ТММ – 2 (2008) Mark: 4 points

PROBLEM

Make kinematic synthesis of four-bar linkage on the base of three given positions

of crank OA (link1) and three corresponding positions of rocker CD (link 3).

332211DADADA

Use Fig.1 for graphical solution.

Fig.1

SOLUTION

1. We note that from the data given А1D1 ≠ А2D2 ≠ А3D3 . So we can guess

scheme of mechanism shown in Fig. 2.

Fig. 2

2. To locate positions of point A with respect to line segment СD we draw

triangles Δ А1D1С, Δ А2D2С, Δ А3D3С (see Fig. 3).

3. We fix link 3 in the first position and use method of inverse motion to

1

3

O

С

ААА

D

D

D

2

31

1

2

3α1α2α3

φ1φ

2φ3

locate new positions of points А (321

А,А,А ) with respect to this link.

For the second position, to find point 2

А we draw triangle Δ А'2D1С (equal

to Δ А2D2С) taking line segment СD1 for side of the triangle.

The similar operations are made for the third position.

If point B of link 3 does not move and connecting rod AB is of constant

length, then point B should be the centre of circle passing through points321

А,А,А .

4. We find centre В1 of a circle passing through points321

А,А,А . Thus, centre

В1 locates the position of axis of turning pair in the first position, while А1В1 gives

length of the connecting rod.

Fig.3

ω1φ1

О В

Е

DA

С

a

b

α1

2

3

4

5

ТММ – 4 (2008) Mark: 5 points

PROBLEM

Fig. 4

Given:

lOA = 0.3 m; lAC = 0.4 m; lCD = 0.35 m; α = 40˚;

a = 0.5 m; b = 0.9 m; ω1 = 7 1/s; φ1 = 45˚.

Determine velocity and acceleration of point D which belongs to link 5. Kinematic

scheme of the mechanism is shown in Fig. 5.

This problem worths 3 points if being solved in a general form (without

calculations).

Fig. 5

SOLUTION

Fig.6

1. Velocity analysis

1.1 Point A belongs to link 1 rotating about O:

VA = lOA · ω1 = 7 · 0.3 = 2.1 m/s.

1.2 The motion of point B2 can be seen from points A and B3 (the latter

belonging to link 3). Hence, we construct a velocity diagram (see Fig. 6) and

solve the set of vector equations

.

Here , .

All vectors are laid to scale .

1.3 To locate point c on velocity diagram we note, that

Whence,

· 56 = 30.3 mm.

1.4 For point D2:

One should note that:

.

In the velocity diagram we go to point d2 following two paths: one starting at

point b2 and another passing trough point c. Hence, triangle Δ b2cd2 should

appear. All sides of it are proportional to sides of triangle Δ B2СD on the

scheme of mechanism; more other, appropriate sides of the triangles are

mutually perpendicular.

It follows from the speculations above that triangles Δ b2cd2 and Δ B2СD are

similar:

Δ B2СD Δ b2cd2.

So, we find graphically Vd2 = Vd4 = 49·KV = 1.63 m/s.

1.5 To determine angular velocity of link 5 we are to investigate point d5 :

.

Here because of rotation with link 5; due to

relative motion.

Readings values from polygon are:

VB2A = 55· KV = 1.83 m/s VD4D5 = 35· KV = 1.17 m/s

VB2B3 = 29· KV = 0.97 m/s VD4D5 = 35· KV = 1.17 m/s.

With given sizes lAB = 0.74 m, lDE = 0.79 m we get:

ω2 = 1.83/0.74 = 2.47 1/s,

ω5 = 1.17/0.79 = 1.48 1/s.

2. Acceleration analysis

2.1 Because of uniform rotation about O:

aA = ,

2.2 Acceleration of point B2 is expressed in two ways:

(*)

. (**)

Here normal component ,

Coriolis acceleration ,

.

The two sequences (*), (**) are laid off from reference point in acceleration

diagram with scale coefficient . It’s known that

; ; relative acceleration ; direction of

Coriolis component is by definition.

2.3 For acceleration of point C we have:

,

With = + , , making counterclockwise

angle α = arctg with a line parallel to AB.

Let’s compare these results with ones obtained for point B2 :

,

With = + , , making

counterclockwise angle α = arctg with a line parallel to AB.

It is obvious that in the acceleration diagram point c lies in line ab2 and divides

this line segment in the ratio ac : ab2 = AC : AB.

So, point c is located.

2.4 Now we investigate acceleration of point D2 (and D4 ).

,

.

Here

, making counterclockwise angle α =

arctg with a line parallel to DC;

, making counterclockwise angle α =

arctg with a line parallel to DB.

Coming back to point C we find

,

, making counterclockwise angle α =

arctg with a line parallel to CB.

Thus, we again can use similarity of triangles Δ b2cd2 and Δ B2СD to find

graphically cd2 = 24 mm in the acceleration diagram (see Fig. 6).

2.5 With no relative motion, .

2.6 Finally,

. (***)

Here

normal component ,

Coriolis acceleration

Vector sequence in the left-hand side of equation (***) are laid off to scale

from reference point π with , . Vector sum in the right-hand

side starts from point d2,4 with upwards (by the definition of

Coriolis acceleration), .

Intersection of lines representing closing components and provides

desired point d5.

Answer: acceleration of point D5 equals

= 45 · 0.1 = 4.5 m/s2.

Table of readings in

velocity diagram

d2

pb3

Table of readings in

acceleration diagram

d2

ТММ – 3 (2007) Mark: 4 points

PROBLEM

A crank-slider mechanism (see the figure on the

left) has the following dimensions: lOA= 100 mm,

h = 15 mm, BS3 = 10 mm.

