Question 2 of 6: Question 2 (paper Question II) — Truss Member Forces (Part A · Statics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination December 2016 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Separately, Question 1's bracket figure (Figure 1) gives only one anchor point's coordinates (E) plus two angle pairs for the other two cables; every printed dimension (50 mm×2, 100 mm×2, 75 mm, the two 45° arcs at C, the 30°/60° pair at D) was used, but the exact 3-D position of points B and C required one reasonable reconstruction, disclosed at first use below. All other figures (2–6) reconstruct cleanly and unambiguously from the printed dimensions, confirmed by internal checks that return clean values (member lengths BD = 5 ft and DE = 10 ft in Question 6) — strong evidence the readings are correct.
Question 2 (paper Question II) — Truss Member Forces (Part A · Statics, equal value)
Given. Joint coordinates A(0,0), B(3,3), C(5,0), D(5,4), E(7,3), F(10,0) m; pin support at A, roller support at F (vertical reaction only); loads 15 kN ↓ at D, 7 kN ↓ at E, 5 kN → at B; members AB, AC, BC, BD, CD, CE, DE, EF, CF (9 members, m + r = 9 + 3 = 12 = 2j — determinate).
Given data
Joint
x (m)
y (m)
Load / support
A
0
0
Pin support
B
3
3
5 kN →
C
5
0
—
D
5
4
15 kN ↓
E
7
3
7 kN ↓
F
10
0
Roller (vertical)
Find. The axial force in every member, and whether it is tension (T) or compression (C).
Figure 2 — truss geometry, supports and applied loads, with the solved member forces (kN) labelled (red = compression, blue = tension).
Approach. Find the two support reactions from global equilibrium, then solve the nine member forces joint-by-joint (method of joints), always starting at a joint with only two unknowns.
Reactions. $\sum F_x=0$: $A_x=-5\ \text{kN}$ (5 kN, pointing left, balancing the applied load at B). Taking moments about A ($\curvearrowleft{+}$, loads at D, E and B):
$$\sum M_A=0:\ -15(5)-7(7)-5(3)+F_y(10)=0\ \Rightarrow\ F_y=\boxed{13.9\ \text{kN}\uparrow}$$
$$\sum F_y=0:\ A_y=15+7-13.9=\boxed{8.1\ \text{kN}\uparrow}$$
Joint B (2 unknowns: BC, BD; AB known). Unit vectors B→A $=(-0.7071,-0.7071)$, B→C $=(0.5547,-0.8321)$, B→D $=(0.8944,0.4472)$, with the 5 kN load.
$$\sum F_x=0:\ 5+8.10+0.5547F_{BC}+0.8944F_{BD}=0$$
$$\sum F_y=0:\ 8.10-0.8321F_{BC}+0.4472F_{BD}=0$$
Solving simultaneously: $F_{BC}=\boxed{1.40\ \text{kN (T)}}$, $F_{BD}=\boxed{-15.51\ \text{kN (C)}}$.
Joint D (2 unknowns: CD, DE; BD known). Unit vectors D→B $=(-0.8944,-0.4472)$, D→C $=(0,-1)$, D→E $=(0.8944,-0.4472)$, with the 15 kN load.
$$\sum F_x=0:\ 13.875+0.8944F_{DE}=0\ \Rightarrow\ F_{DE}=\boxed{-15.51\ \text{kN (C)}}$$
$$\sum F_y=0:\ -15+6.94-F_{CD}-0.4472(-15.51)=0\ \Rightarrow\ F_{CD}=\boxed{-1.13\ \text{kN (C)}}$$
Joints E, F and C close out the same way (2, then 2, then a full equilibrium check), giving $F_{EF}=-19.66\ \text{kN}$, $F_{CF}=13.90\ \text{kN}$ and $F_{CE}=-0.04\ \text{kN}$ (a near-zero brace force, not an error — it is the small member that absorbs the 5 kN horizontal load's asymmetry). All nine equations close within rounding.
