Unit III(KINEMATICS OF CAMS)

16 Marks:

1. 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 position during 120° of cam rotation

(iv) Follower to dwell for the remaining period.

The minimum radius of the cam is 25 mm. The outstroke of the

follower is performed with simple harmonic motion and the return

stroke with equal uniform acceleration and retardation (may/june 2007)

SUDHARSAN ENGINEERING COLLEGE SATHYAMANGALAM.PUDUKKOTTAI

AN ISO 9001:2008 Certified In s t itu t io n (Approved by AICTE, New Delhi, Affiliated to Anna University, Chennai)

2. Construct a tangent cam and mention the important terminologies on it. Also derive the

expression for displacement, velocity and acceleration of a reciprocating roller follower when

the roller has contact with the nose. (may/june 2007)

3. What i s t a n g e n t c a m ? Derive t h e e x p r e s s i o n s for t h e v e l o c i t y a n d

acceleration of a roller follower in the tangent cam. (nov/dec 2009)

4. A symmetrical circular cam operating a flat faced follower has the following particulars:

Minimum radius of the cam=30mm; total lift = 20mm; angle of lift = 75◦

Nose radius = 5mm; speed = 600rpm.

Determine:

i) The principal dimensions of the cam.

ii) Acceleration of the follower at the beginning of lift, at the end of contact with the

circular flank, at the beginning of contact with the nose and at the apex of the

nose.

5. From the following data draw the profile of a cam in which the follower moves with SHM

during ascent while it moves with uniformly accelerated and decelerated motion during

descent.

Least radius of the cam = 50mm

Angle of ascent = 48◦

Angle of dwell = 42◦

Angle of descent = 60◦

Lift of the follower = 40mm

Diameter of the roller = 30mm

If the cam rotates at 360rpm anticlockwise find the maximum velocity and acceleration

of the follower during descent.

6. A symmetrical tangent cam with a least radius of 25 mm operates a roller follower of radius

10mm. the angle of ascent is 60◦ and total lift is 15mm. if the speed of the cam is 400rpm,

then calculate:

The principal dimensions of the cam(i.e) the distance between the cam center and nose

center; nose radius and angle of control of cam with straight flank.(may/june 2010)

Answer: similar to 4 question.

7. Draw the displacement, velocity and acceleration diagrams for a follower when it moves with

uniform acceleration and uniform retardation. Derive the expression for velocity and

acceleration during out stroke and return stroke of the follower. (nov/dec 2009)

8. Draw the profile of an cam operating a knife-edge follower when the axis of the follower

passes through the axis of cam shaft from the following data.

(a) Follower to move outwards through 40mm during 60◦ of cam rotation

(b) Follower to dwell for the next 45◦

(c) Follower to return to its original position during next 90◦

(d) Follower to dwell for the rest of the cam rotation.

The displacement of the follower is to take place with simple harmonic motion during both

the outward and the return strokes. The least radius of cam is 50mm. if the cam rotates at

300rpm, determine the maximum velocity and acceleration of the follower during the

outward stroke and return strokes. (nov/dec 2009)

Similar to this model answer but value is different.

9. 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. (nov/dec 2008)

Similar to this model answer but value is different.

10. 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.(may/june 2009), (nov/dec

2012)

Similar to this model answer but value is different.

Unit IV-(Gears)

16 marks:

1. What do you mean by pitch point, circular pitch, module, and addendum and pressure angle?

Explain in detail. (apr/may 2008)

2. State and prove the law of gear tooth action for constant velocity ratio and show how the

involute teeth profile satisfies the condition. Derive an expression for the velocity of sliding

between a pair of involutes teeth. State the advantages of involute profile as a gear tooth

profile. (apr/may 2008)

3. Two 20◦ involute spur gear mesh externally and give a velocity ratio of 3. Module is 3mm

and the addendum is equal to 1.1 modules. If the pinion rotates at 120rpm, determine

(a) The minimum number of teeth on each wheel to avoid interference

(b) The number of pairs of teeth in contact. (nov/dec 2009)

4. In an epicyclic gear train, an arm carries two gears A and B having 24 and 30 teeth

respectively. The arm rotates at 100rpm. In the clockwise direction. Find the speed of gear B

on its own axis, when the gear A is fixed. If instead of being fixed, the wheel A rotates at

200rpm in the counter clock wise direction, what will be the speed of B?

