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Force AnalysisForce AnalysisPrinciples of Dynamics:Newton's Laws of motion :Newton s Laws of motion. :
1. A body will remain in a state of rest, or of uniform motion in a straight line unless it is acted by external forces to change its state.is acted by external forces to change its state.
2. The rate of change of momentum of a body acted upon by an external Force (or forces) is proportional to the resultant external force F and in the direction of that f ) p p f fforce:
( )d mvFd
=
Where m is the mass and v is the velocity of the body. For constant mass; m:F = m a
dt
Where a is the acceleration of the body.3. To every action of a force there is an equal and opposite reaction.
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Force is a vectorial quantityForce is a vectorial quantity
For several concurrent forces
Paralelogram law of addition
For several concurrent forces∑=
iiFF
F Fi= ∑ F Fi= ∑ F Fi= ∑Paralelogram law of addition F Fx ixi
∑ F Fy iyi
∑ F Fz izi
∑
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couplep
a) The moment vector does not have a point of application. Hence, it is a free vectorvector.b) The relative position vector, r, is in between any two points on the lines of action of the forces forming the couple.c) The force couple that creates a moment M is not unique, e.g. there are other force couples that may create the same momentforce couples that may create the same moment.
Moment of a force
M = r x F
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Resultant of nonconcurrent forcesF Fi= ∑ M r xFi i= ∑ ( )F Fi
i∑ i
∑
or
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Forces in Machine SystemsForces in Machine SystemsReaction Forces: are commonly called the joint forces in machine systemsReaction Forces: are commonly called the joint forces in machine systems since the action and reaction between the bodies involved will be through the contacting kinematic elements of the links that form a joint. The joint forces are along the direction for which the degree-of-freedom is restricted.a e a o g t e d ect o o c t e deg ee o eedo s est cted
Fij= - Fjj
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REACTION FORCES AT KINEMATIC PAIRS (JOINT FORCES)
AL
GR
EE O
F EE
DO
M
TATI
ON
AL
EED
OM
NAMEAN
SLA
TIO
NA
EED
OM
JOINT JOINT
5 3
DEG
FRE
RO
TFR
E
Sphere between parallel planes2
NAME
TRA
FRE JO N
SHAPEy
Fx
JOINT FORCE
Sphere in a cylinder
43
p p
1y
z
z
x
x
Fx
Fy
x
42 Cylinder between
parallel planes2x
z
y
x
F
My
F
3 Spherical pair (Ball joint)
0y
z
y
xx
Fz
yF
Fy
F
1 Pl j i t2
3 2 Slotted sphere in a cylinder1
My y
zx xFMz
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1 Plane joint2z
xMz Fx
NA
L
L
REACTION FORCES AT KINEMATIC PAIRS (JOINT FORCES)
DEG
REE
OF
FREE
DO
M
RA
NSL
ATI
OR
EED
OM
RO
TATI
ON
AL
REE
DO
M
NAME JOINT SHAPE
JOINT FORCED F T FRR F SHAPE FORCE
2 0 Slotted spherey
zx
y
Fz
yF M
xF
022
Torus
z x
z xFz
yxM
xF
Fy
211 Cylindrical joint
xz
y
F
xMx
FMy y
F
11 Slotted cylinder
F
z x
y
zF
y
MxxF
My
Fy
0
0
1
11
Revolute pair (turning joint)
Prismatic pair
z x
y
Fz Mx
Fx
yMy
y yFyM
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0 1 Prismatic pair (sliding joint) z
xMxzM
Fx
Forces in Machine SystemsForces in Machine Systemsa) Joint Forces
b) Physical Forcesc) Friction or Resisting Force:
a) Joint Forces
) g
d) Inertial Forces.
