Ejercicios resueltos por método de tres momentos (resistencia de materiales)

EJERCICIOS RESUELTOS POR EL MÉTODO DE TRES MOMENTOS

1. Universidad Nacional San Cristóbal de Huamanga
Facultad de Ingeníeria Minas , Geología y Civil
Escuela de Ingeniería Civil
PROBLEMAS RESUELTOS DE
RESISTENCIA DE MATERIALES
I - II
Autor:
Calderón Quispe, Gilmer
( [email protected],[email protected] )
Estudiante de Ingeniería Civil Gilmer Calderón Quispe

2. Capítulo
3
Método de tres Momentos
3.1 DeniciónCG CG
L
1 L
2
I
1 I
2
A B C
m
1 n
1
m
2 n
2
MA

L1
I1

2MB

L1
I1

L2
I2

MC

L2
I2

6

A1m1
L1I1

A2n2
L2I2

6E

BA
L1

BC
L2

Ecuación 1Ec. los 3 momentos
Para em marco mostrado en la gura, por el método de tres momentos calcular
1.
2.
Problema N°1
5 Gilmer Calderón Quispe

3. Ingeniería Civil 3 Método de tres Momentos
3.2I
6m 2m
2m
2m
2000kgf/m
2000kgf-m
4000kgf
2m
3I
2I
h=9000
1
A B
C
D
A B C D
-
+
-
+
+
4877.934
7122.066
2873.239
7122.066
2285.211
5633.803
6732.394
2986.258
2760.563
1619.718
1809.859
2760.563
968.258
+
-
-
+
ejercio N° 1
continuacion Ejercicio 1
V
M
(Kgf.m)
(Kgf)
2I
3m 2m
2m
2m
2000kgf/m
2000kgf-m
4000kgf
2m
3I
2I
A B C
D
4877.934 Kgf 9995.305 Kgf
2285.211 Kgf
2873.239 Kgf
1714.789 Kgf
1619.718 Kgf.m
D
A1
2
3
lh36000A2A31000A48000
De la gura (a) aplicando la ecuación de los tres momentos
Solución:2I
6m 2m
2m
2m
2000kgf/m
2000kgf-m
4000kgf
2m
3I
2I
h=9000
1
A B
C
D
A B C D
-
+
-
+
+
4877.934
7122.066
2873.239
7122.066
2285.211
5633.803
6732.394
2986.258
2760.563
1619.718
1809.859
2760.563
968.258
+
-
-
+
ejercio N° 1
continuacion Ejercicio 1
V
M
(Kgf.m)
(Kgf)
2I
3m 2m
2m
2m
2000kgf/m
2000kgf-m
4000kgf
2m
3I
2I
A B C
D
4877.934 Kgf 9995.305 Kgf
2285.211 Kgf
2873.239 Kgf
1714.789 Kgf
1619.718 Kgf.m
D
MA

6
2

2MB

6
2

4
3

MC

4
3

6

36000p3q
62

1000
43

1
3
p2q 2


1000
43

2
3
p2q

MB

4
3

2MC

4
3

4
2

MD

4
2

6

1000
43

2
3
p2q


1000
43

1
3
p2q 2


8000
24
p2q

MC

4
2

2MD

4
2
0

MEp0q 6

8000
24
p2q

Resistencia de Materiales I-II
pagina 6 Gilmer Calderón Quispe

4. Ingeniería Civil 3 Método de tres Momentos
Para:MAME0
8:667MB1:333MC 54666:667 (I)
1:333MB6:667MC2MD12666:667 (II)
2MC4MD 12000 (III)
Resolviendo (I) ,(II)y (III)
MB 6732:394Kgm M C2760:563Kgm MD1619:718Kgm
Hallando las cortantes isostáticas
AB:
"
VAB6000
VBA 6000
BC:
"
VBC500
VCB500
CD:
"
VCD 2000
VDC2000
Factor de corrección de las cortantes
C1
0 p6732:394q
6
1122:066C2
6732:3942760:563
4
2373:239
C3
2760:563 p1619:781q
4
285:211
Hallando las cortantes nales
VVisotC
VAB60001122:0664877:934Kgf V BA 60001122:066 7122:066Kgf
VBC5002373:2392873:239Kgf V CB5002373:2392873:239Kgf
VCD 2000285:211 2285:211Kgf VDC2000285:2111714:789Kgf
Hallando momento ector máximo(Donde la cortante es cero)
enx6mñMmaxpq6732:394Kgm2I
3m 2m
2m
2m
2000kgf/m
2000kgf-m
4000kgf
2m
3I
2I
h=9000
1
A B
C
D
A B C D
-
+
-
+
+
4877.934
7122.066
2873.239
7122.066
2285.211
5633.803
6732.394
2986.258
2760.563
1619.718
1809.859
2760.563
968.258
+
-
-
+
ejercio N° 1
continuacion Ejercicio 1
V
M
(Kgf.m)
(Kgf)
Resistencia de Materiales I-II
pagina 7 Gilmer Calderón Quispe

5. Ingeniería Civil 3 Método de tres Momentos2I
3m 2m
2m
2m
2000kgf/m
2000kgf-m
4000kgf
2m
3I
2I
h =9000
1
A B
C
D
A B C D
-
+
-
+
+
4877.934
7122.066
2873.239
7122.066
2285.211
5633.803
6732.394
2986.258
2760.563
1619.718
1809.859
2760.563
968.258
+
-
-
+
ejercio N° 1
continuacion Ejercicio 1
V
M
(Kgf.m)
(Kgf) 2I
3m 2m
2m
2m
2000kgf/m
2000kgf-m
4000kgf
2m
3I
2I
h=9000
1
A B
C
D
A B C D
-
+
-
+
+
4877.934
7122.066
2873.239
7122.066
2285.211
5633.803
6732.394
2986.258
2760.563
1619.718
1809.859
2760.563
968.258
+
-
-
+
ejercio N° 1
continuacion Ejercicio 1
V
M
(Kgf.m)
(Kgf)
2I
3m 2m
2m
2m
2000kgf/m
2000kgf-m
4000kgf
2m
3I
2I
A B
C
D
4877.934 Kgf 9995.305 Kgf
2285.211 Kgf
2873.239 Kgf
1714.789 Kgf
1619.718 Kgf.m
MAB
2EI
30
p2AB0q 266:667íMAB0:0667EIB266:667
MAB
2EI
30
p2AB0q 266:667
Resolver la viga sabiendo que el apoyo B sufrió un asentamiento de12mm, considerar
I80x10
7
mm
4
yE200KN{mm
2Problema N°2Resistencia de Materiales I-II
pagina 8 Gilmer Calderón Quispe

