A shaft is to be designed to support the spur pinion and hel

A shaft is to be designed to support the spur pinion and helical gear shown in the figure on two hearings spaced 700 mm center-to-center. Bearing A is a cylindrical roller and is to take only radial load: bearing B is to take the thrust load of 900 N produced by the helical gear and its share of the radial load. The bearing at B can be a ball bearing. The radial loads of both gears are in the same plane, and are 2.7 kN for the pinion and 900 N for the gear. The shaft speed is 1200 rev/min. Design the shaft.. Make a sketch to scale of the shaft showing all fillet sizes, key ways, shoulders, and diameters. Specify the material and its heat treatment.

Solution

solution:

1)here design of shaft as per ASME code is as follows

2)here let consider spur gear of 20 degree pressure angle with minimum number of teeth is 18 and for helical gear we choose helix angle to be 23 degree and normal pressure angle to be 20 degree .

3)hence face width of spur gear gives diameter of pinion sas follows

b=10 m

module=100/10=10 mm

where Zp=18 teeth

dp=m*Zp=180 mm

4)for helical gear face width is given by

b=1.15*pi*m/sin23

b=9.24 mn=10 mn

m=50/10= 5 mm

wher diametr is given by

dg=m*Zg/cos23=5*18/cos23=97.77 mm

5)here let consider radial load on shaft,which gives bending moment ofr shaft as follows

here support reactio are

ma=0

2700*175-900*575=Rb*700

Rb=1414.28 N

Mb=0

Ra=2700*525+900*125/700=2185.71 N

hence maximum bending moment is at spur gear as shear force diagram cut axis,maximum bending moment

M1=Ra*175=382499.25 Nmm

6)here for spur gear

F1=resultant force=2700 N act in radial direction hence tangential force is givenby

Ft1=F1cos20

Ft1=2537.17 N

T1=Ft1*dp/2=228345.3 N mm

7)for helical gear it is given that

F2=resultant load in radial plane as resultant of just radial and tangential force =F2=900 N

hence tangential force can be calculated from thrust load as follows

Fa=Ft2*tan23

Ft2=2120.26 N

where torque is given by

T2=Ft2*dg/2=103649.25 Nmm

8)hence resultant torque is sum of both torque

T=T1+T2

T=331994.55 N mm

9)hence by ASME code equivalent torque is

for Kb=1.5,Kt=1

Te=((KB.M)^2+(Kt*T)^2)^.5

Te=662878.68 N mm

10)hence material for shaft is 45C8 steel with

Sut=600 MPA and Syt=380 MPa

here tall=.18*Sut=108 MPa

here reduction by 25% for keyway

tall=.75*108=81 MPa

tall=16*Te/pi*d^3

hence on putting Te and tall we get diameter of shaft as

d=34.67 mm

11)in this way for this system neglecting moment due to thrust load diameter of shaft for minimum number theethed spur and helical gear with standard parameter comes to be d=34.67 mm