Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination — Statics and Dynamics (04-BS-3), December 2014. 3 hours, closed book (one 8.5"×11" self-prepared note sheet permitted). Candidates were required to complete 2 of the 3 questions in Part A (Statics) and 2 of the 3 questions in Part B (Dynamics); every question in both parts is solved below.
Check — figure readings (Questions 1 and 3): (1) In Figure 1's truss only three members meet at joint D (C–D, D–B, D–E) — there is no separate C–E member, and the truss coordinates are A(0,0), B(4,0), C(1,2), D(3,2.6), E(7,4) m. (2) Figure 3A's axial dimension stack (2+4+2+4+2 = 14 in) does not sum to the labelled overall height of 16 in unless the hub's "2 in." label is read as a half-thickness measured from the shaft centreline (giving a full hub thickness of 4 in, which closes the stack exactly: 2+4+4+4+2 = 16 in). Both readings are disclosed here and used consistently in the solutions below.
Given. Plane truss with joints A, B, C, D, E; A is a pin support, B is a roller support (vertical reaction only). A 500 N downward load and a 175 N rightward load act at E. Grid coordinates (measured directly from the source drawing, 1 division = 1 m):
Joint coordinates
Joint
x (m)
y (m)
A (pin)
0
0
B (roller)
4
0
C
1
2
D
3
2.6
E
7
4
Members: AB, AC, CB, CD, DB, DE, EB (7 members; joints j=5, reactions r=3, so m+r=10=2j — statically determinate).
Find. The axial force in every member, and whether it is tension (T) or compression (C).
Figure 1 — truss geometry (A pin, B roller).
Approach. Find the support reactions from global equilibrium, then solve the joints in the order E → A/B (moment) → D → C → B, using $\Sigma F_x=0$, $\Sigma F_y=0$ at each pin.
Support reactions. Summing moments about A (roller at B gives a vertical-only reaction $B_y$):
$$\Sigma M_A = B_y(4) - 500(7) - 175(4) = 0 \implies B_y = \frac{3500+700}{4} = 1050\ \text{N (up)}$$
Then $\Sigma F_y=0$: $A_y + B_y - 500 = 0 \implies A_y = 500-1050=-550\ \text{N}$ (550 N acting downward). $\Sigma F_x=0$: $A_x+175=0 \implies A_x=-175\ \text{N}$ (175 N acting to the left).
$$\boxed{A_x = 175\ \text{N} \leftarrow,\quad A_y = 550\ \text{N}\downarrow,\quad B_y = 1050\ \text{N}\uparrow}$$
Joint E (two unknowns: $F_{DE}$, $F_{EB}$). Unit vectors from E: $\hat u_{ED}=(-0.944,-0.330)$ (length $\sqrt{4^2+1.4^2}=4.238$ m), $\hat u_{EB}=(-0.6,-0.8)$ (length 5 m).
$$\Sigma F_x:\ -0.944F_{DE}-0.600F_{EB}+175=0$$
$$\Sigma F_y:\ -0.330F_{DE}-0.800F_{EB}-500=0$$
Solving simultaneously:
$$\boxed{F_{DE} = +790.1\ \text{N (T)}, \qquad F_{EB} = -951.3\ \text{N (C)}}$$
Joint D (unknowns $F_{CD}$, $F_{DB}$; $F_{DE}$ known). Unit vectors from D: $\hat u_{DC}=(-0.944,-0.330)$, $\hat u_{DB}=(0.164,-0.986)$ (length $\sqrt{1^2+6^2}=6.083$ m), $\hat u_{DE}=(0.944,0.330)$.
$$\Sigma F_x:\ -0.944F_{CD}+0.164F_{DB}+0.944(790.1)=0$$
$$\Sigma F_y:\ -0.330F_{CD}-0.986F_{DB}+0.330(790.1)=0$$
$$\boxed{F_{CD} = +792.0\ \text{N (T)}, \qquad F_{DB} = +35.8\ \text{N (T)}}$$
Joint C (unknowns $F_{AC}$, $F_{CB}$; $F_{CD}$ known). Unit vectors from C: $\hat u_{CA}=(-0.447,-0.894)$ (length $\sqrt{1^2+2^2}=2.236$ m), $\hat u_{CB}=(0.857,-0.514)$ (length $\sqrt{3^2+1.8^2}=3.5$ m... using exact B$-$C$=(3,-2)$, length $\sqrt{13}=3.606$ m so $\hat u_{CB}=(0.832,-0.555)$), $\hat u_{CD}=(0.944,0.330)$.
$$\Sigma F_x:\ -0.447F_{AC}+0.832F_{CB}+0.944(792.0)=0$$
$$\Sigma F_y:\ -0.894F_{AC}-0.555F_{CB}+0.330(792.0)=0$$
$$\boxed{F_{AC} = +614.9\ \text{N (T)}, \qquad F_{CB} = -581.2\ \text{N (C)}}$$
Joint A / member AB (check). With $F_{AC}$ known, horizontal equilibrium at A gives $F_{AB}$ directly:
$$\Sigma F_x:\ A_x + F_{AB} + F_{AC}\cos(63.43^\circ)=0 \implies -175+F_{AB}+0.447(614.9)=0$$
$$\boxed{F_{AB} = -100.0\ \text{N (C)}}$$
Checking joint B with all six connected-member forces confirms $\Sigma F_x=\Sigma F_y=0$ to within rounding — the solution is self-consistent.
