Dummy

Question 1 Solution
According to Equilibrium conditions,
∑FY=0
RH + RC = 50 + 35(sin(30))
RH + RC = 50 + 17.5= 67.5 kN(equation 1)
∑FX=0
HH + HC = 35(cos(30))
HH + HC = 30.31 kN (equation 2)
∑MH=0
50(3) + 17.5(6) = HC ( 3( tan(50))
150 + 105 = HC ( 3.57)
HC = 71.328 kN
Putting value of Hc into equation 2
HH = -41.018 Kn
Rc
RH
Hc
HH
At joint F
∑FY=0
FFA = 17.5 kN (Tension)
∑FX=0
FFG = 30.31 kN (Tension)
At joint A
∑FY=0
FFA + FAD (cos (40)) = 0
FAD (cos (40)) = -17.5
FAD = -22.845 kN (Compression)
∑FX=0
FAB + FAD sin (40) = 0
FAB = 22.845 sin (40) ( we know FAD value)
FAB = 14.684 kN (Tension)
FFA
FFG 30.31
17.5
F
FAF
FAD
FAB A
40°
At joint H
∑FX=0
FHG = HH
FHG = -41.018 kN (Compression)
∑FY=0
RH = 0
At joint C
∑FY=0
RC = FCE cos (40)
67.5 = FCE cos (40) ( from equation 1 RC = 67.5 kN as the value of RH =0)
FCE = 88.12 Kn (Tension)
∑FX=
FCB + FCE sin (40) = HC
FCB + (88.12) sin (40) = 71.38
FCB = 14.686 kN (Tension)
FHG HH
RH
FCB
FCE RC
HC
40°
H
C
At joint D
∑FX=0
FDA cos (50) = FDG (cos (50)) + FDB (cos (40))
-14.684 = FDG (0.642) + FDB (0.766) ( As the value of FDA is -22.845 kN)
FDB = (-14.684 - FDG (0.642)) / 0.766 ( equation 3)
∑FY=0...