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UNIVERSIDAD MAYOR DE SAN ANDRES
FACULTAD DE INGENIERIA
INGENIERIA CIVIL
AUX. APAZA MACUCHAPI CESAR ISRAEL
HORMIGÓN ARMADO I
APLICACIÓN:ARMADURA SIMETRICA
Ejemplo No. 6: Calcular la armadura de la viga en el tramo de manera que sea
simetrica
Datos:
*Sección: ≔
b 0.20 m
≔
h 0.4 m
*Materiales:
Hormigón H30 ≔
fck 30 MPa
*Longitudes: ≔
L1 5.0 m
,
≔
Lv1 2.0 m ≔
Lv2 2.0 m
Acero B400S ≔
fyk 400 MPa
≔
Es 200000 MPa
*Recubrimiento mecánico:
≔
d1 =
⋅
0.08 h 0.032 m
*CNC
*Cargas: ≔
qk 57 ――
kN
m
≔
γHA 25 ――
kN
m3
1.) Materiales:
Hormigon H30 =
fck 30 MPa ≔
γc 1.5 ≔
fcd =
――
fck
γc
2 ――
kN
cm2
Acero B400S: =
fyk 400 MPa ≔
γs 1.15 ≔
fyd =
――
fyk
γs
34.783 ――
kN
cm2
=
Es 200000 MPa ≔
εyd =
――
fyd
Es
⋅
1.739 10-3
2.) Cargas:
≔
gp_viga =
⋅
⋅
γHA b h 2 ――
kN
m
Carga Permanente.
=
qk 57 ――
kN
m
Carga Variable.
Carga Desfavorable: ,
≔
γg 1.5 ≔
γq 1.6
≔
CD =
+
⋅
γg gp_viga ⋅
γq qk 94.2 ――
kN
m
Carga Favorable ,
≔
γg 0.9 ≔
γq 0
≔
CF =
+
⋅
γg gp_viga ⋅
γq qk 1.8 ――
kN
m
3.) Estados de Carga:
* Momento máximo Apoyo "A"
≔
MA =
⋅
⋅
CD Lv1 ――
Lv1
2
188.4 ⋅
kN m
=
Lv1 2 m =
L1 5 m =
Lv2 2 m
09/11/21
Página 1 de 5
UNIVERSIDAD MAYOR DE SAN ANDRES
FACULTAD DE INGENIERIA
INGENIERIA CIVIL
AUX. APAZA MACUCHAPI CESAR ISRAEL
HORMIGÓN ARMADO I
APLICACIÓN:ARMADURA SIMETRICA
* Momento máximo Apoyo "B"
≔
MB =
⋅
⋅
CD Lv2 ――
Lv2
2
188.4 ⋅
kN m
=
Lv1 2 m =
L1 5 m =
Lv2 2 m
* Momento máximo Tramos "A-B"
≔
M'A =
⋅
CF ――
Lv1
2
2
3.6 ⋅
kN m
≔
M'B =
⋅
CF ――
Lv2
2
2
3.6 ⋅
kN m
=
Lv1 2 m =
L1 5 m =
Lv2 2 m
M'A M'B
≔
R'A =
⋅
CF Lv1 3.6 kN ≔
R'B =
⋅
CF Lv2 3.6 kN
=
L1 5 m
≔
Q 0
≔
RA =
⋅
―
1
L1
⎛
⎜
⎜
⎝
+
-
+
⋅
CD ――
L1
2
2
M'A M'B ⋅
R'A L1
⎞
⎟
⎟
⎠
239.1 kN ≔
RB =
⋅
―
1
L1
⎛
⎜
⎜
⎝
+
-
+
⋅
CD ――
L1
2
2
M'B M'A ⋅
R'B L1
⎞
⎟
⎟
⎠
239.1 kN
=
Q -
+
⋅
-CD x RA R'A ---> ≔
x =
―――
-
RA R'A
CD
2.5 m
≔
MAB =
-
+
-
⋅
-CD ―
x2
2
M'A ⋅
RA x ⋅
R'A x 290.775 ⋅
kN m ≔
M0 =
-
⋅
――
CD
8
L1
2
