First find the support reaction which can be determined by considering equilibrium of the truss.
∑V = 0
RA + RD = 50 ...(i)
Taking moment about point A,
∑MA = 0
– RD X 9 + 20 X 6 + 30 x 3 = 0
RD = 23.33KN ...(ii)
Now, from equation (i); we get
RA = 26.67KN ...(iii)
Let draw a section line 1-1 which cut the member BC, BE, FE, and divides the truss in two parts RHS and LHS as shown in fig(a). Make the direction of forces only in those members which cut by the section line.
Choose any one part of them, Since both parts are separately in equilibrium. Let we choose right hand side portion (as shown in fig(b). And the Right hand parts of truss is in equilibrium under the action of following forces,
1. Reaction RD = 23.33KN
2. 20KN load at joint C
3. Force TBC in member BC (From C to B)
4. Force TBE in member BE (From E to B)
5. Force TFE in member FE (From E to F)
All three forces are assumed to be tensile
Now we take moment of all these five forces only from any point of the truss for getting the answers quickly
Taking moment about point E, of all the five forces given above
∑ME = 0
(Moment of Force TBE,TEF and 20KN about point E is zero, since point E lies on the line of action of that forces)
– RD x ED + TEF x 0 + TBE x 0 – TBC x CE + 20 x 0 = 0
– RD x 3 – TBC x 3 = 0
TBC = – 23.33KN ... (iv)
TBC = 23.33KN (Compressive)
Taking moment about point B, of all the five forces given above
∑MB = 0
(Moment of Force TBE, TBC force about point B is zero)
– RD x FD + TBC X 0 + TBE X 0 + TFE X CE + 20 X BC = 0
– RD X FD + TFE X CE + 20 X BC = 0
–23.33 X 6 + TFE X 3 + 20 X 3 = 0
TFE = 26.66KN ...(v)
TFE = 26.66KN (Tensile)
Taking moment about point F, of all the five forces given above
∑MF = 0
(Moment of Force TFE about point B is zero)
– RD x FD – TBC x EC – TBE cos45º x FE + TFE x 0 + 20 x FE = 0
–23.33 x 6 + 23.33 x 3 – TBE cos 45º x 3 + 20 x 3 = 0
TBE = 4.71KN
TBE = 4.71KN (Tensile)
