Moment about an Axis - Memphis about an Axis.pdf3 5 Moments Along an Axis, Couples Monday, September 24, 2012 Moment about an Axis ! There are times that we are interested in the moment

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Moment of a force along an axis Couples

“If you find yourself in a hole, stop digging.” –Will Rogers

Monday, September 24, 2012 Moments Along an Axis, Couples 2

Objectives

¢ Understand the vector formulation for finding the component of a moment along an axis

¢ Understand the idea of a couple and the moment it produces

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Tools

¢  Basic Trigonometry ¢  Pythagorean Theorem ¢  Algebra ¢  Visualization ¢  Position Vectors ¢  Unit Vectors ¢  Reviews ¢  Cross Products ¢  Dot Products


Review

¢ A moment is the tendency of a force to cause rotation about a point or an axis

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Moment about an Axis

¢ There are times that we are interested in the moment of a force that produces some component of rotation about (or along) a specific axis

¢ We can use all the we have learned up to this point to solve this type of problem


Moment about an Axis ¢ First select any point on the axis of

interest and find the moment of the force about that point

¢ Using the dot product and multiplication of the scalar times the unit vector of the axis, the component of the moment about the axis can be calculated

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Moment about an Axis ¢  If we have an axis a-a we can find the

component of a moment along that axis by

Ma−a

= ua−a

ua−a

M ( )

where M

is the moment about any point on a-a

Problem F4-17


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Example ¢ Given: System as Shown


Example ¢ Required: Moment of force F along axis

a-a

6


Example ¢ Solution:

Ma−a

= ua−a

ua−a

M ( )



Ma−a

= ua−a

ua−a

M ( )

need ua−a

and M

where M

is the moment of the forceabout some point on a-a

7


rOA


ua−a

=

r0 A

r0 A

r0 A

= A{ }− O{ }

A{ } = −3,−2,6{ } ft

O{ } = 0,0,0{ } ft

r0 A

= −3i

− 2 j+ 6k( ) ft

r0 A = 7 ft

ua−a

= − 3

7i− 2

7j+ 6

7k


rOA

Example ¢ Solution: Now we have to select some

point on a-a and find the moment of the force F about that point

¢ Please don’t mistake rOA for the moment arm

¢  rOA is just the position vector that we used to generate the unit vector along a-a

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rOA

Example ¢ The origin is on a-a (not always but in

this problem it is) so we can find the moment of F about the origin


rOA

Example ¢ First we define a moment arm from the

origin to the line of action of F

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rOA

rOB B

C

Example ¢ We can use rOB as our moment arm

rOB

= B{ }− O{ }

B{ } = 0,4,0{ } ft

O{ } = 0,0,0{ } ft

rOB

= 4 j( ) ft


Example ¢ Now we have to describe F as a

Cartesian vector

F= FuF

uF

= uCB

uCB

=

rCB

rCB

rCB

= B{ }− C{ }

B{ } = 0,4,0{ } ft

C{ } = 4,0,−2{ } ft

rCB

= −4i

+ 4 j+ 2k( ) ft

rCB = 6 ft

uF

= − 4

6i+ 4

6j+ 2

6k

F= −400i

+ 400 j

+ 200k

( )lb

rOA

rOB B

C

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rOA

rOB B

C

Example ¢ Taking the moment of F about the origin

(O)

M

= rOB

⊗ F

M

= 4 j( ) ft ⊗ −400i

+ 400 j

+ 200k

( )lbM

= 4 j⊗−400i

( ) + 4 j⊗ 400 j

( ) + 4 j⊗ 200k

( )( ) ft ⋅ lb

M

= 1600k+800i

( ) ft ⋅ lb


rOA

rOB B

C

Example ¢ Finding the magnitude of the projection of

M along a-a

ua−a

M

ua−a

= − 3

7i− 2

7j+ 6

7k

M

= 1600k+800i

( ) ft ⋅ lb

ua−a

M

= − 37

⎛⎝⎜

⎞⎠⎟

800 ft ⋅ lb( ) + 67

⎛⎝⎜

⎞⎠⎟

1600 ft ⋅ lb

ua−a

M

= 1028.57 ft ⋅ lb

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rOA

rOB B

C

Example ¢ Converting the magnitude to a Cartesian

vector

ua−a

ua−a

M ( )

