analisis matricial

Ing. Carlos Elmer Cruz Salazar

2016

Autor: ING. CIVIL I 7º “K”

2016

Au

tor:

ING

. CIV

IL I

7º

“K”

ANALISIS ESTRUCTURAL

AVANZADO

ANALISIS ESTRUCTURAL AVANZADO

ING. CIVIL I 7º “K” 2

CUADERNILLO DE EJERCICIOS UNIDAD 3

INGENIERIA CIVIL | 7° “K”

INTEGRATES

SANTOS NATAREN HUGO ALBERTO

SANTOS CAMACHO MANUEL JULIO CESAR

VILLALOBOS HERNANDEZ FCO. GABRIEL

SANCHEZ ZARATE ENRIQUE

TRINIDAD FIGUEROA GUADALUPE

VELAZQUEZ ORTIZ LUIS ENRIQUE

ANALISIS ESTRUCTURAL

AVANZADO



14-1 DETERMINE LA MATRIZ DE RIGIDEZ K P ARA EL ENSAMBLE. CONSIDERE

𝒒𝒖𝒆 𝑨 = 𝟎. 𝟓 𝒑𝒖𝒍𝒈𝟕 Y QUE 𝑬 = 𝟐𝟗(𝟏𝟎𝟑) KSI PARA CADA ELEMENTO.

Barra 1

𝜆𝑥 = cos 36.87 = 0.8

𝜆𝑦 = 𝑠𝑒𝑛 36.87 = 0.6

𝐿 = 60

5 6 1 2

154.667 116 -154.667 -116 5

116 87 -116 -87 6

-154.667 -116 154.667 116 1

-116 -87 116 87 2

K11

Barra 2 𝜆𝑥 = 1 𝜆𝑦 = 0 𝐿 = 72

1 2 3 4

K12

201.389 0 -201.389 0 1

0 0 0 0 2

-201.389 0 201.389 0 3

0 0 0 0 4

Barra 3 𝜆𝑥 = cos 36.87 = 0.8 L=60

𝜆𝑦 = 𝑠𝑒𝑛 36.87 = −0.6

7 8 1 2

154.667 -116 -154.667 116 7

-116 87 116 -87 8

-154.667 116 154.667 -116 1

116 -87 -116 87 2

K13



Ensamble para la determinación de la matriz de rigidez K

1 2 3 4 5 6 7 8

510.723 0 -201.39 0 -154.667 -116 -154.667 116 1

0 174 0 0 -116 -87 116 -87 2

-201.389 0 201.389 0 0 0 0 0 3

0 0 0 0 0 0 0 0 4

-154.667 -116 0 0 154.667 116 0 0 5

-116 -87 0 0 116 87 0 0 6

-154.667 116 0 0 0 0 154.667 -116 7

116 -87 0 0 0 0 -116 87 8

KG

14-2 DETERMINE LOS DESPLAZAMIENTOS HORIZONTALES Y VERTICAL EN LA

JUNTA (3) DEL ENSAMBLE DEL PROBLEMA.

𝐷𝑘 =

[ 00000]

𝑄𝐾 = [0−4]

1 2 3 4 5 6 7 8

0 510.723 0 -201.39 0 -154.667 -116 -154.667 116 D1

-4 0 174 0 0 -116 -87 116 -87 D2

Q3 -201.389 0 201.389 0 0 0 0 0 0

Q4 0 0 0 0 0 0 0 0 0

Q5 -154.667 -116 0 0 154.667 116 0 0 0

Q6 -116 -87 0 0 116 87 0 0 0

Q7 -154.667 116 0 0 0 0 154.667 -116 0

Q8 116 -87 0 0 0 0 -116 87 0

KG

0 = 510.72𝐷1 + 0𝐷2

−4 = 0𝐷1 + 172𝐷2

𝐷1 = 0

𝐷2 = −0.0230𝑖𝑛



14-3 DETERMINE LA FUERZA E N CADA ELEMENTO DEL ENSAMBLE DEL

PROBLEMA 14-1

𝐷1 = 𝐷2 = 𝐷3 = 𝐷4 = 𝐷5 = 𝐷6 = 𝐷7 = 𝐷8 = 0 𝐷2 = −0.230

Barra 1

𝑄1 =0.5(29(103))

60[−0.8 −0.6 0.8 0.6] [

000

−0.230

]

𝑄1 =0.5(29(103))

60(0.6)(−0.230) = −𝟑. 𝟑𝟑𝑲

Barra 2

𝑄2 =0.5(29(103))

72[−1 0 1 0] [

0−0.0230

00

]

𝑄2 = 0

Barra 3

𝑄3 =0.5(29(103))

60[−0.8 0.6 0.8 −0.6] [

000

−0.230

]

𝑄3 =0.5(29(103))

60(0.6)(−0.230) = −𝟑. 𝟑𝟑𝑲



14-4 DETERMINE LA MATRIZ DE RIGIDEZ K PARA LA ARMADURA. CONSIDERE QUE

𝑨 = 𝟎 𝟕 𝟓 𝒑𝒖𝒍𝒈𝟕 Y QUE 𝑬 = 𝟐𝟗(𝟏𝟎𝟑)𝒌𝒔𝒊.

Barra 1

𝜆𝑥 = sen450 = −0.707

𝜆𝑦 = cos 450 = −0.707

𝐿 = 12√32

1 2 3 4

160.155 160.155 -160.155 -160.155 1

160.155 160.155 -160.155 -160.155 2

-160.155 -160.155 160.155 160.155 3

-160.155 -160.155 160.155 160.155 4

K11

Barra 2

𝜆𝑥 = 0 𝜆𝑦 = −1 𝐿 = 48

1 2 5 6

0.000 0.000 0.000 0.000 1

0.000 453.125 0.000 -453.125 2

0.000 0.000 0.000 0.000 5

0.000 -453.125 0.000 453.125 6

K12



Barra 3

𝜆𝑥 = sen36.870 = 0.6

𝜆𝑦 = cos 36.87 = −0.8

𝐿 = 60

1 2 7 8

130.50 -174.00 -130.50 174.00 1

-174.00 232.00 174.00 -232.00 2

-130.50 174.00 130.50 -174.00 7

174.00 -232.00 -174.00 232.00 8

K13


1 2 3 4 5 6 7 8

290.655 -13.845 -160.155 -160.155 0.000 0.000 -130.50 174.00 1

-13.845 845.280 -160.155 -160.155 0.000 -453.125 174.00 -232.00 2

-160.155 -160.155 160.155 160.155 0 0 0 0 3

-160.155 -160.155 160.155 160.155 0 0 0 0 4

0.000 0.000 0.000 0.000 0.000 0.000 0 0 5

0.000 -453.125 0.000 0.000 0.000 453.125 0 0 6

-130.500 -130.500 0.000 0.000 0.000 0.000 130.50 -174.00 7

174.000 174.000 0.000 0.000 0.000 0.000 -174.00 232.00 8

K

14-5 DETERMINACIÓN DEL DESPLAZAMIENTO HORIZONTAL DE LA JUNTA 1 Y LA

FUERZA EN EL ELEMENTO 2.



