UNIVERSIDAD NACIONAL DE MISIONES
FACULTAD DE INGENIERIA
Departamento de Ingeniería Civil
Cátedra: ESTRUCTURAS
Ing. GOLEMBA, Jose Luis
Departamento de Ingeniería Civil
Tema: Método de las Deformaciones
Ejemplo de Aplicación
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
𝜔𝑖 = 𝑅𝑜𝑡𝑎𝑐𝑖𝑜𝑛 𝑑𝑒 𝑁𝑢𝑑𝑜
𝜑𝑖𝑘 = 𝑅𝑜𝑡𝑎𝑐𝑖𝑜𝑛 𝐶𝑢𝑒𝑟𝑑𝑎 𝑑𝑒 𝐵𝑎𝑟𝑟𝑎
𝜔4 = 𝜔6 = 0
(1)
(6)
(2)
(3)
(5)
(4)
Analicemos los posibles desplazamientos
∆𝒗
∆𝒗
∆𝒗
𝝋𝟏𝟐
𝝋𝟔𝟓
𝝋𝟒𝟑
Desplazamiento Vertical:
∆𝒗 = 𝝋𝟏𝟐. 𝒍𝟏𝟐
∆𝒗 = 𝝋𝟓𝟔. 𝒍𝟓𝟔
∆𝒗 = −𝝋𝟑𝟒. 𝒍𝟑𝟒
∆𝒗 = 𝝋𝟏𝟐. 𝒍𝟏𝟐 = 𝝋𝟓𝟔. 𝒍𝟓𝟔 = −𝝋𝟑𝟒. 𝒍𝟑𝟒
𝝋𝟓𝟔. =𝝋𝟏𝟐. 𝒍𝟏𝟐𝒍𝟓𝟔
=𝟑
𝟓𝝋𝟏𝟐
𝝋𝟑𝟒. = −𝝋𝟏𝟐. 𝒍𝟏𝟐𝒍𝟑𝟒
= −𝟑
𝟒𝝋𝟏𝟐
𝝋𝟔𝟓 = 𝝋𝟓𝟔𝝋𝟒𝟑 = 𝝋𝟑𝟒𝝋𝟏𝟐 = 𝝋𝟐𝟏
(1)
(6)
(2)
(3)
(5)
(4)
∆𝒉
𝝋𝟐𝟑
Desplazamiento Horizontal:
∆𝒉 = 𝝋𝟐𝟑. 𝒍𝟐𝟑
(1)
(6)
(2)
(3)
(5)
(4)
Resumiendo:
∆𝒗
∆𝒗
∆𝒗
𝝋𝟏𝟐
𝝋𝟓𝟔
𝝋𝟑𝟒
𝝎𝟐
𝝋𝟓𝟔. =𝝋𝟏𝟐. 𝒍𝟏𝟐𝒍𝟓𝟔
=𝟑
𝟓𝝋𝟏𝟐
𝝋𝟑𝟒. = −𝝋𝟏𝟐. 𝒍𝟏𝟐𝒍𝟑𝟒
= −𝟑
𝟒𝝋𝟏𝟐
∆𝒉
𝝋𝟐𝟑
Incognitas:
𝝎𝟑𝝎𝟓𝝋𝟏𝟐 ; 𝝋𝟑𝟒; 𝝋𝟓𝟔𝝋𝟐𝟑
𝝎𝟏
Ecuación Fundamental del Método de las Deformaciones:
𝑀𝑖𝑘 = 𝑀°𝑖𝑘 +2𝐸𝐽𝑖𝑘𝑙𝑖𝑘2𝜔𝑖 + 𝜔𝑘 − 3𝜑𝑖𝑘
𝑀°𝑖𝑘 = 𝑀𝑜𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝐸𝑚𝑝𝑜𝑡𝑟𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑃𝑒𝑟𝑓𝑒𝑐𝑡𝑜
Tablas M°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
𝑀°12 = 0
𝑀°21 = −𝑞𝑙2
8= −20. 32
8= −22,50 𝑘𝑁.𝑚
3I0
(2)
(3)
q =
20
kN
/m
3,00 𝑀°23 = −𝑀°32= −𝑞𝑙2
12= −20. 32
12= −15 𝑘𝑁.𝑚
4I0(3)
2,00 2,00
20 kN/m 10 kN/m
𝑀°34 =𝑞𝑙2
12+𝑞′𝑐
𝑙2𝑎. 𝑏2 +
𝑐2
12𝑙 − 3𝑏 =
20. 42
12+10.2
423.12 +
22
124 − 3.1 =
185
6𝑘𝑁.𝑚
𝑀°43 = −𝑞𝑙2
12−𝑞′𝑐
𝑙2𝑎2. 𝑏 +
𝑐2
12𝑙 − 3𝑎 = −
20. 42
12−10.2
4232. 1 +
22
124 − 3.3 = −
215
6𝑘𝑁.𝑚
50 kN
5I0
(6)
(5)
3,20 1,80
𝑀°56 = −𝑃𝑎2𝑏
𝑙2= −50. 3,202. 1,80
52= −4608
