Question

A pin jointed cantilever frame is hinged to a vertical wall at A and E, and is loaded as shown in fig. Determine the forces in the member CD, CG and FG.

Answer 1

First find the angle HDG 

Let Angle HDG = θ

 tan θ = GH/HD = 2/2 = 1

Draw a section line which cut the member CD, CG and FG, Consider RHS portion of the truss as shown in fig

Taking Moment about point G, we get

∑M_G = 0

{Moment of force T_FG and T_CG is zero}

– T_CD x 2 + 2 x 2 = 0 

T_CD = 2KN (Tensile)

Since Angle HDG = DGH = HCG = HGC = 45º 

Now for angle GEK; tan¸ = 2/8 =1/4

Angle GEK = 14º

Angle EGK = 76º 

Now resolved force T_FG and T_CG as

Taking Moment about point C, we get

∑M_C = 0

{Moment of force T_CD is zero } 

– T_CG cos 45º x 2 + T_CG sin 45º x 2 – T_FG sin 76º x 2 – T_FG sin 76º x 2 + 2 x 4 = 0

– 2 T_FG (sin 76º + cos 76º) + 8 = 0

T_FG = 3.29KN (Tensile)

Taking Moment about point E, we get

∑M_E = 0

{Moment of force T_FG is zero}

 – T_CD x 4 + 2 x 10 – T_CG cos 45º x 8 – T_CG sin 45º x 2 = 0

– 2 x 4 + 20 -10 T_CG cos 45º = 0

T_CG = 1.69KN (Tensile).