ME ME ME ME –––– 414414414414 MACHINE DESIGN 1MACHINE DESIGN 1MACHINE DESIGN 1MACHINE DESIGN 1

PLATE NO.4PLATE NO.4PLATE NO.4PLATE NO.4 KEYS AND COUPLINGKEYS AND COUPLINGKEYS AND COUPLINGKEYS AND COUPLING

NAME: ODILON CUENCONAME: ODILON CUENCONAME: ODILON CUENCONAME: ODILON CUENCO INSTRUCTOR: ENGR. MICHAEL BURGOSINSTRUCTOR: ENGR. MICHAEL BURGOSINSTRUCTOR: ENGR. MICHAEL BURGOSINSTRUCTOR: ENGR. MICHAEL BURGOS

Design a typical rigid flange coupling for connecting a motor and a centrifugal pump shaft. The coupling

needs to transmit 15 KW at 1000 rpm. The allowable shearing stress of the shaft, key and bolts material

are 60 MPa, 50 MPa, and 25 MPa, respectively. Assume key crushing failure at 100 MPa.

� = 15 �� = 15000 �

= 1000 !"# = 1006 !"%

&'()*+ = 60 ,�-

&./0 = 50 ,�-

1./0 = 100 ,�-

&234+ = 25 ,�-

I. Design for Shaft Diameter 6789

a. Torque

: = �2 × < × = 15000 �

2 × < × =1006 !"%>

= 143.2394488 # = 143239.4488 ##

b. Shaft Diameter

60 ,�- = &'()*+ = 16:<6B'9C = 16 × 6143239.4488 ##9

< × 6B'9C

78 = 22.99 ## ≈ EF. G HH

II. Design for Key 6I, K, LM9

a. Key selection

&./0 = NOPQRST

=UVWP

X×Y - 1./0 = NOZRST[\]

= NX×=^

U >

: = &./0 =_P` > 6a × �9 b = 1./0 ca × =d

`>e

!'()*+ = B'2

: = b × !

&./0 fB'2 g 6a × �9 = 1./0 ha × fi

2 gj fB'2 g

2k&./0l1./0

= i�

1 = i�

K = I

∴ nop-!q �qr stuu vq p%qw.

b. Key Dimensions

� = B'4 = 23 ##

4 = 5.75## ≈ 6 ##

K = I = y HH

- In order for the key to transmit full power from the shaft, the torque that is transmitted

by the shaft is equal to the torque received by the key.

&./0 = `z_P6X{×Y9 &'()*+ = |}z

~6_P9�

: = : & � = B'4

&./0 hB'2 j fa. × B'

4 g = &'()*+ c <16e 6B'9C

a. = &'()*+&./0

=<2> B' = 60 ,�-

50 ,�- =<2> 623 ##9 = 43.354 ##

- Smallest permissible length of key.

- For convenience, we let length of key as:

LM ≈ �� HH

c. Key design failure check

&./0 = `z_P6X×Y9 1./0 = `z

_PfX×=^U >g

50 ,�- = `×|�C`C�.���� ���`C ��×6�� ��×} ��9 100 ,�- = `×|�C`C�.���� ���

`C ��×f�� ��×=� ��U >g

50 ,�- > 46.132 ,"- 100 ,�- > 92.26 ,�-

- Induced shear and crushing stresses in the key are less than the allowable stresses;

therefore the design for key is safe.

III. Design for Hub

a. Hub diameter 6B(9

- Since we are not given with the type of material and strength of the hub, it is safe to

assume that 6B( = 2B'9

7� = 2 × 23## = �y HH

b. Hub length 6a(9

- To avoid any failure at the key or in the hub, let a. = a(

L� = �� HH

c. Hub allowable shear strength

- Given that we’ve assumed B( = 46 ##, we can now calculate for the allowable shear

stress in the hub. Considering the hub as a hollow shaft, and transmitting the same

torque as that of the main solid shaft, we have:

&(�2 = 16:6B(9<66B(9� − 6B'9�9 = 16 × 143239.4488 ## × 46 ##

<6646 ##9� − 623 ##9�9

&(�2 = 7.99 ,�-

���� = � ���

IV. Design for Flange Coupling thickness k��l and diameter 67�9

a. Flange thickness

- Because the hub and flange are of the same material, and the flange at the junction is

under shear with the hub while transmitting torque, we have:

&*4)��/ = 8 ,�-

8 ,�- = b�'(/)�

=2:B(

6< × B(9 × �*

�* = 2:6< × 6B(9`9 × 68 ,�-9 = 2 × 143239.4488 ##

6< × 646 ##9`9 × 8 ,�-

�* = 5.387 ##

- Minimum permissible thickness of flange.