Crankshaft OA rotates at constant angular speed

ω1 = const. Slider В, of weight G = 49 N,

encounters a resistance force Pres = 1000 N.

The mechanism is situated in a vertical plane.

Forces of friction are to be neglected.

Determine what moment M1 should be applied to crankshaft О in order to force the

crank move with acceleration aB = 10 m/s2. Find also reactions in kinematic pairs.

SOLUTION

I. Finding M1

Method 1

Kinematical analysis shows that

.

Inertia force acting on slider has the value of

.

Fig. 8

We reduce to crank OA forces acting on slider B.

–Pred · VA = Pin3 · VB –Pres · VB

– –= = 1097 N.

Crank OA applies the force

Py = –Pred

and couple with moment

M1 = Py·lOA =1097·0.1= 109.7 N·m.

Method 2 (method of Zshukovsky’s lever)

The method is applied in the following steps.

a) Draw velocity diagram.

b) Apply all forces turned through 90º at the appropriate points (see Fig. 9).

c) Sum of moments of all applied forces about reference point of velocity

diagram P should be equal to zero.

d) Find Py and M1 from the equation for moments.

Py·pa + Pin3 · pb – Pred · pb = 0

Py =

M1 = 1097 · lOA = 109.7 N·m

Fig. 9

II. Finding reactions in kinematic pairs

For equilibrium of connecting rod AB and slider B (see force polygon in Fig. 10 ,

scale coefficient KP = 10 N/mm):

+

From the diagram:

R12 = 1100 N

R03 = 600 N.

For the equilibrium of link 3

we have

.

Fig. 10

Considering crank OA (see Fig.11), one should note that

M = 1100 · 0.1 = 110 Nm.

Fig.11

ТММ – 4 (2007) Mark: 5 points PROBLEM

Determine what torque MO1 should be applied to

the camshaft O1 in order to overcome forces of

resistance Q = 730 N and inertia force Fin = 450

N, both exerted on the follower. The dimensions

of cam are: b = 45 mm, l = 30 mm, y = 28 mm.

Coefficient of friction in all kinematical pairs is f

= 0.2.

Friction moment of Mfr = 3 Nm resists rotation of

shaft O1. Ignore gravitational force for both cam

and follower. Cam press against follower at the

angle of α = 20˚ (angle of pressure).

Determine: reactions in kinematic pairs, set up

the condition of no self locking of the

mechanism.

Fig. 12

Solution

1. Finding reactions (see Fig. 13 )

Fig. 13

Consider formal equilibrium of link 2.

= 0 (1)

= 0 (2)

(3)

We express from (3):

(4)

And then substitute the value in (1):

(5)

Since and , then = and = .

Now we put (4) and (5) in (2) to yield

(6)

After putting (5) in (4) we have

(7)

Consider now equilibrium of link 1. Since gravitational force is ignored, and the

link 1 rotates with constant angular speed ω1 = const, it is concluded that cam 1 is

in equilibrium under the action of only two forces.

and

1. A reduced moment is found by calculating powers of couples and forces

(8)

2. 2. The condition of no self locking

When denominator in expression (6) becomes zero, self locking of the

mechanism occurs. Magnitude of driving force tends to infinity no matter

value of force Q2.

Hence, no self locking is granted if

(9)

Now we insert numerical data into formulas (5), (7), (6), (8), (9), take

for angle of friction, and get values of forces

in question.

= 3494.8 N

2697.0 N

2690.1 N

699.0 N

419.4 N

3594.0 N

2138.5 N

Fig.14

In formula (9) we have 1.13 , so there is no self locking in the

mechanism.

3.Finding torque MO1

For equilibrium of link 2:

.

Since forces , , intersect in point A, when force must pass through the

same point.

Fig. 15

,

The moment of friction is expressed:

Here f * = (4/π)f = 0.25 , r = 0.01 m, p = Q +F .

The numerical value is then calculated as

Mfr = 0.25· 1180·0.01 = 2.95 Nm.

By law of sines

3. From the condition of equilibrium of cam 1

4.

The denominator in the formula for R12 should not be equal to 0, i.e.

,

.

ТММ – 5 (2007) Mark: 8 points

PROBLEM

Given: In the gearbox shown wheels have the

following number of teeth:

z1 = 20, z2 = 40, , z4 = 60, z5 = 59,

, z7 = 54.

All gears except for gear wheel 5, are cut with

zero shift.

Every gear has the module of m = 1 mm and

angle of pressure α = 20º.

Wheel 1 runs at ω1 = 100 1/s.

Satellites have equal mass of m = 0.2 kg.

Fig. 16

Determine:

1) gear ratio ;

2) angular speed of each wheel;

3) shift coefficient for wheel 5 providing no backlash in engagement of wheels

3 and 5;

4) inertia force of satellites.

SOLUTION

1.Gear ratio

Find gear ratio from wheel 1 to arm H2

= 480

Here

designates gear ratio between arm H1 and wheel 4 provided wheel 5 is

locked;

designates gear ratio between wheel 4´ and arm H2 provided wheel 7

is locked.

2. Angular speed

Use velocity ratio to find ω2 = ω1/2 = 50 1/s = ωH1, ω2 = ωH1.

For the motion transmission between arm H1 and wheel 4 with wheel 5 kept fixed:

.

Hence,

The further calculations yield:

3.95

= 0.208

The latter result can be obtained alternatively by

.

.

3. Shift coefficient

If the condition of coaxiality is satisfied then there will be no

backlash in the engagement of wheels 3 and 5.

=

The shift coefficient is

4.Inertia forces acting on satellites

,

where m3 and m6 are masses of satellites.

From coaxiality condition: ,

.

Hence,

.