Prova Resistência dos Materiais II Gilfran 2015.1

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SOLUÇÃO
1) σY 345MPa:= E 200GPa:= G 77GPa:= ν 0.3:= CS 2:=
W1 30
kN
m
:= W2 90
kN
m
:= L1 3m:= L2 3m:= L L1 L2+ 6m=:=
RV W1 L1× W2
L2
2
×+ 225 kN×=:= MR_B W1 L1×
L1
2
L2+






× W2
L2
2
×
L2
3






×+ 540 kN m××=:=
(I)
ΣFX 0= => RAx RBx= 0=
(II)
ΣFY 0= => RAy RBy+ RV=
ΣMB 0= => L RAy× MA− MB+ MR_B= (III)
x 0m 0.1m, L..:= S x a, ( ) if x a≥ 1, 0, ( ):= (FUNÇÃO SINGULAR)
M x( ) RAy x× MA− W1
x
2
2
×− W1
x L1−( )2
2
× S x L1, ( )×+
W2
L2
x L1−( )3
6
× S x L1, ( )×−= E I×
d
2.
y
dx
2
×=
E I×
d y
dx
× E I× θ x( )×=
E I× θ x( )× RAy
x
2
2
× MA x×− W1
x
3
6
×− W1
x L1−( )3
6
× S x L1, ( )×+
W2
L2
x L1−( )4
24
× S x L1, ( )×− C1+=
E I× y x( )× RAy
x
3
6
× MA
x
2
2
×− W1
x
4
24
×− W1
x L1−( )4
24
× S x L1, ( )×+
W2
L2
x L1−( )5
120
× S x L1, ( )×− C1 x×+ C2+=
Condições de Contorno: x=0 => y(0)=0 => C2 0:=
x=0 => θ(0)=0 => C1 0:=
x=L => y(L)=0 => Aux1 x( ) W1−
x
4
24
× W1
x L1−( )4
24
× S x L1, ( )×+
W2
L2
x L1−( )5
120
× S x L1, ( )×−:=
θ L( ) 0= => Aux2 x( ) W1−
x
3
6
× W1
x L1−( )3
6
× S x L1, ( )×+
W2
L2
x L1−( )4
24
× S x L1, ( )×−:=
Aux1 L( ) 1579500− L kN×=
Aux2 L( ) 1046.25− m
2
kN×=
k1
L( )
3
6
36 m
3
×=:= k3 L 6m=:=
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−
85.5 kN m××=:=
RAy
Aux1 L( )− k2 MA×+
k1
86.63 kN×=:= L RAy× MA− MB+ MR_B=
RBy RV RAy− 138.38 kN×=:= MB MR_B L RAy×− MA+ 105.75 kN m××=:=
M x( ) RAy x× MA− W1
x
2
2
×− W1
x L1−( )2
2
× S x L1, ( )×+
W2
L2
x L1−( )3
6
× S x L1, ( )×−:=
0 2 4 6
150−
100−
50−
0
50
M x( )
kN m×
x
m
M 0( ) 85.5− kN m××=
M L1( ) 39.38 kN m××=
M L( ) 105.75− kN m××=
Mmax max M 0( ) M L1( ), M L( ), ( ) 105.75 kN m××=:=
V x( ) RAy W1 x×− W1 x L1−( )× S x L1, ( )×+
W2
L2
x L1−( )2
2
× S x L1, ( )×−:=
0 2 4 6
200−
100−
0
100
V x( )
kN m×
x
m
V 0m( ) 86.63 kN×=
V L1( ) 3.38− kN×=
V L( ) 138.38− kN×=
Vmax max V 0m( ) V L1( ), V L( ), ( ) 138.38 kN×=:=
 a) CS 2= σadm
σY
CS
:= σadm 172.5 MPa×=
σmáx
Mmáx
Wn
= σadm= => Wn
Mmax
σadm
:= Wn 613 10
3
mm
3
×=
Perfil escolhido: W360 x 44, 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 86.25 MPa×= τmáx
Vmax
Aalma
:= τmáx 60.3 MPa×= < τadm=>OK
b) y x( )
1
E Ix×
RAy
x
3
6
× MA
x
2
2
×− W1
x
4
24
×− W1
x L1−( )4
24
× S x L1, ( )×+
W2
L2
x L1−( )5
120
× S x L1, ( )×−








×:=
0 2 4 6
4−
3−
2−
1−
0
y x( )
mm
x
m
yC y L1( ) 3.94− mm×=:=
2) P1 100kN:= e1 200mm:= P2 200kN:= e2 0mm:= Le 6m:=
E 200GPa:= σY 345MPa:= σadmf 180MPa:=
Cc
2 π
2
× E×
σY
107=:=
P P1 P2+ 300 kN×=:= Mx P1 e1× 20 kN m××=:= My P2 e2× 0=:=
Adotando: λe 160:= => ry
Le
λe
37.5 mm×=:=
1ª Tentativa: Perfil W360 x 44 com:
A 5730mm
2
:= Wx 693 10
3
× mm
3
:= Wy 95.7 10
3
× mm
3
:= ry 37.8mm:=
Cálculo dos Indices de Esbeltez: λe
Le
ry
158.73=:=
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 40.8 MPa×=
K1
P
A
σadmc
Mx
Wx
σadmf
+
My
Wy
σadmf
+ 1.44=:= Resultado_1 if K1 1≤ "OK", "SUBDIMENSIONADA", ( ):=
Resultado_1 "SUBDIMENSIONADA"=
fa_g
1
4
K1
0.91=:= => λe λe fa_g× 144.8=:= => ry
Le
λe
41.4 mm×=:=
2ª Tentativa: Perfil W360 x 64 com:
A 8140mm
2
:= Wx 1030 10
3
× mm
3
:= Wy 186 10
3
× mm
3
:= ry 48.2mm:=
Cálculo dos Indices de Esbeltez: λe
Le
ry
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
+ 0.66=:= Resultado_2 if K1 1≤ "OK", "SUBDIMENSIONADA", ( ):=
Resultado_2 "OK"=
Logo, o perfil escolhido é o W 360 x 64