static force analysis

Theory of Machines

LECTURE NOTES- MECE 303 Theory of

Machines

9- Static Force Analysis

Fall Semester 2010/2011

Halil Orhan YILDIRAN, MS

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Static Force Analysis

.

-A mechanism is a device that transmits/ transform motion and in doing so transmits power by means of forces and torques in its elements. -In designing a mechanism it is necessary to know the manner in which forces (force or torque) are transmitted from input to output so that components of the mechanism can be properly sized to withstand the stresses developed.

FORCE ANALYSIS

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FORCE ANALYSIS

DYNAMIC FORCE ANALYSIS (Inertia forces and torques due to acceleration are taken into consideration)

STATIC FORCE ANALYSIS


Consider a planar (2D) mechanism:

-Given input motion (displacements, velocities and accelerations)

-Find:

*Motion of the rest of the mechanism

* The forces necessary to generate the motion of the mechanism ,

(# independent forces required to generate the motion of the (# independent forces required to generate the motion of the

mechanism is equal to the dof of that mechanism)

* Joint (reaction) forces

ASSUMPTIONS

-Rigid body assumptions of the components are used

-Gravity may or may not be considered

-Frictional forces at the joints will be assumed to be negligible3


.Force is a vector quantity ( has magnitude and direction). Also it has a line of action (LA) and point of application.

θ

F

4

jFiFF yxˆˆ += : Cartesian

)sinˆcosˆ( θθ jiFF += , FF =

θiFeF = complex

θ∠= FF polar

θ

i

j


.

Force: action of one body on the other

F : force on body j due to action of body i

5

ijF : force on body j due to action of body i

jiF : force on body i due to action of body j

Newton’s third law: ijF =- jiF


In order to calculate forces and torques in a mechanism, it is

necessary to isolate each element of the mechanism as a free

body (which is called Free Body Diagram -FBD) and then apply

equilibrium conditions.

'

1iF

'

1iF

6

FBD of link i

kiF

1iF

"

1iF

jiM

jiF

"

1iF


Same force with different LA is not equivalent as below (because

rotation effect is different)

F

F

7

LA LA


But in a rigid body mechanics the point of application is not

important if our concern is only the equilibrium of the body.

Hence we can slide a force on its LA.

F LA

8

F

LA F

LA


.*Moment of a force about a point.

F

9

FxrFxrM A '==

FhFrFrMhh

A === 321321 'sin'sin αα

(Sense is obtained by right hand rule)

r'

r h

α'

α


.*Moment of a couple (two equal, opposite and non collinear force)

F

CBA MMFxrMrr

===

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Moment vector does not have point of application. It is a free vector, because a rigid body rotates as a whole. Hence the place of a moment on a rigid body is not important.

F−

r

CBA MMFxrM ===


Forces (Moments) in Mechanisms and Machine systems

*Applied forces: Input (Driving) forces , weights, spring forces, etc.

*Joint forces: (Reaction forces and Constraint forces): action and

reactions at the joints, friction forces (resistance to motion)

JOINT FORCES.

Consider frictionless joints.

At a joint, the contact of kinematic elements are such that some

of the possible motions (3 in planar, 6 in spatial) are prevented.

At the directions for which the motion is prevented, there are

forces developed due to contacts.

(See Table 5.2 in the Text Book)

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.Revolute joint:

*λ=3 (3 possible motion: 2 translation, 1 rotation) *2 translation is prevented ⇒ there are joint forces in those directions *Since rotation is not prevented⇒ no joint moment

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FBD

jiF

ijF

θ

θ+π

OR x

jiF

x

ijF

y

jiF

y

ijFi j


.Prismatic joint

*Translation along y axis and rotation are prevented ⇒There are joint forces in y direction and joint moment . *Translation along x axis is allowed⇒ no force in x direction

*LA of force jiF is arbitrary (eg. Thru

midpoint)

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FBD

midpoint)

ijF

jiFjiM

ijM


.

Cam joint:

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Only 1 translation along n is prevented ⇒ 1 force


.

Cylinder in a slot

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Only 1 translation along n is prevented ⇒ 1 force


.STATIC FORCE ANALYSIS In machinery or mechanisms all members have mass. If these members are accelerating or decelerating, inertial forces will be associated with this motion. If magnitude of these forces are small relative to the externally applied loads, they can be neglected in the force analysis. Such an analysis is referred to as STATIC FORCE ANALYSIS.

