Question 2 of 6: Question 2 (paper Question II) — Plane Truss, Method of Joints (Part A · Statics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2019-Dec. Candidates were required to complete 2 questions from PART A (Statics) and 2 questions from PART B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1-3 = paper Part A, I-III; Questions 4-6 = paper Part B, IV-VI).
Check — source reconstruction notes. (2) Question 1's isometric figure gives explicit 1 m / 1.5 m dimensions along the rod, but plotting them on a single straight line makes the six rigid-body equilibrium equations inconsistent (no solution exists). Reading the printed figure, the horizontal run bends twice: A→C is 2 m along the first horizontal direction (with the vertical 1.5 m branch C→D), C→E is 1 m along the perpendicular horizontal direction (this is where the smooth journal bearing sits), and E→F is a further 1.5 m continuing the first horizontal direction to the point that carries the 250 N cylinder. This reading is the only one of the plausible alternatives that returns a statically-determinate, non-degenerate system, and it lands on the clean values reported below — strong corroborating evidence for the reconstruction. (3) Question 4's dimensions place the pin O exactly 90 mm above the floor, equal to the sphere radius (90 mm R); the sphere therefore contacts the floor exactly when the rod is horizontal, which is used as the impact configuration. (4) Question 6 numbers the four-bar linkage O-A-B-C from the source's own 100 mm / 250 mm / 75 mm / 50 mm dimensions; O and C are the fixed pivots, OA and CB are the driven/driving cranks and AB is the coupler.
Question 2 (paper Question II) — Plane Truss, Method of Joints (Part A · Statics, equal value)
Given. Plane truss, pin support at A(0,0), roller support at B(6,0). Joints C(0,3), D(6,3), E(3,7), F(9,7). A vertical 500 N load acts downward at F. Members: AC, CD, AB, CB, CE, ED, DB, DF, EF (9 members, 6 joints, pin+roller = 3 reactions; $m+r=2j=12$, statically determinate).
Find. The force in every member, and whether it is tension (T) or compression (C).
Figure 2. Six-joint, nine-member truss; grid units = 1 m; 500 N load at F.
Approach. Find the two support reactions from global equilibrium, then analyze each joint in turn (starting where only two unknown member forces meet), applying $\Sigma F_x=0,\ \Sigma F_y=0$ at every joint.
Support reactions. Taking moments about A: $B_y(6)-500(9)=0\Rightarrow B_y=750$ N. Then $\Sigma F_y=0$: $A_y+750-500=0\Rightarrow A_y=-250$ N (i.e. 250 N downward), and $\Sigma F_x=0\Rightarrow A_x=0$.
Joint A (members AC vertical, AB horizontal): $\Sigma F_x=0\Rightarrow F_{AB}=0$; $\Sigma F_y=0\Rightarrow F_{AC}+A_y=0\Rightarrow F_{AC}=250$ N (T).
Joint B (members AB horizontal, DB vertical, CB diagonal to (0,3), reaction $B_y=750$ up): with $F_{AB}=0$, resolving the diagonal CB (direction cosines $6/6.708,\,3/6.708$) and DB against $B_y$ gives $F_{CB}=0$ and $F_{DB}=-750$ N (i.e. 750 N C).
Joint C (AC, CD, CB, CE meeting): with $F_{AC}=250$ (T) and $F_{CB}=0$, resolving CD (horizontal) and CE (direction cosines to E(3,7): $3/6.083,\,4/6.083$) gives $F_{CE}=312.5$ N (T) and $F_{CD}=-187.5$ N (i.e. 187.5 N C).
Joint D (CD, ED, DB, DF meeting): with $F_{CD}=-187.5$, $F_{DB}=-750$ known, resolving ED (to E(3,7): dir. $-3/5,\,4/5$) and DF (to F(9,7): dir. $3/5,\,4/5$) gives $F_{ED}=-312.5$ N (312.5 N C) and $F_{DF}=-625$ N (625 N C).
Joint F (EF horizontal, DF diagonal, 500 N load) — check: $\Sigma F_y=0$: $-0.8F_{DF}-500=0\Rightarrow F_{DF}=-625$ N ✓; $\Sigma F_x=0$: $-F_{EF}-0.6F_{DF}=0\Rightarrow F_{EF}=375$ N (T) ✓, matching the independent joint-E check.
Final Results — Question 2 (member forces, T = tension, C = compression)
