Page: Design of Coupling
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Q.1 It is required to design a bushed pin type. Flexible coupling to connect the output shaft of an electric motor to the shaft of a centrifugal pump. The motor delivers $20 KN$ power at $720\ rpm$. The starting torque of the motor can be assumed to 150% of the rated torque. Design the coupling and specify the dimensions of its components with suitable material.

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SELECTION OF MATERIAL FOR SHAFT, BOLT and KEY :-

As the above parts are subjected to torsional shear stress and direct-shear stress, then on the basis of strength.

Let us select C-30 as material for above parts:

PSG 1.9

$\sigma_{\gamma} = 300 N/mm^2$

Selection Of F.O.S:

As the shaft is subjected to torsional shear stress and also there is a stress concentration due to keyway in shaft 50 selecting higher F.O.S $n=4$ based on $\sigma_{\gamma}$ .

Permissible stresses for C-30 :-

$\sigma_t = \frac{\sigma_y}{n} = 300/4 = 75 N/mm^2$

By max. shear stress theory

$t = \frac{0.56y}{F.O.S} = 37.5 N/mm^2$

Selection of material for hub and flanges :-

Select C.I. as material for hub and flange because the flanges are having complex shape. So, they can be manufactured easily with C.I. without involving costly machining operations.

C.I. is having ability to damp the vibrations so they can damp the input vibrations.

C.I. is cheap and it reduces the overall cost of coupling.

Slecting GCI - 25-------- PSG 1.4

$\sigma_u$ = 250 N/$mm^2$

Selecting of F.O.S :

As C.I. material is brittle and also there is stress concentration due to keyways in hub. So consider higher F.O.S $n=5$ based on $\sigma_u$

Permissible Stress :

Consider $\tau= \sigma_t = \frac{\sigma_u}{F.O.S} = 50N/mmm^2$

Selection of material of bush:

Considering rubber as material for bush so that flexibility is maintained and no shock take place.

considering bearing pressure

$P_b = 1 N/mm^2$

Selection of standard coupling :-

From PSG 7.108 -> at last

Rated power = $\frac{\text{KW power application * service factor * 100}}{\text{RPM of application}}$

In this problem consider motor as a prime move to drive the centrifugal pump. Service factor for considered drive and driven combination is selecred from PSG 7.109

Consider service factor for above case = 1.5

Rated power at 100 rpm = $\frac{20\times1.5\times100}{720} = 4.16 KW$

From PSG 7.108 ->

Selecting coupling No.6 whose dimensions are as follows.

A= dia. of shaft $d=75mm$

B= outer dia of flange $D_2 = 200mm$

C= dia of hub =$D = 100mm$

E = Length of hub = 56 mm

G= Length of flange = $l =40mm$

H = protected length of flange $t_f = 15mm$

D= PCD of bolt = $D_1 = 150mm$

n= no of bolts, $n=4$

F= dia of bolt = $d_1 = 12mm$

$d_b$ = dia of rubber bush =30 mm

t = clearance between two flanges = 4 mm

Let us check the indirect stresses in different component of coupling

Let T be the mean torque of coupling

$P= \frac{2 \pi NT}{60}$

$20\times10^3 = \frac{2\times\pi\times720\times T}{60}$

$T= 65.25 N-m$

As starting torque of the motor is 150% of rated torque.

$T_{max} = 1.5\times T_{mean} = 1.5 \times 265.25 = 397.87 N-m$

$T_{max} = 397.87 \times 10^3 N-mm$

I) Shear failure at shaft :

