mechanical 431 project

8/12/2019 Mechanical 431 Project

1/17

Using Simulink to analyze two degrees of freedom system

Nasser M. Abbasi, Spring 2009

Abstract

A two degrees of freedom system consisting of two masses connected by springs and subject to

3 different type of input forces is analyzed and simulated using Simulink

Contents

1 Introduction and Theory 1

2 Analytical solution 3

2.1 Finding the homogenous solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Finding particular solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.1 Finding the particular solution for unit impulse input . . . . . . . . . . . . . . 62.2.2 Finding the particular solution for unit step input . . . . . . . . . . . . . . . . 62.2.3 Finding the particular solution forsin t . . . . . . . . . . . . . . . . . . . . . . 7

3 Simulink simulation and block diagrams 8

3.1 Unit step simulink diagram and output . . . . . . . . . . . . . . . . . . . . . . . . . . 93.1.1 Verification of result from Simulink by Numerically solving the differential equations 10

4 Unit impulse simulink diagram and output 12

4.0.2 Verification of result from Simulink by Numerically solving the differential equations 124.1 sin(t)input simulink diagram and output . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Discussion 17

1 Introduction and Theory

The system that is being analyzed is show in the following diagram

1


2/17

In the above, F(t)is to be taken as each of the following

1. Unit impulse force.

2. Unit step force.

3. sin t

It is required to findx1(t)and x2(t)analytically and then to use Matlabs Simulink software for theanalysis.

The mathematical model of the system is first developed and the equation of motions obtained usingLagrangian formulation then the analytical solution is found by solving the resulting coupled secondorder differential equations for m1 and m2. Next, a simulink model is developed to implement thedifferential equations and the output x1(t) and x2(t) from Simulink is shown and compared to the

output from the analytical solution.

2


3/17

2 Analytical solution

The following is the free body diagram of the above system

Assuming positive is downwards and that x2 > x1, force-balance equations for m1 results in

m1x1= F(t) + k1(x2 x1)

And force-balance equations for m2 results in

m2x2 = k1(x2 x1) k2x2

Hence the EQM for the system become

m1x1

k1x2+ k1x1= F(t)m2x2+ (k1+ k2) x2 k1x1= 0

Or in matrix form m1 00 m2

x1x2

+

k1 k1k1 (k1+ k2)

x1x2

=

F(t)

0

The above can be written in matrix form as

Mx + Kx= F

3


4/17

Where x,x,F are 2 by 1 vectors and M and K are the mass and stiffness matrices. The solution tothe above is

x= xh+ xp (1)

2.1 Finding the homogenous solution

We start by finding xh from the followingm1 00 m2

x1x2

+

k1 k1k1 (k1+ k2)

x1x2

=

00

Now assume x1(t) =A1cos (t + )and x2(t) =A2cos (t + ), hence x1(t) = A1sin (t + )andx2(t) = A2sin (t + )and x1(t) =

2A1cos (t + )and x2(t) = 2A2cos (t + ). Substitut-

ing the above values in the above system results in

m1 00 m2

2A1cos (t + )2A2cos (t + )

+

k1 k1k1 (k1+ k2)

A1cos (t + )A2cos (t + )

=

00

Divide bycos (t + )since not zero (else no solution exist) we obtainm1 00 m2

2A12A2

+

k1 k1k1 (k1+ k2)

A1A2

=

00

Rewrite the above as 2m1A12m2A2

+

k1A1 k1A2

k1A1+ (k1+ k2) A2

=

00

2m1A1+ k1A1 k1A22m2A2 k1A1+ (k1+ k2) A2

=

00

(2m1+ k1) A1 k1A2

2m2A2+ (k1+ k2) A2 k1A1

=

00

2m1+ k1 k1

k1 2m2+ (k1+ k2)

A1A2

=

00

(2)

From the last equation above, we see that to obtain a solution we must have

2m1+ k1 k1k1

2m2+ (k1+ k2)

