static or process force analysis - yıldız teknik ...cdemir/statik_kuv.pdf · 3 static or process...

1

Static or Process Force Analysis

• Based on the equations from statics, i.e. SF = 0 in any direction and SM = 0 about any point which must be satisfied at every position.

• Must know the magnitude and direction of all external loads which are applied to each link at each position to produce the process/function the mechanism is carrying out, e.g. cylinder pressure in an engine. These loads are independent of speed.

• The results yields information about how much energy must be applied at a specific connection to produce the desired function/process and the loads each member and connection in the mechanism must support at each position of operation.

2


For lower pair connections without friction or where friction can be ignored, the load is transferred over an area at the pin connection. As can be seen below, the load is evenly distributed over the area of the pin where the load is normal to the surface and all the lines of action intersect at the center of the pin. Therefore, in moving from link to link the load is transferred through the center of the pin.

3


For higher pair elements such as a ball or roller bearing, the load is transferred at a point between the inner and outer races and the rolling elements. Each force is normal to the surface and all lines of action pass through the center of the bearing.

Therefore, in moving from link to link the load is transferred through the center of the pin.

4


5

Static Equilibrium

From Newton’s first law, a body is in static equilibrium if the

resultant of all the forces acting on a rigid body is zero. This

condition can be expressed mathematicaly as:

In space, these two vector equation yields six scalar equations:

In planar, there are three scalar equations:

0 0and F M

0; 0

0; 0

0; 0

x x

y y

z z

F M

F M

F M

0; 0;

0

x y

z

F F

M

6


• To perform this analysis one must be able to visually recognize which members satisfy the criteria to be either a two force or three force member or are carrying a couple. If this is done incorrectly, it will appear that there are too many unknowns to solve the problem.

• Two force member - member can only support tension or compression, i.e. force line of action must be along the member.

7

Force Systems

In coplanar force system under which the body acted on is in

equilibrium, we can have following force systems:

1. Two force member : If there are only two forces acting on a

body, it is called a “ two force member”.

To satisfy sum of torque is equal to zero distance between the forces

must be zero. This means that the forces are collinear.

F1

F2

To satisfy sum of forces equal to zero, the two forces should be

equal and opposite.

8


• Couple – the lines of action of two forces are parallel, in opposite directions, and separated by a non-zero distance, h, so that a torque is created.

9


• Three force member – the lines of action of all three forces must pass through the same point which means that if the directions of two of the forces are known, the direction of the third force is established.

10

Force Systems

2. Three force member : If there are only three forces acting on a

body, it is called a “ three force member”.

To satisfy sum of torque is equal to zero, the lines of application

of all the three forces intersects at one single point. This point

is called the point of concurrency.

F1

F2

To satisfy sum of forces equal to zero, the vectors must form a

closed polygon and coplanar.

F3

11

Solving static force problems

Graphical Approach : We draw straight lines to represent vectors which are in proper directions and lengths proportional to the magnitudes of the vectors and in an articulated manner as depicted in Figure. Vectors form a closed polygon called a “vector loop”

F1

F2

F3

12


Arithmetical Approach : The simplest arithmetical approach is to separate vector equation into components.

F1

F2

F3

q1

q3

q2

These two component equations are not no longer vector equations. They are scalar and can be simultanously solved to find max two of the following;

1 1 2 2 3 3cos cos cos 0F F Fq q q

1 1 2 2 3 3sin sin sin 0F F Fq q q

1F2F3F

13


• Separate the mechanism into its links, considering each a free body with all the acting external and constraint forces on it.

• Apply the rules of statics each free body which are

Solution of vector equations can be by arithmetical and or graphical.

0 0and F M

14

Örnek Bir üç çubuk mekanizmasında sarkaç kola yatay olarak 10 N

etki etmektedir, 30 mm D noktasından mesafesi. Sistemi

statik dengede tutacak AB koluna etki eden Torku bulunuz.

A

1

B

2

D

43

C

a1

a2

a3 a4

30

10 N

q

a1 = 80 mm

a2 = 30 mm

a3 = 70 mm

a4 = 50 mm

89.86

20,06

60

q

15

Example Arithmetical method:

• Separate the mechanism into free bodies of links,

• Put all the acting and interacting forces,

• Then, apply the law of statics for each free body.