⋅
3.6 kN m 290.775 ⋅
kN m
4.) Diseño:
1er Tanteo Dominio 3lim
≔
α3_lim =
―――
3.5
+
3.5 εyd
0.668
≔
d =
-
h ⋅
0.08 h 36.8 cm
≔
β3_lim =
-
1 ⋅
0.4 α3_lim 0.733
≔
μ3_lim =
⋅
⋅
0.8 α3_lim β3_lim 0.392
09/11/21
Página 2 de 5
UNIVERSIDAD MAYOR DE SAN ANDRES
FACULTAD DE INGENIERIA
INGENIERIA CIVIL
AUX. APAZA MACUCHAPI CESAR ISRAEL
HORMIGÓN ARMADO I
APLICACIÓN:ARMADURA SIMETRICA
≔
M3_lim =
⋅
⋅
⋅
μ3_lim fcd b d2
212.143 ⋅
kN m
≔
εs2 =
⋅
―――――
-
α3_lim ―――
0.08 h
d
α3_lim
3.5 3.044 =
εyd 1.739
≔
z2 =
-
d 0.08 h 33.6 cm ≔
zc_3lim =
-
d 0.4 ⎛
⎝ ⋅
α3_lim d⎞
⎠ 0.27 m
=
=
=
As2 ――
Ns2
σs2
―――
δM
⋅
z2 σs2
―――――
-
MAB M3_lim
⋅
z2 fyd
≔
As2 =
―――――
-
MAB M3_lim
⋅
z2 fyd
6.728 cm2
=
=
=
As1 ――
Ns1
fyd
―――
+
Ns2 Nc
fyd
――――――――
+
―――――
-
MAB M3_lim
z2
―――
M3_lim
zc_3lim
fyd
≔
As1 =
――――――――
+
―――――
-
MAB M3_lim
z2
―――
M3_lim
zc_3lim
fyd
29.346 cm2
2do Tanteo Dominio 2Blim
≔
α2B_lim =
―
7
27
0.259
≔
β2B_lim =
-
1 ⋅
0.4 α2B_lim 0.896
≔
μ2B_lim =
⋅
⋅
0.8 α2B_lim β2B_lim 0.186
≔
M2B_lim =
⋅
⋅
⋅
μ2B_lim fcd b d2
100.7 ⋅
kN m
=
=
=
As2 ――
Ns2
σs2
―――
δM
⋅
z2 σs2
―――――
-
MAB M2B_lim
⋅
z2 fyd
≔
As2 =
―――――
-
MAB M2B_lim
⋅
z2 fyd
16.264 cm2
≔
εs2 =
⋅
――――――
-
α2B_lim ―――
0.08 h
d
α2B_lim
3.5 2.326 =
εyd 1.739
≔
zc_2Blim =
-
d 0.4 ⎛
⎝ ⋅
α2B_lim d⎞
⎠ 0.33 m
=
=
=
As1 ――
Ns1
fyd
―――
+
Ns2 Nc
fyd
――――――――
+
―――――
-
MAB M2B_lim
z2
―――
M2B_lim
zc_2Blim
fyd
≔
As1 =
――――――――
+
―――――
-
MAB M2B_lim
z2
―――
M2B_lim
zc_2Blim
fyd
25.041 cm2
09/11/21
Página 3 de 5
UNIVERSIDAD MAYOR DE SAN ANDRES
FACULTAD DE INGENIERIA
INGENIERIA CIVIL
AUX. APAZA MACUCHAPI CESAR ISRAEL
HORMIGÓN ARMADO I
APLICACIÓN:ARMADURA SIMETRICA
3er Tanteo Dominio 2A lim
≔
α2A_lim =
―
1
6
0.167
≔
β2A_lim =
-
1 ⋅
0.4 α2A_lim 0.933
≔
μ2A_lim =
⋅
⋅
0.8 α2A_lim β2A_lim 0.124
≔
M2A_lim =
⋅
⋅
⋅
μ2A_lim fcd b d2
67.411 ⋅
kN m
≔
εs2 =
⋅
――――――
-
α2A_lim ―――
0.08 h
d
α2A_lim
2 0.957 =
εyd 1.739 ≔
σs2 =
⋅
Es ――
εs2
1000
191.304 MPa
=
=
=
As2 ――
Ns2
σs2
