Question 2 (paper Question II) — Plane Truss Member Forces (Part A · Statics, equal value): 04-BS-3, May 2018, Q2

Question 2 (paper Question II) — Plane Truss Member Forces (Part A · Statics, equal value)

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.

Given. Joint coordinates (grid units = 1 m): A(2,0), B(0,2), C(2,2), D(4,2), E(2,4), F(4,4), G(6,4). Members: BA, BC, BE, AC, CE, CD, DE, DF, EF, FG, DG. Supports: pin at B, roller at A. Load: 150 kN downward at G.

Given data
Item	Value
Joints (m)	A(2,0) B(0,2) C(2,2) D(4,2) E(2,4) F(4,4) G(6,4)
Members (11)	BA, BC, BE, AC, CE, CD, DE, DF, EF, FG, DG
Supports	Pin at B; roller at A (vertical reaction)
Applied load	150 kN down at G

Find. The force (tension or compression) in every member.

Figure 2 (reconstructed). Truss geometry, supports, and the 150 kN load at G.

Approach. Check static determinacy ($m+r=2j$), find the reactions from global equilibrium, then solve the joints in sequence (starting from a joint with only two unknown members) using $\sum F_x=0$, $\sum F_y=0$ at each joint.

Determinacy check. $m=11$ members, $r=3$ reaction components (pin B: 2, roller A: 1), $j=7$ joints. $m+r=14=2j=14$ — statically determinate.
Support reactions. Summing moments about B: $A_y(2)-150(6)=0\Rightarrow\boxed{A_y=450\text{ kN}}$. Then $\sum F_y=0$: $B_y=150-450=-300$ kN (i.e. 300 kN downward), and $\sum F_x=0$ gives $B_x=0$ (no horizontal load).
Joint A (2 unknowns: BA, AC). Member AC runs vertically from A(2,0) to C(2,2); member BA runs from A(2,0) to B(0,2), at 45°. With $A_y=450$ kN up and no horizontal reaction: $\sum F_x=0$ and $\sum F_y=0$ give $F_{BA}=0$ and $F_{AC}=-450$ kN (compression).
Joint B, C, D, E, F (method of joints, solved as a linear system). Proceeding joint-by-joint (B→C→D→E→F→G as a check) gives $F_{BC}=-300$, $F_{BE}=+300\sqrt2=424.3$, $F_{CE}=-450$, $F_{CD}=-300$, $F_{DE}=+150\sqrt2=212.1$, $F_{DF}=0$, $F_{EF}=+150$, $F_{FG}=+150$, $F_{DG}=-150\sqrt2=-212.1$ (all kN; + = tension, − = compression).
Check at joint G. Members FG (horizontal) and DG (diagonal) plus the 150 kN applied load balance exactly: $F_{FG}=150$ kN tension supplies the horizontal pull, $F_{DG}=150\sqrt2$ kN compression supplies the vertical support — confirming the solution.

Final results — member forces (kN)
Member	Force	Sense
BA	0	zero-force member
BC	300	compression
BE	424.3	tension
AC	450	compression
CE	450	compression
CD	300	compression
DE	212.1	tension
DF	0	zero-force member
EF	150	tension
FG	150	tension
DG	212.1	compression

Topic: Plane truss analysis — method of joints.
Key relations: determinacy $m+r=2j$; joint equilibrium $\sum F_x=0$, $\sum F_y=0$ at every pin, solved in an order that never leaves more than two unknown member forces at a joint.
Why this works: each two-force member's line of action is fixed by its end joints, so once a joint has only two unknown members its two scalar equilibrium equations solve them directly; the solved members then reduce the unknown count at neighbouring joints.
Common pitfall: mislabelling a zero-force member as "missing" — BA and DF genuinely carry no load at this loading, which is a legitimate result of the geometry (each sits at a joint where the other two members are collinear or where no other force needs balancing), not a sign of a solving error.
Source: Hibbeler, Engineering Mechanics: Statics, Ch. 6 (Method of joints; zero-force members).

Question 2 of 6: Question 2 (paper Question II) — Plane Truss Member Forces (Part A · Statics, equal value)

Question 2 (paper Question II) — Plane Truss Member Forces (Part A · Statics, equal value)