Let the reaction at ***** A and E are RAV and REV.
First we calculate the support reaction,
RH = 0, RAH = 0 ...(i)
RV = 0, RAV + REV – 10 – 15 – 20 – 10 = 0
RAV + REV = 55 ...(ii)
Taking moment about point A and equating to zero; we get
15 X 3 + 10 X 3 + 20 X 6 – REV X 6 = 0 ...(iii)
REV = 32.5 KN ...(iv)
Value of (iv) putting in equation (ii)
We get, RBV = 22.5 KN ...(v)
Consider FBD of ***** B, as shown in fig(a)
∑H = 0; TBC = 0
∑V = 0; TBA + 10 = 0; TBA = –10KN(C)
Consider FBD of ***** F, as shown in fig (b)
∑H = 0; TFC = 0
∑V = 0; TFE + 20 = 0; TFE= –20KN(C)
Consider FBD of ***** A, as shown in
∑H = 0; TAD + TAC cos 45 = 0
TAD = –TAC cos 45 ...(i)
∑V = 0; TCA sin 45 + 22.5 – 10 = 0
TCA = –17.67KN(C)
Putting this value in equation (i), we get
TAD = 12.5KN(T)
Consider FBD of ***** D, as shown in fig (c)
∑H = 0;
– TAD + TDE = 0
TAD = TDE = 12.5KN (T)
∑V = 0; TDC – 10 = 0
TDC= 10KN (T)
Consider FBD of ***** E, as shown in fig(d)
∑H = 0;
– TED – TECn cos 45 = 0
– 12.5 – TEC cos 45 = 0
TEC= – 17.67KN(C)
∑V = 0; TFE + TEC sin 45 + 32.5 = 0
TFE + (–17.67) sin 45 + 32.5 = 0
TFE = – 20KN(C)
Forces in all the members can be shown as in fig below.
