resmat2007a

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250
500
60
Figura 3.75: Figura do Exerc´\u131cio 7
9. Uma viga de concreto armado tem sec¸a\u2dco retangular 200 mm × 400 mm. A armadura
e´ constitu´\u131da por tre\u2c6s barras de ac¸o de 22mm de dia\u2c6metro, cujos centros esta\u2dco a
50mm da face inferior da viga. Calcular o momento fletor positivo ma´ximo que a
viga pode suportar, dados: Ec = 21Gpa, Ea = 210Gpa, \u3c3c = 9.3Mpa, \u3c3a = 138Mpa
(Resposta: 42, 03kNm)
10. A figura 3.76representa um trecho de uma laje de concreto armado, com armadura
longitudinal de barras de ac¸o de 16 mm de dia\u2c6metro a cada 150 mm. Calcular a
tensa\u2dco ma´xima no concreto e a tensa\u2dco me´dia no ac¸o para um momento fletor positivo
de 4 kNm a cada 300mm de largura da laje. Dados: Ec = 21 GPa, Ea = 210 GPa,
(Resposta: 7,65 MPa e 114, 8 MPa)
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100mm 120mm
Figura 3.76: Figura do Exerc´\u131cio 10
11. Uma laje de concreto com 150mm de espessura e´ reforc¸ada longitudinalmente com
barras de ac¸o de 25mm de dia\u2c6metro a cada 80mm de largura, cujos centros esta\u2dco
a 10mm da face inferior da laje. Determinar o momento fletor ma´ximo admiss´\u131vel
por metro da laje.
Adotar n = 12 e tenso\u2dces admiss´\u131veis 150 MPa para o ac¸o e 8Mpa para o concreto.
(Resposta: 37,1 kNm/m)
85
3.3.8 Flexa\u2dco Inela´stica
Refere\u2c6ncia a R.C. HIBBELER. Resiste\u2c6ncia dos Materias. 5o Edic¸a\u2dco
As equac¸o\u2dces para determinar a tensa\u2dco normal provocada pela flexa\u2dco, desenvolvidas
anteriormente, sa\u2dco va´lidas apenas se o material comporta-se de maneira linear-ela´stica.
Se o momento aplicado provocar escoamento do material, deve-se enta\u2dco usar uma ana´lise
pla´stica para determinar a distribuic¸a\u2dco de tensa\u2dco. No entanto, as tre\u2c6s condic¸o\u2dces para
flexa\u2dco de elementos retos, (exemplo: vigas, colunas), tanto no caso ela´stico como no
pla´stico, devem ser satisfeitas.
1. Distribuic¸a\u2dco da Deformac¸a\u2dco Normal Linear - ²x. Com base em condic¸o\u2dces geome´tricas,
mostramos na sec¸a\u2dco anterior que as deformac¸o\u2dces normais que se desenvolvem no ma-
terial variam sempre linearmente, de zero, no eixo neutro da sec¸a\u2dco transversal, ate´
o ma´ximo no ponto mais afastado deste eixo neutro.
2. O Esforc¸o Normal e´ Nulo. Como somente o momento interno resultante atua sobre
a sec¸a\u2dco transversal, a forc¸a resultante provocada pela distribuic¸a\u2dco de tensa\u2dco deve
ser nula. E, uma vez que \u3c3x cria uma forc¸a sobre a a´rea dA de dF = \u3c3xdA (figura
3.77), para toda a´rea da sec¸a\u2dco transversal A temos:
..
x
y
z
M
Figura 3.77:
N =
\u222b
A
\u3c3xdA = 0 (3.83)
A equac¸a\u2dco 3.83 nos permite obter a localizac¸a\u2dco do eixo neutro.
3. Momento Resultante. O momento resultante na sec¸a\u2dco deve equivaler ao momento
provocado pela distribuic¸a\u2dco de tensa\u2dco em torno do eixo neutro. Como o momento
da forc¸a dFx = \u3c3xdA em torno do eixo neutro e´ dMz = y(\u3c3xdA) o somato´rio dos
resultados em toda a sec¸a\u2dco transversal sera´:
Mz =
\u222b
A
y\u3c3dA (3.84)
Essas condic¸o\u2dces de geometria e carregamento sera\u2dco usadas agora para mostrar como
determinar a distribuic¸a\u2dco de tensa\u2dco em uma viga submetida a um momento interno
resultante que provoca escoamento do material. Suporemos, ao longo da discurssa\u2dco,
que o material tem o mesmo diagrama tensa\u2dco-deformac¸a\u2dco tanto sob trac¸a\u2dco como
sob compressa\u2dco. Para simplificar, comec¸aremos considerando que a viga tenha a´rea
de sec¸a\u2dco transversal com dois eixos de simetria; nesse caso, um reta\u2c6ngulo de altura
h e largura b, como o mostrado na figura3.78. Sera\u2dco considerados tre\u2c6s casos de
carregamento que te\u2c6m interesse especial. Sa\u2dco eles: Momento Ela´stico Ma´ximo;
Momento Pla´stico e Momento Resistente.
86
M E
Figura 3.78:
h
2
h
2
y2
y2
 
