Diseño de una zapata aislada

enseña a diseñar una zapata estructural

1. UNIVERSIDAD
NACIONAL DE JAÉN
CARRERA PROFESIONAL DE INGENIERÍA CIVIL

TEMAS

ZAPATA AISLADA
PREDIMENSIONAMIENTO DE VIGAS Y COLUMNAS

GRUPO N° 1:

ABAD FALLA, Dante Michael.
GUERRERO SÁNCHEZ, Anthony.
ESTELA HORNA, Ronald.
QUIÑONEZ ARÉVALO, Kevin.
VÁSQUES SÁNCHES, Andy.
VÁSQUES QUISPE, Vento.



DOCENTE:

ING. CÉSAR JESÚS DÍAZ CORONEL .



JAÉN – PERÚ
2016

2. DISEÑO DE UNA ZAPATA AISLADA

o Diseñar la zapata aislada
o Presentar las dimensiones
o Distribuir el acero correspondiente en ambos sentidos (parrilla)
o Graficarla

Datos













1. CÁLCULO DE ÁREA DE ZAPATA
Hay que encontrar el esfuerzo neto




??????�= 170 ��
??????�= 35 ��
�
�������
= 1700 ��/�
3

�
�
= 1.65 ��/��
2

��= 1.45 �
����������=400 ��/�
2

�= 0.40 �
�= 0.40 �
�� �= 8 ∅ 5/8"
�´�= 210 ��/��
2

��= 4200 ��/��
2

������= 3"
�
����
= �
�
− �
�������
×��−����������

3. Determinaremos el Área de la zapata requerida:






� ≠�; La columna es rectangular, entonces la zapata también será
rectangular.








2. CÁLCULO DEL PERALTE DE LA ZAPATA



�
����
= 1.65
��
��
2
− 1700
��
�
3
×1.45�−400
��
�
2

�
����
= 16.5
��
�
2
− 2.465
��
�
2
−0.4
��
�
2

�
����
= 13.635
��
�
2

�
��??????�??????�
=
??????
�
����
=
??????�+??????�
�
����

�
��??????�??????�=
(170+35) ��
13.635
��
�
2

�
��??????�??????�=15.03 �
2

�=
�
�
=
0.4
0.4
=1
B
A
t=0 .4 0 m

S =0 .4 0 m
n
m
�= √�
��??????�??????�+(
�−�
2
)= √15.03+(
0.4−0.4
2
)=3.88 �
∴ �=3.90 � ���??????�����
�= √�
��??????�??????�−(
�−�
2
)= √15.03−(
0.4−0.4
2
)=3.88 �
∴ �=3.90 � ���??????�����



�= �+
∅
2
+������??????�??????����

4. Dimensionamos la elevación H, ésta se halla cuando determinamos el
peralte efectivo “d”, mediante la verificación por:
i. LONGITUD DE DESARROLLO
ii. CORTANTE POR PUNZONAMIENTO
iii. CORTANTE POR FLEXIÓN

2.1. LONGITUD DE DESARROLLO POR COMPRESIÓN(cm) ??????�=�
�












2.2. EL ESFUERZO CORTANTE POR PUNZONAMIENTO, SE CALCULA
CON:

Calcularemos la reacción última del suelo (�
�)










�
�=
??????
�
(��)

Dónde:
�
�=�����??????ó� ��� �����
SEGÚN (RNE)
??????
�=1.5 ??????�+1.8 ??????�
??????
�=1.5(170)+1.8(35)
??????
�=318 ��
�
�=
318
(3.90×3.90)

�
�=20.91 ��/�
2

a)��=0.08 ×
�?????? ��
√�´�

��=
0.08 ×4200×
5
8
×2.54
√210
=36.81 �� ≅37 ��
b)��=0.004����
��=0.004×1.5875×4200=26.67 �� ≅27 ��
c)��=20 ��
∴��= �
1=37 ��
Tomamos
el mayor