―――
δM
⋅
z2 σs2
―――――
-
MAB M2A_lim
⋅
z2 σs2
≔
As2 =
―――――
-
MAB M2A_lim
⋅
z2 σs2
34.75 cm2
≔
zc_2Alim =
-
d 0.4 ⎛
⎝ ⋅
α2A_lim d⎞
⎠ 0.343 m
=
=
=
As1 ――
Ns1
fyd
―――
+
Ns2 Nc
fyd
――――――――
+
―――――
-
MAB M2A_lim
z2
―――
M2A_lim
zc_2Alim
fyd
≔
As1 =
――――――――
+
―――――
-
MAB M2A_lim
z2
―――
M2A_lim
zc_2Alim
fyd
24.755 cm2
4to Tanteo Dominio 2B
≔
α2B 0.19098157
≔
β2B =
-
1 ⋅
0.4 α2B 0.924
≔
μ2B =
⋅
⋅
0.8 α2B β2B 0.141
≔
M2B =
⋅
⋅
⋅
μ2B fcd b d2
76.441 ⋅
kN m
≔
εc =
⋅
―――
α2B
-
1 α2B
10 2.361
≔
εs2 =
⋅
―――――
-
α2B ―――
0.08 h
d
α2B
εc 1.286 =
εyd 1.739
≔
σs2 =
⋅
――
εs2
1000
Es 257.164 MPa
09/11/21
Página 4 de 5
UNIVERSIDAD MAYOR DE SAN ANDRES
FACULTAD DE INGENIERIA
INGENIERIA CIVIL
AUX. APAZA MACUCHAPI CESAR ISRAEL
HORMIGÓN ARMADO I
APLICACIÓN:ARMADURA SIMETRICA
≔
zc =
-
d 0.4 ⎛
⎝ ⋅
α2B d⎞
⎠ 0.34 m
≔
As1 =
――――――
+
――――
-
MAB M2B
z2
――
M2B
zc
fyd
24.805 cm2
≔
As2 =
――――
-
MAB M2B
⋅
z2 σs2
24.805 cm2
09/11/21
Página 5 de 5

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Armadura Simetrica.pdf

  • 1. UNIVERSIDAD MAYOR DE SAN ANDRES FACULTAD DE INGENIERIA INGENIERIA CIVIL AUX. APAZA MACUCHAPI CESAR ISRAEL HORMIGÓN ARMADO I APLICACIÓN:ARMADURA SIMETRICA Ejemplo No. 6: Calcular la armadura de la viga en el tramo de manera que sea simetrica Datos: *Sección: ≔ b 0.20 m ≔ h 0.4 m *Materiales: Hormigón H30 ≔ fck 30 MPa *Longitudes: ≔ L1 5.0 m , ≔ Lv1 2.0 m ≔ Lv2 2.0 m Acero B400S ≔ fyk 400 MPa ≔ Es 200000 MPa *Recubrimiento mecánico: ≔ d1 = ⋅ 0.08 h 0.032 m *CNC *Cargas: ≔ qk 57 ―― kN m ≔ γHA 25 ―― kN m3 1.) Materiales: Hormigon H30 = fck 30 MPa ≔ γc 1.5 ≔ fcd = ―― fck γc 2 ―― kN cm2 Acero B400S: = fyk 400 MPa ≔ γs 1.15 ≔ fyd = ―― fyk γs 34.783 ―― kN cm2 = Es 200000 MPa ≔ εyd = ―― fyd Es ⋅ 1.739 10-3 2.) Cargas: ≔ gp_viga = ⋅ ⋅ γHA b h 2 ―― kN m Carga Permanente. = qk 57 ―― kN m Carga Variable. Carga Desfavorable: , ≔ γg 1.5 ≔ γq 1.6 ≔ CD = + ⋅ γg gp_viga ⋅ γq qk 94.2 ―― kN m Carga Favorable , ≔ γg 0.9 ≔ γq 0 ≔ CF = + ⋅ γg gp_viga ⋅ γq qk 1.8 ―― kN m 3.) Estados de Carga: * Momento máximo Apoyo "A" ≔ MA = ⋅ ⋅ CD Lv1 ―― Lv1 2 188.4 ⋅ kN m = Lv1 2 m = L1 5 m = Lv2 2 m 09/11/21 Página 1 de 5
