Lecture 18 and 19-SFA-DFA

BITS PilaniPilani Campus

Dynamic Force Analysis

Lecture-18 and 19 J S Rathore

BITS Pilani, Pilani Campus

2

Static Force analysis:

cgamF

cgJM

0F

0 M

Dynamic analysis has to be carried out

1) Situations, where inertial forces are comparable

with magnitudes of external forces

2) Operating speeds are high

Two or three force members FBD of links Pin reactions Stress analysis and dimensions of links


3

Coupled Wheel Locomotion: Double crank mechanism

X

X

1. What would be maximum stress in

coupler AB?

2. Safe x-sectional dimension (X-X) ?

3. OR for given dimension and material

properties of AB, what would be

maximum possible speed of

crank/locomotive?

Locomotive is running at 120 Km/hr:


4

Dynamic Force analysis:

To complete analysis

1) Variation in velocity and acceleration

2) DAlembert principle

FBD Two or three force members Pin reactions Stress analysis and dimensions of links


F2

Fn

F1

5

Motion of a rigid body subjected to

a system of forces:

cgamRF

JM

Newtons equation

R

e

CG

cga

Resultant force

Resultant torque


6

Motion of a rigid body subjected to

a system of forces:

0)( cgamF

0)( cgJM

D Alemberts Principle

Newtons equation

0 F

0 M

F2

Fn

F1CG

cgam

J

cga

cgamRF

JM

Resultant force

Resultant torque


7

Analytical method:

G3

aG3

3

3GamF

3

JM

0)( 3 GamF

0)( 3

JM

0 F

0 M

Newtons eq D Alemberts eq

W3

PA

PB


8

D Alembert Principle:

Dynamic problem is converted into static problem by having

Fictitious force (-macg) Inertia Force

Fictitious torque (-J) Inertia Torque

W3

PA

PB

aG3

3

- maG3

- J3

G3


9

Four bar mechanism used in coupled wheel locomotion is as shown in Fig. The

crank and follower, O2A and O4B, are of equal length of 600 mm but of negligible

mass. The coupler AB is a rigid uniform circular rod of length 1000 mm, its total

mass being 12 kg. A torque M acts on the crank O2A, causing coupler AB to move

with a linear acceleration aG3 = 25 45o m/s2 and an angular acceleration AB = 60

rad/sec2 (CCW). Find:

1. The FBD of link AB, O4B & O2A.

2. Pin reactions PA and PB.

3. The minimum diameter of the

coupler required, if the allowable

bending stress of the material of the

coupler is 35 MPa.

4. Shifted inertia force in coupler AB.

Ex 1:


10

Soln: FBD

aG3

3

A

Torque M acts on the crank O2A, causing coupler AB to move with a linear

acceleration aG3 = 25 45o m/s2 and an angular acceleration AB = 60 rad/sec

2

(CCW).


11

Soln: Pin reactions at A and B

A

From FBD of coupler AB

0xF

0AM Gives PBY = 224.93 N in positive y-direction

So PBX = 224.93 N in positive X-direction

So PAX = 12.8 N in negative x-direction

PBY

PBXPAX

PAY

300 N

0BM

So PAY = 104.925 N

05.83tan 1

AX

AYA

P

P


12

Soln: FBD

A


13

For minimum diameter of coupler AB, draw BMD of AB

3

32

d

Mbending

mm 32 d

Soln: Diameter of coupler AB


14

Soln: Shifted inertia force in AB

A

Shifted inertia force 300 N force at a distance

45cos3G

AB

ma

Je

me 2828.045cos300

60 Right from the centre of mass

e

300 N


15

Objectives:

Dynamic force analysis of 4-bar mechanism

aG3

3

Crank and follower with negligible mass External force/moment is acting only on crank

DAlembert principle: Dynamic problem is converted into static problem by having

Pseudo force (-macg) Pseudo torque (-J)

F2M2


16

Dynamic Force Analysis:

F3M3

F4F2M2

M4

aG22

DAlembert principle: Dynamic problem is converted into static problem by having

Pseudo force (-macg) Pseudo torque (-J)


17

FBD

Fj,x = External force in the x-direction applied at CGj

Fj,y = External force in the y-direction applied at CGj

Mj = External Moment on the jth link applied at CGj

Lj = dj + fj

aG22

aG33

aG4

4


18

Link 2

222

2

2

22

2

222222222

2

2

cosf sinf- cosd- sind

1 0 1 0

0 1 0 1

k

a

a

m

P

P

P

P

M

F

F

yg

xg

Ay

Ax

yo

xo

y

x

aG2

2

Fj,x = External force in the x-direction applied at CGj

Fj,y = External force in the y-direction applied at CGj

Mj = External Moment on the jth link applied at CGj

Lj = dj + fj


19

Link 3

233

3

3

3

333333333

3

3

cosf- sinf cosd- sind

1 0 1 0

0 1 0 1

k

a

a

m

P

P

P

P

M

F

F

yg

xg

By

Bx

Ay

Ax

y

x

aG33


20

Link 4

244

4

4

4

4

4444444444

4

4

cosf- sinf cosd- sind

1 0 1 0

0 1 0 1

K

a

a

m

P

P

P

P

M

F

F

yg

xg

yo

xo

By

Bx

y

x

aG4

4


21


22

Assuming gravitational and frictional effects are negligible, Determine all constraint forces

and driving torque (on link 2) required to produce the velocity and acceleration conditions

specified.

m3 = 1.5 kg, m4 = 5 kg

JG2= 0.025 kg m2, JG3

= 0.012 kg m2, JG4= 0.054 kg m2

2 = 0, 3 = 119k rad/s2, 4 = 625k rad/s

2

AG3= [email protected] m/s2, AG4

= 104@233 m/s2

FC = 0.8 j kN.

RAO2= 60 mm

RO4O2= 100 mm

RBA = 220 mm

RBO4= 150 mm

RCO4= RCB =120 mm

RG3A= 90 mm

RG4O4= 90 mm

Ex 1: Problem 14.4 (Uicker-Shigley) p 536


23

Four bar mechanism used in coupled wheel locomotion is shown in

Fig. The crank and follower, O2A and O4B, are of equal length of

600 mm but of negligible mass. The coupler AB is a rigid uniform

circular rod of length 1000 mm and its total mass being 15 kg. If

the locomotive is running at constant speed of 120 km/hr then

find:

1. The linear acceleration (aG3) and angular acceleration (AB) ofcoupler AB.

2. The inertial loading on coupler AB.

Ex 2: Home assignment

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Lecture 18 and 19-SFA-DFA