Chapter 1 Static Force Analysis

When the inertia forces are neglected in comparison to the externally applied load, one may go for static

force analysis. If the body is under equilibrium condition, then this equilibrium is known as static

equilibrium and this condition is applicable in many machines where the movement is relatively slow.

These include clamps, latches, support linkages, and many hand operated tools, such as pliers and cutters.

In case of lifting cranes also, the bucket load and the static weight loads may be quite high relative to any

dynamic loads due to accelerating masses and hence one may go for static force analysis.

When the inertia effect due to the mass of the components is also considered, it is called dynamic force

analysis.

Applied and Constraint forces:

When two or more bodies are connected together to form a group or system, the pair of action

and reaction forces between any two of the connecting bodies is called constrained forces.

These forces constrain the connected bodies to behave in a specific manner defined by the nature

of the connection.

Forces acting on this system of bodies from outside the system are called applied forces.

Electric, Magnetic and gravitational forces are example of forces that may be applied without actual

physical contact. But most of the forces we are concerned in mechanical equipment occur through direct

physical or mechanical contact.

External force

Constraint forces

T

F4

Figure 1: Four bar mechanism showing external and constraint forces

Constraint forces of action and reaction at a mechanical contact occur in pairs and thus have no net force

effect on the system of bodies being considered.

When a part of the body is considered in isolation the effect of such force is considered by using the

freebody diagram.

Characteristics of a force are its magnitude, its direction and its point of application

1

Two equal and opposite forces along two parallel but noncollinear straight lines in a body cannot

be combined to constitute a single force and they constitute a couple. The arm of the couple is the

perpendicular distance between their lines of action and the plane of the couple is the plane

containing the two lines of action.

The moment of the couple M is a vector directed normal to the plane of the couple and the sense

of M is in accordance to the right-hand rule for rotation.

The moment of couple

= BAM R F

The value of M is independent of the choice of the reference point about which the moments

are taken, because the vector is the same for all positions of the origin. BAR

As the moment vector M is independent of any particular origin or line of application, hence it is a free vector.

Figure 2

B

A

F

F

Free-body diagram A free body diagram is a sketch or drawing of the body, isolated from the rest of the machine and its

surroundings, upon which the forces and moments are shown in action. In case of the four bar mechanism

shown in figure 1 the free body diagram of link 3 is as shown below.

43F 23F C

B

Free body diagram of link 3

2

When a link or body is subjected to only two forces it is called a two-force member and when it is subjected to 3 forces it is called a three-force member. Similarly one may consider multi-force member also.

Static equilibrium: A body is in static equilibrium if

the vector sum of the forces acting on the body is zero i.e., = 0F

the vector sum of all the moments about any arbitrary point is zero i.e., = 0M

Hence a two force member as shown in figure 3(a) will be in equilibrium if (i) both forces are equal and opposite and (b) their line of action coincide. If the forces are equal and opposite but not collinear as shown in Figure 3(b) they will form a couple and body will start to rotate. Hence these two forces should be equal, opposite and collinear.

1F2F

1F1F

2F

(c) (b) (a)

2F

Figure 3. Equilibrium of a two force member

Similarly a three force member will be in equilibrium if the vector sum of all these forces equal to zero

and to satisfy the vector sum of all the moments about any arbitrary point equal to zero, their line of

action should meet at a point.

1F2F

3F (b)

O

(a) 3F

2F 1F

O

2F

1F

3F

(c)

Figure 4: Equilibrium of three-force member

3

Figure 4(a) shows a body subjected to three forces Also the line of action of coincide at point O. Hence the resultant of must pass through point O and it should

be equal and opposite to force . Hence for equilibrium, line of action of should pass through point

O as shown in Figure 4(b). In figure 4(c) the forces are shown to form a close polygon (triangle) and one may use Lamis theorem (sine rule of tringle) to find the unknown forces if atleast one force is known both in magnitude and direction and the line of action of one more force is known. According to this

theorem

1 2 3, and .F F F

1 and F 2F 2F1 and F

3F 3F

31 2

sin sin sinFF F

= =

where , and are angle as shown in figure 4(c).

For more than three forces one may draw force vector polygon or resolve the forces and moments to get

the required force components.

To find the constraint forces in a mechanism one may either go for analytical or graphical method of

solution if the maximum number of forces in a member is limited to three and if the system has more than

three force members one should go for analytical methods.

Example 1: Find the bearing forces and the torque required for static equilibrium of the four bar

mechanism shown in fig 1.

Solution:

Analytical: For Planar mechanism , 0= XF 0= YF and 0= ZM .

Step1: Let us first write all the quantities in vector form

jABiABRAB sincos 22 +=K

23F 43F C

BjBciBCRCB sincos 33 +=

K

jCDiCDRDC sincos 44 +=K

Free body diagram of link 3

jDQiDQRQD sincos 44 +=K

Similarly

jPiPP sincos +=G

4

Here link 3 is a two-force member and at this stage we know only the line of action of the forces

which should be along the line BC. 23 43and F F

34F p

D

14FG

C

Free body diagram of Link 4

Link 4 is a three-force member in whih force P is completely known and the line of action of force

which is equal and opposite to is known. Only the point of application of force , which is at point

D, is known. As link 4 is a three force member, taking moment about D,

34F

43F 14F

0= ZM 034 =+ PRFR QDCDGGGG

As P is completely known one may obtain 34FG

One may note that link 3 is a two-force member, so 344323 FFFGGG

==

Link 2 which is acted upon by two forces i.e., 12FG

and 32FG

, and the external applied torque, will be in

equilibrium only if = - 12FG

32FG

, i.e., these forces are equal and opposite and the resulting moment of the

couple is equal to the applied torque.

Also one may find the torque by taking moment about point A.

Graphical method

As link 4 is a three force member, the line of action of 14FG

should pass through the intersection of the

line of action of P and 34FG

.

Taking proper scale and by drawing the force polygon one may obtain the magnitude of 34FG

and 14FG

.

Then considering equilibrium of link 3, force 23FG

can be determined.

