ENGR-36_Lec-08_Moments_Math_H13e

Report
Engineering 36
Chp 4: Moment
Mathematics
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Moments are VECTORS
 As Described Last Lecture a Moment
is a measure of “Twisting Power”
 A Moment has Both MAGNITUDE &
Direction and can be Represented as a
Vector, M, with Normal
Vector properties
M  M xiˆ  M y ˆj  M z kˆ
M M  M M M
2
x
Engineering-36: Engineering Mechanics - Statics
2
2
y
2
z
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Moments are VECTORS
 Describe M in terms of a unit vector, û,
directed along the LoA for M

M  M uˆ  Muˆ  M iˆ cos  x  ˆj cos  y  kˆ cos  z
 Find the θm by Direction CoSines
My
Mx
Mz
cos x 
cos y 
cos z 
M
M
M
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt

M=rXF
1. Magnitude of M measures the
tendency of a force to cause rotation
of a body
about an
Axis thru
the
pivot-Pt O
d
M  d  F   r sin  F   r  F sin
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
M=rXF
2. The sense of the moment may be
determined by the right-hand rule
• If the fingers of the
RIGHT hand are
curled from the
direction of r toward
the direction of F,
then the THUMB
points in the
Moment
direction of the Direction
Moment
Engineering-36: Engineering Mechanics - Statics
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M
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
M=rXF
c  a b
 Combining (1) & (2) yields
the Definition of the
vector CROSS
PRODUCT (c.f. MTH3)
 Engineering Mechanics uses the Cross
Product to Define the Moment Vector
M  r  F   r  F sin  uˆ
• û is a unit vector directed by the Rt-Hand Rule
• θ is the Angle Between the LoA’s for r & F
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
M = r X F → θ by Tail-toTail
 When Finding Moment
Magnitudes using:
M  r  F  sin 
 The Angle θ MUST be
determined by placing
Vectors r & F in the
TAIL-to-TAIL
Orientation
• See Diagram at Right
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Cross Product Math Properties
 Recall Vector ADDITION Behaved As
Algebraic Addition
– BOTH Commutative and Associative.
 The Vector PRODUCT Math-Properties do
NOT Match Algebra - Vector Products:
• Are NOT Commutative
• Are NOT Associative
• ARE Distributive
P Q
Q  P  P  Q  P  Q
NONcommutative
P  Q S  P  Q  S
NONassociative
P  Q1  Q 2   P  Q1  P  Q 2 Distributive
Engineering-36: Engineering Mechanics - Statics
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Q P
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Vector Prod: Rectangular Comps
 Vector Products Of Cartesian Unit Vectors
 
i i  0
  
i  j k
 

i k   j
   
 
j  i  k k  i  j
 
 

j  j  0 k  j  i
 
  
j k  i
k k  0
 Vector Product In Terms Of
Rectangular Coordinates









V  P  Q  Px i  Py j  Pz k  Qx i  Q y j  Qz k




 Py Qz  Pz Qy i  Pz Qx  PxQz  j  PxQy  Py Qx k
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
rxF in 3D  Deteriminant Notation
 Consider 3D versions of r & F
 Taking the Cross Product Yields M
 Determinant Notation provides a
convenient Tool For the Calculation
iˆ
r  F  rx
Fx
ˆj
ry
Fy
kˆ
rz  ry Fz  rz Fy iˆ  rx Fz  rz Fx  ˆj  rx Fy  ry Fx kˆ
Fz
• Don’t Forget the MINUS sign in the Middle
(j)Term
– See also TextBook pg123
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Varignon’s Theorem
 The Moment About a Point O
Of The Resultant Of Several
Concurrent Forces Is Equal
To The Sum Of The Moments
Of The Various Forces
About The Same Point O


  
   
• Stated Mathematically r  F1  F2    r  F1  r  F2  
 Varignon’s Theorem Makes It Possible To
Replace The Direct Determination Of The
Moment of a Force F By The Moments of Its
Components (which are concurrent)
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
rxF in 3D  Vector Properties
M x  ry Fz  rz Fy
M y  rx Fz  rz Fx 
M z  rx Fy  ry Fx
 Cartesian CoOrds
for a 3D M vector
 The Magnitude of a 3D M vector
M M 

r F  r F   r F  r F   r F
2
y
z
z
x
z
z
M x ry Fz  rz Fy
Direction cos x  M  M
CoSines
M y  rx Fz  rz Fx 
cos y 

