Prova Resistência dos Materiais II Gilfran 2014.2

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SOLUÇÃO
1) σY 345MPa:= E 200GPa:= CS 2:= L1 3m:= L2 3m:=
W1 30
kN
m
:= W2 60
kN
m
:= MC 20kN m×:= L L1 L2+ 6 m=:=
WR W2 W1− 30
kN
m
×=:=
RV W1 L× WR
L1
2
×+ WR
L2
2
×+ 270 kN×=:=
MR_B W1 L×
L
2
× WR
L1
2
×
L1
3
L2+






×+ WR
L2
2
×
2L2
3
×+ MC+ 830 kN m××=:=
(I)
=>ΣFX 0= RAx 0=
(II)
=>ΣFY 0= RAy RBy+ RV=
=>ΣMB 0= L RAy× MA− MB+ MR_B= (III)
M x( ) RAy x× MA− W1
x
2
2
×−
WR
L2
x
3
6
×− 2
WR
L2
x L1−( )3
6
×+ MC x L1−( )0×−= E I×
d
2.
y
dx
2
×=
E I×
d y
dx
× E I× θ x( )×= RAy
x
2
2
× MA x×− W1
x
3
6
×−
WR
L2
x
4
24
×− 2
WR
L2
x L1−( )4
24
×+ MC x L1−( )×− C1+=
E I× y x( )× RAy
x
3
6
× MA
x
2
2
×− W1
x
4
24
×−
WR
L2
x
5
120
×− 2
WR
L2
x L1−( )5
120
×+ MC
x L1−( )2
2
×− C1 x×+ C2+=
x 0m 0.1m, L..:= S x a, ( ) if x a≥ 1, 0, ( ):= (FUNÇÃO SINGULAR)
Condições de Contorno: x=0 => y(0)=0 => C2 0:=
x=0 => θ(0)=0 => C1 0:=
x=L => y(L)=0 =>
RAy
x
3
6
× MA
x
2
2
×− W1
x
4
24
×−
WR
L2
x
5
120
×− 2
WR
L2
x L1−( )5
120
× S x L1, ( )×+ MC
x L1−( )2
2
× S x L1, ( )×− 0=
Aux1 x( ) W1−
x
4
24
×
WR
L2
x
5
120
×− 2
WR
L2
x L1−( )5
120
× S x L1, ( )×+ MC
x L1−( )2
2
× S x L1, ( )×−:=
Aux1 L( ) 2317.5− m
3
kN××=
θ L( ) 0= => Aux2 x( ) W1−
x
3
6
×
WR
L2
x
4
24
×− 2
WR
L2
x L1−( )4
24
× S x L1, ( )×+ MC x L1−( )× S x L1, ( )×−:=
Aux2 L( ) 1612.5− m
2
kN×=
k3 L 6 m=:=
k1
L( )
3
6
36 m
3
×=:= k2
L( )
2
2
18m
2
=:=
k1 RAy k2 MA×− Aux1 L( )−= (IV) => RAy
Aux1 L( )− k2 MA×+
k1
=
k2 RAy k3 MA×− Aux2 L( )−= (V) => MA
Aux2 L( )−
k2
k1
Aux1 L( )×+
k2
2
k1
k3−
151.25 kN m××=:=
RAy
Aux1 L( )− k2 MA×+
k1
140 kN×=:= L RAy× MA− MB+ MR_B=
RBy RV RAy− 130 kN×=:= MB MR_B L RAy×− MA+ 141.25 kN m××=:=
M x( ) RAy x× MA− W1
x
2
2
×−
WR
L2
x
3
6
×− 2
WR
L2
x L1−( )3
6
× S x L1, ( )×+ MC x L1−( )0× S x L1, ( )×−:=
0 2 4 6
200−
100−
0
100
M x( )
kN m×
x
m
∆ 10
10−
m:=
M 0( ) 151.25− kN m××=
M L1 ∆−( ) 88.75 kN m××=
M L1 ∆+( ) 68.75 kN m××=
M L( ) 141.25− kN m××=
Mmax max M 0( ) M L1( ), M L( ), ( ) 151.25 kN m××=:=
V x( ) RAy W1 x×−
WR
L2
x
2
2
×− 2
WR
L2
x L1−( )2
2
× S x L1, ( )×+:=
0 2 4 6
200−
100−
0
100
200
V x( )
kN m×
x
m
V 0m( ) 140 kN×=
V L1 ∆−( ) 5 kN×=
V L1 ∆+( ) 5 kN×=
V L( ) 130− kN×=
Vmax max V 0m( ) V L1( ), V L( ), ( ) 140 kN×=:=
 a) CS 2= σadm
σY
CS
:= σadm 172.5 MPa×=
σmáx
Mmáx
Wn
= σadm= => Wn
Mmax
σadm
:= Wn 877 10
3
mm
3
×=
Perfil escolhido: W360 x 57,8, com as seguintes propriedades:
A 7220mm
2
:= Ix 161 10
6
× mm
4
:= Wx 899 10
3
× mm
3
:= d 358mm:= tf 13.1mm:= tw 7.9mm:=
Aalma d 2 tf×−( ) tw×:= Aalma 2621.22 mm
2
×=Tensão de Cisalhamento:
τadm
σY
2
CS
86.25 MPa×=:= τmáx
Vmax
Aalma
53.41 MPa×=:= < τadm=>OK
b)
y x( )
1
E Ix×
RAy
x
3
6
× MA
x
2
2
×− W1
x
4
24
×−
WR
L2
x
5
120
×−
2
WR
L2
x L1−( )5
120
× S x L1, ( )× MC
x L1−( )2
2
× S x L1, ( )×−+
...