Then determine the torque taking moment about A.

5

When multiple forces act on a mechanism, one may use superposition theory, which states that in a

linear system, the net effect (e.g., bearing forces or torque) due to all the forces taken simultaneously will be equal to the summation of the effects due to individual forces taken one at a time.

If one wishes to find only the torque acting on the mechanism, the method of virtual work may be used. It

states the work performed during a virtual displacement from equilibrium is equal to zero. The

virtual displacement is defined as an imaginary infinitesimal displacement of the system that is consistent

with the constraints on the system. For example, the constraints on the slider-crank mechanism are that all

members including the frame are rigid and all joints maintain contact

Example 2. Calculate the torque required (assuming no friction in the bearing) for static equilibrium of an in-line reciprocating engine in the position when crank angle = 45 deg (from inner dead center). The dimensions are crank length r =30 mm, connecting rod length L = 70mm, and the piston force is P = 40

N.

B

A

O

Lr

P

XSolution Here OB is link 1, crank OA is the 2nd link, connecting rod AB is the 3rd link and the piston is the 4th link.

Crank radius r =30 mm, Length of connecting rod =70 mm

Letting

It may be observed that link 3 is a two force member and subjected to forces 23FThe free-body diagram of link 4, i.e., that of piston is shown below. For the present case, it is a three-

force member subjected to a force due to gas pressure, vertical reaction force and force of connecting rod on piston ( ) at the gudgeon pin. Force is known completely both in magnitude and

direction and the line of action and point of application of force is known. Now drawing the force

polygon as shown in Figure (b) one will be able to find the unknown forces and .

P 14F

34F P

34F

14F 34F

P

14F

34F

14F 34F

P

Figure 5 (a) Free-body diagram of link 4 (b) force polygon

Now one may use a vector method or use simple algebraic calculation using Lamis theorem to find the

forces. Also one may use graphical method to fid the same. All these methods are described briefly below

Vector method

As , 14 340, hence, 0F P F F= + + =

=34 14 So, (0.953 0.303 ) 0i j F Pi F j +

Equating the ith and jth compoment of the forces one may obtain

34

14 34

40 41.973N0.9530.304 12.72N

F

F F

= =

= =

Hence and 34 41.973 342.35 NF = 14 12.72 90 NF = .

Using Lamis formula from the force diagram shown in Figure (b) 34 14

sin 90 sin sin(90 )F F P

= =

7

Hence

3440 41.974 N

sin(90 17.64)F = = and

8

1440sin(17.64) 12.72 N

sin(90 17.64)F = = .

23 43F F=

23F

B

A

43F

Now considering free-body diagram of link 3

But, 43 34 41.974 342.35F F= =

So 23 43 41.974 342.35F F= =

Considering equilibrium of link 2

Link 2 is subjected to forces . For equilibrium these two forces must be equal and opposite. But as they are acting at A and O respectively they will form a couple which will try to rotate the link OA

in anti-clock wise direction. Hence for static equilibrium a torque T must be applied in clockwise

direction whose magnitude should be equal to the couple formed by these forces.

32 12 and FF

32F

12F

T

Now 32 23 41.974 342.35 40 12.7265F F i= = = + jj32 ( ) (21.213 21.213 ) ( 40 12.73 )

= -1118.56

AOT R F i j ik

= = + +

Negative sign indicate the applied torque should be applied in clock-wise direction.

Static force analysis with friction As we are considering only simple mechanisms with prismatic and revolute joints, the effect due to dry or Coulomb friction and greasy friction at the journals are discussed. Consider a pair of sliding surfaces as shown in figure **. When a force is applied on the block to move it towards right, a friction force is

generated which oppose this motion. According to Coulombs law, the magnitude of this force for

impending motion is

F

R , where R is the reaction force due to weight . WW

F

R

R

Figure 6

Greasy friction at a journal

Generally greasy or boundary lubrication type friction force occurs in heavily loaded, slow running

bearings. Figure 7 (a) shows a journal inside a bearing during static condition. Here A is the contact point and the weight of the journal W

and the reaction force R act in the vertical directions as shown in the

figure. Now let us consider a torque T is applied to the journal in the clockwise direction. The friction force will now oppose this motion and so the contact point between the bearing and the journal shifts to point B as shown in figure (b). The resultant ( R ) of the normal reaction force ( nR ) and the friction force

( nR ) at B should be equal and opposite to the weight as the journal is under static equilibrium condition. These two forces will form a couple in anticlockwise direction, which will oppose the applied

torque.

Let OC be the perpendicular distance between . If one draw a circle with radius OC and center

at O, the reaction force will be tangent to that circle. This circle is known as friction circle. Now to find

the radius of the friction circle, consider the triangle OBC. Here OC = OB sin

and W R

where is the angle

between the resultant and normal reaction force. Also the coefficient of friction tan = . Hence radius

of the friction circle = 2/( (1 )).fr r = + where r is the radius of the journal. For small value of ,

.fr r= Friction couple = 21f

Wrr W Wr

=

+ .

A

W

T

C r

nR R

B

R

W

n(a)

O O

Lubricant

Journal

Bearing

R

(b)

Figure 7 (a) Journal in static condition (b) Journal when a torque is applied to start the motion.

9

Example 3: Determine the torque required at the crank and also bearing forces in a slider-crank mechanism when the inertia forces are neglected. Also develop a matlab code for the same.

User Specified Parameters

Crank Length = R1Connecting Rod Length = L1Radius of Journal = R

Coefficient of friction =

Piston Force = P

Angle of the crank =

Calculations:

Radius of friction circle = R2 = 2/ (1 )R + To find angle that connecting rod makes with horizontal 3

L1/ sin=R1/sin(180-3) Performing Force analysis on the connecting rod :

10

Free body diagram of crank:

Free Body diagram of slider:

Force of reaction by connecting rod on slider = F (Combination of forces F5 & F6) Reaction force because of friction between ground and slider = Rxn (combination of Normal reaction NN and friction f) The following figures show the forces acting on the connecting rod, a combination of forces F5 and F6

11

Out of these only the figure 1 shows the correct direction of forces. This can be understood by consulting

the initial figure. The tendency of the friction in the bearing is to increase 3 .the tendency of the friction

in bearing connecting crank and connecting rod would be to increase 2.thus the force direction of the

link reaction force can be found out. The angle made by reaction force with connecting rod can be

found out using friction circle radius.

tan = R2/(L1/2) Angle made by reaction force with horizontal = 3-

Force on bearing connecting crank and connecting rod F (Vector) From the code we find the magnitude of the link force and its direction.

