ME
3491-THEORY OF MACHINES
UNIT – I KINEMATICS OF MECHANICS
- Explain the three inversions of double slider-crank chain with
suitable example. (APR/MAY 2023)
The following three inversions of a double
slider crank chain are important from the subject point of view:
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1.Elliptical trammels. It is an instrument used for drawing ellipses. This inversion
is obtained
by fixing the slotted plate (link 4), as shown in
Fig. The fixed plate or link 4 has two
straight grooves cut in it, at right angles to each other.
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2.Scotch yoke mechanism.
This mechanism is used for
converting rotary motion into a reciprocating motion. The inversion is
obtained by fixing either the link 1 or link 3. In Fig. link 1 is fixed.
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3.
Oldham’s coupling.
An Oldham’s coupling is used for connecting two parallel shafts whose
axes are at a small distance apart. The shafts are coupled in such a way that
if one shaft rotates, the other shaft also rotates at the same speed. This
inversion is obtained by fixing the link 2, as shown in Fig
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- The lengths of crank and connecting rod of a horizontal
reciprocating engine are 125 mm and 500 mm respectively. The crank is
rotating at 600 rpm. When the crank has turned 45° from inner dead centre,
find analytically, (1) the velocity and acceleration of the slider and (2)
the angular velocity and angular acceleration of the connecting rod.
(APR/MAY
2023)
Solution. Given : NBO = 600
r.p.m. or ωBO
= 2x π × 600/60 = 62.83 rad/s; OB = 125 mm =0.125 m ; B A = 600 mm = 0.6 m
We know that linear velocity of B with
respect to O or velocity of B,
vBO = vB = ωBO × OB = 62.83 × 0.125 = 7.85 m/s
1.
Velocity and acceleration of the slider
We know that ratio of the length of
connecting rod and crank,
n = l / r = 0.5 / 0.125 = 4
∴ Velocity of
the slider,
Acceleration of the slider,
Angular
velocity and angular acceleration of the connecting rod
We know that
angular velocity of the connecting rod,
angular acceleration
of the connecting rod,
- Derive expressions for displacement, velocity and acceleration for
a tangent cam operating radial-translating roller follower;
(i)
When the contact is on straight flank, and
(ii)
When the contact is on circular nose.
1.
When the roller has contact with the straight flanks ; and
2.
When the roller has contact with the nose. (APR/MAY 2023)
Let r1 = Radius of the base circle or
minimum radius of the cam,
r2 = Radius of the roller,
r3 = Radius of nose,
α = Semi-angle of action of cam or angle of
ascent,
θ = Angle turned by the cam from the
beginning of the roller displacement,
φ = Angle turned by the cam for contact of
roller with the straight flank, and
ω = Angular velocity of the cam.
1. When the roller has contact with
straight flanks. A roller having contact with straight flanks is shown in Fig. .
The point O is the centre of cam shaft and the point K is the centre of nose.
EG and PQ are straight flanks of the cam. When the roller is in lowest
position,
2.
When the roller has contact with the nose.
A roller having contact with the circular
nose at G is shown in Fig 20.41. The centre of roller lies at D on the pitch
curve. The displacement is usually measured from the top position of the
roller, i.e. when the roller has contact at the apex of the nose (point H) and
the centre of roller lies at J on the pitch curve.
- What do you mean by Grashoff chain? Draw the inversions of a four
bar mechanism and explain clearly how they are differing from each other.
The type of motion is a function of the
link lengths. Grashof's theorem (or Grashof’s rule) gives the criteria for
these various conditions as follows:
Let us identify the link lengths in a
four-bar chain as:
l
+ s < p + q
l= length of the longest link
s= length of the shortest link
p,q = length of the two intermediate links
The following statements are valid (stated
without proof. One can prove these statements by using the input-output
equation of a four-bar See Appendix AIII for the proof of the theorem).
