Lista de exercício Concreto Armado 1 Resolvida ATUALIZADA 2

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RESOLUGIO DA LISTA
SOWGIO QUESFAO 1
1)
ESFORQ DE Cfi LWLO
Md = ff . MK = 1,4 . 90=126 KN . m = 12.6001W . cm
2)
ESTIMATNA DA ALTURA VTIL ( d )
D= h - ( Cnom + 0++0121
D= 50 - (2,5+0,5+1,2) = 46,5 cm3)A' LWLO DOS COEFICIENTES Kc E KSKc =bw=d2= 2046,5-2= 3,43 cm2/kNMd 12600PARA CONCRETO C 20 E Act CA - 50 ( TABELA A - 2)Kc = 3,4 ⇒ ks= 0,027=D Id = 0,35 ⇒ DOMINIO 3Id = 0,35 < 0,45 , LOGO RESPEITA 0 LIMITE DE DUCTIBILIDADE Ok4)POSIQIO Da LINHA NEUTRA REGIIO COMPRIMIDA1=0,35 { . x= 16,28 amd REGIIO TRACIONADA× = 0,35 . D= 0,35-46,5=16,28 am5)
DEFORMAOIO DO CONCRETO E DO Act
DOMIN 103 : CONCRETO ( Gc = 3,5%0 FIXO )
a @ ( 6yd < Eos < 90% . VARIAVEL )
§t=6sd
- ×
0,0035 = Gs
-6,28 46,5 - 16,28 Gc= 3,5%0
Gs =
0,0035-(46,5-16,28)=16,28
6s= 6,5%0
GS = 0,0065 = 6,5%0
6) A' LWLODA AREA DE AGO ( As )
As
=KsgMd_
=
0,027.1-2600=7,32
am
'
=D 9010mm ( As ,ef= 7,20am
' ) + 1AM
46,5 ↳ TOLERENCIA 5% 6012,5mm ( As ,ef= 7,50mi ) + 1 AM
6,95cm
'
4016mm ( As ,ef= 8,0cm
'
) 1AM
3020mm ( AS ,ef= 9,45am
'
) 1AM
810cm
'
MA 'S PRJXIMO DE 7,32am
'
2022mm (AS ,ef= 7,60cm
'
) 0 NEO COM .
DO OWE 9,45cm
'
,
LOGO ESWLHO
4016mm
7) ARMADURA MINIMA DE TRAEAO ( As , min )
AS
,
min =L
,
min . AC
PARAWNCRETOS ( 20,025 E C 30 : fs , min = 0,15%
As ,mim= 0,15% . ( bw . h ) = 0,15--20.50
= 1,5cm
'
100
8,0cm
'
> 1,5 am
'
( Ok )
8) ARMADURA MAXIMA DE TRACEO ( As , mdx )
As ,mdx= 4% . Ac
AS ,moix=hfj ( bw
. h ) =L
. ( 20.50 ) = 40cm '
8,0cm
'
< 40mi ( Ok )
9) ARMADURA DE PELE ( ASPELE )
AS
,
PELE = 0,10% . AC
, ALMA
As
,
PELE = %kf
. ( bwh )= Gq
. (20-50)=1 oni =D 208,0mm
10 ) DETALHAMENTO
h=50 205mm ( PORTA
- ESTRBO OU ARMADURA CONSTRUTWA )
⇐- 5mm → 2×2.0 8mm PELE
ESPAEAMENTO VERTICAL DA ARMADURA
( nc ) - 4016mm
DE PELE
bw : 20
a > (20£70, 8am
0,5 . OAGREGADO = 0,5 . 1,9cm - -0,95cm
ev = 2am
SOWETO OWESTAT 2
1) ESFORGO DE ofLWLO
Md = ?
