Question 1 of 6: Truss Member Forces (Part A — Statics)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination — May 2013 — 04-BS-3 Statics and Dynamics. Three-hour, closed-book exam (one 8.5"×11" self-prepared aid sheet permitted; Casio or Sharp approved calculator). Format: Part A (Statics) offers 3 questions, answer any 2; Part B (Dynamics) offers 3 questions, answer any 2. All six are solved below for completeness.
Given. Pin joints at grid coordinates (metres) A(0,0), B(0,4), C(3,6), D(5,2), E(8,7). A and B are pinned directly to a rigid wall (each supplies its own pair of reactions; the drawn A–B edge is the wall outline, not a load-carrying member). Members: AD, BC, BD, CD, CE, DE. Applied loads at E: 50.0 kN downward and 75.0 kN to the right.
Given data
Joint
x (m)
y (m)
A
0
0
B
0
4
C
3
6
D
5
2
E
8
7
Figure 1 — truss geometry (1 grid division = 1 m); loads applied at E.
Find. The axial force in each of the six members (AD, BC, BD, CD, CE, DE) and whether each is in tension (T) or compression (C).
Approach. With m + r = 6 + 4 = 10 = 2j (j = 5 joints), the truss is statically determinate as a system. Starting at the only joint with two unknowns first (E, loaded, connects only to C and D), solve joint equilibrium in the sequence E → C → D, then read the wall reactions at B and A from their own joint equations.
Joint E — two unknowns (FCE, FDE). Unit vectors from E: $\hat u_{EC}=(-0.5547,-0.8321)$, $\hat u_{ED}=(-0.7682,-0.6402)$. Equilibrium $F_{CE}\hat u_{EC}+F_{DE}\hat u_{ED}+(75.0,-50.0)=0$ gives two scalar equations in two unknowns. Solving,
$$F_{CE}=\boxed{121.68\text{ kN (T)}}, \qquad F_{DE}=\boxed{86.14\text{ kN (C)}}$$
Joint C — two unknowns (FBC, FCD). The force member CE exerts on C is now known ($121.68$ kN pulling C toward E). With $\hat u_{CB}=(-0.5547,-0.8321)$, $\hat u_{CD}=(0.5547,-0.8321)$, solving $F_{BC}\hat u_{CB}+F_{CD}\hat u_{CD} = -F_{CE}\hat u_{CE}$ gives
$$F_{BC}=\boxed{118.31\text{ kN (T)}}, \qquad F_{CD}=\boxed{46.69\text{ kN (C)}}$$
Joint D — two unknowns (FAD, FBD). FCD and FDE are now known. With $\hat u_{DA}=(-0.9285,-0.3714)$, $\hat u_{DB}=(-0.9285,0.3714)$, solving the joint-D equilibrium gives
$$F_{AD}=\boxed{168.29\text{ kN (C)}}, \qquad F_{BD}=\boxed{143.04\text{ kN (T)}}$$
Wall reactions (check, not asked for). Joint B: $R_B = -(F_{BC}\hat u_{BC}+F_{BD}\hat u_{BD}) = (-231.25,-12.50)$ kN. Joint A: $R_A=-F_{AD}\hat u_{AD}=(156.25,62.50)$ kN. Summing $R_A+R_B+(75.0,-50.0)=(0,0)$ confirms global equilibrium, and taking moments about A also closes to zero.
