156

Figure XP3.3 (a). Spring-mass-damper system with an appliedforce, (b). Applied force definition

Lecture 10. Transient Solutions 3 — More base ExcitationExample Problem XP3.3.

Figure XP3.3a illustrates a spring-mass-damper systemcharacterized by the following parameters: ,

, , with the forcingfunction illustrated. The mass system start from rest with thespring undeflected. The force increases linearly until

when it reaches a magnitude equal to . For , .Tasks: Determine the mass’s position and velocity at

. Plot for and .

Solution. This Example problem has the same stiffness and mass

157

(ii)

(i)

as Example problem 9.1; hence,

The equation of motion for is

For , the equation of motion is We will use the solution for the first equation for todetermine and , which we will then use as initialconditions in stating the solution to the second equation ofmotion for .

The applicable homogeneous solution is

From Table B.2, the particular solution for , is

158

(iii)

(iv)

Hence, the particular solution for is

and the complete solution is

The velocity is

Solving for the constants from the initial conditions gives

The complete solution satisfying the initial conditions is

159

(v)

Y versus t

0

0.002

0.004

0.006

0.008

0 0.05 0.1 0.15

t (seconds)

Y (m

eter

s)

Figure XP3.3 b. Solution for0#t #t1.

Y dot versus t

0.00E+00

2.00E-02

4.00E-02

6.00E-02

8.00E-02

1.00E-01

1.20E-01

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14

t (seconds)

Y do

t (m

/sec

)

The complete solution for t # 0 # t1 is illustrated below

The final conditions from figure XP3.3b are, and at

.

For , the equation of motion has the

160

solution

and derivative

We could solve for the unknown constants A and B via and . We are going

to take an easier tack by restarting time via the definition. Since,

the equation of motion is unchanged. The solution in terms of is

with the derivative

161

where , , and . Again, we are basically

restarting time to simplify this solution. Solving for the constantsin terms of the initial conditions gives

and the complete solution in terms of the initial conditions is

The solution is plotted below

162

Y versus t'

-2.00E-03

-1.00E-03

0.00E+00

1.00E-03

2.00E-03

3.00E-03

4.00E-03

5.00E-03

6.00E-03

7.00E-03

8.00E-03

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

t' (seconds)

Y (m

eter

s)

Y dot versus t'

-1.50E-01

-1.00E-01

-5.00E-02

0.00E+00

5.00E-02

1.00E-01

1.50E-01

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

t' (seconds)

Y do

t (m

`/`se

c)

163

Example Problem Xp3.5

Figure XP3.5a illustrates a cart that is connected to a movablebase via a spring and a damper. The cart’s change of position isdefined by x. The movable base’s location is defined by xb.

First, we need to derive the cart’s equation of motion, startingwith an assumption of the relative positions and velocities. The

Figure Xp3.5 Movable cart withbase excitation. (a). Coordinates,(b). Free-body diagram for

164

(i)

free-body diagram of XP 3.4a is drawn assuming that (placing the spring in tension) and (placing the damper in tension). Applying Newton’s 2nd law of motion nets

In Eq.(i), the spring and damper forces are positive because theyare in the + x direction.

The model of Eq.(i) can be used to approximately predict themotion of a small trailer being towed behind a much largervehicle with the spring and damper used to model the hitchconnection between the trailer and the towing vehicle. (Note: Tocorrectly model two vehicles connected by a spring and damper,we need to write for both bodies and account for theinfluence of the trailer’s mass on the towing vehicle’s motion. We will consider motion of connected bodies following the nexttest.)

Assume that the cart and the base are initially motionless with thespring connection undeflected. The base is given the constantacceleration . Complete the following engineeringanalysis tasks:

a. State the equation of motion.

165

b. State the homogeneous and particular solutions, and statethe complete solution satisfying the initial conditions.c. For , plot thecart’s motion for two periods of damped oscillations.

Stating the equation of motion simply involves proceeding from to find and plugging them into Eq.(i), via

Substituting these results into Eq.(i) produces

The homogeneous solution is

We can pick the particular solutions from Table B.2 below asfollows

166

Table B.2. Particular solutions for .

Excitation,

h = constant

a t

167

The complete solution is

The constant A is obtained via,

To obtain B, first

168

(ii)

Hence,

The complete solution satisfying the boundary conditions is

and Eq.(ii) completes Task b.

Moving towards Task c,.

169

Two cycles of damped oscillations will be completed inseconds, where

. The solutionsare shown below.

170

x versus t

-1012345678

0 0.1 0.2 0.3 0.4 0.5

t (seconds)

x (in

ches

)

x_b -x versus t

00.010.020.030.040.050.060.070.080.090.1

0 0.1 0.2 0.3 0.4 0.5

t (seconds)

( x_b

-x )

( in

ches

)

(x dot_b-x dot) versus t

-0.4-0.20

0.20.40.60.81

1.21.4

0 0.1 0.2 0.3 0.4 0.5

t (seconds)

x do

t_b-

x do

t ( in

/ se

c )

Discussion. After an initial transient, the relative displacement

171

approaches a constant value , while therelative velocity approaches zero; i.e., . The scalingfor hides the initial transient, but the plot shows quadraticincrease with time, which is consistent with its constantacceleration at . The spring force

, while the forcerequired to accelerate the mass at g/5 is

Hence, the asymptotic value for creates the spring forcerequired to accelerate the mass at .

172

(iii)

Example XP3.6Think about an initially motionless cart being “snagged” by amoving vehicle with velocity through a connection consistingof a parallel spring-damper assembly. The connecting spring isundeflected prior to contact. That circumstance can be modeledby giving the base end of the model in figure XP3.5 the velocity

; i.e., , and the governingequation of motion is

with initial conditions . The engineering analysistasks for this example are:

a. Determine the complete solution that satisfies the initialconditions.

b. For the data set, , , and, determine .

c. For produce plots for the mass displacement, relative displacement , and the mass

velocity for two cycles of motion.

173

Solution. From Table B.2, the particular solutions correspondingto the right-hand terms are:

The complete solution is

Solving for A ,

Solving for B from

174

(iv)

gives

The complete solution satisfying the initial conditions is

which completes Task a.

For Task b,

175

x versus t

0

1

2

3

4

5

6

7

0 0.2 0.4 0.6 0.8 1 1.2

t ( seconds )

x (

met

ers

)

x_b - x versus t

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 0.2 0.4 0.6 0.8 1 1.2

t ( seconds )

( x_b

- x)

( m

eter

s )

From the last of these results, the period for a damped oscillationis .

For Task c,. The

figures below illustrate the solution for two cycles of motion.

176

x dot verus t

012345678

0 0.2 0.4 0.6 0.8 1 1.2

t (seconds)

x do

t ( m

/ se

c)

Discussion. The first and last figures show the mass rapidlymoving towards the towing velocity . The secondfigure shows the relative position approaching zero. Note that atabout 0.3 seconds, the towed cart appears to actually passes thetowing vehicle when becomes negative. However, thepresent analysis does not account for the initial spring length. Athe towed vehicle could “crash” into the back of the towingvehicle, but it’s not likely. You could plot to find therelative velocity if impact occurs.