5y}=52.4$ kN ↑.
Joint equilibrium (method of joints), solved as a full linear system for all 13 bar forces. The three requested bars are:
$L_1U_1 = \boxed{85.7\ \text{kN (C)}}$ — the steep left end-post is in compression as it carries the panel loads down to the pin.
$U_1L_2 = \boxed{39.5\ \text{kN (C)}}$ (the vertical hanger/post below $U_1$).
$U_1L_3 = \boxed{12.5\ \text{kN (T)}}$ (the interior diagonal).
4(b) — Forces in L₂L₃, M₁L₃ and M₁M₂
Truss 4(b): pin at L₁, roller at L₃; 24 kN down at U₁, 24 kN horizontal at U₂ and at M₂. Members solved are shown in red.
Given. Bottom chord $L_1(0,0)\ldots L_4(12,0)$ at 4 m spacing; mid nodes $M_1(4,3),M_2(12,3)$; top $U_1(4,6),U_2(12,6)$. Pin at $L_1$, roller at $L_3$. Loads: 24 kN down at $U_1$; 24 kN horizontal (→) at $U_2$ and at $M_2$.
Find. Forces in $L_2L_3$, $M_1L_3$, $M_1M_2$ (T/C).
Reactions. $L_{1x}=48$ kN (←, balancing the two 24 kN horizontals), $L_{1y}=15$ kN ↓, $L_{3y}=39$ kN ↑.
Joint/section equilibrium yields:
$L_2L_3 = \boxed{28.0\ \text{kN (T)}}$ (bottom chord in tension).
$M_1L_3 = \boxed{50.0\ \text{kN (C)}}$ (mid-to-bottom diagonal in compression).
$M_1M_2 = \boxed{60.0\ \text{kN (T)}}$ (the mid chord in tension).
