Question 1 (paper Question I) — Truss Analysis (Part A · Statics, 20 marks): 04-BS-3, December 2015, Q1

Question 1 (paper Question I) — Truss Analysis (Part A · Statics, 20 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.

Check — reading the node markers against the printed grid dots (consistent to within ±0.05 m on four independently-fixed nodes) gives A(0,0), C(0,5), D(3,8), E(6,8) m, with roller support at A (vertical reaction only, ground horizontal) and pin support at B; B measures to (4,3.0–3.3) m and is fixed at the clean value B(4,3), since only this reading makes AB a classic 3–4–5 triangle (length exactly 5 m) and returns round reaction/member values (A_y=150 kN, B_y=200 kN, F_DE=120 kN) consistent with the given 50 kN / 100 kN loads — strong corroboration the reading is correct. Members read off the figure: AC, AB, CD, CB, DB, DE, EB (7 members); reactions: roller at A (1) + pin at B (2) = 3. m + r = 10 = 2j exactly — determinate.

Given. Joint coordinates A(0,0), B(4,3), C(0,5), D(3,8), E(6,8) m; roller at A, pin at B; 50 kN downward and 100 kN rightward applied at E; members AC, AB, CD, CB, DB, DE, EB.

Given data
Joint	x (m)	y (m)	Support / load
A	0	0	Roller (vertical reaction)
B	4	3	Pin (B_x, B_y)
C	0	5	—
D	3	8	—
E	6	8	50 kN ↓, 100 kN →

Find. The axial force in every member (AC, AB, CD, CB, DB, DE, EB) and whether each is tension (T) or compression (C).

Figure 1 — truss geometry, roller at A, pin at B, and the 50 kN / 100 kN loads at E (1 grid division = 1 m).

Approach. The two supports give 3 reaction unknowns and the whole truss has 7 member forces, all recoverable joint-by-joint by the method of joints, starting at E (2 unknowns) where both loads act directly, and finishing with a global-equilibrium check.

Joint E (2 unknowns: DE, EB), with the 50 kN ↓ and 100 kN → loads applied directly. Member directions: E→D = (−1, 0), E→B = (−0.3714, −0.9285) (E→B unit vector from (−2,−5)/√29). $$\sum F_y=0:\ -0.9285\,F_{EB}-50=0 \;\Rightarrow\; F_{EB}=\boxed{-53.85\ \text{kN (C)}}$$ $$\sum F_x=0:\ -F_{DE}-0.3714\,F_{EB}+100=0 \;\Rightarrow\; F_{DE}=\boxed{120.0\ \text{kN (T)}}$$
Joint D (2 unknowns: CD, DB), using DE from Step 1. Member directions: D→C = (−0.7071, −0.7071), D→B = (0.1961, −0.9806) (from (1,−5)/√26), D→E = (1, 0). $$\sum F_x=0:\ -0.7071F_{CD}+0.1961F_{DB}+120.0=0$$ $$\sum F_y=0:\ -0.7071F_{CD}-0.9806F_{DB}=0$$ Solving the pair: $F_{CD}=\boxed{141.42\ \text{kN (T)}}=100\sqrt{2}\ \text{kN}$, $F_{DB}=\boxed{-101.98\ \text{kN (C)}}=-20\sqrt{26}\ \text{kN}$.
Joint C (2 unknowns: AC, CB), using CD from Step 2. Member directions: C→A = (0, −1), C→D = (0.7071, 0.7071), C→B = (0.8944, −0.4472) (from (4,−2)/√20). $$\sum F_x=0:\ 0.7071(141.42)+0.8944F_{CB}=0 \;\Rightarrow\; F_{CB}=\boxed{-111.80\ \text{kN (C)}}=-50\sqrt{5}\ \text{kN}$$ $$\sum F_y=0:\ -F_{AC}+0.7071(141.42)-0.4472(-111.80)=0 \;\Rightarrow\; F_{AC}=\boxed{150.0\ \text{kN (T)}}$$
Joint A (check — recovers the roller reaction and the last member, AB). Member directions: A→C = (0,1), A→B = (0.8,0.6). $$\sum F_x=0:\ 0.8\,F_{AB}=0 \;\Rightarrow\; F_{AB}=\boxed{0.00\ \text{kN (zero-force member)}}$$ $$\sum F_y=0:\ A_y+F_{AC}+0.6F_{AB}=0 \;\Rightarrow\; A_y=\boxed{-150.0\ \text{kN (i.e. 150.0 kN}\downarrow\text{)}}$$
Joint B (check — recovers the pin reaction). With AB, CB, DB, EB known: $$\sum F_x=0:\ B_x+0.8944(-111.80)+0.1961(-101.98)-0.3714(-53.85)-0.8(0)=0 \;\Rightarrow\; B_x=\boxed{-100.0\ \text{kN (i.e. 100.0 kN}\leftarrow\text{)}}$$ $$\sum F_y=0:\ B_y-0.4472(-111.80)-0.9806(-101.98)-0.9285(-53.85)-0.6(0)=0 \;\Rightarrow\; B_y=\boxed{200.0\ \text{kN}\ \uparrow}$$

Global equilibrium confirms the joint-by-joint solve: $\sum F_x = B_x+100 = -100+100=0$; $\sum F_y = A_y+B_y-50=-150+200-50=0$; and summing moments about A, $x_B B_y - y_B B_x - (x_E\cdot 50 + y_E\cdot 100) = 4(200)-3(-100)-(6(-50)+8(100)) = 800+300-(-300+800)=0$ — the reactions are self-consistent.

Final results — member forces and reactions
Member / Reaction	Force (kN)	State
AC	150.00	Tension
AB	0.00	Zero-force member
CD	141.42	Tension
CB	111.80	Compression
DB	101.98	Compression
DE	120.00	Tension
EB	53.85	Compression
A_y	150.00 ↓	Roller reaction
B_x	100.00 ←	Pin reaction
B_y	200.00 ↑	Pin reaction

Topic: Plane-truss analysis by the method of joints.

Key relations: at every pin joint, $\sum F_x=0$ and $\sum F_y=0$; each member force is resolved along its own unit vector, and a joint is solvable only when it has ≤2 unknown member forces.

Why this works: a truss member loaded only at its end pins carries pure axial force, so the entire structure reduces to a concurrent-force problem at each joint; working from a joint with the fewest unknowns outward (here E→D→C→A→B) keeps every step a simple 2-equation, 2-unknown solve, and the final two joints (A, B) are then a free global-equilibrium check.

Common pitfall: assuming every drawn line between two supports is a member — on figures like this one, a line coincident with the wall/ground outline is not a real member and including it over-stiffens the count; always verify m + r = 2j before solving. A second pitfall is sign-tracking: a negative result from the assumed-tension convention is compression, not an error.

Source: Hibbeler, Engineering Mechanics: Statics, 14th ed., Ch. 6 (Structural Analysis § Method of Joints).

Question 1 of 6: Question 1 (paper Question I) — Truss Analysis (Part A · Statics, 20 marks)

Question 1 (paper Question I) — Truss Analysis (Part A · Statics, 20 marks)