A building frame is subjected to wind loads as shown in Figure. Determine the resultant of the loads.

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.