resmat2007a
135 pág.

resmat2007a


DisciplinaResistência dos Materiais II4.580 materiais117.035 seguidores
Pré-visualização27 páginas
m; a = 0,35m e P = 120 kN, calcular \u3c3max. Resposta: 27,38 MPa.
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a L a
A
 P P
B
Figura 3.50: Exerc´\u131cio 1
2. A viga representada na fig3.51 tem sec¸a\u2dco constante, retangular com h = 2b. Cal-
cular as dimenso\u2dces h e b para as tenso\u2dces admiss´\u131veis 12 MPa a` trac¸a\u2dco e 10 MPa a`
compressa\u2dco, de um certa qualidade de madeira. Resposta: m\u131´nimo 132 x 264 mm.
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A B
1m
25 kN 10 kN10 kN
2m 2m 1m
Figura 3.51: Exerc´\u131cio 2
3. Calcular o valor ma´ximo admiss´\u131vel de q na viga da fig3.52 , para tenso\u2dces admiss´\u131veis
140 MPa a` trac¸a\u2dco e 84 MPa a` compressa\u2dco, sendo a sec¸a\u2dco transversal constante
mostrada (dimenso\u2dces em cm). Resposta: 21,3 kN/m
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2,54
10,16
2,54 25,4
1,2m 2m 2m 1,2m
B DEAC
Figura 3.52: Exerc´\u131cio 3
4. A viga da fig3.53 tem sec¸a\u2dco constante em duplo T assime´trico (mom. de ine´rcia em
relac¸a\u2dco a` LN 7570 cm4), que pode ser colocado na posic¸a\u2dco 1 ( T ) ou 2 ( L ). Dados
\u3c3t =150 MPa e \u3c3c = 120 MPa, calcular qadm na posic¸a\u2dco mais eficiente (aquela que
suporta maior carga). Resposta: 18,55 kN/m na posic¸a\u2dco 2.
5. Dimensionar um eixo de ac¸o (\u3c3 =120 MPa, E=210 GPa ) de sec¸a\u2dco circular cheia para
suportar um momento flexa\u2dco de 60 kNm. Calcular o a\u2c6ngulo de rotac¸a\u2dco espec´\u131fica
da sec¸a\u2dco. Resposta: Dia\u2c6metro 172 mm; Rotac¸a\u2dco 0,00665 rd/m.
73
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3m
A B
q
G
.
7,65cm
13,60cm
Figura 3.53: Exerc´\u131cio 4
6. Em uma sec¸a\u2dco anular (coroa circular) a raza\u2dco entre os dia\u2c6metros externo interno e´
D/d = 1,5. Pede-se dimensiona-la para suportar um momento fletor de 32 kNm,
para uma tensa\u2dco admiss´\u131vel de 80 MPa. Resposta: D = 172 mm.
7. Uma viga tem momento fletor ma´ximo 18 kNm. Para ama sec¸a\u2dco transversal con-
stante e retangular a x 2a, vazada por um retangulo 0,6 a x a (conservada a
simetria), dimensiona´-la para uma tensa\u2dco admiss´\u131vel 10MPa. Resposta: a = 143
mm
8. Calcular as tenso\u2dces normais extremas da viga abaixo, dado P = 7 kN, representada
a sec¸a\u2dco transversal constante. Resposta: comp. 153,2 MPa nas fibras sup; trac¸a\u2dco
88,7 nas fibras inf.
A B
P P
100cm 50cm50cm
3cm 3cm 3cm
2cm
4cm
Figura 3.54: Exerc´\u131cio 8
9. Calcular o valor m\u131´nimo de a na sec¸a\u2dco transversal da viga da fig3.55/ para \u3c3t =100MPa
e \u3c3c =60 MPa. Resposta: a = 41 mm.
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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
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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
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2m2m 4m
40 kN 100 kN 100 kN 40 kN
a
9a
0,8a
3,6a 3,6a
2m 2m
Figura 3.55: Exerc´\u131cio 9
74
10. A viga abaixo e´ constitu´\u131da por duas pec¸as de madeira de sec¸a\u2dco 300 mm x 100 mm,
conforme mostra a figura. Dadas as tenso\u2dces admiss´\u131veis 12 MPa a` compressa\u2dco e
18 MPa a` trac¸a\u2dco, calcular Padm e representar o diagrama de tenso\u2dces da sec¸a\u2dco E.
Resposta: P = 102 kN.
