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!!!