School

Texas A&M UniversityDepartment

Aerospace EngineeringCourse Code

AERO 306Professor

John WhitcombStudy Guide

FinalThis

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p. 1 of 9

Dealing with Complex Kinematics

When the relationships between various displacements are not very simple, it is useful to

get some help from your calculus background. Several examples are illustrated in this

Maple worksheet: AnimatedMechanisms\__mechanismCollection.mw

The next section introduces the “variational operator”, which is closely related to the

differential operator that you saw in your first calculus course. You will find that using

the variational operator greatly simplifies many problems.

Pre-calculus approach (the hard way)

When the kinematics are not trivial, it is useful to understand something called the

“variational operator”. Up until now, we have used

as a label to indicate a virtual

displacement. It actually means more than that… it is referred to as the variational

operator. Before explaining what this new operator is, let’s first solve the following

problem without using it. This example will help motivate the need for the variational

operator.

P, u

a

M

Inextensible bar of

length L

Suppose the current equilibrium state is as shown

0

0

a a u

=+

=+

0

0

where initial value of u=translation in direction of P

a initial value of a α=rotation in direction of

=

=

Now impose virtual displacements

,u

VW P u M 0

= + =

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p. 2 of 9

This is the virtual work, but

u and

are not independent. To proceed further we must

find the relationship between

u and

. If we assume we do not know about calculus,

we can proceed as follows:

If we assume the bar is inextensible and it's length = L, then simple trigonometry will

give us a relationship between

andu

.

( )

0

0

a u u

cos L

++

+ + =

This can be simplified by recognizing that

u and

are infinitesimal.

1 2 1 2 1 2

0 0 0

Recall the formula: cos( ) cos cos sin sin

cos( ) cos( ) cos( ) sin( ) sin( )

But is extremelysmall sin cos 1and

+ = −

+ + = + − +

= =

0

0 0 0

0

0

0

a u u

cos( ) cos( ) sin( ) L

au

but cos( ) L

sin( )

Therefore

uL

++

+ + = + − + =

+

+=

= − +

Now substitute this relationship into the expression for VW and solve to obtain

M PLsin

=

There is an easier way. Regardless of the particular values of

00

,,a and

, the

relationship between u and

is given by

( )

0 0 0

cos a u where L

+ = + + =

Take the differential of both sides, and we obtain:

( )

0

sin duLd

− + =

. Hence,

( )

0

sindu L d

= − +

, which is very close to what we obtained above. The difference

is "d" instead of

, e.g.

.du vs u

. Both refer to shall changes in "u". The

differential operator "d" has specific behavior that you have studied before in math

classes. The

was just used to indicate some small change in a displacement… up until

now. Now we are going to define

to be the variational operator and it will have

properties that are very similar to the differential operator, as described in the next

section. I should point out that

( )

0

d d d

+ = =

.

Variational Operator

Suppose we have a function F = F (x,u,u’), where u(x) is a displacement field. The strain

energy density is an example of such a function. The differential of the function is

F F F

dF d x d u d u

xuu

= + +

In contrast, the variation of F… i.e.

F

is

FF

Fuu

uu

=+

du and u

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p. 3 of 9

The term

Fdx

x

does not appear in the variation of F. The differential is meant to

describe changes in the function due to all possible effects: changes in position x, u, and

u'. The variation only describes changes that are related to displacement. In this class a

simple rule will suffice to be able to take a variation of a function: Take the differential of

the function, set all coordinate differentials (e.g. dx and dy) to zero, and then simply

replace the "d" with a

. The following simply illustrate the properties of the variational

operator. No new properties are stated.

• Suppose u=u(x). We will define

( )

to be the variation (small change) of u xu

(see

plot below).

u u+

( )

u x

• Suppose ci = a list of constants. Then

ci

= changes in the values of those constants.

• At points where u(x) is fixed (specified),

u=0

. Similarly, if the constant c2 is

specified,

c20=

• The variational operator behaves almost exactly like the differential operator, as

shown below.

( )

( )

( )

( ) ( )

1nn

bb

aa

uv u v v u

u v u v

u n u u

d dx u du dx

udx udx

−

=+

+ = +

=

=

=

Example

In the leaning bar problem earlier, we had the relationship

( )

0

0

au

cos L

+

+=

.

According to the rules above

( )

( )

( )

0

00

00

au

cos sin L

The variations of , a , and L are all zero, since all are fixed values.

u

and L

+

+ = − + =

Hence, we obtain the relationship

( )

0

sinuL

=− +

.

In this case, the "thing" being varied was a single number. Suppose the displacement of a

beam was being approximated by the polynomial function

23

()v x a bx cx dx= + + +

,

where a, b, c, and d are all constants which are not specified, but will be determined

somehow. The variation of v(x) is

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