universidad nacional de misiones facultad de ingenieria
TRANSCRIPT
UNIVERSIDAD NACIONAL DE MISIONES
FACULTAD DE INGENIERIA
Departamento de IngenierΓa Civil
CΓ‘tedra: ESTRUCTURAS
Ing. GOLEMBA, Jose Luis
Departamento de IngenierΓa Civil
Tema: MΓ©todo de las Deformaciones
Ejemplo de AplicaciΓ³n
Sistema HIPERESTATICO Grado 4
q = 20 kN/m
50 kN
3I0
4I0
3I0
5I0
(1)
(6)
(2)
(3)
(5)
(4)
2I0
20 kN
3,20 1,80
3,00
2,00 2,00
3,00
2,00
q = 20 kN/mq = 30 kN/mq = 20 kN/m
20 kN
1,20
ππ = π ππ‘πππππ ππ ππ’ππ
πππ = π ππ‘πππππ πΆπ’ππππ ππ π΅ππππ
π4 = π6 = 0
(1)
(6)
(2)
(3)
(5)
(4)
Analicemos los posibles desplazamientos
βπ
βπ
βπ
πππ
πππ
πππ
Desplazamiento Vertical:
βπ = πππ. πππ
βπ = πππ. πππ
βπ = βπππ. πππ
βπ = πππ. πππ = πππ. πππ = βπππ. πππ
πππ. =πππ. ππππππ
=π
ππππ
πππ. = βπππ. ππππππ
= βπ
ππππ
πππ = ππππππ = ππππππ = πππ
(1)
(6)
(2)
(3)
(5)
(4)
βπ
πππ
Desplazamiento Horizontal:
βπ = πππ. πππ
(1)
(6)
(2)
(3)
(5)
(4)
Resumiendo:
βπ
βπ
βπ
πππ
πππ
πππ
ππ
πππ. =πππ. ππππππ
=π
ππππ
πππ. = βπππ. ππππππ
= βπ
ππππ
βπ
πππ
Incognitas:
πππππππ ; πππ; ππππππ
ππ
EcuaciΓ³n Fundamental del MΓ©todo de las Deformaciones:
πππ = πΒ°ππ +2πΈπ½πππππ2ππ + ππ β 3πππ
πΒ°ππ = ππππππ‘π ππ πΈππππ‘πππππππ‘π πππππππ‘π
Tablas MΒ°ik
i) Momentos de empotramiento perfectoq = 20 kN/m
50 kN
3I0
4I0
3I0
5I0
(1)
(6)
(2)
(3)
(5)
(4)
2I0
20 kN
3,20 1,80
3,00
2,00 2,00
3,00
2,00
q = 20 kN/mq = 30 kN/mq = 20 kN/m
20 kN
1,20
q = 20 kN/m
3I0(1)
3,00
πΒ°12 = 0
πΒ°21 = βππ2
8= β20. 32
8= β22,50 ππ.π
3I0
(2)
(3)
q =
20
kN
/m
3,00 πΒ°23 = βπΒ°32= βππ2
12= β20. 32
12= β15 ππ.π
4I0(3)
2,00 2,00
20 kN/m 10 kN/m
πΒ°34 =ππ2
12+πβ²π
π2π. π2 +
π2
12π β 3π =
20. 42
12+10.2
423.12 +
22
124 β 3.1 =
185
6ππ.π
πΒ°43 = βππ2
12βπβ²π
π2π2. π +
π2
12π β 3π = β
20. 42
12β10.2
4232. 1 +
22
124 β 3.3 = β
215
6ππ.π
50 kN
5I0
(6)
(5)
3,20 1,80
πΒ°56 = βππ2π
π2= β50. 3,202. 1,80
52= β4608
125ππ.π
πΒ°65 =ππ2π
π2= β50. 1,802. 3,20
52=2592
125ππ.π
πΒ°5π£ = π. π = 20.1,20 = 24 ππ.π
ii) Momentos en barras ikq = 20 kN/m
50 kN
3I0
4I0
3I0
5I0
(1)
(6)
(2)
(3)
(5)
(4)
2I0
20 kN
3,20 1,80
3,00
2,00 2,00
3,00
2,00
q = 20 kN/mq = 30 kN/mq = 20 kN/m