D

C

10 N

89,86°

B

3

C

20,06°

A

B

2

60°

T

4

FDx

FDy

FCy

FCy

FCx

FCx

FBx

FBy

FBx

FBy

FAx

FAy

16

D

C

10 N

89,86°

4

FDx

FDy

FCy

FCx

10010;0 CxDxCxDxx FFFFF

0; 0y Dy Cy Dy CyF F F F F 08568950868950868930100 ,cos**F,sin**F,sin**;M CyCxD

99,299*12,0*99,49 CyCx FF

4 numaralı uzuv için statik denklem;

17

B

3

C

20,06°

FCy

FCx

FBx

FBy

3 numaralı uzuv için statik denklem;

CxBxCxBxx FFFFF 0;0

CyByCyByY FFFFF 0;0

006,20cos*70*06,20sin*70*;0 CyCxB FFM

0*75,60*01,24 CyCx FF

18

A

B

2

60°

T

FBx

FBy

FAx

FAy

2 nolu uzuv için statik denklem;

BxAxBxAxx FFFFF 0;0

ByAyByAyy FFFFF 0;0

TFFM BxByB 00,60sin*30*00,60cos*30*;0

TFF BxBy *98,25*00,15

19

D

C

10 N

89,86°

B

3

C

20,06°

A

B

2

60°

T

4

FDx

FDy

FCy

FCy

FCx

FCx

FBx

FBy

FBx FBy

FAx

FAy

9 adet denklemi beraber çözersek ; FAx, FAy, FBx, FBy, FCx, FCy, FDx, FDy, and T.

N,FCy 372

N,FCx 995

N,FFFF AyByDyCy 372

NFFFF DxAxCxBx 99,510

CWmm.N,T 07120 cvp.

10010;0 CxDxCxDxx FFFFF

CyDyCyDyy FFFFF 0;0

0856,89cos*50*86,89sin*50*86,89sin*30*10;0 CyCxD FFM

99,299*12,0*99,49 CyCx FF

CxBxCxBxx FFFFF 0;0

CyByCyByY FFFFF 0;0

006,20cos*70*06,20sin*70*;0 CyCxB FFM

0*75,60*01,24 CyCx FF

BxAxBxAxx FFFFF 0;0

ByAyByAyy FFFFF 0;0

TFFM BxByB 00,60sin*30*00,60cos*30*;0

TFF BxBy *98,25*00,15

20

Örnek Grafik Yöntem:

• Mekanizmayı ölçekli çiz,

• Bilinmeyen değerler ölçkelenmiş çizimden okunur(açılar),

• Mekanizmayı serbest cisim diyagramlarını ölçekli çıkart,

• Sistemi iki kuvvet veya üç kuvvet etkisinde olduğunu belirle iç kuvvetler sistem üzerine yerleştir,

• Statik kuralları vektörel uygula.

21

D

C

10 N

89,86°

B

3

C

20,06°

A

B

2

60°

T

4

3 nolu uzuv 2 kuvvet 4 nolu uzuv 3 kuvvetli eleman

F23

F43

F34

F32

F14

F12 F14 F34

10 N

F14

22

D

C

10 N

B

3

C

A

B

2

4

F43

F34

F23

F14

F32

d1

T12

OF

10 N

F14 F34

F12

F14 ve F34 kuvvet poligonundan direkt olarak ölçülür. 10 N : 50 mm F14 : 22.5 mm F34 : 32.5 mm

NF 5.450

5.22*1014

NF 5.650

5.32*1034

N..*.F*dT 751265651932112

23


As an example consider the slider-crank below. The externally applied force, FC, is shown acting below the centerline of the slider as it represents a cutting force instead of a pressure force. In order to proceed we must determine if there are any 2 or 3-force members. Our goal is to find the couple that must act about point d to hold the mechanism in place. The negative of this couple is the torque required to drive the cutter forward. Link a is then rotated to the next position, FC changed to the next value, and the process repeated until all positions have been analyzed.

24


To begin, we draw free body diagrams of each link. The solid arrows represent forces where magnitude and direction are known. The double headed dashed arrows represent forces where the direction only is known. The curved dashed arrows represent forces where neither direction nor magnitude is known.