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∑ = GamF . ,

∑ = α.GIM If the RHS are small ⇒ can be neglected

G denotes center of mass, a denotes acceleration, I denotes mass moment of inertia In static analysis inertia forces and torgues are neglected to yield

∴∑ = 0F , ∑ = 0M static equilibrium equations


.

For the equilibrium of a body we can solve for 3 unknowns using 3 equations

∑ = 0xF , ∑ = 0xF , ∑ = 0AMr

∑ ∑ ∑

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∑ = 0yF ,or ∑ = 0yF , or ∑ = 0BM

∑ = 0AMr

∑ = 0BM ∑ = 0CM


.These equilibrium equations are simplified in some special cases

Two Force member (2F)

AF

BF

FF B

r=

18

∑ = 0F ⇒ BA FF −= : forces must be equal and opposite

∑ = 0M ⇒ AF and BF must be collinear

FF A

r−=

⇒


.Eg: a binary link with R joints, no other force on it. Note: A two force member is either in tension or compression

kjF

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The line of action (LA) is drawn connecting the point of application of forces, forces are shown as equal and opposite.

ijF


.

Two Force and a Moment Member (2F+M)

FF A

r=

Tr

α

β

r

20

∑ = 0F ⇒ BA FF −= : force must be equal and opposite

∑ = 0M ⇒ gives the relation of moment M (T) to the forces.

∴ once the forces are shown equal and opposite in //LA’s, only the moment equation is needed.

∑ = 0M : 0ˆ. =+ FxrkT

0)sinˆcosˆ()sinˆcosˆ(ˆ. =+++ ββαα jixFjirkT

0)sin( =−+ αβFrT

FF B

r−=


.*Three Force Member (3F)

A rigid body under the action of three forces is in static equilibrium only if the forces intersect at one point and the RESULTANT OF THE 3 FORCES IS ZERO.

3 forces AF , BF and CF acting on the body

Let LA of the two forces intersect at O. As ∑ = 0M ⇒LA of the third force must also pass

thru O. ie. 3 forces must be CONCURRENT ∴ If the forces are shown concurrent in the FBD, moment equation is not necessary.

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FA

FB

FC


.Ex. Slider crank Mechanism

θ

θ13

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Given force P, find T12 as a function of position variables Assume inertial forces are small compared to other forces and gravity and friction forces are negligible.

θ12


.

FBD’s

θ13

θ13+π

θ13 θ12

23

θ13+π

θ12


.∑ = 0xF ⇒ -P+F34 cos(θ13+π)=0 ⇒

)cos( 13

34πθ +

=P

F

∑ = 0yF ⇒R14+F34sin(θ13+π)=0 ⇒ R14= -F34sin (θ13+ π )=)cos( 13 πθ +

−P

sin(θ13+ π )

= -Ptan (θ13+π)

Link 3 (Link 3 is two force member)

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Link 3 (Link 3 is two force member) F43= F23 (in magnitude) Link 2 (Link 2 is two force and a moment member) F23= F32=R12(in magnitude)

∑ = 0oM

F32 r2sin(θ13- θ12)+T12=0 ⇒ T12=- F32 r2sin(θ13- θ12)


.

Ex: Four Bar Mechanism

Given P14 (external applied load) and α. Find T12 as a function of position variables. (B0C=b4) Link length; A0A=r2, AB=r3, B0B=r4

π+α

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θ12

θ13

θ14

α


.

FBD’s

θ13 α

π+α π+θ13 θ14

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θ12

θ13

θ14

α


.

Link 4 (3F)

∑ = 0xF ⇒R14x+F34cos(θ13+π)+P14 cos(α +π)=0⇒R14x =F34cos θ13- P14 cos α =0 (A)

∑ = 0yF ⇒R14y+F34sin(θ13+π)+P14 sin(α +π)=0⇒R14y =F34sin θ13- P14 sin α =0 (B)

∑ = 0oBM ⇒F34r4sin(θ13+π-θ14)+ P14b4 sin(α+ π-θ14)=0

⇒ -F34r4sin(θ13-θ14)- P14b4 sin(α-θ14)=0 (C)

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Using (C) F 34 can be found. Then from (A) R14x and from (B) R14y can be found

If we want to find R14 and α

2

14

2

1414 yx RRR += ,

=

x

y

R

Ra

14

14tanα


.

