Taller Sistemas Dinamicos Segundo Corte

Taller Sistemas Dinamicos Segundo Corte

Diego Andrés Espinel Hernández, Daniel Felipe Gualdron Orjuela

1 Ejercicio Sistema Translacional

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Fk_{1}=k_{1}X_{1}

Fk_{2}=k_{2}X_{2}

Fb{}_{1}=b_{1}\mathring{X_{1}}

Fb_{2}=b_{2}(\mathring{X_{1}}+\mathring{X_{2}})

1.1 Diagrama de cuerpo libre para masa 1

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f_{1}(t)-Fk_{1}-Fb_{1}-Fb_{2}=m_{1}\ddot{X}_{1}

f_{1}(t)=m_{1}\ddot{X}_{1}+Fk_{1}+Fb_{1}+Fb_{2}

f_{1}(t)=m_{1}\ddot{X_{1}}+(b_{1}+b_{2})\mathring{X_{1}}+k_{1}X_{1}+b_{2}\mathring{X_{2}}

1.2 Diagrama de cuerpo libre para masa 2

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f_{2}(t)-Fb_{2}-Fk_{2}=m_{2}\ddot{X_{2}}

f_{2}(t)=m_{2}\ddot{X_{2}}+Fb_{2}+Fk_{2}

f_{2}(t)=m_{2}\ddot{X_{2}}+b\mathring{X_{2}}+k_{2}X_{2}+b_{2}\ddot{X_{2}}

1.3 Funciones de transferencia

Aplicamos tansformada de Laplace en f1 y f2

f_{1}(s)=x_{1}(s)[m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}]+x_{2}(s)b_{2}s

Ecuacion 1

f_{2}(s)=x_{2}(s)[m_{2}s^{2}+b_{2}s+k_{2}]+x_{1}(s)b_{2}s

Ecuacion 2

Para las fdt se aplica superposición. Iniciando ahora con f_{2}(s)=0

DespejamosX_{2}(s)

de la Ecuación 2

x_{2}(s)=\frac{-b_{2}s}{m_{2}s^{2}+b_{2}s+k_{2}}x_{1}(s)

Se reemplaza X_{2}(s)

en la Ecuación 1

f_{1}(s)=x_{1}(s)[m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}]+[\frac{-b_{2}s}{m_{2}s^{2}+b_{2}s+k_{2}}x_{1}(s)]*b_{2}s\}

f_{1}(s)=x_{1}(s)[m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}]-\frac{(b_{2}s)^{2}}{m_{2}s^{2}+b_{2}s+k_{2}}x_{1}(s)

f_{1}(s)=\frac{[m_{2}s^{2}+b_{2}s+k_{2}]*[m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}]-(b{}_{2}s){}^{2}}{m_{2}s^{2}+b_{2}s+k_{2}}x_{1}(s)

\frac{x_{1}(s)}{f_{1}(s)}=\frac{m_{2}s^{2}+b_{2}s+k_{2}}{[m_{2}s^{2}+b_{2}s+k_{2}]*[m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}]-(b{}_{2}s){}^{2}}

DEN=s^{4}(m_{1}m_{2})+s^{3}[(m_{2}b_{1})+(m{}_{2}b_{2})+(m{}_{1}b_{2})]+s^{2}(m_{2}k_{1}+b{}_{1}b_{2}+m_{1}k_{2})+s[b_{2}k_{1}+k_{2}(b_{1}+b_{2})]+k_{1}k_{2}

Primera funcion de transferencia \frac{X_{1}(s)}{f_{1}(s)}

\frac{x_{1}(s)}{f_{1}(s)}=\frac{m_{2}s^{2}+b_{2}s+k_{2}}{DEN}

De la Ecuación 2 despejamos X_{1}(s)

x_{1}(s)=-\frac{m_{2}s^{2}+b_{2}s+k_{2}}{b_{2}s}x_{2}(s)