6. Ingeniería Civil 3 Método de tres Momentos80KN
60KN
20KN/m
26KN/m
3m 3m 4m 2m 3m
3m 2m 7/8m
A B C D
305.295
245.295
165.295
105.295
150.842
254.842
100.704
40.704
743.923
487.847
328.521
122.113
+
-
+
-
+
-
M
P
W
aw
W
wb
x
y
x
y
M
P
a
b
c
d
e
f
p
q
A
C D B
2m 2m 2m 2m
26KN/m
2KN/m 2KN/m
5KN-m 5KN-m
A
C D B
2m 2m 2m 2m
2KN/m 2KN/m
5KN-m 5KN-m
4KN 4KN
80KN
60KN
20KN/m
26KN/m
A
B
C D
A1360 ;A2
2
3
p6q p90q 360 ;A3138:667 ;A4180Solución:80KN
60KN
20KN/m
26KN/m
3m 3m 4m 2m 3m
3m 2m 7/8m
A B C D
305.295
245.295
165.295
105.295
150.842
254.842
100.704
40.704
743.923
487.847
328.521
122.113
+
-
+
-
+
-
M
P
W
aw
W
wb
x
y
x
y
M
P
a
b
c
d
e
f
p
q
A
C D B
2m 2m 2m 2m
26KN/m
2KN/m 2KN/m
5KN-m 5KN-m
A
C D B
2m 2m 2m 2m
2KN/m 2KN/m
5KN-m 5KN-m
4KN 4KN
80KN
60KN
20KN/m
26KN/m
A
B
C D
M0p0q 2MA

0
6
2I

MB

6
2I

6

0
360p3q
62I

360p3q
62I

6E

0
1210
3
6

6MA3MB 6

360p3q
62

360p3q
62

6EI

1210
3
6

6MA3MB 3000 :::::::::::::::::::::::::::::::::::::::::::::(I)
MA

6
2I

2MB

6
2I

4
I

MC

4
I

6

360p3q
62I

360p3q
62I

138:667p2q
4I


6E

1210
3
6

1210
3
4

3MA14MB4MC3304::::::::::::::::::::::::::::::::::::::::::::pIIq
MB

4
I

2MC

4
I

5
I

MD

5
I

6

138:667p2q
4I

180p8{3q
5I

6E

1210
3
4

4MB18MC 3872 ::::::::::::::::::::::::::::::::::::::::::::::::pIIIq
Resolviendo las ecuacionespIq;pIIqypIIIq
MA 743:923KN:m M B487:847KN:m M C 323:521KN:m
Hallando las cortantes isostáticas
AB:
#
VAB100
VBA 100
BC:
#
VBC52
VCB 52
CD:
#
VCD36
VDC 24
pKNq
Resistencia de Materiales I-II
pagina 9 Gilmer Calderón Quispe

7. Ingeniería Civil 3 Método de tres Momentos
Hallando las correcciones
Ci
MizqMder
L
C1 205:295KN C2202:842KN C3 64:704KN
Cortantes nales
VAB305:295KN VBC 150:842KN VCD100:704KN
VBA105:295KN VCB 254:842KN VDC40:704KN80KN
60KN
20KN/m
26KN/m
3m 3m 4m 2m 3m
3m 2m 7/8m
A B C D
305.295
245.295
165.295
105.295
150.842
254.842
100.704
40.704
743.923
487.847
328.521
122.113
+
-
+
-
+
-
M
P
W
aw
W
wb
x
y
x
y
M
P
a
b
c
d
e
f
p
q
A
C D B
2m 2m 2m 2m
26KN/m
2KN/m 2KN/m
5KN-m 5KN-m
A
C D B
2m 2m 2m 2m
2KN/m 2KN/m
5KN-m 5KN-m
4KN 4KN
80KN
60KN
20KN/m
26KN/m
A
B
C D
M
(KN.m)
V
(KN)
Resolver la viga mostrada en la que los asentamientos de los apoyos son enA10mm
,enC65mm, enE40mmy enG25mm E200Gpa,I500x10
6
mm
4Problema N°3Resistencia de Materiales I-II
pagina 10 Gilmer Calderón Quispe

8. Ingeniería Civil 3 Método de tres Momentos10
15.536
19.464
21.698
18.302
50
89.286
55.357
105.357
27.678
175 100
10ft
-
+
-
-
-
+ +
-
-
-
+
+
V
(K)
M
(K.ft)
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
A
1 A
2
A
3 A
4
A
5
288 288
300
6m 4m 6m 4m 4m 4m
A C E
G
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
45.877KN 100.504KN 198.296KN 45.323KN
45.877
26.381
104.677
45.323
93.619
74.123
21.230
237.417
275.262
137.289
181.291
300
288
288
+
+
+
-
-
-
-
+
+ +
V
(KN)
M
(KN.m)
50KN 50KN
30KN
A
B C
2m 3m 2m
1.5m
2.5m
Viga: 300mmx400mm
colum: 300mmx500mm
2
E=20KN/mm
A
B
B
C
C
30KN
1.5m
2.5m
50KN 50KN
30KN
A
B C
2m 3m 2m
1.5m
2.5m
56.7KN
121.97KN.m30KN
43.3KN
A1
1
2
p6q p288q 864 ;A2576 ;A3864 ;A4576 ;A51200 ;Solución:10
15.536
19.464
21.698
18.302
50
89.286
55.357
105.357
27.678
175 100
10ft
-
+
-
-
-
+ +
-
-
-
+
+
V
(K)
M
(K.ft)
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
A
1 A
2
A
3 A
4
A
5
288 288
300
6m 4m 6m 4m 4m 4m
A C E
G
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
45.877KN 100.504KN 198.296KN 45.323KN
45.877
26.381
104.677
45.323
93.619
74.123
21.230
237.417
275.262
137.289
181.291
300
288
288
+
+
+
-
-
-
-
+
+ +
V
(KN)
M
(KN.m)
50KN 50KN
30KN
A
B C
2m 3m 2m
1.5m
2.5m
Viga: 300mmx400mm
colum: 300mmx500mm
2
E=20KN/mm
A
B
B
C
C
30KN
1.5m
2.5m
50KN 50KN
30KN
A
B C
2m 3m 2m
1.5m
2.5m
56.7KN
121.97KN.m30KN
43.3KN
MA

