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 45^o  + 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,

f_EDcos 45^o  = f_CD  ⇒      f_CD= 2   √2 × (1/√2) = 2 t (tension)

Now letting the equilibrium to the right of section (2) - (2) (let the forces f_BC, f_FC, and f_FE 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 45^o = 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 + f_GF  × 3 = 0  (tension)

 (Taking moments about B).

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

Resolving the forces vertically at B

2 + 2 + 2 + f_GB  cos 45^o = 0

∴          f_GB  =- 6√2 t   ⇒ f_GB  = 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)