Torque due to this force = R1 x F (Vector Product) Matlab code:

R1=0;

while (R1

end

end;

L1=R1;

while (L1=(R1/10)) fprintf ('not acceptable value, enter agn') else break

end

end;

C=1

while (C>=1) C=input ('enter the coefficient of friction='); if (C>=1) fprintf ('not acceptable value, enter agn') else break

end

end;

P=input ('enter the piston force='); Tht= input ('Enter the value of angle considered(degrees)='); %R2=radius of journal bearing Tht = Tht*pi/180;

R2= (C*R)/ sqrt (1+C^2) A= atan ( (2*R2)/L1) Tht3= asin ((R1/L1)*sin(Tht)) %Angle made with horizontal B

fprintf('the angle made by rxn force with horizontal') B= 180*(Tht3-A)/pi fprintf('the value of link rxn force=')

13

F=P/(cos(B)+sin(B)*C) Rxn = (F*sin (Tht))/cos (atan(C)); fprintf('the value of torque=') T= F*R1*sin (Tht + B)

Example 4. Calculate the torque required (assuming no friction in the bearing) for the static equilibrium of an in-line slider crank mechanism in the position when crank angle (from the inner dead center). The dimensions are, Crank length =30 cm, Connecting rod length=70 cm and the piston force = 40N. Also find the torque required assuming that the co-efficient for all bearing is 0.1. The three journal bearings all have radii of 10 mm, and the crank is rotating in the clockwise.

045=

Solutions: Given data:--

045= Crank length (link OA) =30 cm. Connecting rod length (link AB)=70 cm. And the piston force (P)= 40N.

045=

From the figure, and using the sine rule. We can write,

0

0

sin sin30

sin sin 45 0.30370

17.64

AB OA

=

= =

=

Case (I):----

14

Without friction Considering the link4, and using static force analysis,

34

34 14

34 14

cos 40& sin

,

41.97 ,& 12.12

F PF F

ThereforeF N F N

= =

=

= =

Also, 34 43F F= (equal and opposite reaction) 43 4341.97F N= = F

Considering the link 3.

Since link 3 is a 2-force member,

23 43

23

,

41.97

ThereforeF F

F N

=

=

And also, we can write,(By equal and opposite reaction)

15

32 23

32 41.97F F

F N=

=

Considering link 2.

Now the torque due to reaction force is given by, 32

0 0 0.03(cos 45 sin 45 ) 41.97( cos sin )

1.112

T r F

T i j iT kN m

j =

= + +

=

Case (II):-- With friction,

Consediring the link 4.

Radius of friction circle, is given by21

fr

r

=

+

Where, r= 10, and =0.1

2

0.1 10 0.991 (0.1)f

r

=

+

the angle is given the angle by which the reaction force hift and is determined by =tan-1=5.71 Now, Since the rotation of the crank is clockwise direction, thus the angle r will decreased and simultaneously, angle , angle will increased and deceased. Also the piston (link 4)

16

Will move toward the X-axis (to the right). Thus the direction will be towards left & thus =+5.71(according to the figure) Also the force F34 can act in two ways shown in below.

Since the angle is the decreasing and friction will tends to oppose it. Thus force F34 will act in the orientation (i).

Considering the link (3) Since the link 3 is a 2-force member, then this link can be showed the four possible force situations.

Same way as above, the forces F43 and F23 will act in the orientation (iii). Similarly, we can write for the link 2.

A

D

// B

Now we have to find

17

1 1

0

0.9917.64 tan ( ) 17.64 tan ( ) 1635

16

fr

DB0

=

= = = =

=

Now considering link .4. 34 14

34 14

34

14

cos sin&

sin cos,

42.8411.87

F P F

F FThereforeF NF N

= +

=

=

=

34 43 23 32 42.84F F F F= = = = N as in case (I) considering no friction. Considering link.2.

Torque 32

0 0 0 0.03(cos 45 sin 45 ) 42.84 ( cos16 sin16 )

1.124

T r F

T i j iT kN m

=

= + +

=

0

Example 5: Determine the driving torque available on the crank of a slider-crank mechanism, if a force of 2000 N pointing towards the main bearing is applied horizontally to the piston. Length of the crank and the connecting rod are 10 and 30 cm respectively. At the instant considered the crank has rotated 60 degree (CCW) from the inner dead center. Take coefficient of friction between all the pairing surfaces as 0.13. The diameter of the main bearing, crank pin and piston pin are respectively 10, 6 and 6 cm. Also find the driving torque in the absence of friction using virtual work principle.

Solution: Considering the friction in all turning and sliding joints:

18

l=300 mm

r=10

0 mm

A

B2000 N

Impending motion

32

42O

Figure 1

Now angle of friction in joints, = tan-1() = tan-1(0.13) = 7.407 o. Radius of friction circle, sinhr f =

For the crank pin and gudgeon pin, rf = 60 sin = 7.735 mm

For the main bearing of diameter 100 mm, rf = 12.89 mm.

figure 2

Figure 3

mmr f 735.7= 34Fo72.19

dN2000

23F43F

mml 300=

13F

Consider the link 2 (figure 2), 300/(sin) = 100/(sin)

= 16.778 o. Now, from the figure, d = tan-1(7.735/150) = 2.95 o

Now the Free body diagram of the piston is shown in figure 3, The force equations are (F14 can be divided into their frictional and normal components),

0.13 N +F34 cos19.72o = 2000 where N is the normal force acting and F34 sin19.72o = N F34 =2030 N

19

We know, = 180-(+) = 103.23o. (103.23+2.95) From the figure 4,

Now the resisting couple, equal to F32 d having a clockwise direction. Torque = F32 d = 2030 0.10355 = 210.21 Nm.

