resmat2007a
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resmat2007a


DisciplinaResistência dos Materiais II4.834 materiais119.145 seguidores
Pré-visualização27 páginas
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\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
e e
b
Figura 3.117: Exemplo 2
2. Calcular o valor ma´ximo admiss´\u131vel de P na viga da figura 3.118 (dimenso\u2dces em m),
de sec¸a\u2dco retangular 100 mm × 150 mm, de madeira com \u3c3traca\u2dco e comp. =10 MPa e \u3c4
=1,4 MPa
Resposta: P = 8,333kN
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2.10 0.450.45
P P
Figura 3.118: Figura do exerc´\u131cio 2
3. Calcular o valor ma´ximo admiss´\u131vel de uma carga P na extremidade livre de uma
viga em balanc¸o (figura 3.119) de 0,9 m, constitu´\u131da por tre\u2c6s ta´buas de madeira de
sec¸a\u2dco 100 mm × 50 mm, se a \u3c4uniao =350 kPa, e calcular o valor de \u3c3.
Resposta: P = 3937,5 N e \u3c3 = 9,45 MPa
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\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
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Figura 3.119: Figura do exerc´\u131cio 3
4. Calcular os valores ma´ximos da tensa\u2dco normal e da tensa\u2dco tangencial na viga da
figura 3.120 conhecida sua sec¸a\u2dco transversal (dimenso\u2dces em mm).
Resposta: \u3c3 = 7,872 MPa e \u3c4 = 929,6 kPa
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50
50
100
50
100
1 m2 m
2kN/m
6kN
5,36kN
Figura 3.120: Figura do exerc´\u131cio 4
5. A figura 3.121 (dimenso\u2dces em mm) mostra a sec¸a\u2dco transversal de uma viga de 4
m de comprimento, simplesmente apoiada nos extremos, que suporta uma carga
106
uniformemente distribu´\u131da de 4 kNm sobre todo seu comprimento. Em uma sec¸a\u2dco
a 0,5 m da extremidade esquerda e em um ponto desta sec¸a\u2dco a 40 mm abaixo da
superf´\u131cies neutras, calcular a tensa\u2dco normal e a tensa\u2dco tangencial.
Resposta: \u3c3 = 1,402 MPa,trac¸a\u2dco; \u3c4 = 925,5 kPa
120
40
40
70 40 70
Figura 3.121: Figura do exerc´\u131cio 5
6. A figura 3.122 (dimenso\u2dces em mm) mostra a sec¸a\u2dco transversal de um trecho de uma
viga. Na sec¸a\u2dco A o momento fletor e´ - 4 kNm e o esforc¸o cortante e´ 5 kN. Calcular
a tensa\u2dco normal e a tensa\u2dco de cisalhamento na camada situada 40 mm da LN, na
sec¸a\u2dco B.
Resposta: \u3c3 = -3,505 MPa e \u3c4 = 1,084 MPa
120
40
40
4040 40
6kN/m
A B
2 m
Figura 3.122: Figura do exerc´\u131cio 6
7. Calcular os tenso\u2dces ma´ximas de trac¸a\u2dco, compresa\u2dco e cisalhamento em uma viga en-
gastada e livre de comprimento 0,38 m que suporta uma carga concentrada transver-
sal de 6,7 kN na extremidade livre. A figura 3.123 mostra a sec¸a\u2dco transversal da
viga (dimenso\u2dces em mm).
Resposta: \u3c3t = 92,58 MPa; \u3c3c = 277,75 MPa e \u3c4 = 16,45 MPa
100
45
10
45
50
10
Figura 3.123: Figura do exerc´\u131cio 7
8. Uma viga de sec¸a\u2dco \u201c T \u201d (dimenso\u2dces em mm). Suporta cargas indicadas. Calcular
a tensa\u2dco:
(a) tangencial ma´xima.
107
(b) normal de flexa\u2dco ma´xima de compressa\u2dco.
(c) tangencial vertical a 3,4 m da extremidade esquerda e 60 mm acima da base.
(d) normal de flexa\u2dco a 1,5 m da extremidade direita e 50 mm acima da base.
Resposta: 8a) 694 kPa; 8b) 11,73 MPa de compressa\u2dco; 8c) 148,1 kPa e 8d) 6,17MPa
de trac¸a\u2dco
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200
50
200
75
2kN/m
R
15 kN
2 m 2 m
1
2 m
R2
3 m
Figura 3.124: Figura do exerc´\u131cio 8
9. Verificar a estabilidade da viga 3.125 (dimenso\u2dces em mm na sec¸a\u2dco transversal). Para
\u3c3traca\u2dco = 160MPa, \u3c3compressa\u2dco = 110MPa e \u3c4 = 14MPa.