10
I

2MC

10
I

10
2I

ME

10
2I

6

864
10I

2
3
p6q


576
10I

6
1
3
p4q


864
10p2Iq

4
1
3
p6q


576
10p2Iq

2
3
p4q

6E

0:0650:01
10

0:0650:04
10

MA0
30MC5ME 1824:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::piq
MC

10
2I

2ME

10
2I

8
I

MG

8
I

6

864
10p2Iq

2
3
p6q


576
10p2Iq

6
4
3


1200
8I
p4q

6E

0:040:065
10

0:040:025
8

5MC26ME 6279::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::piiq
Resolviendo las ecuaciones (i) y (ii)
MC 21:230 ;ME 237:417 ;MAMG0rKN:ms
Calculando las cortantes isostaticas
AB:
"
VAC48
VCA 72
CE:
"
VCE48
VEC 72
EG:
"
VEG75
VGE 75
rKNs
Calculo de las correcciones
C1
0 p21:230q
10
2:123 ;C2
21:230 p237:417q
10
21:619
C3
237:4170
8
29:677rKNs
Resistencia de Materiales I-II
pagina 11 Gilmer Calderón Quispe

9. Ingeniería Civil 3 Método de tres Momentos
Fuerzas cortantes nales
VAC482:12345:877 VCE4821:61926:381
VCA 722:123 74:123 VEC 7221:619 93:619
VEG75 p29:677q 104:667VGE 75 p29:667q 45:323rKNs10
15.536
19.464
21.698
18.302
50
89.286
55.357
105.357
27.678
175 100
10ft
-
+
-
-
-
+ +
-
-
-
+
+
V
(K)
M
(K.ft)
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
A
1 A
2
A
3 A
4
A
5
288 288
300
6m 4m 6m 4m 4m 4m
A C E
G
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
45.877KN 100.504KN 198.296KN 45.323KN
45.877
26.381
104.677
45.323
93.619
74.123
21.230
237.417
275.262
137.289
181.291
300
288
288
+
+
+
-
-
-
-
+
+ +
V
(KN)
M
(KN.m)
50KN 50KN
30KN
A
B C
2m 3m 2m
1.5m
2.5m
Viga: 300mmx400mm
colum: 300mmx500mm
2
E=20KN/mm
A
B
B
C
C
30KN
1.5m
2.5m
50KN 50KN
30KN
A
B C
2m 3m 2m
1.5m
2.5m
56.7KN
121.97KN.m30KN
43.3KN 10
15.536
19.464
21.698
18.302
50
89.286
55.357
105.357
27.678
175 100
10ft
-
+
-
-
-
+ +
-
-
-
+
+
V
(K)
M
(K.ft)
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
A
1 A
2
A
3 A
4
A
5
288 288
300
6m 4m 6m 4m 4m 4m
A C E
G
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
45.877KN 100.504KN 198.296KN 45.323KN
45.877
26.381
104.677
45.323
93.619
74.123
21.230
237.417
275.262
137.289
181.291
300
288
288
+
+
+
-
-
-
-
+
+ +
V
(KN)
M
(KN.m)
50KN 50KN
30KN
A
B C
2m 3m 2m
1.5m
2.5m
Viga: 300mmx400mm
colum: 300mmx500mm
2
E=20KN/mm
A
B
B
C
C
30KN
1.5m
2.5m
50KN 50KN
30KN
A
B C
2m 3m 2m
1.5m
2.5m
56.7KN
121.97KN.m30KN
43.3KN
Aplicando el teorema de los tres momentos hallar:
1.
2.
3.
4.
Problema N°4Resistencia de Materiales I-II
pagina 12 Gilmer Calderón Quispe

10. Ingeniería Civil 3 Método de tres Momentos
5.
6. 30x50cm
2
yE2x10
6
MPacalcule la deformacion de
esta.A B C D E
F
G
2.5T
10T.m
3T/m
I 2I
I
2I
2I I
4m 2m 2m 5m 2m
5m
3m
0.507T.m
0.38T
0.189T
2.779T
0.676T
0.676T.m
2.222T 6.665T
1.668T
0.487T
0.812T.m
0.507
1.015
3.99
1.353
7.276
2.724
6
4.142
0.812
0.676
1.623
-
+
-
+
+
-
-
-
0.38
0.487
4.067
1.57
0.665
6
0.676
+
+
-
+
-
-
-
V
(T)
M
(T.m)
2.5
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
A
0A B C D
A
1
A
2
A
3
8
16
45
2m 2m 6m 6m
A
B
C DB
C D
8KN
2m
2m
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
7.46KN
8KN
14.174KN.m
25.507KN
41.957KN
-
+
8
7.464
34.493
25.507
7.46
26.956
1.826
17.826
31.522
14.174
45
18
-
-
-
+
+
+
+
-
-
+
V
(KN)
M
(KN.m)
N
(KN)
3m
A1
1
2
p4q p8q 16A2
1
2
p6q p16q 48A3
2
3
p6q p45qSolución:A B C D E
F
G
2.5T
10T.m
3T/m
I 2I
I
2I
2I I
4m 2m 2m 5m 2m
5m
3m
0.507T.m
0.38T
0.189T
2.779T
0.676T
0.676T.m
2.222T 6.665T
1.668T
0.487T
0.812T.m
0.507
1.015
3.99
1.353
7.276
2.724
6
4.142
0.812
0.676
1.623
-
+
-
+
+
-
-
-
0.38
0.487
4.067
1.57
0.665
6
0.676
+
+
-
+
-
-
-
V
(T)
M
(T.m)
2.5
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
A
0A B C D
A
1
A
2
A
3
8
16
45
2m 2m 6m 6m
A
B
C DB
C D
8KN
2m
2m
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
7.46KN
8KN
14.174KN.m
25.507KN
41.957KN
-
+
8
7.464
34.493
25.507
7.46
26.956
1.826
17.826
31.522
14.174
45
18
-
-
-
+
+
+
+
-
-
+
V
(KN)
M
(KN.m)
N
(KN)
3m A B C D E
F
G
2.5T
10T.m
3T/m
I 2I
I
2I
2I I
4m 2m 2m 5m 2m
5m
3m
0.507T.m
0.38T
0.189T
2.779T
0.676T
0.676T.m
2.222T 6.665T
1.668T
0.487T
0.812T.m
0.507
1.015
3.99
1.353
7.276
2.724
6
4.142
0.812
0.676
1.623
-
+
-
+
+
-
-
-
0.38
0.487
4.067
1.57
0.665
6
0.676
+
+
-
+
-
-
-
V
(T)
M
(T.m)
2.5
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
A
0A B C D
A
1
A
2
A
3
8
16
45
2m 2m 6m 6m
A
B
C DB
C D
8KN
2m
2m
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
7.46KN
8KN
14.174KN.m
25.507KN
41.957KN
-
+
8
7.464
34.493
25.507
7.46
26.956
1.826
17.826
31.522
14.174
45
18
-
-
-
+
+
+
+
-
-
+
V
(KN)
M
(KN.m)
N
(KN)
3m Resistencia de Materiales I-II
pagina 13 Gilmer Calderón Quispe

