ME – 414 MACHINE DESIGN 1
PLATE NO.4 KEYS AND COUPLING
NAME: ODILON CUENCO INSTRUCTOR: 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.
= 60 = 50 = 100 100 = 25
= 15 = 15000 = 1000 J = 100 6 J I.
Design for Shaft Diameter a.
Torque
= 2 × × = b.
II.
{
15000 = 143.23 143.23944 94488 88 = 143239. 143239.4488 4488 100 2 × × 6 J
Shaft Diameter
16 × {143239.4488 60 = = 16 = {% × {% = 22.99 ≈ . Design for Key {,, a. Key selection
= = = × = × = $ { × = × $ J = 2 =×J 2F { × = @ × 2 F 2F 2 = 1 = = ∴ J J .
b. Key Dimensions
= 4 = 23 4 = 5.75 ≈ 6 = = -
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.
= {$× ×
= # {
= & = 4 @2 × 4F = 16 {% = 60 {23 = = 43.3 43.354 54 2 50 2 c.
Smallest permissible length of key. For convenience, we let length of key as:
≈
Key design failure check
= {$× × 50 = $%$×#%$%. ×{ ×{' × 50 >46.132 -
III.
$ = × F
100 100 = $%$×#%$%. ×' × F 100 >92.26
Induced shear and crushing stresses in the key are less than the allowable stresses; therefore the design for key is safe.
Design for Hub
a. Hub diameter -
Since we are not given with the type of material and strength of the hub, it is safe to assume that
{ = 2
b. Hub length -
{
{
X = 2 × 23 = = X =
To avoid any failure at the key or in the hub, let
c.
Hub allowable shear strength -
= 46
Given that we’ve assumed , 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:
16{ = 16×143239.4488 ×46 = {{{{16{ {{{{46 46 − {23 − { = 7.99 99 X = IV.
Design for Flange Coupling thickness
and diameter {
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, w e have:
= 8 2 8 = = { × × 2×143239.4488 = { × { $2 × {8 = { × {46 $ × 8 = 5.387 387 -
- Minimum permissible thickness of flange. For better geometry/proportion with respect to the shaft:
=0.5
= 0.5 × 23 = . b. Flange Coupling Diameter -
= 4
Since there is no direct load at the outer diameter of the flange, we can let , provided that there will be an allowance left for the diameter of the bolt circle and for the diameter of the bolts.
= 4 × 23 =
c.
Flange design failure check
8 = {2×143239.4488 × {46 $ × 11.5 11.5 8 > 3.75 7 5 -
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
{ and the Bolt Circle {
a. Bolt Diameter -
Bolts are subject to shearing and bearing loads while transmitting torque, and we let the number of bolts to be . 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 .
{ = 4
-
$ = = {{ { = { $
{ =0.5 = = { { = {{
J = 2 =×J { × { $ = {{ $ = 4{50 { {11.5 11.5 = 4{ {{{ {25 {4 = 7.32 32
- Minimum permissible diameter of bolt Next standard diameter of a hexagonal-head he xagonal-head thorough bolt is 8 mm. So,
-
=
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,
-
c.
= {8{{$ = {8{{$ = 8×143239.4488 {25 { {88 ${4 = 57 - Minimum permissible diameter of the bolt circle. To give some space for the bolt head and nut, we can let = 3 = 3 × 23 =
Bolt design failure check
25 = 8×143239.4488 {69 { {88 ${4 25 > 20.6 20.655 -
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.
V = . = . 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 BSME – IV 1.
ME – 414 ASSIGNMENT
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?
a.
= 4 J 550 I− J = 66000 I−J = 1000 H × 1 H × 12 660 0000 00 1 1 = 100 =1000 × 60 6 6600000 I−J = 63025 =2×× = 63025.3.357 5746 46 I − J 2 × × 100 6
Given:
= 1500 150000
= 30000 30000
Solution:
$ = × = ×× = ×$××
=
×
& = ××
= ×&××
= × × × = × × H × 2 4 2 = 2{15000 = H 30000 H= = 4 = 44J = X = X × 6302 63025.5.3574 3 57466 II − J = . = ×2 × = 2150 15000 × 4 J J × 1 J J . X
b.
Given:
= 5000 5000 Solution:
= 20000 20000
2 = 2{5000 = H 20000 0.5=H = 4 = 44J = X = . X × 6302 63025.5.3574 3 57466 II − J = . X = ×2 × = 25000 5000 × 4 J J × 1 J J
2. A flange coupling is to connect two 57 mm shaft, the hubs of the coupling are each 111 mm in diameter and 92 mm thick and the flanges webs are 19 mm thick. Six 16 mm bolts in a 165 mm diameter bolt circle connected the flanges. The key is 6 mm shorter than the hub thickness and the key is 14 mm x 14 mm. Coupling is to transmit 45 KW at 165 RPM. For all the parts, yield point in shear is one-half the yield point in tension or compression which is 448 MPa. Find the following stress and factor of safety based on yield point: Torque:
= 45 = 45000 = 165 = 2.75 = 2604.35 = 2 × × = 2 × 45000 36144 = 260435 2604353.6 3.614 14 × {2.75 2604.35361 Material:
= 448 448 =0.5 = 224 224
Hub and flange:
= 57
= 111 111 = 92 = 19
Key:
Bolt circle and Bolts:
Shaft:
= 14 H = 14 = 92 − 6 = 86 a.
= 165 165 J = 6 I = 16
Shear in the key
2 = 75.8977 = × = ×2 × = 572×2604353.614 8977 × 14 × 86 = 224 = 2.951 = = 75.8977 951 ∴ J J ℎ ℎ . . W W J . .
b. Bearing in the key
2 = 151.7954 = H = ×4H × = 574×2604353.614 7954 × 14 × 86 2 × = 448 = 2.951 = = 151.7954 951 ∴ J IJ IJ ℎ . . W W J . .
c.
Shear in the bolts
2 = 26.1 = {$J = × 8× { $J = 1658×2604353.614 68 × × {16 $ × 6 26.168 4 = 224 =8.56 = = 26.168 ∴ J ℎJ ℎJ ℎ I I . . W W J . .