5. 2.3. ESFUERZO CORTANTE POR FLEXIÓN









�
��������=�
�×
[�×�−(�+�)×(�+�)]
2�×(�+�+2�)

El que tendrá que ser menor o igual que el esfuerzo
cortante admisible
�
���??????�??????���=∅×0.27(2+
4
�
)×√�´�
�
���??????�??????���=0.85×1.1√�´�

�=
�
�
=
0.4
0.4
=1
∅=0.85
�
���??????�??????���=0.85×0.27 2+
4
1
×√210=19.95 ��/��
2

�
���??????�??????���=0.85×1.1√210=13.549 ��/��
2
=135.49 ��/�
2

Tomamos el
menor valor
Haciendo: �
���??????�??????���=�
��������
135.49=20.91×
[3.90×3.90−(0.40+�)×(0.40+�)]
2�×(0.40+0.40+2�)

∴ �
2=0.57� ≅57 ��
CORTANTE ACTUANTE:
�������� ��������=�
�=�
�(�−�)×�
ESFUERZO CORTANTE ACTUANTE:
�������� ��������= �
�=
�
�
��
=
�
�(�−�)×�
��
=
�
�(�−�
3)
�
3

Igualando:
�
�(�−�
3)
�
3
=�
�×
[�×�−(�+�
2)×(�+�
2)]
2�
2×(�+�+2�
2)

(1.75−�
3)
�
3
=
[3.90
2
−(0.40+0.57)
2
]
2×0.57×(0.80+2×0.57)

∴ �
3=0.23 � ≅23 ��
�=
�−�
2
=
3.90−0.40
3
=1.75 �
�=
�−�
2
=
3.90−0.40
3
=1.75 �

6. 3. CÁLCULO DEL ACERO DE LA PARRILLA





3.1. PRIMERA FORMA DE DISEÑO














De los tres peraltes �
1,�
2 � �
3 se escoge el mayor:
�
1=37 ��
�
2=57 �� ∴�=57��
�
3=23 ��
Hallamos el peralte de la zapata: H
�= �+
∅
2
+������??????�??????���� ����??????���� ∅=3/4"
�= 57 ��+
1.905 ��
2
+7.5��=65.45 ��
∴�=70 �� ���??????�����

�
�=
20.91×1.75
2
×3.90
2

�
�=124.87 ��.�
Según Morales Morales se
toma la longitud transversal
de la zapata mas no A
�
�=
�
��
2
�
2

�
�=
�
�
��
2
=
124.87×10
5
390×57
2
=9.85 ��/��
2

 ??????=??????����í� �� �����
�
�=∅×??????×��(1−0.59×
??????��
�´�
)
9.85=0.90×??????×4200(1−
0.59×??????×4200
210
)
9.85=0.90×??????×4200(1−
0.59×??????×4200
210
)
9.85=3780×??????−44604×??????
2

44604×??????
2
−3780×??????+9.85=0
∴ ??????=0.27%

7.  AREA DEL ACERO : As



Asumo:
∅ 3/4" = 2.84 ��
2

 NÚMERO DE VARILLAS
# ��������=
��.��
�.��
=��
# ��������= ??????� ����á �� ��������

 SEPARACIÓN
�=
(�−��−2×������??????�??????����)
(# ���??????����−1)

�=
(390−1.905−7.5×2)
(27−1)

�= 14 ��
3.2. SEGUNADA FORMA DE DISEÑO (iterando)




Según el código ACI

0.8√�´�
�??????
=0.00276 ∴ ??????
�í�=0.34 %
??????
&#3627408474;í&#3627408475;??????&#3627408474;&#3627408462;= ??????<??????
&#3627408474;í&#3627408475;

14.1
&#3627408467;??????
=0.0034 ∴ ??????=??????
&#3627408474;í&#3627408475;=0.34 %

