Question

Determine the forces in the members FH, HG and GI in the truss shown in Fig.(a). Each load is 10 kN and all triangles are equilateral with sides 4 m.

Answer 1

Due to symmetry,

R_A = R₀ = 1/2 × 10 × 7 = 35 kN

Take section (A)–(A), which cuts the members FH, GH and GI and separates the truss into two parts. Consider the equilibrium of left hand side part as shown in Fig. (b) (Prefer part in which number of forces are less).

ΣMG = 0, gives 

F_FH × 4 sin 60° – 35 × 12 + 10 × 10 + 10 × 6 + 10 × 2 = 0 

F_FH = 69.2820 kN (Comp.) 

∑V = 0, gives 

F_GH sin 60° + 10 + 10 + 10 – 35 = 0 

F_GH = 5.7735 kN (Comp.) 

∑ H = 0, gives

F_GI – F_FH – FGH cos 60° = 0 

F_GI = 69.2820 + 5.7735 cos 60° 

= 72.1688 kN (Tension)