(nov/dec 2009)

5. (i) Derive an expression for minimum number of teeth on the wheel in

Order to avoid interference.

(i) Two mating gears have 20 and 40 involute teeth of module 10mm and 20 pressure

angle. The addendum on each wheel is to be made of such a length that the line of

contact on each side of the pitch point has half of the maximum possible length.

Determine the addendum height for each gear wheel, length of the path of contact,

arc of contact and contact ratio.

(nov/dec 2008)

This sum is similar to solved sum below here:

6. In an epicyclic gear train shown in fig. the pinion A has 15 teeth and is rigidly fixed in the

motor shaft. The wheel B has 20 teeth and gears with A, and also with annular fixed wheel

D. pinion C has 15 teeth and is integral with B(C,B being a compound gear wheel). Gear C

meshes with annular wheel E, which is keyed to the machine shaft. The arm rotates about the

same shaft on which A is fixed and carries the compound wheel B-C. if the motor runs at

1000rpm. Find the speed of the machine shaft. (may/jun 2009)

7. Anepicyclic train of gear is arranged as shown in fig. Determine the revolution of the arm, to

which the pinions B and C are attached, when (i) A makes one revolution clockwise and D

makes half a revolution anticlockwise and (ii) A makes one revolution clockwise and D is

stationary.

DA/2 = DB+DC/2

TA/2 = TB + TC/2

TB = (TA – TC)/2 = (54 -24)/2 = 15

8. An epicylic gear train is shown in fig. the number of teeth on A &B are 100 and 220

respectively. Determine the speed of the arm a

(a) If A rotates at 100rpm clockwise and B at 50 rpm counter clockwise

(a) If A rotates at 100 rpm clockwise and B is stationary. (May/Jun 2011)

9. (a) (i) State and prove “Law of gearing”, and thus derive the expression for “Velocity

of sliding”.

(ii) Prove that the maximum length of arc of contact between a pair of gear tooth to

avoid interference is (r + R)tan . (Apr/May 2010)

10. (i) Diagrammatically show the following with reference to a spur gear

tooth:

Face width, Pitch circle, Clearance, Tooth thickness,Addendum and Dedendum.

(ii) An epicyclic gear train is shown in the following figure. How many

revolutions does the arm makes when (1) A makes one revolution in

clockwise and D makes 1/2 a revolution in the opposite sense and

(2) A makes one revolution in clockwise and D remains stationary?

The number of teeth in gears A and D are 40 and 90 respectively.

11. Two planet gear B and C having 30 teeth each are attached to the arm E as shown in fig. and

gear A is having 40 teeth instead of 50, then find the number of revolution made by the arm

when:

(i) Gear A makes one revolution clockwise and D makes half a revolution

anticlockwise.

(ii) Gear A makes one revolution clockwise and D is stationary.

(nov/dec 2011)

Unit V- (Friction)

1. Sketch and explain any three inversion of a double slider crank chain.(nov/dec 2008)(nov/dec

2011)(nov/dec 2010)

Many a times mechanisms are designed to perform repetitive operations. During these

operations for a certain period the mechanisms will be under load known as workingstroke and the remaining period is known as the return stroke, the mechanism

returnsto repeat the operation without load. The ratio of time of working stroke to that of

thereturn stroke is known a time ratio. Quick return mechanisms are used in machinetools to give a slow cutting stroke and a quick return stroke. The various quick returnmechanisms commonly used are i) Whitworth ii) Drag link. iii) Crank and slottedlever mechanism

1. Whitworth quick return mechanism:

Whitworth quick return mechanism is an application of third inversion of the single

slider crank chain. This mechanism is shown in the figure below. The crank OC isfixed and OQ rotates about O. The slider slides in the slotted link and generates acircle of radius CP. Link 5 connects the extension OQ provided on the opposite

sideof the link 1 to the ram (link 6). The rotary motion of P is taken to the ram R

whichreciprocates. The quick return motion mechanism is used in shapers and slottingmachines.