Elemento 2

𝜆𝑥 = 0 𝜆𝑦 = 1,−1

𝐷1 = −500

𝐷2 = 0

𝐷𝑘 =

000000

Qk=⌈5000⌉

1 2 3 4 5 6 7 8

-500 290.655 -13.845 -160.155 -160.155 0.000 0.000 -130.50 174.00 1 D1

0 -13.845 845.280 -160.155 -160.155 0.000 -453.125 174.00 -232.00 2 D2

Q3 -160.155 -160.155 160.155 160.155 0 0 0 0 3 0

Q4 -160.155 -160.155 160.155 160.155 0 0 0 0 4 0

Q5 0.000 0.000 0.000 0.000 0.000 0.000 0 0 5 0

Q6 0.000 -453.125 0.000 0.000 0.000 453.125 0 0 6 0

Q7 -130.500 -130.500 0.000 0.000 0.000 0.000 130.50 -174.00 7 0

Q8 174.000 174.000 0.000 0.000 0.000 0.000 -174.00 232.00 8 0

K

[−5000

] = [0.003443191 5.63967E − 05

5.63967E − 05 0.001183964] [𝐷1𝐷2]

−500 = 𝐴𝐸(0.003443191D1 − 5.63967E − 05D2)

0 = 5.63967E − 05D1 − 0.001183964D2

𝐷2 =-1.72159573 𝐷 = -0.02819834

ELEMENTO 2

𝜆𝑥 = 0 𝜆𝑦 = −1 𝐿 = 48

𝑄2 =0.75(29(103))

48[0 1 0 −1] [

−1.7116−0.0282

00

]

𝑄2 =0.75(29(103))

48(1)(−0.082) = −𝟏𝟐. 𝟕𝟖𝒍𝒃



14-7 DETERMINE LA MATRIZ DE RIGIDEZ K ARA LA ARMADURA. CONSIDERE QUE

A = 0.0015 𝑴𝟐 Y QUE E = 200 G P A P ARA CADA ELEMENTO.

Barra 1

𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 2

5 6 7 8

150000000 0 -150000000 0 5

0 0 0 0 6

-150000000 0 150000000 0 7

0 0 0 0 8

K11

Barra 2 𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 2

1 2 5 6

150000000 0 -150000000 0 1

0 0 0 0 2

-150000000 0 150000000 0 5

0 0 0 0 6

K12

Barra 3

𝜆𝑥 = sen45 = −0.707

𝜆𝑦 = cos 45 = −0.707

𝐿 = 2√2

1 2 3 4

53020000 -53020000 -53020000 53020000 1

-53020000 53020000 53020000 -53020000 2

-53020000 53020000 53020000 -53020000 3

53020000 -53020000 -53020000 53020000 4

k13



Barra 4 𝜆𝑥 = 0 𝜆𝑦 = 1 𝐿 = 2

5 6 3 4

0 0 0 0 5

0 150000000 0 -150000000 6

0 0 0 0 3

0 -150000000 0 150000000 4

k14

Barra 5 𝜆𝑥 = −0.7071 𝜆𝑦 = −0.7071 𝐿 = 2.8284

3 4 7 8

53020000 53020000 -53020000 -53020000 3

53020000 53020000 -53020000 -53020000 4

-53020000 -53020000 53020000 53020000 7

-53020000 -53020000 53020000 53020000 8

k15

Barra 6 𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 2

3 4 9 10

150000000 0 -150000000 0 3

0 0 0 0 4

-150000000 0 150000000 0 9

0 0 0 0 10

k16


1 2 3 4 5 6 7 8 9 10

203020000 -53020000 -53020000 53020000 -150000000 0 0 0 0 0 1

-53020000 53020000 53020000 -53020000 0 0 0 0 0 0 2

-53020000 53020000 256040000 0 0 0 -53020000 -53020000 -150000000 0 3

53020000 -53020000 0 256040000 0 -150000000 -53020000 -53020000 0 0 4

-150000000 0 0 0 300000000 0 150000000 0 0 0 5

0 0 0 -150000000 0 150000000 0 0 0 0 6

0 0 -53020000 -53020000 -150000000 0 203020000 53020000 0 0 7

0 0 -53020000 -53020000 0 0 53020000 53020000 0 0 8

0 0 -150000000 0 0 0 0 0 150000000 0 9

0 0 0 0 0 0 0 0 0 0 10

KG



14-8 DETERMINE EL DESPLAZAMIENTO VERTICAL EN LA JUNTA 2 Y LA FUERZA EN

LA JUNTA EL ELEMENTO [5]. CONSIDERE QUE 𝑨 = 𝟎. 𝟎𝟎𝟏𝟓 Y QUE 𝑬 = 𝟐𝟎𝟎 𝑮𝑷𝒂.

𝑄𝑘 =

[

0−30000

0000 ]

123456

𝐷𝑘 = [

0000

]

78910

1 2 3 4 5 6 7 8 9 10

203020000 -53020000 -53020000 53020000 -150000000 0 0 0 0 0 D1

-53020000 53020000 53020000 -53020000 0 0 0 0 0 0 D2

-53020000 53020000 256040000 0 0 0 -53020000 -53020000 -150000000 0 D3

53020000 -53020000 0 256040000 0 -150000000 -53020000 -53020000 0 0 D4

-150000000 0 0 0 300000000 0 150000000 0 0 0 D5

0 0 0 -150000000 0 150000000 0 0 0 0 D6

KG

0 = [203.033 𝐷1 − 53.033𝐷2 − 53.033𝐷3 + 53.033𝐷4 − 150𝐷5](106) − 30(103)

= [−53.033𝐷1 + 53.033𝐷2 + 53.033𝐷3 − 53.033𝐷4](106)

0 = [−53.033𝐷1 + 53.033𝐷2 + 256.066𝐷3](106)

0 = [53.033𝐷1 − 53.033𝐷2 + 256.066𝐷4 − 150𝐷6](106)

0 = [−150𝐷4 + 300𝐷5](106)

0 = [−150𝐷4 + 150𝐷6](106)

𝐷1 = −0.0004𝑚 𝐷2 = −0.0023314𝑚 𝐷3 = 0.0004𝑚 𝐷4 = −0.00096569𝑚

𝐷5 = −0.0002𝑚 𝐷6 = 0.00096569𝑚 = 0.000966𝑚

𝜆𝑦 = −cos 45 . 𝜆𝑥 = −𝑠𝑒𝑛 45

𝐿 = 2.8284𝑀

(𝑞5)𝐹 =0.0015[200(109)]

2.8284[. 707 .707 − .707 − .707]= -42.4 kN



14.9 DETERMINE LA MATRIZ DE RIGIDEZ K PARA LA ARMADURA. CONSIDERE QUE

A= 0.0015 𝑴𝟐 Y QUE E = 200 GPA PARA CADA ELEMENTO.

