Download - Abaco Para Columnas-final
UNIVERSIDAD PRIVADA DEL NORTE
DIAGRAMA DE ITERACCIÓN - VARIAS CUANTIAS
1.-) Una ρ=1%
Se establecen los siguientes datos a considerar en el problema:
f'c= 210fy= 4200
Recubrimiento= 5 cm
Datos asumidos para el problema:
3/8 pulgᴓ Acero 5/8 cm
d1 = 6.75 cm
60 d2 = 30 cm
d3 = 53.25 cm
30
2000000
Areas:
5.97
3.98
5.97Ast= 8Ø5/8'' (Area total)
Distancias:
6.75
30
53.25
Ag= 1800Ast= 15.92Yo= 30 cm
kg/cm2
kg/cm2
ᴓ
Coeficiente de Elasticidad
kg/cm2
As1=
As2=
As3=
d1=
d2=
d3=
cm2
cm2
a) carga concentrica
388.164 tn
b)Punto 2 del diagrama
Ey = f´c/2x10^6 0.003 Cb =
Ey (def.acero) = 0.0021 cb Cb= 40.00Ec (def.concreto) = 0.003
Ey+Ec = 0.0051
0.0021 32.00
C 40 cmCb 31.32 cmC2 20 cmC3 15 cm
4.9875
1.5
1.9875
Cálculo de fuerzas: CONSIDERANDOf'c= 2.1
Cc= 171.36 Tn
12.537 Tn
5.970 Tn
11.865 Tn
Haciendo uso de las formulas:
178.00 Tn
Pn0=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
kg/cm2
Cs1=
Cs2=
Ts3=
Pn=
𝑃𝑛o=0.85∗𝑓^′ 𝑐∗𝐴𝑔+𝐴𝑠𝑡∗𝑓𝑦
𝑃_𝑛𝑏=𝐶𝑐+𝐶𝑠_1+𝐶𝑠_2−T𝑠_3𝑀_𝑛𝑏=𝐶𝑐(𝑦_0−𝑎/2)+𝐶𝑠_1 (𝑦_0−𝑑_1)+𝐶𝑠_2 (𝑦_0−𝑑_2)+T𝑠_3 ( 〖𝑑 _3−𝑦〗_0)
〖𝑓𝑠〗 =6* (c − )/𝑏 𝑑 𝑐𝑏
29.66 tn-m
0.17 m
c) condicion balanceada
Cb= 31.32 cm
25.06 cm
4.71
0.253
4.20
Cálculo de fuerzas: CONSIDERANDOf'c= 2.1
Cc= 134.17488 Tn
12.537 Tn
1.006 Tn
12.537 Tn
135.18 Tn
29.27 tn-m
0.22
d) falla en fluencia
condicion C<Cb
C= 20 cm
16 cm
3.975
3
9.975
CONSIDERANDO
Mn=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
kg/cm2
Cs1=
Cs2=
Ts3=
Pnb=
Mnb=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
f'c= 2.1Cc= 85.68 Tn
12.537 Tn
8.358 Tn
12.537 Tn
77.322 tn
24.679 tn-m
0.32
e) punto 4 del diagrama de iteraccion
condicion C<Cb
C= 15.0 cm
12 cm
3.300
6.000
15.300
CONSIDERANDOf'c= 2.1
Cc= 64.26 Tn
12.537 Tn
8.358 Tn
12.537 Tn
55.902 0 Tn
21.252 Tn-m
f) traccion puraPn= -66.864 Tn
resultadocarga momento
388.164 0178.00 29.66135.18 29.27377.322 24.679-66.864 0
kg/cm2
Cs1=
Ts2=
Ts3=
Pn=
Mn=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
kg/cm2
Cs1=
Ts2=
Ts3=
Pn=
Mn=
ACERO COMERCIAL
3/8'' 0.71
1/2'' 1.29
5/8`` 1.993/4'' 2.841'' 5.1
1 3/8'' 10.06
cm2
cm2
cm2
cm2
cm2
cm2
1.-) Una ρ=2%
