Question 2 of 6: Question 2 (paper Question II): Truss member forces
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3 (Statics and Dynamics) — 2016-May. 3 hours, closed book (one 8.5″×11″ formula sheet permitted). Part A (Statics) offers 3 questions, answer any 2; Part B (Dynamics) offers 3 questions, answer any 2. Every printed question is solved below (all 6), even though the exam instructions require only 4.
Given. A pin-jointed plane truss with joints A–F pinned into a rigid wall at A and B, and members AC, BC, BD, CD, CE, DE, DF, EF (8 members). A vertical load of 22.5 kN acts downward at F. All dimensions are in metres, read from the figure's grid: A(0,0), B(0,2), C(2,1), D(2,4), E(4,3), F(5,5).
Joint coordinates (m)
Joint
x
y
A
0
0
B
0
2
C
2
1
D
2
4
E
4
3
F
5
5
Find. The axial force (magnitude and sense – tension or compression) in every member: AC, BC, BD, CD, CE, DE, DF, EF.
Fig. 2 – truss geometry with the solved member forces (T = tension, C = compression).
Check: the pin symbols at A and B (both directly into the rigid wall) are taken as two independent pin supports (2 reaction components each). This gives 8 members + 4 reactions = 12 unknowns against $2j=12$ joint equations – exactly determinate – confirming the reading (a single rigid connection spanning A–B would over-constrain the truss by one degree).
Approach. Method of joints. Joint F carries the only external load and only 2 members (DF, EF) – solve it first, then work back through E, D, C, and finally the two supports B and A, using the unit vector along each member for the equilibrium equations $\Sigma F_x=0$, $\Sigma F_y=0$ at each joint (tension assumed positive).
Joint E (members CE, DE, EF; $F_{EF}$ known from Step 1). Unit vectors from E: $\hat{u}_{ED}=(-0.8944,0.4472)$, $\hat{u}_{EC}=(-0.8944,-0.4472)$. Substituting the known $F_{EF}$ and solving the remaining $2\times2$ system:
$$\boxed{F_{DE}=10.06\text{ kN (T)}},\qquad \boxed{F_{CE}=-31.82\text{ kN (C)}}$$
Joint D (members BD, CD, DE, DF; $F_{DE},F_{DF}$ known). Unit vectors from D: $\hat{u}_{DB}=(-0.7071,-0.7071)$, $\hat{u}_{DC}=(0,-1)$. Solving:
$$\boxed{F_{BD}=31.82\text{ kN (T)}},\qquad \boxed{F_{CD}=-22.50\text{ kN (C)}}$$
Joint C (members AC, BC, CD, CE; $F_{CD},F_{CE}$ known). Unit vectors from C: $\hat{u}_{CA}=(-0.8944,-0.4472)$, $\hat{u}_{CB}=(-0.8944,0.4472)$. Solving:
$$\boxed{F_{AC}=-62.89\text{ kN (C)}},\qquad \boxed{F_{BC}=37.73\text{ kN (T)}}$$
Joint B (reactions $B_x,B_y$; members BC, BD known). $\Sigma F_x=0,\ \Sigma F_y=0$ give
$$\boxed{B_x=-56.25\text{ kN}},\qquad \boxed{B_y=-5.63\text{ kN}}$$
Joint A (reactions $A_x,A_y$; member AC known, this is the only member at A). $\Sigma F_x=0,\ \Sigma F_y=0$ give
$$\boxed{A_x=56.25\text{ kN}},\qquad \boxed{A_y=28.13\text{ kN}}$$
Check – whole-truss equilibrium: $A_x+B_x=0$ and $A_y+B_y-22.5=0$, both satisfied.