&#3627408436;
&#3627408480;=(0.34 %)(390)(57)
&#3627408436;
&#3627408480;=75.58 &#3627408464;&#3627408474;
2

 MECHA
&#3627408448;=4.5475 &#3627408464;&#3627408474;

⟹&#3627409360;&#3627409365; ∅ &#3627409361;/&#3627409362;" @ &#3627409358;.&#3627409359;&#3627409362; &#3627408526;
&#3627408448;
&#3627408482;=124.87 &#3627408481;&#3627408475;.&#3627408474;
&#3627408436;
&#3627408480;=
&#3627408448;
&#3627408482;
∅×&#3627408467;&#3627408486;×(&#3627408465;−
&#3627408462;
2
)

&#3627408462;=
&#3627408436;
&#3627408480;×&#3627408467;&#3627408486;
0.85×&#3627408467;´&#3627408464;×&#3627408463;

∴ &#3627408436;
&#3627408480;=59.85 &#3627408464;&#3627408474;
2
⟹&#3627408462;=3.61 &#3627408464;&#3627408474;

8. 4. DIMENSIONAMIENTO DE LA ZAPATA














Según el código ACI

0.8 × √&#3627408467;´&#3627408464; × &#3627408463; × &#3627408465;
&#3627408467;??????
=61.36 &#3627408464;&#3627408474;
2
∴ &#3627408436;&#3627408480;
&#3627408474;í&#3627408475;=74.63 &#3627408464;&#3627408474;
2

&#3627408436;&#3627408480;
&#3627408474;í&#3627408475;??????&#3627408474;&#3627408462;= &#3627408436;&#3627408480;<&#3627408436;&#3627408480;
&#3627408474;í&#3627408475;

14.1 × &#3627408463; ×&#3627408465;
&#3627408467;??????
=74.63 &#3627408464;&#3627408474;
2
∴ &#3627408436;&#3627408480;=&#3627408436;&#3627408480;
&#3627408474;í&#3627408475;=74.63 &#3627408464;&#3627408474;
2


# &#3627408535;&#3627408514;&#3627408531;&#3627408522;&#3627408525;&#3627408525;&#3627408514;&#3627408532;=&#3627409360;&#3627409365; &#3627408535;&#3627408514;&#3627408531;&#3627408522;&#3627408525;&#3627408525;&#3627408514;&#3627408532; ∅=&#3627409361;/&#3627409362;"
??????=&#3627409359;&#3627409362; &#3627408516;&#3627408526;
??????&#3627408518;&#3627408516;&#3627408521;&#3627408514;=&#3627409362;.&#3627409363;&#3627409362;&#3627409365;&#3627409363;

⟹ ∅ / " @ .

10. PREDIMENSIONAMIENTO DE VIGAS Y COLUMNAS

PREDIMENSIONAMIENTO DE COLUMNAS

En una edificación que presenta un sistema aporticado con 6 pisos, cuya planta
típica es la que se presenta. Se solicita predimensionar las columnas C1, C2 y
C3. Cualquier dato(s) faltante(s) asúmalo e indique su criterio para dicho fin.











Datos:
&#3627408449;=&#3627408449;ú&#3627408474;&#3627408466;&#3627408479;&#3627408476; &#3627408465;&#3627408466; &#3627408477;??????&#3627408480;&#3627408476;&#3627408480; = 6
&#3627408447;&#3627408476;&#3627408480;&#3627408462; &#3627408462;&#3627408473;??????&#3627408468;&#3627408466;&#3627408479;&#3627408462;&#3627408465;&#3627408462; = 20
&#3627408436;&#3627408464;&#3627408462;&#3627408463;&#3627408462;&#3627408465;&#3627408476; = 100 &#3627408472;&#3627408468;/&#3627408474;
2
&#3627408441;´&#3627408464; =280 &#3627408472;&#3627408468;/&#3627408464;&#3627408474;
2
&#3627408441;&#3627408486; = 4200 &#3627408472;&#3627408468;/&#3627408464;&#3627408474;
2
&#3627408456;&#3627408480;&#3627408476; = &#3627408439;&#3627408466;&#3627408477;&#3627408462;&#3627408479;&#3627408481;&#3627408462;&#3627408474;&#3627408466;&#3627408475;&#3627408481;&#3627408476;