11. Ingeniería Civil 3 Método de tres Momentos
MA0p0q 2MA

0
4
I

MB

4
I

6

0
16p2q
4I

6E

0
0
4

8MA4MB1:5EI 48 (i)
MA

4
I

2MB

4
I

6
I

MC

6
I

6

16p2q
4I

48p4q
6I

6E


4
0

4MA20MB6MC1:5EI 240 (ii)
MB

6
I

2MC

6
I

6
I

MD

6
I

6

48p2q
6I

180p3q
6I

6MB24MC 636 (iii)
De la gura se observa
VA8KN
Haciendo equilibrio
MAMBVAp4q 160

1
4
pMAMB16q VA

1
4
pMAMB16q 8
MAMB 16 (iv)
Resolviendo las ecuaciones
MA 14:174 ;MB1:826 ;MC 26:956rKN:msEI38:725

KN m
3

Cálculo de las cortantes isostaticas
AB:
"
VAB4
VBA 4
;BC:
"
VBC2:667
VCB 2:667
;CD:
"
VCD30
VDC 30
rKNs
Cálculo de las correcciones
C1
14:1741:826
4
4C2
1:826 p26:956q
6
4:797 ;C3
26:956
6
4:493
Cortantes nales
VAB4 p4q 8
VBA 4 p4q 0
VBC 2:6674:797 7:464
VCB 2:6674:797 7:464
VCD30 p4:493q 34:493
VDC 30 p4:493q 25:507rKNs
Resistencia de Materiales I-II
pagina 14 Gilmer Calderón Quispe

12. Ingeniería Civil 3 Método de tres Momentos
Hallando desplazamiento
IC
30p50q
3
12
312500cm
4
E2x10
6
MPa0:2x10
6KN
L
cm
2
EI38:725

38:725x10
6
312500x0:2x10
6
0:0006196cmA B C D E
F
G
2.5T
10T.m
3T/m
I 2I
I
2I
2I I
4m 2m 2m 5m 2m
5m
3m
0.507T.m
0.38T
0.189T
2.779T
0.676T
0.676T.m
2.222T 6.665T
1.668T
0.487T
0.812T.m
0.507
1.015
3.99
1.353
7.276
2.724
6
4.142
0.812
0.676
1.623
-
+
-
+
+
-
-
-
0.38
0.487
4.067
1.57
0.665
6
0.676
+
+
-
+
-
-
-
V
(T)
M
(T.m)
2.5
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
A
0A B C D
A
1
A
2
A
3
8
16
45
2m 2m 6m 6m
A
B
C DB
C D
8KN
2m
2m
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
7.46KN
8KN
14.174KN.m
25.507KN
41.957KN
-
+
8
7.464
34.493
25.507
7.46
26.956
1.826
17.826
31.522
14.174
45
18
-
-
-
+
+
+
+
-
-
+
V
(KN)
M
(KN.m)
N
(KN)
3m A B C D E
F
G
2.5T
10T.m
3T/m
I 2I
I
2I
2I I
4m 2m 2m 5m 2m
5m
3m
0.507T.m
0.38T
0.189T
2.779T
0.676T
0.676T.m
2.222T 6.665T
1.668T
0.487T
0.812T.m
0.507
1.015
3.99
1.353
7.276
2.724
6
4.142
0.812
0.676
1.623
-
+
-
+
+
-
-
-
0.38
0.487
4.067
1.57
0.665
6
0.676
+
+
-
+
-
-
-
V
(T)
M
(T.m)
2.5
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
A
0A B C D
A
1
A
2
A
3
8
16
45
2m 2m 6m 6m
A
B
C DB
C D
8KN
2m
2m
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
7.46KN
8KN
14.174KN.m
25.507KN
41.957KN
-
+
8
7.464
34.493
25.507
7.46
26.956
1.826
17.826
31.522
14.174
45
18
-
-
-
+
+
+
+
-
-
+
V
(KN)
M
(KN.m)
N
(KN)
3m
Resistencia de Materiales I-II
pagina 15 Gilmer Calderón Quispe

