6161103 4.9 further reduction of a force and couple system
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Transcript of 6161103 4.9 further reduction of a force and couple system
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� Consider special case where system of forces and moments acting on a rigid body reduces at point O to a resultant force FR = ∑F and MR = ∑ MO, which are perpendicular to one anotherperpendicular to one another
� Further simplify the system by moving FR to another point P either on or off the body
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� Location of P, measured from point O, can be determined provided FR and MR known
� P must lie on the axis bb, which is perpendicular to the line of action at FR and the aa axisthe line of action at FR and the aa axis
� Distance d satisfies MRo = Frd or d = MRo/Fr
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� With FR located, it will produce the same resultant effects on the body
� If the system of forces are concurrent, coplanar or parallel, it can be reduced to a single resultant force actingparallel, it can be reduced to a single resultant force acting
� For simplified system, in each case, FR and MRo will always be perpendicular to each other
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Concurrent Force Systems
� All the forces act at a point for which there is no resultant couple moment, so that point P is no resultant couple moment, so that point P is automatically specified
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Coplanar Force Systems � May include couple moments directed perpendicular
to the plane of forces
� Can be reduced to a single resultant force
When each force is moved to any point O in the x-y � When each force is moved to any point O in the x-y plane, it produces a couple moment perpendicular to the plane
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Coplanar Force Systems
� For resultant moment,
MRo = ∑M + ∑(r X F)
� Resultant moment is perpendicular to resultant force � Resultant moment is perpendicular to resultant force
� FR can be positioned a distance d from O to create this same moment MRo about O
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Parallel Force Systems � Include couple moments that are perpendicular to the
forces
� Can be reduced to a single resultant force� Can be reduced to a single resultant force
� When each force is moved to any point O in the x-y plane, it produces a couple moment components only x and y axes
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Parallel Force Systems � For resultant moment,
MRo = ∑MO + ∑(r X F)
� Resultant moment is perpendicular to resultant force � Resultant moment is perpendicular to resultant force
� FR can be positioned a distance d from O to create this same moment MRo about O
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4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� Three parallel forces acting on the stick can be replaced a single resultant force FR acting at a distance d from the grip
� To be equivalent,
F = F + F + FFR = F1 + F2 + F3
� To find distance d,
FRd = F1d1 + F2d2 +F3d3
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Procedure for Analysis � Establish the x, y, z axes and locate the resultant
force an arbitrary distance away from the origin of the coordinates
Force Summation
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Force Summation� For coplanar force system, resolve each force
into x and y components� If the component is directed along the positive x
or y axis, it represent a positive scalar� If the component is directed along the negative x
or y axis, it represent a negative scalar
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Procedure for Analysis Force Summation
� Resultant force = sum of all the forces in the system
Moment Summation
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Moment Summation
� Moment of the resultant moment about point O = sum of all the couple moment in the system plus the moments about point O of all the forces in the system
� Moment condition is used to find location of resultant force from point O
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Reduction to a Wrench � In general, the force and couple moment system
acting on a body will reduce to a single resultant force and a couple moment at o that are not
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
force and a couple moment at o that are not perpendicular
� FR will act at an angle θ from MRo
� MRo can be resolved into one perpendicular M┴and the other M║ parallel to line of action of FR
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Reduction to a Wrench � M┴ can be eliminated by moving FR to point P
that lies on the axis bb, which is perpendicular to both MRo and FR
� To maintain equivalency of loading, for distance
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� To maintain equivalency of loading, for distance from O to P, d = M┴/FR
� When FR is applied at P, moment of FR tends to cause rotation in the same direction as M┴
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Reduction to a Wrench � Since M ║ is a free vector, it may be moved to P so that it
is collinear to FR
� Combination of collinear force and couple moment is called a wrench or screw
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
called a wrench or screw
� Axis of wrench has the same line of action as the force
� Wrench tends to cause a translation and rotation about this axis
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Reduction to a Wrench
� A general force and couple moment system acting on a body can be reduced to a wrench
� Axis of the wrench and the point through which
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� Axis of the wrench and the point through which this axis passes can always be determined
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Example 4.16
The beam AE is subjected to a system of coplanar forces. Determine the magnitude, direction and location on the beam of a resultant force which is
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
location on the beam of a resultant force which is equivalent to the given system of forces measured from E
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Solution
Force Summation
Σ=→+ FF xRx ;
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
↓=−=
+−=
Σ=→+
→==
+=
NN
NNF
FF
NN
NNF
Ry
yRy
Rx
0.2330.233
20060sin500
;
0.3500.350
10060cos500
o
o
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Solution
� For magnitude of resultant force
5.420
)0.233()0.350()()( 2222
=
+=+= RyRxR
N
FFF
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� For direction of resultant force
o7.33
0.350
0.233tantan
5.420
11
=
=
=
=
−−
Rx
Ry
F
F
N
θ
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Solution
- Moment Summation
� Summation of moments about point E,MM ERE ;Σ=
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
md
mNmN
mNmNNdN
MM ERE
07.50.233
1.1182
)5.2)(200()5.0)(100(
)0)(60cos500()4)(60sin500()0(350)(0.233
;
==
−−
+=+
Σ=oo
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Example 4.17
The jib crane is subjected to three coplanar
forces. Replace this loading by an equivalent
resultant force and specify
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
resultant force and specify
where the resultant’s line of
action intersects the column
AB and boom BC.