Se establecen los siguientes datos a considerar en el problema:
f'c= 210fy= 4200
Recubrimiento= 5 cm
Datos asumidos para el problema:
3/8 pulgᴓ Acero 5/8 cm
60
30
Elasticidad 2000000
Ast= 23.88Ast= 12Ø5/8''
Areas:
5.97 1.8
3.98
3.98
3.98
5.97
Distancias:
6.75
13.87
30
46.08
53.25
Ag= 1800Ast= 23.88
kg/cm2
kg/cm2
ᴓ
kg/cm2
cm2
As1=
As2=
As3=
As4=
As5=
d1=
d2=
d3=
d4=
d5=
cm2
cm2
Yo= 30.00 cm
a) carga concentrica
421.596 Kg
b) Punto 2 del diagrama
0.0051Ey (def.acero) = 0.0021 Cb= 40
Ec (def.concreto) = 0.003 cb
Ey+Ec = 0.0051
320.003
C= 40 cmCb= 31.32 cmC2= 20 cmC3= 15 cm
4.9875
3.9189375
1.5
0.9114375
1.9875
Cálculo de fuerzas: CONSIDERANDOCc= 171.36 Tn f'c= 2.1
12.537 Tn
8.358 Tn
5.970 Tn
8.358 Tn
12.537 Tn
Haciendo uso de las formulas:
Pn0=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Cs3=
Ts4=
Ts5=
𝑃𝑛_0=0.85∗𝑓^′ 𝑐∗𝐴𝑔+𝐴𝑠𝑡∗𝑓𝑦
𝑃_𝑛𝑏=𝐶𝑐+𝐶𝑠_1+𝐶𝑠_2+𝐶𝑠_3−"T" 𝑠_4−T𝑠_5𝑀_𝑛𝑏=𝐶𝑐(𝑦_0−𝑎/2)+𝐶𝑠_1 (𝑦_0−𝑑_1)+𝐶𝑠_2 (𝑦_0−𝑑_2)+T𝑠_3 ( 〖𝑑_3−𝑦〗 _0)
〖𝑓𝑠〗 =6* (c − )/𝑏 𝑑 𝑐𝑏
194.05 Tn
32.508 tn-m
0.17 m
c) condicion balanceada
Cb= 31.32 cm
25.056 cm
4.71
3.34
0.25
2.83
4.20
CONSIDERANDOCálculo de fuerzas: f'c= 2.1
Cc= 134.17488 Tn
12.537 Tn
8.358 Tn
1.006 Tn
8.358 Tn
12.537 Tn
135.18 Tn
31.962 tn-m
0.24 m
d) falla en fluencia
condicion C<Cb
C= 20 cm
16 cm
3.98
Pn=
Mn=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Cs3=
Ts4=
Ts5=
Pnb=
Mnb=
eb=
ab=BC=
ab=
fs1= Tn/cm2
𝑀_𝑛𝑏=𝐶𝑐(𝑦_0−𝑎/2)+𝐶𝑠_1 (𝑦_0−𝑑_1)+𝐶𝑠_2 (𝑦_0−𝑑_2)+T𝑠_3 ( 〖𝑑_3−𝑦〗 _0)
1.84
3.00
7.824.2
9.98
Cálculo de fuerzas: CONSIDERANDOCc= 85.68 Tn f'c= 2.1
12.537 Tn
7.315 Tn
8.358 Tn
8.358 Tn
12.537 Tn
76.279 TnMn= 27.201 Tn-m
0.36 m
e) punto 4 del diagrama de iteraccion
condicion C<Cb
C= 15 cm
12 cm
3.30
-0.45
6.00
12.43
15.30
Cálculo de fuerzas:Cc= 64.26 Tn CONSIDERANDO
19.701 Tn f'c= 2.1
fs1= Tn/cm2
fs2= Tn/cm2
fs2= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Ts3=
Ts4=
Ts5=
Pn=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Cs1= Tn/cm2
8.358 Tn
8.358 Tn
8.358 Tn
12.537 Tn
46.350 TnMn= 25.608 Tn-m
f) traccion puraPn= -100.296 tn
resultadocarga momento
421.596 0194.05 32.508135.18 31.96276.279 27.201
-100.296 0
Ts2=
Ts3=
