First find the support reaction which can be determined by considering equilibrium of the truss. Let RAH & RAV be the support reaction at hinged support A and RDV be the support reaction at roller support D.
∑H = 0
RAH + 10 = 0
RAH = – 10 KN
∑V = 0
RAV + RDV = 40 ...(i)
Taking moment about point A,
∑MA = 0
40 x 2 – 10 x 2 – RDV x 6 = 0
RDV = 10KN
From equation (i); RAV = 30KN
Let draw a section line 1-1 which cut the member BC, EC, FE, and divides the truss in two parts RHS and LHS as shown in fig. Make the direction of forces only in those members which cut by the section line. i.e. in BC, EF and EC, Since the question ask the forces in the member BC and EF, but by draw a section line member EC is also cut by the section line, so we consider the force in the member EC.
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 force
1. Reaction RDV = 10KN at the joint D
2. 10KN load at joint F
3. Force TBC in member BC
4. Force TCE in member CE
5. Force TFE in member FE
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 C, of all the five forces given above
∑MC = 0
(Moment of Force TBC,TCE about point C is zero, since point C lies on the line of action of that forces)
– RD x CD + TEF x CF – 10x CF = 0
–10 x 2 + TEF x 2 – 10x 2 = 0
TEF = 20KN
TEF = 20KN (Tensile) ...(iii)
Taking moment about point E, of all the five forces given above
∑ME = 0
(Moment of Force TEF,TEC and 10KN about point E is zero, since point E lies on the line of action of that forces).
– RD x BD – TBC x CF = 0
–10 x 4 – TBC x 2 = 0
TBC = –20KN
TBC = 20KN(Compressive).
