The shaft shown in the figure below receives P hp from a win

The shaft shown in the figure below receives P hp from a wind turbine output shaft through a V-belt sheave at point C. The gear pair at E delivers P_1 hp to an electric generator. The V belt sheave at A delivers P_2 to a bucket elevator to carry food grains. The shaft rotate at N rpm. The sheavs and gear are located axially by retaining rings. Assume suitable key sets to relain the sheaves and gear on the the shaft. Use AISI 1050 cold drawn steel for the shaft. Compute the minimum expectable diameters D_1 through D_2 as explained in the figure. Assume P = (100 + x + y)hp P_1 = (65 + y)hp P_2 = (P + P_1)hp N = (1500 + 10X+ Y)rpm Orientation of elements A, C, and E as viewed from right end of by shaft.

Solution

solution:

1)here design of shaft of diameter D2

power=P2=41*746=(F1-F2)*V

we get,D=L=.1524 m

F1-F2=2446.07 N

where for v belt with 180 wrap angle and groove angle 35 degree force ratio is 2.56

F1=2.56F2

on putting we get

F2=1567.99 N

F1=4014.06 N

torque is=T=(F1-F2)*.1524/2=186.39 Nm

for ces in vertical and horizontal plane are

Fv1=F1sin60=1357.9 n

Fv2=F2sin60=3476.27 N

Frv=4834.18 N

Fh1=F1cos60=783.9 N

Fh2=f2cos60=2007.03 N

Frh=2791.02 N

bending moment in vertiacal and horizontal plane is

Mv=Frv*L=736.72 Nm

Mh=Frh*L=425.35 Nm

M=850.7 Nm

equavalent torque

Te=((KbM)^2+(KtT)^2)^.5

Kb=1.5

Kt=1

we get Te=1289.59 Nm

t allowable=.18 Sut

Sut=790 N/mm2

t all=.18*Sut=142.2

but reduction 25%

t all=106.65 N/mm2

by ASME code

Tall=16*te/pi*d^3

D2=15.54 in

2)design for diameter D4

here design of shaft of diameter D4

power=P2=113*746=(F1-F2)*V

we get,D=.2032,L=.1524 m

F3-F4=5056.12 N N

where for v belt with 180 wrap angle and groove angle 35 degree force ratio is 2.56

F3=2.56F4

on putting we get

F4=3241.16 N N

F3=8297.3 N

torque is=T=(F1-F2)*.2032/2=513.71 Nm

for ces in vertical and horizontal plane are

Fv1=F3sin40=2088.37 N

Fv2=F4sin40=5333.4 N

Frv=7415.4 N

Fh1=F3cos40=2482.87 N

Fh2=f4cos40=6356.1 N

Frh=8838.97 N

where reaction at B and D in vertical and horizontal plane are

Rbv=Rdv=3707.7 N

Rbh=Rdh=4419.48 N

Rbr=Rdr=5768.78 N

M=Rbr*L=879.16 Nm

equavalent torque

Te=((KbM)^2+(KtT)^2)^.5

Kb=1.5

Kt=1

we get Te=1415.26 Nm

t allowable=.18 Sut

Sut=790 N/mm2

t all=.18*Sut=142.2

but reduction 25%

t all=106.65 N/mm2

by ASME code

Tall=16*te/pi*d^3

D4=16.03 in

2)design for diameter D6

1)here design of shaft of diameter D6

power=P1=72*746=T*pi*2*N/60

torque is=T=P/w=327.32 Nm

for ces in tangential and radial direction are

Ft=T*2/D=2147.76 N

Fr=Fttan20=781.72 N

f=(Ft^2+Fr^2)^.5

M=F*L=348.32 Nm

M=348.32 Nm

equavalent torque

Te=((KbM)^2+(KtT)^2)^.5

Kb=1.5

Kt=1

we get Te=616.54 Nm

t allowable=.18 Sut

Sut=790 N/mm2

t all=.18*Sut=142.2

but reduction 25%

t all=106.65 N/mm2

by ASME code

Tall=16*te/pi*d^3

D6=12.15 in