y 
y 
1
1
\u3b5
\u3b5
\u3b5
\u3b5
\u3b5
\u3b5
E
E
1
1
2
2
Figura 3.79: Diagrama de deformac¸a\u2dco
Momento Ela´stico Ma´ximo.
Suponhamos que o momento aplicado Mz = ME seja suficiente apenas para produzir
deformac¸o\u2dces de escoamento nas fibras superiores e inferiores da viga, conforme mostra
a figura 3.79. Como a distribuic¸a\u2dco de deformac¸a\u2dco e´ linear, podemos determinar a dis-
tribuic¸a\u2dco de tensa\u2dco correspondente usando o diagrma tensa\u2dco-deformac¸a\u2dco (figura 3.80).
Vemos aqui que a deformac¸a\u2dco de escoamento ²E causa o limite de escoamento \u3c3E, en-
quanto as deformac¸o\u2dces intermediarias ²1 e ²2 provocam as tenso\u2dces \u3c31 e \u3c32, respectiva-
mente. Quando essas tenso\u2dces, e outras como elas, te\u2c6m seus gra´ficos montados nos pontos
y = h/2, y = y1, y = y2, etc., tem-se a distribuic¸a\u2dco de tensa\u2dco da figura 3.81 ou 3.82.
Evidentemente, a linearidade de tensa\u2dco e´ conseque\u2c6ncia da Lei de Hooke.
\u3c3
\u3c3
\u3c3
\u3c31
2
E
\u3b5 \u3b5 \u3b5
\u3b5
1 2 E
Figura 3.80: Diagrama tensa\u2dco-deformac¸a\u2dco
Agora que a distribuic¸a\u2dco de tensa\u2dco foi estabelecida, podemos verificar se a equac¸a\u2dco
3.83 foi satisfeita. Para isso, calculemos primeiro a forc¸a resultante de cada uma das
duas partes da distribuic¸a\u2dco de tensa\u2dco (figura 3.82). Geometricamente, isso equivale a
87
calcular os volumes de dois blocos triangulares. Como mostrado, a sec¸a\u2dco transversal
superior do elemento esta´ submetida a` compressa\u2dco, enquanto a sec¸a\u2dco transversal inferior
esta´ submetida a` trac¸a\u2dco.
Temos:
T = C =
1
2
(
h
2
\u3c3E
)
b =
1
4
bh\u3c3E (3.85)
Como T e´ igual mas oposta a C, a equac¸a\u2dco 3.83 e´ satisfeita e, de fato, o eixo neutro
passa atrave´s do centro´ide da a´rea da sec¸a\u2dco transversal.
O momento ela´stico ma´ximo ME e´ determinado pela equac¸a\u2dco 3.84, que o declara
equivalente ao momento da tensa\u2dco de distribuic¸a\u2dco em torno de um eixo neutro. Para
aplicar essa equac¸a\u2dco geometricamente, temos de determinar os momentos criados por T
e C em torno do eixo neutro (figura 3.82). Como cada forc¸a atua atrave´s do centro´ide do
volume do seu bloco de tensa\u2dco triangular associado, temos:
ME = C
(
2
3
)
h
2
+ T
(
2
3
)
h
2
ME = 2
(
1
4
)
bh\u3c3E
(
2
3
)
h
2
ME =
1
6
bh2\u3c3E (3.86)
Naturalmente, esse mesmo resultado pode ser obtido de maneira mais direta pela
fo´rmula da flexa\u2dco, ou seja, \u3c3E =ME(h/2)/[bh
3/12], ou ME = bh
2\u3c3E/6
h
2
h
2
y2
y2
 
y 
y 
1
1
2
1
1
2
E
E
\u3c3
\u3c3
\u3c3
\u3c3
\u3c3
\u3c3
Figura 3.81: Diagrama de tensa\u2dco
b
N
A
2
h
2
h
\u3c3
\u3c3
E
E
C
T
ME
Figura 3.82:
88
Momento Pla´stico
Alguns materiais, tais como ac¸o, tendem a exibir comportamento ela´stico perfeita-
mente pla´stico quando a tensa\u2dco no material exceder \u3c3E. Considereremos, por exemplo, o
elemento da figura 3.83. Se o momento interno M > ME, o material comec¸a a escoar nas
partes superior e inferior da viga, o que causa uma redistribuic¸a\u2dco de tensa\u2dco sobre a sec¸a\u2dco
transversal ate´ que o momento internoM de equilibrio seja desenvolvido. Se a distribuic¸a\u2dco
da deformac¸a\u2dco normal assim produzida for como a mostrada na figura 3.79, a distribuic¸a\u2dco
de tensa\u2dco normal correspondente sera´ determinada pelo diagrama tensa\u2dco-deformac¸a\u2dco da
mesma maneira que no caso ela´stico. Usando esse diagrama para material mostrado na
figura 3.84, temos que as deformac¸o\u2dces ²1, ²2 = ²E, ²2 correspondem, respectivamente, a`s
tenso\u2dces \u3c31, \u3c32 = \u3c3E, \u3c3E (essas e outras tenso\u2dces sa\u2dco mostradas na figura 3.85 ou na 3.86).
Nesse caso, os so´lido de tenso\u2dces de esforc¸os de compressa\u2dco e trac¸a\u2dco sa\u2dco parte retangulares
e parte triangulares, observa-se na figura 3.86:
M > ME
Figura 3.83:
\u3c3
\u3c3
\u3c3
\u3b5 \u3b5
\u3b5
\u3b5E
E
1
1 2
Figura 3.84:
h
2
h
2
\u3c3
\u3c3
\u3c3
\u3c3
\u3c3
\u3c3
1
1
2
2
E
E
y
y E
E
Figura 3.85: Diagrama de tensa\u2dco
89
T1 = C1 =
1
2
yE\u3c3Eb