7.00 m
7.40 m
7.20 m
5.80 m 6.20 m 5.80 m

11. Metrado de Cargas:
??????.&#3627408462;&#3627408473;??????&#3627408468;&#3627408466;&#3627408479;&#3627408462;&#3627408465;&#3627408476; = 300 &#3627408472;&#3627408468;/&#3627408474;
2
&#3627408455;&#3627408462;&#3627408463;??????&#3627408478;&#3627408482;&#3627408466;&#3627408479;í&#3627408462; = 120 &#3627408472;&#3627408468;/&#3627408474;
2
(??????&#3627408532;&#3627408534;&#3627408526;&#3627408522;&#3627408526;&#3627408528;&#3627408532;)
&#3627408436;&#3627408464;&#3627408462;&#3627408463;&#3627408462;&#3627408465;&#3627408476; = 100 &#3627408472;&#3627408468;/&#3627408474;
2
??????&#3627408466;&#3627408480;&#3627408476; &#3627408465;&#3627408466; &#3627408483;??????&#3627408468;&#3627408462;&#3627408480; = 100 &#3627408472;&#3627408468;/&#3627408474;
2
(??????&#3627408532;&#3627408534;&#3627408526;&#3627408522;&#3627408526;&#3627408528;&#3627408532;)
??????&#3627408466;&#3627408480;&#3627408476; &#3627408465;&#3627408466; &#3627408464;&#3627408476;&#3627408473;&#3627408482;&#3627408474;&#3627408475;&#3627408462;&#3627408480; = 60 &#3627408472;&#3627408468;/&#3627408474;
2
(??????&#3627408532;&#3627408534;&#3627408526;&#3627408522;&#3627408526;&#3627408528;&#3627408532;)
&#3627408454;/&#3627408438; = 250 &#3627408472;&#3627408468;/&#3627408474;
2
??????.&#3627408474;&#3627408482;&#3627408466;&#3627408479;&#3627408481;&#3627408462; = 680 &#3627408472;&#3627408468;/&#3627408474;
2
??????.&#3627408483;??????&#3627408483;&#3627408462; = 250 &#3627408472;&#3627408468;/&#3627408474;
2

??????&#3627408442; = ??????.&#3627408474;&#3627408482;&#3627408466;&#3627408479;&#3627408481;&#3627408462; + ??????.&#3627408483;??????&#3627408483;&#3627408462; = 680 +250 = 930 &#3627408472;&#3627408468;/&#3627408474;
2

Columna C3 (interior)
Área tributaria
&#3627408436; =
(7+7.20)
2
×
(5.30+6.20)
2
= 42.6 &#3627408474;
2
?????? = 930 &#3627408472;&#3627408468;/&#3627408474;
2
∗ 42.6&#3627408474;
2
= 39618 &#3627408472;&#3627408468;

&#3627408463;∗ &#3627408439; =
1.10 × ??????
&#3627408475; × &#3627408467;′&#3627408464;
&#3627408475;=0.30
Considerando que &#3627408463; = &#3627408439; = &#3627408455;
&#3627408463; .&#3627408439; =
1.10 × 39618 × 6
0.30 ×280

&#3627408455; = 55.79 cm por lo tanto usamos &#3627408455; = 60&#3627408464;&#3627408474;
∴&#3627408438;3∶ 0.60 ×0.60 &#3627408474;
2

