z3 prz1 2

1

5kN

2,25 kN

	0
	Jl
zy = o	Ni 2,25 = 0 => 2 "
M 1! O	4i -5 + 3,18—+ M =0 2

=> Ni = 2,75 [kN]

ł

Rys. 1.124

N₂ = 2,25^2 =3,!8[kN]

węzeł I

i

W₃=O

i

2,75 kN

N.

O

Rvs. 1.125

ZA' = O, Ną — 2,75 fkN];

węzeł II

Zy = 0,    = O [kN]

3.18 k

V

\

ir = o

LX - O

2

"3 fj

N₆ -3,18 + —— = O

2

N₅ — -2,12 fkN]

N₆ = 1,06 [kN]

- węzeł III

i

O

f~

IX = O

zy = o

-2,75 + 2,12 — + N* = 0

2

Jl

7/7-2,12 — + 6-0

2

7/₈ -1,25 [kN]

N₇ = -4,5 [kN]

węzeł B

i

O

0

- 0

1,25 + 5 V2

Jl

N,

il

o

N_l} = 3,75 V2 = 5.3 [kN]

Z )■' - 0

-- 1

- 8,75 + 5/2 — + /V, | + 3,75/i •

/z

= 0    => A5 1 = 0 [kN]