Barra 1

𝜆𝑥 = cos 36.87 = − 0.8

𝜆𝑦 = 𝑠𝑒𝑛 36.87 = 0.6

𝐿 = 5𝑚

Barra 2 𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 4𝑚

1 2 3 4

75000000 0 -75000000 0 1

0 0 0 0 2

-75000000 0 75000000 0 3

0 0 0 0 4

k12

Barra 3 𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 4𝑚

3 4 9 10

75000000 0 -75000000 0 3

0 0 0 0 4

-75000000 0 75000000 0 9

0 0 0 0 10

k13

Barra 4 𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 3𝑚

3 4 5 6

0 0 0 0 3

0 100000000 0 -100000000 4

0 0 0 0 5

0 -100000000 0 100000000 6

k14

1 2 5 6

38400000 28800000 -38400000 -28800000 1

28800000 21600000 -28800000 -21600000 2

-38400000 -28800000 38400000 28800000 5

-28800000 -21600000 28800000 21600000 6

k11



Barra 5 𝜆𝑥 = cos 36.87 = − 0.8 𝜆𝑦 = 𝑠𝑒𝑛 36.87 = 0.6 𝐿 = 5𝑚

5 6 9 10

38400000 28800000 -38400000 -28800000 5

28800000 21600000 -28800000 -21600000 6

-38400000 -28800000 38400000 28800000 9

-28800000 -21600000 28800000 21600000 10

k15

Barra 6 𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 4𝑚

5 6 8 7

75000000 0 -75000000 0 5

0 0 0 0 6

-75000000 0 75000000 0 8

0 0 0 0 7

k16

Barra 7 𝜆𝑥 = 0 𝜆𝑦 = 1 𝐿 = 3𝑚

8 7 9 10

0 0 0 0 8

0 100000000 0 -100000000 7

0 0 0 0 9

0 -100000000 0 100000000 10

k17




1 2 3 4 5 6 7 8 9 10

113400000 28800000 -75000000 0 -38400000 -28800000 0 0 0 0 1

28800000 21600000 0 0 -28800000 -21600000 0 0 0 0 2

-75000000 0 150000000 0 0 0 0 0 0 0 3

0 0 0 100000000 0 -100000000 0 0 0 0 4

-38400000 -28800000 0 0 84600000 57600000 0 -75000000 0 0 5

-28800000 -21600000 0 -100000000 57600000 143200000 0 0 0 0 6

0 0 0 0 0 0 100000000 0 0 -100000000 7

0 0 0 0 -75000000 0 0 75000000 0 0 8

0 0 -75000000 0 -75000000 -28800000 0 0 0 0 9

0 0 0 0 -28800000 -28800000 -100000000 0 0 100000000 10

KG

14.10, 14-11 DETERMINE LA FUERZA EN EL ELEMENTO (5). CONSIDERE QUE A =

00015𝐌𝟐 Y QUE E = 200 GPA PARA CADA ELEMENTO.

1 2 3 4 5 6 7

0 113400000 28800000 -75000000 0 -38400000 -28800000 0 D1 0

-20000 28800000 21600000 0 0 -28800000 -21600000 0 D2 -20000

0 -75000000 0 150000000 0 0 0 0 D3 0

0 0 0 0 100000000 0 -100000000 0 D4 0

0 -38400000 -28800000 0 0 84600000 57600000 0 D5 0

0 -28800000 -21600000 0 -100000000 57600000 143200000 0 D6 0

0 0 0 0 0 0 0 100000000 D7 0

2.66667E-08 -3.55556E-08 1.33333E-08 -2.34968E-23 -3.1329E-23 -2.34968E-23 0 D1 0.000711111

-3.55556E-08 0.00000014 -1.77778E-08 4.62963E-08 2.09832E-22 4.62963E-08 0 D2 -0.0028

1.33333E-08 -1.77778E-08 1.33333E-08 -1.18424E-23 -1.18424E-23 -1.18424E-23 0 D3 0.000355556

0 4.62963E-08 0 2.84217E-07 -1.7094E-07 2.74217E-07 0 D4 -0.000925926

0 1.15348E-22 0 -1.7094E-07 1.28205E-07 -1.7094E-07 0 D5 -2.30696E-18

0 4.62963E-08 0 2.74217E-07 -1.7094E-07 2.74217E-07 0 D6 -0.000925926

0 0 0 0 0 0 0.00000001 D7 0

M INVERSA



0.8 -0.6 -0.8 0.6 -2.30696E-18

-0.000925926

0

0

𝑸𝟓 = 𝟑𝟑. 𝟑𝟑

14.12 DETERMINE LA MATRIZ DE RIGIDEZ K PARA LA ARMADURA. CON SIDERE QUE

𝑨 = 𝟐 𝒑𝒖𝒍𝒈 𝟐 Y QUE 𝑬 = 𝟐 𝟗 ( 𝟏𝟎𝟑) KSI.

Barra 1 𝜆𝑥 = 0 𝜆𝑦 = −1 𝐿 = 72𝑖𝑛

1 2 5 6

0 0 0 0 1

0 805.556 0 -805.56 2

0 0 0 0 5

0 -805.56 0 805.556 6

K11

Barra 2 𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 96𝑖𝑛

5 6 3 4

604.167 0 -604.17 0 5

0 0 0 0 6

-604.17 0 604.167 0 3

0 0 0 0 4

K12



Barra 3 𝜆𝑥 = −1 𝜆𝑦 = 0 𝐿 = 96𝑖𝑛

1 2 7 8

604.167 0 -604.17 0 1

0 0 0 0 2

-604.17 0 604.167 0 7

0 0 0 0 8

K13

Barra 4 𝜆𝑥 = 0 𝜆𝑦 = −1 𝐿 = 72𝑖𝑛

3 4 7 8

0 0 0 0 3

0 805.556 0 -805.56 4

0 0 0 0 7

0 -805.56 0 805.556 8

K14

Barra 5 𝜆𝑥 = −0.8 𝜆𝑦 = −0.6 𝐿 = 120𝑖𝑛

1 2 3 4

309.333 232 -309.33 -232 1

232 174 -232 -174 2

-309.33 -232 309.333 232 3

-232 -174 232 174 4

K15

Barra 6 𝜆𝑥 = −0.8 𝜆𝑦 = 0.6 𝐿 = 120𝑖𝑛

5 6 7 8

309.333 -232 -309.33 232 5

-232 174 232 -174 6

-309.33 232 309.333 -232 7

232 -174 -232 174 8

K16

ENSAMBLE PARA LA DETERMINACIÓN DE LA MATRIZ DE RIGIDEZ K

1 2 3 4 5 6 7 8

913.5 232 -309.333 -232 0 0 -604.167 0 1

232 979.556 -232 -174 0 -805.556 0 0 2

-309.333 -232 913.5 232 -604.167 0 0 0 3

-232 -174 232 979.556 0 0 0 -805.556 4

0 0 -604.167 0 913.5 -232 -309.333 232 5

0 -805.556 0 0 -232 979.556 232 -174 6

-604.167 0 0 0 -309.333 232 309.333 -232 7

0 0 0 -805.556 232 -174 -232 979.556 8

KG



14.13 DETERMINE E L DESPLAZAMIENTO HORIZONTAL DE LA JUNTA Y LA FUERZA

E N E L ELEMENTO [5]. CONSIDERE QUE A -2 PULG2 Y QUE E = 29(𝟏𝟎𝟑) KSI. NO TOME

EN CUENTA EL ESLABÓN CORTO EN (2).