The angle covered during cutting stroke from P1 to P2 in counter clockwise direction

is α or 360 -2θ. During the return stroke, the angle covered is 2θ or β.

2. Drag link mechanism :

This is four bar mechanism with double crank in which the shortest link is fixed. Ifthe crank AB rotates at a uniform speed, the crank CD rotate at a non-uniform

speed.This rotation of link CD is transformed to quick return reciprocatory motion of theram E by the link CE as shown in figure. When the crank AB rotates through an

angleα in Counter clockwise direction during working stroke, the link CD rotates

through180. We can observe that / α >/ β. Hence time of working stroke is α /β times

more orthe return stroke is α /β times quicker. Shortest link is always stationary link.

Sum ofthe shortest and the longest links of the four links 1, 2, 3 and 4 are less than the

sum ofthe other two. It is the necessary condition for the drag link quick return

mechanism.

3. Crank and slotted lever mechanism:It is an application of second inversion. The crank and slotted lever mechanism isshown in figure below.

2. Explain the working of a quick return motion mechanism. Also derive an eqution for the ratio of time taken for return stroke and forward strokes. (may/june 2009)Many a times mechanisms are designed to perform repetitive operations. During theseoperations for a certain period the mechanisms will be under load known as workingstroke and the remaining period is known as the return stroke, the mechanism returnsto repeat the operation without load. The ratio of time of working stroke to that of thereturn stroke is known a time ratio. Quick return mechanisms are used in machinetools to give a slow cutting stroke and a quick return stroke. The various quick returnmechanisms commonly used are i) Whitworth ii) Drag link. iii) Crank and slottedlever mechanism

1. Whitworth quick return mechanism:

Whitworth quick return mechanism is an application of third inversion of the singleslider crank chain. This mechanism is shown in the figure below. The crank OC isfixed and OQ rotates about O. The slider slides in the slotted link and generates acircle of radius CP. Link 5 connects the extension OQ provided on the opposite sideof the link 1 to the ram (link 6). The rotary motion of P is taken to the ram R whichreciprocates. The quick return motion mechanism is used in shapers and slottingmachines.

The angle covered during cutting stroke from P1 to P2 in counter clockwise directionis α or 360 -2θ. During the return stroke, the angle covered is 2θ or β.

2. Drag link mechanism :

This is four bar mechanism with double crank in which the shortest link is fixed. Ifthe crank AB rotates at a uniform speed, the crank CD rotate at a non-uniform speed.This rotation of link CD is transformed to quick return reciprocatory motion of theram E by the link CE as shown in figure. When the crank AB rotates through an angleα in Counter clockwise direction during working stroke, the link CD rotates through180. We can observe that / α >/ β. Hence time of working stroke is α /β times more orthe return stroke is α /β times quicker. Shortest link is always stationary link. Sum ofthe shortest and the longest links of the four links 1, 2, 3 and 4 are less than the sum ofthe other two. It is the necessary condition for the drag link quick return mechanism.

3. Explain the working of a toggle mechanism and its application with a neat sketch.

(may/june 2009)1. Toggle Mechanism:

In slider crank mechanism as the crank approaches one of its dead centre position, theslider approaches zero. The ratio of the crank movement to the slider movementapproaching infinity is proportional to the mechanical advantage. This is the principleused in toggle mechanism. A toggle mechanism is used when large forces act througha short distance is required. The figure below shows a toggle mechanism. Links CDand CE are of same length. Resolving the forces at C vertically F Sin α =P Cos α 2Therefore, F = P . (because Sin α/Cos α = Tan α) 2 tan α Thus for the given value ofP, as the links CD and CE approaches collinear position (αO), the force F risesrapidly.

12. Diagrammatically show the following with reference to a spur gear

tooth:

Face width, Pitch circle, Clearance, Tooth thickness,Addendum and Dedendum.