First we calculate the support reaction, Draw FBD as shown in fig(a)
RH = 0,
RAH – RBV cos 60 = 0 ...(i)
RV = 0, RAV + RBV sin 60 – 24 – 7 – 7 – 8 = 0 ...(ii)
Taking moment about point
B, RAV X 6 – 24 X 3 – 7 X 6 – 8 X 3 = 0 ...(iii)
RAV = 23KN ...(iv)
Value of (iv) putting in equation (ii) We get,
RBV = 26.6KN ...(v)
Value of (v) putting in equation (i) We get,
RAH =13.3KN ...(vi)
***** E, Consider FBD as shown in fig(b) From article 8.8.2,
TED = 0,...(vii)
And, TAE = –7KN...(viii)
TAE = 7KN (compression)
Joint C: Consider FBD as shown in fig (c)
TCD = 0, ...(ix)
TCD = 0
And, TBC = – 7KN ...(x)
TBC = 7KN (compression)
Note: Since for perfect frame the condition n = 2j – 3 is necessary to satisfied.
Here Point F is not a , if we take F as a then,
Number of *****(j) = 6 and No. of member (n) = 7
n = 2j – 3, 7 = 2 X 6 – 3
≠ 9 i.e
i.e if F is not a *****, then j = 5
7 = 2 X 5 – 3
= 7, i.e. F is not a . But at F a force of 8KN is acting. Which will effect on A and B, Since 8KN is acting at the middle point of AB, So half of its magnitude will equally effect on A and B. i.e. 4KN each acting on ***** A and B downwards.