14.14 DETERMINE LA FUERZA EN EL ELEMENTO [3] SI EL ELEMENTO ERA 0.025

PULGADAS MÁS CORTO DE LO ESPERADO ANTES DE AJUSTARSE EN LA

ARMADURA. CONSIDERE QUE A = 2 PULG2 Y QUE E=29(𝟏𝟎𝟑) KSI. NO TOME EN

CUENTA EL ESLABÓN CORTO E N (2).

913.5 232 -309.333 -232 0 D1 0

232 979.556 -232 -174 0 D2 0

-309.333 -232 913.5 232 -604.167 D3 3

-232 -174 232 979.556 0 D4 0

0 0 -604.167 0 913.5 D5 0

CONOCIDOS

0.00140996 -0.00013793 0.00072414 0.00013793 0.00047893 D1 0.00217241

-0.00013793 0.00116379 0.00040733 7.7586E-05 0.0002694 D2 0.00122198

0.00072414 0.00040733 0.00274946 -0.00040733 0.00181843 D3 0.00824839

0.00013793 7.7586E-05 -0.00040733 0.00116379 -0.0002694 D4 -0.00122198

0.00047893 0.0002694 0.00181843 -0.0002694 0.00229736 D5 0.00545529

M INVERSA



16-1 DETERMINAR LA MATRIZ DE LAS RIGIDECES K DE LA ESTRUCTURA PARA EL

MARCO.

DATOS

𝐴 = 10 𝐸3 𝑚𝑚4. = 0.01𝑚2

𝐼 = 300 𝐸6 𝑚𝑚4 = 0.0003𝑚4.

𝐸 = 200 𝐺𝑝𝑎.= 200𝐸9𝑁

𝐿 = 4𝑚

Componentes de la matriz de rigideces.

𝐴𝐸

𝐿=(0.01𝑚)(200𝐸9𝑁)

4= 500𝐸6 𝑁/𝑚

12𝐸𝐼

𝐿3=12(200𝐸9𝑁)(300 𝐸−6)

42= 11.25𝐸6 𝑁/𝑚

6𝐸𝐼

𝐿2=6(200𝐸9𝑁)(0.0003𝑚4)

42= 22.5𝐸6 𝑁

4𝐸𝐼

𝐿=4(200𝐸9𝑁)(0.0003𝑚4)

4= 60𝐸6 𝑁.𝑚

2𝐸𝐼

𝐿=2(200𝐸9𝑁)(0.0003𝑚4)

4= 30𝐸6 𝑁.𝑚

Elemento 1 (VIGA)

𝜆𝑋 =4−0

4= 1 𝜆𝑌 =

0−0

4= 0

7 8 9 1 2 3 500 0 0 -500 0 0 7

0 11.25 22.5 0 -11.25 22.5 8 0 22.5 60 0 -22.5 30 9

-500 0 0 500 0 0 1 0 -11.25 -22.5 0 11.25 -22.5 2

0 22.5 30 0 -22.5 60 3

E6



7

8

9

1

2

3

500 0 0 -500 0 0 7 0 11.25 22.5 0 -11.25 22.5 8 0 22.5 60 0 -22.5 30 9

-500 0 0 500 0 0 1 0 -11.25 -22.5 0 11.25 -22.5 2 0 22.5 30 0 -22.5 60 3

Elemento 2 (COLUMNA)

𝜆𝑋 =4−4

4= 0 𝜆𝑌 =

−4−0

4= −1

Matriz de rigideces del marco

1 2 3 4 5 6 7 8 9 511.25 0 22.5 -11.25 0 22.5 -500 0 0 1

0 511.25 -22.5 0 -500 0 0 -11.25 -22.5 2

22.5 -22.5 120 -22.5 0 30 0 22.5 30 3

-11.25 0 -22.5 11.25 0 -22.5 0 0 0 4 0 -500 0 0 500 0 0 0 0 5

22.5 0 30 -22.5 0 60 0 0 0 6

-500 0 0 0 0 0 500 0 0 7 0 -11.25 22.5 0 0 0 0 11.25 22.5 8 0 -22.5 30 0 0 0 0 25.5 60 9

E6

1 2 3 4 5 6 11.25 0 22.5 -11.25 0 22.5 1

0 500 0 0 -500 0 2

22.5 0 60 -22.5 0 30 3

-11.25 0 -22.5 11.25 0 -22.5 4

0 -500 0 0 500 0 5

22.5 0 30 -22.5 0 60 6

E6



16-2 DETERMINE LAS REACCIONES EN LOS SOPORTES FIJOS 1 Y 2. CONSIDERE

QUE E = 200 GPA, I = 300(106) MM4 Y A =10(103) MM2 PARA CADA ELEMENTO.

Conociendo las cargas en los nodos y las deflexiones

𝑄𝑘 [

−5(103)

−24(103)

11(103)

]123 𝐷𝑘

[ 000000] 456789

Relación de cargas-desplazamiento Q=KD

[ −5(103)

−24(103)

11(103)𝑄4𝑄5𝑄6𝑄7𝑄8𝑄9 ]

=

[ 511.25 0 22.50 511.25 −22.522.5−11.25022.5−50000

−22.50

−50000

−11.25−22.5

120−22.5030022.530

−11.25 0 22.50 −500 0

−22.511.50

−22.5000

005000000

30−22.5060000

−500 0 00 −11.25 −22.5000050000

22.50000

11.2522.5

30000022.560 ]

(106)

[ 𝐷1𝐷2𝐷3000000 ]



Qu = K11Du + K12DK

−5(103) = (511.25𝐷1 + 22.5𝐷3)(106) (1)

−24(103) = (511.25𝐷2 − 22.5𝐷3)(106) (2)

11(103) = (511.25𝐷1 − 22.5𝐷2 + 120𝐷3)(106) (3)

Resolviendo las ecuaciones 1, 2 y 3

𝐷1 = 13.57(10−6)𝑚

𝐷2 = −43.15(10−6)𝑚

𝐷3 = 86.12(10−6)𝑟𝑎𝑑

Usando estos resultados y aplicándolos en Qu = K21Du + K22Dk

𝑄4 = −11.25(106)(−13.57)(10−6) + (−22.5)(106)(86.12)(10−6) = −1.785𝐾𝑁

𝑄5 = −500(106)(−43.15)(10−6) = 21.58𝐾𝑁

𝑄6 = 22.5(106)(−13.57)(10−6) + 30(106)(86.12)(10−6) = 2.278𝐾𝑁.𝑚

𝑄7 = −500(106)(−13.57)(10−6) = 6.785𝐾𝑁

𝑄8 = −11.25(106)(−43.15)(10−6) + 22.5(106)(86.12)(10−6) = 2.423𝐾𝑁

𝑄9 = −22.5(106)(−43.15)(10−6) + 30(106)(86.12)(10−6) = 3.555𝐾𝑁.𝑚

Reacciones.

