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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