Engineering Mechanics

(^276) „„„„„ A Textbook of Engineering Mechanics
Now let us pass section (1-1) cutting the truss into two parts as shown in Fig 13.51.
Fig. 13.51.
Now consider equilibrium of the left part of the truss. Let the directions of the forces PBD, PCE
and PCD be assumed as shown in Fig 13.51. Taking moments about the joint C and equating the same,
PBD × 2 = 14.9 × 4 = 59.6
∴
59.6
29.8 kN (Compression)
BD 2
P ==
Similarly taking moments about the joint D and equating the same,
PCE × 6 tan 30° = 14.9 × 6 = 89.4
∴
89.4 89.4
25.8 kN (Tension)
6tan 30 6 0.5774
PCE== =
°×
Now for finding out PCD, we shall take moments about the A (where the other two members
meet). Since there is no force in the lift of the truss (other than the reaction VA, which will have zero
moment about A), therefore the value of PCD will be zero.
Note: The force PCD may also be found out as discussed below :
At joint B, the force in member BC is zero, as there is no other member to balance the force (if
any) in the member BC. Now at joint C, since the force in member BC is zero, therefore the force in
member CD is also equal to zero.
Example 13.16. A truss hinged at A and supported on rollers at D, is loaded as shown in
Fig. 13.52.
Fig. 13.52.
Find the forces in the members BC, FC, FE of the truss.