𝑅4 = −1.785 + 5 = 3.214𝐾𝑁 = 3.21KN

𝑅5 = 21.58 + 0 = 21.58𝐾𝑁 = 21.6𝑁

𝑅6 = 2.278 − 5 = −2.722𝐾𝑁.𝑚 = −2.722𝐾𝑁.𝑚

𝑅7 = 6.785 + 0 = 6.785𝐾𝑁 = 6.79𝐾𝑁

𝑅8 = 2.423 + 24 = 26.42𝐾𝑁 = 26.4𝐾𝑁

𝑅9 = 3.555 + 16 = 19.55𝐾𝑁.𝑚 = 19.6𝐾𝑁.𝑚



16-3 DETERMINE LA MATRIZ DE RIGIDEZ DE ESTRUCTURA K PARA EL MARCO.

SUPONGA QUE (3). ESTÁ ARTICULADO Y QUE (1) ESTÁ FIJO. CONSIDERE QUE E =

200 MPA, I= 300(10)6 MM4 Y A=21(10)3 MM2 PARA CADA ELEMENTO.

ELEMENTO 1:

𝜆𝑥 =5−0

5= 1 𝜆𝑦 = 0

𝐴𝐸

𝐿=(0.021)(200𝐸8)

5= 840000

6𝐸𝐼

𝐿2=6(200𝐸6)(300𝐸−6)

52= 14000

4𝐴𝐸

𝐿=4(200𝐸6)(300𝐸−6)

5= 48000

12𝐸𝐼

𝐿3=12(200𝐸6)(300𝐸−6)

53 = 5760

2𝐸𝐼

𝐿=2(200𝐸6)(300𝐸−6)

5= 24000

K1=

840000 0 0 −840000 0 00 5760 14400 0 −5760 144000 14400 48000 0 −14400 24000

−840000 0 0 840000 0 00 −5760 −14400 0 5760 −144000 14400 24000 0 −14400 48000



ELEMENTO 2:

𝜆𝑌 =0 − (−4)

4= 1 𝜆𝑋 = 0

𝐴𝐸

𝐿=(0.021)(200𝐸6)

4= 1050000

6𝐸𝐼

𝐿2=6(200𝐸6)(300𝐸−6)

42= 22500

4𝐴𝐸

𝐿=4(200𝐸6)(300𝐸−6)

4= 60000

12𝐸𝐼

𝐿3=12(200𝐸6)(300𝐸−6)

43 = 11250

2𝐸𝐼

𝐿=2(200𝐸6)(300𝐸−6)

4= 30000

K2=

11250 0 −22500 −11250 0 −225000 1050000 0 0 −1050000 0

−22500 0 60000 22500 0 30000−11250 0 22500 11250 0 22500

0 −1050000 0 0 1050000 0−22500 0 30000 22500 0 60000

𝐾 =

851250 0 22500 22500 −11250 0 −840000 0 00 1055760 −14400 0 0 −1050000 0 −5760 −14400

22500 −14400 108000 30000 −22500 0 0 14400 2400022500 0 30000 60000 −22500 0 0 0 0−11250 0 −22500 −22500 11250 0 0 0 0

0 −1050000 0 0 0 1050000 0 0 0−840000 0 0 0 0 0 840000 0 0

0 −5760 14400 0 0 0 0 5760 144000 −14400 24000 0 0 0 0 14400 48000



16-4 DETERMINE LAS REACCIONES CON LOS SOPORTES 1 Y 3 CONSIDERE QUE

E= 200 MPA, I=300 (10⁶) MM⁴ Y A= 21(10³) MM² PARA CADA ELEMENTO.

7 8 9 1 2 3

𝐾1 =

[ 840000

00

−84000000

05760144000

−576014400

014400480000

−1440024000

−84000000

84000000

0−5760−14400

05760−14400

014400240000

−1440048000 ]

789123

1 2 3 5 6 4

𝐾2 =

[ 112500

−22500−11250

0−22500

01050000

00

−10500000

−225000

60000225000

30000

−112500

22500112500

22500

0−1050000

00

10500000

−225000

30000225000

60000 ] 123564



MATRIZ GLOBAL

1 2 3 4 5 6 7 8 9

𝐊 =

[ 851250

02250022500−11250

0−840000

00

01055760−14400

00

−10557600

−5760−14400

22500−1440010800030000−22500

00

1440024000

225000

3000060000−22500

0000

−112500

−22500−22500112500000

0−1050000

000

1050000000

−84000000000

84000000

0−5760144000000

57601440

0−14400240000000

1440048000 ]

𝟏𝟐𝟑𝟒𝟓𝟔𝟕𝟖𝟗

𝑫𝑲 =

[ 𝟎𝟎𝟎𝟎𝟎]

𝑸𝑲 = [

𝟎𝟎𝟑𝟎𝟎𝟎

]

[ 𝟎𝟎𝟑𝟎𝟎𝟎𝑸𝟓𝑸𝟔𝑸𝟕𝑸𝟖𝑸𝟗 ]

=

[ 851250

02250022500−11250

0−840000

00

01055760−14400

00

−10557600

−5760−14400

22500−1440010800030000−22500

00

1440024000

225000

3000060000−22500

0000

−112500

−22500−22500112500000

0−1050000

000

1050000000

−84000000000

84000000

0−5760144000000

57601440

0−14400240000000

1440048000 ]

[ 𝑫𝟏𝑫𝟐𝑫𝟑𝑫𝟒𝟎𝟎𝟎𝟎𝟎 ]

CALCULO DE LAS REACCIONES

[

003000

] = [

8512500

2250022500

01055760−14400

0

22500−1440010800030000

225000

3000060000

] [

D1D2D3D4

] + [

0000

]

0 = 851250 𝐷1 + 22500 𝐷3 + 22500 𝐷4

0 = 1055760 𝐷2 +−14400 𝐷3

300 = 22500 𝐷1 +−14400 𝐷2 + 108000 𝐷3 + 30000 𝐷4

0 = 22500 𝐷1 + 30000 𝐷3 + 60000 𝐷4



SOLUCION

𝐷1 = −0.00004322 𝑚 𝐷2 = 0.00004417 𝑚 𝐷3 = 0.00323787 𝑟𝑎𝑑 𝐷4 = −0.00160273 𝑟𝑎𝑑

[ Q5Q6Q7Q8Q9]

=

[ −11250

0−840000

00

0−1050000

0−5760−14400

−2250000

1440024000

−225000000

]

[

−0.000043220.000044170.00323787−0.00160273

] +

[ 00000]

𝐐𝟓 = −𝟑𝟔. 𝟑 𝑲𝑵 𝐐𝟔 = −𝟒𝟔. 𝟒 𝑲𝑵 𝐐𝟕 = 𝑵𝟑𝟔. 𝟑 𝑲𝑵 𝐐𝟖 = 𝟒𝟔. 𝟒 𝑲𝑵 𝐐𝟗 = 𝟕𝟕. 𝟏 𝑲𝑵 ∗𝒎



16-5 DETERMINE LA MATRIZ DE RIGIDEZ DE LA ESTRUCTURA K PARA EL MARCO.