$T_{max}$ = $\frac{\pi}{16}$

$z_{ind} = 4.80 N/mm^2 \lt \tau = 37.5 N/mm^2$

$\therefore$ Shaft is safe

II) Shear failure of hub

$T_{max} = \frac{\pi}{16} \tau_{ind}\times[\frac{D^4 - d^4}{D}]$

$397.87\times10^3 = \frac{\pi}{16} \tau_{ind}\times(\frac{100^4-75^4}{100})$

$397.87\times10^3 = \frac{\pi}{16} \tau_{ind}\times683.59 \times10^3$

$\tau_{ind} 22.96 N/mm^2 \lt \tau = 50 N/mm^2$

$\therefore$ Hub is safe is shear

SELECTION OF KEY :-

PSG 5.21

Slecting rectangular taper key for diameter $d=75mm$

Width of key

$b= 20mm$

Height of key $h= 12mm$

Depth in shaft $t_1 = 7.5mm$

Depth in hub $t_2 = 3.8mm$

I) Considering crushing failure of key in hub :-

$T_{max}$ = $(t_2\times l)\times \sigma_{crushing}\times \frac{d}{2}$

$397.87\times10^3 = (3.8 \times l)\times 75 \times 75/2$

$l= 37.22 mm$

But minimum length of key = length of hub + 5mm

$= 56+5 = 61mm$

II) Checking shear stress developed in key

$T_{max} =(b\times l)\times \tau_{ind}\times d/2$

$397.87\times 10^3 = (20 \times 61) \times \tau_{ind}\times 75/2$

$t_{ind} = 8.69 N/mm^2 \lt \tau = 37.5 N/mm^2$

$\therefore$ Key is safe in shear stress

III) Checking the key in crushing in shaft

$T_{max} = (t_1\times l)\times \sigma_{crushing}\times d/2$

$397.87 \times 10^3 = (7.5 \times 61) \times \sigma_{crushing} \times 75/2$

$\sigma_{crushing} = 23.19 N/mm^2 \lt 75 N/mm^2$

$\therefore$ It is safe in crushing

DESIGN OF BOLT :-

Checking of induced shear stress in bolt.

$T_{max} =n\times(\frac{\pi}{4}(d_1)^2)\times \tau_{ind}\times\frac{D_1}{2}$

$397.87 \times 10^3 = 4 \times (\pi/4 \times 12^2)\times \tau_{ind}\times 150/2$

$t_{ind}= 11.72 N/mm^2 \lt \tau = 37.5 N/mm^2$

$\therefore$ Bolts are safe in shear.

Let dp be the dia. of pin which is usually

$dp= d_1 + 4 =12+4$

$dp = 16mm$

Let $F_b$ be the bearing load acting on all pins

$F_B$ = no. of plns X bearing pressure X bearing area

$F_b= 4 \times P_b \times [t+(l-2/3d_1)]\times d_p$

$F_B= 4\times1\times[4+(40-2/3\times12)]\times16$

$= 2304 N$

From PSG 7.106

$M_{max} = \frac{F_B[t+1/2(l-2/3d_1)]}{n}$

$=\frac{2304[4+1/2(40-2/3\times 12)]}{4}$

$= 11520\ N-mm$

$\therefore$ Induced bending stress

$\sigma_{induced} = \frac{M_{max}}{z} =\frac{11520}{\frac{\pi}{32}\times(d_1)^3} = \frac{11520}{\frac{\pi}{32}\times(12)^3}$

$\sigma_{b_{induced}}$ = $67.90 N/mm^2 \lt \sigma_t = 75 N/mm^2$

Principle stress developed in bolt

$\sigma = \frac{1}{2} [\sigma_{induced} + \sqrt{(\sigma^2_{bind} + 4\tau^2_{ind})}]$

$= \frac{1}{2}[67.90+\sqrt{(67.90)^2+4(11.72)^2}]$

$= 69.86 N/mm^2 \lt \sigma_t =75 N/mm^2$

$\therefore$ Bolts are safe.

DESIGN OF FLANGE :-

Consider the shear failure of flange at junction of flange and hub.

$T = [\pi \times D \times (2-t_F)]\times \tau_{ind}\times \frac{D}{2}$

$397.87 \times 10^3 = [\pi \times 100 \times (40-15)]\times \tau_{ind} \times 100/2$

$\tau_{ind} = 1.013 N/mm^2 \lt \tau = 50 N/mm^2$

$\therefore$ Flange is safe in shear.

  • Checking the bearing pressure on pin from PSG 7.106

$P_b = \frac{F_n}{d_b[D-2/3d_1]\times n}$

$= \frac{2304}{30[40-2/3\times12]\times 4}$

$= 0.6N/mm^2 \lt P_b = 1 N/mm^2$


Q.2 Design a flexible coupling of pin-bush types construction for connecting reduction gear shaft to a stone crushes shaft. The unit is driven by 30kw & 720 rpm motor through 5:1 reduction. Choose suitable materials and the design stresses for the parts of coupling.

enter image description here

$i= \frac{n_1}{n_2}$

$n_2 = 144 rpm$

Selection of standard coupling:from PSG 7.108

Rated power = $\frac{\text{KW power application} \times service factor \times 100}{\text {RPM of application}}$

In this problem, consider motor as a prime mover to drive stone crusher

service factors for considered driven and driven combination is selected from PSG 7.109

Rated power at 100 rpm= $\frac{30\times2.5\times100}{144} = 52.08 KW$

from PSG 7.108

Selecting coupling no. 9

Then design the coupling as per previous problem.


Q.3: A protective type rigid flanged coupling required to transmit $25 KW$ at $100 rpm$.

1) Select suitable material for the shaft and design the shaft

2) select suitable material for the key select cross section of the key.

3) Select no of bolts, select bolt material design P.C.D. for bolts and bolt size. check the bolt for safety.

