Question 1 of 6: Question 1 (paper Question I) — Truss Analysis (Part A · Statics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, December 2013 — Statics and Dynamics (04-BS-3). Closed book (one self-prepared 8½×11 in. sheet of notes permitted). Candidates were instructed to complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); every question is solved below as a complete study resource.
Check — the vision-extracted figure lists 8 members (AC, AB, CB, CD, AD, DE, DF, EF) on 6 joints with 3 reaction components (pin A + roller B), i.e. m+r = 11 ≠ 2j = 12: one degree of freedom short of a stable, determinate truss. Geometrically, A(0,0), C(2,4) and E(4,8) are exactly collinear (y = 2x), as are A(0,0), D(4,4) and F(8,8) (y = x) — a straight continuation of member AC to E (member CE) is easy for an automated figure-caption pass to miss because it looks like one long line, not two segments. Adding CE gives m + r = 2j = 12 exactly, and the resulting 9-equation system solves EXACTLY (residual ≈ 10⁻¹⁴) with clean closed-form values — strong corroborating evidence that CE is the missing member. Solved with CE included; the reconstruction is disclosed here rather than hidden.
Given. Truss with joints A(0,0), B(4,0), C(2,4), D(4,4), E(4,8), F(8,8) m; pin support at A, roller support at B; downward loads of 300 kN at E and 200 kN at F; members AC, AB, CB, CD, AD, CE, DE, DF, EF.
Given data
Joint
x (m)
y (m)
Support / load
A
0
0
Pin
B
4
0
Roller (vertical)
C
2
4
—
D
4
4
—
E
4
8
300 kN ↓
F
8
8
200 kN ↓
Find. The axial force in every member (AC, AB, CB, CD, AD, CE, DE, DF, EF) and whether each is in tension (T) or compression (C).
Figure 1 — truss geometry, supports, applied loads, and the reconstructed member CE (dashed).
Approach. Find the support reactions from global equilibrium, then solve the joints in an order that never leaves more than two unknown member forces (F, then E, then D, then C, then A/B as a check).
Support reactions — moments about A.
$$\sum M_A = 0:\quad B_y(4) - 300(4) - 200(8) = 0$$
$$B_y = \dfrac{1200+1600}{4} = \boxed{700\ \text{kN} \uparrow}$$
Joint F (2 unknowns: DF, EF). With θ = 45° on member DF:
$$\sum F_y=0:\ -DF\sin45^\circ - 200 = 0 \Rightarrow DF = \boxed{-282.84\ \text{kN (C)}}$$
$$\sum F_x=0:\ -DF\cos45^\circ - EF = 0 \Rightarrow EF = \boxed{200.00\ \text{kN (T)}}$$
Joint E (2 unknowns: DE, CE), using EF = 200 kN (T) from Step 3:
$$\sum F_x=0:\ -0.4472\,CE + 200 = 0 \Rightarrow CE = \boxed{447.21\ \text{kN (T)}}$$
$$\sum F_y=0:\ -DE - 0.8944\,CE - 300 = 0 \Rightarrow DE = \boxed{-700.00\ \text{kN (C)}}$$
Joint D (3 members carry known force from DE, DF; solve CD, AD):
$$\sum F_x=0,\ \sum F_y=0 \Rightarrow CD = \boxed{700.00\ \text{kN (T)}}, \qquad AD = \boxed{-1272.79\ \text{kN (C)}}$$
Joint C (solve CB using CE, CD, AC known from the remaining equations):
$$CB = \boxed{-782.62\ \text{kN (C)}}$$
Joint B (check, and solve AB):
$$AB = \boxed{350.00\ \text{kN (T)}}, \qquad AC = \boxed{1229.84\ \text{kN (T)}}$$
Substituting all nine member forces back into every joint's ΣFx and ΣFy equations closes to zero — the solution is self-consistent.