CONSIDERE QUE E= 200 GPA, I= 350 (𝟏𝟎𝟔 ) 𝒎𝒎𝟒 Y 𝑨 = 𝟏𝟓(𝟏𝟎𝟑)𝒎𝒎𝟐 PARA CADA ELEMENTO. LAS JUNTAS EN 1 Y 3 ESTÁN ARTICULADOS.

𝐸 = 200 𝐺𝑃𝑎 = 200 ∗ 109𝑁/𝑚

𝐼 = 350 ∗ 106𝑚𝑚4 = 350 ∗ 10−6𝑚4

𝐴 = 15 ∗ 103𝑚𝑚2 = 0.015 𝑚2 𝐴𝐸

𝐿=(0.015)(200𝐸9)

4= 750𝐸6𝑁/𝑚

12𝐸𝐼

𝐿3=12(200𝐸9)(350𝐸−6)

43 = 13.125𝐸6𝑁/𝑚

6𝐸𝐼

𝐿2=6(200𝐸9)(350𝐸−6)

42= 26.25𝐸6𝑁

4𝐸𝐼

𝐿=4(200𝐸9)(350𝐸−6)

4= 70𝐸6𝑁 ∙ 𝑚

2𝐸𝐼

𝐿=2(200𝐸9)(350𝐸−6)

4= 35𝐸6𝑁 ∙ 𝑚

Por miembro 1 𝜆𝑥 =4−0

4= 1 𝜆𝑦 =

0−0

4= 0

𝐾1 =

0080 00009 0000500 00010000 020000 03000

[ 750 0 0 −750 0 00 13.125 26.25 0 −13.125 26.250 26.25 70 0 −26.25 35

−750 0 0 750 0 00 −13.125 −26.25 0 13.125 −26.250 26.25 35 0 −26.25 70 ]

(106)

Por miembro 2 𝜆𝑥 =4−4

4= 0 𝜆𝑦 =

−4−0

4= −1

𝐾2 =

0080 00009 0000500 00010000 020000 03000

[ 750 0 0 −750 0 00 13.125 26.25 0 −13.125 26.250 26.25 70 0 −26.25 35

−750 0 0 750 0 00 −13.125 −26.25 0 13.125 −26.250 26.25 35 0 −26.25 70 ]

(106)

Matriz general 9X9

𝐾2 =

000010 00002000 0000300 0004000000 0050000 60000 7 08000 9

[ 763.125 0 26.25 26.25 0 −13.125 0 0 0

0 763.125 −26.25 0 −26.25 0 −750 −750 −13.12526.25 −26.25 140 35 35 −26.25 0 0 26.2526.25 0 35 70 0 −26.25 0 0 00 −26.25 35 0 70 0 0 0 26.25

−13.125 0 26.25 −26.25 0 13.125 0 0 00 −750 0 0 0 0 750 750 0

−750 0 0 0 0 0 0 0 00 −13.125 26.25 0 26.25 0 0 0 13.125 ]

(106)



16-6 DETERMINE LAS REACCIONES EN LOS SOPORTES ARTICULADOS 1 Y 3

CONSIDERE QUE E=200 GPA, I=350 (𝟏𝟎𝟔 ) MM4 Y A=15 (𝟏𝟎𝟑) MM2 PARA CADA ELEMENTO.

𝑄𝑘 =

[

0 −41.25(103)

45(103) 0 0 ]

1 234 5

𝐷𝑘 = [

0 0 0 0

]

6 7 8 9

[

0 −41.25(103)

45(103) 0 0𝑄6𝑄7𝑄8𝑄9 ]

=

[ 763.125

026.25 26.25 0

−13.1250

−7500

0 763.125−26.250

−26.250

−7500

−13.125

26.25 −26.2514035 35

−26.2500

26.25

26.25 03570 0

−26.25000

0 −26.25350 70000

26.25

−13.125 0

−26.25−26.25 0

13.125000

0 −75000 0075000

−750 000 0007500

0 −13.12526.250

26.25000

13.125

]

106

[ 𝐷1 𝐷2 𝐷3 𝐷4𝐷50000 ]

Desde la partición de matriz, Qk= K11Du + K12Dk

0= (763.125D1 + 26 .25D3 + 26.25D4) ( 106 ) (1)

-41.25 (103)= 763.125D1 + 26 .25D3 + 26.25D5) ( 106 ) (2)

45(103)= (26.25D1 - 26.25D2 + 140D3 + 35D4 + 35D4) ( 106 ) (3)

0= (26.25D1 + 35D3 + 70D4) (106) (4)

0= (26.25D1 + 35D3 + 70D4) (106) (5)

Resultados ecuación. (1) a (5)

D1= -7.3802(10−6) D2= -47.3802(10−6) D3= 423.5714(10−6)

D5= -209.0181(10−6) D5= -229.5533(10−6)

Utilizando estos resultados y aplicando Qu= K21Du +K22Dk



Q6= (-13.125) (10−6) – 7.3802(10−6) – 26.25(10−6)423.5714(10−6) – 26.25(106) - 209.0181(10−6) + 0 = -

5.535 kN

Q7= -750(106) – 47.3802(10−6) + 0 = 35.535 kN

Q8= -750(106) – 7.3802(10−6) + 0 = 5.535 kN

Q9= -13.125(106) – 47.3802(10−6) + 26.25(106) + 423.5714(10−6) + 26.25(106) – 229.5533(10−6) + 0 =

5.715 kN

Superponer estos resultados a los de la figura (a),

R6 = - 5. 535 kN + 0 = 5.54 kN

R7= 35.535 + 0 = 35.5 kN

R8= 5.535 + 0 = 5.54 kN

R9= 5.715 + 18.75 = 24.5 kN



16.7 DETERMINE LA MATRIZ DE RIGIDEZ DE LA ESTRUCTURA K, PARA EL MARCO.

CONSIDERE QUE 𝑬 = 𝟐𝟗 × 𝟏𝟎𝟑 𝒌𝒔𝒊 , 𝑰 = 𝟔𝟓𝟎𝒑𝒖𝒍𝒈𝟒 , 𝑨 = 𝟐𝟎𝒑𝒖𝒍𝒈𝟐 , PARA CADA

ELEMENTO.