4) Select a suitable standard fit for the diameter and the hole in the flange. Note down the corresponding tolerance on the bolt diameter and the hole diameter in the flange.

enter image description here

ELECTION OF MATERIAL FOR and BOLTS :

As the above parts are subjected to torsional stresses and to account for good shear strength. Let us select general material i.e. mild steel.

Let us select C-30 for above components from PSG 1.9

$\sigma_y = 300 N/mm^2$

As the shaft is subjected to torsional as well reverse stresses and also in the bolt there is stress concentration due to the threads hence let us select higher FOS $n=4$ based on $\sigma_{\gamma}$

Permissible Stress in C-30 :-

$\sigma_L = \sigma_C = \frac{\sigma_{\gamma}}{n} = \frac{300}{4} = 75N/mm^2$

By max shear stress theory,

$\tau = 0.5\frac{\sigma_y}{n} = \frac{0.5 \times300}{4}= 87.5 N/mm^2$

Selection of material for hub and flange

selecting CL material

Taking GCI - 25 from PSG 1.04

$\sigma_u = 250N/mm^2$

Selecting of FOS :-

Selecting FOS $n=5$ for C.I. to take into account the effect of high torsional stresses then permissible stresses

$\tau = \sigma_t = \frac{\sigma_u}{FOS} = 250/5= 50N/mm^2$

DESIGN OF SHAFT:-

By using relation

P =$\frac{2 \pi N T}{60}$

$25\times10^3 = \frac{2 \pi \times200\times T}{60}$

$T = 1193.66 N-m= 1193.66 \times 10^3 N-mm$

using the relation

$T =\frac{\pi z d^3}{16}$

$1193.66 \times10^3 = \frac{\pi}{16} \times 37.5\times d^3$

$d = 54.52 mm$

Selecting standard dia of shaft $[d=55mm]$

SELECTION OF KEY :-

Form PSG 5.21 selecting taper key where width of key

$b=16mm$

Height $h = 10mm$

depth on shaft $t_1 = 6mm$

depth on shaft $t_2 = 3.4mm$

CALCULATION OF LENGTH OF KEY :-

1) Consider shear failure of key

$T= (b\times l)\times (\tau \times d/2)$

$1193.66 \times 10^3 = (16 \times l) \times (37.5 \times 55/2)$

$l=72.34 mm$

2) Consider the crushing failure of key shaft

$T = (t_1\times2)\times(\sigma_c\times d/2)$

$1193.66 \times 10^3 = (6\times l) \times (75 \times 55/2)$

$l= 96.45 mm$

3) consider the crushing failure of key in hub :-

$T= (t_2\times l)\times\sigma_c\times d/2$

$1193.6 \times 10^3 = (3.4 \times l) \times (75\times 55/2)$

$l = 170.21 mm$

Length of hub = $L = 1.5d = 1.5 \times 55$

$= 82.5 mm$

say $L= 85 mm$

DESIGN OF BOLTS :-

Shaft dia No. of Bolts

Upto 40 mm = 3

Upto 100 mm = 4

Upto 180 mm = 6

Upto 250 mm = 8

Upto 250 mm = 10

As shaft dia is $55mm$, so select no of bolt $n=4$

Pitch circle dia = $D_1 = 3d = 3\times 55$

$= 165 mm$

$D_1 = 165 mm$

using the relation,

$T=n\times [\frac{\pi}{4} \times d_c^2] \times \tau_{max} \times (\frac{D_1}{2})$

$1193.66 \times 10^3 = 4 \times (\frac{\pi}{4}\times d_c^2)\times 37.5 \times 165/2$

$d_c = 11.08$

PSG 5.42 sketching standard bolt is M16

Let us check the crushing failure of bolt,

$T=(n\times d_c \times t_f)\sigma_c\times \frac{D_1}{2}$

consider $t_f = 0.5d = 0.5 \times 55$

$= 27.5 mm$

$1193.66 \times 10^3 = (4\times16\times28)\times\sigma_c\times 165/2$

$\sigma_c = 8.07 N/mm^2$

As $\sigma_c (8.07) \lt \sigma_c (75 N/mm^2)$

$\therefore$ Bolt is soft in crushing

DESIGN OF HUB :-

Selecting Hub dia $D= 2d = 2 \times 55$

$= 110 mm$

Let us check shear stress induced,

$T = \frac{\pi}{16}\times \tau_{induced}\times(\frac{D^4 - d^4}{D})$

$1193.66 \times 10^3 = \frac{\pi}{16}\times \tau_{induced} \times (\frac{110^4 - 55^4}{110})$

$1193.66 \times 10^3 = \tau_{induced} \times 245 \times \ 10^3$

$\tau_{induced}$ = 4.87 $N/mm^2 \lt 50 N/mm^2$

$\therefore$ Design of Hub is soft.

