This is the question of multiple beam (i.e., beam on a beam). In this type of question, first consider the top most beam, then second last beam as, In this problem on point E and F, there are roller support, and this support give reaction to both up and down beam. Consider FBD of top most beam EB as shown in fig(a)
∑V = 0;
RE + RB – 10 – 4 – 6 – 8 = 0
RE + RB = 28KN ........(i)
10 × 1 + 4 × 2 + 6 × 3 + 8 × 4 – RB × 6 = 0
RB = 11.33 KN
Putting the value of RB in equation (i)
RE = 16.67 KN
Consider FBD of second beam AF as shown in fig(b)
∑V = 0;
RA + RF – 6 – 8 – RE = 0
RA + RF = 30.67 KN........(ii)
Now taking moment about point ‘A’
6 × 1 + 8 × 2 + 16.67 × 3 – RF × 6 = 0
RF = 12 KN .
Putting the value of RF in equation (ii)
RA = 18.67 KN
Consider FBD of third beam CD as shown in fig(c)
∑V = 0;
RC + RD – RF = 0
RC + RD = 12 KN ........(iii)
Now taking moment about point ‘C’
– RD × 5 + 12 × 3 = 0
RD = 7.2 KN
Putting the value of RD in equation (iii)
RC = 4.8 KN.
