Prova Resistência dos Materiais II Gilfran 2011.2

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SOLUÇÃO
 -DADOS:: kN 1000N:= MPa 10
6
Pa:= σY 280MPa:= E 200000MPa:=
1)
(I)
ΣFX 0= => RAx 0:=
(II)
ΣFY 0= => RAy RBy+ 180=
ΣMB 0= => 6 RAy× MA− 585= (III)
M x( ) RAy x× MA− 30
x
2
2
×− 30
x 3−( )
2
2
×+ 60
x 3−( )
2
2
×− 20
x 3−( )
3
6
×+= E I×
d
2.
y
dx
2
×=
E I×
d y
dx
× E I× θ x( )×= RAy
x
2
2
× MA x×− 30
x
3
6
×− 30
x 3−( )
3
6
×+ 60
x 3−( )
3
6
×− 20
x 3−( )
4
24
×+ C1+=
E I× y x( )× RAy
x
3
6
× MA
x
2
2
×− 30
x
4
24
×− 30
x 3−( )
4
24
×+ 60
x 3−( )
4
24
×− 20
x 3−( )
5
120
×+ C1 x×+ C2+=
Condições de Contorno: x=0 => y(0)=0 => C2 0:=
x=0 => θ(0)=0 => C1 0:=
x=6 => y(6)=0 => RAy
x
3
6
× MA
x
2
2
×− 30
x
4
24
× 30
x 3−( )
4
24
×− 60
x 3−( )
4
24
×+
x 3−( )
5
120
+=
36RAy 18 MA×− 1680.75= (IV)
36− RAy 6 MA×+ 3510−= (III x -6)
12− MA 1829.25−=
MA
1829.25
12
:= MA 152.4= kN m×
RAy
585 152.4+
6
:= RAy 122.9= kN
RBy 180 RAy−:= RBy 57.1= kN
x 0 1, 6..:= S x a, ( ) if x a≥ 1, 0, ( ):= (FUNÇÃO SINGULAR)
M x( ) RAy x× S x 0, ( )× MA S x 0, ( )×− 30
x
2
2
× S x 0, ( )×− 30
x 3−( )
2
2
× S x 3, ( )×+ 60
x 3−( )
2
2
× S x 3, ( )×−
20
x 3−( )
3
6
× S x 3, ( )×+
...:=
Mmáx 152.4kN m×:=
M 0( ) 152.438−= M 3( ) 81.262= M 6( ) 0−=
V x( ) RAy S x 0, ( )× 30 x× S x 0, ( )×− 30 x 3−( )× S x 3, ( )×+ 60 x 3−( )× S x 3, ( )×− 20
x 3−( )
2
2
× S x 3, ( )×+:=
V 0( ) 122.9= V 3( ) 32.9= V 6( ) 57.1−= Vmáx 122.9kN:=
 a) CS 2:= σadm
σY
CS
:= σadm 140 MPa×=
σmáx
Mmáx
Wn
= σadm= => Wn
Mmáx
σadm
:= Wn 1.089 10
6
× mm
3
×=
Perfil escolhido: W360 x 79, com as seguintes propriedades:
A 10100mm
2
:= Ix 225 10
6
× mm
4
:= Wx 1271 10
3
× mm
3
:=
Aalma 354 2 16.8×−( ) 9.4× mm
2
:= Aalma 3.012 10
3
× mm
2
×=Tensão de Cisalhamento:
τadm
σY
2
CS
:= τadm 70 MPa×= τmáx
Vmáx
Aalma
:= τmáx 40.8 MPa×= < τadm=>OK
b)
y x( )
1000 m×
200 10
9
× 225 10
6−
××
122.9
x
3
6
× S x 0, ( )× 152.4
x
2
2
× S x 0, ( )×− 30
x
4
24
× S x 0, ( )×−
30
x 3−( )
4
24
× S x 3, ( )× 60
x 3−( )
4
24
× S x 3, ( )×− 20
x 3−( )
5
120
× S x 3, ( )×++
...












×:=
y 3.5( ) 5.4− 10
3−
× m= y 3.5( ) 5.4− mm×=
2) P1 200kN:= P2 100kN:= e 300mm:= L 4m:= Le_xz 2 L×:= Le_xz 8m=
E 200GPa:= σY 280MPa:= σadmf 180MPa:= Le_yz 0.7 L×:= Le_yz 2.8m=
P P1 P2+:= P 300 kN×= Mx P2 e×:= Mx 30 kN m××=
Cc
2 π
2
× E×
σY
:= Cc 118.7=
λe_yz
Le_yz
rx
=
λe_xz
Le_xz
ry
=
λe_xz 170:=Como Le_xz Le_yz> e rx ry> o maior indice de esbeltez ocorre no plano x-z. Adotando:
ry
Le_xz
λe_xz
:= ry 47.059 mm×=
Vamos testar incialmente o perfil W360 x 64:
ry 48mm:= => λe_xz
Le_xz
ry
:=
A 8130mm
2
:= Wx 1027 10
3
× mm
3
:=
λe_xz 166.7=
σadmc
π
2
E×
1.92 λe_xz
2
×
:= σadmc 37.011 MPa×=
K
P
A
σadmc
Mx
Wx
σadmf
+:= K 1.159= Subdimensionada
fa
1
4
K
:= fa 0.96= λe_xz fa λe_xz×:= λe_xz 160.62=
ry
Le_xz
λe_xz
:= ry 49.8 mm×=
Vamos testar agora o perfil W360 x 79:
A 10100mm
2
:= Wx 1271 10
3
× mm
3
:= ry 48.8mm:= λe_xz
Le_xz
ry
:= λe_xz 163.9==>
σadmc
π
2
E×
1.92 λe_xz
2
×
:= σadmc 38.255 MPa×=
K
P
A
σadmc
Mx
Wx
σadmf
+:= K 0.91= "OK"
Logo, o perfil escolhido é o W360 x 79