Prova Resistência dos Materiais II Gilfran 2013.2

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SOLUÇÃO
1) σY 320MPa:= E 200GPa:= CS 2:=
W 60
kN
m
:= L1 3m:= L2 2m:=
L L1 L2+ 5 m=:=
RV W
L1
2
× W L2×+ 210 kN×=:=
MR_B W
L1
2
×
L1
3
L2+






× W L2×
L2
2






×+ 390 kN m××=:=
(I)
ΣFX 0= => RAx RBx= 0=
(II)
ΣFY 0= => RAy RBy+ RV=
ΣMB 0= => L RAy× MA− MB+ MR_B= (III)
M x( ) RAy x× MA−
W
L1
x
3
6
×−
W
L1
x L1−( )
3
6
× S x L1, ( )×+= E I×
d
2.
y
dx
2
×=
E I×
d y
dx
× E I× θ x( )×= RAy
x
2
2
× MA x×−
W
L1
x
4
24
×−
W
L1
x L1−( )
4
24
× S x L1, ( )×+ C1+=
E I× y x( )× RAy
x
3
6
× MA
x
2
2
×−
W
L1
x
5
120
×−
W
L1
x L1−( )
5
120
× S x L1, ( )×+ 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 => Aux1 x( )
W
L1
−
x
5
120
×
W
L1
x L1−( )
5
120
× S x L1, ( )×+:= Aux1 L( ) 515500− L kN×=
θ L( ) 0= => Aux2 x( )
W
L1
−
x
4
24
×
W
L1
x L1−( )
4
24
× S x L1, ( )×+:= Aux2 L( ) 507.5− m
2
kN×=
k1
L( )
3
6
20.83 m
3
×=:= k2
L( )
2
2
12.5 m
2
=:= k3 L 5 m=:=
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−
79.28 kN m××=:=
RAy
Aux1 L( )− k2 MA×+
k1
72.31 kN×=:= L RAy× MA− MB+ MR_B=
RBy RV RAy− 137.69 kN×=:= MB MR_B L RAy×− MA+ 107.72 kN m××=:=
M x( ) RAy x× MA−
W
L1
x
3
6
×−
W
L1
x L1−( )
3
6
× S x L1, ( )×+:=
0 1 2 3 4 5
150−
100−
50−
0
50
100
M x( )
kN m×
x
m
M 0( ) 79.28− kN m××=
M L1( ) 47.66 kN m××=
M L( ) 107.72− kN m××=
Mmax max M 0( ) M L1( ), M L( ), ( ) 107.72 kN m××=:=
V x( ) RAy
W
L1
x
2
2
×−
W
L1
x L1−( )
2
2
× S x L1, ( )×+:=
0 1 2 3 4 5
200−
100−
0
100
V x( )
kN m×
x
m
V 0m( ) 72.31 kN×=
V L1( ) 17.69− kN×=
V L( ) 137.69− kN×=
Vmax max V 0m( ) V L1( ), V L( ), ( ) 137.69 kN×=:=
 a) CS 2= σadm
σY
CS
:= σadm 160 MPa×=
σmáx
Mmáx
Wn
= σadm= => Wn
Mmax
σadm
:= Wn 673 10
3
mm
3
×=
Perfil escolhido: W360 x 64, com as seguintes propriedades:
A 5730mm
2
:= Ix 122 10
6
× mm
4
:= Wx 693 10
3
× mm
3
:= d 352mm:= tf 9.8mm:= tw 6.9mm:=
Aalma d 2 tf×−( ) tw×:= Aalma 2293.56 mm
2
×=Tensão de Cisalhamento:
τadm
σY
2
CS
:= τadm 80 MPa×= τmáx
Vmax
Aalma
:= τmáx 60 MPa×= < τadm=>OK
b) y x( )
1
E Ix×
RAy
x
3
6
× MA
x
2
2
×−
W
L1
x
5
120
×−
W
L1
x L1−( )
5
120
× S x L1, ( )×+








×:=
0 1 2 3 4 5
4−
3−
2−
1−
0
y x( )
mm
x
m
yC y L1( ) 2.94− mm×=:=
2) P 400kN:= ey 100mm:= ex 0mm:= L 5m:=
Le_zy 2 L×:= Le_zx 0.7 L×:= Le_x Le_zy 10 m=:= Le_y Le_zx 3.5 m=:=
E 200GPa:= σY 320MPa:= σadmf 180MPa:=
Cc
2 π
2
× E×
σY
111.1=:=
Mx P ey× 40000 N m××=:= My P ex× 0=:=
Adotando: λe 140:= => ry
Le_y
λe
25 mm×=:= rx
Le_x
λe
71.4 mm×=:=
1ª Tentativa: Perfil W360 x 32,9 com:
A 4170mm
2
:= Wx 474 10
3
× mm
3
:= Wy 45.8 10
3
× mm
3
:= rx 141mm:= ry 26.4mm:=
Cálculo dos Indices de Esbeltez: λe_x
Le_x
rx
70.92=:= λe_y
Le_y
ry
132.58=:=
λe max λe_x λe_y, ( ) 132.58=:=
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 58.49 MPa×=
K1
P
A
σadmc
Mx
Wx
σadmf
+
My
Wy
σadmf
+ 2.11=:= Resultado_1 if K1 1≤ "OK", "SUBDIMENSIONADA", ( ):=
Resultado_1 "SUBDIMENSIONADA"=
fa_g
1
4
K1
0.83=:= => λe λe fa_g× 110=:= => ry
Le_y
λe
31.8 mm×=:= => rx
Le_x
λe
90.9 mm×=:=
2ª Tentativa: Perfil W360 x 44 com:
A 5730mm
2
:= Wx 693 10
3
× mm
3
:= Wy 95.7 10
3
× mm
3
:= rx 146mm:= ry 37.8mm:=
Cálculo dos Indices de Esbeltez: λe_x
Le_x
rx
68.49=:= λe_y
Le_y
ry
92.59=:=
λe max λe_x λe_y, ( ) 92.59=:=
CSc 1.92return λe Cc≥if
5
3
3
8
λe
Cc
×+
1
8
λe
Cc






3
×−return λe Cc<if
:=
CSc 1.91=
σadmc
π
2
E×
CSc λe
2
×
return λe Cc≥if
σY
CSc
1
1
2
λe
Cc






2
−








×return λe Cc<if
:=
σadmc 109.5 MPa×=
K1
P
A
σadmc
Mx
Wx
σadmf
+
My
Wy
σadmf
+ 0.96=:= Resultado_1 if K1 1≤ "OK", "SUBDIMENSIONADA", ( ):=
Resultado_1 "OK"=
Logo, o perfil escolhido é o W 360 x 44