Question 1 of 6: Question 1 (paper Question I) — Truss Analysis (Part A · Statics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination May 2015 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Check — the printed figure has no coordinate table. Reading the grid directly gives joints A(0,0), B(3,3), C(6,5), D(0,6), E(3,6) m, with A and D both pinned to a wall along x = 0. The drawn segment D–A runs along that same wall line; it is taken as the wall outline, not a load-carrying member. Excluding it leaves 6 members (DE, DB, EB, EC, BC, AB) and 4 reaction components (two pins) against 5 joints: m + r = 6 + 4 = 10 = 2j exactly — determinate. The resulting 10-equation joint system solves to a self-consistent set (global equilibrium residual < 10⁻&sup9; kN), corroborating the reading.
Given. Truss with joints A(0,0), B(3,3), C(6,5), D(0,6), E(3,6) m; pin supports at A and D; downward load 25 kN at C; members DE, DB, EB, EC, BC, AB.
Given data
Joint
x (m)
y (m)
Support / load
A
0
0
Pin
B
3
3
—
C
6
5
25 kN ↓
D
0
6
Pin
E
3
6
—
Find. The axial force in every member (DE, DB, EB, EC, BC, AB) and whether each is in tension (T) or compression (C).
Figure 1 — truss geometry, pin supports at A and D, and the 25 kN load at C (1 grid division = 1 m).
Approach. With this truss's two pin supports, the 4 reaction components cannot be found from global equilibrium alone (3 equations, 4 unknowns) — solve directly by the method of joints, starting at C (2 unknown members), then E, then B, then use joints D and A as a check that recovers the pin reactions.
Joint C (2 unknowns: EC, BC), with the 25 kN load applied directly at C. Member directions: C→E = (−0.9487, 0.3162), C→B = (−0.8321, −0.5547).
$$\sum F_x=0,\ \sum F_y=0 \;\Rightarrow\; F_{BC} = \boxed{-30.05\ \text{kN (C)}}, \qquad F_{EC} = \boxed{26.35\ \text{kN (T)}}$$
Joint E (2 unknowns: DE, EB), using EC from Step 1. Member directions: E→D = (−1, 0), E→B = (0, −1).
$$\sum F_x=0,\ \sum F_y=0 \;\Rightarrow\; F_{DE} = \boxed{25.00\ \text{kN (T)}}, \qquad F_{EB} = \boxed{-8.33\ \text{kN (C)}}$$
Joint B (2 unknowns: DB, AB), using BC and EB already found. Member directions: B→D = (−0.7071, 0.7071), B→A = (−0.7071, −0.7071).
$$\sum F_x=0,\ \sum F_y=0 \;\Rightarrow\; F_{DB} = \boxed{0.00\ \text{kN (zero-force member)}}, \qquad F_{AB} = \boxed{-35.36\ \text{kN (C)}}$$
Joint D (check — recovers the pin reaction at D). With DE and DB known:
$$\sum F_x=0:\ 25.00(1)+0+D_x=0 \Rightarrow D_x=\boxed{-25.00\ \text{kN (i.e. 25.00 kN}\leftarrow\text{)}}$$
$$\sum F_y=0:\ 0+0+D_y=0 \Rightarrow D_y=\boxed{0.00\ \text{kN}}$$
Joint A (check — recovers the pin reaction at A). With AB known:
$$\sum F_x=0,\ \sum F_y=0 \;\Rightarrow\; A_x=\boxed{25.00\ \text{kN}\ \rightarrow},\qquad A_y=\boxed{25.00\ \text{kN}\ \uparrow}$$
Global check: $A_x+D_x=25.00-25.00=0$ and $A_y+D_y-25=25.00+0-25=0$ — self-consistent.
