Mechanics of Machines MCB3043

Lecture 18

T. Nagarajan

Outline

Inertial Force

Dynamic Force Analysis

Inertial Force

maF

0 maF

maFi

0 iFF

Inertial force is defined as:

d’Alembert’s Principle

By considering inertial force as one of the forces in the

system, the system is considered to achieve dynamic

equilibrium.

Inertial Force

Inertial force is not a real force (Same

concept as a centrifugal ‘force’)

Inertial ‘force’ opposes acceleration

Example:

As a car is accelerated, the head of the

passenger appeared to be lurched

backward,

As the car is braked, the head of the

passenger appeared to be lurched forward.

Topics for Self-Revision

Mass & Weight

Centre of Gravity

Mass Moment of Inertia (inc radius of

gyration & parallel axis theorem)

Dynamic Force Analysis

Dynamic force analysis should be

performed to determine the forces on

each link of a linkage as it is moving.

Static force analysis should be done

prior to DFA

Need info on the mass, centre of gravity,

mass moment of inertia of each link.

Example 1

The compressor mechanism is driven clockwise by a dc electric motor at a constant rate of 600 rpm.

The piston weighs 3 N and the coefficient of friction between the piston and the compressor cylinder is 0.1. The weight of all other links is negligible.

At the position shown, as the cylinder pressure is 300 kPa, determine the torque required from the motor to operate the compressor.

200 mm 40 mm

50 mm

300 kPa

Procedure For Solution`

200 mm 40 mm

50 mm

300 kPa A

B

C

Torque

required to

turn link AB

Force exerted by

link BC on link

AB at joint B

Force exerted by

the piston on link

BC at joint C

Analyse forces by

drawing FBD and

apply Eq of M to

each link.

Inertial force

Frictional force

HP Gas force

Fi =mCaC

Ff =μNC

FG =pGAC

FBD of All Links

F23

F21Y

F21X

F32

F34

F43

Fi Ff

F41=NC

FG

mCg

200 mm 40 mm

50 mm

300 kPa

T21

`Kinematic Diagram & Velocity

Analysis

vC = vB + vC/B

vB

ωAB= 600 rpm cw = 62.8 rad/s cw

vB = ωABrBA = 62.8 x 0.050 = 3.14 m/s

Measuring

vC = 2.00 m/s

vC/B = 2.08 m/s

vB

ωBC = vC/B / rCB = 2080 / 200 = 10.4 rad/s cw

A

B

C

aC = aB

n + aC/B

n + aC/Bt

aC/Bn = ω2

2rBC=10.42 x 0.2 = 21.6 m/s2

aBn = ωAB

2rAB= 62.82 X 0.05 = 197.2 m/s2

`Acceleration Analysis

aBn

A

B

C

aC = aB

n + aC/B

n + aC/Bt

aC/Bn = 21.6 m/s2

aBn = 197.2 m/s2

Measuring

aC = 135.2 m/s

aC/Bt = 147.2 m/s

aBn

aC

aC/Bt

aC/Bn

aC/Bn

Force Analysis on the Piston

F43

Fi Ff

NC

FG

mCg Σ F = 0

Σ M = 0

Note: d’Alembert’s

principle applies here.

Σ FX = F43cos θ - Fi - FG-Ff =0

Σ FY = -F43sin θ - mCg +NC =0

Fi = -mC aC = -3/9.8 x 135.2 = -41.4 N = 41.4 N

Ff =μNC = 0.1 NC

FG =pGAC = (300 x 103) (π x ¼ x 0.042 ) = 377.0 N

From the kinematic diagram, = 11º

F43cos 11º - 41.4 - 377-0.1 NC =0

-F43sin 11º - 3 +NC =0 F43 = 435.0 N

Force Analysis of the Crank and

Connecting Arm

F32

F34

F34= -F43= 435 N

F32= -F34= 435 N

F21Y

F21X

T21

F23= -F32 = 435 N

50º

F23 11º

-T21 + F23 sin(50º-11º) x 0.050 = 0

T21 = 13.7 Nm cw

Inertial Torque

IM

0 IM

ITi

0 iTM

Inertial torque is defined as:

By considering the inertial torque as one of the moments in

the system, the system is considered to achieve dynamic

equilibrium. The inertial torque direction is the opposite

direction to the angular acceleration, .