103.55 mm

3

100 mm

12.89mm

Figure 4

32F

12F

In the absence of friction using virtual work principle:

od 23=

o60F

According to principle of virtual work, F dx = T d T = (F dx) / d From the figure, dx= 42.93 mm and d = 23o = 23/180 rad. Now, T = (20000.042) / (23/180) = 209.254 Nm. (answer)

20

Example 6: Determine the required input torque T1 for the static equilibrium of the four bar mechanism shown in the figure. Forces F2 and F3 have magnitudes of 50 N and 75 N, respectively. Forces F2 acts in the horizontal direction. Use both graphical and analytical methods. AB=30 cm, BC=40 cm, CD=50 cm and the fixed link AD=75 cm and CE=15 and CF= 20cm.

D

C E

A

B F

450

450

Solution: Given data: -- AB=30 cm, F2=50 N, F3=75 N, BC=40 cm, CD=50 cm, AD=75 cm, CE=15 and CF= 20cm

Let link AB is rotating in counterclockwise with rad/s Then, Vb=30 cm/s. One may obtain the velocity of different points on the mechanism by using , graphical method, vector method, complex number methods etc. Here,

0

0

20

3

13.33 284.28 26

5075 45

E

c

VVF iF

=

=

=

=

Resolving the velocities at E and F, parallel and perpendicular to the applied forces at these positions respectively.

1 11.67 ,eV = Parallel to the F2 1 3.81 ,fV = Parallel to the F3.

Assuming T to be counter clockwise and applying principle of virtual work 3 3

3 3

3.81 3.81 03.81 3.81

297.75 297.75Nm

T F FT F FT Nm

+ =

=

= =

Alternatively, one may use dot product to find the virtual work done as follows.

2 3. .

( 50 ).(13.33 cos 28 13.33 sin 28 )

(75 4.28 )(cos 45 sin 45 ).(cos 26 sin 26 )

E cT F V F V

i i ji j i j

= +

= +

+ + +

21

588.484 303.51284.97Nm

TT =

=

Analytical Method:---

0 0 023 43 2

cos 20 sin 20 cos 64 sin 64 0F i j F i j F i + + =

F

Equating the i component of the equations, we have, 0 0

23 43 2cos 20 cos 64F F+ = Equating the j part of the equations:-

0 023 43sin 20 sin 64F F=

0 0 023

23

cos 20 sin 20 cot 64 5045.187

F

F

+ =

=

1 1

2 2r i = +

j

23aT r F=

( ) 0 00 0

30 45.187( cos 20 sin 20 )

230 30( 45.187sin 20 ) 45.187cos 20

2 2572.9055

i j i j= +

= +

=

0 0 0 034 3 14

cos33 sin 33 cos 45 sin 45 cos58.5 sin 58.5 0F i j F i j F i j + + + =

0 0334 14

0 0334 14

0 0 034 14

14 34

cos33 cos58.52

sin 33 sin 58.52

(cos33 sin 33 ) (sin 58.5 cos58.5 ),

0.8906

FF F

FF F

F FThereforeF F

=

=

=

=

0

0 034

34 32

,

75cos33 0.8906cos58.5

240.6689 40

Therefore

F

F N F

+ =

= =

We know that,

22

0 032

0 0

30 ( )

2,

30 ( ) 40.7 cos33 sin 33

230 40.7 sin 33 cos33 253.86

2

b

b

r i jTherefore

T r F i j

T N

= +

= = + +

= =

By superposition principle, ( ) 572.9055 253.68 319.04a bTotal T T T= + = = , Ans.

Gear force Analysis

The fundamental law of gearing states that in order to obtain a constant velocity ratio, the common normal to the tooth profile at the point of contact should always pass through a fixed point, called the pitch point. Thus the point of contact of the two gears has the same velocity. Applying Newtons third law, the force exerted by one gear to the other at the point of contact is same in magnitude but opposite in direction. In this section the forces in spur and helical gears are discussed. Spur Gear In figure 8(a) shows the pitch circles of a pair of spur gears with center at a and b and rotating with angular velocities 2 and 3 . The line of action and pressure angle are

clearly shown in this figure. In figure 8(b) the pair of constraint forces ( 23F d 32F cting at the pitch point along the line of action are shown. Considering the freebody diagram of gear 2 as

shown in figure 8(c), the force is balanced by the reaction force acing at the bearing. As these two forces are equal and opposite, they will form a couple. To overcome this reaction couple, the prime mover (say motor) should provide a torque equal in magnitude but opposite in direction, which is represented by in the figure.

an a)

32F 12F

a2T

23

(a) (c ) (b)

Figure 8 : Force analysis of Spur gear

One may resolve these forces in tangential and radial direction. Clearly, the tangential force is responsible for rotating the gear. Now let us derive an expression for the gear forces form the given power and speed of operation. Let be the speed of rotation (in rpm) of the gear with module m and number of teeth z. The value of the pressure angle(in degrees) is and the power (in KW) it transmits be P. The diameter D of the gear can be calculated as D = m z (a) and the torque T transmitted by the gears is

660 102

PT

= (b)

From Figure 8(c), we see that the force that is responsible for transmitting the torque T is the tangential component (Ft32 =Ft23 =Ft). The radial component (Fr32 =Fr23 =Fr) is the separating force, which always acts towards the center of the gear. Thus we get

t t

D 2T=T, or, F

2 D=F (c)

So, using equation (c) one may obtain the tangential force Ft from known value of D and T. From Figure 8(c) the radial component can be obtained as Fr = Ft tan (d)

Hence the resultant force acting on the gear or on the bearing equals to 2 2t rF = (F F )+

Also one may note that t rF =Fcos and F Fsin =

24

Thus the net force F can be easily calculated using the above expressions. This analysis of the gear tooth force is based on certain assumptions which are as follows

1. As the point of contact moves, the value of the resultant force F changes, which is

neglected in the above analysis. 2. It is assumed that only one pair of teeth take the entire load. At times there are two pairs

which simultaneously are in contact and share the load. This aspect is also neglected in this case.

