A shaft is needed to support two pulleys as shown. The shaft rotates at 900 RPM

Question

A shaft is needed to support two pulleys as shown. The shaft rotates at 900 RPM and 15 hp is transmitted between the two pulleys. The driven and slack belt sides of each pulley all lie in the same plane and act in the same direction. Both pulleys have a tension ratio of 3. The ID of each bearing is fixed at 1 in. Each pulley should be seated securely against a solid shoulder, and attached to the shaft by an end-milled key seat. You are not required to size the keys. The inner diameter of each pulley may be machined as needed to match your calculated shaft diameters. You are to use a machined surface finish, and 99.9% reliability as needed. All diametric dimensions should be kept in increments of 0.1 in. The material is to be 4140 steel with a heat treatment of your choosing. Design the shaft, such that the bearing deflections do not exceed .002 rad, and n_infinity 3.5. Your solution will consist of a properly dimensioned drawing of the shaft, along with all calculations necessary to show your design meets the above criteria. Also, calculate and report the weight of your shaft design.

Explanation / Answer

solution:

1)here we have to design a shaft to support two pullies which are transferring power of 15 hp so we can write as

P=T*w

15*746=T*(2*pi*900/60)

T=118.7295 N m

2)where this torque is transferred through pulley so we can write as here we consider F1 and F2 as force on left small pulley and F3 and F4 are as force on large pulley so we get relation as

T=(F1-F2)*d1/2=(F3-F4)*d2/2=118.7295

d1=8'' or .2048 m

d2=12'' or .3072 m

F1/F2=F3/F4=3

so we get on solving as

F2=579.79 N

F1=1739.20 N

Fr1=F1+F2=2318.99 N

F4=386.489 N

F3=1159.467 N

Fr2=F3+F4=1545.95 N

3)here let draw bending moment diagram ,first reaction at bearing A and B are

Ma=0

Rb*.6144-Fr1*.256-Fr2*(.6144-.2176)=0

Rb=1964.67 N

Mb=0 gives

Ra=1900.2681 N

so we get moment as follows

ma=0

Mb=0

M1=486.46 N m

M2=427.51 N m

so maximum bending moment occure at small pulley and maximim bending moment as

M1=mmax=486.46 N,m

where maximum torque as

Tmax=118.7295 N,m

4)here let calculate endurance limit of shaft of AISI4140 steel

Sut=655 MPa

Syt=415 MPa

G=80 GPa

density=7850 kg/m3

i)for surface finish factor,as it is machined

Ka=aSut^b=4.51(655)^-.265=.8088

Kg=.753 for 99.9% reliabilty

Se'=.5Sut

so we get final endurance limit of shafts

Se=ka*kg*Se'=199.456 MPa

5)here bending moment acting on shaft is completely reversed and hence

Sba=32Mmax/pi*d^3=4955.04/d^3 N/m2

Sbm=0

similarly as torque is uniform so we can write

tm=16Tmax/pi*d^3=604.68/d^3 N/m2

ta=0

so we get resultant mean and amplitude bending moment as

Sm=(Sbm^2+3*tm^2)^.5=1047.33/d^3

Sa=(Sba^2+3*ta^2)^.5=4955.04/d^3

here factor of safety as Nf=3.5

so we can write equation for modified goodman line as

Sm/Sut+Sa/Se=1/Nf

[1047.33/655+4955.04/199.456]*(1/d^3*10^6)=1/3.5

d=.0452327 m

d=45.2327 mm

6)on checking for deflection by formula we get that

T/J=G*theta/L

T=118.7295 N m

J=pi*d^4/32

G=80 GPa

L=.6144

putting value on solving we get that

theta=2.2187*10^-3 as this greater than .002 so we choose diameter of shaft as per requirement to be

d=47 mm and it gives deflection of theta=1.9033*10^-3

so design of shaft is safe for

d=47 mm and L=.6144 m

7)weight of shaft as follows and heat treatment is hardening by quenching in oil

W=density*(pi*d^2*L/4)=7850*1.06594*10^-3=8.3677 kg

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