M
M
M z rx Fy  ry Fx
cos z 

M
M
Engineering-36: Engineering Mechanics - Statics
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y
2
x
x
y
 ry Fx 
2
 Unit Vector
uˆ 
Mx ˆ M y ˆ Mz ˆ
i
j
k
M
M
M
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
rxF in 2D  r & F in XY Plane
 If r & F Lie in the XY Plane, then
rz = Fz = 0. Thus the rxF Determinant
iˆ
r  F  rx
Fx
ˆj
ry
Fy
kˆ
0  ry 0  0Fy iˆ  rx 0  rz 0 ˆj  rx Fy  ry Fx kˆ
0
 So in this case M is confined to the
Z-Direction:
r  F 
XY
 M z  M z kˆ  rx Fy  ry Fx kˆ
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Direction for r in rF
 Consider the
CrowBar Below
 We Want to find the
Torque (Moment)
About pt-B due to
Pull, P, applied at
pt-A using rP
 We have Two
Choices for r:
• r points A→B
• r points B→A
 Which is Correct?
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Direction for r in rF
 We can find the
Direction for r by
considering the
SIGN of the Moment
y
x
Engineering-36: Engineering Mechanics - Statics
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 In this case it’s
obvious (to me,
anyway) that P will
cause CLOCKwise
Rotation about Pt-B
 In the x-y Plane
ClockWise Rotation
is defined as
NEGATIVE
 Test rP
and rP
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Direction for r in rF
 Put r and r into
Component form
• Equal but Opposite
ycomp  36" sin50  27.58"
xcomp  36" cos50  23.14"
 Then the two r’s
rAB  r AB  23.14iˆ  27.58 ˆj
rB A  r BA  23.14iˆ  27.58 ˆj
y
 Now let
x
Engineering-36: Engineering Mechanics - Statics
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P  10 lb iˆ
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Direction for r in rF
^
^

  ^
rA B  P   23.14 i  27.58 j   10 i 
 then the rxP

 

calculations noting
 ^
^ ^
^ ^
^
 0  275.8  k  in - lb
i i  0
j i   k
 

^
^

  ^
rB A  P    23.14 i  27.58 j   10 i 

 

 ^
 0  275.8  k  in - lb
 
y

x
Engineering-36: Engineering Mechanics - Statics
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 Thus rB→A is the
CORRECT
position vector
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Direction for r in rF
 To Calc the Moment
about pt-B use:
 The position Vector
points FROM the
PIVOT-point TO the
Force APPLICATIONpoint on the Force LoA
 Summarize this as
FROM the
PIVOT TO
the FORCE
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Unit Vector Notation: u ≡ λ
 Our Text uses u to denote the unit
vector
 While u is quite popular as the unit
vector notation, other symbols are often
used (kind of like θ & φ for angles)
 On Occasion I will use λ to represent
the unit vector
u λ
• This is usually apparent
from the problem
or situation context
Engineering-36: Engineering Mechanics - Statics
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AB
AB
uˆ AB  ˆAB
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Example: 3D Moment
 A Rectangular Plate Is Supported
By The Brackets At A and B and
By A Wire CD. Knowing That The
Tension In The Wire is 200 N,
Determine The Moment About A
Of The Force Exerted By The Wire
At connection-point C.
 Solution Plan
• The Moment MA Of The Force F
Exerted By The Wire Is Obtained By
Evaluating The Vector Product
M A  r AC  F
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Example 3D Moment - Solution
 Resolve Both F and rAC into
Cartesian Components
 Take Cross-Product Using Determinant

 
rAC  rC  rA  0.3 mi  0.4 mk  0.32 mk   0.3 mi  0.08 mk
rAC

CD
F  Fuˆ  200 N 
CD



 0.3 m i  0.24m  j  0.32 m k
 200 N 
0.5 m



 120 N  i  96 N  j  128 N k
iˆ
ˆj
kˆ
M A  rAC  Fuˆ  0.3
0 0.08
 120 96  128
Which Moment will
Most Likely Cause
DEFORMATION?



M A  7.68 N  mi  28.8 N  m j  28.8 N  mk
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Moment About an Axis (§4.5)
 Moment MO Of A Force F ,
Applied at The Point A,
About a Point O, Recall

 
MO  r  F
 Scalar Moment MOL About An
AXIS OL Is The Projection Of
The Moment Vector MO Onto
The OL Axis using the Dot Product
M OL

Engineering-36: Engineering Mechanics - Statics
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
  

   MO    r  F

MOL it the tendency
of the applied force
to cause a rotation
about the AXIS OL
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Moment About an Axis – cont.
 Moments of F About The CoOrd Origin
M x  yFz  zFy
as
r  xi  y j  z k
M y  zFx  xFz
M z  xFy  yFx
 Moment Of A Force About
An Arbitrary
Axis BL
 
M BL    M B
 

   rBA  F

 
rBA  rA  rB

r BA  r A  r B

• Similar Analysis for CL, Starting With MC, Shows
That MCL = MBL; i.e., the Result is Independent of
the Location of the Point ON the Line
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Moment About an Axis – cont.
 Since the moment, ML,
about an arbitrary axis
r
is INDEPENDENT of
position vector, r, that
r
runs from ANY Point
on the axis to ANY
pointon the LoA of the force we can
choose the MOST CONVENIENT
Points on the Axis and the Force LoA to
determine ML
BA
CA
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
MOL Physical Significance
 MOL Measures the Tendency of an
Applied Force to Impart to a Rigid Body
Rotation about a fixed Axis OL
• i.e., How Much will the Applied Force
Cause The body to Rotate about an AXLE
• MOL can be
Considered as
the Component
of M directed
along “axis” OL
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Example: MOL

