Question 2 of 6: Question 2 (paper Question II) — Forces in All Members of a Space Frame (Part A · Statics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2017-December. Candidates were required to answer any 2 of 3 questions in Part A (Statics) and any 2 of 3 in Part B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1–3 = paper Part A, I–III; Questions 4–6 = paper Part B, IV–VI).
Check — figure reconstruction. Three figures needed a reconstructed reading, each corroborated by an independent numerical check rather than assumed: (1) Question 1's wall cable anchors C and D — the only geometry (C on the +x side, D on the −x side of the pipe) that returns positive (physically valid, tension-only) cable forces for both cables; any same-side reading returns a negative "tension" in one cable, which is impossible. (2) Question 2's overall frame height is not printed directly; it follows from the stated 30° leg angle and the 4 m offset of the lower joints, and the resulting member forces close exactly by symmetry (equal reactions, two exact zero-force members) — strong corroborating evidence the reconstruction is right. (3) Question 5's second suspension cable (block A) is read as the same 6 ft length as sphere B's cable, since the figure's one printed "6 ft" dimension spans a horizontal reference line common to both hanging positions.
Question 2 (paper Question II) — Forces in All Members of a Space Frame (Part A · Statics, equal value)
Given. A symmetric pin-jointed frame: outer legs A–E (vertical) and B–C (vertical), diagonal legs A–F and B–G at 30° from vertical, a top chord E–D–C, a lower cross-tie F–G, and four web diagonals E–F, F–D, D–G, G–C. A is a pin support, B is a roller. Top-chord spacing is 4 m between each of E, the point above F, D, the point above G, and C (16 m total, E to C); F and G sit 4 m below the top chord, directly under those quarter points. Downward loads of 1200 N at E, 2500 N at D and 1200 N at C.
Given data — joint coordinates (m), A at the origin
Joint
x
y
Support / load
A
0
0
pin
B
16
0
roller (vertical)
E
0
10.928
1200 N ↓
D
8
10.928
2500 N ↓
C
16
10.928
1200 N ↓
F
4
6.928
—
G
12
6.928
—
Find. The axial force (tension or compression) in every member: AE, AF, EF, ED, FD, FG, DG, DC, GC, GB, CB.
Figure 2. Blue members are in tension, red in compression, grey (AF, GB) are zero-force members.
Approach. Confirm determinacy, find the two vertical reactions from overall equilibrium (symmetric loading), then solve every joint by the method of joints, starting from the ends where only two unknown members meet.
Geometry and determinacy. The 30° leg angle with a 4 m horizontal offset from A to F gives the height from the ground to F, G: $4/\tan30^\circ=4\sqrt3=6.928\ \text{m}$, so the top chord sits $6.928+4=10.928\ \text{m}$ above the ground. Members: AE, AF, EF, ED, FD, FG, DG, DC, GC, GB, CB $\;(m=11)$; reactions $r=3$ (pin + roller); joints $j=7$: $m+r=14=2j$ — determinate.
Support reactions. The frame and loading are symmetric about $x=8$ (under D), so by inspection $A_x=0$ and $$\boxed{A_y=B_y=\dfrac{1200+2500+1200}{2}=2450\ \text{N}}$$
Method of joints, starting at A. At A, members AE (vertical) and AF (at 30° from vertical) meet the pin reaction $(0,2450)$: resolving horizontally forces $\boxed{F_{AF}=0}$ (a zero-force member), then vertically $\boxed{F_{AE}=-2450\ \text{N}}$ (2450 N compression). By the mirror symmetry of the frame, $F_{GB}=0$ and $F_{CB}=-2450\ \text{N}$ (compression) follow immediately at joint B.
Remaining joints (E, F, D, G, C). Proceeding joint by joint (E → F → D, with G and C following by symmetry) and solving each joint's two scalar equilibrium equations gives the complete member set below; every value satisfies both joint equilibrium at all seven joints and the global check $\sum F_x=\sum F_y=\sum M_A=0$.
Final results — Question 2 (T = tension, C = compression)