If l
+ s < p + q (if the sum of the
lengths of the shortest and the longest links is less than the sum of the two
intermediate links)
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a, Two different crank-rocker mechanisms are
possible. In each case the shortest link is the crank, the fixed link is
either of the adjacent links
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b) Two different crank-rocker mechanisms are
possible. In each case the shortest link is the crank, the fixed link is
either of the adjacent links
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c. One double-crank (drag-link) is possible when the
shortest link is the frame.
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One double-rocker mechanism is possible when the
link opposite the shortest link is the frame.
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- Explain crank and slotted lever quick return motion mechanism with
a neat diagram.
This mechanism is mostly used in shaping
machines, slotting machines and in rotary internal combustion engines.
In this mechanism, the link AC (i.e. link 3) forming the
turning pair is fixed, as shown in Fig. 5.26. The link 3 corresponds to the
connecting rod of a reciprocating steam engine. The driving crank CB revolves
with uniform angular speed about the fixed centre C. A sliding block attached
to the crank pin at B slides along the slotted bar AP and thus causes AP to
oscillate about the pivoted point A. A short link PR transmits the motion from
AP to the ram which carries the tool and reciprocates along the line of stroke
R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC
produced.
In
the extreme positions, AP1 and AP2 are tangential to the circle and the cutting
tool is at the end of the stroke. The forward or cutting stroke occurs when the
crank rotates from the position CB1 to CB2 (or through an angle β) in the
clockwise direction. The return stroke occurs when the crank rotates from the
position CB2 to CB1 (or through angle α) in the clockwise direction. Since the
crank has uniform angular speed,
therefore,
Since the tool travels a distance of R1 R2
during cutting and return stroke, therefore travel of the tool or length of
stroke
- A crank and slotted lever
mechanism used in a shaper has a centre distance of 300 mm between the
centre of oscillation of the slotted lever and the centre of rotation of
the crank. The radius of the crank is 120 mm. Find the ratio of the time
of cutting to the time of return
stroke.
Solution. Given : AC = 300 mm ; CB1 = 120
mm
The extreme positions of the crank are
shown in Fig. We know that
- The link lengths of a quadratic cycle chain are shown in Fig. below
Find all the inversions of the given chain and classify them.
s = length of shortest link = 10 cm
l = length of longest link = 40 cm
p & q = length of other
two links
= 25 + 30 = 55 cm
Here s + l = 10+40 = 50 cm
p + q = 55 cm
since, (i) s + l <= p +q
(ii) any one of the link adjacent to the shortest link is fixed therefore the
give mechanism is a crank-rocker
mechanism.
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Any one of the link
adjacent to the shortest link is fixed
therefore the give mechanism is a crank-rocker
mechanism.
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any one of the link adjacent to the shortest link is fixed therefore the
give mechanism is a crank-rocker
mechanism.
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The shortest link is fixed, then a double-crank
mechanism obtained.
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The link opposite to the shortest link is fixed,
then the a double-rocker mechanism is obtained.
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- A crank-rocker mechanism has a 70 mm fixed link, a 20 mm crank, 50
mm coupler and a 70 mm rocker. Determine the maximum and minimum transmission
angles.
A crank-rocker mechanism has a 70 mm fixed link, a 20
mm crank, 50 mm coupler and a 70 mm rocker.
First let us identify the type of mechanism using grashofs law.
Here s = 20 mm, l = 70 mm, and p = 70, & q =50 mm
Since s + l <= p +q
20+70 <= 70+50
90 <120 – Grashofs law satisfied
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b2 + c2 - 2bc cos μmax =(a+d)2
702 + 502 - 2 x70x50 cos μmax =(20+70)2
7400-7000 cos μmax = 8100
cos μmax = 0.1
μmax = cos -1 0.1 = 93.6230
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b2 + c2 - 2bc cos μmax =(a-d)2
702 + 502 - 2 x70x50 cos μmax =(20+70)2
7400-7000 cos μmax = 8100
cos μmax = 0.1
μmax = cos -1 0.1 = 93.6230
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- A mechanism, as shown in
Fig. 6.15, has the following dimensions:
OA
= 200 mm; AB = 1.5 m; BC = 600 mm; CD = 500 mm and BE = 400 mm. Locate all the
instantaneous centres. If crank OA rotates uniformly at 120 r.p.m. clockwise,
find 1. the velocity of B, C and D, 2. the angular velocity of the links AB, BC
and CD.
