Find out the forces in the members of the tower:
Find out the forces in the members of the tower.
Solution
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 f_{AB} and f_{AC} 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 f_{AC} is under compression and f_{AB} = + 11.1806 t (that means tension).
Joint B
Suppose both f_{BC} and f_{CD} to be both tensile; and resolving forces at right-angles to AD:
F_{BC }sin θ= 0 ⇒ ∴ f _{BC} = 0
Hence, f _{AB} = f _{BD} = 11.1806 t(i.e. tensile).
Joint C
Suppose forces f_{CD} and f_{CE} to be both tensile, and resolving all the forces at right angles to AE (taking ∠ DCE = α).
f_{CD} sin α= 0 ⇒ ∴ f_{CD} = 0
Since both f_{BC} and f_{CD} are equal to zero:
f_{CE} = f _{AC} = 11.1806 t(comp.)
Joint D
Suppose f_{DE }and f_{DF} both to be tensile and resolving out forces at right-angles to BF:
f _{DE} sin θ= 0 ⇒ ∴ f _{DE} = 0
As f_{CD} = f _{DE } = 0 ⇒ ∴ f _{DF } = f _{BD} = 11.1806 t
Joint E
Assume f_{EF} and f_{EG} to be both tensile, and resolving forces at right-angles to CG :
f _{EF} sin α= 0 ⇒ f _{EF } = 0
∴ f _{EG } = f_{CE} = 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.