Chapter 4 Rigid Bodies Equivalent Force/Moment …cac542/L4.pdfChapter 4 Rigid Bodies Equivalent...
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Transcript of Chapter 4 Rigid Bodies Equivalent Force/Moment …cac542/L4.pdfChapter 4 Rigid Bodies Equivalent...
2
MEM202 Engineering Mechanics - Statics MEM
Equilibrium of Rigid Bodies
0=−=∑ PPFrrr
Pr
Pr
0=∑Fr
Pr
Pr
0≠∑Mr
0 0 == ∑∑ MFrr
Equilibrium for non-concurrent force systems
Force systems to be studied:1. Coplanar, parallel force systems2. Coplanar, non parallel, non-concurrent force systems3. Non-coplanar, parallel force systems4. Non-coplanar, nonparallel, non-concurrent force systems
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MEM202 Engineering Mechanics - Statics MEM
4.2 Moments and Their Characteristics
OMr
O
FrMFdM OO
rrr×== or
AAF −about ofMoment r
Fr
A
A
Orr
Fr
O
drr
• The magnitude of a moment is the product of the magnitude of theforce (F) and the perpendicular distance (d) from the line of action of the force to the axis of rotation (A-A).
• The units for moments are in⋅lb (U.S.) and N⋅m (SI).• A moment is a vector. Thus, parallelogram law of addition applies.
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MEM202 Engineering Mechanics - Statics MEM
4.2 Moments and Their CharacteristicsSign Convention for Moments
Right-hand Rule :When fingers of right hand curl from positive r to positive F, the thumb is pointing in the direction of positive moment.
In 2-D, a counter-clockwise moments ( ) is positive and a clockwise moments ( ) is negative.
x
y
x
y
+ −
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MEM202 Engineering Mechanics - Statics MEM
4.2 Moments and Their CharacteristicsExamples
BCd
Moment of FA about point E( ) (ccw) lbin 100010100 ⋅=== AEAE dFM
r
Moment of FE about point A( ) (ccw) lbin 240012200 ⋅=== EAEA dFM
r
Moment of FD about point B( ) (cw) lbin 420014300 ⋅=== DBDB dFM
r
Moment of FC about point B( ) (ccw) mN 2.2512.016.090 ⋅=+== CBCB dFM
r
Moment of FD about point A( ) (ccw) mN 0.6620.035.0120 ⋅=+== DADA dFM
r
Moment of FB about point C( )[ ] (cw) mN 4.3630cos12.016.0150 ⋅=+== o
rBCBC dFM
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MEM202 Engineering Mechanics - Statics MEM
Moment of FB about point C( ) (cw) lbin 60015400 ⋅=== BCBC dFM
r
Moment of FC about point B( ) (ccw) lbin 360012300 ⋅=== CBCB dFM
r
Moment of FD about point B( ) (ccw) lbin 7503250 ⋅=== DBDB dFM
r
o13.531520tan 1 == −φ
in. 251520 22 =+=ACL
in. 0.313.53sin1515
in. 0.1213.53sin15
in. 0.1510
=−=
==
=−=
o
o
DB
CB
ACBC
d
d
Ld
4.2 Moments and Their CharacteristicsExamples
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MEM202 Engineering Mechanics - Statics MEM
Varignon’s Theorem: The moment of the resultant of a system of forces with respect to any axis or point is equal to the vector sum of the moments of the individual forces of the system with respect to the same axis or point.