EA B
60cm60cm 60cm 60cm
C
 P
D
P
Figura 3.56: Exerc´\u131cio 10
11. Dimensionar a viga abaixo a` flexa\u2dco (a=?) e representar o diagrama de tenso\u2dces da
sec¸a\u2dco C. A viga tem sec¸a\u2dco constante de ferro fundido com tenso\u2dces admiss´\u131vel 35
MPa a` trac¸a\u2dco e 140 MPa a` compressa\u2dco. Escolher a mais favora´vel entre as posic¸o\u2dces
1 (T ) e ( L ) da sec¸a\u2dco. Resposta: a = 4,2 cm, posic¸a\u2dco 2
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2,2m2,2m 2,2m
30 kN30 kN
2a 2aa
a
7a
BA
C D
Figura 3.57: Exerc´\u131cio 11
75
3.3.4 Va´rias formas da sec¸a\u2dco transversal
Sec¸o\u2dces sime´tricas ou assime´tricas em relac¸a\u2dco a` LN
Com o objetivo de obter maior eficie\u2c6ncia (na avaliac¸a\u2dco) ou maior economia (no dimen-
sionamento) devemos projetar com \u3c3max = \u3c3, onde \u3c3max e´ a tensa\u2dco maxima na sec¸a\u2dco e \u3c3
e´ a tensa\u2dco maxima admissivel(propriedade do material).
Levando-se em conta que
\u3c3s
\u3c3i
=
ds
di
ha´ dois casos a considerar:
1. Se o material e´ tal que \u3c3s 6= \u3c3i enta\u2dco e´ indicada a forma assime´trica em relac¸a\u2dco a`
LN, ficando esta mais pro´xima da fibra de menor \u3c3, sendo ideal ds
di
= \u3c3s
\u3c3i
, pois neste
caso podemos projetar \u3c3s = \u3c3s e \u3c3i = \u3c3i por exemplo, para M > 0 e
\u3c3c
\u3c3t
= 0, 5, o
ideal e´ ds
di
= 0, 5
\u3c3i
s\u3c3 \u3c3c
\u3c3t
ds=h/3
di=2h/3
=
=
Figura 3.58:
2. Se o material e´ tal que \u3c3c = \u3c3t, enta\u2dco e´ indicada a sec¸a\u2dco sime´trica em relac¸a\u2dco a LN:
ds = di = h/2. O projeto pode contemplar a situac¸a\u2dco ideal: \u3c3max = \u3c3 (trac¸a\u2dco ou
compressa\u2dco).
\u3c3i
s\u3c3
=
= \u3c3
\u3c3
h/2
h/2
M>0
Figura 3.59:
Sec¸o\u2dces sime´tricas a` LN - Sec¸o\u2dces I
Maior a´rea A da sec¸a\u2dco transversal na\u2dco significa maior mo´dulo de resiste\u2c6ncia a flexa\u2dco W ,
pois este depende da forma da sec¸a\u2dco.
1. Entre duas sec¸o\u2dces de mesma W, a mais econo\u2c6mica e´ a de menor A
2. Entre duas sec¸o\u2dces de mesma A, a mais eficiente e´ a de maior W
Sejam va´rias sec¸o\u2dces sime´tricas a` LN, com a mesma a´rea A.
\u2022 Reta\u2c6ngular b × h: W = bh2/6 e A = bh \u2192 W = Ah/6 = 0, 167Ah. (sec¸o\u2dces
reta\u2c6ngulares de mesma a´rea \u2192 maior eficie\u2c6ncia = maior h)
76
\u2022 Circular, dia\u2c6metro D: W = piD3/32 e A = piD2/4 \u2192 W = AD/8 = 0, 125AD.
\u2022 Quadrada,lado L (mesma a´rea L2 = piD2/4 \u2192 L = 0, 886D):
W = 0, 167AL = 0, 167 A 0, 886D \u2192 W = 0, 148 A D
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Eficiencia crescente^
A A A A
Figura 3.60:
Concluimos que, para obter maior eficiencia, devemos dispor a maior massa do material
(a´rea de sec¸a\u2dco) o mais afastado poss´\u131vel da LN.
A situac¸a\u2dco ideal e´ mostrada na figura 3.61
/2\u3b4
/2\u3b4
\ufffd\ufffd
\ufffd\ufffd