Columna C1
Área tributaria

12. &#3627408436; =
(5.80)
2
×
(7.00)
2
= 10.15 &#3627408474;
2
?????? = 930 &#3627408472;&#3627408468;/&#3627408474;
2
∗ 10.15&#3627408474;
2
= 9439.5 &#3627408472;&#3627408468;
&#3627408463;∗&#3627408439; =
1.50∗??????
&#3627408475;∗&#3627408467;´&#3627408464;
&#3627408475;=0.20
Considerando que &#3627408463; = &#3627408439; = &#3627408455;
&#3627408463; .&#3627408439; =
1.50 × 9439.5 × 6
0.20 ×280

&#3627408455; = 38.9 &#3627408464;&#3627408474; por lo tanto usamos &#3627408455; = 40&#3627408464;&#3627408474;
∴&#3627408438;1=0.40 ×0.40 &#3627408474;
2

Columna C2
Área tributaria
&#3627408436; =
(7+7.20)
2
×
(5.80)
2
= 20.59 &#3627408474;
2

?????? = 930 &#3627408472;&#3627408468;/&#3627408474;
2
∗ 20.59&#3627408474;
2
= 19148.7 &#3627408472;&#3627408468;
&#3627408463;∗&#3627408439; =
1.25∗??????
&#3627408475;∗&#3627408467;´&#3627408464;
&#3627408475;=0.25
Considerando que &#3627408463; = &#3627408439; = &#3627408455;
&#3627408463; .&#3627408439; =
1.25 × 19148.7 × 6
0.25 ×280

&#3627408455; = 45.30 &#3627408464;&#3627408474; por lo tanto usamos &#3627408455; = 50&#3627408464;&#3627408474;
∴&#3627408438;2=0.50 ×0.50 m
2

13. PREDIMENSIONAMIENTO DE VIGAS













&#3627408535;−&#3627409359;&#3627409358;&#3627409359;
&#3627408463;=
&#3627408437;
20
=
2.2
20
=0.11 0.5∗0.11=0.25&#3627408485;
ℎ=
??????&#3627408475;
??????
=
5.5
11
=0.5 &#3627408485;=0.22

&#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.25
Usando el criterio de igualdad de rigidez &#3627408463;
0=0.25
 Zona de alto riesgo sísmico

&#3627408463;ℎ
3
=&#3627408463;
0ℎ
0
3

ℎ=0.5 &#3627408463;=0.11
&#3627408463;
0=0.25

0.11∗0.5
3
=0.25ℎ
0
3

ℎ
0=0.38
4.40 m
4.40 m
4.90 m
5.50 m
5.90 m 5.50 m 1.80 m

14. &#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.40&#3627408474;
2

Usando criterios de igualdad de cuantía
 Zona mediamente sísmica

&#3627408463;ℎ
2
=&#3627408463;
0ℎ
0
2


&#3627408463;=0.11 0.11∗0.5
2
=0.25ℎ
0
2

ℎ=0.5 ℎ
0=0.33
&#3627408463;
0=0.25
&#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.35&#3627408474;
2


Volado V-101  &#3627408447;&#3627408483;=1.6&#3627408474;
&#3627408463;=
&#3627408437;
20
=
2.2
20
=0.11
ℎ=1.4ℎ
&#3627408480;=1.4
??????&#3627408475;
??????
=1.4
2∗1.6
10
=0.448
0.11∗0.448=0.25∗ℎ
ℎ=0.20
&#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.25&#3627408474;
2


&#3627408535;−&#3627409359;&#3627409358;&#3627409360;−&#3627409359;&#3627409358;&#3627409361;
&#3627408463;=
&#3627408437;
20
=
4.65
20
=0.2325
ℎ=
??????&#3627408475;
??????
=
5.5
11
=0.50
&#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.50&#3627408474;
2

Usando el criterio de igualdad de rigidez &#3627408463;
0=0.25
 Zona de alto sísmico