Solution.
Given : NOA = 120 r.p.m.
or ωOA = 2 π × 120/60 = 12.57 rad/s
Since the length of crank OA = 200 mm = 0.2
m, therefore linear velocity of crank OA,
vOA = vA = ωOA × OA = 12.57 × 0.2 = 2.514
m/s
Location of instantaneous centres
The
instantaneous centres are located as discussed below:
- Since the mechanism consists of six links
(i.e. n = 6), therefore the number of instantaneous centres,
- N = n(n-1)/2 = 6(6-1)/2 =15
- Make a list of all the instantaneous
centres in a mechanism. Since the mechanism has 15 instantaneous centres,
therefore these centres are listed in the following book keeping table.
3. Locate the fixed and permanent instantaneous centres by
inspection. These centres are I12 I23, I34, I45, I56, I16 and I14 as shown in
Fig. 6.16.
4. Locate the remaining neither fixed nor permanent instantaneous
centres by Aronhold Kennedy’s theorem. Draw a circle and mark points equal to
the number of links such as 1, 2,
3, 4, 5 and 6 as shown in Fig. 6.17. Join the points 12,
23, 34, 45, 56, 61 and 14 to indicate the centres I12, I23, I34, I45, I56, I16 and
I14 respectively.
5. Join point 2 to 4 by a dotted line to form the
triangles 1 2 4 and 2 3 4. The side 2 4, common to both triangles, is responsible
for completing the two triangles. Therefore the instantaneous centre I24 lies
on the intersection of I12 I14 and I23 I34 produced if necessary. Thus centre I24
is located. Mark number 8 on the dotted line 24 (because seven centres have
already been located).
Answer : TOM -KURMI Page No:138 (130)
- A four bar mechanism has
the following dimensions :
DA
= 300 mm ; CB = AB = 360 mm ; DC = 600 mm. The link DC is fixed and the angle
ADC is 60°. The driving link DA rotates uniformly at a speed of 100 r.p.m.
clockwise and the constant driving
torque has the magnitude of 50 N-m. Determine the velocity of the point B and
angular velocity of the driven link CB.
Also find the actual mechanical advantage and the resisting torque if the efficiency of the mechanism is 70 per
cent.
Solution.
Given : NAD = 100 r.p.m.
or ωAD = 2 π × 100/60 = 10.47 rad/s ; TA = 50 N-m
Since the length of driving link, DA = 300
mm = 0.3 m, therefore velocity of A with respect to D or velocity of A (because
D is a fixed point),
vAD = vA = ωAD × DA = 10.47 × 0.3 = 3.14
m/s . . . (Perpendicular to DA)
Velocity of point B
First of all draw the space diagram, to
some suitable scale, as shown in Fig. 7.25 (a). Now the velocity diagram, as
shown in Fig. 7.25 (b), is drawn as discussed below :
1. Since the link DC is fixed, therefore
points d and c are taken as one point in the velocity diagram. Draw vector da
perpendicular to DA, to some suitable scale, to represent the velocity of A with
respect to D or simply velocity of A (i.e. vAD or vA) such that
vector da = vAD = vA = 3.14
m/s
2. Now from point a, draw vector ab
perpendicular to AB to represent the velocity of B with respect to A (i.e.
vBA), and from point c draw vector cb perpendicular to CB to represent the
velocity of B with respect to C or simply velocity of B (i.e. vBC or vB). The
vectors ab and cb intersect at b.