4.2 Moments and Their Characteristics
2α
1α
2αα
Rr
1Fr
2Fr
1α
α
1d
d2d
( )αcos :about ofMoment hRRdMOR O ==r
21
2211
2211
2211
coscoscoscoscoscos
MMMdFdFRd
hFhFRhFFR
R +=⇒+=⇒
+=⇒+=
αααααα
nnO
n
dFdFdFRdMFFFR+++==⇒
+++=L
rL
rrr
2211
21 general In11 cosαF22 cosαF
αcosR
O
( )111111 cos :about ofMoment αhFdFMOF ==r
h
( )222222 cos :about ofMoment αhFdFMOF ==r
21 FFRrrr
+=
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MEM202 Engineering Mechanics - Statics MEM
4.2 Moments and Their CharacteristicsExample of Varignon’s Theorem
ft 83.230sin330cos5 =−= ood lb 0.43330cos50030cos === ooFFx
lb 0.25030sin50030sin === ooFFylbft 1415)83.2(500 ⋅−=−=−= FdMO
( ) ( ) lbft 1415534 ⋅−=−= xO FFM
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MEM202 Engineering Mechanics - Statics MEM
lb 8.25930cos30030cos === ooFFx
lb 0.15030sin30030sin === ooFFy
4.2 Moments and Their CharacteristicsExample of Varignon’s Theorem and Principle of Transmissibility
( ) ( ) mN 0.9520.025.0 ⋅−=−−= yxB FFM
( ) m 366.030tan2.025.02 =+= od( ) mN095366.082592 ⋅−=−=−= ..dFM xB
( ) m 633.030cot25.020.03 =+= od( ) mN095633.00.1503 ⋅−=−=−= .dFM yB
Move F to D (aligned horizontally with B)
FdM B = But d is not easy to determine
d
Determine Moment of F about B
Move F to C (aligned vertically with B)Find the components of F first
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a MomentDot and Cross Products of Vectors (Appendix A)
( ) ( ) ( )kBABAjBABAiBABA
BBBAAAkji
eABBAC
xyyxzxxzyzzy
zyx
zyx
rrr
rrr
rrrr
−+−+−=
=
=×= sinθ
kAjAiAA zyx
rrrr++=
kBjBiBB zyx
rrrr++=
zzyyxx BABABAABBA ++==⋅ θcos
BAe
BeAe
C
CCrrr
rrrr
and containing plane
,
⊥
⊥⊥Czyx eCkCjCiCC rrrrr
=++=
y
z
Ar
BrC
rPlane containing A and B
θ
x OCer
Dot (Scalar) product of two vectors yields a scalarrr
Cross (Vector) product of two vectors yields a vector
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a MomentCross Product of Two Vectors
kBjBiBBkAjAiAA zyxzyx
rrrrrrrr++=++= ,
kCjCiCBBBAAAkji
BAC zyx
zyx
zyx
rrr
rrr
rrr++==×=
yxzyx
yxzyx
zyx
zyx
BBBBBAAAAAjikji
BBBAAAkji
rrrrrrrr
=
+ + +− − −( ) ( ) ( )kBABAjBABAiBABA xyyxzxxzyzzy
rrr −+−+−=
CkCjCiC
eCCCC zyxCzyx
rrrr ++
=++= ,222
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a MomentCross Product of Two Vectors
kjiBkjiArrrrrrrr
352 ,243 :Example ++−=+−=
( ) ( ) ( )( ) ( )( ) ( ) ( )( )
kji
kji
kBABAjBABAiBABA
jikjikji
BBBAAAkji
BAC
xyyxzxxzyzzy
zyx
zyx
rrr
rrr
rrr
rrrrrrrrrrr
rrr
7 13 22
2453 3322 5234
5235243243
352243
+−−=
−×−−×+×−−×+×−×−=
−+−+−=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−−−−=
−−==×=
( ) ( )( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( ) 0264.03491.05830.02
0264.02491.04830.03
264.0491.0830.071322
71322 :NOTE222
=+−+−−=⋅
=+−−+−=⋅
+−−=+−+−
+−−=
C
C
C
eB
eA
kjikjie
rr
rr
rrrrrr
r
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a MomentMoment of a Force About A Point
FdMO =
2-D 3-D
Fr
d x
y
O
FdMO =
x
y
z Fr
rr
Od
α
α
rF
erF
eFrFrMO
rr
r
rr
rrrr
and containing plane the to
larperpendicur unit vecto A : and between angle The :
sin
α
α=×=
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MEM202 Engineering Mechanics - Statics MEM
αsinrd =
eMeFd
eFrFrM
O
Orr
rrrr
==
=×=
sinα
4.3 Vector Representation of a MomentMoment of a Force About A Point
Frrr
O
α
1α2α
1rr
2rr
rF rr and
containing Plane
d
Moment of a force about O is equal to the cross product of a vector drawn from O to any point along the force and the force itself.22
11
sin
sin
α
α
r
rd
=
=
eMeFd
eFrFr
eFrFrM
O
O
rr
rrr
rrrr
==
=×=
=×=
sin
sin
222
111
α
α
Direction of M is determined by using the right-hand rule
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a MomentBF point about of Moment :Example
r
x y
zFr
O
B
ABArr rr
=
Brr
Arr
• Select any point (A) on the line of action of F.