Ts4=
Ts5=
Pn=
0 5 10 15 20 25 30 35
-200
-100
0
100
200
300
400
500
ρ2%
Momento
Carg
a ax
ial
areas de acero
3/8'' 0.71
1/2'' 1.29
5/8`` 1.993/4'' 2.841'' 5.1
1 3/8'' 10.06
cm2
cm2
cm2
cm2
cm2
cm2
1.-) Una ρ=3%
Se establecen los siguientes datos a considerar en el problema:
f'c= 210fy= 4200
recubrimiento 5 cm
Datos asumidos para el problema: 3/8 pulg
ᴓ Acero 5/8 cm
60
30
Elasticidad 2000000
Ast= 35.82Ast= 12 Ø 5/8''
Areas: areas de acero
9.95 3/8''
5.97 1/2''
3.98 5/8``
5.97 3/4''
9.95 1''1 3/8''
Distancias:
6.75
13.87
30
46.08
53.25
Ag= 1800
kg/cm2
kg/cm2
ᴓ
kg/cm2
cm2
As1=
As2=
As3=
As4=
As5=
d1=
d2=
d3=
d4=
d5=
cm2
Ast= 35.82Y0= 30.00 cm
a) carga concentrica
471.744 Kg
b) Punto 2 del diagrama
0.003
Ey (def.acero) = 0.0021 cb Cb= 40Ec (def.concreto) = 0.003
Ey+Ec = 0.0051
0.0021 32
C = 40 cmCb= 31.32 cmC2= 20 cmC3= 15 cm
4.988
3.919
1.500
0.911
1.988
Cálculo de fuerzas: CONSIDERANDOCc= 171.36 Tn f'c= 2.1
20.895 Tn
12.537 Tn
5.970 Tn
cm2
Pn0=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Cs3=
𝑃𝑛_0=0.85∗𝑓^′ 𝑐∗𝐴𝑔+𝐴𝑠𝑡∗𝑓𝑦
〖𝑓𝑠〗 =6* (c − )/𝑏 𝑑 𝑐𝑏
12.537 Tn
20.895 Tn
Haciendo uso de las formulas:
202.40 Tn
37.739 tn-m
0.19 m
c) condicion balanceada
Cb= 31.32 cm
25.056 cm
4.706896551724
3.34219348659
0.252873563218
2.826867816092
4.201149425287
Cálculo de fuerzas: CONSIDERANDOCc= 134.17488 Tn f'c= 2.1
20.895 Tn
12.537 Tn
1.006 Tn
12.537 Tn
20.895 Tn
135.18 Tn
37.193 tn-m
0.28 m
Ts4=
Ts5=
Pn=
Mn=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Cs3=
Ts4=
Ts5=
Pnb=
Mnb=
eb=
𝑀_𝑛𝑏=𝐶𝑐(𝑦_0−𝑎/2)+𝐶𝑠_1 (𝑦_0−𝑑_1 )+𝐶𝑠_2 (𝑦_0−𝑑_2 )+𝐶𝑠_3 (𝑦_0−𝑑_3 )+"T" 𝑠_4 ( 〖𝑑 _4−𝑦〗 _0)+T𝑠_5 ( 〖𝑑 _5−𝑦〗 _0)𝑃_𝑛𝑏=𝐶𝑐+𝐶𝑠_1+𝐶𝑠_2+𝐶𝑠_3−"T" 𝑠_4−T𝑠_5
d) falla en fluencia
condicion C<Cb
C= 20 cm
16 cm
3.98
1.84
3.00
7.82
9.98
Cálculo de fuerzas: CONSIDERANDOf'c= 2.1
Cc= 85.68 Tn
20.895 Tn
10.972 Tn
8.358 Tn
12.537 Tn
20.895 Tn
75.757 tnMn= 32.348 tn-m
0.43
e) punto #4 del diagrama de iteraccion
condicion C<Cb
C= 15
12
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Ts3=
Ts4=
Ts5=
Pn=
eb=
ab=BC=
ab=
0.45
0.45
12.43
12.43
15.30
Cálculo de fuerzas: CONSIDERANDOCc= 64.26 Tn f'c= 2.1
4.482 Tn
2.689 Tn
