Given the frame shown, determine the internal loadings at D
The Attempt at a Solution
This should be a simple problem.....but I cannot see where I am making a mistake.
If I approach it by examining the entire structure, I get a wrong answer:
ΣFx: Ax + (cos60)IB = 0
ΣFy: Ay - (sin60)IB – 150 = 0
ΣM about I I: 3.44Ay – (7.44)(150) = 0
The 3.44 comes from determining, by trig, that IA is 3.44; the 150LB load is therefore 7.44 ft away from I
This yields Ay = 324.4
Plug back into sum of forces in y-direction; that yields IB = -201
OK now examine the section BDHC:
ΣFx: -(cos60)(201) + (cos45)GH=0
This yields GH= 142, which is wrong.
However, if I start instead with just the structure BDHC, I get:
sum of moments about B: (2)[(sin45)GH] – (4)(150)=0 then GH=424 which agrees with the book
Where did I make a wrong turn? Any help greatly appreciated. Thank you.