Design of Transmission Systems 6

7/29/2019 Design of Transmission Systems 6

http://slidepdf.com/reader/full/design-of-transmission-systems-6 1/7

SCHEME OF VALUATION

Q.P.Set No. Q.P. Serial No.

Total Number of Pages : Sub. Code

------------------------ DEGREE EXAMINATION, March/April/October/November ------------

Branch --------------------------------------

Title of the Paper ------------------------------

PART – A 10 X 2 = 20 marks

Answer ALL questions

1 Disadvantages of Rolling contact bearing over sliding contact bearing: (any four)- 4 x ½ = 2 marks

i. More noisy at high speeds

ii. Low resistance to shock loading

iii. More initial cost

iv. Design of housing is complicated

v. Less Reliable in service

2 Factors influencing the selection of belt drive(any four)

i – Power to be transmitted

ii - Space availability

iii - Required speed reduction ratio

iv – Distance between the centers of rotating shafts

v – Service conditions. - 4 x ½ = 2 marks

3 Desirable characteristics of friction surfaces: (any four)

i. Should have high and uniform coefficient of friction

ii. Should have the ability to withstand high temperature

iii. Should not have moisture absorption characteristics

iv. Should have high heat conductivity

v. Should have high resistance to wear and scoring. - 4 x ½ = 2 marks

4 Undercutting will occur if large lift is attempted with small cam rotation in a plate cam

with flat face follower

5 Backlash is the difference between the tooth space and the tooth thickness as measured

on the pitch circle.

6 Formative number of teeth of a helical gear can be defined as the number of teeth that

can be generated on the surface of the cylinder having radius equal to the radius of

curvature at a point at the tip of the minor axis of the ellipse obtained by taking a

section of the gear in the normal plane.7 When the bevel gears connect two shafts whose axis intersects at an angle greater than

a right angle and one of the bevel gears has a pitch angle of 900, then it is known as a

crown gear or crown wheel

8 The worm gear box is provided with an oil bath to minimize the effect of friction

9 Components of a speed reducer: (any four)

i. Shaft ii. Bearing iii. Housing

iv. Spacers v. Oil level indicator, drain plug

vi. Cooling fins - 4 x ½ = 2 marks

10The spindle speed in a gear box is arranged in geometric progression and the ratio of

1



adjacent speeds will be a constant and is called as progression ratio.

Mathematically,1

min

max −= n

N

N φ , where is the progression ratio.

Part – B 5 *12 = 60 marks

Answer any FIVE questions, choosing ONE question from each unit

UNIT - I

11 Equivalent load P = (XFr + YFa)K s

For the above application, since Fa / Fr is less than 1.14 , X = 1 and Y = 0 - 1 mark

Life of the bearing = 624 mrev - 1 mark

Service factor Ks = 1.5

Hence, P = 3.75 KN - 2 marks

Dynamic capacity C = Pk

1

10L

L

= 31.364 KN - 2 marks

For the above rating, SKF 7310 B is suitable - 2 marks

Bearing specifications are - 2 marks

Bore diameter = 50mm

Outer diameter = 110 mm

Breadth = 27 mm

Rated dynamic capacity = 53 KN

Tolerance for shaft is m6 and its dimensions are 000.50

009.0

025.0

+

+

mm - 1 mark Tolerance for the housing is J7 and its dimensions are 000.110 013.0

022.0

−

+

mm - 1 mark

OR

12 Determination of the transmission ratio

i= 970/330 = 2.94

Selection of teeth on the driver sprocket

Z1 = 25 (for i= 2 to 3) &

Number of teeth on the driven sprocketZ2 = iZ1 = 74 - 1 mark

Determination of optimum centre distance & pitch - 1 mark

a = 500 mm ; pmax = a/30 = 16.6 mm

pmin = a/50 = 10 mm; p = 15.875 mm is chosen.

Selection of chain & chain number - 1 mark

“Duplex” 10A-2/DR50

Determination of total load on the driving side of the chain - 1 mark

PT = Pt + Pc + Ps = 1876.4 N

Determination of Design Load = T s P k × - 1

mark

k s = k 1 . k 2 . k 3 . k 4 . k 5 . k 6 = 1.6

Design load = 1.6*1876.4 = 3002 N

Determination of Factor of safety - 1 mark

Breaking load / Design load = 14.79

Check for induced Bearing stress on Rollers - 1 mark

2



Induced stress = 97.19=×

= A

k P st σ N/mm2 < 4.22][ =σ N/mm2

Determination of Corrected Chain Length - 2 marks

p p pa

z z z z a I /

22

2

2

1221

−

+

+

+= π = 116 links.

= p×116 = 1.842 m.