Link 3 (2F) F43= F34=F23

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Link 2 (2F+M) F32= R12

∑ = 0AoM ⇒T12+F32r2sin(θ13-θ12)=0 ⇒ T12 can be found


.Ex: Quick Return Mechanism

θ15

29

θ12

θ14


.

(a)Using static equilibrium find joint forces and input torque T12 to drive the slider (link 6) against load P for a given crank angle. (b) Find forces on the links and input torque T12 for the given conditions below

jP ˆ800= N

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jP ˆ800= N

A0A=r2=22.5 cm B0B=r4=100 cm BC=r5=40 cm r1=60 cm, b1=110 cm θ12=120°


.Solution: (a) FBD’s

θ15

31

θ12

θ14

θ14


.

Link 6

∑ = 0yF ⇒ F56 sin(θ15+3π/2)+P=0 ⇒ )2/3sin( 15

56πθ +

−=P

F

∑ = 0xF ⇒F56cos(θ15+3π/2)+F16=0 ⇒ F16= P cot (θ15+3π/2)

F56= F65= F45= F54

Link 4

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Link 4

∑ = 0BoM ⇒ -F34 s43+F54 r4 sin (θ15+π/2- θ14) ⇒43

141545434

)2/sin(

s

rFF

θπθ −+=

∑ = 0yF ⇒R14y+F34 sin(θ14-π/2)+F54sin( θ15+ π/2)=0

⇒ R14y=-F34 sin(θ14-π/2)-F54sin( θ15+ π/2)=0

∑ = 0xF ⇒F34cos(θ14-π/2)+F54cos( θ15+ π/2) - R14x=0

⇒ R14x= F34 cos(θ14-π/2)+ F54cos( θ15+ π/2)


.

To find polar form

2

14

2

1414 yx RRR += ,

=

x

y

R

Ra

14

14tanα

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F43= F34= F23= F32 Link 2

∑ = 0oAM ⇒T12+F32r2sin(θ14+ π/2 -θ12)=0 ⇒ T12 is found


.

(b)Numerical values From position analysis when θ12=120° θ14=164.70°, θ15=19.79°, s43=73.87 cm Substitute the numerical values in the equations above we get: F56=850.21 N

F16= -287.84 N

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F16= -287.84 N

F65=F56=F45=F54=850.21 N

F34=-941.9 N

R14x= - 536.3 N, R14y=108.5 N , R14=547.2 N

F43= F34= F23= F32= - 941.9 N T12=15062.2 N.cm (CCW) ANS


.FBD with correct force directions

θ15

35

θ12

θ14

θ14


.Principle of superposition

For linear systems the individual responses to several inputs can be superposed on each other to obtain total system response. Ex: P and Q are given. Find T12

36


.We consider each force separately and then add them

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Then

"

12

'

1212 TTT +=

"'

ijijij FFF +=

Static Force Analysis-Resistive Forces

.1-Static Frictional Force

This is the force that is necessary to slide one body relative to the other. This force is along the contact surface and it is related by the normal force “N” acting at the contacting surfaces.

f = - µN where µ= the coefficient of static friction.

2-Sliding Frictional Forces

This is the force necessary to keep the relative motion of two sliding bodies continuously. It is also given by the equation:

f = - µN where

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where µ=coefficient of sliding friction, which is less than the coefficient of static friction. Sliding friction is also known as Coulomb friction.

3-Viscous Damping Force

This is the force that is proportional to the relative velocity of the two contacting surfaces. f = - cv where c=coefficient of viscous friction v=relative velocity. Viscous friction assumes that there is a fluid film between the two surfaces in contact.

Static Force Analysis-Resistive Forces

. Prismatic Joints

In prismatic joints, the reaction force between two links in contact will not be perpendicular to the slider axis if resisting force is taken into account. The resultant reaction force will be inclined. The angle Φ which is known as friction angle.

Revolute Joints

In revolute joints, the line of

ji

ij

F

f== µφtan

Ø

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In revolute joints, the line of action of the reaction force will not pass through the center of the joint but will be tangent to a circle, which is known as the friction circle. The radius, rf, of this circle is : rf = rij sinΦ , where “rij” is the radius of revolute joint. For small angles sinΦ≈tanΦ. Therefore: rf = rij sinΦ ≈ rij tanΦ≈µrij

ωij

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