Question :- A building frame is subjected to wind loads as shown in Figure. Determine the resultant of the loads.
Solution:
The roof is inclined at 45° to horizontal and loads are at 90° to the roof. Hence, the loads are also inclined at 45° to vertical/horizontal.
Now,
R = ΣFx = (5+10+10+5+5+10+10 + 5) cos 45°
= 60× 1/√2= 42.426 kN
ΣFy = -(5 + 10 + 10 + 5) sin 45° + (5 + 10 + 10 + 5) sin 45°
y = 0
R = Fx = 42.426 kN
and its direction is horizontal.
Let R be at a distance d from the ridge A.
Then,
Rd = ΣΜΑ
60×1/√2×d=5x 3/√2+10x 2/√2+10x 1/√2+10x1/√2 +10x 2/√2 +5x3√2
d = 1.5 m
... Resultant is a horizontal force of magnitude 42.426 kN at 1.5 m below A.