125𝑘𝑁.𝑚
𝑀°65 =𝑃𝑏2𝑎
𝑙2= −50. 1,802. 3,20
52=2592
125𝑘𝑁.𝑚
𝑀°5𝑣 = 𝑃. 𝑙 = 20.1,20 = 24 𝑘𝑁.𝑚
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
M°12 =0
M°21 =-22,50
M°23 =-15
M°32 =15
M°34 =185/6
M°43 =-215/6
M°56 =-4608/125
M°65 =2592/125
𝝋𝟓𝟔. =𝟑
𝟓𝝋𝟏𝟐𝝋𝟑𝟒. = −
𝟑
𝟒𝝋𝟏𝟐
𝑀𝑖𝑘 = 𝑀°𝑖𝑘 +2𝐸𝐽𝑖𝑘𝑙𝑖𝑘2𝜔𝑖 + 𝜔𝑘 − 3𝜑𝑖𝑘
𝑀12 = 0 +2𝐸3𝐽032𝜔1 + 𝜔2 − 3𝜑12 = 4𝜔1 + 2𝜔2 − 6𝜑12
𝑀21 = −45
2+2𝐸3𝐽032𝜔2 + 𝜔1 − 3𝜑12 = −
45
2+ 4𝜔2 + 2𝜔1 − 6𝜑12
𝑀23 = −15 +2𝐸3𝐽032𝜔2 + 𝜔3 − 3𝜑23 = −15 + 4𝜔2 + 2𝜔3 − 6𝜑23
𝑀32 = 15 +2𝐸3𝐽032𝜔3 + 𝜔2 − 3𝜑23 = 15 + 4𝜔3 + 2𝜔2 − 6𝜑23
𝑀34 =185
6+2𝐸4𝐽042𝜔3 + 𝜔4 − 3(−
3
4𝜑12) =
185
6+ 4𝜔3 +
9
2𝜑12
𝑀43 = −215
6+2𝐸4𝐽042𝜔4 + 𝜔3 − 3(−
3
4𝜑12) = −
215
6+ 2𝜔3 +
9
2𝜑12
𝑀35 = 0 +2𝐸2𝐽022𝜔3 + 𝜔5 − 3𝜑35 = 4𝜔3 + 2𝜔5
𝑀53 = 0 +2𝐸2𝐽022𝜔5 + 𝜔3 − 3𝜑35 = 4𝜔5 + 2𝜔3
𝑀56 = −4608
125+2𝐸5𝐽052𝜔5 + 𝜔6 − 3(
3
5𝜑12) = −
4608
125+ 4𝜔5 −
18
5𝜑12
𝑀65 =2592
125+2𝐸5𝐽052𝜔6 + 𝜔5 − 3(
3
5𝜑12) =
2592
125+ 2𝜔5 −
18
5𝜑12
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
𝑁𝑢𝑑𝑜 1 → 𝑀1 = 0 → 𝑀12 = 0 4𝜔1 + 2𝜔2 − 6𝜑12 = 0 𝐼
𝑁𝑢𝑑𝑜 2 → 𝑀2 = 0 → 𝑀21 +𝑀23 = 0
−45
2+ 4𝜔2 + 2𝜔1 − 6𝜑12 + −15 + 4𝜔2 + 2𝜔3 − 6𝜑23 = 0
2𝜔1 + 8𝜔2 + 2𝜔3 − 6𝜑12 − 6𝜑23 =75
2𝐼𝐼
𝑁𝑢𝑑𝑜 3 → 𝑀3 = 0 → 𝑀32 +𝑀34 + 𝑀35 = 0
2𝜔2 + 12𝜔3 + 2𝜔5 +9
2𝜑12 − 6𝜑23 = −
275
6𝐼𝐼𝐼
15 + 4𝜔3 + 2𝜔2 − 6𝜑23 +185
6+ 4𝜔3 +
9
2𝜑12 + 4𝜔3 + 2𝜔5 = 0
𝑁𝑢𝑑𝑜 5 → 𝑀5 = 0 → 𝑀53 +𝑀56 +𝑀5𝑣 = 0
4𝜔5 + 2𝜔3 + −4608
125+ 4𝜔5 −
18
5𝜑12 + 24 = 0 2𝜔3 + 8𝜔5 −
18
5𝜑12 =
1608
125𝐼𝑉
iv) Ecuación de piso
𝐹𝐻 = 0
Desplazamiento Horizontal
(1)
(6)
(2)
(3)
(5)
(4)
∆𝒉
𝐹𝐻 = 0 → 𝑄23 = 0
iv) Ecuación de piso3I0
(1)
(2)
(3)
q =
20
kN
/m
QM
M/L
Q°qq
. L
/2
M23
M32
Q23
