Question

Design a speedreducer that is capable of driving two mixers at different rotational velocities. The first mixer requires 10 kW output power, and the second one requires 5 kW output power. The first mixer shouldbe driven at 250 ± 5 rpm The second mixer should be driven at 80 ± 5 rpm. The axis ofinput and output shafts areparallel. Input shaft Gear Box Output shaft Select the type of gear set. Determine the number of teeth for each gear. ords

Answer 1

Speed Reducers are mechanical devices generally used for 2 purposes. The primary use is to multyply the amount of Torque generated by an input power source to increse the amount of usable work. They also reduce the input power source speed to achive desired output speeds.

DESIGNE PROCESS

Given :

Power (P) = 10 kW

1st mixer speed (n1)= 225 (tol. 5) rpm

2nd mixer speed (n2)=60 (tol. 5) rpm

y (psy) = 150

x (phy)= 200

1.) GEARS

Type of gear set selected for this speed reducer is HELICAL GEAR

• FORCE ANALYSIS :

Torque (Mt) = (60*10^6*P)/2*pi*n1 ....................... for 1st mixer

put P= 10 and n1= 230, we get

Mt = 415397.39 N-mm

Similarly for 2nd mixer, using the above formula,

put P=10 and n2= 65, we get

Mt= 1469867.712 N-mm

Now, diameter d1 for 1st mixer could determined as,

d1 = (z1*m)/cos y

where z1 is no. of teeth =20 (assumed as per standard data)

and m is module = 5mm (assumed as per standard data)

y is 150 (given)

d1 = (20*5)/cos150

d1= 116.2 mm

Now, calculating loads

1.) Transmitted load (Wt) = 2*Mt/d1

= 2*415397.39/116.2

= 7149.69 N

2.) Axial load (Wa) = Wt*tan(y)

= 7149.69*tan150

=4127.8 N

3.) Radial Load (Wr) = Wt[tan(x)/cos(y)]

2931.37 N

Similarly for 2nd mixer diameter will be

d2 = (Z2*m)/cos(y)

d2 = (30*5)/cos150 .............. here value of Z2 amd m are asumed as per the standard data.

d2 = 174.4 mm

1.) Transmitted load (Wt) = 2*Mt/d2

= 16856.28 N

2.) Axial load (Wa) = Wt*tan(y)

= 9731.97 N

3.) Radial load (Wr) = Wt[tan(x)/cos(y)]

= 6911.07 N

• STRESS ANALYSIS

By stress analysis, let the material of gear is hardened steel (Sut=750)

stress = Sut/f.s ........... here Sut is ultimate strength and f.s is factor of safety.

stress(Ts) = 750/2

stress(Ts)= 375 N/mm2

By lewis equations .. Sb= m*b*Ts*Y  

also Sb= Peff *(f.s)

Peff = 859.61/m .......... by standard data ................ (1)

and Sb = 1350*m^2 .................................................... (2)

equating (1) and (2) we get

1350*m^2 = 859.61/m

m = 1.6 mm

b (thickness) = 10*m

= 16 mm

2.) DESIGN OF SHAFT :

Designing of shaft could be done by the following steps

• Permisible shear stress

T(shear stress) = Ssy/ f.s  

= 0.5*Syt/ f.s  

here, Ssy is yeild Strength and Syt is tesile strength and f.s is factor of safety

• Torsional moment

P1 = P2*e^(uo)

where, P1 and P2 are maximum and minimum tension

u is coefficient of friction and o is angle

Now, torque supplied will be

Mt (torque) = (P1 - P2)*R ........... here, R is radius of shaft

Hence, by this way optimim disigning could be achived