&#3627408463;ℎ
3
=&#3627408463;
0ℎ
0
3

&#3627408463;=0.2325
ℎ=0.50 0.2325∗0.50
3
=0.25ℎ
0
3

&#3627408463;
0=0.25
ℎ
0=0.49&#3627408474;
Asumimos 10 porque
la S/C = 400kg/&#3627408474;
2

15. &#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.50&#3627408474;
2

Usando criterios de igualdad de cuantía
 Zona mediamente sísmica

&#3627408463;ℎ
2
=&#3627408463;
0ℎ
0
2

&#3627408463;=0.2325
ℎ=0.50 0.2325∗0.50
2
=0.25ℎ
0
2

&#3627408463;
0=0.25
ℎ
0=0.48
&#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.50&#3627408474;
2


Volado V-102-103  &#3627408447;&#3627408483;=1.6&#3627408474;
&#3627408463;=
&#3627408437;
20
=
4.65
20
=0.2325
ℎ=1.4ℎ
&#3627408480;=1.4
??????&#3627408475;
??????
=1.4
2∗1.6
10
=0.448
&#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.45&#3627408474;
2




V-104

&#3627408463;=
&#3627408437;
∗
20
&#3627408437;
∗
=&#3627408437;+&#3627408437;&#3627408462;&#3627408465;
ℎ=
??????&#3627408475;
??????
&#3627408437;&#3627408462;&#3627408465;=
??????&#3627408482;
&#3627408484;&#3627408482;
=
??????
&#3627408484;


&#3627408458;=(??????.??????.&#3627408462;&#3627408473;??????&#3627408468;)+(??????.&#3627408462;&#3627408464;&#3627408462;&#3627408463;)+(??????.&#3627408481;&#3627408462;&#3627408463;)+(
&#3627408480;
&#3627408464;
⁄)
P.P.aligerado de 20&#3627408464;&#3627408474;=300 &#3627408472;&#3627408468;/&#3627408474;
2

??????.&#3627408462;&#3627408464;&#3627408462;&#3627408463;&#3627408462;&#3627408465;&#3627408476; =100&#3627408472;&#3627408468;/&#3627408474;
2

??????.&#3627408481;&#3627408462;&#3627408463;??????&#3627408478;&#3627408482;&#3627408466;&#3627408479;??????&#3627408462;=
50&#3627408472;&#3627408468;
&#3627408474;
2
….(&#3627408514;&#3627408532;&#3627408534;&#3627408526;&#3627408522;&#3627408517;&#3627408528;)
&#3627408480;
&#3627408464;
⁄=200&#3627408472;&#3627408468;/&#3627408474;
2

Asumimos 10 porque
la S/C = 400kg/&#3627408474;
2

Las dimensiones de la viga V-103 Y V-104 y sus respectivos volados
van hacer iguales debido a que tienen el mismo ancho tributario y
longitud libre

16. &#3627408458;=300+100+50+20=650&#3627408472;&#3627408468;/&#3627408474;
2

??????=0.25∗2.3∗1800=
1035&#3627408472;&#3627408468;
&#3627408474;

1800&#3627408472;&#3627408468;
&#3627408474;
3
….(&#3627408529;&#3627408518;&#3627408532;&#3627408528; &#3627408534;&#3627408527;&#3627408522;&#3627408533;&#3627408514;&#3627408531;&#3627408522;&#3627408528; &#3627408517;&#3627408518; &#3627408514;&#3627408525;&#3627408515;&#3627408514;ñ&#3627408522;&#3627408525;&#3627408518;&#3627408531;&#3627408522;&#3627408514; &#3627408517;&#3627408518; &#3627408514;&#3627408531;&#3627408516;&#3627408522;&#3627408525;&#3627408525;&#3627408514; &#3627408516;&#3627408528;&#3627408516;&#3627408522;&#3627408517;&#3627408514; &#3627408532;&#3627408528;&#3627408525;&#3627408522;&#3627408517;&#3627408514;)
&#3627408437;=
4.4
2
=2.2&#3627408474;
&#3627408437;&#3627408462;&#3627408465;=
1035
650
=1.592&#3627408474;
&#3627408437;
∗
=2.2+1.592=3.792