By measurement, we find that velocity of
point B,
vB = vBC = vector cb = 2.25 m/s Ans.
Angular velocity of the driven link CB
Since CB = 360 mm = 0.36 m, therefore
angular velocity of the driven link CB,
Actual mechanical advantage
We know that the efficiency of the
mechanism, η = 70% = 0.7 . . . (Given)
∴Actual
mechanical advantage,
Resisting torque
Let TB = Resisting torque.
We know that efficiency of the mechanism
(η),
Answer : TOM -KURMI Page No:170(162)
- It is required to set out
the profile of a cam to give the following motion to the reciprocating
follower with a flat mushroom contact face :
(i)
Follower to have a stroke of 20 mm during 120° of cam rotation;
(ii)
Follower to dwell for 30° of cam rotation;
(iii)
Follower to return to its initial position during 120° of cam rotation; and (iv)
Follower to dwell for remaining 90° of cam rotation.
The
minimum radius of the cam is 25 mm. The out stroke of the follower is performed
with simple harmonic motion and the return stroke with equal uniform
acceleration and retardation.
Answer: TOM -KURMI Page No:812(804)
- A symmetrical cam with convex flanks operates a flat-footed
follower. The lift is 8 mm, base circle radius 25 mm and the nose radius
12 mm. The total angle of the cam action is 120°. 1. Find the radius of convex flanks, 2.
Draw the profile of the cam, and 3. Determine the maximum velocity and the
maximum acceleration when the cam shaft rotates at 500 r.p.m.
Solution.
Given : x = JK = 8 mm ; r1
= OE = OJ = 25 mm ; r2 = QF = QK = 12 mm ;
2α = ∠EOG =120° or
α = ∠EOK = 60° ; N = 500 r.p.m. or ω = 2π × 500/60 = 52.37 rad/s
1. Radius of
convex flanks
Let R = Radius
of convex flanks = PE = P G′
A symmetrical
cam with convex flanks operating a flat footed follower is shown in Fig.
20.47. From
the geometry of the figure,
OQ =OJ + JK
−QK = r1 + x − r2
= 25 + 8 – 12
= 21 mm
PQ = PF −QF =
PE −QF = (R – 12) mm
and OP = PE
−OE =(R − 25) mm
Now consider
the triangle OPQ. We know that
(PQ)2 = (OP)2
+ (OQ)2 − 2OP×OQ×cosβ
(R −12)2 = (R
− 25)2 + (21)2 − 2(R − 25)21cos (180° − 60°)
R2 − 24R +144
= R2 −50 R + 625 + 441+ 21R −525
– 24 R + 144 =
– 29 R + 541 or 5 R = 397
∴ R = 397/5 = 79.4 mm Ans.
We know that
maximum velocity,
vmax = ω(R −
r1)sin φ = 52.37(79.4 − 25)sin15.65° = 770 mm/s
= 0.77 m/s
Ans.
and maximum
acceleration,
amax = ω
2 (R − r1) = (52.37) 2 (79.4 − 25) = 149200 mm/s2= 149.2
m/s2 Ans.
Answer: TOM -KURMI Page No:829(821)
- Layout the profile of a cam operating a roller reciprocating
follower for the following data. Lift of follower = 30mm; angle during the
follower rise period = 120◦Angle during the follower after rise = 30◦ ;
Angle during the follower return period = 150◦ ; angle during which follower dwell
after return = 60◦ ; minimum radius of cam = 25mm; roller diameter = 10mm.
the motion of follower is uniform
acceleration and deceleration during the rise and return period.
- A cam is to designed for a
knife edge follower with the following data: Cam lift = 40mm during 90◦ of
cam rotation with SHM, dwell for the next 30◦, during the next 60◦ of cam
rotating, the follower returns to its originals position with SHM, dwell
during the remaining 180◦. Draw the profile of the cam when the line of
stroke is offset 20mm from the axis of the cam shaft. The radius of the
base circle of the cam is 40mm.