DESIGN OF FLANGE:

Consider $t_f = 0.5d = 0.5\times55 = 27.5 mm$

Take = $t_f = 28 mm$

Consider $t_p = 0.25d = 0.25\times55$

$= 13.75 mm$

Protective circumferential flange $t_p = 14mm$

Let us check the shear stress induced the junction of flange.

$T= (\pi \times D \times t_f) \times \tau_{induced} \times D/2$

$1193.66 \times 10^3 = (\pi \times 110 \times 28)\times \tau_{ind} \times 110/12$

$t_{induced} = 2.24 N/mm^2 \lt 50 N/mm^2$

$\therefore$ Safe

Height of Nut which is to be protected

$=0.8 \times d_1 bolt= 0.8\times16$

$= 12.8 mm$

The fit between the bolt and the hole in the flange is non running, normal location fit.

Hence fir sketched is of type locational and assembly fit quality of fit is normal location.

i.e. for combination of hole and shaft a fit $H_g h_g$

Tolerances are specified as

Basic size tolerance on hole(mm) Tolerance on bolt (mm) H8 h8

16 -> +0.027 -> -0.027


Q.4 Design muff coupling to transmit 20 KW @720 rpm.

Data: $P=20\times 10^3 watts$

$N: 720\ rpm$

Calculation of torque to be transmitted

Using P = $\frac{2\pi NT_m}{60}$

$20 \times 10^3 = \frac{2\pi \times 720 \times T_m}{60}$

$T_m = 265.26 N-m$

$T_m = 265.26 \times 10^3 N-mm$

For 50% overload

$T_{max} = 1.5\times T_m = 397.89 \times 10^3 N-mm$

enter image description here

Selection of materials FOS and calculation of design stress.

1] Shaft of keys:

Mat 1 : C-45

Justification:

i) High strength

ii) High toughness

iii) low in coast compared to allow

iv) readily available

$\sigma_yt$ = 360 $N/mm^2$ (PSG 1.9)

Let FOS = 3 from V.M fairs for ductile material under uniform bad selecting a higher value to assant for stress concentration,

[$\sigma_t$] = $\frac{\sigma_{v}}{F.O.S} = 360/3 = 120 N/mm^2$

Also, $\tau = 0.5 \times \sigma_t = 60 N/mm^2$

Accounting weakening effect due to keyway present,

$\tau = 0.75 \times [\tau] = 45 N/mm^2$

2] Sleeve or muff :-

mate : GCI-25

Justification :

i) Low in cost

ii) Adequate strength

Let; FOS = 8

$[\sigma_t] = \frac{\sigma_{ut}}{F.OS.} = 235/8 = 29.375 N/mm^2$

C.I are equally strong in tension and shear.

$[\tau] = [\sigma_t] = 29.375 N/mm^2$

Accounting for the weaking effect because of keyway

$\tau = 0.75 \times [\tau] = 22.03 N/mm^2$

1] Design of shaft :

Considering torsional shear of shaft

$T_{max} = \frac{\pi}{16} [\tau]$

$397.89 \times 10^3 = \frac{\pi}{16} \times 45 \times d^3$

$d = 35.58 mm \approx 40mm$

2] Design of key :

From PSG 5.21 ; selecting a std. surf taper key

$b= 12 mm$

$h = 8 mm$

$t_1 =5mm$

$t_2 =2.5mm$

i) considering shear failure of key :-

$T_{max} = {(b\times l)\times(\tau)}\times \frac{d}{2}$

$397.89 \times 10^3 = (12 \times l) \times 60 \times 40/20$

$l=27.63 mm$

ii) considering crushing of key portion in hub:-

$T_{max} = {(t_2\times l)\times(\sigma_d)}\times\frac{d}{2}$

$397.89 \times 10^3 = {(25 \times l)\times 192 \times 40/2}$

$l=41.46 mm$

sleecting a higher value, $L=2l = 84 mm$,

Total length of key $=84+20=104 mm$

3] design of muff ;-

Empirically, i) $D=2d+13 mm = 93 mm$

ii) $L= 3.5d (mm) = 140 mm$

considering torsional shear, of muff(which is like a hollow shaft)

$T_{max} = \frac{\pi}{16}(\tau)$

$397.89 \times 10^3 = \frac{\pi}{16}\tau\times \frac{93^4 - 40^4}{93}$

$(\tau) = 2.61 N/mm^2$

summary:-

$d=40 mm$

$D=93 mm$

$L= 140mm$

$b= 18mm$

$h=8 mmm$

length of key $= 140+ 20=160 mm$

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modified 19 days ago by gravatar for Sanket Shingote Sanket Shingote250 written 4 months ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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