8.acceleration analysis

Kinematic Analysis: Scope

•Need to know the dynamic forces to be able to compute stresses in the components

• Dynamic forces are proportional to acceleration (Newton second law)

• Goal shifts to finding acceleration of all the moving parts in the assembly

•In order to calculate the accelerations:

• need to find the positions of all the links , for all increments in input motion

• differentiate the position eqs. to find velocities, diff. again to get accelerations

Acceleration analysis

Acceleration: Rate of change of velocity with respect to time

The change of velocity, as the body moves from P to Q can be determined by drawing the

vector triangle opq, in which op and oq represent the velocities at P and Q, respectively.

pq represents the change of velocity in time δt.

pq can be resolved into two components, namely,

- px (parallel to op), and

- xq (perpendicular to op).

Acceleration analysis

The change of velocity (pq) has two components (px and xq) which are mutually

perpendicular, hence, the rate of change of velocity, that is, acceleration will also have two

mutually perpendicular components.

AT = r α AN = r ω2 = V2

T

r

Acceleration analysis

The change of velocity (pq) has two components (px and xq) which are mutually

perpendicular, hence, the rate of change of velocity, that is, acceleration will also have two

mutually perpendicular components.

If the body moves with a uniform velocity, then dv/dt=d(w)/dt=α=0; and the body has only AN

If the body moves on a straight path, r is infinite, and v2/r=0; and the body has only AT

Straight path Uniform velocity

Only tangential acceleration Only centripetal acceleration

Acceleration analysisGraphical method Vector loop method

Consider a rigid link AB. If A is assumed to be fixed then the only possible motion for B is rotation

about A (as the centre).

Acceleration analysis: Graphical Method

Acceleration of a point on a link

2 things are known:

I . the acceleration of A.

ii. Direction of VB

The acceleration of B can be determined

in magnitude and direction, graphically

Completely known

Just direction is known, perpendicu-

lar to the radial component, so….

…..just draw the direction

Just direction is known, parallel to

the path of B ….so just draw aB…….

…………..parallel to VB

Acceleration analysisAcceleration in the slider crank mechanism

Uniform ω

MD2

MD1D1

D2

Acceleration analysis: Coriolis component

When a point on one link slides along another rotating link, then a component of

acceleration, called Coriolis component of acceleration comes into play

Acceleration analysis: Coriolis component

Change of velocity: Along & perpendicular to OA due to

linear (sliding) velocity tangential (rotational) velocity

Acceleration analysis: Coriolis component

Acceleration components

of the slider (B)

wrt the coincident point (C) on the rotating link (2)

Acceleration components of the coincident point (C) on the rotating link (2)

wrt the origin (pivot:O)

Acceleration components of the slider (B)

wrt the origin (pivot:O)

minus

Stage-I

Stage-II

Acceleration analysis: Coriolis componentChange of velocity: Along & perpendicular to OP due to

linear (sliding) velocity tangential (rotational) velocity

Along OAPerpendicular to OA

Stage-I

Acceleration analysis: Coriolis componentChange of velocity: Along & perpendicular to OP due to

linear (sliding) velocity tangential (rotational) velocity

Along OA Perpendicular to OA

Stage-I

Acceleration analysis: Coriolis component

Change of velocity: Along & perpendicular to OP due to

Along OP Perpendicular to OP

linear (sliding) velocity

tangential (rotational) velocity

Total change in velocity

along radial direction (along OP)

Total change in velocity in

tangential direction (perp. to OP)

Stage-I

Acceleration analysis: Coriolis component

Summary: Acceleration components of the slider (B) wrt the origin (pivot:O)

Stage-I

Acceleration analysis: Coriolis component

Summary: Acceleration components of the coincident point (C) on the rotating link (2) wrt O

Stage-II

Acceleration analysis: Coriolis componentStage-I – Stage II

Acceleration components

of the slider (B)

wrt the coincident point (C) on the rotating link (2)

Acceleration analysis: Coriolis component

The tangential component of acceleration of the slider (B) with respect to the coincident point (C) on the

rotating link is known as coriolis component of acceleration and is always perpendicular to the link

Direction of coriolis component

of acceleration is obtained by

rotating V at 90º, about its origin,

in the direction of ω

Acceleration analysis: Problem-IA mechanism of a crank and slotted lever quick return motion is shown in the figure. If the

crank rotates counter clockwise at 120 r.p.m, then for the configuration shown, determine:

- the velocity and acceleration of the ram.

- the angular acceleration of the slotted lever. Crank AB=50mm, link CD=200mm

Slotted arm OC=700mm

Velocity AB BB’ B’ O CD OD

Radial

Tangential

Acceleration AB BB’ B’O CD OD

Radial

Tangential

___

VBA

perp. to AB

___

___ ___ ___VBB’

parallel to BB’

VB’O

perp. to B’O

VCD

perp. to CD

VD=VDo

parallel to

motion of D

___No α

M+D___

B:slider

B’: Coincident point

Acceleration analysis: Problem-IA mechanism of a crank and slotted lever quick return motion is shown in the figure. If the

crank rotates counter clockwise at 120 r.p.m, then for the configuration shown, determine:

- the velocity and acceleration of the ram.

- the angular acceleration of the slotted lever. Crank AB=50mm, link CD=200mm

Slotted arm OC=700mm

Velocity AB BB’ B’O CD OD

Radial

Tangential

Acceleration AB BB’ B’O CD OD

Radial

Tangential

___

VBA

perp. to AB

___

___ ___ ___VBB’

parallel to BB’

VB’O

perp. to B’O

VCD

perp. to CD

VD=VDo

parallel to

motion of D

___No α

M+D ___

B:slider

B’: Coincident point

Acceleration analysis: Problem-IDetermine: the velocity and acceleration of the ram;

: the angular acceleration of the slotted lever.

1

1

Locating B’

2(D/b)

3(D/O)

2

3

B’ is under

pure rotation

about O: VB’O

is perpendicular

to OB’

Path of B’ along OC

2

3

Locating C

4

5

4

5

Locating D

4(D/C)

5(D/O)

D is under pure rotation about C:

VDC is perpendicular to CD

Horizontal

path of D

Acceleration analysis: Problem-IFind: the velocity and acceleration of the ram;

-the angular acceleration of the slotted lever.

Acceleration analysis: Problem-I

Acceleration analysis: Problem-I

Coriolis

direction

Direction of coriolis component

of acceleration is obtained by

rotating V at 90º, about its origin,

in the direction of ω

Mag not

known

Accn. of B w.r.t the

coincident point B’

Locating image of B’, i.e., b’’

Accn. of B’ w.r.t O

Locating the image

of D wrt the image

of C

C lies on

OB’ produced

Acceleration analysis: Problem-I

Locating the image

of D wrt the image

of C and O

Acceleration analysis: Problem-I

Acceleration analysis: Complex No. Notation; Vector Loop Equation

90º rotation of RPA

in the direction of α

Opposite

to RPA

AP=APA, since, point A

Is the origin of the GCS

Acceleration analysis: Complex No. Notation; Vector Loop Equation

Acceleration analysis: 4-bar pin jointed linkage

Acceleration analysis: 4-bar slider crank

Acceleration analysis: Coriolis component of acceleration

Acceleration analysis: Coriolis

Acceleration analysis: Coriolis