2) ESTIMATIVA DA ALTURA JTIL ( d )
D= h - ( Cnomt OH + 0£ )
D= 40 - ( 3+0,5 + !z )
D= 36cm
3) 0 MAXIMO MOMENTO FLETOR SUPORTADO POR ESTA VIGA SERA
'
NO
DOMINIO 3
,
RESPEITANDO 0 LIMITE DE DUCTIBILIDADE ( Xg= 0,45
)
PARA ONCRETO C25 E CA - 50 OWANDO Id
= 0,45 TEMOS :
14=2,2 =D KS = 0,028=D Dominic 3 =D Id = 0,45
Kc = bw#
Md
z
2,2 =
20€MdMd =2036=2= 11781,82 KN . am
2,2
Md = ff . MK
MK
=M¥
=
1178,1yd
= 8415,59 KN . on = 84,15 KN . m
4) POSIQIO DAUNHA NEUTRA ( × )
REGIIO COMPRIMIDA
1=0,45 { . x= 16,2 am÷
- 0,45 . D= 0,45 . 36=16,2 am
REGIIOTRACIONADA
5) DEFORMAEAO DO CONCRETO EDO AGO
DOMINO 3 : CONCRETO ( Gc = 3,5%. FIXO )
A @ ( 6yd < Gs < 10%0 VANAVEL)
§t=±d- x
0*0,35=62 Go . 3,5%0
36 - 16,2
6s= 0,0035 .(36-16,2)=0,0042
= 4,2%
. 6s= 4,2%0
-16,2
6) A' LWLO DA AREA DE AGO ( As )
AS = KS . Md = 0,028 . 11781,82 = 9,16 am
'
=D 7012,5mm ( As ,ef= 8,75cm
'
) + 1AM-d 36 ↳ TOLERANCIA 5% 5016mm ( As ,ef= 10mi ) + 1AM
8,70mi 3020mm ( As ,ef= 9,45mi ) 1 AM
2025mm ( As ,ef= 9,45cm
'
) ON Com .
LOGO ESCOLHO 3020mm ( AS
,
ef = 9,45 am
'
)
7) ARMADURA MINIMA DE TRAEAO ( As , min )
As
,
min = I
,
min . AC
PARA WNCRETOS C 20,025 E C 30 : fs , min = 0,15%
As
,
min = 0,15% . ( bw . h ) =
0,15--20.40
=
1,2cm
'
100
9,45 am
'
>
1,2cm
'
( Ok )
8) ARMADURA MAXIMA DE TRACEO ( AS , mdx )
As ,mdx= 4% . Ac
As
,
moix = ⇒ ( bw . h ) = ¥
.
(20-40)=32
cni
9,45 am
'
< 32 cmi ( Ok )
9) DETALHAMENTO
205mm ( PORTA - ESTRIBO OU ARMADURA CONSTRUTIVA )
⇐-5mm h= 40
( k ) -
3020mm
bw= 20
SOWETO OWESTAJ 3
1) ESFOR .WS DE A' LWLO
Md = ff . MK = 1,4 . 55=77 KN . m = 77001W . cm
2) ESTIMATNA DA ALTURA VTIL
D= ?
3) of LWLO DO
'
d
'
PARA 0 DOMINIO 3 RESPEITANDO 0 UMITE DEDUCTIBILIDADE( 1=0,45)
d
PARA CONCRETO C25 E AGO A -50
Kc= 2,2 ; ks= 0,028 ; DOMINO 3 ; Id
= 0,45
Kc = bw . d2
Td
2,2=1212
7700
D= way = 37,57cm
.
4) A' LWLODA UNHA NEUTRA ( × )
X = 0,45
- REGIIO COMPRIMIDA
d {
. x= 16,90 cm
X = 0,45 . D= 0,45 . 37,57=16,90 am
REGIEOTRACIONADA
5) of LWLO DADEFORMAEAO DO CONCRETO E DO A ,O
DOMINI 03 : CONCRETO ( Gc = 3,5%. FIXO )
A @ ( 6yd < Gs < 10%0 VARIAVEL)
E×=±d- x
0,0035 = GS
- -
16,90 37,57 - 16,90
Eos = 0,0035 . (
37,57
- 16,90 ) Gc= 3,5%0
16,9065-0,0042=4,2
%o 65-4,2%0
6) A' LWLO DA AREA DE AGO ( As )
As=KsgMd_
= 0,028.7700 = 5,74cm
'
=D 7010mm ( As ,ef= 5,60cm
' ) team
-37,57
↳ TOLERANCIA 5% 5012,5mm ( AS ,ef= 6,25mi ) + 1AM
5,45 am
'
3016mm ( As ,ef= 6,0cm
'
) + 1AM
2020mm ( As ,ef= 6,30am
'
) + 1AM
ESWLHO 3016mm Pas As = 6,0cm
'
E
'
Mais PRJXIMO DE 5,74 an
'
PORTANTOTEREMOS 2 GMADAS a = {¥[j|a%mp€%Yj%Yqmm=
9,5mm
7) A' LWLO DO ' h ' ↳ 0,95cm
