THEORY OF MACHINES 16 MARKS SOLVED Q&A

 ME 3491-THEORY OF MACHINES

UNIT – I  KINEMATICS OF MECHANICS

 

1. 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:

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.
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.
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

2. 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 : N_BO = 600 r.p.m. or ω_BO = 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,

v_BO = v_B = ω_BO × 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,

3. 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.

4. 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)

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	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
c. One double-crank (drag-link) is possible when the shortest link is the frame.	One double-rocker mechanism is possible when the link opposite the shortest link is the frame.

 

5. 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

 

6.  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

 

7. 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.

 

Any one of the link adjacent  to the shortest link is fixed therefore the give mechanism is a crank-rocker mechanism.	any one of the link adjacent  to the shortest link is fixed therefore the give mechanism is a crank-rocker mechanism.
The shortest link is fixed, then a double-crank mechanism obtained.	The link opposite to the shortest link is fixed, then the a double-rocker mechanism is obtained.

 

8. 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

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

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

 

 

9.  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:

1. Since the mechanism consists of six links (i.e. n = 6), therefore the number of instantaneous centres,
2. N = n(n-1)/2 = 6(6-1)/2 =15
3. 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)

 

10.  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)

 

11.  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)

 

12. 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)

 

13. 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.                                               

14.   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.

 

15. 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.

16. 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.

17.  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.

 

18. 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.

19. 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.

20. 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.

 

21. 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.

 

22.  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.

23. 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.

24.   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 :

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.
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.
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

 

25.  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.

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.
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.
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.

 

 

 

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.

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.