20 kN
1,20
MΒ°12 =0
MΒ°21 =-22,50
MΒ°23 =-15
MΒ°32 =15
MΒ°34 =185/6
MΒ°43 =-215/6
MΒ°56 =-4608/125
MΒ°65 =2592/125
πππ. =π
πππππππ. = β
π
ππππ
πππ = πΒ°ππ +2πΈπ½πππππ2ππ + ππ β 3πππ
π12 = 0 +2πΈ3π½032π1 + π2 β 3π12 = 4π1 + 2π2 β 6π12
π21 = β45
2+2πΈ3π½032π2 + π1 β 3π12 = β
45
2+ 4π2 + 2π1 β 6π12
π23 = β15 +2πΈ3π½032π2 + π3 β 3π23 = β15 + 4π2 + 2π3 β 6π23
π32 = 15 +2πΈ3π½032π3 + π2 β 3π23 = 15 + 4π3 + 2π2 β 6π23
π34 =185
6+2πΈ4π½042π3 + π4 β 3(β
3
4π12) =
185
6+ 4π3 +
9
2π12
π43 = β215
6+2πΈ4π½042π4 + π3 β 3(β
3
4π12) = β
215
6+ 2π3 +
9
2π12
π35 = 0 +2πΈ2π½022π3 + π5 β 3π35 = 4π3 + 2π5
π53 = 0 +2πΈ2π½022π5 + π3 β 3π35 = 4π5 + 2π3
π56 = β4608
125+2πΈ5π½052π5 + π6 β 3(
3
5π12) = β
4608
125+ 4π5 β
18
5π12
π65 =2592
125+2πΈ5π½052π6 + π5 β 3(
3
5π12) =
2592
125+ 2π5 β
18
5π12
iii) Equilibrio de Nudosq = 20 kN/m
50 kN
3I0
4I0
3I0
5I0
(1)
(6)
(2)
(3)
(5)
(4)
2I0
20 kN
3,20 1,80
3,00
2,00 2,00
3,00
2,00
q = 20 kN/mq = 30 kN/mq = 20 kN/m
20 kN
1,20
ππ’ππ 1 β π1 = 0 β π12 = 0 4π1 + 2π2 β 6π12 = 0 πΌ
ππ’ππ 2 β π2 = 0 β π21 +π23 = 0
β45
2+ 4π2 + 2π1 β 6π12 + β15 + 4π2 + 2π3 β 6π23 = 0
2π1 + 8π2 + 2π3 β 6π12 β 6π23 =75
2πΌπΌ
ππ’ππ 3 β π3 = 0 β π32 +π34 + π35 = 0
2π2 + 12π3 + 2π5 +9
2π12 β 6π23 = β
275
6πΌπΌπΌ
15 + 4π3 + 2π2 β 6π23 +185
6+ 4π3 +
9
2π12 + 4π3 + 2π5 = 0
ππ’ππ 5 β π5 = 0 β π53 +π56 +π5π£ = 0
4π5 + 2π3 + β4608
125+ 4π5 β
18
5π12 + 24 = 0 2π3 + 8π5 β
18
5π12 =
1608
125πΌπ
iv) EcuaciΓ³n de piso
πΉπ» = 0
Desplazamiento Horizontal
(1)
(6)
(2)
(3)
(5)
(4)
βπ
πΉπ» = 0 β π23 = 0
iv) EcuaciΓ³n de piso3I0
(1)
(2)
(3)
q =
20
kN
/m
QM
M/L
QΒ°qq
. L
/2
M23
M32
Q23
πΉπ» = 0 β π23 = 0
π
Desplazamiento Horizontal
π23 = ππ + π0
π23 =π23 +π32π23
βπ. π
2
π23 =1
3β15 + 4π2 + 2π3 β 6π23 + 15 + 4π3 + 2π2 β 6π23 β
20.3
2
π23 = 2π2 + 2π3 β 4π23 β 30
πΉπ» = 0 β π23 = 02π2 + 2π3 β 4π23 = 30
iv) EcuaciΓ³n de piso
πΉπ = 0 β π21 + π56 + π34 + π5π + π = 0 (ππΈπΆπππ πΌπ΄πΏ)
Desplazamiento Vertical
(1)
(6)
(2)
(3)
(5)
(4)
βπ
βπ
βπ
iv) EcuaciΓ³n de pisoq = 20 kN/m
50 kN
(2)
(5)
20 kN
q = 20 kN/mq = 30 kN/m
20 kN
M21
QM M/L
M12
QΒ°qq. L/2
Q23
(3) M12 M21
QM M/L
