resmat2007a
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resmat2007a


DisciplinaResistência dos Materiais II4.852 materiais119.208 seguidores
Pré-visualização27 páginas
da viga da direc¸a\u2dco longitudinal.
A aplicac¸a\u2dco destas u´ltimas equac¸o\u2dces segue enta\u2dco um procedimento bastante parecido
com o exposto no item anterior. Deve-se ter bastante atenc¸a\u2dco no ca´lculo do momento
esta´tico, identificando corretamente qual a a´rea a ser considerada no seu ca´lculo. Obvia-
mente a aplicac¸a\u2dco destas equac¸o\u2dces podem ser extendidas a vigas de um so´ elemento.
Observa-se que das equac¸o\u2dces 3.106 e 3.107 que o fluxo de cisalhamento e´ uma grandeza
vetorial e define a direc¸a\u2dco das tenso\u2dces as tenso\u2dces
Observa-se tambe´m que na aba do perfil o fluxo de cisalhamento na direc¸a\u2dco vertical
provoca tenso\u2dces \u3c4xy de baixa magnitude pois a espessura t e´ relativamente grande. Por um
outro lado o fluxo de cisalhamento na direc¸a\u2dco horizontal provoca tenso\u2dces de cisalhamento
\u3c4xz de altas magnitudes pois a espessura e e´ relativamente pequena. Assim sendo, e´
comum analisarmos o fluxo de cisalhamento somente nas direc¸o\u2dces paralelas aos lados da
sec¸a\u2dco: direc¸a\u2dco horizontal na(s) aba(s) e direc¸a\u2dco vertical na alma.
O sentido do fluxo de cisalhamento e, consequ¨entemente das tenso\u2dces cisalhantes, nas
abas sa\u2dco mostrados na figura 3.130 e sa\u2dco obtidos pela simetria do tensor de tenso\u2dces
(equac¸o\u2dces de equil´\u131brio). Ja´ na alma a direc¸a\u2dco do fluxo e´ a mesma direc¸a\u2dco do cortante
atuante na sec¸a\u2dco.
110
M
\u3c4zx
\u3c4xz
\u3c4zx
\u3c4xz
\u3c4zx
\u3c4xz
dx
dF F + dF
F
\u3c4xz
\u3c4zx
dx
t
e
M + dMx
y
z
F
dx
F + dF
dF
F
dx
dF F + dF
dx
F + dF
F
dF
Figura 3.130: Fluxo de cisalhamento num perfil I
A figura 3.131 resume as direc¸o\u2dces de fluxo de cisalhamento considerados na ana´lise de
uma viga I, bem como suas intensidades. Estas u´ltimas sera\u2dco discutidas logo a seguir.
f
max
aba
f
max
aba
f
max
alma
max
aba2f
Figura 3.131: Fluxo de cisalhamento num perfil I
No que se refere a` intensidade do fluxo de cisalhamento tem-se para uma viga I, tem-se:
\u2022 Para as abas do perfil: (ver figura 3.132)
y
z
d/2
e
b/2
x
Q
dz
Figura 3.132: Fluxo de cisalhamento nas abas de um perfil I
111
f =
QMs
I
=
Qd
2
( b
2
\u2212 z)e
I
=
Qed
2I
(
b
2
\u2212 z
)
(3.109)
Verifica-se que o fluxo de cisalhamento varia linearmente com z conforme mostra
figura 3.131.
A forc¸a total desenvolvida em cada trecho das abas pode ser obtida pela integrac¸a\u2dco
que segue e as resultantes sa\u2dco mostradas na figura 3.133
Faba =
\u222b
dF =
\u222b
fdz =
\u222b
0
b/2
Qed
2I
(
b
2
\u2212 z
)
dz =
Qedb2
16I
(3.110)
Faba Faba
FabaFaba
Falma = Q
Figura 3.133: Resultantes do fluxo de cisalhamento num perfil I
Observa-se facilmente que a resultante de forc¸as na horizontal e´ nula, como era de
se esperar ja´ que para o caso analisado so´ existe cortante na direc¸a\u2dco y.