8.358 Tn
12.537 Tn
20.895 Tn
24.263 0 tnMn= 23.771 tn-m
f) traccion puraPn= -150.444 tn
resultadocarga momento
471.744 0202.40 37.739135.18 37.19375.757 32.348
-150.444 0
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Ts2=
Ts3=
Ts4=
Ts5=
Pn=
0 5 10 15 20 25 30 35 40
-200
-100
0
100
200
300
400
500
600
ρ3%
Momento
Carg
a ax
ial
1.-) Una ρ=3%
Se establecen los siguientes datos a considerar en el problema:
f'c= 210fy= 4200
Recubrimiento 5 cm
Datos asumidos para el problema: 3/8 pulg
ᴓ Acero 5/8 cm
60
30Elasticidad 2000000
Ast= 49.75Ast= 25Ø5/8''
Areas:
9.95
9.95
9.95
9.95
9.95
Distancias:
6.75
13.87
30
46.08
53.25
Ag= 1800Ast= 49.75
kg/cm2
kg/cm2
ᴓ
kg/cm2
cm2
As1=
As2=
As3=
As4=
As5=
d1=
d2=
d3=
d4=
d5=
cm2
cm2
Yo= 30.00 cm
a) carga concentrica
530.25 Kg
b) Punto 2 del diagrama0.003
Ey (def.acero) = 0.0021 cb Cb= 40Ec (def.concreto) = 0.003
Ey+Ec = 0.0051
0.0021 32
C 40 cmCb 31.32 cmC2 20 cmC3 15 cm
4.99
3.92
1.50
4.51
1.99
Cálculo de fuerzas: CONSIDERANDOCc= 171.36 Tn f'c= 2.1
20.895 Tn
99.002 Tn
14.925 Tn
20.895 Tn
20.895 Tn
Pn0=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Cs3=
Ts4=
Ts5=
𝑃𝑛_0=0.85∗𝑓^′ 𝑐∗𝐴𝑔+𝐴𝑠𝑡∗𝑓𝑦
〖𝑓𝑠〗 =6* (c − )/𝑏 𝑑 𝑐𝑏
306.18 Tn
53.021 tn-m
0.17 m
c) condicion balanceada
Cb= 31.32 cm
25.056 cm
4.71
3.34
0.25
2.83
4.20
Cálculo de fuerzas: CONSIDERANDOCc= 134.17488 Tn f'c= 2.1
20.895 Tn
20.895 Tn
2.516 Tn
20.895 Tn
20.895 Tn
136.69 Tn
39.883 tn-m
0.29 m
d) falla en fluencia
Pnb=
Mnb=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Cs3=
Ts4=
Ts5=
Pnb=
Mnb=
eb=
𝑀_𝑛𝑏=𝐶𝑐(𝑦_0−𝑎/2)+𝐶𝑠_1 (𝑦_0−𝑑_1 )+𝐶𝑠_2 (𝑦_0−𝑑_2 )+𝐶𝑠_3 (𝑦_0−𝑑_3 )+"T" 𝑠_4 ( 〖𝑑 _4−𝑦〗 _0)+T𝑠_5 ( 〖𝑑 _5−𝑦〗 _0)𝑃_𝑛𝑏=𝐶𝑐+𝐶𝑠_1+𝐶𝑠_2+𝐶𝑠_3−"T" 𝑠_4−T𝑠_5
condicion C<Cb
C= 20 cm
16 cm
3.975
1.838
3.000
7.823
9.975
Cálculo de fuerzas: CONSIDERANDOCc= 85.68 Tn f'c= 2.1
20.895 Tn
20.895 Tn
20.895 Tn
20.895 Tn
20.895 Tn
47.865 tnMn= 31.357 tn-m
0.66 m
e) punto 4 del diagrama de iteraccion
condicion C<Cb
C= 15 cm
12 cm
3.300
0.451
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Ts3=
Ts4=
Ts5=
Pn=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
6.000
12.431
15.300
