ROZCI
GANIE PR
TÓW PROSTYCH - uzupełnienia 3u
Pierwszy wiersz macierzy naprężeń
x
3
Ã13
x
Ã11 2
Ã12
r
O
II
r '
q
I
x
1
Ã11 dA = q dA
+"+" +"+"
AA
S WII = S Z Ã12 dA = 0 dA
( ) ( )
I
+"+" +"+"
AA
Ã13 dA = 0 dA
+"+" +"+"
AA
M WII = M Z r × p dA = r' × q dA
( ) ( )
I
+"+" +"+"
AA
r 0,x2 ,x3 r' L - x1,x2 ,x3
( ) ( )
p Ã11,Ã12 ,Ã13 q q,0,0
( ) ( )
+"+"( x2 Ã13 - x3 Ã12) dA = +"+"( x2 0 - x3 0) dA
AA
x3 Ã11 dA = x3 q dA
+"+" +"+"
AA
- x2 Ã11 dA = - x2 q dA
+"+" +"+"
AA