3. The analysis is valid under static conditions, when the gears are running at very low velocities. In practice there are dynamic forces also due to the power transmission. The effect of these dynamic forces are neglected in the analysis.

Helical gears Like the spur gears, the helical gears also connect parallel shafts. But a major difference between a helical gear and a spur gear is that the teeth in case of a helical gear are cut in the form of a helix on the pitch cylinder. In these types of gears the contact between the meshing teeth begins with a point on the leading edge of the tooth and gradually extends along the diagonal line across the tooth. There is a gradual pick up of the load by the tooth resulting in a smooth engagement and a quiet operation even at very high speeds.

nF

tF

rF

aF n

In a helical gear n =normal pressure angle

=helix angle

Figure 9: Force analysis of helical gear

In helical gear the normal force consists of three components viz., tangential component ,

radial component and the axial or thrust component as shown in the figure 9. They are

related as follows.

nF tF

rF aF

Tangential force cos cost n nF F = (a) Radial force sinr nF F n= (b)

25

Thrust or axial force = cos sin tana n n tF F F = = (c) Let N be the speed of rotation in rpm of the gear with module m and number of teeth z, is the transverse pressure angle and is the helix angle. Now the diameter D of the gear can be determined from the relation D = m z (d)

The angular velocity of the gear = 260

N = (e)

The torque T transmitted by the gears can be calculated from the power P from the relation

P T= (f)

From Figure 9, we see that the force that is responsible for transmitting the torque T is the tangential component Ft. The radial component Fr is the separating force, which always acts towards the center of the gear, and Fa is the axial or thrust component. The direction of this axial component depends upon whether the gear is left or right handed, the direction of rotation and on whether the driving or driven gear is under consideration. Thus we get

2tDT F= or 2t

TFD

= (g)

It may be recalled that the normal pressure angle n , helix angle and transverse pressure angle

are related by

tancos

tann

= (h)

Once we have calculated the value of the normal pressure angle from (h), we can calculate the tangential force from equations (f and g), axial or the thrust force using equation (c) and radial component from expression (b). The resultant force on the helical gear can now be calculated using the three components as

2 2 2( )n t rF F F F= + + a (i) The following guidelines should be followed while calculating the axial or thrust component Fa

1. Select the driving gear from the pair. 2. Use right hand for right-handed helix and left hand for left handed helix. 3. Keep the fingers in the direction of rotation of the gear and the thumb will indicate the

direction of the thrust component of the driving gear.

26

4. The direction of the thrust component of the driven gear will be the opposite to that for the driving gear.

Example 7 Two helical gears on the parallel shafts have a normal pressure angle of 20 degrees and a normal module of 6 mm. The centre distance is 200 mm and the assembly has 20 and 40 teeth. The gear set transmits 50 KW at a pinion speed of 1200 rev/min. Determine the tangential, radial and thrust loads on the gear teeth, and show these forces on the gears. The pinion is handed and rotates clockwise. Solution: Given data: ---

Normal pressure angle = n =200

Normal module = =6mm. == 21 mm nm

Center distance=C=200mm No of teeth of the pinion=20,

No of the teeth of the gear=40.

212

1 221

4020

rrr

r===

1 1 1 1

2

2002 200 3 200. Hence, .3

400Therefore, 3

r r r r

r

+ = = =

=

Now as C= 21 rr +

27

1 2 1 1 2 2 1 2

0

1 1 1( ) ( ) (2 2 2 cos

So, cos 25.8419

nmc d d mT m T T T

= + = + = +

=

)

1 1 1

50kw,12002 ( ) 125.6rad/sec

6050 1000 398.089Nm

125.6

P

w

T w P T

=

= =

= = =

211

398.089 1000 5971.335(200 / 3)T TF N

r

= = =

We know that

12 21 21 12,Hence, 5971.335NT T T TF F F F= = =

cos , cos

sin , sin

T Tn n n n n

r n n a n

F F F FF F F F

n

= =

= =

012 12

21 12

tan 5971.335 tan 25.8419 2892.04532N,

a T

a a

F FF F

= = =

=

012 12 sin 5971.335 tan 20 2173.388Nr T

nF F = = = Ans.

Summery The following points are learned in this chapter

Classification of forces: external and constraint forces Determination of moments Freebody diagram Two and three force members Static force analysis using graphical method Static force analysis using analytical method (vector method) Use of superposition theory for multiple external forces acting on a mechanism Static force analysis with sliding and grease friction (concept of friction circle) Application of virtual work principle for static force analysis.

28

Exercise Problems 1. Draw the constraint forces in all the six types of lower pairs, viz., (i) revolute or turning

pair (ii) prismatic or sliding pair, (iii) cylindrical pair, (iv) screw or helical pair, (v) planar or flat pair, and (vi) globular or spherical pair.

2. Explain with neatly drawn free-body diagram the effect of friction in the bearings on the torque required by the crankshaft in a slider-crank mechanism when the crank is rotating in (i) clock wise direction (ii) anti-clock wise direction.

3. Calculate the torque required for static equilibrium of an in-line slider crank mechanism in the position when crank angle = 60 deg (from inner dead center). The dimensions are crank length r = 100 mm, connecting rod length L = 175 mm, and the piston force is P = 50 N. Assume crank to be rotating in anticlockwise direction. Use, graphical, analytical and virtual work principle to find the result.

4. Taking same data as in problem 3, also find the torque required assuming that the coefficient of friction for all bearings is 0.1. The three journal bearings all have radii of 20 mm, and the crank is rotating in the clockwise direction.

