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Machine Design, by U. C. Jindal, Pearson Education, India. 

1. Conceptual design is the description of how a new product will work and meet its performance requirements. In conceptual design, product is created instead of a visual representation—which would directly be used in a final product, e.g. a film, animation or video game.

2. Tolerance is a valuable tool for reducing manufacturing costs by improving productivity. Tolerance specification is an important link between engineering and manufacturing. It can become a common ground on which to build an interface between the two, to open a dialog based on common interests and competing requirements.

3. A key is device, which is used for connecting two machine parts for preventing relative motion of rotation with respect to each other.

4. Shafts are rotating members which are subjected to bending moments and twisting moments and sometimes to axial loads. It twists and transmits power.Axles are rotating or non-rotating members which are subjected to only bending moments due to members supported by it. It does not transmit torque. In other words, an axle is not twisted it only bends.

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  6.

11a.

PART –B - 16 Marks

11a. Importance and types of theories of failure - 8 Marks Explanation of any 2 types of theories - 8 Marks

11b. Diameter of bar-8 Marks, Diameter of pin – 8 Marks

12a. Advantages of shaft couplings - 4 Marks Problem - 12 Marks

Solution:

The torque transmitted by the shaft, T = 60P 2πN

= 60x35x10 3 2πx400

= 835.563 N-m = 835563 N-mm And also , Torque , T = (T1 - T2) R

(T1 - T2) = T/R = 835563/450 = 1856.807 N ----(1)

W.K.T, Ratio of tensions, T1/T2 = eµθ

Assume, θ = π radians and µ = 0.25

T1/T2 = 2.1932

T1 = 2.1932 T2 -------(2)

By solving (1) and (2), we get T1 = 1556.15 N, T2 = 3407.9 N Total load on the shaft , W = T1+ T2 + weight of pulley = 5764.05 N

B.M due to the above load, M = WL =5764.05 x 250 = 1441012.5 N-mm

Equivalent Torque, Te = √(Kb x M)2+(Kt x T)2

Assume, Kb = 2 Kt = 1.5

Te = 3142760.003 N-mm

And also w.k.t Te = π s ( 1- k4 )do3 N-mm 16

K = di / do = 0.5

By substituting, we get, do = 44.03 mm = 45 mm (R20 series)

And di = 22.5 mm = 25 mm (R20 series)

12 b. Design Calculation - 12 Marks Drawing - 6 Marks

Solution: Power (P) = 12000W, N = 300 rpm, Ks= 1.25, c = 100 N/mm2

Shear stress for shaft & key (sh or k = 50N/mm2 ,

Shear stress for sleeve(s = 10 MPa =10N/mm2

Torque, T = 60P 2πN

=381.97 N-m = 381970 N-mm

(i) Design of Shaft:

We know that, Torque, T = π sh d3 N-mm 16 Diameter of shaft (d) = 33.88 mm say d = 34 mm.

(ii) Design of sleeve:

Diameter of sleeve = D = 2d +13 = 81 mm

Length of Sleeve = L = 3.5d=119 mm

Check for induced shear stress in muff

Torque, T = π s ( D4- d4 ) N-mm 16D

s = 3.77 N/mm2

which is less than the permissible value (10 N/mm2). Hence the design is safe.

(iii) Design for keys:

From the PSG DDB for the shaft diameter d = 34 mm,

Width of the key , b = 10 mm

Height of the key, h = 8 mm

Length of each key, l = L/2 = 119/2 = 59.5 mm.

Check for shearing:

Torque, T = l b k d/2 by substituting the values

We get k = 37.76 N/mm2

which is less than the allowable value (50 N/mm2). Hence the design is safe.

Check for crushing:

Torque, T = c l h/2 d/2 by substituting the values

We get c = 94.40 N/mm2

which is less than the allowable value (100 N/mm2). Hence the design is safe.

14a.

Given data: N= 300 rpm, n = 2, µ = 0.28, d1 = 220 mm, r1 = 110 mm, d2 = 140 mm,

r2 = 70 mm, Pmax = 0.1 MPa = 10000 N/m2 , I = 7.2 kg-m2

Solution:

(i) The time to attain the full speed by the machine ( with uniform wear)

Since the intensity of pressure(p) is maximum at the inner radius (r2),

Pmax r2 = C or C= 7000 N/m

Axial thrust exerted, W = 2πC (r1-r2) = 1759.29 N

Torque , T = n µ W ( r1 + r2 )/2 = 88.6682 N-m

Power, P = 2 π N T / 60 =2785.59 Watts

w.k.t Torque , T = I α

Angular acceleration, α = T / I = 12.315 rad/s2

And also w.k.t α = ω / t

Time , t = ω/α = 2.55 sec

(ii) The energy lost in slipping of the clutch

Energy lost in friction = T ( θ1- θ2)

Where θ1 = ω t = 80.108 rad

θ2 = ωo t + ½ α t2 = 0 + 40.039 = 40.039 rad

Energy lost in friction = 88.6682 ( 80.108 – 40.039 )

= 3552.846 N-m

14b.