- A cam is to give the following motion to a knife-edged
follower:
1.
Outstroke during 60° of cam rotation ; 2. Dwell for the next 30° of cam
rotation ; 3. Return stroke during next 60° of cam rotation, and 4. Dwell for
the remaining 210° of cam rotation. The stroke of the follower is 40 mm and the
minimum radius of the cam is 50 mm. The follower moves with uniform velocity
during both the outstroke and return strokes. Draw the profile of the cam when
(a) the axis of the follower passes through the axis of the cam shaft, and(b) the axis of the follower is
offset by 20 mm from the axis of the cam shaft.
- 13. A cam is to be designed for a knife edge follower with the
following data :1. Cam lift = 40 mm during 90° of cam rotation with simple
harmonic motion. 2. Dwell for the next 30°. 3. During the next 60° of cam
rotation, the follower returns to its original position with simple
harmonic motion. 4. Dwell during the remaining 180°.
Draw
the profile of the cam when (a) the line of stroke of the follower passes
through the axis of the cam shaft, and
(b)
the line of stroke is offset 20 mm from the axis of the cam shaft.
The
radius of the base circle of the cam is 40 mm. Determine the maximum velocity
and acceleration of the follower during its ascent and descent, if the cam
rotates at 240 r.p.m.
- A cam, with a minimum
radius of 25 mm, rotating clockwise at a uniform speed is to be designed
to give a roller follower, at the end of a valve rod, motion described
below : 1. To raise the valve through 50 mm during 120° rotation of the
cam ; 2. To keep the valve fully raised through next 30°; 3. To lower the
valve during next 60°; and 4. To keep the valve closed during rest of the
revolution i.e. 150° ; The diameter of the roller is 20 mm and the
diameter of the cam shaft is 25 mm. Draw the profile of the cam when (a)
the line of stroke of the valve rod passes through the axis of the cam
shaft, and (b) the line of the stroke is offset 15 mm from the axis of the
cam shaft. The displacement of the valve, while being raised and lowered,
is to take place with simple harmonic motion. Determine the maximum
acceleration of the valve rod when the cam shaft rotates at 100 r.p.m.
Draw the displacement, the velocity and the acceleration diagrams for one
complete revolution of the cam.
- A cam drives a flat reciprocating follower in the following manner
: During first 120° rotation of the cam, follower moves outwards through a
distance of 20 mm with simple harmonic motion. The follower dwells during
next 30° of cam rotation. During next 120° of cam rotation, the follower
moves inwards with simple harmonic motion. The follower dwells for the
next 90° of cam rotation. The minimum radius of the cam is 25 mm. Draw the
profile of the cam.
- Design a cam for operating the exhaust valve of an oil engine. It
is required to give equal uniform acceleration and retardation during
opening and closing of the valve each of which corresponds to 60° of cam
rotation. The valve must remain in the fully open position for 20° of cam
rotation. The lift of the valve is 37.5 mm and the least radius of the cam
is 40 mm. The follower is provided with a roller of radius 20 mm and its
line of stroke passes through the axis of the cam.
- A cam rotating clockwise at a uniform speed of 1000 r.p.m. is
required to give a roller follower the motion defined below : 1. Follower
to move outwards through 50 mm during 120° of cam rotation,
2.
Follower to dwell for next 60° of cam rotation, 3. Follower to return to its
starting position during next 90° of cam rotation, 4. Follower to dwell for the
rest of the cam rotation. The minimum radius of the cam is 50 mm and the
diameter of roller is 10 mm. The line of
stroke of the follower is off-set by 20 mm from the axis of the cam
shaft. If the displacement of the follower takes place with uniform and equal
acceleration and retardation on both the outward and return strokes, draw
profile of the cam and find the maximum velocity and acceleration during out
stroke and return stroke.
