resmat2007a
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resmat2007a


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Figura 3.61:
\u2022 Perfis I ou S te\u2c6m altura bem maior que a largura.
\u2022 Perfis H ou WF (abas largas) te\u2c6m largura mais pro´xima da altura.
Os produtores de perfis fornecem tabelas com as caracteristicas geome´tricas (dimensso\u2dces,
a´rea, momento de ine´rcia...) necessa´rias ao projeto. Na Resiste\u2c6ncia dos Materiais I va-
mos usar as tabelas do livro \u201cResiste\u2c6ncia dos Materiais\u201d de Beer e Johnston, que esta\u2dco
reproduzidas em anexo.
Os perfis sa\u2dco designados pela letra S(perfil I) ou W(perfil H) seguida da altura nominal
(mm) e da sua massa em kg por metro (kg/m). Encontram-se em ordem decrescente de
altura e, em cada grupo de mesma altura, em ordem decrescente de peso.
3.3.5 Exerc´\u131cios
1. Calcular o valor ma´ximo admissivel da carga P, na viga na figura 3.62 para uma
\u3c3 = 140Mpa, se a viga e´ um perfil W150× 37, 1. Na\u2dco desprezar o peso pro´prio do
perfil.
Resposta: 14, 88 kN
2. Escolher o perfil I mais econo\u2c6mico para a viga da figura 3.63, para \u3c3 = 140Mpa
Resposta: S 510× 97, 3
77
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P
2,5m
Figura 3.62: Exerc´\u131cio 1
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8m
BA
27kN/m
Figura 3.63: Exerc´\u131cio 2
3. Duplicando a carga da viga do exerc´\u131cio 2 (q\u2032 = 54 kN/m) e conservando o perfil
adotado, para se obter resiste\u2c6ncia sa\u2dco soldados duas chapas (mesma \u3c3 = 140 MPa)
sobre as mesas, de espessura do reforc¸o igual a espessura da mesa. Determine a
largura das chapas e o trecho da viga em que e´ necessa´rio usa´-las. Desprezar os
pesos pro´prios.
Resposta: largura 121 mm, reforc¸o nos 5,0 m centrais da viga
4. A viga da figura 3.64 e´ contituida de um perfil W 200 × 86, de ac¸o com \u3c3 = 130
MPa). Calcular o valor ma´ximo admissivel de P desprezando o peso pro´prio.
Resposta: 59, 57 kN/m
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5,4m
A B
Figura 3.64: Exerc´\u131cio 4
5. Calcular as tenso\u2dces extremas na viga da figura 3.65, indicando a sec¸a\u2dco onde ocorrem.
A viga e´ constitu´\u131da por um perfil W130×28, 1. Considerar o efeito do peso pro´prio,
ale´m da sobrecarga.
Resposta: ±66, 1 MPa
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5,0m
1,5kN
Figura 3.65: Exerc´\u131cio 5
78
6. Idem para a viga da figura 3.66 constitu´\u131da por um perfil W150x37, 1
Resposta: ±10, 77 MPa
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5,0m
1,5kN
Figura 3.66: Exerc´\u131cio 6
7. Escolher o perfil mais econo\u2c6mico (I ou W, conforme indicado) para cada uma da
figura 3.67, desconsiderando o efeito do peso pro´prio,ale´m da sobrecarga represen-
tada. A tensa\u2dco admissivel e´ dada.
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\u3c3 \u3c3
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\u3c3 = 140Mpa
d) Perfil W,\u3c3
c) Perfil W,
 
 = 120Mpa
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12kN
0,8m
(S 130 x 15 )
30kN
10kN/m
2,0m 2,0m
(S 310 x 47,3)
a) Perfil I, b) Perfil I,
 = 120Mpa
 = 140Mpa
65kN 65kN
1,0m 0,6m0,6m
(W 250 x 32,7 ou W 310 x 32,7)
3,0m
25kN/m
(W 460 x 52)
Figura 3.67: Exerc´\u131cio 7
8. Para uma tensa\u2dco admiss´\u131vel de 150 MPa, calcular o valor ma´ximo admissivel de q
na viga da figura 3.68, constitudida por duas chapas de ac¸o, 200 mm de largura e 12
mm de espessura, soldadas a dois perfis I (S 180× 30), conforme indicado na figura
3.68.Resposta: q = 27,05 kN/m
3.3.6 Vigas de dois materiais
Sa\u2dco vigas de madeira reforc¸adas por cintas meta´licas, vigas de concreto reforc¸adas com
barras de ac¸o (concreto armado), vigas-sanduiche, etc, genericamente designadas por vigas
armadas.
Estas vigas sa\u2dco constituidas por elementos longitudinais (camadas) de materiais difer-
entes, seguramente aderentes de modo a ter necessa´ria resiste\u2c6ncia a`s tenso\u2dces tangenciais
longitudinais
Sa\u2dco admitidas as mesmas hipo´teses da flexa\u2dco em vigas de um so´ material. Portanto,
para um momento fletor Mz = M , as sec¸o\u2dces permanecem planas e normais ao eixo e a
79
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q(kN/m)
6,0m0,6m 0,6m
Figura 3.68: Exerc´\u131cio 8
deformac¸ ao especifica em uma camada de ordenada y em relac¸a\u2dco a LN (linha neutra) e´
²x = ky (k constante)
A figura 3.69 representam a sec¸a\u2dco transversal, o diagrama de deformac¸o\u2dces espec´\u131ficas e
o diagrama de tenso\u2dces de uma viga constituida de dois materiais com a´reas de sec¸a\u2dco A1 e
A2 e mo´dulos de elasticidade E1 e E2, respectivamente. Nestas figuras adimitimos E1 < E2
e a LN situada acima da superf´\u131cie de contato entre os materiais, mas as concluso\u2dces sa\u2dco
gene´ricas.
A , E
1 1
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A , E
M M
E
2 2
\u3b5x
=ky
\u3c3x2
=E \u3b5x2
\u3c3x =E \u3b5x1 1
y 
L.N. M
Figura 3.69: Viga de dois materiais
Na camada de contato entre os dois materiais ha´ uma descontinuidade no diagrama
de tenso\u2dces, com valores \u3c3x1 = E1²x para o material 1 e \u3c3x2 = E2²x para o material 2.
A posic¸a\u2dco da LN e o valor da constante k sera\u2dco determinados