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Solution
Force Summation
−
−=
Σ=→+
kNkNF
FF xRx
75.13
5.2
;
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
↓=−=
−
−=
Σ=→+
←=−=
−
−=
NkN
kNNF
FF
kNkN
kNkNF
Ry
yRy
Rx
60.260.2
6.054
5.2
;
25.325.3
75.153
5.2
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Solution
� For magnitude of resultant force
16.4
)60.2()25.3()()( 2222
=
+=+= RyRxR
kN
FFF
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� For direction of resultant force
o7.38
25.360.2
tantan
16.4
11
=
=
=
=
−−
Rx
Ry
F
F
kN
θ
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Solution
Moment Summation
Method 1
� Summation of moments about point A,
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� Summation of moments about point A,
my
mkNmkN
mkNmkn
kNykN
MM ARA
458.0
)6.1(54
50.2)2.2(53
50.2
)6.0(6.0)1(75.1
)0(60.2)(25.3
;
=
−
+
−=
+
Σ=
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Solution
� Principle of Transmissibility
xkNmkN
MM ARA
)(60.2)2.2(25.3
;
−
Σ=
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
mx
mkNmkN
mkNmkn
xkNmkN
177.2
)6.1(5
450.2)2.2(
5
350.2
)6.0(6.0)1(75.1
)(60.2)2.2(25.3
=
−
+
−=
−
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Solution
Method 2
� Take moments about point A,
;Σ= MM
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
49.160.225.3
)6.1(5
450.2)2.2(
5
350.2
)6.0(6.0)1(75.1
)(60.2)(25.3
;
=−
−
+
−=
−
Σ=
xy
mkNmkN
mkNmkn
xkNykN
MM ARA
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Solution
� To find points of intersection, let x = 0 then y = 0.458m
� Along BC, set y = 2.2m then x =
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
� Along BC, set y = 2.2m then x = 2.177m
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Example 4.18
The slab is subjected to four parallel forces.
Determine the magnitude and direction of the
resultant force equivalent to the given force
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
resultant force equivalent to the given force
system and locate its point of application on
the slab.
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Solution
Force Summation
−−+−=
Σ=↑+
NNNNF
FF
R
R
500400100600
;
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
↓=−=
−−+−=
NN
NNNNFR
14001400
500400100600
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Solution
Moment Summation
NmNmNNyN
MM xRx
)0(500)10(400)5(100)0(600)(1400
;
−=−
−−+=−
Σ=
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
mx
x
NNmNmNxN
MM
my
y
yRy
00.3
42001400
)0(500)0(400)6(100)8(600)(1400
;
50.2
35001400
=
=
++−=
Σ=
=
−=−
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Solution
� A force of FR = 1400N placed at point P (3.00m, 2.50m) on the slab is equivalent to the parallel force system acting on the slab
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
system acting on the slab
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Example 4.19
Three parallel bolting forces act
on the rim of the circular cover
plate. Determine the magnitude
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
plate. Determine the magnitude
and direction of a resultant
force equivalent to the given
force system and locate its
point of application, P on the
cover plate.
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Solution
Force Summation
FFR ;rrrr
rrΣ=
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Nk
kkkF
FF
R
R
}650{
150200300
;
r
rrrr
−=
−−−=
Σ=
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Solution
Moment Summation
)150()200()300(
;
−+−+−=
Σ=
kXrkXrkXrFXr
MM
CBAR
ORo
rrrrrrr
rrrrrrrr
rr
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
Equating i and j components,
85.84160650
85.84240650
85.8485.84160240650650
)150()45cos8.045sin8.0(
)20()8.0()300()8.0()650()(
)150()200()300(
−=−
−=
−−+=−
−+−+
−−+−=−+
−+−+−=
y
x
ijijjyjx
kXji
kXjkXikXjyix
kXrkXrkXrFXr CBAR
rrrrrr
rrr
rrrrrrr
oo
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Solution
� Solving, x = 0.239m and y = -0.116m
� Negative value of y indicates that the +ve direction is wrongly assumed
4.9 Further Reduction of a Force and Couple System4.9 Further Reduction of a Force and Couple System
wrongly assumed
� Using right hand rule,
)45cos8.0(150)8.0(200650
;
)45sin8.0(150)8.0(300650
;
mNmNx
MM
mNmNx
MM
xRx
yRy
o
o
−=−
Σ=
−=
Σ=