- For better geometry/proportion with respect to the shaft: k�* = 0.5B'l

�� = 0.5 × 23 ## = ��. � HH

b. Flange Coupling Diameter

- Since there is no direct load at the outer diameter of the flange, we can let B� = 4B',

provided that there will be an allowance left for the diameter of the bolt circle and for

the diameter of the bolts.

7� = 4 × 23 ## =  E HH

c. Flange design failure check

8 ,�- = 2 × 143239.4488 ##6< × 646 ##9`9 × 11.5 ##

8 ,�- > 3.75 ,�-

- Induced shear stress in the flange is less than the allowable stress; therefore the design

thickness for the flange is safe.

V. Design for Bolts 67�9 and the Bolt Circle 67��9

a. Bolt Diameter

- Bolts are subject to shearing and bearing loads while transmitting torque, and we let the

number of bolts to be6¡ = 49.

- We are given with the allowable shear of bolts at 25 Mpa, it is therefore safe to assume

that the crushing strength of a bolt is at 50 Mpa 6&234+ = 0.51234+9.

1234+ =UV

WZ¢_Zk+£l = `z

6_Z¢96_Z9k+£l &234+ = NOPQRST

= �N~6_Z9U�

: = 1234+6B29 _Z¢` k�*l b = ¤Z¥¦§6~96_Z9U�

�

! = B2�2

: = b × !

1234+6B29 _Z¢` k�*l = =¤Z¥¦§6~96_Z9U�

� > × =_Z¢` >

B2 = 461234+9k�*l&234+6<96¡9 = 4650 ,�-9611.5 ##9

625 ,"-9<649

B2 = 7.32 ##

- Minimum permissible diameter of bolt

- Next standard diameter of a hexagonal-head thorough bolt is 8 mm. So,

7� = � HH

b. Bolt Circle

- We already have the number of bolts, and the diameter of each bolt. So we can now

calculate for the design diameter of the bolt circle,

&234+ = 8:<6B2�96B29`¡

B2� = 8:<6&234+96B29`¡ = 8 × 143239.4488 ##

<625 ,�-968 ##9`649

B2� = 57 ##

- Minimum permissible diameter of the bolt circle.

- To give some space for the bolt head and nut, we can let B2� = 3B'

B2� = 3 × 23 ##

7�� = y  HH

c. Bolt design failure check

25 ,�- = 8 × 143239.4488 ##<669 ##968 ##9`649

25 ,�- > 20.65 ,�-

- Induced shearing stress in the bolts is less than the allowable stress; therefore the

design diameter for the bolts is safe.

VI. Spigot Depth (A & B)

a. Protrusion and Recess

- The spigot depth on one flange and recess on the opposing face (A and B) is for the

purpose of alignment and ease of assembly.

- The protrusion and corresponding recess of the flanges which is mainly provided for

location may be taken as 1.75 mm.

¨ = �. ©� HH

ª = �. ©� HH

VII. Design sketch

a. Orthographic design sketch

- Illustrates side and cross-sectional view of each machine part, which also includes their

respective dimensions.

b. Isometric Design sketch

- Illustrates isometric assembly of the coupling.

Odilon Cuenco ME – 414

BSME – IV ASSIGNMENT

1. Select a key for a 4-inch diameter shaft transmitting 1000 HP at 1000 rpm, the allowable shear stress

in the key is 15000 PSI and an allowable compressive stress is 30000 PSI. What type of key should be

used if allowable shearing stress is 5000 PSI and the allowable compressive stress is 20,000 PSI?