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Free Body DiagramFree-Body Diagram
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Static Equilibrium ∑ 0F ∑ = 0MStatic Equilibrium ∑ = 0F ∑ = 0M
In space, these two vector equations yield six scalar equations:
∑ ∑Fx =∑ 0 Fy =∑ 0 Fz =∑ 0
Mx =∑ 0 My =∑ 0 Mz =∑ 0x∑
In case of coplanar force systems, there are three scalar equations:
Fx =∑ 0 Fy =∑ 0 Mz =∑ 0
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Two Force MemberTwo Force Member
In a Two-Force member, the forces must be equal and opposite and must have the same line of action
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Two force and one moment member:
Two forces form a couple whose moment is equal in magnitude but in opposite sense to the applied moment
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Three Force MemberThree Force Member
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STATIC FORCE ANALYSIS OF MACHINERYSystems without Resisting ForceSystems without Resisting Force
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Link 2
F32y + Gl2y = 0 (∑Fy= 0)For link 4:
F34x + Gl4x - Fl4 cosη = 0 (∑Fx= 0)F32x + Gl2x = 0 (∑Fx= 0)
F32y a2 cosθl2 - F32x a2 sinθl2+ T12= 0 (∑MAo = 0)F34y + Gl4y - Fl4 sinη = 0 (∑Fy= 0)
F34x
a4
sinθ14
+ F34y
a4
cosθ14
+ F14
r4
(cosηsinθ14
-sinηcosθ14
)=0 (∑MBo
=0)
For link 3:
F23x + F43x = 0 (∑Fx= 0)
F23y + F43y =0 (∑Fy= 0) F32y = - F23y
Due to Third Law:
F23y F43y 0 (∑Fy 0)
F43x a3sinθl3+ F43y a3 cosθl3= 0 (∑MA = 0)
y y
F32x = - F23x
F43y = - F34y
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F43x = - F34x
F23=F ∠ θ1323 13
F43
= -F32
=-F34
= -G12
= -F23
- F14 cosη + F34 cosθ13 + G14 cosφ = 0 or -F14cosη + F34cosθ13 + G14x = 0
- F14 sinη + F34 sinθ13 + G14 sinφ = 0 or - F14 sinη + F34 sinθ13 + G14y =0
r F (cosη sinθ sinη cosθ )+a F (sinθ cosθ cosθ sinθ ) 0r4F14(cosη sinθ14 - sinη cosθ14)+a4F34(sinθ13 cosθ14 - cosθ13 sinθ14)=0
And
a2F32(cosθ13sinθ12 -sinθ13 cosθ12) + T12 = 0
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(F32 = - F34)
a) For two force members you don't have to write the equilibrium equations You
In general:
a) For two-force members you don't have to write the equilibrium equations. You can simply state that the forces are equal and opposite and their line of action coincides with the line joining the points of applications.
b) For two-force plus a moment members you must write the moment equilibrium equation only. The two forces are equal and opposite and they form a couple equal and opposite to the moment applied.
c) In case of three or four force members, the three equilibrium equations ( ΣFx = 0, ΣFy = 0, Σ M = 0) must be written.
F = Fv r = ruF = Fv r ru
M = rxF = ruxFv = rF(uxv)
u = 1 ∠ α = cos α i + sin α j
v = 1 ∠ β = cos β i + sin β j
u x v = (sin β cos α - cos β sin α) k
Or: u x v = sin (β - α) k
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M = rxF = r F sin (β - α) k
F + ( F14) (ΣM 0 )a4u4 x F34u3 + r4u4 x (-F14)v1 (ΣMBo = 0 )
a2u2 x (-F32)u3 + T12 k = 0 (ΣMAo = 0 )
Which result in:Which result in:
a4 F34 sin(θ13-θ14) - r4F14 sin( η−θ14) = 0
a F sin(θ θ ) + T =0
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-a2 F32 sin(θ13-θ12) + T12 =0
Graphical SolutionGraphical Solution
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Graphical SolutionGraphical Solution
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Modes of Contact in Prismatic Joints
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Principle of SuperpositionPrinciple of Superposition
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Systems with Resisting ForceSystems with Resisting Force1. Static Frictional Force :
R32= - μF32
μ, is known as the coefficient of static friction.2. Sliding Frictional Forceg
R32= - μF32,
μ is the coefficient of sliding friction which is less than the coefficient of static friction Slidingμ is the coefficient of sliding friction, which is less than the coefficient of static friction. Slidingfriction is also known as Coulomb friction.
– Experiments have shown that the static or sliding friction force does not depend on the area of contact.It depends on the types of materials in contact, on the surface quality in contact and the type of filmf d b t th t ti f Th ffi i t f t ti f i ti i li htl l th thformed between the contacting surfaces. The coefficient of static friction is slightly larger then thecoefficient of sliding friction.
3. Viscous Damping ForceR = cvR32 = -cv2/3
Where c is the coefficient of viscous friction and v3/2 is the relative velocity. Viscous friction assumes that there is a fluid film between the two surfaces in contact.
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friction angle : friction circle.
F R FR23 23 23= +
R32
r r r rf = ≅ =sin tanφ φ μ
tan φ = μ = M r FR F i ti T26
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F32
32
tan φ = μ = M r FfR
32 32= Friction Torque
θ12=550
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-F34cos(φ)-μG14-F14=0 ( ∑Fx=0)
F sin(φ)+G = 0 ( ∑F =0)-F34sin(φ)+G14 = 0 ( ∑Fy=0)
μr34F34-M14-μG14a=0 ( ∑MB=0)
-F34cos(φ)-0.1G14=100 (i)
-F34sin(φ)+G14 = 0 (ii)
5F34-M14-5G14 =0 (M in Nmm) (iii)
Unknowns are F34, φ , G14 and M14.
