Find out the forces in the members of the tower, Mechanical Engineering

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.

∴   22_Find out the forces in the members of the tower.png

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).

Joint B

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).

Joint C

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.)

Joint D

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

Joint E

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.

Posted Date: 1/29/2013 7:43:06 AM | Location : United States

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