  • 2. UNIVERSIDAD MAYOR DE SAN ANDRES FACULTAD DE INGENIERIA INGENIERIA CIVIL AUX. APAZA MACUCHAPI CESAR ISRAEL HORMIGÓN ARMADO I APLICACIÓN:ARMADURA SIMETRICA * Momento máximo Apoyo "B" ≔ MB = ⋅ ⋅ CD Lv2 ―― Lv2 2 188.4 ⋅ kN m = Lv1 2 m = L1 5 m = Lv2 2 m * Momento máximo Tramos "A-B" ≔ M'A = ⋅ CF ―― Lv1 2 2 3.6 ⋅ kN m ≔ M'B = ⋅ CF ―― Lv2 2 2 3.6 ⋅ kN m = Lv1 2 m = L1 5 m = Lv2 2 m M'A M'B ≔ R'A = ⋅ CF Lv1 3.6 kN ≔ R'B = ⋅ CF Lv2 3.6 kN = L1 5 m ≔ Q 0 ≔ RA = ⋅ ― 1 L1 ⎛ ⎜ ⎜ ⎝ + - + ⋅ CD ―― L1 2 2 M'A M'B ⋅ R'A L1 ⎞ ⎟ ⎟ ⎠ 239.1 kN ≔ RB = ⋅ ― 1 L1 ⎛ ⎜ ⎜ ⎝ + - + ⋅ CD ―― L1 2 2 M'B M'A ⋅ R'B L1 ⎞ ⎟ ⎟ ⎠ 239.1 kN = Q - + ⋅ -CD x RA R'A ---> ≔ x = ――― - RA R'A CD 2.5 m ≔ MAB = - + - ⋅ -CD ― x2 2 M'A ⋅ RA x ⋅ R'A x 290.775 ⋅ kN m ≔ M0 = - ⋅ ―― CD 8 L1 2 ⋅ 3.6 kN m 290.775 ⋅ kN m 4.) Diseño: 1er Tanteo Dominio 3lim ≔ α3_lim = ――― 3.5 + 3.5 εyd 0.668 ≔ d = - h ⋅ 0.08 h 36.8 cm ≔ β3_lim = - 1 ⋅ 0.4 α3_lim 0.733 ≔ μ3_lim = ⋅ ⋅ 0.8 α3_lim β3_lim 0.392 09/11/21 Página 2 de 5
  • 3. UNIVERSIDAD MAYOR DE SAN ANDRES FACULTAD DE INGENIERIA INGENIERIA CIVIL AUX. APAZA MACUCHAPI CESAR ISRAEL HORMIGÓN ARMADO I APLICACIÓN:ARMADURA SIMETRICA ≔ M3_lim = ⋅ ⋅ ⋅ μ3_lim fcd b d2 212.143 ⋅ kN m ≔ εs2 = ⋅ ――――― - α3_lim ――― 0.08 h d α3_lim 3.5 3.044 = εyd 1.739 ≔ z2 = - d 0.08 h 33.6 cm ≔ zc_3lim = - d 0.4 ⎛ ⎝ ⋅ α3_lim d⎞ ⎠ 0.27 m = = = As2 ―― Ns2 σs2 ――― δM ⋅ z2 σs2 ――――― - MAB M3_lim ⋅ z2 fyd ≔ As2 = ――――― - MAB M3_lim ⋅ z2 fyd 6.728 cm2 = = = As1 ―― Ns1 fyd ――― + Ns2 Nc fyd ―――――――― + ――――― - MAB M3_lim z2 ――― M3_lim zc_3lim fyd ≔ As1 = ―――――――― + ――――― - MAB M3_lim z2 ――― M3_lim zc_3lim fyd 29.346 cm2 2do Tanteo Dominio 2Blim ≔ α2B_lim = ― 7 27 0.259 ≔ β2B_lim = - 1 ⋅ 0.4 α2B_lim 0.896 ≔ μ2B_lim = ⋅ ⋅ 0.8 α2B_lim β2B_lim 0.186 ≔ M2B_lim = ⋅ ⋅ ⋅ μ2B_lim fcd b d2 100.7 ⋅ kN m = = = As2 ―― Ns2 σs2 ――― δM ⋅ z2 σs2 ――――― - MAB M2B_lim ⋅ z2 fyd ≔ As2 = ――――― - MAB M2B_lim ⋅ z2 fyd 16.264 cm2 ≔ εs2 = ⋅ ―――――― - α2B_lim ――― 0.08 h d α2B_lim 3.5 2.326 = εyd 1.739 ≔ zc_2Blim = - d 0.4 ⎛ ⎝ ⋅ α2B_lim d⎞ ⎠ 0.33 m = = = As1 ―― Ns1 fyd ――― + Ns2 Nc fyd ―――――――― + ――――― - MAB M2B_lim z2 ――― M2B_lim zc_2Blim fyd ≔ As1 = ―――――――― + ――――― - MAB M2B_lim z2 ――― M2B_lim zc_2Blim fyd 25.041 cm2 09/11/21 Página 3 de 5
  • 4. UNIVERSIDAD MAYOR DE SAN ANDRES FACULTAD DE INGENIERIA INGENIERIA CIVIL AUX. APAZA MACUCHAPI CESAR ISRAEL HORMIGÓN ARMADO I APLICACIÓN:ARMADURA SIMETRICA 3er Tanteo Dominio 2A lim ≔ α2A_lim = ― 1 6 0.167 ≔ β2A_lim = - 1 ⋅ 0.4 α2A_lim 0.933 ≔ μ2A_lim = ⋅ ⋅ 0.8 α2A_lim β2A_lim 0.124 ≔ M2A_lim = ⋅ ⋅ ⋅ μ2A_lim fcd b d2 67.411 ⋅ kN m ≔ εs2 = ⋅ ―――――― - α2A_lim ――― 0.08 h d α2A_lim 2 0.957 = εyd 1.739 ≔ σs2 = ⋅ Es ―― εs2 1000 191.304 MPa = = = As2 ―― Ns2 σs2 ――― δM ⋅ z2 σs2 ――――― - MAB M2A_lim ⋅ z2 σs2 ≔ As2 = ――――― - MAB M2A_lim ⋅ z2 σs2 34.75 cm2 ≔ zc_2Alim = - d 0.4 ⎛ ⎝ ⋅ α2A_lim d⎞ ⎠ 0.343 m = = = As1 ―― Ns1 fyd ――― + Ns2 Nc fyd ―――――――― + ――――― - MAB M2A_lim z2 ――― M2A_lim zc_2Alim fyd ≔ As1 = ―――――――― + ――――― - MAB M2A_lim z2 ――― M2A_lim zc_2Alim fyd 24.755 cm2 4to Tanteo Dominio 2B ≔ α2B 0.19098157 ≔ β2B = - 1 ⋅ 0.4 α2B 0.924 ≔ μ2B = ⋅ ⋅ 0.8 α2B β2B 0.141 ≔ M2B = ⋅ ⋅ ⋅ μ2B fcd b d2 76.441 ⋅ kN m ≔ εc = ⋅ ――― α2B - 1 α2B 10 2.361 ≔ εs2 = ⋅ ――――― - α2B ――― 0.08 h d α2B εc 1.286 = εyd 1.739 ≔ σs2 = ⋅ ―― εs2 1000 Es 257.164 MPa 09/11/21 Página 4 de 5
  • 5. UNIVERSIDAD MAYOR DE SAN ANDRES FACULTAD DE INGENIERIA INGENIERIA CIVIL AUX. APAZA MACUCHAPI CESAR ISRAEL HORMIGÓN ARMADO I APLICACIÓN:ARMADURA SIMETRICA ≔ zc = - d 0.4 ⎛ ⎝ ⋅ α2B d⎞ ⎠ 0.34 m ≔ As1 = ―――――― + ―――― - MAB M2B z2 ―― M2B zc fyd 24.805 cm2 ≔ As2 = ―――― - MAB M2B ⋅ z2 σs2 24.805 cm2 09/11/21 Página 5 de 5