Question 2 (paper Question II) — Plane Truss Member Forces (Part A · Statics, equal value): 04-BS-3, December 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 (m, each grid division = 1 m): A(0,0), B(4,2), C(0,4), D(4,6), E(9,6). Members: A–C, A–B, A–D, C–D, B–D, B–E, D–E. Pin support at A; roller at C bearing against the vertical wall (horizontal reaction only). Applied at E: 1.5 kN down and 2 kN right.

Given data
Item	Value
Joints (m)	A(0,0) B(4,2) C(0,4) D(4,6) E(9,6)
Members (7)	AC, AB, AD, CD, BD, BE, DE
Supports	Pin at A (2 reactions); roller at C on the vertical wall (1 horizontal reaction)
Applied load at E	1.5 kN down, 2 kN right

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

Figure 2. Truss with pin support at A, roller (horizontal reaction) at C; 1.5 kN down + 2 kN right applied at E.

Approach. Confirm static determinacy, find the three support reactions from global equilibrium, then solve member forces joint-by-joint starting at the joints with only two unknown members.

Determinacy check. $m+r=7+3=10=2j=2(5)$ — the truss is statically determinate.
Support reactions. Summing moments about A (roller reaction $C_x$ is horizontal, acting at $(0,4)$): $\sum M_A=0$: $-4C_x+[9(-1500)-6(2000)]=0\Rightarrow \boxed{C_x=-6375\ \text{N}}$ (6375 N pointing into the wall, $-x$). $\sum F_x=0$: $A_x+C_x+2000=0\Rightarrow \boxed{A_x=4375\ \text{N}}$. $\sum F_y=0$: $A_y-1500=0\Rightarrow \boxed{A_y=1500\ \text{N}}$.
Joint E (members BE, DE only). $\overrightarrow{EB}=(-5,-4)$, $|EB|=\sqrt{41}=6.403$; $\overrightarrow{ED}=(-5,0)$, $|ED|=5$. $\sum F_y=0$: $F_{BE}(-4/6.403)-1500=0\Rightarrow \boxed{F_{BE}=-2401\ \text{N (2401 N C)}}$. $\sum F_x=0$: $F_{BE}(-5/6.403)+F_{DE}(-1)+2000=0\Rightarrow \boxed{F_{DE}=3875\ \text{N (T)}}$.
Joint C (members AC, CD, plus reaction $C_x$). $\overrightarrow{CA}=(0,-4)$; $\overrightarrow{CD}=(4,2)$, $|CD|=\sqrt{20}=4.472$. $\sum F_x=0$: $C_x+F_{CD}(4/4.472)=0\Rightarrow F_{CD}=-6375/0.8944=7127\ \text{N}\Rightarrow \boxed{F_{CD}=7127\ \text{N (T)}}$. $\sum F_y=0$: $-F_{AC}+F_{CD}(2/4.472)=0\Rightarrow \boxed{F_{AC}=3188\ \text{N (T)}}$.
Joint B (members AB, BD, BE — $F_{BE}$ already known). $\overrightarrow{BA}=(-4,-2)$, $|BA|=\sqrt{20}=4.472$; $\overrightarrow{BD}=(0,4)$; $\overrightarrow{BE}=(5,4)$, $|BE|=6.403$. $\sum F_y=0$: $F_{AB}(-2/4.472)+F_{BD}+F_{BE}(4/6.403)=0$. $\sum F_x=0$: $F_{AB}(-4/4.472)+F_{BE}(5/6.403)=0\Rightarrow \boxed{F_{AB}=-2096\ \text{N (2096 N C)}}$; substituting, $\boxed{F_{BD}=563\ \text{N (T)}}$.
Joint A (check). With $F_{AC}$, $F_{AB}$ known, equilibrium at A gives $\boxed{F_{AD}=-4507\ \text{N (4507 N C)}}$, and back-substituting into $\sum F_x, \sum F_y$ at A reproduces $A_x=4375$ N and $A_y=1500$ N exactly, confirming the solution.

Final results (T = tension, C = compression)
Member	Force
AC	3188 N (T)
AB	2096 N (C)
AD	4507 N (C)
CD	7127 N (T)
BD	563 N (T)
BE	2401 N (C)
DE	3875 N (T)
Reactions	$A_x=4375$ N, $A_y=1500$ N, $C_x=6375$ N (into wall)

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)