Find out the forces in cantilevers frame, Mechanical Engineering

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Find out the forces in cantilevers frame:

Find out the forces in the members of the cantilevers frame by the method of sections.

Solution

Let the equilibrium of the frame to the right of section (1) - (1).

Joint D

f ED  sin 45o  + 2 = 0   ⇒ f ED  = - 2 √2 t

The - ve sign define that the member is in compression as against tension as supposed above.

∴          f ED  = 2 √2 t  (comp.)

and,

fEDcos 45o  = fCD  ⇒      fCD= 2   √2 × (1/√2) = 2 t (tension)

Now letting the equilibrium to the right of section (2) - (2) (let the forces fBC, fFC, and fFE be taken as tensile); and taking moment about F (where two of these three forces intersect) :

f BC  × 3 ← = 2 × 6 → + 2 × 3 →

∴          f BC  = 6 t               (tensile)

Taking moment about C

2 × 3 + f FE  × 3 = 0

 (that means the sum of moments has to the zero, for equilibrium, no matter whatever the sign of forces in assumed.)

∴          f EF = - 2 t          

∴          f EF = 2 t   (comp.)

Further,

2 × 2 + f FC cos 45o = 0

∴          f FC  =- 4 × √2 t   ⇒ ∴ f FC  = 4  √2 t      (comp.)

Likewise, with respect to section (3) - (3)

2 × 9 + 2 × 6 + 2 × 3 - f AB × 3 = 0

(Taking moments about G).

 ∴         f AB  = 12 t

2 × 6 + 2 × 3 + fGF  × 3 = 0  (tension)

 (Taking moments about B).

∴          fGF = - 6 t   ⇒  ∴ fGF = 6 t   (comp.)

Resolving the forces vertically at B

2 + 2 + 2 + fGB  cos 45o = 0

∴          fGB  =- 6√2 t   ⇒ fGB  = 6√2 t  (comp.)

Letting the right-hand portion of the frame, with reference to section (4) - (4), and resolving the forces vertically.

f EC   = 2 t

Likewise, we get

f FB  = 4 t   (tension)


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