\u2022 Para a alma do perfil: (ver figura 3.134)
Similarmente, observando figura 3.134 faz-se a ana´lise da alma.
y
z
e
Q
b
t
dy
y
e
d/2
Figura 3.134: Fluxo de cisalhamento na alma de um perfil I
Pode-se escrever para o ca´lculo do momento esta´tico na alma:
Ms = bed/2 + t(d/2\u2212 e/2\u2212 y)(d/2\u2212 e/2\u2212 y)1/2 (3.111)
Considerando que d/2 À e/2 (paredes finas), pode-se simplificar a equac¸a\u2dco 3.111
por:
Ms ' bed/2 + t(d/2\u2212 y)(d/2\u2212 y)1/2 = bed/2 + t/2(d2/4\u2212 y2) (3.112)
Resultando para o fluxo de cisalhamento, para o caso de t = e:
112
f =
QMs
I
=
Qt
I
[
db
2
+
1
2
(
d2
4
\u2212 y2
)]
(3.113)
Neste caso, conforme mostrado na figura 3.131 o fluxo de cisalhamento varia de
modo parabo´lico, de f = 2fabamax = Qtdb/(2I) em y = d/2 ao ma´ximo de f =
falmamax = (Qtd/I)(b/2 + d/8) em y = 0.
A forc¸a total desenvolvida na alma pode ser obtida pela integrac¸a\u2dco que segue e a
resultante e´ mostrada na figura 3.133
Falma =
\u222b
dF =
\u222b
fdy (3.114)
Desenvolvendo a integral da equac¸a\u2dco 3.114, sendo o valor do fluxo de cisalhamento
dado por 3.113, pode-se mostrar que:
Falma =
\u222b
dF =
\u222b d/2
\u2212d/2
fdy = Q (3.115)
ou seja, que a resultante vertical e´ igual ao cortante que atua na sec¸a\u2dco, conforme
era esperado. (ver figura 3.133)
3.4.6 Exerc´\u131cios
1. Um esforc¸o cortante vertical de 18 kN atua na sec¸a\u2dco transversal de uma viga con-
stitu´\u131da de quatro pec¸as de madeira 50 mm × 200 mm (veja figura 3.135 Determinar:
(a) a tensa\u2dco tangencial horizontal ma´xima, indicando onde ela ocorre na sec¸a\u2dco
transversal.
(b) a tensa\u2dco tangencial vertical a 80 mm abaixo do topo.
Resposta: 821,7 kPa e 706,7 kPa
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Figura 3.135: Figura do exerc´\u131cio 1
2. Uma viga caixa e´ formada por quatro ta´buas de madeira 25 mm × 150 mm, unidas
com parafusos. O esforc¸o cortante de 4 kN e´ constante ao longo do comprimento.
Calcular o espac¸amento entre os parafusos, no comprimento, se cada um suporta
uma forc¸a de cisalhamento de 1 kN.
Resposta: 110 mm, no ma´ximo.
113
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Figura 3.136: Figura do exerc´\u131cio 2
3. Uma viga caixa\u2dco de madeira sec¸a\u2dco quadrada 250mm × 250 mm externamente,
espessura de 50 mm, e´ formada por quatro pec¸as de madeira pregadas de uma das
tre\u2c6s formas indicadas. O esforc¸o cortante de 3,02 kN ao longo do comprimento e
cada prego resiste a uma forc¸a cortante de 240 N. escolher a soluc¸a\u2dco que exige menor
nu´mero de pregos e calcular o espac¸amento entre os pregos para esta soluc¸a\u2dco.
Resposta: (b) 60 mm (para (a) 36 mm e para (c) 45 mm)
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(a) (b) (c)
Figura 3.137: Figura do exerc´\u131cio 3
4. A sec¸a\u2dco transversal AB da viga da figura 3.138, e´ constitu´\u131da por va´rias pec¸as de
madeira (dimenso\u2dces em mm), conforme a figura 3.138. O momento de ine´rcia em
relac¸a\u2dco a` LN e´ igual a 2360×106 mm4. Cada parafuso representado e´ capaz de
resistir a uma forc¸a de cisalhamento longitudinal de 2 kN. Pede-se o espac¸amento,
ao longo do comprimento, dos parafusos necessa´rios a` ligac¸a\u2dco:
(a) nos trechos AC e DB.
(b) no trecho CD.
Resposta: 120 mm e 240mm
3.4.7 Centro de cisalhamento
Seja uma sec¸a\u2dco com perfil U como a mostrada na figura 3.139. que esta´ em balanc¸o em
um apoio fixo e submetida a` forc¸a P. Se a forc¸a for aplicada ao longo do eixo vertical
assime´trico que passa pelo centro´ide C da a´rea da sec¸a\u2dco transversal,