13. Ingeniería Civil 3 Método de tres MomentosA B C D E
F
G
2.5T
10T.m
3T/m
I 2I
I
2I
2I I
4m 2m 2m 5m 2m
5m
3m
0.507T.m
0.38T
0.189T
2.779T
0.676T
0.676T.m
2.222T 6.665T
1.668T
0.487T
0.812T.m
0.507
1.015
3.99
1.353
7.276
2.724
6
4.142
0.812
0.676
1.623
-
+
-
+
+
-
-
-
0.38
0.487
4.067
1.57
0.665
6
0.676
+
+
-
+
-
-
-
V
(T)
M
(T.m)
2.5
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
A
0A B C D
A
1
A
2
A
3
8
16
45
2m 2m 6m 6m
A
B
C DB
C D
8KN
2m
2m
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
7.46KN
8KN
14.174KN.m
25.507KN
41.957KN
-
+
8
7.464
34.493
25.507
7.46
26.956
1.826
17.826
31.522
14.174
45
18
-
-
-
+
+
+
+
-
-
+
V
(KN)
M
(KN.m)
N
(KN)
3m
Resolver la estructura mostrada utilizando la ecuación de los tres momentos.5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
1055.195
535.714
3944.805
2288.961
876.623
2483.766
2483.766
5405.844
876.623
+
+
-
-
-
-
+
V
(Kgf)
M
(Kgf.m)
5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
535.714
535.714
-
3944.805
-
-
N
(Kgf )
Problema N°5Solución:5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
1055.195
535.714
3944.805
2288.961
876.623
2483.766
2483.766
5405.844
876.623
+
+
-
-
-
-
+
V
(Kgf)
M
(Kgf.m)
5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
535.714
535.714
-
3944.805
-
-
N
(Kgf ) Resistencia de Materiales I-II
pagina 16 Gilmer Calderón Quispe

14. Ingeniería Civil 3 Método de tres Momentos
MA0

0
8

2MA

0
8

3
I

MB

3
I

6E

0

3

6MA3MB2EI0 (i)
MA

3
I

2MB

3
I

3
I

MC

3
I

6E


3

3MA12MB3MC2EI0 (ii)
MB

3
I

2MC

3
I

2
I

MD

2
I

6E

0

2

3MB10MC2MD3EI0 (iii)
MC

2
I

0
8

2MD

2
I

0
8

MD0

0
8

6E



2

2MC4MD3EI0 (iv)
De la gura obtenemos una ecuacion mas para resolver
MAMBVABp3q 0 MCMDVCDp2q 0
VAB
1
3
pMAMBq VCD
1
2
pMCMDq
¸
Fy0 VABVDC50000

1
3
pMAMBq


1
2
pMCMDq

50000

MA
3

MB
3

MC
2

MD
2
50000 (v)
Resolviendo las ecuaciones
MA 2288:961MB876:623 ;MC2483:766 ;MD 5405:844rkgfms
Fuerzas cortantes
VABVBA0
1
3
p2288:961876:623q 1055:195
VBCVCB0
1
3
p876:6232483:766q 535:714
VCDVDC0
1
2
p2483:7665405:844q 3944:805rkgfs
Resistencia de Materiales I-II
pagina 17 Gilmer Calderón Quispe

15. Ingeniería Civil 3 Método de tres Momentos5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
1055.195
535.714
3944.805
2288.961
876.623
2483.766
2483.766
5405.844
876.623
+
+
-
-
-
-
+
V
(Kgf)
M
(Kgf.m)
5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
535.714
535.714
-
3944.805
-
-
N
(Kgf ) 5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
1055.195
535.714
3944.805
2288.961
876.623
2483.766
2483.766
5405.844
876.623
+
+
-
-
-
-
+
V
(Kgf)
M
(Kgf.m)
5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
535.714
535.714
-
3944.805
-
-
N
(Kgf )
Resistencia de Materiales I-II
pagina 18 Gilmer Calderón Quispe

16. Ingeniería Civil 3 Método de tres Momentos5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
1055.195
535.714
3944.805
2288.961
876.623
2483.766
2483.766
5405.844
876.623
+
+
-
-
-
-
+
V
(Kgf)
M
(Kgf.m)
5000kgf
3m 2m
3m
A
B
C D
A
B
C D
B
C
3m
2m
5000kgf
3m 2m
3m
A
B
C D
1055.195kgf
2288.961kgf.m
535.714kgf
3944.805kgf
535.714kgf
5405.844kgf.m
535.714
535.714
-
3944.805
-
-
N
(Kgf )
Para la vigamostrada en la gura determinar las reacciones en los apoyos, el diagrama
de fuerza cortante y el diagrama de momento exionante EI=cte8.682
4.885
4.885
4.264
4.264
6.932
4.8 16.616
9
6.791
6.40
1.521
9.375
0.843 12.317
8.841
2.239
3.2
N
(T )
M
(T.m)-
+
+
+
- --
+
+
-
-
-
-
-
+
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
A E
G
C
0.01m
0.055m
0.025m
0.015m
30KN/m
25KN/m
100KN
60KN
A B C D E
2m 3m 6m 2.5m 2.5m 4.5m
60KN.m 60KN.m
800N 900N
600N/m
2m2m4m2m
800N
900N
600N/m
160N.m
A B C D
1200
900
-
+
1600
42.429
464.286
457.143
189.286
A
1
A
2
+
-
800
1494.643
905.357
439.286
460.759
Problema N°6Solución:8.682
4.885
4.885
4.264
4.264
6.932
4.8 16.616
9
6.791
6.40
1.521
9.375
0.843 12.317
8.841
2.239
3.2
N
(T )
M
(T.m)-
+
+
+
- --
+
+
-
-
-
-
-
+
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
A E
G
C
0.01m
0.055m
0.025m
0.015m
30KN/m
25KN/m
100KN
60KN
A B C D E
2m 3m 6m 2.5m 2.5m 4.5m
60KN.m 60KN.m
800N 900N
600N/m
2m2m4m2m
800N
900N
600N/m
160N.m
A B C D
1200
900
-
+
1600
42.429
464.286
457.143
189.286
A
1
A
2
+
-
800
1494.643
905.357
439.286
460.759 Resistencia de Materiales I-II
pagina 19 Gilmer Calderón Quispe