𝐹𝐻 = 0 → 𝑄23 = 0
𝑉
Desplazamiento Horizontal
𝑄23 = 𝑄𝑀 + 𝑄0
𝑄23 =𝑀23 +𝑀32𝑙23
−𝑞. 𝑙
2
𝑄23 =1
3−15 + 4𝜔2 + 2𝜔3 − 6𝜑23 + 15 + 4𝜔3 + 2𝜔2 − 6𝜑23 −
20.3
2
𝑄23 = 2𝜔2 + 2𝜔3 − 4𝜑23 − 30
𝐹𝐻 = 0 → 𝑄23 = 02𝜔2 + 2𝜔3 − 4𝜑23 = 30
iv) Ecuación de piso
𝐹𝑉 = 0 → 𝑄21 + 𝑄56 + 𝑄34 + 𝑄5𝑉 + 𝑃 = 0 (𝑉𝐸𝐶𝑇𝑂𝑅𝐼𝐴𝐿)
Desplazamiento Vertical
(1)
(6)
(2)
(3)
(5)
(4)
∆𝒗
∆𝒗
∆𝒗
iv) Ecuación 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
Q°qq. L/2
Q23
(3) M12 M21
QM M/L
Q°q; q'
Q34
M56
QM M/L
M65
Q°P
Q56
Q°P
Q5V
𝐹𝑉 = 0 → 𝑄21 + 𝑄56 − 𝑄34 − 𝑄5𝑉 − 𝑃 = 0
Desplazamiento Vertical
𝑄21 = 𝑄𝑀 + 𝑄0
𝑄21 =𝑀12 +𝑀21𝑙12
−𝑞. 𝑙
2
𝑄21 =1
34𝜔1 + 2𝜔2 − 6𝜑12 + −
45
2+ 4𝜔2 + 2𝜔1 − 6𝜑12 −
20.3
2
𝑄21 = 2𝜔1 + 2𝜔2 − 4𝜑12 −75
2
𝑄56 = 𝑄𝑀 + 𝑄0𝑝 𝑄56 =
𝑀56 +𝑀65𝑙56
−𝑃. 𝑎
𝑙
𝑄56 =1
5−4608
125+ 4𝜔5 −
18
5𝜑12 +
2592
125+ 2𝜔5 −
18
5𝜑12 −
50.3,20
5
𝑄56 =6
5𝜔5 −36
25𝜑12 −
22016
625
iv) Ecuación 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
Q°qq. L/2
Q23
(3) M12 M21
QM M/L
Q°q; q'
Q34
M56
QM M/L
M65
Q°P
Q56
Q°P
Q5V
Desplazamiento Vertical
𝑄34 = 𝑄𝑀 + 𝑄0𝑞 + 𝑄0𝑞’
𝑄34 =𝑀34 +𝑀43𝑙34
+𝑞. 𝑙
2+𝑞′. 𝑐. 𝑏
𝑙
𝑄34 =1
4
185
6+ 4𝜔3 +
9
2𝜑12 + −
215
6+ 2𝜔3 +
9
2𝜑12 +
20.4
2+10.2.1
4
𝑄34 =3
2𝜔3 +9
4𝜑12 +
175
4
𝑄5𝑉 = 𝑄0𝑝
𝑄5𝑉 = 20
iv) Ecuación 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
Q°qq. L/2
Q23
(3) M12 M21
QM M/L
Q°q; q'
Q34
M56
QM M/L
M65
Q°P
Q56
Q°P
Q5V
𝐹𝑉 = 0 → 𝑄21 + 𝑄56 − 𝑄34 − 𝑄5𝑉 − 𝑃 = 0
Desplazamiento Vertical
𝑄21 = 2𝜔1 + 2𝜔2 − 4𝜑12 −75
2
𝑄56 =6
5𝜔5 −36
25𝜑12 −
22016
625
𝑄34 =3
2𝜔3 +9
4𝜑12 +
175
4
𝑄5𝑉 = 20
𝐹𝑉 = 0 → 2𝜔1 + 2𝜔2 − 4𝜑12 −75
2+6
5𝜔5 −36
25𝜑12 −
22016
625−3
2𝜔3 +9
4𝜑12 +
175
