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_{ED}cos 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)