Prova Resistência dos Materiais II Gilfran 2018.2

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SOLUÇÃO
1) σY 345MPa E 200GPa CS 2
W 60
kN
m
 P 30kN M0 20kN m L1 2m L2 3m
L L1 L2 5 m
RV P W L1 W
L2
2
 240 kN
MR_B P L2 W L1
L1
2
L2






 W
L2
2

2L2
3






 M0 730 kN m
ΣFX 0= => RAx RBx= 0= (I)
ΣFY 0= => RAy RBy RV= (II)
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 W
x
2
2

W
L2
x L1 3
6
 S x L1  P x L1  S x L1 
M0 x L1 0 S x L1 
= E I
d
2.
y
dx
2
=
v x( ) RAy W x
W
L2
x L1 2
2
 S x L1  P x L1 0 S x L1 =
E I
d y
dx
 E I θ x( )= RAy
x
2
2
 MA x W
x
3
6

W
L2
x L1 4
24
 S x L1  P
x L1 2
2
 S x L1 
M0 x L1  S x L1  C1
=
E I y x( ) RAy
x
3
6
 MA
x
2
2
 W
x
4
24

W
L2
x L1 5
120
 S x L1  P
x L1 3
6
 S x L1 
M0
x L1 2
2
 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( ) W
x
4
24

W
L2
x L1 5
120
 S x L1  P
x L1 3
6
 S x L1  M0
x L1 2
2
 S x L1 
Aux1 L( ) 1567000
m
4
kg
s
2

θ L( ) 0= =>
Aux2 x( ) W
x
3
6

W
L2
x L1 4
24
 S x L1  P
x L1 2
2
 S x L1  M0 x L1  S x L1 
Aux2 L( ) 1.26 10
6

m
3
kg
s
2

k1
L
3
6
20.83 m
3
 k2
L
2
2
12.5m
2
 k3 L 5m
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
126.92 kN m
RAy
Aux1 L( ) k2 MA
k1
151.37 kN RBy RV RAy 88.63 kN
L RAy MA MB MR_B= MB MR_B L RAy MA 100.08 kN m
M x( ) RAy x MA W
x
2
2

W
L2
x L1 3
6
 S x L1  P x L1  S x L1 
M0 x L1 0 S x L1 

0 1 2 3 4 5
200
100
0
100
M x( )
kNm
x
m
Δ 10
15
m
M 0( ) 126.92 kN m
M L1 Δ  55.82 kN m
M L1 Δ  75.82 kN m
M L( ) 100 kN m
Mmax max M 0( ) M L1  M L( ) 
Mmax 126.92 kN m
V x( ) RAy W x
W
L2
x L1 2
2
 S x L1  P x L1 0 S x L1 
0 1 2 3 4 5
100
0
100
200
V x( )
kNm
x
m
V 0( ) 151.37 kN
V L1 Δ  31.37 kN
V L1 Δ  1.37 kN
V L( ) 88.63 kN
Vmax max V 0( ) V L1  V L( ) 
Vmax 151.37 kN
 a) CS 2 σadm
σY
CS
 σadm 172.5 MPa
σmáx
Mmáx
Wn
= σadm= => Wn
Mmax
σadm
 Wn 735.77 10
3
 mm
3

Escolhendo: 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 bf 172mm
tf 13.1mm tw 7.9mm
Tensão de Cisalhamento: Aalma d 2 tf  tw Aalma 2621.2 mm
2

τadm
σY
2
CS
86.25 MPa τmáx
Vmax
Aalma
57.75 MPa
τmáx 57.7 MPa Resul_1 if τmáx τadm "OK" "REDIMENSIONAR"  "OK"
Perfil Escolhido: W360 x 57,8
b) y x( )
1
E Ix
RAy
x
3
6
 MA
x
2
2
 W
x
4
24

W
L2
x L1 5
120
 S x L1  P
x L1 3
6
 S x L1 
M0
x L1 2
2
 S x L1  C1 x C2
















0 1 2 3 4 5
4
3
2
1
0
1
y x( )
mm
x
m
y L1  2.86 mm
2) P1 100kN P2 300kN s 100mm e1 0mm e2 s Le 6m
E 200GPa σY 345MPa σadmf 180MPa
Cc
2 π
2
 E
σY
107
P P1 P2 400 kN Mx P1 e1 P2 e2 30 kN m My 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
 1.95 Resultado_1 if K1 1 "OK" "SUBDIMENSIONADA" 
Resultado_1 "SUBDIMENSIONADA"
fa_g
1
4
K1
0.85 => λe λe fa_g 134.3 => ry
Le
λe
44.7 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
 0.9 Resultado_2 if K1 1 "OK" "SUBDIMENSIONADA" 
Resultado_2 "OK"
Logo, o perfil escolhido é o W 360 x 64