Cálculo de fuerzas: CONSIDERANDOCc= 64.26 Tn f'c= 2.1
20.895 Tn
4.482475 Tn
20.895 Tn
20.895 Tn
20.895 Tn
17.988 0 tnMn= 29.218 tn-m
f) traccion puraPn= -208.950 Tn
resultadocarga momento
530.25 0306.18 53.021136.69 39.88347.865 31.3570.000 29.218
-208.950 0
fs3= Tn/cm2
fs4= Tn/cm2
fs5= Tn/cm2
Tn/cm2
Cs1=
Ts2=
Ts3=
Ts4=
Ts5=
Pn=
0 10 20 30 40 50 60
-300
-200
-100
0
100
200
300
400
500
600
ρ4%
Momento
Carg
a ax
ial
areas de acero3/8'' 0.711/2'' 1.29
5/8`` 1.99
3/4'' 2.84
1'' 5.1
1 3/8'' 10.06
cm2
cm2
cm2
cm2
cm2
cm2
1.-) Una ρ=3%
Se establecen los siguientes datos a considerar en el problema:
f'c= 210fy= 4200
recubrimiento 5 cm
3/8 pulgᴓ Acero 5/8 cm
60
30
Elasticidad 2000000
Ast= 49.75Ast= 30Ø5/8''
Areas:
9.95
9.95
9.95
9.95
9.95
9.95
Distancias:
6.75
12.33
16.92
6.75
4.38
12.00
Ag= 1800Ast= 59.7
kg/cm2
kg/cm2
ᴓ
kg/cm2
cm2
As1=
As2=
As3=
As4=
As5=
As6=
d1=
d2=
d3=
d4=
d5=
d6=
cm2
cm2
Yo= 21.17 cm
a) carga concentrica
572.04 Kg
b) Punto #2 del diagrama 0.003
Ey (def.acero) = 0.0021 cb Cb= 7.06Ec (def.concreto) = 0.003
Ey+Ec = 0.0051
0.0021 5.648
C 10 cmCb 7.06 cmC2 5 cmC3 3 cm
0.26
4.48
8.38
0.26
2.28
4.20
Cálculo de fuerzas: CONSIDERANDOCc= 30.24504 Tn f'c= 2.1
2.621 Tn
44.595 Tn
83.388 Tn
2.621 Tn
22.705 Tn
41.773 Tn
Pn0=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
f5= Tn/cm2
fs6= Tn/cm2
Tn/cm2
Cs1=
Cs2=
Cs3=
Cs4=
Cs5=
Cs6=
𝑃𝑛_0=0.85∗𝑓^′ 𝑐∗𝐴𝑔+𝐴𝑠𝑡∗𝑓𝑦
𝑃_𝑛=𝐶𝑐+𝐶𝑠_1+𝐶𝑠_2+𝐶𝑠_3−𝑇𝑠_4−T𝑠_5−T𝑠_6−"T" 𝑠_7−"T" 𝑠_8−"T" 𝑠_9𝑀_𝑛𝑏=𝐶𝑐(𝑦_0−𝑎/2)+𝐶𝑠_1 (𝑦_0−𝑑_1 )+𝐶𝑠_2 (𝑦_0−𝑑_2 )+𝐶𝑠_3 (𝑦_0−𝑑_3 )+𝐶𝑠_4 (𝑦_0−𝑑_4 )+"C" 𝑠_5 ( 〖𝑑 _5−𝑦〗 _0)+T𝑠_6 ( 〖𝑑 _6−𝑦〗 _0)+T𝑠_7 ( 〖𝑑 _7−𝑦〗 _0 )+T𝑠_8 ( 〖𝑑 _8−𝑦〗 _0)+T𝑠_9 ( 〖𝑑 _9−𝑦〗 _0)
〖𝑓𝑠〗 =6* (c − )/𝑏 𝑑 𝑐𝑏
144.40 Tn
5.388 tn-m
0.04 m
c) condicion balanceada
Cb= 7.06 cm
5.648 cm
0.26
4.48
8.38
0.26
2.28
4.20
Cc= 30.245 Tn CONSIDERANDO
2.621 Tn f'c= 2.1
44.595 Tn
83.388 Tn
2.621 Tn
22.705 Tn
41.773 Tn
93.75 Tn
5.388 tn-m
0.06 m
d) falla en fluencia
condicion C<Cb
C= 5 cm
Pnb=
Mnb=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
f5= Tn/cm2
fs6= Tn/cm2
Cs1= Tn/cm2
Cs2=
Cs3=
Ts4=
Ts5=
Ts6=
Pnb=
Mnb=
eb=
