VIETAB CIMENTACION
Páginas: 10 (2423 palabras)
Publicado: 21 de octubre de 2015
DISEÑO
DE
CONTRATRABE:
DATOS:
σu
:
Longitud
de
contratrabe:
Rt
:
B:
4.13 ton/m2
10 m
3 ton/m2
1.0 m
(considerado)
w
=
σu
x
B
=
4.13
x
1.0
= 4.13 ton/m
1
2
4
6
w
= 4.13 ton/m
D
40 cm
A
B
C
2.5
4.3 m
k = 3EI/L = 160000
K=4EI/L =
A
FD
FT
ME
1D
1T
2D
2T
3D
3T
4D
4T
5D
5T
6D
6T
7D
7T
8D
8T
9D
SUMA
=
D3.2 m
124031.01
C
BA
-‐0.56
0
-‐3.23
-‐1.77
0.00
-‐0.15
0.00
-‐0.10
0.00
-‐0.01
0.00
-‐0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-‐5.26
1
I
= b
x
h3/12
= 133333.33 cm4
E=
1.0
kg/cm2
k = 3EI/L = 125000.00
B
AB
0
0
0
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
25
BC
-‐0.44
0.5
6.37
-‐1.37
0.27
-‐0.12
0.17
-‐0.07
0.01-‐0.01
0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
5.26
D
CB
-‐0.50
0.5
-‐6.37
0.54
-‐0.69
0.34
-‐0.06
0.03
-‐0.04
0.02
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-‐6.23
2
CD
-‐0.50
0
5.29
0.54
0.00
0.34
0.00
0.03
0.00
0.02
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
6.23
DC
0
0
0
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.000.00
4
6
w
= 4.13 ton/m
A
B
C
2.5 m
VI
VH
∑V
AB
5.17
-‐2.10
3.06
D
4.3 m
BA
5.17
2.10
7.27
BC
8.88
-‐0.22
8.66
8.66 ton
3.2 m
CB
8.88
0.22
9.11
CD
6.61
1.95
8.56
8.56 ton
DC
6.61
-‐1.95
4.67
3.06 ton
9.07 ton-‐m
1.13 ton-‐m
1.76
8.86 ton-‐m
2.20
0.74
1.13
2.10
6.39 ton-‐m
0
ton
2.07
10.04 ton-‐m
2.63 ton-‐m
-‐7.27 ton
-‐4.67 ton
-‐9.11 ton
3.82 ton-‐m2.63 ton-‐m
1.13 ton-‐m
0.0
0.0
-‐5.26 ton-‐m
-‐6.23 ton-‐m
DIMENSIONAMIENTO POR FLEXIÒN:
250
Acero
fy
=
4200
f*c
=
200
fIIc
=
170
Pb
=
0.0190
Pmax
=
0.75
Pb
=
Pmin
=
0.0026
qmax
=
0.3529
qmin
=
0.0651
kumax
=
44.47
kumin
=
9.64
Concreto
f`c
=
kg/cm2
kg/cm2
kg/cm2
kg/cm2
0.0143
MR
= FRbd2f`IIcq(1 -0.5q). Por lo tantoMR
6.23 x 105
bd2
=FRfIIcq(1-‐0.5q)
bd
=0.9
x
170
x
0.3529
x
(1-‐0.5
x
0.3529)
2
bd2
=13999
considerando
d/b
=
2.
Por
lo
tanto
b
=
d/2,
sustituyendo
en
bd2
se
obtiene
d3/2.
d3/2
=13999
por
lo
tanto
d
=
3√
2
x
13999
Datos
calculados:
d
=
b
=
h
=
h
=
Datos propuesto:
30 cm
15 cm
d
+
r
+
Ø/2
34 cm
r
= 3 cm
Proponiendo
var
#
4,Ø
=1.3 cm
d
= 36 cm
b
= 25 cm
h
= 40 cm
Nota:La
sección
propuesta
para
la
contratrabe
es
correcta
por
lo
q
se
adoptara
sus
dimensiones
REVISIÒN POR CORTANTE DE LA SECCIÒN TRANSVERSAL
Se
debe
cumplir;
Vu
<
VR
VR
=2FRbd√f*c
= 2
x 0.8
x
25
x
36
x
√200
VR
=20.36 ton > Vu =
9.11
Por
lo
tanto
cumple.
CALCULOS CON RESPECTO AL MU:
Mu = 1.13 ton-m
Q
=
Mu
FRbd2fIIc
1.13
x105
0.9
x
25
x
362
x170
=
=
0.0229
q
= 1
-‐
√1
-‐
2Q
= 1
-‐
√1
-‐
2(0.0229)
=
0.0232
p
= q
x
fIIc/fy
= 0.0232
x
170/4200
=
0.00094
As
= pbd
=
0.00094
x
25
x
36
=
0.84 cm2
Mu = 5.26 ton-m
Q
=
Mu
FRbd2fIIc
5.26
x105
0.9
x
25
x
362
x170
=
=
0.106
q
= 1
-‐
√1
-‐
2Q
= 1
-‐
√1
-‐
2(0.106)
=
0.112
p
= q
x
fIIc/fy
=
0.112
x 170/4200
=
0.0045
As
= pbd
=
0.0045
x
25
x
36
=
4.09 cm2
Mu = 3.82 ton-m
Q
=
Mu
FRbd2fIIc
3.82
x105
0.9
x
25
x
362
x170
=
=
0.077
q
= 1
-‐
√1
-‐
2Q
= 1
-‐
√1
-‐
2(0.077)
=
0.080
p
= q
x
fIIc/fy
=
0.080
x...
