[email protected] engr-36_lec-08_moments_math.ppt 1 bruce mayer, pe engineering-36:...
TRANSCRIPT
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt1
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 36
Chp 4: Moment
Mathematics
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt2
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
222zyx
zyx
MMMM
kMjMiM
M
ˆˆˆM
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt3
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Moments are VECTORS
Describe M in terms of a unit vector, û, directed along the LoA for M
zyx kjiuMu cosˆcosˆcosˆMˆˆMM
Find the θm by Direction CoSines
Mcos
Mcos
Mcos z
zy
yx
x
MMM
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt4
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
M = r X F
1. Magnitude of M measures the tendency of a force to cause rotation of a body about an Axis thru the pivot-Pt O
sinFrFsinrM Fd
d
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt5
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
M = r X F
2. The sense of the moment may be determined by the right-hand rule
M
MomentDirection
• 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 direction of the Moment
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt6
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
M = r X F
Combining (1) & (2) yields the Definition of the vector CROSS PRODUCT (c.f. MTH3)
bac
Engineering Mechanics uses the Cross Product to Define the Moment Vector
usinFrFrM • û is a unit vector directed by the Rt-Hand Rule• θ is the Angle Between the LoA’s for r & F
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt7
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
M = r X F → θ by Tail-toTail When Finding Moment
Magnitudes using:
The Angle θ MUST be determined by placing Vectors r & F in the TAIL-to-TAIL Orientation• See Diagram at Right
sin FrM
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt8
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
veDistributi
tiveNONassocia
tiveNONcommuta
2121 QPQPQQP
SQPSQP
QPQPPQ
QP
PQ
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt9
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Vector Prod: Rectangular Comps Vector Products Of Cartesian Unit
Vectors
0
0
0
kkikjjki
ijkjjkji
jikkijii
kQjQiQkPjPiPQPV zyxzyx
kQPQPjQPQPiQPQP xyyxzxxzyzzy
Vector Product In Terms Of Rectangular Coordinates
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt10
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
• Don’t Forget the MINUS sign in the Middle (j)Term– See also TextBook pg123
kFrFrjFrFriFrFr
FFF
rrr
kji
Fr xyyxxzzxyzzy
zyx
zyxˆˆˆ
ˆˆˆ
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt11
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Varignon’s Theorem
The Moment About a Point OOf The Resultant Of SeveralConcurrent Forces Is Equal To The Sum Of The MomentsOf The Various Forces About The Same Point O• Stated Mathematically
2121 FrFrFFr
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)
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt12
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
rxF in 3D Vector Properties
Cartesian CoOrds for a 3D M vector
xyyxz
xzzxy
yzzyx
FrFrM
FrFrM
FrFrM
The Magnitude of a 3D M vector
222xyyxxzzxyzzy FrFrFrFrFrFrM M
DirectionCoSines
M
M
M
xyyxzz
xzzxyy
yzzyxx
FrFr
M
M
FrFr
M
M
FrFr
M
M
cos
cos
cos Unit Vector
kM
Mj
M
Mi
M
Mu zyx ˆˆˆˆ
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt13
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
rxF in 2D r & F in XY Plane
If r & F Lie in the XY Plane, then rz = Fz = 0. Thus the rxF Determinant
kFrFrjrriFr
FF
rr
kji
Fr xyyxzxyy
yx
yxˆˆˆ
ˆˆˆ
0000
0
0
So in this case M is confined to the Z-Direction:
kFrFrkMFr xyyxzzXYˆˆM
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt14
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Direction for r in rF Consider the CrowBar Below We Want to find the
Torque (Moment) About pt-B due to Pull, P, applied at pt-A using rP
We have Two Choices for r:• r points A→B
• r points B→A
Which is Correct?
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt15
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Direction for r in rF We can find the Direction for r by
considering the SIGN of the Moment
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 rP and rP
x
y
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt16
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Direction for r in rF Put r and r into Component form
• Equal but Opposite
x
y
".cos"
".sin"
14235036
58275036
comp
comp
x
y
Then the two r’s
jir
jir
BAAB
ABBA
ˆ.ˆ.r
ˆ.ˆ.r
58271423
58271423
Now let
ilb 10P
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt17
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Direction for r in rF then the rxP calculations noting
x
y
lb-in k8.2750
i10j58.27i14.23
^
^^^
Pr BA
Thus rB→A is the CORRECT position vector
lb-in k8.2750
i10j58.27i14.23
^
^^^
Pr AB
^^^^^
kij0ii
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt18
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Direction for r in rF To Calc the Moment about pt-B use: The position Vector
points FROM the PIVOT-point TO the Force APPLICATION-point on the Force LoA
Summarize this as
FROM the PIVOT TO the FORCE
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt19
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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• This is usually apparent
from the problem or situation context ABAB
ABAB
u ˆ
λu
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt20
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
FrM ACA
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt21
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example 3D Moment - Solution Resolve Both F and rAC into
Cartesian Components Take Cross-Product Using Determinant
kjiM A
mN 8.82mN 8.82mN 687 .