𝑃_𝑛=𝐶𝑐+𝐶𝑠_1+𝐶𝑠_2+𝐶𝑠_3−𝑇𝑠_4−T𝑠_5−T𝑠_6−"T" 𝑠_7−"T" 𝑠_8−"T" 𝑠_9𝑀_𝑛𝑏=𝐶𝑐(𝑦_0−𝑎/2)+𝐶𝑠_1 (𝑦_0−𝑑_1 )+𝐶𝑠_2 (𝑦_0−𝑑_2 )+𝐶𝑠_3 (𝑦_0−𝑑_3 )+𝐶𝑠_4 (𝑦_0−𝑑_4 )+"C" 𝑠_5 ( 〖𝑑 _5−𝑦〗 _0)+T𝑠_6 ( 〖𝑑 _6−𝑦〗 _0)+T𝑠_7 ( 〖𝑑 _7−𝑦〗 _0 )+T𝑠_8 ( 〖𝑑 _8−𝑦〗 _0)+T𝑠_9 ( 〖𝑑 _9−𝑦〗 _0)
4 cm
2.10
8.80
14.31
2.10
0.75
8.40
Cc= 21.42 Tn CONSIDERANDO
20.895 Tn f'c= 2.1
20.895 Tn
20.895 Tn
20.895 Tn
7.462 Tn
20.895 Tn
-6.938 tnMn= 3.671 tn-m
-0.53 m
e) punto 4 del diagrama de iteraccion
condicion C<Cb
C= 3 cm
2.4 cm
7.50
18.67
27.84
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
f5= Tn/cm2
fs6= Tn/cm2
Cs1= Tn/cm2
Cs2=
Ts3=
Ts4=
Ts5=
Ts6=
Pn=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
7.50
2.75
18.00
Cc= 12.852 Tn CONSIDERANDO
20.895 Tn f'c= 2.1
20.895 Tn
20.895 Tn
20.895 Tn
20.895 Tn
20.895 Tn
-28.938 0 tnMn= -0.125 tn-m
f) traccion puraPn= -250.740 tn
resultadocarga momento
572.04 0144.40 5.38893.75 5.388-6.938 3.6710.000 -0.125
-250.740 0
fs4= Tn/cm2
f5= Tn/cm2
fs6= Tn/cm2
Cs1= Tn/cm2
Cs2=
Cs3=
Ts4=
Ts5=
Ts6=
Pn=
-1 0 1 2 3 4 5 6
-300
-200
-100
0
100
200
300
400
500
600
700
ρ5%
Momento
Carg
a ax
ial
areas de acero
3/8'' 0.71
1/2'' 1.29
5/8`` 1.99
3/4'' 2.84
1'' 5.1
1 3/8'' 10.06
cm2
cm2
cm2
cm2
cm2
cm2
1.-) Una ρ=3%
Se establecen los siguientes datos a considerar en el problema:
f'c= 210fy= 4200
recubrimiento: 5 cm
3/8 pulgᴓ Acero 5/8 cm
40
30
Elasticidad 2000000
Ast= 71.64Ast= 30Ø5/8''
Areas:
11.94
11.94
11.94
11.94
11.94
11.94
Distancias:
6.75
12.33
16.92
23.08
28.67
33.25
kg/cm2
kg/cm2
ᴓ
kg/cm2
cm2
As1=
As2=
As3=
As4=
As5=
As6=
d1=
d2=
d3=
d4=
d5=
d6=
Ag= 1200Ast= 71.64Yo= 20.10 cm
a) carga concentrica
515.088 Kg
b) Punto 2 del diagrama
Def acero 0.0021 Cb= 30 cmDef concreto 0.003
Acum 0.0051
24 cmC 30 cm
Cb 19.56 cmC2 15 cmC3 13.9 cm
4.65
3.53
2.62
1.38
0.27
0.65
Cc= 128.52 Tn CONSIDERANDO
25.074 Tn f'c= 2.1
25.074 Tn
25.074 Tn
16.528 Tn
3.185 Tn
cm2
cm2
Pn0=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
f5= Tn/cm2
fs6= Tn/cm2
Cs1= Tn/cm2
Cs2=
Cs3=
Cs4=
Cs5=
𝑃𝑛_0=0.85∗𝑓^′ 𝑐∗𝐴𝑔+𝐴𝑠𝑡∗𝑓𝑦
7.761 Tn
215.69 Tn
18.283 tn-m
0.08 m
c) condicion balanceada
Cb= 19.56 cm
15.648 cm
3.93
2.22
0.81
1.08
2.79
4.20
Cc= 83.79504 Tn CONSIDERANDO
25.074 Tn f'c= 2.1
25.074 Tn
9.665 Tn
12.888 Tn
25.074 Tn
25.074 Tn
80.57 Tn
21.715 tn-m
0.27 m
Cs6=
Pnb=
Mnb=
eb=
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
f5= Tn/cm2
fs6= Tn/cm2
Cs1= Tn/cm2
Cs2=
Cs3=
Ts4=
Ts5=
Ts6=
Pnb=
Mnb=
eb=
𝑃_𝑛=𝐶𝑐+𝐶𝑠_1+𝐶𝑠_2+𝐶𝑠_3−𝑇𝑠_4−T𝑠_5−T𝑠_6−"T" 𝑠_7−"T" 𝑠_8−"T" 𝑠_9𝑀_𝑛𝑏=𝐶𝑐(𝑦_0−𝑎/2)+𝐶𝑠_1 (𝑦_0−𝑑_1 )+𝐶𝑠_2 (𝑦_0−𝑑_2 )+𝐶𝑠_3 (𝑦_0−𝑑_3 )+𝐶𝑠_4 (𝑦_0−𝑑_4 )+"C" 𝑠_5 ( 〖𝑑 _5−𝑦〗 _0)+T𝑠_6 ( 〖𝑑 _6−𝑦〗 _0)+T𝑠_7 ( 〖𝑑 _7−𝑦〗 _0 )+T𝑠_8 ( 〖𝑑 _8−𝑦〗 _0)+T𝑠_9 ( 〖𝑑 _9−𝑦〗_0)
d) falla en fluencia
condicion C<Cb
C= 15 cm
12 cm
3.30
1.07
0.77
3.23
5.47
7.30
Cc= 64.26 Tn CONSIDERANDO
25.074 Tn f'c= 2.1
12.734 Tn
9.176 Tn
25.074 Tn
25.074 Tn
25.074 Tn
17.670 tnMn= 19.880 tn-m
1.13 m
e) punto 4 del diagrama de iteraccion
condicion C<Cb
C= 13.9 cm
11.12 cm
ab=BC=
ab=
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
f5= Tn/cm2
fs6= Tn/cm2
Cs1= Tn/cm2
Cs2=
Ts3=
Ts4=
Ts5=
Ts6=
Pn=
eb=
ab=BC=
ab=
3.09
0.68
1.30
3.96
6.37
8.35
Cc= 59.5476 Tn CONSIDERANDO
25.074 Tn f'c= 2.1
8.072 Tn
15.571 Tn
25.074 Tn
25.074 Tn
25.074 Tn
1.901 tnMn= 19.319 tn-m
f) traccion puraPn= -300.888 tn
resultadocarga momento
515.088 0215.69 18.28380.57 21.715
17.670 19.8800.000 19.319
-300.888 0
fs1= Tn/cm2
fs2= Tn/cm2
fs3= Tn/cm2
fs4= Tn/cm2
f5= Tn/cm2
fs6= Tn/cm2
Cs1= Tn/cm2
Cs2=
Cs3=
Ts4=
Ts5=
Ts6=
Pn=
areas de acero3/8'' 0.711/2'' 1.295/8`` 1.99
3/4'' 2.84
1'' 5.1
1 3/8'' 10.06
cm2
cm2
cm2
cm2
cm2
cm2
DIAGRAMA DE ITERACCIÓN
Cuantía de 1% Cuantía de 2%388.164 0 421.596 0.000
178.001625 29.6639522 194.046 32.508135.181 29.273 135.181 31.962
77.322 24.679 76.279 27.201#NAME? #NAME? #NAME? #NAME?
Cuantía de 3% Cuantía de 4%471.744 0.000 530.250 0.000202.404 37.739 306.183 53.021135.181 37.193 136.691 39.883
75.757 32.348 47.865 31.357#NAME? #NAME? 0.000 29.218
Cuantía de 5% Cuantía de 6%572.040 0.000 515.088 0.000144.402 5.388 215.694 18.283
93.750 5.388 80.572 21.715-6.938 3.671 17.670 19.8800.000 -0.125 0.000 19.319
Una vez obtenidos los diagramas de iteracción para 6 cuantías, sobreponemos las gráficas y obtenemos:
-10 0 10 20 30 40 50 60-100
0
100
200
300
400
500
600
700
ρ1%ρ2%ρ3%ρ4%ρ5%ρ6%
Momentos
Fuer
za