25


When analyzing any mechanism, one should consider what is to be determined. If only Ma is needed, take links b and c together as a free body. Summing moments about the pin between links a and b will yield Fd/c. Then taking the entire linkage as one free body and summing moments about the ground pin on link a will yield Ma. This approach avoids the need to find Fb/c and Fa/b if they are not of interest in the solution or they can be determined later.

26


Reviewing each link yields that link b is a 2-force member (pinned ends and no externally applied load) and the slider is a 3-force member. This establishes the direction of the force b exerts on c (Fb/c) and allows the force polygon on link c to be constructed (3 directions and 1 magnitude are known). Fc is drawn to scale and

SF = 0 is solved graphically. Where the two directions intersect deter- mines the magnitudes of Fb/c and Fd/c. As Fb/c = -Fc/b and Fa/b = -Fc/b, the torque needed to hold the slider in place is

a ba

M F x

O2A = 6 in. AB = 9 in. AC = 15 in. O4C = 8.25 in. O2O4 = 17.13 in.

For the mechanism shown below, determine the value of the torque T2 needed to hold the mechanism in place.

O2A = 6 in. AB = 9 in. AC = 15 in. O4C = 8.25 in. O2O4 = 17.13 in.

The free body diagram of each link is shown below. Because a torque acts on links 2 and 4, only link 3 can be identified as a 3-force member. As the magnitudes and directions of all pin forces are unknown, use the free body diagram of link 4 to determine the transverse component of F34.

F12

F32

F23

F43

F14

F34

FR34

FT34

F34 bileşenlerine ayıralım 4 nolu uzuv boyunca, FR34, ve

4 nolu uzva dik, FT34.

F14

F34

As SMO4 = 0 = FT34*O4C - T4, we can find FT

34 from FT

34 = T4/ O4C FT

34 = 72.7 lbf

F23

To determine FR43 , use SMA = FR

43*hR43 - FT

43*hT43 - P*hP = 0

-OR-

FR43 = (P*hP + FT

43*hT43)/hR

43 = 69.4 lbf F43 = [(FR

43)2 + (FR43)2]0.5 = 100.5 lbf

A force polygon is used to find F23 = 155 lbf (1 in. = 50 lbf)

+ OF

F23

F43

P

FT43

FR43

hT43

hR43

hP

F23

F32

F12

h32

The value of the torque T2 is obtained from

T2 = -F32*h32

and F12 = -F32

F12 F14

A force polygon for the entire mechanism is used to determine F14 = 81.3 lbf (1 in. = 50 lbf)

P +OF

F12 F14

34

METHOD of SUPERPOSITION

Sometimes the number external forces and inertial forces acting on a mechanism is too much for graphical solution.

A

D(6cm, 3cm)

B

E

C1

2 3

4

100 N

60o

x

y

F1

F2

t1

F3

In this case we apply the method of superposition. Using superposition the entire system is broken up into (n-1) problems, where n is the number of forces, by considering the external and inertial forces of each link individually.

35

METHOD of SUPERPOSITION

A

D(6cm, 3cm)

B

E

C1

2 3

4

100 N

60o

x

y

F1

F2

t F3

Response of a linear system to several forces acting simultaneously is equal to the sum of responses of the system to the forces individually.

A

D(6cm, 3cm)

B

E

C1

2 3

4

100 N

60o

x

y

F1 t1

A

D(6cm, 3cm)

B

E

C1

2 3

4

100 N

60o

x

y

F2

t2

A

D(6cm, 3cm)

B

E

C1

2 3

4

100 N

60o

x

y

t3

F3 321 tttt

36

Example 1 Onto crank CD of the mechanism shown, a CCW torque of 100 N.m is acting. Onto the

slider of link 6 a 50 N force is acting horizontally rightward. Calculate the amount of torque required on crank AB to keep the mechanism in static equilibrium using the

method of “superposition”. AB = BE=CE = 7 cm. CD = EF = 6 cm.AD = 10 cm.e = 8 cm.

A

1

B

2

3

D

E

100 N.m

60°

T

5

6

4

50 N

C

F

80

120

37

Example 1 This problem can not be solved graphically. We have to use the method of

superposition. • Assume, only 50 N force is acting onto mechanism. Then, Link 6, 3 are three force member. Link 5, 4 are two force member.