= 0

since if we had

A1A2

= 0 then no solution will exist. Therefore, taking the determinant and setting it

to zero results in

2m1+ k1

2m2+ (k1+ k2)

k21 = 0

k21+ k1k2 2k1m1

2k1m2 2k2m1+

4m1m2 k21 = 0

4m1m2+ 2 (k1m1 k2m1 k1m2) +

k21+ k1k2 k

21

= 0

4


5/17

Let 4 =2, hence the above becomes

2m1m2+ (k1m1 k2m1 k1m2) +

k21+ k1k2 k21

= 0

Solving for gives

1 = 12m1m2 k1m1+ k1m2+ k2m1+ k21m21+ k21m22+ k22m21+ 2k1k2m21+ 2k21m1m2 2k1k2m1m2

2= 12m1m2

k1m1+ k1m2+ k2m1

k21m

21+ k

21m

22+ k

22m

21+ 2k1k2m

21+ 2k

21m1m2 2k1k2m1m2

(3)

For each of the above solutions, we obtain a different

A1A2

from equation (2) as follows

For 1, (2) becomes 1m1+ k1 k1

k1 1m2+ (k1+ k2)

A1A2

=

00

(1m1+ k1) A1 k1A2

k1A1

1m2A2+ (k1+ k2) A2=

0

0

From the first equation above, we have

(1m1+ k1) A1 k1A2 = 0

1m1+ k1k1

=

A2

A1

(1)

Similarly for 2,2m1+ k1

k1=

A2

A1

(2)

Let

r1 =1m1+ k1

k1=

A2

A1

(1)(4)

r2 =2m1+ k1

k1=

A2

A1

(2)

Hence now xh can be written asx1x2

h

=

A

(1)1 cos(1t + 1) + A

(2)1 cos(2t + 2)

A(1)2 cos(1t + 1) + A

(2)2 cos(2t + 2)

But A(1)2 =r1A

(1)1 and A

(2)2 =r2A

(2)1 , hence the above becomes

x1x2

h

=

A

(1)1 cos(1t + 1) + A

(2)1 cos(2t + 2)

r1A(1)1 cos(1t + 1) + r2A

(2)1 cos(2t + 2)

(5)

Now, given numerical values fork1, k2, m1, m2we can find1, 2from (3) above, and next find r1, r2from

(4). Hence (5) contains 4 unknowns, A(1)1 , A

(2)1 , 1, 2 which now can be found from initial conditions

(after we find the particular solution) which we will now proceed to do.

5


6/17

2.2 Finding particular solutions

There are 3 different F(t)which we are asked to consider

1. Unit impulse force.

2. Unit step force.

3. sin t

For each of the above, we find xp and then add it to xh found above in (5) to obtain (1).

2.2.1 Finding the particular solution for unit impulse input

Using the standard response for a unit impulse which for a single degree of freedom system is x (t) =1

mnsin nt, then we write xp as

xp=

x1x2

p

=

1m1

sin 1t + 1m2

sin 2t

0

Hence, the general solution becomesx1x2

=

A

(1)1 cos(1t + 1) + A

(2)1 cos(2t + 2)

r1A(1)1 cos(1t + 1) + r2A

(2)1 cos(2t + 2)

+

1m1

sin 1t + 1m2

sin 2t

0

(6)

2.2.2 Finding the particular solution for unit step input

Since unit step is1for t >0, then, using convolution we write

xp(t) = t0

f() h (t ) d

= t0

h (t

) d

= t0

1

mnsin n(t ) d

= 1

mn

cos(n(t ))

n

t0

= 1

m2n[cos(n(t ))]

t

0

= 1

m2n[1 cos(nt)]

Then, since now we have 2 natural frequencies, we can write xp as

xp=

x1x2

p

=

1m2

1

[1 cos(1t)] + 1m2

2

[1 cos(2t)]

0

Hence, the general solution becomesx1x2

=

A

(1)1 cos(1t + 1) + A

(2)1 cos(2t + 2)

r1A(1)1 cos(1t + 1) + r2A

(2)1 cos(2t + 2)

+

1m2

1

[1 cos(1t)] + 1m2

2

[1 cos(2t)]

