a x x a b m pl m m pl a a a c y y p y p a a - lem.ep.usp.br · 1 0) 0) 0) 0 2 ( ) ( ) ( ) ( ) ( ) =...
TRANSCRIPT
Exercício 1
∑∑∑
=⇒−==
=⇒+−==
==
PYPYYc
PlMMPlMb
XXa
AA
AAA
A
0)
0)
0)
)(
PlPxMMPxPlMf
PVVPYe
NXd
XXS
XX
X
−=⇒+−==
=⇒−==
==
∑∑∑
)()(
)()(
)(
0)
0)
0)
Diagramas:
Exercício 2
PxVPxMPxVPxV
N
XX
XX
X
=⇒=+
=⇒=−
=
)()(
)()(
)(
0)3
0)2
0)1
Diagramas:
Exercício 3
lPxMx
lxPM
lPxV
lPxV
NxP
lP
XX
XX
X
X
60
32)4
20
2)3
0)2
)1
3
)(
2
)(
2
)(
2
)(
)(
−=⇒=
+
=⇒=−
=
=
Diagramas
Exercício 4 :
lPbYA =
PblYMlPaY
lYPaMYPY
X
AB
B
BA
BA
A
=−⇒=
=
−⇒=
=+−=
∑
∑
00)4
0)30)2
0)1
)(
)(
ISeção)5 )0( ax ≤≤
x
lPbMMx
lPbc
lPbVb
Na
XX
X
X
=⇒=−
=
=
)()(
)(
)(
0)
)
0)
IISeção)6 ( ) lxa ≤≤
( ) ( )
( )
( )( ) ( )
( ) ( )
01
0)
0)
0)
2
)(
)(
)(
)(
)(
=⇒=
=−
=+−
=⇒=
−=
+−=+−=
=−−−
−=−
=−
=−=
=−−
=
MxlPab
lalPa
lPalPaMax
lPaxPal
lPaxlbPPaPx
lPbxM
MaxPlPbxc
lPa
llbP
lPlPbP
lPbV
VPlPbb
Na
X
X
X
X
X
Diagramas :
Exercício 6 :
∑
∑
=⇒−==
=
−
==
=
620)3
3
32
20)2
0)1
)(
)(
PlYPllYM
PlY
lYlPlM
X
AAB
B
BA
A
( )
( )
33
32626
363
61
660
32
6)
260
26)
0)
)4
2222
2
32
22
lxxll
PxPll
PxPlV
PlVlPllPlPlVx
lPxPlxMM
l
xPxPlxc
lPxPlVV
lPxPlb
NalPxP
lP
xP
XX
=====−=
−=⇒−
=−=⇒=
−=⇒=−−
−=⇒=−−
=
=⇒=
Diagramas :
Exercício 7
PxPPP X 2=⇒=
( ) lx X
21
≤≤
20 lx
lPxPlxMM
lxPxPlxc
lPxPlVV
lPxPlb
Na
340
34)
40
4)
0)
32
22
−=⇒=−−
−=⇒=−−
=
Diagramas :
Exercício 8 :
lMYlYMM
lMYlYMM
X
AAB
oBBA
A
00)(
0)(
0)3
0)2
0)1
−=⇒+==
=⇒−==
=
∑
∑
( ) lx ≤≤0
xlMMx
lMMc
lM
lMVb
Na
00
00
0)
0)
0)
=⇒=−
−⇒=+
=
Diagramas :
Exercício 9 : Ex5 Ex8
lMPlYMLPllYM
lMPlYMlYlPlM
Y
AAB
BBA
A
00)(
00)(
220)3
2220)2
0)1
−=⇒+−==
+=⇒+−==
=
∑
∑
22
02
2)
20
2)
0)
20
0
00
PxxlMPlM
MxPxx
lMPl
c
PxlMPlVVPx
lMPlb
Na
−
−=
=−−
−
−−=⇒=−−−
=
Diagramas :
Exercícios 10 :
2
2)2
2
20)1
)(
)(
PllPbY
lPlPblYM
PllPaY
PllYPaM
A
AB
B
BA
+=
−−=
+=
+−==
∑
∑
( )ax ≤≤0
22
)
20
2)
0)
2
)(
PxxPll
Pbc
PxPblPbVVPxPl
lPbb
Na X
=
+
−+=⇒=−−+
=
( )( )
lPaPaPb
lPaPalPblPPaPb
lPbPV
PaPbaPalPbPaaPl
lPbM
PaPblPbPaPl
lPbV
A
A
A
−−=−−
=−−+=−
−++=−+=
−+=−+=
2222
22
22222
222
)(
222
)(
)(
Diagramas :