Resposta: As tenso\u2dces ma´ximas sa\u2dco 15,35 MPa; 9,43 MPa e 1,27 MPa
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\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
175
25
25
100
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
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\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
1.3 m 4.5 m 1.3 m
2kN/m
Figura 3.125: Figura do exerc´\u131cio 9
10. Calcular os valores ma´ximo admiss´\u131vel da carga q na viga da figura 3.126, sec¸a\u2dco \u201c
T \u201d constitu´\u131da por suas pec¸as de madeira 40 mm × 120 mm, para \u3c3 = 9 MPa (de
flexa\u2dco, trac¸a\u2dco ou compressa\u2dco) e \u3c4 = 0,7 MPa (tangencial horizontal).
Resposta: q = 1,741 kN/m; \u3c4max = 0,6 MPa; \u3c3
T
max = 9,0 MPa e \u3c3
c
max = 5,4 MPa.
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2 m 2 m
q
Figura 3.126: Figura do exerc´\u131cio 10
11. Calcular os valores ma´ximo admiss´\u131vel da carga P na viga da figura 3.127, de modo
que a sec¸a\u2dco longitudinal de trac¸a\u2dco na\u2dco exceda 12 MPa e a tensa\u2dco tangencial hori-
zontal na\u2dco ultrapasse 0,7 MPa. Na figura as dimenso\u2dces sa\u2dco dadas em mm.
Resposta: 14,58 kN
108
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\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
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\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
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\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
2 m
P
3 m
75
200
50
200
Figura 3.127: Figura do exerc´\u131cio 11
12. Uma viga bi-apoiada nos extremos, de 6 m de comprimento, suporta uma carga uni-
formemente distribu´\u131da de 5 kN/m em todo o seu comprimento. A sec¸a\u2dco transversal
e´ mostrada na figura 3.128 (dimenso\u2dces em mm)
(a) a tensa\u2dco tangencial horizontal ma´xima, indicando onde ela ocorre na sec¸a\u2dco
transversal.
(b) a tensa\u2dco tangencial vertical a 0,5 m da extremidade direita e a 100 mm abaixo
do topo.
Resposta: 931 kPa e 751 kPa
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\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
60
140
60
160 60
Figura 3.128: Figura do exerc´\u131cio 12
3.4.5 Fluxo de cisalhamento
Seja a figura 3.129 uma pec¸a constitu´\u131da de dois va´rios elementos (viga em madeira no
caso).
dF
F
F + dF
dx
A\u2019
\u3c4zx
\u3c4xz
M
dx
M + dMx
y
z
e
t
Figura 3.129: Viga de madeira composta por va´rios elementos
Analisando-se o equil´\u131brio do elemento destacado na figura tem-se uma situac¸a\u2dco ana´loga
a` estudada no item anterior, considerando \u3c4zx = \u3c4xz constante na espessura e:
109
Portanto:
dF = \u3c4xzedx =
\u222b
Ai
d\u3c3xdA
\u2032 =
\u222b
Ai
dM
I
ydA\u2032 (3.102)
obte´m -se:
\u3c4xz = \u3c4 =
1
Ize
dM
dx
\u222b
A\u2032
ydA\u2032\ufe38 \ufe37\ufe37 \ufe38
Ms
(3.103)
lembrando que dM
dx
= Q (esforc¸o cortante Q = Qy) tem-se enta\u2dco:
\u3c4xz =
QMs
Ize
(3.104)
A equac¸a\u2dco 3.104 mostra que e´ poss´\u131vel se calcular tenso\u2dces longitudinais (direc¸a\u2dco x )
num plano paralelo ao plano xz em vigas sujeitas a flexa\u2dco (\u3c4xz) com a mesma equac¸a\u2dco que
vem sendo usada ate´ o momento no ca´lculo das tenso\u2dces longitudinais no plano xy (\u3c4xy),
calculando-se o momento esta´tico para a a´rea A\u2032 e substituindo-se a espessura \u201ct\u201d por
\u201ce\u201d.
Definindo o fluxo de cisalhamento\u201cf \u201d como sendo o valor da forc¸a dF por unidade de
comprimento, ou seja:
f =
dF
dx
=
\u3c4xzedx
dx
=
QMs
Iz
(3.105)
Consequ¨entemente,
\u3c4xy =
f
t
(3.106)
e
\u3c4xz =
f
e
(3.107)
A forc¸a F a ser transmitida de um elemento para outro fica enta\u2dco:
F = fL =
QMs
Iz
L (3.108)
onde L e´ o comprimento