Given data: TB = 30 N-m = 30 x 103 N-mm, d = 300 mm, r = 150 mm,

2θ = 90º = 90 x π/180 = 1.57 rad, µ = 0.4 , p = 0.3 N / mm2

Solution: (i) To find the spring force (S)

Torque, TB = ( Ft1 + Ft2 ) r --------(1)

Ft1, Ft2 ---- Braking force on the R.H.S, L.H.S of shoe.

RN1, RN2 -----Normal reaction on the R.H.S, L.H.S of shoe.

Taking moment about fulcrum on R.H.S,

S x 475 = RN1 x 225 + Ft1 x 150

Where RN1 = Ft1 / µ’ and

µ’ = 4 µ sin θ/ (2 θ +sin 2θ) = 0.4402

By substituting we get, Ft1 = 0.718 S

Taking moment about fulcrum on L.H.S,

S x 475 + Ft2 x 150 = RN2 x 225

Where RN2 = Ft2/ µ’ and

µ’ = 4 µ sin θ/ (2 θ +sin 2θ) = 0.4402

By substituting we get, Ft2 = 1.315 S

From equation (1), i.e 30 x103 = (0.718 S + 1.315 S) 150

Spring force (S) = 98.376 N

(iii) To find the width of the shoe (b)

Let, the projected bearing area of shoe , Ab = b (2r sinθ) = 212.132 b mm2

w.k.t RN1 = Ft1 / µ’ = 0.718 S/ µ’ = 160.458 N and

RN2 = Ft2 / µ’ = 1.315 S/ µ’ = 293.876 N (max)

Therefore, take pb = RN2 / Ab

Ab = 1049.55

212.132b = 1049.55

Width, b= 4.947 mm

15a.

Given data: For centrifugal pump journal bearing, Load W =13.25 x 103 N, D = 80mm,

n = 1440 rpm, p = 0.7- 1.4 N/mm2, Atm.temp = 30ºC

Solution:

Step-1

From PSG DDB Select L/D ratio = 1 to 2

Let us take L/D = 1.5

Length of journal, L = 1.5 D = 120 mm

Step-2

Pressure developed , p = W/ LD = 1.380 N/mm2

This pressure is within the safe limit (0.7- 1.4 N/mm2).

Step-3

Selection of lubricating oil

The minimum value of Zn/p from PSGDDB is,

(Zn/p)min = 2844.5

Zmin= 2844.5x13.8/1440 = 27.638 Centipoise say 30CP

From PSGDDB, for 30CP and 60ºC, Select SAE 30 oil

Step-4

Calculation of coefficient of friction, µ = 33.25/1010 (Zn/p)(D/C) + k

Where C = Diameteral Clearance = 150 microns (assume) = 150x10-3mm

And k = 0.002 for 0.75<L/D<2.8

µ = 0.00255

Step-5

Heat Calculation,

Heat generated Hg = µW v

Where , v = π D n/60 = π x 0.08 x1440/ 60 = 6.03 m/sec

Hg = 203.801 watts

Heat dissipated, Hd = (Δt + 18)2LD/K

Where, Δt = ½ (to – ta) = ½ (60 -30) = 15º

and K = 0.484 (medium construction)

Therefore, Hd = 21.6 watts

Since Hg > Hd , artificial cooling arrangement must be provided by using cooling fans or by circulating water.

Diameter of bearing, Db = D + C = 80.15 mm

Material: From PSGDDB select, Rubber or Modulated plastic laminate.

15 b.

Solution:

Load on the spring = 2P =140 x 103

But, 2P = 140 x 103 / Number of springs

P = 17500 N

Permissible stress, σ = 6PL/nbt2

600= 6x17500x500 / 10xbt2

bt2= 8750 -------------(1)

Deflection of the spring, y = 6PL3/Enbt3

80 = 6x17500x5003 / 200x103x10xbt3

bt3= 82031.25

bt2.t= 82031.25 -------------(2)

From equations (1) & (2), 8750 t = 82031.25

t = 82031.25/8750 = 9.375mm

Say thickness of the leaves, t = 10mm

b = 8750/t2 = 8750/102 = 87.5mm

Width of the leaves, b = 87.5mm

The nearest standard size, b = 90mm