Determination of Exact Centre Distance - 1 mark

p M ee

a ×−+

=4

82

= 512.91 mm

Determination of Sprocket diameters - 2 marks

Pitch circle diameter of smaller sprocket,)/180sin( 1

1 z

pd = = 126.66 mm

Outside diameter of smaller sprocket, d01 = d1 + 0.8dr = 134.788 mm

Pitch circle diameter of larger sprocket,)/180sin( 2

2 z

pd = = 374.034 mm

Outside diameter of larger sprocket d02 = d2 + 0.8 dr = 382.162 mm

UNIT – II

13 Calculation of Braking torque = load x barrel radius

= 11250Nm - 2 marks

Using, T1-T2 = 37,500 & µθ e

T

T =

2

1, T1=51.38 kN & T2=13.88 kN - 3 marks

Calculation of band width - 2 marks

Bt

T t

1=σ ⇒ B=350mm

Calculation of band thickness t = 0.01D = 6mm - 1 mark

Check for bearing pressure - 2 marks

Br T P 1

max = = 0.489 N/mm2

Calculation of the force to be applied at the end of the lever - 2 marks

Taking moment about fulcrum, P=480.2N

OR

14Determination of mean diameter of the screw

20

pd d −= = 50 mm - 1 mark

Determination of force required at the circumference of the screw

P= ( )φ α +tanW = 86.4 N - 2 marks

Determination of mean radius of flat surface R = 37.5 mm - 1 mark

Determination of torque required T = WRd

P 12

* µ + = 4410 N-mm - 2

marks

Determination of power required to operate the nut = Tω = 277 Watts - 2

marks

Determination of ideal effort To =2

*tand

W α = 635 N-mm - 2 marks

Determination of efficiency of the screw = 14.4% - 2 marks

3



UNIT – III

15 Determination of pitch line velocity v = 0.236m m/sec. - 2 marks

Determination of velocity factor v

C v +=3

3=

m236.03

3

+- 1

mark

Determination of tangential tooth load,m

C v

P W sT

84746== N - 1

mark

Determination of form factor for the pinion p Z

y912.0

154.0 −= = 0.0932 - 1

mark

Determination of form factor for the gear

G

Z y

912.0154.0 −= = 0.1337 - 1

mark

Based on the product of operating stress and form factor,

Decision about weaker section – Pinion - 1 mark

Using the Lewis equation, formulation of 0.0174m3 -0.236m – 3 = 0 - 1 mark

Solving the above, Determination of standard module m = 8 mm. - 1 mark

Determination of face width b = 112 mm - 1 mark

Determination of pitch circle diameter of the pinion, D p = 120 mm - 1 mark

Determination of pitch circle diameter of the gear, DG = 360 mm - 1 mark

OR

16

Calc. of the distance between crank and wheel centre, e=

Z

Sin

r

π = 100 mm. – 2

marks

Calculation of radius of Geneva wheel, R = e

Z

Sin

−2

1π

= 86.67 mm. - 3 marks

Calculation of slot length, h =

+

Z Cos

Z Sine

π π

= 36.6 mm. - 3 marks

Calculation of crank shaft diameter, dc =

−

Z Cose

π 12 = 25 mm. - 2 marks

Calculation of Geneva wheel shaft diameter, dg =

−

Z Sine

π 12 = 35 mm. -

2 marks

UNIT – IV

17 1. Selection of materials

a) Worm - steel

b) Wheel – bronze (sand cast)

[ ] 2

2

/50

/390

mm N

mm N

b

u

=

<

σ

σ

2. Calculation of center distance - 2 marks

4





Average module, mav = 1.26[ ]

[ ]3

vmbv

t

Z y

M

ψ σ = 2.7 mm. - 1 mark

Standard module, m = mav ( )b R

R

5.0− =4 mm. - 1 mark

Diameter of pinion, d1 = 80mm. - 2 marks

Diameter of gear, d2 = 240mm.

Actual cone radius, R = 12.65 cm.

Face width, b = 4.25 cm

Check for induced surface compressive stress - 1 mark

ib

M E i

b R

t c

][)1(

)5.0(

72.0 32 +−

=σ = 3870 kg/cm2 < [ ]cσ

Check 2. Check for induced bending stress - 1 mark

[ ]

( )v

t

bbmY b R

M i R2

2

5.0

1

−

+=σ = 410 kg/cm2< [ ]bσ

UNIT – V

19Calculation of speed ratio, Nmax / Nmin = 55.1,

5.31

1050 19 == − φ φ - 2 marks

The speeds are 31.5, 48.8, 75.7, 117.3, 181.8, 281.8, 436.8, 677, 1049.5 rpm.– 2 marks

Construction of Ray diagram - 4 marks

Construction of kinematic arrangement - 4 marks

Calculation of number of teeth & % of error - 4 marks

Z7 = 25 (given)

3.117

5.31

8

7 = z

z , Z8 = 93

Using Z9+Z10 = Z7+Z8 = 118 &

3.117

5.31

8

7 = z

z , Z9 = Z10 = 59

Using Z11 + z12 = 118

72.33.117

8.436

12

11 == z

z , Z12 =25, Z11 = 93

Z1 = 25;8.281

3.117

2

1 = z

z , Z2 = 60

Using Z1+Z2 = Z3+Z4 = 85 &

8.281

8.181

4

3 = z

z , Z3 = 33, Z4 = 52

Using Z5+Z6 = 85 & Z5/Z6 = 1, Z5 = 42, Z6 = 43

Output speed & % of error calculation

Speed

Number

Output speed Standard

speed

% error

1 31.6 31.5 0.317

2 48.1 48.8 -1.433 74 75.7 -2.24

4 117.4 117.3 0.08

5 178.8 181.8 -1.65

6



6 275.2 281.8 -2.34

7 436.8 436.8 08 665.3 677 -1.72

9 1023.9 1049.5 -2.43

The permissible deviation = ( ) %5.5%110 =−± φ

∴The deviations are well within the permissible limits.OR

20 Calculation of progression ratio as Ф = 1.2535 ≈ 1.25 - 1 mark

The intermediate speeds are

100, 125, 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250 - 2 marks

The possible arrangement is 1x2x2x3 - 1 mark

Ray diagram - 4 marks

Kinematic arrangement - 4 marks

Documents

Design of Transmission Systems 6