A Cube With Side Length a
is Acted On By a Force P
as Shown

Determine The Moment Of P:
a) About Pt A
b) About The Edge (Axis) AB
c) About The Diagonal (Axis) AG
of The Cube

For Lines AG and FC
d) Determine The Perpendicular
Distance Between them
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Example MOL - Solution
a) Moment
of P about A


M A  rAF  P




 

 
rAF  rF  rA  a i  a k  a j  a k  a i  a j  ai  j 



FC
a j  ak
a j k  P  
PP
P
P

 j k
2
2
2
FC

a 2
a a


  

  



aP
MA 
M A  ai  j   P  j  k
 i  j k
2
2




r AF




 

b) Moment of P about AB

3aP
MA 
 1.225aP
2
M AB

27

 



 
  AB  M A  1i  0 j  0k  M A  i  M A
  P   
 i a
 i  j k
2


M AB  aP
Engineering-36: Engineering Mechanics - Statics
 
2

 0.707aP
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt

Example MOL - Solution
a) Alternative Moment of P about A



M A  rAF  P

 
rAF  rF  rA  xiˆ  yˆj  zkˆ
 aiˆ   a  ˆj  0kˆ  aiˆ  aˆj  a iˆ  ˆj

FC
a ˆj  akˆ
a ˆj  kˆ  P
 ˆj  kˆ
PP
P
P


2
2
2
FC


a 2
a a

r AF
iˆ
ˆj
rAF  P  a
a
0 P 2
kˆ

0
 aP
P 2


M A  aP
Engineering-36: Engineering Mechanics - Statics
28
 
2 iˆ   aP





 
2 ˆj  aP
  
2 i  j k

Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt


2 kˆ
Example MOL - Solution
c) Moment of P About Diagonal AG
M AG
ˆAG 

MA 
M AG 

M AG







aP

 0.408aP (dir. is OPPOSITE to  )
6
Engineering-36: Engineering Mechanics - Statics
29

AG  M A

 

AG rAG ai  aj  ak
1   



i  j k
AG rAG
a 3
3
aP   
i  j k
2
1    aP   
i  j k 
i  j k
3
2
aP
1  1  1
6
 ˆ
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt

Example MOL - Solution
 Perpendicular distance between
AG and FC
• Notice That Plane OFC Appears To
Be  to Line AG, And FC Resides In
this Plane
– Since P Has Line-of-Action FC We Can
Test Perpendicularity with Dot Product
  P   1   
P
0  1  1
P 
j k 
i  j k 
2
3
6
0





• Then the Moment (or twist) Caused by
P About AG = Pd; Thus
M AG 
Engineering-36: Engineering Mechanics - Statics
30
aP
 Pd
6
d
a
 0.408a
6
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Mixed Triple Product
 Do Find MOL we used the Qty
û•(r x F). Formalize this
Operation as the Mixed Triple
Product for vectors S, P, & Q


  
S  P  Q  scalar result
 Associativity and Communtivity for the Mixed
Triple Product Of Three Vectors
  
  
  
S  PQ  P  Q S  Q  S  P
 
  
  
  S  Q  P   P  S  Q  Q  P  S




Engineering-36: Engineering Mechanics - Statics
31







Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt

Evaluate the Mixed Triple Prod
 Let V = PxQ, Then


S  P  Q  S  V  S xVx  S yV y  S zVz
 And
Vx  PyQz  PzQy SimilarlyforVy andVz
 Thus


  
S  P  Q  S x Py Qz  Pz Qy   S y Px Qz  Pz Qx 
Sx
  
S  P  Q  Px
Sy
Py
S z • Determinant
Notation
Pz
Qx
Qy
Qz


Engineering-36: Engineering Mechanics - Statics
32
 S z Px Qy  Py Qx 
Yet Again
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Mixed Triple Product vs MOL
 The Mixed Triple Product can be used
to find the Magnitude of the Moment
about an Axis.
MOL  uˆol  r  F
uˆol , x
uˆol , y
uˆol , z
MOL  rx
ry
rz
Fx
Fy
Fz
Engineering-36: Engineering Mechanics - Statics
33
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
WhiteBoard Work
Let’s Work
This Nice
Problem
 Determine MA as caused by application of
the 120 N force
Engineering-36: Engineering Mechanics - Statics
34
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Engineering-36: Engineering Mechanics - Statics
35
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Engineering-36: Engineering Mechanics - Statics
36
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Vector/Cross Product
 TWISTING Power of a Force
 MOMENT of the Force
• Quantify Using VECTOR
PRODUCT or CROSS PRODUCT
 Vector Product Of Two Vectors
P And Q Is Defined As The
Vector V Which Satisfies:
• Line of Action of V Is Perpendicular To Plane
Containing P and Q.
– Rt Hand Rule Determines Direction for V
• |V| =|P|•|Q|•sin
Engineering-36: Engineering Mechanics - Statics
37
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt
Engineering-36: Engineering Mechanics - Statics
38
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt

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