A

1

B

2

3

D

E

100 N.m

60°

T

5

6

4

50 N

C

F

80

38

Example 1 Link 6, 3 are three force member. Link 5, 4 are two force member.

5

50 NF

F

B

3E

A

B

2T 4

C

N

N FF

FF=59.8N

FF

FE

FE

FC

FC

FD

FB

FB

FA FF=59.8N FC

FB=49N

CCWNm.*T 0375049

39

Example 1

• Assume, only 100 N.m torque is acting onto mechanism. Then, Link 3 is two force member. There is no force and torque on link 5 and 6.

A

1

B

2

3

D

E

100 N.m

60°

T

5

6

4

50 N

C

F

80

40

Example 1 Link 3 is two force member.

There is no force and torque on link 5 and 6.

5

FF

B

3E

A

B

2T 4

C

N FF

FF

FE

100 Nm

41

Example 1 Link 3 is two force member.

There is no force and torque on link 5 and 6.

5

FF

B

3E

A

B

2T 4

C

N FF

FF

FE

FC

FC

FD

FB

FB

FA

N.F.*F CC 1417240580100

100 Nm

NmCW..*..d*FT B 27548141724702603

CWm.N,,, 4924578422754821 TTT

42


QR = 140 in.; QFc = 75 in.; PR = 285 in.

PFb = 126 in.; OP = 94 in.; Fb = 90 lb

Fc = 85 lb; Mc = 1,000 in-lb

43


44


hFc

/

/

/

*

85 *61.875 . 1,000

140 .

44.7

c F ct cb c

tb c

tb c

F h MF

QR

lb in in lbF

in

F lb

45


hFb

hFn

c/b

hFtc/b

//

/

/

* *90 *125.7 . 44.7 *145.1 .

19.7244.8 .

tb F c b tn Fb c b

c bnFc b

F h F hlb in lb in

F lbh in

Use a force polygon for link b to find Fa/b

46

Static or Process Force Analysis +

Fb

Fn

c/b

Ft

c/b

Fa/b

If the assumed magnitudes for Fb, Fc, and Mc were correct,

this would be the magnitude and direction of Fa/b.

47


For illustration purposes only – NOT to scale

48


49


50


51


52


53


AB = 9.375 in. BD = 5.25 in. DO4 = 5 in. BO4 = 10 in. AC = 7 in. AO2 = 5 in.

To determine T2, we first identify links 3 and 4 as 3-force members which specifies the directions of the line of action of the third force once the other two directions are known.

AB = 9.375 in. BD = 5.25 in. DO4 = 5 in. BO4 = 10 in. AC = 7 in. AO2 = 5 in.

FT3/4

FR3/4

hP

hT3/4

F14

F3/4

The free body diagram of link 4 is shown below where the forces F14 and F34 are unknown in magnitude and direction.

To solve for T2 we first replace F34 with the components FT34 and FR

34 so that their directions are known to be along and transverse to O4B.

To determine FR34 in magnitude and direction we must move to the free

body diagram of link 3.

We now want to evaluate SMO4 = 0 = hT

34*FT34 - hP*P or

FT34 = P(hP/hT

34) = 75 lbf

FR43 FT

43

F23

Shown below is the free body diagram of link 3. The forces at B are FT43 (which has the

same magnitude but opposite direction of FT34) and FR

43 which is unknown in magnitude. The force at A (F23 ) is unknown in magnitude and direction.

hT34

AC

A force polygon gives F23 = 445.9 lbf (1 in. = 100 lbf)

To solve for FR43 we use SMA = 0 = FR

43*hR43 - FT

43*hT43 - S*AC or

+ OF

F23

FR43

FT43

S

hR34

FR43 = (FT

43*hT43 + S*AC)/hR

43 = 408.7 lbf

F32

F12

h32

To determine the magnitude and direction of T2 we use T2 = - F32*h32

-OR- T2 = 2229.5 in-lbf CW

From a force polygon for the entire mechanism F14 = 462.2 lbf (1 in. = 100 lbf)

F12

F14

F12 = -F32 = F23

S

OF +

P F14

F23

F34

static or process force analysis - yıldız teknik ...cdemir/statik_kuv.pdf · 3 static or process...

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