5. Figure below shows a mechanism used to crush rocks. The mechanism is moving slowly, so the inertia forces may be neglected. In the position shown, determine the torque required to drive the input link AB when the crushing force acting in the horizontal direction is 5000N. Here, AB = 50 cm, BC=100 cm, CD=120 cm and the fixed link AD=150 cm and CE=25 cm and the angle CED of the ternary link CED is 900. Use (a) graphical method, (b) analytical method and (c) virtual work principle to determine the bearing forces and required torque.

5000 N

D A

B

E

450

C

F

29

CHAPTER 2 DYNAMIC FORCE ANALYSIS

Inertia force and couple

a

Equivalent off-set Inertia force

iF cI

a

a

a

iF

iFcIiF

iF

iF

cI iF

Dynamic force analysis of four bar mechanism

Dynamic Force Analysis of a Four bar Mechanism using Matrix Method

Solution procedure

The MATLAB code

%DYNAMIC FORCE ANALYSIS OF FOUR-BAR MECHANISM %TAKING INPUTS FROM THE USER FOR THE FOUR-BAR MECHANISM a = input('enter the length of the link AB : '); b = input('enter the length of the link BC : '); c = input('enter the length of the link CD : '); d = input('enter the length of the link DA(fixed link) : ');

ma = input('enter the mass of link AB : '); mb = input('enter the mass of link BC : '); mc = input('enter the mass of link CD : '); ka = input('enter the radius of gyration of link AB : '); kb = input('enter the radius of gyration of link BC : '); kc = input('enter the radius of gyration of link CD : '); rga = input('enter the magnitude of the p.v. of the c.g. of link AB from the fixed pivot A : '); rgb = input('enter the magnitude of the p.v. of the c.g. of link BC from the pivot B : '); rgc = input('enter the magnitude of the p.v. of the c.g. of link CD from the fixed pivot D : '); deltaa = input('enter the deviation angle of the p.v. of the c.g. of AB from the p.v. of pivot B : '); deltab = input('enter the deviation angle of the p.v. of the c.g. of BC from the p.v. of pivot C(wrt B) : '); deltac = input('enter the deviation angle of the p.v. of the c.g. of CD from the p.v. of pivot C(wrt D) : '); thetaa = input('enter the input angle (angle between AB and AD) in degrees : '); omegaa = input('enter the angular velocity of link AB : '); alphaa = input('enter the angular acceleration of the link AB : '); Tl = input('enter the load torque : ');

%CONVERTING DEGREES TO RADIANS AND CHECKING FOR FEASIBILITY thetaa = 3.1415926*thetaa/180; K = ((a*a) - (b*b) + (c*c) + (d*d))/2; P = K - (a*(d-c)*cos(thetaa)) - (c*d) ; Q = -2*a*c*sin(thetaa); R = K - (a*(d+c)*cos(thetaa)) - (c*d);

flag=0; if ((Q*Q - 4*P*R)

alphab1 = ((-1*a*alphaa*sin(thetac1 - thetaa)) + (a*omegaa*omegaa*cos(thetac1 - thetaa)) + (b*omegab1*omegab1*cos(thetac1 - thetab1)) - (c*omegac1*omegac1))/(b*sin(thetac1 - thetab1)); alphab2 = ((-1*a*alphaa*sin(thetac2 - thetaa)) + (a*omegaa*omegaa*cos(thetac2 - thetaa)) + (b*omegab2*omegab2*cos(thetac2 - thetab2)) - (c*omegac2*omegac2))/(b*sin(thetac2 - thetab2)); alphac1 = ((-1*a*alphaa*sin(thetab1 - thetaa)) + (a*omegaa*omegaa*cos(thetab1 - thetaa)) + (b*omegab1*omegab1) - (c*omegac1*omegac1*cos(thetac1 - thetab1)))/(c*sin(thetac1 - thetab1)); alphac2 = ((-1*a*alphaa*sin(thetab2 - thetaa)) + (a*omegaa*omegaa*cos(thetab2 - thetaa)) + (b*omegab2*omegab2) - (c*omegac2*omegac2*cos(thetac2 - thetab2)))/(c*sin(thetac2 - thetab2));

%CALCULATION OF THE ELEMENTS OF THE 'b' MATRIX b1 = -1*ma*rga*(alphaa*cos(thetaa + deltaa - (3.1415926/2)) + omegaa*omegaa*cos(thetaa + deltaa)); b2 = ma*9.81 - ma*rga*(alphaa*sin(thetaa + deltaa - (3.1415926/2)) + omegaa*omegaa*sin(thetaa + deltaa)); b3 = ma*ka*ka*alphaa; b4 = -1*mb*rgb*(alphab1*cos(thetab1 + deltab - (3.1415926/2)) + omegab1*omegab1*cos(thetab1 + deltab)); b5 = mb*9.81 - mb*rgb*(alphab1*sin(thetab1 + deltab - (3.1415926/2)) + omegab1*omegab1*sin(thetab1 + deltab)); b6 = mb*kb*kb*alphab1; b7 = -1*mc*rgc*(alphac1*cos(thetac1 + deltac - (3.1415926/2)) + omegac1*omegac1*cos(thetac1 + deltac)); b8 = mc*9.81 - mc*rgc*(alphac1*sin(thetac1 + deltac - (3.1415926/2)) + omegac1*omegac1*sin(thetac1 + deltac)); b9 = mc*kc*kc*alphac1 - Tl;

%CALCULATIONS OF THE ELEMENTS OF THE 'A' MATRIX A31 = rga*sin(thetaa + deltaa); A32 = -1*rga*cos(thetaa + deltaa); A33 = a*sin(thetaa) - rga*cos(thetaa + deltaa); A34 = rga*cos(thetaa + deltaa) - a*cos(thetaa); A39 = 1;

A63 = rgb*sin(thetab1 + deltab); A64 = -1*rga*cos(thetab1 + deltab); A65 = b*sin(thetab1) - rgb*cos(thetab1 + deltab); A66 = rgb*cos(thetab1 + deltab) - b*cos(thetab1);

