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Extraterrestrial Vehicle Two - Modeling of Physical Systems - Home Work, Exercises of Mathematical Modeling and Simulation

The main points in the home work assignment of the Modeling of Physical Systems are:Extraterrestrial Vehicle Two, Spring Radius, Motion, Horizontal, Vehicle, Rotational Dynamics, Longitudinal, Vertical Directions, Loading Conditions., Vehicle Body.

Typology: Exercises

2012/2013

Uploaded on 05/08/2013

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Modeling of Physical Systems: HW 6–SOLUTION Page 1
Problem 1: Work problem 7-2 from Karnopp, et al [1].
The problem statement from KMR is shown below:
Solution: One way to begin building the bond graph is to formulate the expressions for the
xand ydirection forces.
(continued next page)
R.G. Longoria, Fall 2012 ME 383Q, UT-Austin
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Problem 1: Work problem 7-2 from Karnopp, et al [1].

The problem statement from KMR is shown below:

Solution: One way to begin building the bond graph is to formulate the expressions for the

x and y direction forces.

(continued next page)

R.G. Longoria, Fall 2012 ME 383Q, UT-Austin

Now, the constitutive relations:

R.G. Longoria, Fall 2012 ME 383Q, UT-Austin

Problem 3: Work problem 7-9 from Karnopp, et al [1].

The problem statement from KMR is shown below:

Here is solution from KMR:

R.G. Longoria, Fall 2012 ME 383Q, UT-Austin

Problem 4: A suspension for an extraterrestrial vehicle consists of four wheels and semi-

circular springs, readily folded from the body of the vehicle after landing and then clamped

in the position shown. Study the handout from Chapter 10 of Burr [2] that will be posted

on the course log.

1. Show how you can use a two-port C element to represent each semi-circular spring

element, especially so you can answer the question from Burr: “Determine in terms of

EI, vehicle load W , and spring radius r: (a) the downward deflection of the supported

body and (b) the the horizontal motion of a wheel due to vehicle load only.”

Solution: The figure below shows that energy can be stored in the curved beam

suspension either by deflection in the x or the z direction. Assume that axle bearing

prevents any moment from being applied at the tip of the beam, so that effect is

neglected.

The detailed analysis is shown on the attached Curved Beam Vehicle Suspension:

MathCAD Appendix. This two-port C-element has linear constitutive relations

derived in compliance form, x = CF. They can also be expressed in stiffness form,

F = kx. The details of the derivation are provided in the appendix.

2. Develop the bond graph for a quarter-vehicle model under static loading conditions.

Consider the translational mass of the vehicle in the longitudinal and vertical directions

only. A quarter-vehicle model does not include rotational dynamics of the vehicle body.

Solution: Consider a 1/4-vehicle model represented by mass, m

v

; no rotational mo-

tion, only translations in x and z. The tire has mass, m

t

, stiffness, k

t

, and damping,

b

t

The suspension stiffness is represented by the two-port C, which couples into the

translational mass of the vehicle as well as the tire in both directions. The tire is

forced by ground input, v

g

R.G. Longoria, Fall 2012 ME 383Q, UT-Austin

References

[1] Karnopp, D., D. Margolis, R. Rosenberg, System Dynamics, Wiley, New York, (any

edition after 2001).

[2] Burr, A.H. Mechanical Analysis and Design, Elsevier, NY, 1981.

R.G. Longoria, Fall 2012 ME 383Q, UT-Austin

Curved Beam Vehicle Suspension: MathCAD Appendix RGL - 10/28/

As there are no applied moments, we utilize equation 10.10 in Burr (Chapter 10). Find each of

the deflections in the x (horizontal) and z (vertical) directions. Assume Fx and Fz are applied.

First, the moment in the y direction is,

M

y

F

z

⋅ rsin φ ( ) ⋅ F x

⋅r 1 cos φ ( ) − ( ) = − ⋅

Now, apply equation 10.10 to each deflection:

δ x

E I⋅

0

π

F φ z

⋅r sin φ ( ) ⋅ F x

⋅ r 1 cos φ ( ) − ( )

⋅ −r 1 cos φ ( ) − ( )

⎡ ⋅^ ⋅r

= ⋅ d

δ x

E I⋅

r

3 F z

2 E I⋅ ⋅

r

3 F x

= + ⋅ ⋅ ⋅π

δ z

E I⋅

0

π

F φ z

⋅ rsin φ ( ) ⋅ F x

⋅r 1 cos φ ( ) − ( )

r sin φ ( ) ⋅ ( ) ⋅ ⋅r

= ⋅ d

δ z

2 E I⋅ ⋅

r

3 π F z

E I⋅

r

3 F x

Compliance form:

Stiffness form:

δ x

δ z

3 ⋅ πr

3 ⋅

2 E⋅ ⋅I

− 2 r

3 ⋅

E I⋅

− 2 r

3 ⋅

E I⋅

π r

3 ⋅

2 E⋅ ⋅I

F

x

F

z

F

x

F

z

E I⋅

r

3

3 ⋅ π

π

− 1

δ x

δ z

Now, consider case where Fx = 0 (only weight is applied):

δ x

δ z

3 ⋅ πr

3 ⋅

2 E⋅ ⋅I

− 2 r

3 ⋅

E I⋅

− 2 r

3 ⋅

E I⋅

π r

3 ⋅

2 E⋅ ⋅I

F

z

δ x

δ z

E I⋅

r

3 F z

2 E I⋅ ⋅

r

3 π F z

= Let Fz = W/4 to solve for the deflections.