0

6


7/17

2.2.3 Finding the particular solution forsin t

In this case, we guess that x1p= c1cos t + c2sin t, and since there is no forcing function being applieddirectly on m2 then x2p= 0hence

xp=

x1(t)x2(t)

p

=

c1cos t + c2sin t

0

Then x1p(t) =c1sin t+c2cos t and x1p(t) =2c1cos t+

2c2sin t and now we substitutethese into the original ODE for x1 which is

m1x1 k1x2+ k1x1= F(t)

m1x1p k1x2p+ k1x1p= sin t

We obtain the following

m12c1cos t +

2c2sin t k1(0) + k1(c1cos t + c2sin t) = sin t

cos t 2c1

m1+ k

1c1+ sin t m

12c

2+ k

1c2= sin t

Hence by comparing coefficients, we obtain

2c1m1+ k1c1 = 0

m12c2+ k1c2 = 1

or

c12m1+ k1

= 0

c2 m12 + k1= 1

c1 must be zero since k1 2m1 = 0 only when = n and we assume that this is not the case here.

Hence

c1 = 0

c2 = 1

(m12 + k1)

Therefore xp becomes

xp=

x1(t)x2(t)

p

=

1

(m12+k1)sin t

0

And the general solution becomes

x1x2

=

A

(1)1 cos(1t + 1) + A

(2)1 cos(2t + 2)

r1A(1)1 cos(1t + 1) + r2A

(2)1 cos(2t + 2)

+

1

(m12+k1)sin t

0

(8)

7


8/17

3 Simulink simulation and block diagrams

In simulink, we will directly solve the system from the original formulation

m1x1 k1x2+ k1x1= F(t)

m2x2+ (k1+ k2) x2 k1x1= 0

or

x1 k1

m1x2+

k1

m1x1 =

F(t)

m1

x2+(k1+ k2)

m2x2

k1

m2x1 = 0

Hence

x1=F(t)

m1+

k1

m1x2

k1

m1x1

x2=

(k1+ k2)m2

x2+ k1m2

x1

8


9/17

3.1 Unit step simulink diagram and output

The simulink block diagram will be as follows for the unit step input

For an initial run with parameters m1= m2 = k1= k2= 1I get this warning below

EDU>> simulink

Warning: Using a default value of 0.2 for maximum step size. The simulation

step size will be equal to or less than this value. You can disable this

diagnostic by setting Automatic solver parameter selection diagnostic to

none in the Diagnostics page of the configuration parameters dialog.

And this is the output for x1(t)and x2(t)for the unit step response

9


10/17

3.1.1 Verification of result from Simulink by Numerically solving the differential equa-

tions

To verify the above output from Simulink, I solved the same coupled differential equations for zeroinitial conditions numerically (using a numerical differential equation solver) and plotted the solutionfor x1(t)and x2(t)and the result matches that shown above by simulink. Here is the code the plot asa result of this verification

10


11/17

11


12/17

4 Unit impulse simulink diagram and output

And the output for x1(t)and x2(t)is as follows

4.0.2 Verification of result from Simulink by Numerically solving the differential equa-

tions

To verify the above output from Simulink, The same coupled differential equations were solved numer-ically for zero initial conditions numerically and the solution plotted for x1(t)and x2(t)and the resultwas found to match that shown above by simulink. Here is the code used to do the verification

verification

12


13/17

13


14/17

14


15/17

4.1 sin(t) input simulink diagram and output

The simulink block diagram will be as follows for the sin t input

For an initial run with parameters m1 = m2 =k1 = k2 = 1 this is the output for x1(t) and x2(t) andshowing the input signal at the same time

15


16/17

16


17/17

5 Discussion

A coupled system of two masses and springs was analyzed using Simulink. The simulation was donefor one set of parameters (masses and stiffness). Simulink made the simulation of this system underdifferent loading conditions easy to do. The 2 masses response were recorded using simulink scope andthe signals captured on the same plot to make it easy to compare the response of the first mass to thesecond mass.

The analytical analysis was more time consuming than actually making the simulation in simulink.The ability to easily change different sources to the system was useful as well as the ability to changethe frequency of the input and immediately see the effect on the response.

This was my first project using Simulink, and I can see that this tool will be useful to learn more asit allows one to easily analyze engineering problems.

17

mechanical 431 project

Documents