Find out the forces in the members of the tower:
Find out the forces in the members of the tower.
By symmetry, ∠ AFG = ∠ AGF = θ and HG = 7.5 m; while AH = 3 × 5 = 15 m.
cos θ= 7.5 /AG = 7.5 /16.77 = 0.4472
(∴ θ = 63.434 degrees)
and, sin θ= 15/16.77 = 0.8944
Letting the equilibrium of joint A (and, supposing fAB and fAC both to be tensile), and resolving forces horizontally and vertically, we obtain:
10 + f AC cos θ - f AB cos θ = 0 --------- (i)
And f AB sin θ+ f AC sin θ= 0 -------- (ii)
From (ii), f AB = - f AC
and, thus, from (i)
10 + 2 f AC cos θ = 0
∴ f AC =- 5 /0.4472
= - 11.1806 t
(- ve sign indicating fAC is under compression and fAB = + 11.1806 t (that means tension).
Suppose both fBC and fCD to be both tensile; and resolving forces at right-angles to AD:
FBC sin θ= 0 ⇒ ∴ f BC = 0
Hence, f AB = f BD = 11.1806 t(i.e. tensile).
Suppose forces fCD and fCE to be both tensile, and resolving all the forces at right angles to AE (taking ∠ DCE = α).
fCD sin α= 0 ⇒ ∴ fCD = 0
Since both fBC and fCD are equal to zero:
fCE = f AC = 11.1806 t(comp.)
Suppose fDE and fDF both to be tensile and resolving out forces at right-angles to BF:
f DE sin θ= 0 ⇒ ∴ f DE = 0
As fCD = f DE = 0 ⇒ ∴ f DF = f BD = 11.1806 t
Assume fEF and fEG to be both tensile, and resolving forces at right-angles to CG :
f EF sin α= 0 ⇒ f EF = 0
∴ f EG = fCE = 11.1806 t (comp.)
(a) A frame illustrated in Figure is supported by a hinge at A and a roller at E. Compute the horizontal & vertical components of hinge forces at B and C as they act upon member AC.