×:=
yC y L1( ) 5.35− mm×=:=
0 2 4 6
6−
4−
2−
0
y x( )
mm
x
m
2) P1 250kN:= P2 100kN:= e1 200mm:= e2 100mm:= L 3m:=
Le_zx 2 L×:= Le_zy 2 L×:= Le_x Le_zy 6 m=:= Le_y Le_zx 6 m=:=
E 200GPa:= σY 345MPa:= σadmf 180MPa:=
Cc
2 π
2
× E×
σY
107=:=
Mx P1 e1× P2 e2×− 40000 N m××=:= My 0:= P P1 P2+ 350 kN×=:=
Adotando: λe 145:= => ry
Le_y
λe
41.4 mm×=:= rx
Le_x
λe
41.4 mm×=:=
1ª Tentativa: Perfil W360 x 57,8 com:
A 7220mm
2
:= Wx 899 10
3
× mm
3
:= Wy 129 10
3
× mm
3
:= rx 149mm:= ry 39.2mm:=
Cálculo dos Indices de Esbeltez: λe_x
Le_x
rx
40.27=:= λe_y
Le_y
ry
153.06=:=
λe max λe_x λe_y, ( ) 153.06=:=
CSc 1.92return λe Cc≥if
5
3
3
8
λe
Cc
×+
1
8
λe
Cc






3
×−return λe Cc<if
:=
CSc 1.92=
σadmc
π
2
E×
CSc λe
2
×
return λe Cc≥if
σY
CSc
1
1
2
λe
Cc






2
−








×return λe Cc<if
:=
σadmc 43.88 MPa×=
K1
P
A
σadmc
Mx
Wx
σadmf
+
My
Wy
σadmf
+ 1.35=:= Resultado_1 if K1 1≤ "OK", "SUBDIMENSIONADA", ( ):=
Resultado_1 "SUBDIMENSIONADA"=
fa_g
1
4
K1
0.93=:= => λe λe fa_g× 141.9=:= => ry
Le_y
λe
42.3 mm×=:= => rx
Le_x
λe
42.3 mm×=:=
2ª Tentativa: Perfil W360 x 64 com:
A 8140mm
2
:= Wx 1030 10
3
× mm
3
:= Wy 186 10
3
× mm
3
:= rx 148mm:= ry 48.2mm:=
Cálculo dos Indices de Esbeltez: λe_x
Le_x
rx
40.54=:= λe_y
Le_y
ry
124.48=:=
λe max λe_x λe_y, ( ) 124.48=:=
CSc 1.92return λe Cc≥if
5
3
3
8
λe
Cc
×+
1
8
λe
Cc






3
×−return λe Cc<if
:=
CSc 1.92=
σadmc
π
2
E×
CSc λe
2
×
return λe Cc≥if
σY
CSc
1
1
2
λe
Cc






2
−








×return λe Cc<if
:=
σadmc 66.35 MPa×=
K1
P
A
σadmc
Mx
Wx
σadmf
+
My
Wy
σadmf
+ 0.86=:= Resultado_1 if K1 1≤ "OK", "SUBDIMENSIONADA", ( ):=
Resultado_1 "OK"= Logo, o perfil escolhido é o W 360 x 64