Engineering Mechanics

(^298) „„„„„ A Textbook of Engineering Mechanics
Solution. Taking moments about A and equating the same,
RD × 15 = (3 × 5) + (6 × 12.5) = 90
∴
90
6kN
D 15
R ==
and RA = (3 + 6) – 6 = 3 kN
First of all, draw the space diagram and name all the members of the truss and forces according
to Bow’s notations as shown in Fig. 14.14 (a).
Fig. 14.14.
Now draw vector diagram as shown in Fig. 14.14 (b). Measuring various sides of the vector
diagram, the results are tabulated here :
S. No. Member Magnitude of force in kN Nature of force
1 AG (1-5) 3.5 Compression
2 FG (1-6) 3.2 Compression
3 FE (1-8) 3.2 Compression
4 ED (2-9) 7.0 Compression
5 AB (4-5) 1.7 Tension
6 BG (5-6) 3.0 Tension
7 BF (6-7) 0.5 Tension
8 BC (3-7) 3.0 Tension
9 CF (7-8) 0.5 Tension
10 CE (8-9) 0.5 Compression
11 CD (3-9) 3.5 Tension