Miembro 1

λx =10 − 0

10= 1 λy = 0

𝐴𝐸

𝐿=(20)(29 × 103)

10(12)= 4833.33

6𝐸𝐼

𝐿2=6(29 × 103)(650)

(10)2(12)2= 7854.17

2𝐸𝐼

𝐿=2(29 × 103)(650)

10(12)= 314166.67

12𝐸𝐼

𝐿3=12(29 × 103)(650)

(10)3(12)3= 130.90

4𝐸𝐼

𝐿=4(29 × 103)(650)

10(12)= 628333.33

𝐾1 =

[ 4833.33

00

−4833.3300

0130.907854.17

0−130.907854.17

07854.17628333.33

0−7854.17314166.67

−4833.3300

4833.3300

0−130.90−7854.17

0130.90−7854.17

07854.17314166.6

0−7854.17628333.33]



Miembro 2

λY =12 − 0

12= 1 λX = 0

𝐴𝐸

𝐿=(20)(29 × 103)

(12)(12)= 4027.78

6𝐸𝐼

𝐿2=6(29 × 103)(650)

(12)2(12)2= 5454.28

2𝐸𝐼

𝐿=2(29 × 103)(650)

(12)(12)= 261805.55

12𝐸𝐼

𝐿3=12(29 × 103)(650)

(12)3(12)3= 75.75

4𝐸𝐼

𝐿=4(29 × 103)(650)

(12)(12)= 523611.11

𝐾2 =

[ 75.750

5454.28−75.750

5454.28

04027.78

00

−4027.780

5454.280

523611.11−5454.28

0261805.55

−75.750

−5454.2875.750

−5454.28

0−4027.78

00

4027.780

5454.280

261805.55−5454.28

0523611.11]

MATRIZ DE RIGIDEZ ESTRUCTURAL

𝐾:

[ 4833.33

00

−4833.33 0 0000

0 130.907854.17

0−130.907854.17

000

0 7854.17628333.33

0−7854.17314166.67

000

−4833.33 00

4909.08 0

5454.28−75.750

5454.28

0 −130.90−7854.17

0 4158.68−7854.17

0−4027.78

0

0 7854.17314166.675454.28−7854.171151944.44−5454.28

0261805.55

0 00

−75.75 0

−5454.2875.750

−5454.28

0 000

−4027.7800

4027.780

0 00

5454.28 0

261805.55−5454.28

0523611.11

]



16-8 DETERMINE LOS COMPONENTES DE LOS DESPLAZAMIENTOS EN 1

CONSIDERE QUE E =29(𝟏𝟎𝟑 ) KSI, I=650𝒑𝒖𝒍𝒈𝟒 Y A=20 𝒑𝒖𝒍𝒈𝟐PARA CADA ELEMENTO.

𝐷𝑘=|000| 𝑄𝑘=

|

|

−4−60000

|

|

[ −4 −6 0 0 00𝑄7𝑄8𝑄9 ]

=

[ 4833.33

00

−4833.33 0 0000

0 130.907854.17

0−130.907854.17

000

0 7854.17628333.33

0−7854.17314166.67

000

−4833.3300

4909.08 0

5454.28−75.750

5454.28

0 −130.90−7854.17

04158.68−7854.17

0−4027.78

0

0 7854.17314166.675454.28−7854.171151944.44−5454.28

0261805.55

0 00

−75.75 0

−5454.2875.750

−5454.28

0 000

−4027.7800

4027.780

0 00

5454.28 0

261805.55−5454.28

0523611.11

]

[ 𝐷1 𝐷2 𝐷3 𝐷4𝐷5𝐷6000 ]

[ −4 −6 0 0 00 ]

+

[ 4833.33

00

−4833.33 0 0

0 130.907854.17

0−130.907854.17

0 7854.17628333.33

0−7854.17314166.67

−4833.3300

4909.08 0

5454.28

0 −130.90−7854.17

04158.68−7854.17

0 7854.17314166.675454.28−7854.171151944.44

]

[ 𝐷1 𝐷2 𝐷3 𝐷4𝐷5𝐷6]

+

[ 0 00 000]



-4 = 4833.33D1 - 4833.33D4

-6 = 130.90D2 + 7854.17D3 - 130.90D5 + 7854.17D6

0 = 7854.17D2 + 628333.33D3 - 7854.17D5 + 314166.67D6

0 = -4833.33D1 + 4909.08D4 + 5454.28D6

0 = -130.90D2 - 7854.17D3 + 4158.68D5 - 7854.17D6

0 = 7854.17D2 + 314166.67D3 + 5454.28D4 – 7854.17D5 + 11519944.44D6

Resolviendo los rendimientos de las ecuaciones anteriores

D1= -0.608 in

D2= -1.12 in

D3= 0.0100 rad

D4= -0.6076 in

D5= -0.001490 in

D6= 0.007705 rad



16.9 DETERMINE LA MATRIZ DE RIGIDEZ K PARA EL MARCO. CONSIDERE 𝑬 =

𝟐𝟗 × 𝟏𝟎𝟑 𝒌𝒔𝒊, 𝑰 = 𝟑𝟎𝟎𝒑𝒖𝒍𝒈𝟒, 𝑨 = 𝟏𝟎𝒑𝒖𝒍𝒈𝟐, PARA CADA ELEMENTO.

Miembro 1

λy =10 − 0

10= 1 λx = 0

𝐴𝐸

𝐿=(10)(29 × 103)

10(12)= 2416.67 𝑘/𝑖𝑛

6𝐸𝐼

𝐿2=6(29 × 103)(300)

(10)2(12)2= 3625 𝑘

2𝐸𝐼

𝐿=2(29 × 103)(300)

10(12)= 145000 𝑘. 𝑖𝑛

12𝐸𝐼

𝐿3=12(29 × 103)(300)

(10)3(12)3= 60.4167 𝑘/𝑖𝑛

4𝐸𝐼

𝐿=4(29 × 103)(300)

10(12)= 290000 𝑘. 𝑖𝑛

𝐾1 =

[

860.4167

0−3625−60.4167

0−3625

90

2416.6700

−2416.670

5−36250

29000036250

145000

1−60.4167

0362560.4167

03625

20

−2416.6700

2416.670

3−36250

14500036250

290000] 895123



Miembro 2

λx =20 − 0

20= 1 λy =

10 − 10

20= 0

𝐴𝐸

𝐿=(10)(29 × 103)

(120)(12)= 1208.33 𝑘/𝑖𝑛

6𝐸𝐼

𝐿2=6(29 × 103)(300)

(20)2(12)2= 906.25 𝑘

2𝐸𝐼

𝐿=2(29 × 103)(300)

(20)(12)= 72500 𝑘. 𝑖𝑛

12𝐸𝐼

𝐿3=12(29 × 103)(300)

(20)3(12)3= 7.5521 𝑘/𝑖𝑛

4𝐸𝐼

𝐿=4(29 × 103)(300)

(20)(12)= 145000 𝑘. 𝑖𝑛

𝐾2 =

[

11208.33

00

−1208.3300

20

7.5521906.250

−7.5521906.25

30

906.25145000

0−906.2572500

6−1208.33

00

1208.3300

70

−7.5521−906.25

07.5521−906.25

40

906.25725000

−906.25145000 ]

123684

Matriz de rigidez estructural

𝐾:

[ 1268,75

036250

3625−1208.33

0−60.4167

0

0 2424.22906.25906.2500

−7.55210

−2416.67

3625 906.2543500072500145000

0−906.25−36250

0 906.2572500145000 00

−906.2500

36250

1450000

29000000

−36250

−1208.33 0000

1208.33000

0 −7.5521−906.25−906.25

00

7.552100

−60.4167 0

−36250

−362500

60.41670

0 −2416.67

00 0000

2416.67

]



16-10 DETERMINE LA REACCIONES DE SOPORTE EN (1) Y (3). TOMAR E = 2911032 KSI, I = 300 IN4, A = 10 IN2 PARA CADA MIEMBRO.