A95 = rgc*cos(thetac1 + deltac) - c*sin(thetac1); A96 = c*cos(thetac1) - rgc*cos(thetac1 + deltac); A97 = rgc*sin(thetac1 + deltac); A98 = -1*rgc*cos(thetac1 + deltac);

B = [b1 b2 b3 b4 b5 b6 b7 b8 b9]; A = [1 0 -1 0 0 0 0 0 0;0 1 0 -1 0 0 0 0 0;A31 A32 A33 A34 0 0 0 0 1;0 0 1 0 -1 0 0 0 0;0 0 0 1 0 -1 0 0 0;0 0 A63 A64 A65 A66 0 0 0;0 0 0 0 1 0 1 0 0;0 0 0 0 0 1 0 1 0;0 0 0 0 A95 A96 A97 A98 0];

%CALCULATION OF THE FIRST SOLUTION X = (inv(A))*B'; theta112 = (atan(X(2,1)/X(1,1)))*180/3.1415926; if(X(1,1)

theta134 = (atan(X(6,1)/X(5,1)))*180/3.1415926; if(X(5,1)

disp('Set I : '); disp('X = [F12x F12y F23x F23y F34x F34y F14x F14y Ts] ');disp(X); disp('F12 = ');disp(sqrt(X(1,1)^2 + X(2,1)^2)); disp('theta_F12 = ');disp(theta112); disp('F23 = ');disp(sqrt(X(3,1)^2 + X(4,1)^2)); disp('theta_F23 = ');disp(theta123); disp('F34 = ');disp(sqrt(X(5,1)^2 + X(6,1)^2)); disp('theta_F34 = ');disp(theta134); disp('F14 = ');disp(sqrt(X(7,1)^2 + X(8,1)^2)); disp('theta_F14 = ');disp(theta114);

disp('Set II : '); disp('Y = [F12x F12y F23x F23y F34x F34y F14x F14y Ts] ');disp(Y); disp('F12 = ');disp(sqrt(Y(1,1)^2 + Y(2,1)^2)); disp('theta_F12 = ');disp(theta212); disp('F12 = ');disp(sqrt(Y(3,1)^2 + Y(4,1)^2)); disp('theta_F23 = ');disp(theta223); disp('F12 = ');disp(sqrt(Y(5,1)^2 + Y(6,1)^2)); disp('theta_F34 = ');disp(theta234); disp('F12 = ');disp(sqrt(Y(7,1)^2 + Y(8,1)^2)); disp('theta_F14 = ');disp(theta214);

flag=flag+2; end %END OF CODE

Dynamic analysis of Slider Crank Mechanism

Example1:

Solution:

l=450 mm

r=150 m

m

A

B

B

AG

l=300 mm

Inertia forces due to reciprocating masses:

Torque to consider the correction couple:

Torque due to mass at A:

Total torque on the crankshaft:

(answer)

Example 2:

Solution:

Inertia forces due to reciprocating mass:

l=2m

r=0.4 m

A

B

O

Torque to consider the correction couple:

Torque due to weight of mass at A:

Torque due to weight of mass at B:

Total inertia torque on the crankshaft:

Graphical method:

Inertia torque

QC

O

N

D

Ix

Y

Z

G

E

dg

K

L

P

2

0.4

QC

O

N

D

E

dg

K

L

Summery

Exercise Problems

1.

CAM DYNAMICS Mechanism provides a non-linear I/O relationship. Different mechanism like single or multi-

degrees of freedom, intermittent motion mechanisms and linkages etc. have different I/O

relationship. When we can not obtain a certain functions from the well known mechanisms,

we use a cam mechanism. It is a one degree of freedom mechanism of two moving links. One

is cam and the other is follower.

There are limitations of its use. After some critical

speed the nett acceleration of the follower system may

be in a direction to separate it from the cam surface. So

the follower does not follow the surface profile any

more. We have to be able to calculate this critical

speed and must not operate the mechanism beyond it.

Critical speed depends on the profile of the cam, the stiffness and pretension of any restoring

spring if exist at all or the mass of the follower. So, a differential analysis requires for each

widely used cam profile. One example will be done here concerning an eccentric cam. An

eccentric cam has a circular profile, but rotates about an axis which is displaced from the cam

center by e amount, called the "eccentricity".

Follower displacement becomes;

coseey =

where t = , cam is rotating with constant angular velocity, . taking first and second time derivative of follower displacement equation we will get velocity

and acceleration, respectively;

mass

xSliding cam

Cam profileFollower

y=f(x)

e

k

y

m

y

m

NA NB

Fc Fs=k(y+d)

mg

tey sin=

And tey cos2=

There will be inertia forces acting onto the follower proportional to this acceleration.

Differential motion equation for the follower is;

= ymF

From the free body diagram of the follower, nett force in y direction becomes;

)( +== ykFFFF csc

Where, cF is contact force

is spring pre-compression

Substituting, y ,

y and F into differential equation of motion yields;

)(cos)( 2 ++= ektkmeFc

0=cF is the critical condition. It must always have positive value. When 0=cF , contact is

lost and hence cam and follower are no longer touching each other. It means that follower is

no more following the profile of the cam. It is free and floating in the air.

Example An uniform steel bar shown in the figure is used as an oscillating cam follower and

its mass moment of inertia is CGI as 0.65 kg.m. The cam follower is driven by an eccentric

cam to give a rise of 4 cm. Assuming that cam system is working on the horizontal plane and

there is no friction between the cam and follower.

k=2kN/m

2 cm 25 cm 50 cm 25 cm

5 cm

y

xO

a) Draw the freebody diagram of the follower, showing all the forces,

OCG

zy,y,y

l/4 l/4l/4l/4

. ..

Fs=kz Fc

manFoy

Fox

b) Drive the cam contact force equation,

teyandteyteey cos,sin,cos 2...===

Tangential acceleration at cam contact point is equal to the ..y , then

ltetelyat

cos2cos2

*2

2..

===

Equation of motion of follower is;

= oo IM

= 4**

2* lzklFM co

Substituting angular acceleration, mass moment of inertia and moment equation into EQM,

+=

ltewlmlzklFc

cos***2*48

47*4

**2

*222

In this equation we need a relationship between y and z. This comes from similar triangles.