QΒ°q; q'
Q34
M56
QM M/L
M65
QΒ°P
Q56
QΒ°P
Q5V
πΉπ = 0 β π21 + π56 β π34 β π5π β π = 0
Desplazamiento Vertical
π21 = ππ + π0
π21 =π12 +π21π12
βπ. π
2
π21 =1
34π1 + 2π2 β 6π12 + β
45
2+ 4π2 + 2π1 β 6π12 β
20.3
2
π21 = 2π1 + 2π2 β 4π12 β75
2
π56 = ππ + π0π π56 =
π56 +π65π56
βπ. π
π
π56 =1
5β4608
125+ 4π5 β
18
5π12 +
2592
125+ 2π5 β
18
5π12 β
50.3,20
5
π56 =6
5π5 β36
25π12 β
22016
625
iv) EcuaciΓ³n de pisoq = 20 kN/m
50 kN
(2)
(5)
20 kN
q = 20 kN/mq = 30 kN/m
20 kN
M21
QM M/L
M12
QΒ°qq. L/2
Q23
(3) M12 M21
QM M/L
QΒ°q; q'
Q34
M56
QM M/L
M65
QΒ°P
Q56
QΒ°P
Q5V
Desplazamiento Vertical
π34 = ππ + π0π + π0πβ
π34 =π34 +π43π34
+π. π
2+πβ². π. π
π
π34 =1
4
185
6+ 4π3 +
9
2π12 + β
215
6+ 2π3 +
9
2π12 +
20.4
2+10.2.1
4
π34 =3
2π3 +9
4π12 +
175
4
π5π = π0π
π5π = 20
iv) EcuaciΓ³n de pisoq = 20 kN/m
50 kN
(2)
(5)
20 kN
q = 20 kN/mq = 30 kN/m
20 kN
M21
QM M/L
M12
QΒ°qq. L/2
Q23
(3) M12 M21
QM M/L
QΒ°q; q'
Q34
M56
QM M/L
M65
QΒ°P
Q56
QΒ°P
Q5V
πΉπ = 0 β π21 + π56 β π34 β π5π β π = 0
Desplazamiento Vertical
π21 = 2π1 + 2π2 β 4π12 β75
2
π56 =6
5π5 β36
25π12 β
22016
625
π34 =3
2π3 +9
4π12 +
175
4
π5π = 20
πΉπ = 0 β 2π1 + 2π2 β 4π12 β75
2+6
5π5 β36
25π12 β
22016
625β3
2π3 +9
4π12 +
175
4β 20 β 20 = 0
2π1 + 2π2 β3
2π3 +6
5π5 β769
100π12 β 156,476
πI
v) Sistemas de ecuaciones
2π1 + 2π2 β3
2π3 +6
5π5 β769
100π12 β 156,476
πI
4π1 + 2π2 β 6π12 = 0 πΌ
2π1 + 8π2 + 2π3 β 6π12 β 6π23 =75
2πΌπΌ
2π2 + 12π3 + 2π5 +9
2π12 β 6π23 = β
275
6πΌπΌπΌ
2π3 + 8π5 β18
5π12 =
1608
125πΌπ
2π2 + 2π3 β 4π23 = 30 π
Resolviendo
π1 = β63,0985
π2 = β38,1490
π3 = 19,2500
π5 = β27,8564
π12 = β54,7820
π23 = β16,9495
vi) Reemplazando en Mik
π12 = 4π1 + 2π2 β 6π12 = 0
π21 = β45
2+ 4π2 + 2π1 β 6π12 = 27,39
π23 = β15 + 4π2 + 2π3 β 6π23 = β27,39
π32 = 15 + 4π3 + 2π2 β 6π23 = 117,39
π34 =185
6+ 4π3 +
9
2π12 = β138,68
π43 = β215
6+ 2π3 +
9
2π12 = β243,85
π35 = 4π3 + 2π5 = 21,28
π53 = 4π5 + 2π3 = β72,92
π56 = β4608
125+ 4π5 β
18
5π12 = 48,92
π65 =2592
125+ 2π5 β
18
5π12 = 162,23
π1 = β63,0985
π2 = β38,1490
π3 = 19,2500
π5 = β27,8564
π12 = β54,7820
π23 = β16,9495
π5π = 24
(1) (2)
(3) (4)
(5)(6)