Cargas y deflexiones normales conocidas. Las cargas normales que actúan sobre el grado de libertad.

20 = - 2416.67D2 D2 = - 8.275862071 (10-3) 5 = - 7.5521 (- 8.2758)(10- 3) - 906.25D3 - 906.25D4 0 = 906.25 (- 8.2758)(10- 3) + 72500D3 + 145000D4 4.937497862 = - 906.25D3 - 906.25D4



Desde la partición de matriz Qk = K11Du + K12Dk,

0 = 1268.75D1 + 3625D3 + 3625D5 - 1208.33D6 (1) -25 = 2424.22D2 + 906.25D3 + 906.25D4 (2) -1200 = 3625D1 + 906.25D2 + 435000D3 + 72500D4 + 145000D5 (3) 0 = 906.25D2 + 72500D3 + 145000D4 (4) 0 = 3625D1 + 145000D3 + 290000D5 (5) 0 = -1208.33D1 + 1208.33D6 (6)

Resolución de La ecu. (1) a (6) D1 = 1.32 D2 = -0.008276 D3 = -0.011 D4 = 0.005552 D5 = -0.011 D6 = 1.32

Usando estos resultados y aplicando Qk = K21Du + K22Dk

Q7 = -7.5521 (-0.008276) - 906.25(-0.011) - 906.25(0.005552) = 5

Q8 = 60.4167 (1.32)-3625(-0.011)-3625(-0.011) = 0

Q9 = -2416.67 (-0.008276) = 20

Superponer estos resultados a los de FEM mostrados

R7 = 5 + 15 = 20 k Ans.

R8 = 0 + 0 = 0 Ans.

R9 = 20 + 0 = 20 k Ans.



16.-11 DETERMINE LA MATRIZ DE RIGIDEZ DE LA ESTRUCTURA K PARA EL MARCO,

CONSIDERE QUE E =29(103) KSI. 1= 700 PULG4 Y A = 20 PULG2 PARA CADA

ELEMENTO.

𝐴𝐸

𝐿=20((29(102))

24(12)= 2013.89𝐾/𝑖𝑛

13𝐸𝐼

𝐿2=12(29(103)(700)

(24(12))2= 10.197𝑘/𝑖𝑛

6𝐸𝐼

𝐿2=6((29(102))(700)

24(12))2= 1468.46 𝐾

4𝐸𝐼

𝐿=4(29(103)(700)

(24(12)= 281944 𝐾. 𝑖𝑛

2𝐸𝐼

𝐿=2(29(103))(700)

24(12)= 140972 𝐾. 𝑖𝑛



L=16 ft AX = 24−24

16= 0 𝑎𝑛𝑑 𝑥𝑦 =

16−0

16= 1.

𝐴𝐸

𝐿=20((29(103))

16(12)= 3020.83𝐾/𝑖𝑛

12𝐸𝐼

𝐿3=12(29(103)(700)

(16(12))3= 34.4170𝑘/𝑖𝑛

6𝐸𝐼

𝐿=6((29(103))(700)

16(12))2= 3304.04𝐾

4𝐸𝐼

𝐿=4(29(103)(700)

(16(12)= 42217 𝐾. 𝑖𝑛

2𝐸𝐼

𝐿=2(29(103))(700)

16(12)= 211458 𝐾. 𝑖𝑛



16.12 DETERMINE LAS REACCIONES EN LOS SOPORTES FIJOS 1 Y 2. CONSIDERE

QUE E = 200 GPA, I= 300(10^6) MM^4 Y A= 10(10^3) MM^2 PARA CADA ELEMENTO.

CARGAS Y DEFLEXIONES NODALES CONOCIDAS, LAS CARGAS NODALES ACTÚAN

A LIBERTAD.

Qk=

{

0−13.75108000 }

{

12345}

DK={

0000

}{

6789

}

RELACIÓN CARGAS, DESPLAZAMIENTO Y APLICACIÓN DE Q=KD.

[

0−13.759000

− − −−𝑄6𝑄7𝑄8𝑄9 ]

[ 2048.31

0−3304.04−3304.04

0− − − −−34.4170

0−2013. .89

0

]

[

03031. .03−1468.46

0−1468.46− − − −

0−3020.83

0−10.1976]

[ −3304.04−1468.46704861211458140972− − −−3304.04

00

1468.46 ]

[ −3304.04

0211458422917

0− −− −3304.04

000 ]

[

0−1468.46140972

0281944− − − −

000

1468.46 ]

[ −34.4170

03304.043304.04

0− −− −34.4170

000 ]

[

0−3020. .83

000

− − −− −0

−3020.8300 ]

[ −2013.89

0000

− −− −00

2013.890 ]

[

0−10.19761468.46

01468.46− − − −

000

10.1976 ]

[

𝐷1𝐷2𝐷3𝐷4𝐷5

− −− −0000 ]



A PARTIR DE LA PARTICIÓN DE MATRIZ.

Qk= K11Du+K12DK

0=2048.31D1 – 3304.04D3- 3304.04D4 (1)

-13.75=3031.03D2 -1468.46D3 -1468.46D5 (2)

90= -3304.04D1-1468.46D2+704861D3+211458D4+140972D5 (3)

0= -3304.04D1+211458D3+422917D4 (4)

0=-1468.46D2+140972D3+281944D5 (5)

SOLUCION DE ECUACIONES.

D1=0.001668 D2=-0.004052 D3=0.002043 D4=-0.001008 D5=-0.001042

USANDO ESTOS RESULTADOS Y APLICANDO Qu=K21Du+K22Dk

Q6=-34.4170 (0.001668) +3304.04(0.002043) +3304.04(-0.001008) =3.360

Q7= -3020.83 (-0.004052) = 12.24

Q8= -2013.89 (0.001668) = -3.360

Q9= -10.1976 (-0.004052) +1468.46(0.002043) +1468.46(-0.001008) = 1.510

Superponer estos resultados a los de FEM

R6=3.360 + 0 = 3.36K

R7= 12.24 + 0 = 12.2K

R8= -3.360 + 0 = -3.36K

R9=1.510 +6.25 = 7.76K

analisis matricial

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