- Construct the profile of a cam to suit the following
specifications : Cam shaft diameter = 40 mm ; Least radius of cam = 25 mm
; Diameter of roller = 25 mm; Angle
of lift = 120° ; Angle of fall = 150° ; Lift of the follower = 40 mm ;
Number of pauses are two of equal interval between motions.
- It is required to set out
the profile of a cam to give the following motion to the reciprocating
follower with a flat mushroom contact face :
(i)
Follower to have a stroke of 20 mm during 120° of cam rotation ;
(ii)
Follower to dwell for 30° of cam rotation ;
(iii)
Follower to return to its initial position during 120° of cam rotation ; and
(iv)
Follower to dwell for remaining 90° of cam rotation.
The
minimum radius of the cam is 25 mm. The out stroke of the follower is performed
with simple harmonic motion and the return stroke with equal uniform
acceleration and retardation.
- 20. It is required to set
out the profile of a cam with oscillating follower for the following
motion :
(a)
Follower to move outward through an angular displacement of 20° during 90° of
cam rotation ; (b) Follower to dwell for 45° of cam rotation ; (c) Follower to
return to its original position of zero displacement in 75° of cam rotation ;
and (d) Follower to dwell for the remaining period of the revolution of the
cam. The distance between the pivot centre and the follower roller centre is 70
mm and the roller diameter is 20 mm. The minimum radius of the cam corresponds
to the starting position of the follower as given in (a). The location of the
pivot point is 70 mm to the left and 60 mm above the axis of rotation of the
cam. The motion of the follower is to take place with S.H.M. during out stroke
and with uniform acceleration and retardation during return stroke.
- Explain inversion of Four
bar mechanism with neat sketch
Inversions
of Four Bar Chain Though
there are many inversions of the four bar chain, yet the following are
important from the subject point of view :
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1. Beam engine (crank and lever mechanism).
A part of the mechanism of a beam engine (also known as crank and
lever mechanism) which consists of four links, is shown in Fig.
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2.
Coupling rod of a locomotive (Double crank mechanism). The mechanism of a
coupling rod of a locomotive (also known as double crank mechanism) which
consists of four links, is shown in Fig.
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3.
Watt’s indicator mechanism (Double lever mechanism). A *Watt’s indicator
mechanism (also known as Watt's straight line mechanism or double lever
mechanism) which consists of four links
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- Explain inversion of Single
slider crank chain with neat sketch.
A single slider crank chain is a
modification of the basic four bar chain. It consist of one sliding pair and
three turning pairs.
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1.
Pendulum pump or Bull engine. In
this mechanism, the inversion is obtained by fixing the cylinder or link 4
(i.e. sliding pair), as shown in Fig..
In this case, when the crank (link 2) rotates, the
connecting rod (link 3) oscillates about a pin pivoted to the fixed link 4 at
A and the piston attached to the piston rod (link 1) reciprocates.
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2.
Oscillating cylinder engine. The
arrangement of oscillating cylinder engine mechanism, as shown in Fig. 5.24,
is used to convert reciprocating motion into rotary motion. In this
mechanism, the link 3 forming the turning pair is
fixed.
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3.
Rotary internal combustion engine or Gnome engine. Sometimes back, rotary
internal combustion engines were used in aviation. But now-a-days gas
turbines are used in its place. It consists of seven cylinders in one plane
and all revolves about fixed center D, as shown in Fig.
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4. Crank and slotted lever quick
return motion mechanism. This mechanism is mostly used in shaping machines,
slotting machines and in rotary internal combustion engines. In this mechanism, the link AC (i.e. link 3) forming the turning
pair is fixed, as shown in Fig. The link 3 corresponds to the connecting rod
of a reciprocating steam engine.
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5. Whitworth quick return motion mechanism.
This mechanism is mostly used in shaping and slotting machines. In this
mechanism, the link CD (link 2) forming the turning pair is fixed, as shown
in Fig. The link 2 corresponds to a crank in a reciprocating steam engine.
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