17. Ingeniería Civil 3 Método de tres Momentos
MA

4
I

2MB

4
I

4
I

MC

4
I

6

3200p2q
4I

1800p2q
4I

4MA16MB4MC 15000 (i)
MB

4
I

2MC

4
I

0
8

MDp0q 6

1800p2q
4I
0

4MB8MC 5400 (ii)
resolviendo las ecuaciones se tiene
MA 1600MB 421:429 ;MC 464:286rN:ms
Cálculo de cortantes isostaticas
AB:
"
VAB1200
VBA 1200
;BC:
"
VBC450
VCB 450
rNs
Cálculo de las correcciones
C1
1600 p421:429q
4
294:643C2
421:429p464:286q
4
10:759
Cálculo de las cortantes nales
VAB1200p294:843q 1494:643 ;VBA 1200p294:843q 905:357
VBC45010:459439:286 ; VCB 45010:759 460:759rNs8.682
4.885
4.885
4.264
4.264
6.932
4.8 16.616
9
6.791
6.40
1.521
9.375
0.843 12.317
8.841
2.239
3.2
N
(T )
M
(T.m)-
+
+
+
- --
+
+
-
-
-
-
-
+
A B E F
G
DC
120KN 120KN
150KN
I 2I I
6m 4m 6m 4m 4m 4m
A E
G
C
0.01m
0.055m
0.025m
0.015m
30KN/m
25KN/m
100KN
60KN
A B C D E
2m 3m 6m 2.5m 2.5m 4.5m
60KN.m 60KN.m
800N 900N
600N/m
2m2m4m2m
800N
900N
600N/m
160N.m
A B C D
1200
900
-
+
1600
42.429
464.286
457.143
189.286
A
1
A
2
+
-
800
1494.643
905.357
439.286
460.759
M
(N.m)
V
(N)
Resistencia de Materiales I-II
pagina 20 Gilmer Calderón Quispe

18. Ingeniería Civil 3 Método de tres Momentos800N 900N
600N/m
2294.643N 1344.643N 460.714N
464.286N.m
2Tn
2Tn
2Tn/m
3m 4m 3m
A
0 A B C D D
0
6.25
A B
C
D
A
B
C
D
B
C
D
D
10Tn
6Tn
8Tn
M
A
M
B
V
AB
+
-
+
-
3
3
6
6
5
5
+
-
14.985
3
14.985
3.015
3.015
2Tn
2Tn
2Tn/m
A
B
C
D
A
6Tn
3Tn
14.985Tn.m
14.985Tn.m
6Tn
3Tn
N
(Tn)
V
(Tn)
M
(TKn.m)
1m 3m 2m 3m
40Tn/m
20Tn
30Tn.m
A B C DE
A C DB
45
30
40/3
A
1
A
2
A
3 A
4
En la viga quebrada que se muestra calcule las reacciones , DFC, DMF , DFN y la
deexión en el punto B800N 900N
600N/m
2294.643N 1344.643N 460.714N
464.286N.m
2Tn
2Tn
2Tn/m
3m 4m 3m
A
0 A B C D D
0
6.25
A B
C
D
A
B
C
D
B
C
D
D
10Tn
6Tn
8Tn
M
A
M
B
V
AB
+
-
+
-
3
3
6
6
5
5
+
-
14.985
3
14.985
3.015
3.015
2Tn
2Tn
2Tn/m
A
B
C
D
A
6Tn
3Tn
14.985Tn.m
14.985Tn.m
6Tn
3Tn
N
(Tn)
V
(Tn)
M
(TKn.m)
1m 3m 2m 3m
40Tn/m
20Tn
30Tn.m
A B C DE
A C DB
45
30
40/3
A
1
A
2
A
3 A
4
Problema N°7Solución:800N 900N
600N/m
2294.643N 1344.643N 460.714N
464.286N.m
2Tn
2Tn
2Tn/m
3m 4m 3m
A
0 A B C D D
0
6.25
A B
C
D
A
B
C
D
B
C
D
D
10Tn
6Tn
8Tn
M
A
M
B
V
AB
+
-
+
-
3
3
6
6
5
5
+
-
14.985
3
14.985
3.015
3.015
2Tn
2Tn
2Tn/m
A
B
C
D
A
6Tn
3Tn
14.985Tn.m
14.985Tn.m
6Tn
3Tn
N
(Tn)
V
(Tn)
M
(TKn.m)
1m 3m 2m 3m
40Tn/m
20Tn
30Tn.m
A B C DE
A C DB
45
30
40/3
A
1
A
2
A
3 A
4 Resistencia de Materiales I-II
pagina 21 Gilmer Calderón Quispe

19. Ingeniería Civil 3 Método de tres Momentos
La deformación en B y C son iguales por ser simetricas
MA0p0q 2MA

0
3
I

MB

3
I

6p0q 6E


3

6MA3MB2EI0 (i)
MA

3
I

2MB

3
I

5
I

MC

5
I

6

020:833p2:5q
5I

6E


3

3MA16MB5MC2EI 62:499 (ii)
MB

5
I

2MC

5
I

3
I

MD

3
I

6

20:8333p2:5q
5I
0

6EI


3

5MB16MC3MD2EI 62:499 (iii)
MC

3
I

2MD

3
I

MD0p0q 6p0q 6E


3

3MC6MD2EI0 (iv)
haciendo equilibrio
MAMB6p3q 0ñMAMB 18 (v)
VAB6Tn
Resolviendo las ecuaciones
MA 14:985MBMC3:015 ;MD 14:985rTn:ms
Cortantes isostaticos
VABVBAVCDVDC0VBC VCB5Tn
Cortantes nales
VABVBA0
1
3
p14:9853:015q 6Tn
VBCVCB5Tn
VCDVDC0
1
3
p3:01514:985q 6Tn
Resistencia de Materiales I-II
pagina 22 Gilmer Calderón Quispe

20. Ingeniería Civil 3 Método de tres Momentos800N 900N
600N/m
2294.643N 1344.643N 460.714N
464.286N.m
2Tn
2Tn
2Tn/m
3m 4m 3m
A
0 A B C D D
0
6.25
A B
C
D
A
B
C
D
B
C
D
D
10Tn
6Tn
8Tn
M
A
M
B
V
AB
+
-
+
-
3
3
6
6
5
5
+
-
14.985
3
14.985
3.015
3.015
2Tn
2Tn
2Tn/m
A
B
C
D
A
6Tn
3Tn
14.985Tn.m
14.985Tn.m
6Tn
3Tn
N
(Tn)
V
(Tn)
M
(TKn.m)
1m 3m 2m 3m
40Tn/m
20Tn
30Tn.m
A B C DE
A C DB
45
30
40/3
A
1
A
2
A
3 A
4 800N 900N
600N/m
2294.643N 1344.643N 460.714N
464.286N.m
2Tn
2Tn
2Tn/m
3m 4m 3m
A
0 A B C D D
0
6.25
A B
C
D
A
B
C
D
B
C
D
D
10Tn
6Tn
8Tn
M
A
M
B
V
AB
+
-
+
-
3
3
6
6
5
5
+
-
14.985
3
14.985
3.015
3.015
2Tn
2Tn
2Tn/m
A
B
C
D
A
6Tn
3Tn
14.985Tn.m
14.985Tn.m
6Tn
3Tn
N
(Tn)
V
(Tn)
M
(TKn.m)
1m 3m 2m 3m
40Tn/m
20Tn
30Tn.m
A B C DE
A C DB
45
30
40/3
A
1
A
2
A
3 A
4
Resistencia de Materiales I-II
pagina 23 Gilmer Calderón Quispe