DE
CONTRATRABE:
DATOS:
σu
:
Longitud
de
contratrabe:
Rt
:
B:
4.13 ton/m2
10 m
3 ton/m2
1.0 m
(considerado)
w
=
σu
x
B
=
4.13
x
1.0
= 4.13 ton/m
1
2
4
6
w
= 4.13 ton/m
D
40 cm
A
B
C
2.5
4.3 m
k = 3EI/L = 160000
K=4EI/L =
A
FD
FT
ME
1D
1T
2D
2T
3D
3T
4D
4T
5D
5T
6D
6T
7D
7T
8D
8T
9D
SUMA
=
D3.2 m
124031.01
C
BA
-‐0.56
0
-‐3.23
-‐1.77
0.00
-‐0.15
0.00
-‐0.10
0.00
-‐0.01
0.00
-‐0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-‐5.26
1
I
= b
x
h3/12
= 133333.33 cm4
E=
1.0
kg/cm2
k = 3EI/L = 125000.00
B
AB
0
0
0
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
25
BC
-‐0.44
0.5
6.37
-‐1.37
0.27
-‐0.12
0.17
-‐0.07
0.01-‐0.01
0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
5.26
D
CB
-‐0.50
0.5
-‐6.37
0.54
-‐0.69
0.34
-‐0.06
0.03
-‐0.04
0.02
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-‐6.23
2
CD
-‐0.50
0
5.29
0.54
0.00
0.34
0.00
0.03
0.00
0.02
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
6.23
DC
0
0
0
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.000.00
4
6
w
= 4.13 ton/m
A
B
C
2.5 m
VI
VH
∑V
AB
5.17
-‐2.10
3.06
D
4.3 m
BA
5.17
2.10
7.27
BC
8.88
-‐0.22
8.66
8.66 ton
3.2 m
CB
8.88
0.22
9.11
CD
6.61
1.95
8.56
8.56 ton
DC
6.61
-‐1.95
4.67
3.06 ton
9.07 ton-‐m
1.13 ton-‐m
1.76
8.86 ton-‐m
2.20
0.74
1.13
2.10
6.39 ton-‐m
0
ton
2.07
10.04 ton-‐m
2.63 ton-‐m
-‐7.27 ton
-‐4.67 ton
-‐9.11 ton
3.82 ton-‐m2.63 ton-‐m
1.13 ton-‐m
0.0
0.0
-‐5.26 ton-‐m
-‐6.23 ton-‐m
DIMENSIONAMIENTO POR FLEXIÒN:
250
Acero
fy
=
4200
f*c
=
200
fIIc
=
170
Pb
=
0.0190
Pmax
=
0.75
Pb
=
Pmin
=
0.0026
qmax
=
0.3529
qmin
=
0.0651
kumax
=
44.47
kumin
=
9.64
Concreto
f`c
=
kg/cm2
kg/cm2
kg/cm2
kg/cm2
0.0143
MR
= FRbd2f`IIcq(1 -0.5q). Por lo tantoMR
6.23 x 105
bd2
=FRfIIcq(1-‐0.5q)
bd
=0.9
x
170
x
0.3529
x
(1-‐0.5
x
0.3529)
2
bd2
=13999
considerando
d/b
=
2.
Por
lo
tanto
b
=
d/2,
sustituyendo
en
bd2
se
obtiene
d3/2.
d3/2
=13999
por
lo
tanto
d
=
3√
2
x
13999
Datos
calculados:
d
=
b
=
h
=
h
=
Datos propuesto:
30 cm
15 cm
d
+
r
+
Ø/2
34 cm
r
= 3 cm
Proponiendo
var
#
4,Ø
=1.3 cm
d
= 36 cm
b
= 25 cm
h
= 40 cm
Nota:La
sección
propuesta
para
la
contratrabe
es
correcta
por
lo
q
se
adoptara
sus
dimensiones
REVISIÒN POR CORTANTE DE LA SECCIÒN TRANSVERSAL
Se
debe
cumplir;
Vu
<
VR
VR
=2FRbd√f*c
= 2
x 0.8
x
25
x
36
x
√200
VR
=20.36 ton > Vu =
9.11
Por
lo
tanto
cumple.
CALCULOS CON RESPECTO AL MU:
Mu = 1.13 ton-m
Q
=
Mu
FRbd2fIIc
1.13
x105
0.9
x
25
x
362
x170
=
=
0.0229
q
= 1
-‐
√1
-‐
2Q
= 1
-‐
√1
-‐
2(0.0229)
=
0.0232
p
= q
x
fIIc/fy
= 0.0232
x
170/4200
=
0.00094
As
= pbd
=
0.00094
x
25
x
36
=
0.84 cm2
Mu = 5.26 ton-m
Q
=
Mu
FRbd2fIIc
5.26
x105
0.9
x
25
x
362
x170
=
=
0.106
q
= 1
-‐
√1
-‐
2Q
= 1
-‐
√1
-‐
2(0.106)
=
0.112
p
= q
x
fIIc/fy
=
0.112
x 170/4200
=
0.0045
As
= pbd
=
0.0045
x
25
x
36
=
4.09 cm2
Mu = 3.82 ton-m
Q
=
Mu
FRbd2fIIc
3.82
x105
0.9
x
25
x
362
x170
=
=
0.077
q
= 1
-‐
√1
-‐
2Q
= 1
-‐
√1
-‐
2(0.077)
=
0.080
p
= q
x
fIIc/fy
=
0.080
x...
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