vii) Reacciones y Esfuerzos CaracterΓsticos
Ponemos de manifiesto losmomentos en extremo debarra
M12 = 0M21 = 27,39M23 = -27,39M32 = 117,39M34 = -138,68M35 = 21,28M53 = -72,92M56 = 48,92M5v = 24,00M43 = -243,85M65 = 162,23
(1) (2)
(3) (4)
(5)(6)
27.39
27.39
vii) Reacciones y Esfuerzos CaracterΓsticos
Ponemos de manifiesto losmomentos en extremo debarra
M12 = 0M21 = 27,39M23 = -27,39M32 = 117,39M34 = -138,68M35 = 21,28M53 = -72,92M56 = 48,92M5v = 24,00M43 = -243,85M65 = 162,23
(1) (2)
(3) (4)
(5)(6)
27.39
27.39
117.39
243.85138.68
21.28
72.92
24
vii) Reacciones y Esfuerzos CaracterΓsticos
Ponemos de manifiesto losmomentos en extremo debarra
M12 = 0M21 = 27,39M23 = -27,39M32 = 117,39M34 = -138,68M35 = 21,28M53 = -72,92M56 = 48,92M5v = 24,00M43 = -243,85M65 = 162,23
(1) (2)
(3) (4)
(5)(6)
M12 + M21L
q.L2
V1 = 39.13
vii) Reacciones y Esfuerzos CaracterΓsticos
Analizamos los cortes encada barra
M12 = 0M21 = 27,39M23 = -27,39M32 = 117,39M34 = -138,68M35 = 21,28M53 = -72,92M56 = 48,92M5v = 24,00M43 = -243,85M65 = 162,23
(1) (2)
(3) (4)
(5)(6)
M12 + M21L
q.L2
V1 = 39.13
20.87 20
40.87
-40.8
7
vii) Reacciones y Esfuerzos CaracterΓsticos
Analizamos los cortes encada barra
M12 = 0M21 = 27,39M23 = -27,39M32 = 117,39M34 = -138,68M35 = 21,28M53 = -72,92M56 = 48,92M5v = 24,00M43 = -243,85M65 = 162,23
(1) (2)
(3) (4)
(5)(6)
M12 + M21L
q.L2
V1 = 39.13
20.87 20
40.87
-40
.87
M3
2 -
M2
3L q.L 2
60
M34 + M43L
q.L2
q'.c.bL
q'.c.aL
50.63
V4 = 150.63
+9
.76
M5
3 -
M3
5L
25
.82 -85.82
H4 = 85.82
P
M56 + M65L
P.aL
P.bLV6 = 60.23
10.93 20
-25.82H6 = 25.82
60.23
150.63
39.13
vii) Reacciones y Esfuerzos CaracterΓsticos
Analizamos los cortes encada barra
M12 = 0M21 = 27,39M23 = -27,39M32 = 117,39M34 = -138,68M35 = 21,28M53 = -72,92M56 = 48,92M5v = 24,00M43 = -243,85M65 = 162,23
viii) Diagramas de Esfuerzos CaracterΓsticos
Momento Flector
Corte
Normal
M21 = 27,39
M32 = 117,39
M34 = -138,68
M43 = -243,85
M65 = 162,23
-25,82
-85,82
--4
0,8
7
60,23
150,63
39,13
Calcular el momento mΓ‘ximo (Q=0)
M35 = 21,28
M53 = -72,92
Dudas? Consultas?
Se entiende?
Sencillo, no?
Bastaβ¦.basta, por hoy!!!