So,

2cos

224 teezyzy

l

z

l

=== ,

( )teelk

ltewl

lmFc

coscos***2*24

47*222

+

+=

c) Sketch a displacement diagram showing the follower motion for the entire 360 of the

cam rotation without computing the numerical values, superimpose graph of cam contact

force onto the same axis. Show where jump is most likely to begin,

/2 3/2 2

Jump is most likely occurs here

y Fc

d) Calculate the jump speed,

When 0== cFandt

lke

lewl

lm **2**2*

2447*0

222

+

+=

+= 22 37

24*wlm

lk

e) List the things that can be done to increase the critical speed of this system.

In order to increase the speed;

k can be increased. Harder spring can be used.

m can be reduced. Smaller mass means smaller dimensions.

Example The cam-and-follower

system shown in figure has 2=k

kN/m, 1=m kg,

tcosy 1515 = mm, and 50=

rad/sec. The retaining spring is

assembled with a preload of 5 N.

a) Draw the freebody diagram of the follower.

y

FC Fs=k(y+ )

NA NB

mg

b)What is the acceleration of the follower, when 1=t sec.

222 18636223618615050151515

1515

sec/m.sec/mm.)*cos(**tcos**ytsin**ytcosy

====

=

=

DD

D

c) Drive conctact force equation between the cam and follower.

( )C

C

Cy

y

Ftcosk)tcos**(m

F)y(kym)y(kymF)y(kF

ymF

=

+

+

=+++=

++=

=

10001515

100015 2

DDDD

DD

d) Compute the maximum and minimum values of the contact force.

( )

tsin)mk(*.tsin**.*ktsin*.*m)t(d

tcosk)tcos**(md

dtdFC

22

2

0150015001500

10001515

100015

0

+==

+

+

==

210001500 2

,,nwherenttsintsin)mk(*.

===

+=

A B

yk

CF is maximum when 20 == tort and CF is minimum when =t .So,

N.)cos(*.*cos**.(*FN.)cos(.*)cos**.(*F

minC

maxC

52751015020005001501

54250101502000050015012

2

=++=

=++=

e) If the follower is found to jump off the cam, compute the angle t corresponding to the

very beginning of jump.

it is seen from previous answer minCF >0,so there is no jump.

Example A cam of linear displacement

profile is moving leftward with a constant

velocity to actuate the spring-mass system

shown.

a) Draw the freebody diagrams (mass and

spring)

b) Drive differential equation of motion of the follower.

c) Determine the natural frequency of the system in rad/sec

d) Determine the amplitude of the rise motion

a) Draw the freebody diagrams

b) Drive differential equation of motion of the body

= xmFx DD

m x

F=k(x-y)

x

y

F=k(x-y)

F=k(x-y)

= )yx(kFx Then equation of motion becomes;

kykxxm =+DD where taby =

c) Determine the natural frequency of the system in rad/sec

Solution of this equation of motion is;

kykxxm =+DD ymkx

mkx =+ DD

.sec/rad.mk

n 724412000

===

d) Determine the amplitude of the motion

Complementary part of this equation of motion is;

tsinBtcosAx nnc +=

Particular part of this equation of motion is;

tabx np =

Overall solution is

tabtsinBtcosAx nnnc ++=

22 BAX amp +=

A and B is to be found from initial conditions.

0

0000

00

=

++=++=

===

AabsinBcosAt

abtsinBtcosAx

mxx,t

nnnnnn

o

abB

abcosBsinA

abtcosBtsinAx

x,t

nnnnnnnnnn

=

++=++=

==

000

00

D

D

m.abBAX amp 6600

22

22=+

=+=

Example The mass m is driven up and down by the eccentric

cam as shown in figure. The eccentricity is 3 cm. Assume no

friction

a) Derive the equation for the contact force.

b) Find the cam velocity w corresponding to the beginning of

the jump.

tcoseey =

tsiney =D

tcosey 2=DD

FBD of the follower;

ymF DD=

mgFF c =

mgymFymmgF cc == DDDD

mgtcosmeFc += 2 (1)

Jump most probably occurs when 0=cF

To obtain where cF is minimum;

tsinme)t(d

dFc

20 ==

,.......,,,,n,*ntwhere,tsin 432100 ===

obviously when =t , cF is minimum. So, subsituting this into (1);

sec/rad...

egmgcosme 08318

0308190 2 ===+=

my

Fc

my

e

Q-5 (20%) The cam-and-follower system shown in figure has

.sec/radandmme,tcoseey,kgm,m/kNk

501522==

===

Mechanism is working on vertical plane. Assume that friction of the rollers at A

and Bs are negligible small. a) Draw the freebody diagram of the follower.

y

FC

Fs=ky

NA

NB

mg

b) What is the acceleration of the follower, when 1=t sec.

222 18636223618615050151515

1515

sec/m.sec/mm.)*cos(**tcos**ytsin**ytcosy

====

=

=

DD

D

c) Drive contact force equation between the cam and follower.

( )C

C

Cy

y

Fmgtcosk)tcos**(m

FmgkyymmgkyymFmgkyF

ymF

=+

+

=++=

+=

=

10001515

100015 2

DDDD

DD

d) Compute the maximum and minimum values of the contact force. ( )

tsin)mk(*.tsin**.*ktsin*.*m)t(d

mgtcosk)tcos**(md

dtdFC

22

2

0150015001500

10001515

100015

0

=+=

+

+

==

210001500 2

,,nwherenttsintsin)mk(*.

===

=

CF is maximum when 20 == tort and CF is minimum when =t .So,

N..*)cos(*.*cos**.(*FN..*)cos(.*)cos**.(*F

minC

maxC

124281921015020005001501

125781920101502000050015012

2

=++=

=++=

e) If the follower is found to jump off the cam, compute the angle t corresponding to the very beginning of jump. it is seen from previous answer minCF >0,so there is no jump.

A

B

y

k