Reemplazando en FDT1

\frac{-[m_{2}s^{2}+b_{2}s+k_{2}]*x_{2}(s)}{\frac{b_{2}s}{f_{1}(s)}}=\frac{m_{2}s^{2}+b_{2}s+k_{2}}{DEN}

Segunda funcion de transferencia \frac{X_{2}(s)}{f_{1}(s)}

\frac{x_{2}(s)}{f_{1}(s)}=\frac{-b{}_{2}s}{DEN}

Ahora para dejar en términos de f_{2}(s)

despejamos de la

Ecuación 1

Igualamos por superposición f_{1}(s)=0

, y despejamos X_{1}(s)

x_{1}(s)=-\frac{b_{2}s}{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}*x_{2}(s)

Reemplazamos X_{1}(s)

en la Ecuación 2

f_{2}(s)=x_{2}(s)[m_{2}s^{2}+b_{2}s+k_{2}]+[-x_{2}(s)\frac{b_{2}s}{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}]b{}_{2}s

f_{2}(s)=x_{2}(s)[m_{2}s^{2}+b_{2}s+k_{2}]-\frac{x_{2}(s)*(b{}_{2}s){}^{2}}{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}

f_{2}(s)=x_{2}(s)\frac{[m_{2}s^{2}+b_{2}s+k_{2}]*[m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}]-(b{}_{2}s){}^{2}}{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}

\frac{x_{2}(s)}{f_{2}(s)}=\frac{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}{[m_{2}s^{2}+b_{2}s+k_{2}]*[m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}]-(b{}_{2}s){}^{2}}

Tercera funcion de transferencia \frac{X_{2}(s)}{f_{2}(s)}

\frac{x_{2}(s)}{f_{2}(s)}=\frac{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}{DEN}

Ahora despejamos x_{2}(s)

de la Ecuación 2

x_{2}(s)=-x_{1}(s)*\frac{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}{b_{2}s}

Reemplazamos X_{2}(s)

en la FDT 3

-x_{1}(s)\frac{\frac{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}{b_{2}s}}{f_{2}(s)}=\frac{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}{DEN}

Cuarta funcion de transferencia \frac{X_{1}(s)}{f_{2}(s)}

\frac{x_{1}(s)}{f_{2}(s)}=-\frac{b{}_{2}s}{DEN}

1.4 Polos, ceros, tiempo de estabilización y frecuencia de

oscilación

Con m_{1}=10kg

m_{2}=8kg

k_{1}=80N/m

k_{2}=110N/m

b_{1}=8Ns/m

b_{2}=10Ns/m

f_{1}(t)=10N

f_{2}(t)=15N

Polos:

Polo 1 y 2=-1.1942\pm2.7804i

Polo 3 y 4= -0.3308\pm3.4502i

Ceros:

En FDT1

-\frac{5}{8}\pm3.655i

En FDT2

0

En FDT3

-\frac{9}{8}\pm2.6814i

En FDT4

0

El sistema es estable. Tiene un tiempo de estabilización

\tau=\frac{1}{0.3308}=3.02seg

5\tau=15.1seg

Frecuencia

w=2.7804rad/seg

f=0.44Hz

1.5 Deformacion de cada elemento

Para el resorte K_{1}

y amortiguador b_{1}

\frac{x_{1}(s)}{f_{1}(s)}=\frac{m_{2}s^{2}+b_{2}s+k_{2}}{DEN}

Para el amortiguador b_{2}

\frac{x_{1}(s)}{f_{1}(s)}+\frac{x_{2}(s)}{f_{2}(s)}=\frac{(m_{1}+m_{2})s^{2}+(b_{1}+2b_{2})s+k_{1}+k_{2}}{DEN}

Para el resorte K_{2}

\frac{x_{2}(s)}{f_{2}(s)}=\frac{m_{1}s^{2}+(b_{1}+b_{2})s+k_{1}}{DEN}

1.6 Gráficas por Ecuaciones diferenciales y funciones de

transferencia

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2 Ejercicio Sistema Rotacional Mixto

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Hallamos el valor

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