• Determine vector r using the coordinates of points A and B:r = rA/B = rA − rB
• Compute the moment of F about B:MB = r × F = rA/B × F
If the coordinates of points A and B are (xA, yA, zA) and (xB, yB, zB), then ( ) ( ) ( )kzzjyyixxrrrr BABABABABA
rrrrrrr −+−+−=−==
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a Moment2-D Cases
Fr
rr
x
y
O d
αFr
rr x
y
Oyr
xr xF
yF
kFr
kFd
kMM OO
r
r
rr
sin
α=
=
=
( )kFrFr
FrM
xyyx
Or
rrr
−=
×=
Fr
rr
x
y
O
α
rθ
Fθ
rθ
( )rFrF
rF
θθθθθθα
sincoscossinsinsin
−=−=
( )( )kFrFr
kFrFr
kFrM
xyyx
rFrF
O
r
r
rr
sincoscossin
sin
−=
−=
=
θθθθ
α
ry
rx
Fy
Fx
rF
rrrrFFFF
θθθθ
θθα
sincos
sincos
====
−=
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a Moment2-D Cases - Example
1. Moment of F about O2. Perpendicular distance d from O to F
( ) N 60080060.080.01000 jijiFrrrrr
+=+=
( )m 20.010.0 jirr OA
rrrr+==
( ) mN 100
00
⋅−==−=
=×=
kkMkFrFr
FFrr
kjiFrM
zxyyx
yx
yxO
rrr
rrr
rr
1. Moment of F about O
d
BAr
2. Determine the distance d:
m 22.0==F
Md B
r
r
jir BA
rr35.010.0 += mN 220 ⋅−=×= kFrM BAB
rrrr
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a Moment3-D Cases
kFjFiFF
krjrirr
zyx
zyxrrrr
rrrr
++=
++=
( ) ( ) ( )
MO
zyx
xyyxzxxzyzzy
zyx
zyxO
eMkMjMiM
kFrFrjFrFriFrFr
FFFrrrkji
FrM
r
rrr
rrr
rrr
rrr
=
++=
−+−+−=
=×=
xyyxzzxxzyyzzyx FrFrMFrFrMFrFrM −=−=−=
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a MomentMagnitude and Direction of the Resulting Moment
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
++=
++=
−−−
O
zz
O
yy
O
xx
zyxM
zyxO
MM
MM
MM
kjie
MMMM
111
222
coscoscos
coscoscos
θθθ
θθθrrrr
MOzyxO eMkMjMiMM rrrrr =++=
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a MomentForce as a Sliding Vector (Principle of Transmissibility)
CACBABA rrrrr rrrrr+=+=
( )( ) Frr
Frr
FrM
CAC
BAB
AO
rrr
rrr
rrr
×+=
×+=
×=FrAr
r
O
Brr
Crr
AB
C0//
0//
=×⇒
=×⇒
FrFr
FrFr
CACA
BABArrrr
Q
rrrrQ
FrFrFrM CBAO
rrrrrrr×=×=×=∴
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a Moment
( ) ( ) ( )N 7.5608.6004.300
14015075
14015075875222
kji
kjiF
rrr
rrrr
++=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++
++=
kjir OA
rrrr 15.025.020.0 ++=
mN 1.451.671.50
7.5608.6004.30015.025.020.0
⋅+−=
=×=
kji
kjiFrM OAO
rrr
rrr
rrr
1. Moment of F about O2. Perpendicular distance d from O to F
( ) ( ) ( ) mN 95.11.451.671.50 222 ⋅=+−+== OO MMr
mm 108.6m 0.108687595.1
====F
Md O
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a MomentMoment of a Force About A Line (Axis)
MOO eMFrM rrrr=×= ( )
( )[ ] nOBnn
nnOOB
eMeeFr
eeMMrrrrr
rrrr
=⋅×=
⋅=
knjnine zyxn
rrrr++=
( ) ( ) ( ) ( )[ ] [ ]
( ) ( ) ( ) zxyyxyzxxzxyzzy
zyx
zyx
zyx
n
zyx
zyx
zyxxyyxzxxzyzzynnOOB
nFrFrnFrFrnFrFrFFFrrrnnn
eFFFrrrkji
knjninkFrFrjFrFriFrFreFreMM
−+−+−==⋅=
++⋅−+−+−=⋅×=⋅=
r
rrr
rrrrrrrrrrr
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MEM202 Engineering Mechanics - Statics MEM
4.3 Vector Representation of a Moment
( ) ( ) ( )N 9.1829.3189.338
8.458800850
8.458800850500222
kji
kjiF
rrr
rrrr
+−−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+−+−
+−−=
kir OB
rrr 75.085.0 +−=
mN 1.2717.982.239
9.1829.3189.33875.0085.0
⋅+−=
−−−=×=
kji
kjiFrM OBO
rrr
rrr
rrr
mN 2.2399.1829.3189.338
75.0085.0000.1
mN 2.239
⋅−=−−
−−
=⋅×=
⋅−=⋅=
OCOB
OCOOC
eFr
eMM
rrr
rrieOC
rr 0.1−=
Determine moment of F about line OC