ua−a

= − 3

7i− 2

7j+ 6

7k

ua−a

M

= 1028.57 ft ⋅ lb

ua−a

ua−a

M ( ) = −440.82i

− 293.88 j

+881.63k

( ) ft ⋅ lb


rOA

rOB B

C

Example ¢ To find the magnitude of the projection,

we can also use what is known as the mixed product

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rOA

rOB B

C

Example ¢ Again, starting with the force F the

moment arm rOB and the unit vector along the axis of concern uOA


rOA

rOB B

C

Example ¢ Taking advantage of the properties of the

dot product and the cross product, we can set up the problem in a matrix form as

x y z

x y z

x y z

u u ur r rF F F

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rOA

rOB B

C

Example ¢ Notice that since there are no vectors in

this expression, the result of reducing the matric will be a scalar.

x y z

x y z

x y z

u u ur r rF F F


rOA

rOB B

C

Example ¢ We can use the same method that we

used when we calculated the cross product

¢ Here we use it to find the magnitude of the moment along the axis

x y z

x y z

x y z

u u ur r rF F F

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rOA

rOB B

C

Example ¢ Substituting the coefficients of the unit

vector

3 2 67 7 7x y z

x y z

r r rF F F

− −


rOA

rOB B

C

Example ¢ Substituting the coefficients of the

moment arm

−37

−27

67

0 ft 4 ft 0 ftFx Fy Fz

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rOA

rOB B

C

Example ¢ Substituting the coefficients of the force

−37

−27

67

0 ft 4 ft 0 ft−400lb 400lb 200lb


rOA

rOB B

C

Example ¢ Using our method for reducing the matrix

−37

−27

67

0 ft 4 ft 0 ft−400lb 400lb 200lb

−37

0 ft−400lb

−27

4 ft400lb

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rOA

rOB B

C


− 37

− 27

67

0 ft 4 ft 0 ft−400lb 400lb 200lb

− 37

0 ft−400lb

− 27

4 ft400lb

− 37

4 ft( ) 200lb( )− 400lb( ) 0 ft( )( )− 2

70 ft( ) −400lb( )− 200lb( ) 0 ft( )( )

+ 67

0 ft( ) 400lb( )− −400lb( ) 4 ft( )( )


rOA

rOB B

C


− 37

− 27

67

0 ft 4 ft 0 ft−400lb 400lb 200lb

− 37

0 ft−400lb

− 27

4 ft400lb

− 37

800 ft • lb( )− 27

0( ) + 67

1600 ft • lb( )

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rOA

rOB B

C

Example ¢ The magnitude of the moment produced

by F along a-a is then

−37

−27

67

0 ft 4 ft 0 ft−400lb 400lb 200lb

−37

0 ft−400lb

−27

4 ft400lb

−37

800 ft • lb( ) − 27

0( ) + 67

1600 ft • lb( )=

7200 ft • lb7

= 1028.57 ft • lb


rOA

rOB B

C

Example ¢ Take care in identifying the unit vector of

the force and the position vector that is the moment arm

−37

−27

67

0 ft 4 ft 0 ft−400lb 400lb 200lb

−37

0 ft−400lb

−27

4 ft400lb

−37

800 ft • lb( ) − 27

0( ) + 67

1600 ft • lb( )=

7200 ft • lb7

= 1028.57 ft • lb

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rOA

rOB B

C

Example ¢ Remember that the unit vector will not

have units and that the position vector will have units

−37

−27

67

0 ft 4 ft 0 ft−400lb 400lb 200lb

−37

0 ft−400lb

−27

4 ft400lb

−37

800 ft • lb( ) − 27

0( ) + 67

1600 ft • lb( )=

7200 ft • lb7

= 1028.57 ft • lb


Couples ¢ A couple is a system of two forces ¢ The forces must satisfy the following