&#3627408463;=
3.792
20
=0.1896&#3627408474; 0.1896∗0.5=0.25∗&#3627408485;
ℎ=
5.5
11
=0.5&#3627408474; &#3627408485;=0.3792&#3627408474;

&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.40 &#3627408474;
2

Usando el criterio de igualdad de rigidez &#3627408463;
0=0.25
 &#3627408436;&#3627408473;&#3627408481;&#3627408476; &#3627408479;??????&#3627408466;&#3627408480;&#3627408468;&#3627408476; &#3627408480;??????&#3627408480;&#3627408474;??????&#3627408464;&#3627408476;
&#3627408463;∗ℎ
3
=&#3627408463;
0∗ℎ
0
3

0.1896∗0.5
3
=0.25∗ℎ
0
3

ℎ
0=0.46&#3627408474;
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.5&#3627408474;
2


Usando criterios de igualdad de cuantía
 Zona mediamente sísmica

&#3627408463;ℎ
2
=&#3627408463;
0ℎ
0
2

&#3627408463;=0.1896
ℎ=0.50 0.1896∗0.50
2
=0.25ℎ
0
2

&#3627408463;
0=0.25
ℎ
0=0.44

17. &#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.45&#3627408474;
2

Volado V-104  &#3627408447;&#3627408483;=1.6&#3627408474;









V-105
&#3627408463;=
&#3627408437;
20
=
4.55
20
=0.2275
ℎ=
&#3627408447;&#3627408475;
&#3627409148;
=
4.5
11
=0.41
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.45 &#3627408474;
2

Usando el criterio de igualdad de rigidez &#3627408463;
0=0.25

 &#3627408436;&#3627408473;&#3627408481;&#3627408476; &#3627408479;??????&#3627408466;&#3627408480;&#3627408468;&#3627408476; &#3627408480;??????&#3627408480;&#3627408474;??????&#3627408464;&#3627408476; &#3627408463;
0=0.25&#3627408464;&#3627408474;
&#3627408463;∗ℎ
3
=&#3627408463;
0∗ℎ
0
3

0.2275∗0.41
3
=0.25∗ℎ
0
3

ℎ
0=0.40&#3627408474;
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.40&#3627408474;
2


Usando criterios de igualdad de cuantía

 &#3627408487;&#3627408476;&#3627408475;&#3627408462; &#3627408474;&#3627408466;&#3627408465;??????&#3627408462;&#3627408475;&#3627408462;&#3627408474;&#3627408466;&#3627408475;&#3627408481;&#3627408466; &#3627408480;??????&#3627408480;&#3627408474;??????&#3627408464;&#3627408462; &#3627408463;
0=0.25&#3627408464;&#3627408474;
&#3627408463;=
&#3627408437;
∗
20
&#3627408437;
∗
=&#3627408437;+&#3627408437;
&#3627408462;&#3627408465;??????&#3627408464;??????&#3627408476;&#3627408475;&#3627408462;&#3627408473;
&#3627408437;
&#3627408462;&#3627408465;??????&#3627408464;??????&#3627408476;&#3627408475;&#3627408462;&#3627408473;=
??????
??????
=
1035
650
=1.592 &#3627408474;
&#3627408437;
∗
=2.20+1.592=3.792 &#3627408474;
&#3627408463;=
3.792
20
=0.1896 &#3627408474;
ℎ
&#3627408480;=1.4 ℎ=1.4 (
&#3627408447;
&#3627408475;
&#3627409148;
)=1.4(
2×&#3627408447;
&#3627408483;
&#3627409148;
)=
1.4×2×1.6
10

