Draw the Free Body Diagram of the examined system and procee

Draw the Free Body Diagram of the examined system and proceed with a complete static analysis (i.e. calculate forces and moments exerted on the shaft by the pulley/belt systems, calculate the reaction forces at supports, calculate the necessary bending moment and shear force diagrams, calculate the resultant shear stress, define the required diameter of the shaft.

What should P1 and P2 be equal to?

Angular velocity = 1000rpm

AISI 1040 steel Tensile strength = 620 MPA, Yield Strength 415MPa.

300 250 D, 400 50N 27ON AS 150 300 D

Solution

solution:

1)for given shaft which has to drive two pulleys we has equal and opposite torque for two pulleys so we can write that

Ta=Tc

(p1-p2)*da/2=(F1-F2)*dc/2

(p1-p2)=(270-50)*(300/250)=264 N

2)where for belt at c we have force ratio as

F1/F2=270/50=5.4

if we are driving another pulley by same material then coefficent of friction will be same and as angle of wrap is same for both of them that is 180 degree,we can take same pressure ratio for pulley at A so we get that

p1/p2=5.4

putting value in p1-p2=264 we get that

p2=60 N

and p1=324 N

3)here total force at A is given by

p1+p2=384 N

and at C we have that

F1+F2=320 N

so we get reaction for vertical plane as

force at A are,Fva=(p1+p2)sin45

so taking moment at o and E as zero we get reaction as follows

Mov=0=(p1+p2)sin45*300-Re*850

Re=95.8310 N

Mev=0=(p1+p2)sin45*550-Ro850

Ro=175.68 N

for vertical plane moment at A is given by

Mov=0

Ma=Ro*300=52704 N mm

Mev=0

for horizontal plane we have force at A as

Fah=(p1+p2)cos45

Moh=0=(p1+p2)cos45*300-320*700+Reh*850

Reh=167.69 N

Meh=0

Roh=48000-(p1+p2)cos45*550/850=-119.22 N

so moment in horizontal plane is

Moh=0

Mah=35766 N mm

Mch=-25153.5 N mm

Meh=0

as maximum moment in horizontal and vertical plane occure at A so we get maximum resultant moment as

Mmax=(Mah^2+Mav^2)^.5=63693.94 N mm

where torque is given by

Tmax=(F1-F2)*dc/2=33000 N mm

so resultant torque is given by

where take Kb=2 and Kt=1.5 ,for bending load is sudden and torque is gradual

Te=((Kb*Mamax)^2+(Kt*Tmax)^2)^.5=136667.1998 N mm

4)where allowable shaer stess is given by with keyways for pulley

tall=.75*.18*Sut=83.7 N/mm2

tall=.75*.3*Syt=93.375 N/mm2

we choose smallest of them so we choose

tall=83.7 MPa

where our shear stress is given by

tall=16*Te/pi*d^3

83.7=16*136667.1998/pi*d^3

d=20.25 mm

and hence for given shaft diameter to sustain all of this is 20.25 mm