conditions for the force system to be a couple l The magnitudes of the forces must be

the same l The unit vectors of the forces must be

opposite in direction

u

1 = −1( )u

21 2F F=

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Couples ¢ A couple produces pure rotation on a

system ¢ The sum of the two forces will be equal to

0

1 2F F= u

1 = −1( )u

2


Couples ¢ Most importantly, a couple produces the

same amount of rotation about any moment center

1 2F F= u

1 = −1( )u

2

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Couples ¢ We will start with two forces in space who

satisfy the conditions of a couple

u

F1 = −1( )u

F2

1 2F F=F1

F2


Couples ¢ We will then pick some point in space as

the moment center

u

F1 = −1( )u

F2

1 2F F=F1

F2

a

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Couples ¢ We can pick a point on each force’s line

of action and construct moment arms from point a to a point on the life of action on each of these forces

u

F1 = −1( )u

F2

1 2F F=F1

F2

arab

rac

b

c


Couples ¢ The total moment generated by the two

forces about a is equal to

u

F1 = −1( )u

F2 1 2F F=

F1

F2

arab

rac

b

c

Ma

= rab

⊗ F1

+ rac

⊗ F2

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Couples ¢ Substituting the definition of F1 and F2

u

F1 = −1( )u

F2 1 2F F=

F1

F2

arab

rac

b

c

Ma

= rab

⊗ u1

F1( ) + rac

⊗ u2

F2( )


Couples ¢ Using our definition of the magnitudes in

a couple

1 2F F=

F1

F2

arab

rac

b

c

Ma

= rab

⊗ u1

F1( ) + rac

⊗ u2

F1( )

u

F1 = −1( )u

F2

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Couples ¢ And the relationship of the unit vectors

1 2F F=

F1

F2

arab

rac

b

c

Ma

= rab

⊗ u1

F1( ) + rac

⊗ −u1

F1( )

u

F1 = −1( )u

F2


Couples ¢ Manipulating the cross product

1 2F F=

F1

F2

arab

rac

b

c

Ma

= rab

− rac

( )⊗ u1

F1( )

u

F1 = −1( )u

F2

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Couples ¢ Now if we generate another position

vector from c to b and by vector addition we can say

1 2F F= Ma

= rab

− rac

( )⊗ u1

F1( )

F1

F2

arab

rac

b

c

rcb rab

= rac

+ rcb

u

F1 = −1( )u

F2


Couples ¢ Rearranging terms

1 2F F= Ma

= rab

− rac

( )⊗ u1

F1( )

F1

F2

arab

rac

b

c

rcb rab

− rac

= rcb

u

F1 = −1( )u

F2

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Couples ¢ And substituting in the expression for the

moment at a

1 2F F=

Ma

= rcb

( )⊗ u1

F1( ) F1

F2

arab

rac

b

c

rcb

rab

− rac

= rcb

u

F1 = −1( )u

F2


Couples ¢ So no matter where a is, the moment of

the couple will depend on how far apart the forces are

Ma

= rcb

( )⊗ u1

F1( ) F1

F2

b

c

rcb

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Couples ¢ The moment of a couple can be

calculated by taking a moment arm from line of action of one of the forces to the line of action of the other force and forming the cross product of the moment arm and the force that we went to

F1

F2

b

c

rcb


Couples ¢  In a two-dimensional problem, if you can

find the perpendicular distance between the forces, you can calculate the magnitude of the moment by multiplying the perpendicular distance times the magnitude of either one of the forces

¢ The sign will be given by the sense of rotation

F1

F2

b

c

rcb

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Homework ¢ Problem 4-50 ¢ Problem 4-51 ¢ Problem 4-53

Documents

Moment about an Axis - Memphis about an Axis.pdf3 5 Moments Along an Axis, Couples Monday, September 24, 2012 Moment about an Axis ! There are times that we are interested in the moment