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F43= -F34 = -F23
(∑M =0):(∑MA=0):
-⎜AB⎜F43sin(φ−θ13)-μr23F23-μr34F43=0or:
-800F43sin(φ−θ13) - 25F23 -5 F43 =0
since F43 = F23 (in magnitude):
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-800F43sin(φ−θ13) -30 F43 =0
F F F FF32= -F23 = -F34 =F43
G12 = - F32
(∑M =0):(∑MAo=0):
⎜A0A⎜F23sin(φ−θ12)+μr23F23+ μr12G12+T12 = 0
which can be simplified as (noting F43 = F23)
200 F43sin(φ−θ12)+30 F43 +T12= 0 (v)
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When there are several forces acting on different links:
1. Principle of superposition cannot be used.
2 S f th ti t li f th k2. Some of the equations are not linear for the unknowns. Therefore, numerical iterative solutions are required.
In mechanical systems, if the designer has taken some good design measures, friction can usually be neglected in revolute joints with size small compare to link length dimensions (this usually simplifies thesmall compare to link length dimensions (this usually simplifies the solution)
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DYNAMIC FORCE ANALYSISDYNAMIC FORCE ANALYSISCenter of Mass G is commonly known as the center of gravity
mrr i
ii
G
∑=
Center of Mass, G, is commonly known as the center of gravity,
xG = Σ ximi /m
mrG yG = Σ yimi /m
Moment of inertia
∑ ∑=+=i i
i2ii
2i
2i0 mrm)yx(I
mI
k 020 =
∑ += 22 m)vu(I ∑ +=i
iiiG m)vu(I
( ) ( )[ ] ii
iGiGi
iii mvyuxmyxI ∑∑ +++=+= 22220 )(
∑ ∑∑ ∑ +++++=i i
iiGiiGi i
iGGiii mvymuxmyxmvuI 22)()( 22220
)rk(mmrII 2G
2G
2GG0 +=+=
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Parallel Axis Theorem
N t ' S d L f M ti f Ri id B dNewton's Second Law of Motion for a Rigid Body
∑ ==+ 2ii
2
iijii dt)rm(d
amFF ∑j
2iijii dt
Fji is commonly known as the internal force.
∑∑
∑∑∑ ==+j
2i
ii2
iiiji
iii dt
)rm(damFF
For all particles:
j
∑∑ = FFi
isum of all external forces acting.
∑∑ = 0F ii rmrm =∑∑∑ =i j
ji 0F Gi
ii rmrm =∑
Newton's Second Law of motion for Linear momentum
GG
2G
2
amdt
)vm(ddt
)rm(dF ===∑
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∑ =+j
iiijiiii maxrFrFxr ai = a0 + ai/0 = a0 + ai/0 n+ ai/0
tMoment equilibrium
ai/0 t= α x r ai/0
n= -ω2ri
( )[ ]∑∑ ∑∑∑ ω−α+==+i
ii2
i0ij i
iiijiiii
ii mrrxaxrmaxrFrFxr
∑ ∑= MFxr∑ ∑=i
0ii MFxr
∑∑ =i j
jii 0Fxri j
( )[ ] ∑∑∑∑ ω−α+=ω−α+i
iii2
ii
ii0i
iii
ii2
i0i mrxrrxxrmaxrmmrrxaxr
0G0i
ii axrmaxrm =∑∑∑ 2 α=α=α ∑∑ 0i
i
2ii
iii Imrrxxrm
0mrxri
iii2 =ω ∑
α+=∑ 00G0 IaxrmM∑Newton's Second Law of Motion for Angular Momentum
ωθ∑2 )I(dId
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α=ω
=θ
=∑ GG
2G
G Idt
)I(ddtId
M
D'Alambert's PrincipleD Alambert s Principle0F∑ 0amF G =−∑
0IM GG =α−∑
Gi amF −= A nonexistant (fictituous) force inertia force
α−= Gi IT A nonexistant (fictituous) torque inertia torque
Considering Inertia force and torque as if an external force or torque
0F =∑ 0MG =∑
Considering Inertia force and torque as if an external force or torque
D'Alambert's PrincipleIn a body moving with a known angular acceleration and a linear acceleration
of the center of gravity, the vector sum of all the external forces and inertia forces
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and the vector sum of all the external moments and inertia torque are both separately equal to zero.
ExampleExampleiyeax i
3 =+ θ
θ
−=θ
sinayaxcos
3θ
=ωsinax
313
θθ= sinay 3 θω= cosay 133
θω−=α 13 cosx
θ=α 2
313 sina
θα+θω−= cosasinay 1332133
θ+= i33G ea
21xr 3G
i3 amF −=
θω+= i1333G eia
21xv
)i(1 2iθ
=α−= 133Gi
3G IT
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)i(ea21xa 2
1313i
33G ω−α+= θ
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For Link 3F43 - F23x = 0R3
i - F23y = 0
500F43sin(180-113.578) + 250(25.97)sin(90-113.578)-865 = 0
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