ev= 2cm
FORMULA DO
'
d
'
PARA DUAS CAMADAS
d=h - ( cnom + Ott Ol + eg )37,57
 = h - (2,5+0,5+1,6+22)
h=
37,57+5,6=43,17
I 45cm
8) ARMADURA MINIMA DE TRAAIO ( AS , min )
AS ,mim= I. min . Ac
AS ,min= 0,15% . ( bwh )
AS ,min= 0,1 . ( 12.45 ) = 0,81cm
'
100
6,070,81 ( OK )
9) ARMADURA MAXIMA DE TRAETO ( As , moix )
AS ,mdx= 4% . Ac
As ,mdx= 4- . ( bw . h )
100
As
,
mix = 1 . 12.45 = 21,6 cni
100
6,0<21,6 ( ON
10 ) DETALHAMENTO
205mm ( PORTA - ESTRIBO OU ARMADURA CONSTRUTIVA )
⇐-5mm h= 45
- ( 2C ) - 1016mm
ev= 2cm -
( nc ) -2016mm
bw= 12
Soweto OWESTEO 4
1) ESFORGO DE A' LWLO
Md = ff . Mk = 1,4 . C- 91 ) = - 127,4 KN . m = - 12740 KN . am ( MOMENTO NEGATIVE)
2) ESTIMATNA DA ALTURA JTIL ( d )
D= h - ( Cnom + Ot +
OFGPESTIMO
10mm = 1 am
D= 30 - ( 2,5cm + 0,5 + 12 )
D= 26,5cm
3) A' LWLO DO KC E KS
Kc = bing.gl =
20=26,52--1,10
→ PARA CONCRETO C 20 E A .W A -50
12740 ↳ DOMINIO 4 → REDIMENSIONAR A PEEA
4) A' LWLO DO NOVO
'
d
'
PARA 0 DOMINO 3 RESPEITANDO 0 UMITE DE
DUCTIBIUDADE ( Id
= 0,45 )
PARA dI= 0,45 E CONCRETO C 20 E AQ A
- 50 TEMOS :
Kc = 2,8 ; KS = 0,028 ; DOMINIO 3 ; ×g= 0,45
Kc = bw#
Md
2,8=20 . d2
-12740
D= feta20
D= 42,23 am
5) A' LWLO DA Posiofo DA UNHA NEUTRA ( × )
REGIIO TRACIONADA
1=0,45 . x= 19, 0 am:
= 0,45 . 42,23 = 19 am
{ % REGIO COMPRIMIDA
6)
of LWLO DADEFORMAEAO DO CONCRETO E DO AGO
DOMINI 03 : CONCRETO ( Gc = 3,5%. FIXO )
A @ ( 6yd < Gs < 10%0 VARIAVEL)
§t=±d- x
0,0035 = Gs Gs = 4,2% .⇒ 2,23 - 19
Gs = 0,0035 ( 42,23 - 19 ) = 0,0042 = 4,2%0 Gc= 3,5%0
.9
7)A' LWLO DA AREA DE AGO ( As )
As
=KsdMd_
= 0,028 . 12740 = 8,45cm
'
=D 7012,5mm ( AS
, ef= 8,75cm
'
) + 1AM
-42,23 ↳ TOLERANCIA 5% 5016mm ( As , ef : 10mi ) + 1AM
8,02cm
'
3020mm ( As
, ef = 9,45mi )
2025mm ( As ,ef= 9,80cm
'
) NE COM .
ESWLHO 3020mm
8)
of LWLO DO ' h '
UMA
FORMULA DO
'
d
'
PARA
#At
CAMADAS
d=h - ( cnom+0++012)42,23 = h -(2,5+0,5+2-2)h = 42,23+4=46,23 = 50 cm9)ARMADURA MINIMA DE TRAAIO ( AS , min )AS ,mim= I. min . AcAS ,min= 0,15% . ( bwh )
AS ,min= 0,15 .
(20-50)=1,5
cm
'
TOO
9,45 >
1,5
( Ok )
10 ) ARMADURA MAXIMA DE TRAETO ( As , moix )
AS ,mdx= 4% . Ac
As ,mdx= 4- . ( bw . h )
100
AS ,mdx = 1.
(20-50)=40,0
cni
100
9,45<40,0
( ON
10 ) ARMADURA DE PELE ( ASPELE )
AS
,
PELE = 0,10% . AC
, ALMA
As
,
PELE = %kf
. ( bwh )= Gq
. (20-50)=1 oni =D 208,0mm
11 ) DETALHAMENTO
h=5O (
1C )-3020mm⇐- 5mm → 2×2.0 8mm PELE
ESPAEAMENTO VERTICAL DA ARMADURA
205mm
DE PELE
( PORTA - ESTRIBO )
bw : 20
a > (20£70, 8am
0,5 . OAGREGADO = 0,5 . 1,9cm - -0,95cm
ev = 2am
A VIGA FOI REDIMENSIONADA
DE (20×30) PARA (20×50)