�� = 4 ��

� = 1000 �� ×550 �� − ��

�1 �� × 12 ��

1 �� = 6600000 �� − ���

� = 1000 ���� × 1 �

60 � = 1006 ���

� = �2 × � × � =

6600000 �� − ���

2 × � × �1006 ����

= 63025.35746 �� − ��

a. Given:

#$%& = 15000 �'( )$%& = 30000 �'(

Solution:

#$%& =*+,-

.×/ = 012-×.×/ )$%& =

*+,-

�3* �×/

= 412-×5×/

� = 6789×2-×.×/0 � = :789×2-×5×/

4

� = �

#$%& × �� × ; × <2 = )$%& × �� × � × <

4

2#$%&)$%&

= 2=15000 �'(>30000 �'( = �

;

� = ;

? = ��4 = 4 ��

4 = @ AB

C = @ AB

D = 2�#$%& × �� × ; = 2 × 63025.35746 �� − ��

15000 �'( × 4 �� × 1 �� = E. @FFGH AB

b. Given:

#$%& = 5000 �'( )$%& = 20000 �'(

Solution:

2#$%&)$%&

= 2=5000 �'(>20000 �'( = �

;

0.5; = �

? = ��4 = 4 ��

4 = @ AB

C = F. H AB

D = 2�#$%& × �� × ; = 2 × 63025.35746 �� − ��

5000 �'( × 4 �� × 1 �� = I. JFJ AB

2. A flange coupling is to connect two 57 mm shaft, the hubs of the coupling are each 111 mm in

diameter and 92 mm thick and the flanges webs are 19 mm thick. Six 16 mm bolts in a 165 mm

diameter bolt circle connected the flanges. The key is 6 mm shorter than the hub thickness and

the key is 14 mm x 14 mm. Coupling is to transmit 45 KW at 165 RPM. For all the parts, yield

point in shear is one-half the yield point in tension or compression which is 448 MPa. Find the

following stress and factor of safety based on yield point:

Torque:

� = 45 K; = 45000 ;

� = 165 L�M = 2.75 L�'

� = �2 × � × � = 45000 ;

2 × � × =2.75 L�'> = 2604.353614 �� = 2604353.614 ���

Material:

)&N = 448 M�P

#&N = 0.5)&N = 224 M�P

Shaft:

�� = 57 ��

Key:

; = 14 ��

� = 14 ��

< = 92 �� − 6 �� = 86 ��

Hub and flange:

�R = 111 ��

<R = 92 ��

�S = 19 ��

Bolt circle and Bolts:

�TU = 165 ��

� = 6 �V���

�T = 16 ��

a. Shear in the key

#$%& =2���

; × < = 2��� × ; × < = 2 × 2604353.614 ���

57 �� × 14 �� × 86 �� = 75.8977 M�P

#$%& = #&N�

� = #&N#$%&

= 224 M�P75.8977 M�P = 2.951

∴ (�XYZ�X �ℎ�P� ������ P� �ℎ� \�] �� ^H. G_^^ `ab P�X cPZ�V� V� 'P���] �� E. _H@

b. Bearing in the key

)$%& =2���

��2 � × <

= 4��� × � × < = 4 × 2604353.614 ���

57 �� × 14 �� × 86 �� = 151.7954 M�P

)$%& = )&N�

� = )&N)$%&

= 448 M�P151.7954 M�P = 2.951

∴ (�XYZ�X ��P���d ������ P� �ℎ� \�] �� @H@. ^_He `ab P�X cPZ�V� V� 'P���] �� E. _H@

c. Shear in the bolts

#Tfgh� =2��TU

�=�T>0�4

= 8��TU × � × =�T>0� = 8 × 2604353.614 ���

165 �� × � × =16 ��>0 × 6 = 26.168 M�P

#Tfgh� = #&N�

� = #&N#Tfgh�

= 224 M�P26.168 M�P = 8.56

∴ (�XYZ�X �ℎ�P���d ������ P� �ℎ� �V��� �� EI. @IG `ab P�X cPZ�V� V� 'P���] �� G. HI