kji
kji
CD
CDuFF
N 128N 69N 120
m 5.0
m 32.0m 0.24m 3.0N 200
N 200ˆ
kikkirrr ACAC m 080m 30m 320m 40m 30 .....
12896120
08.003.0
ˆˆˆ
ˆ
kji
uFACA rMWhich Moment will Most Likely Cause DEFORMATION?
ACr
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt22
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Moment About an Axis (§4.5)
Moment MO Of A Force F ,Applied at The Point A, About a Point O, Recall
FrMO
FrMM OOL
Scalar Moment MOL About AnAXIS OL Is The Projection OfThe Moment Vector MO OntoThe OL Axis using the Dot Product
MOL it the tendency of the applied force to cause a rotation about the AXIS OL
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt23
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Moment About an Axis – cont. Moments of F About The CoOrd Origin
xyz
zxy
yzx
yFxFM
xFzFM
kzjyixrzFyFM
as
BABA
BA
BBL
rrr
Fr
MM
Moment Of A Force AboutAn Arbitrary Axis BL
• 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
BABA rrr
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt24
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Moment About an Axis – cont.
Since the moment, ML,about an arbitrary axisis INDEPENDENT ofposition vector, r, thatruns from ANY Pointon 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
BAr
CAr
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt25
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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 asthe Componentof M directed along “axis” OL
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt26
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt27
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example MOL - Solutiona) Moment of P about A
kjPjiaM
kjPa
kjaP
aa
kajaP
FC
FCPP
jiajaiakajakaiarrr
PrM
A
AFAF
AFA
2
2222
kjiaPM A
2
kji
Pai
MiMkjiMM AAAABAB
2
001
aPaPM AB 707.02
aPaP
M A 225.12
3
b) Moment of P about AB
AFr
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt28
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example MOL - Solution
kaPjaPiaP
PP
aa
kji
PrAF ˆ2ˆ2ˆ2
220
0
ˆˆˆ
AFr
a) Alternative Moment of P about A
kjPa
kjaP
aa
kajaP
FC
FCPP
jiajaiakjaia
kzjyixrrr
PrM
AFAF
AFA
ˆˆ22
ˆˆˆˆ
ˆˆˆˆˆ0ˆˆ
ˆˆˆ
22
kjiaPM A
2
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt29
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example MOL - Solution
c) Moment of P About Diagonal AG
1116
23
12
3
1
3
aP
kjiaP
kjiM
kjiaP
M
kjia
kajaia
r
r
AG
AG
MM
AG
A
AG
AGAG
AAGAG
ˆ
ˆ
) toOPPOSITE is (dir.408.06
aPaP
M AG
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt30
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
0
11063
1
2
P
kjikjP
P
PdaP
M AG 6
aa
d 408.06
• Then the Moment (or twist) Caused by P About AG = Pd; Thus
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt31
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Mixed Triple Product
Do Find MOL we used the Qtyû•(r x F). Formalize thisOperation as the Mixed TripleProduct for vectors S, P, & Q resultscalar QPS
SPQQSPPQS
PSQSQPQPS
Associativity and Communtivity for the Mixed Triple Product Of Three Vectors
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt32
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Evaluate the Mixed Triple Prod
Let V = PxQ, Then
zzyyxx VSVSVSVSQPS
zyyzzyx VVQPQPV andforSimilarly
zyx
zyx
zyx
xyyxz
xzzxyyzzyx
QQQ
PPP
SSS
QPS
QPQPS
QPQPSQPQPSQPS
And
Thus
• Determinant Notation Yet Again
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt33
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Mixed Triple Product vs MOL
The Mixed Triple Product can be used to find the Magnitude of the Moment about an Axis.
zyx
zyx
zolyolxol
OL
olOL
FFF
rrr
uuu
u
,,, ˆˆˆ
M
FrˆM
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt34
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhiteBoard Work
Let’s WorkThis NiceProblem
Determine MA as caused by application ofthe 120 N force
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt35
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt36
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt37
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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 TheVector 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
[email protected] • ENGR-36_Lec-08_Moments_Math.ppt38
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics