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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,
.
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