4− 20 − 20 = 0
2𝜔1 + 2𝜔2 −3
2𝜔3 +6
5𝜔5 −769
100𝜑12 ≅ 156,476
𝑉I
v) Sistemas de ecuaciones
2𝜔1 + 2𝜔2 −3
2𝜔3 +6
5𝜔5 −769
100𝜑12 ≅ 156,476
𝑉I
4𝜔1 + 2𝜔2 − 6𝜑12 = 0 𝐼
2𝜔1 + 8𝜔2 + 2𝜔3 − 6𝜑12 − 6𝜑23 =75
2𝐼𝐼
2𝜔2 + 12𝜔3 + 2𝜔5 +9
2𝜑12 − 6𝜑23 = −
275
6𝐼𝐼𝐼
2𝜔3 + 8𝜔5 −18
5𝜑12 =
1608
125𝐼𝑉
2𝜔2 + 2𝜔3 − 4𝜑23 = 30 𝑉
Resolviendo
𝜔1 = −63,0985
𝜔2 = −38,1490
𝜔3 = 19,2500
𝜔5 = −27,8564
𝜑12 = −54,7820
𝜑23 = −16,9495
vi) Reemplazando en Mik
𝑀12 = 4𝜔1 + 2𝜔2 − 6𝜑12 = 0
𝑀21 = −45
2+ 4𝜔2 + 2𝜔1 − 6𝜑12 = 27,39
𝑀23 = −15 + 4𝜔2 + 2𝜔3 − 6𝜑23 = −27,39
𝑀32 = 15 + 4𝜔3 + 2𝜔2 − 6𝜑23 = 117,39
𝑀34 =185
6+ 4𝜔3 +
9
2𝜑12 = −138,68
𝑀43 = −215
6+ 2𝜔3 +
9
2𝜑12 = −243,85
𝑀35 = 4𝜔3 + 2𝜔5 = 21,28
𝑀53 = 4𝜔5 + 2𝜔3 = −72,92
𝑀56 = −4608
125+ 4𝜔5 −
18
5𝜑12 = 48,92
𝑀65 =2592
125+ 2𝜔5 −
18
5𝜑12 = 162,23
𝜔1 = −63,0985
𝜔2 = −38,1490
𝜔3 = 19,2500
𝜔5 = −27,8564
𝜑12 = −54,7820
𝜑23 = −16,9495
𝑀5𝑉 = 24
(1) (2)
(3) (4)
(5)(6)
vii) Reacciones y Esfuerzos Característicos
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 Característicos
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 Característicos
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 Característicos
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 Característicos
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 Característicos
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 Característicos
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 máximo (Q=0)
M35 = 21,28
M53 = -72,92
Dudas? Consultas?
Se entiende?
Sencillo, no?
Basta….basta, por hoy!!!
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