Σ M = 0

Example 2

The crank in Example 1 has a weight of 10 N. The centre of

gravity is 20 mm from the fixed pivot A. Recalculate the

torque required to operate the compressor if the crank has

an angular acceleration of 500 rad/s2 and has a speed of

600 rpm at the position shown.

200 mm 40 mm

50 mm

300 kPa

A

B

C

10 N

Ti

`Acceleration Analysis

aBn

A

B

C

aC = aB

n + aB

t + aC/Bn + aC/B

t

aC/Bn = 21.6 m/s2

aBn = 197.2 m/s2

Measuring

aC = 152 m/s

aC/Bt = 130 m/s

aBn

aC

aC/Bt

aC/Bn

aC/Bn

aBt = rAB=500 x 0.05= 25 m/s2

aBt

aBt

Force Analysis on the Piston

F43

Fi Ff

NC

FG

mCg Σ F = 0 Σ M = 0

Σ FX = F43cos θ - Fi - FG-Ff =0

Σ FY = -F43sin θ - mCg +NC =0

Fi = -mC aC = -3/9.8 x 152 = -46.5 N = 46.5 N

Ff = 0.1 NC

FG = 377.0 N

From the kinematic diagram, = 11º

F43cos 11º - 46.5 - 377-0.1 NC =0

-F43sin 11º - 3 +NC =0 F43 = 440.3 N

Force Analysis of the Crank and

Connecting Arm

F32

F34

F34= -F43= 440.3 N

F32= -F34= 440.3 N

F21Y

F21X

T21

F23= -F32 = 440.3 N

50º

F23 11º 0.106+ 10(0.02cos 50º) -T21 +

440.3 sin 39º(0.05)= 0

T21 = 14.1 Nm cw

Σ M = Ti +m21g (rcg cos 50º) -T21 + F23 sin(50º-11º)r21= 0

ccw Nm 106.0Nm 106.050005.0)( 2

8.910

1212

121 mlITi

Ti

Example 3 (Q 14-16)

A

F

E

D

C B

30°

Determine the motor torque T21

required at the instant shown to turn

the crank AB at 120 rpm ccw.

T21

μ = 0.15

4 kg

Weights of all links

are negligible. AB= 0.1 m

BC = 0.4 m

CD = 0.3 m

CE = 0.32 m

DE = 0.6 m

EF = 0.65 m

FAY = 0.4 m

ADY = 0.2 m

ADX = 0.37 m

Example 3: FBD of Box

Fi

Ff

NC

Fbox

mg Fi =-maF

Ff =μNC

Σ FX = Fi +Ff -Fbox = 0

Σ FY = NC - mg =0

FBD of the box

NC = mg = 4 x 9.8 = 39.2 N

Ff =μNC = 0.15 x 39.2 = 5.9 N

-4aF +5.9 -Fbox = 0

To determine Fbox , need to determine aF first, by performing

acceleration analysis (and velocity analysis).

B C

D

A

Kinematic Analysis

F E

vC = vB + vC/B

vF = vE + vF/E

ωCDE = vC / rCD

vE = ωCDErDE

Get vC by graphical method

Get vF by graphical method.

Calculate ω for all links and

calculate the normal

components of acceleration.

αCDE = aCt / rCD

aCn

+aCt = aB

n + aC/B

n + aC/B

t

aEt = α CDErDE

aF = aEn+ aE

t + aF/E

n+ aF/Et

3°

Example 3

A

F

E

D

C B

30°

Measure the angular positions of the links from the kinematic

diagram. These angles are required in constructing free-body

diagrams of all links in analysis of the forces.

T21

μ = 0.15

4 kg

1°

66°

10°

Example 3: FBD links 5 & 6

Fbox

F65 1°

F56

F54 F65 ≈ Fbox

F54 =F65

2-force member

F41Y

F41X

F45

T21

Σ MD = F45 rED cos (10-1)° + F43 rCD cos (90-66+3)° = 0

F43

D

1°

3°

F32

F23

F21X

F21Y

3°

30°

A

E

C B

C

B 10°

F34

66°

F45 =-F54

F32 =F43

Σ MA = T21 - F23 rAB cos (90-30+3)° = 0

Example 3: FBD of Links 2,3 & 4

Tutorial Week 13

14-18

11-12