ℎ
&#3627408480;=0.448
0.1896×0.448=0.25×ℎ
ℎ=0.34
&#3627408456;&#3627408480;&#3627408462;&#3627408479;=0.25∗0.35&#3627408474;
2

18. &#3627408463;∗ℎ
2
=&#3627408463;
0∗ℎ
0
2

0.2275∗0.41
2
=0.25∗ℎ
0
2

ℎ
0=0.39&#3627408474;
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.40&#3627408474;
2

V-106-107
&#3627408463;=
&#3627408437;
20
=
5.7
20
=0.285
ℎ=
&#3627408447;&#3627408475;
&#3627409148;
=
4.5
11
=0.41
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.30∗0.45 &#3627408474;
2


Usando el criterio de igualdad de rigidez &#3627408463;
0=0.25
 &#3627408461;&#3627408476;&#3627408475;&#3627408462;&#3627408480; &#3627408465;&#3627408466; &#3627408462;&#3627408473;&#3627408481;&#3627408476; &#3627408479;??????&#3627408466;&#3627408480;&#3627408468;&#3627408476; &#3627408480;??????&#3627408480;&#3627408474;??????&#3627408464;&#3627408476;
&#3627408463;∗ℎ
3
=&#3627408463;
0∗ℎ
0
3

0.285∗0.41
3
=0.25∗ℎ
0
3

ℎ
0=0.43&#3627408474;
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.45&#3627408474;
2

Usando criterios de igualdad de cuantía

 &#3627408487;&#3627408476;&#3627408475;&#3627408462; &#3627408474;&#3627408466;&#3627408465;??????&#3627408462;&#3627408475;&#3627408462;&#3627408474;&#3627408466;&#3627408475;&#3627408481;&#3627408466; &#3627408480;??????&#3627408480;&#3627408474;??????&#3627408464;&#3627408462; &#3627408463;
0=0.25&#3627408464;&#3627408474;
&#3627408463;∗ℎ
2
=&#3627408463;
0∗ℎ
0
2

0.285∗0.41
2
=0.25∗ℎ
0
2

ℎ
0=0.44&#3627408474;
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.45&#3627408474;
2

V-108
&#3627408463;=
&#3627408437;
20
=
2.75
20
=0.1375 0.1375∗0.41=0.25∗&#3627408485;
ℎ=
&#3627408447;&#3627408475;
&#3627409148;
=
4.5
11
=0.41 &#3627408485;=0.23 &#3627408474;

19. &#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.25 &#3627408474;
2


Usando el criterio de igualdad de rigidez

 &#3627408436;&#3627408473;&#3627408481;&#3627408476; &#3627408479;??????&#3627408466;&#3627408480;&#3627408468;&#3627408476; &#3627408480;??????&#3627408480;&#3627408474;??????&#3627408464;&#3627408476; &#3627408463;
0=0.25&#3627408464;&#3627408474;
&#3627408463;∗ℎ
3
=&#3627408463;
0∗ℎ
0
3

0.1375∗0.41
3
=0.25∗ℎ
0
3

ℎ
0=0.34&#3627408474;
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.35&#3627408474;
2

Usando criterios de igualdad de cuantía

 &#3627408487;&#3627408476;&#3627408475;&#3627408462; &#3627408474;&#3627408466;&#3627408465;??????&#3627408462;&#3627408475;&#3627408462;&#3627408474;&#3627408466;&#3627408475;&#3627408481;&#3627408466; &#3627408480;??????&#3627408480;&#3627408474;??????&#3627408464;&#3627408462; &#3627408463;
0=0.25&#3627408464;&#3627408474;
&#3627408463;∗ℎ
2
=&#3627408463;
0∗ℎ
0
2

0.1375∗0.41
2
=0.25∗ℎ
0
2

ℎ
0=0.30&#3627408474;
&#3627408482;&#3627408480;&#3627408462;&#3627408479; 0.25∗0.30&#3627408474;
2