21. Ingeniería Civil 3 Método de tres Momentos
Empleando la ecuacion de los tres momentos determinar:
1.
2.
3.
4.
EI=cte800N 900N
600N/m
2294.643N 1344.643N 460.714N
464.286N.m
2Tn
2Tn
2Tn/m
3m 4m 3m
A
0 A B C D D
0
6.25
A B
C
D
A
B
C
D
B
C
D
D
10Tn
6Tn
8Tn
M
A
M
B
V
AB
+
-
+
-
3
3
6
6
5
5
+
-
14.985
3
14.985
3.015
3.015
2Tn
2Tn
2Tn/m
A
B
C
D
A
6Tn
3Tn
14.985Tn.m
14.985Tn.m
6Tn
3Tn
N
(Tn)
V
(Tn)
M
(TKn.m)
1m 3m 2m 3m
40Tn/m
20Tn
30Tn.m
A B C DE
A C DB
45
30
40/3
A
1
A
2
A
3 A
4
B
C
Problema N°8
A1
2
3
p3q p45q 90A2
1
2
p3q p30qA3
1
2
p1q

40
3


20
3
A4
1
2
p2q

40
3


40
3
Solución:800N 900N
600N/m
2294.643N 1344.643N 460.714N
464.286N.m
2Tn
2Tn
2Tn/m
3m 4m 3m
A
0 A B C D D
0
6.25
A B
C
D
A
B
C
D
B
C
D
D
10Tn
6Tn
8Tn
M
A
M
B
V
AB
+
-
+
-
3
3
6
6
5
5
+
-
14.985
3
14.985
3.015
3.015
2Tn
2Tn
2Tn/m
A
B
C
D
A
6Tn
3Tn
14.985Tn.m
14.985Tn.m
6Tn
3Tn
N
(Tn)
V
(Tn)
M
(TKn.m)
1m 3m 2m 3m
40Tn/m
20Tn
30Tn.m
A B C DE
A C DB
45
30
40/3
A
1
A
2
A
3 A
4
B
C Resistencia de Materiales I-II
pagina 24 Gilmer Calderón Quispe

22. Ingeniería Civil 3 Método de tres Momentos
MA 20Tn:m
MA

3
I

2MB

3
I

2
I

MC

2
I

6


90p1:5q
3I

20
{
3

2
{
3

3I

40
{
3

1
1
{
3

3I
0


3MA10MB2MC 323:333
10MB2MC 263:333 (i)
MB

2
I

2MC

2
I

3
I

MD

3
I

6

0
45

2
{
3
p3q

3I


2MB10MC3MD 180 (ii)
MC

3
I

2MD

3
I

MD0p0q 6


45

1
{
3
p3q

3I
0


3MC6MD 90 (iii)
Resolviendo las ecuaciones
MB 24:30MC 10:165MD 9:918
Cortantes isostaticas
AB:
"
VAB73:333
VBA 66:667
;BC:
"
VBC0
VCB0
;CD:
"
VCD 10
VDC 10
rTns
C11:433C2 7:068C3 8:23rTns
Cortantes anles
AB:
"
VAB71:90
VBA 68:10
;BC:
"
VBC7:068
VCB7:068
;CD:
"
VCD 9:918
VDC 9:918
rTns
Utilizando la ecuación general de los tres momentos para calcular la echa en el voladizo
ME

1
I

2MA

4
I

MB

3
I

6


10
{
3
p0:5q
I

90p1:5q
3I

20
{
3

1
{
3
2

3I

40
{
3

2
{
3
p2q

3I

6Epq
MA20Tn:m M B 24:30Tn:m

18:961
EI
Ò
Resistencia de Materiales I-II
pagina 25 Gilmer Calderón Quispe

23. Ingeniería Civil 3 Método de tres Momentos-
+
40
71.90
31.90
11.90
68.1
7.068
9.918
-
+
20
24.30
10.165
19.835
9.918
40Tn/m
20Tn
30Tn.m
A B C DE
111.90Tn
75.168Tn 16.986Tn
9.918Tn
9.918Tn.m
A
B C
C
C
1
C C
1
C
B
D
O
18KN/m
16/3m
20/3m
5m
2m 2m
120KN
(2)
18KN/m
5m
4m
D
M
A
M
A
M
B
M
C
M
D
M
A
0
A B C DD
0
36
120
240
96
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D =
V
(Tn)
M
(Tn.m) -
+
40
71.90
31.90
11.90
68.1
7.068
9.918
-
+
20
24.30
10.165
19.835
9.918
40Tn/m
20Tn
30Tn.m
A B C DE
111.90Tn
75.168Tn 16.986Tn
9.918Tn
9.918Tn.m
A
B C
C
C
1
C C
1
C
B
D
O
18KN/m
16/3m
20/3m
5m
2m 2m
120KN
(2)
18KN/m
5m
4m
D
M
A
M
A
M
B
M
C
M
D
M
A
0
A B C DD
0
36
120
240
96
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D =
V
(Tn)
M
(Tn.m)
Resolver el portico con método de los tres momentos56.7
6.7
43.3
30
100
100
46.9
46.9
121.97
66.5
86.6
-
+
-
-
+
+
+
+
-
-
56.7
M
(KN.m)
V
(KN)
N
(KN)
120KN
18KN/m
A
B C
D
2m 2m 3m
4m
Para todos los casos
b=400mm
h=500mm
2
E=200KN/mm
A
B C
C
C
1
C C
1
C
B
D
O
18KN/m
5.333m
6.667m
5m
2m 2m
120KN
(2)
18KN/m
5m
4m
-
+
120KN
18KN/m
A
B C
D
2m 2m 3m
4m
-
+
22.634KN
67.391KN
4.149KN.m
49.366KN
52.609KN
11.589KN.m
- -
-
67.391
49.366
71.707
N
(KN)
28.1
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D =
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D =
Problema N°9Solución:Resistencia de Materiales I-II
pagina 26 Gilmer Calderón Quispe

24. Ingeniería Civil 3 Método de tres Momentos-
+
40
71.90
31.90
11.90
68.1
7.068
9.918
-
+
20
24.30
10.165
19.835
9.918
40Tn/m
20Tn
30Tn.m
A B C DE
111.90Tn
75.168Tn 16.986Tn
9.918Tn
9.918Tn.m
A
B C
C
C
1
C C
1
C
B
D
O
18KN/m
16/3m
20/3m
5m
2m 2m
120KN
(2)
18KN/m
5m
4m
D
M
A
M
A
M
B
M
C
M
D
M
A
0
A B C DD
0
36
120
240
96
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D =
V
(Tn)
M
(Tn.m)
M0A

0
8

2MA

0
4
I

MB

4
I

6

0
96p2q
4I

6EI

0
0
4

8MA4MB1:5EI 288 (i)
MA

4
I

2MB

4
I

4
I

MC

4
I

6

96p2q
4I

240p2q
4I

6E


0
4

0

3
{
4

4


4MA16MB4MC2:625EI 1008 (ii)
MB

4
I

2MC

4
I

5
I

MD

5
I

6

240p2q
4I
0

6E

3
{
4
0
4

5
{
4
0
5

4MB18MC5MD2:265EI 720 (iii)
MC

5
I

2MD

5
I
0

MD0

0
8

6p0q 6


0

3
{
4

5
0


5MC10MD1:5EI0 (iv)
Resistencia de Materiales I-II
pagina 27 Gilmer Calderón Quispe

25. Ingeniería Civil 3 Método de tres Momentos
¸
MO0
MAMDVA

16
3
4

VD

5
20
3

18p4q

16
3
2

120p2q 0
MAMD
28
3
pVAq
35
3
VD2880
Equilibrio en el miembro AB
MAMB18p4q p2q VAp4q 0
VA36
MBMA
4
Equilibrio en el miembro CD
MCVDp5q MD0
VD
MDMC
5
MAMD
28
3

36
MBMA
4


35
3

MDMC
5

2880
MAMD
7
3
pMBMAq
7
3
pMDMCq 48

4
3
MA
7
3
MB
4
3
MD
7
3
MC 48 (v)
Resolviendo las ecuaciones se tiene
MA 4:149 ;MB 57:614 ;MC 28:049 ;MD11:589rKN:ms
EI 16:234rKN:m
3
s
Cálculo de cortantes
VAB36
1
4
p4:149 p37:614qq 22:634
VBA 36
1
4
p4:149 p37:614qq 49:366
VBC60
1
4
p57:614 p28:049qq 67:391
VCB 60
1
4
p57:614 p28:049qq 52:609
VCDVDC0
1
5
p28:04911:589q 7:928rKNs
Resistencia de Materiales I-II
pagina 28 Gilmer Calderón Quispe

26. Ingeniería Civil 3 Método de tres Momentos56.7
6.7
43.3
30
100
100
46.9
46.9
121.97
66.5
86.6
-
+
-
-
+
+
+
+
-
-
56.7
M
(KN.m)
V
(KN)
N
(KN)
120KN
18KN/m
A
B C
D
2m 2m 3m
4m
Para todos los casos
b=400mm
h=500mm
2
E=200KN/mm
A
B C
C
C
1
C C
1
C
B
D
O
18KN/m
5.333m
6.667m
5m
2m 2m
120KN
(2)
18KN/m
5m
4m
-
+
120KN
18KN/m
A
B C
D
2m 2m 3m
4m
-
+
22.634KN
67.391KN
4.149KN.m
49.366KN
52.609KN
11.589KN.m
- -
-
67.391
49.366
71.707
N
(KN)
28.1
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D =
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D = 67.391
49.366
22.634
7.928
52.609
-
+
+
+
-
-
+
V
(KN)
57.614
28.049
57.614
4.149
11.589
28.049
- -
+
-
+
+
-
120
77.169
5.118
2m
M
(KN.m)
36
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
A
B
C DB
C D
8KN
2m
2m
A
B
C D
16KN.m
10KN/m
8KN
6m 6m
2m
2m
7.46KN
8KN
14.17KN.m
25.51KN
41.95KN
8
7.46
34.49
25.51
7.46
26.95
1.82
17.82
31.52
14.17
45
18
-
-
-
+
+
+
+
-
-
+
V
(KN)
M
(KN.m)
N
(KN)
3m
A B C D E
F
G
2.5T
10T.m
3T/m
I 2I
I
2I
2I I
4m 2m 2m 5m 2m
5m
3m
Resistencia de Materiales I-II
pagina 29 Gilmer Calderón Quispe

27. Ingeniería Civil 3 Método de tres Momentos56.7
6.7
43.3
30
100
100
46.9
46.9
121.97
66.5
86.6
-
+
-
-
+
+
+
+
-
-
56.7
M
(KN.m)
V
(KN)
N
(KN)
120KN
18KN/m
A
B C
D
2m 2m 3m
4m
Para todos los casos
b=400mm
h=500mm
2
E=200KN/mm
A
B C
C
C
1
C C
1
C
B
D
O
18KN/m
5.333m
6.667m
5m
2m 2m
120KN
(2)
18KN/m
5m
4m
-
+
120KN
18KN/m
A
B C
D
2m 2m 3m
4m
-
+
22.634KN
67.391KN
4.149KN.m
49.366KN
52.609KN
11.589KN.m
- -
-
67.391
49.366
71.707
N
(KN)
28.1
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D =
1
5 3
' ;'tan
cos 4 4
CC CC b
b
D D D
= = =D =
Resistencia de Materiales I-II
pagina 30 Gilmer Calderón Quispe