Design a speed reducer that is capable of driving two mixers at different rotati

Question

Design a speed reducer that is capable of driving two mixers at different rotational velocities. Both mixers require a 10 kW output power. The first mixer should be driven at 225 ± 5 revolutions per minute. The second mixer should be driven at 60 ± 5 rpm. The axis of input and output shafts are parallel. Design has to be compact with least manufacturing cost.

MAJOR TASKS EXPECTED FROM YOU AS A DESIGNER:

Gears

Select the type of gear set. Determine the number of teeth for each gear.

Analyze the forces on gears and shafts: Force analyses are to be made based on the power transmitted by the gear system. You will need to determine the loads acting on the teeth of each gear pair, i.e. the transmitted load Wt, the axial load Wa and the radial load Wr need to be determined for each gear pair. (You may use n = 200, and = 150 if required)

Determine the stress acting on the gear teeth and select the material to withstand these stresses. Safety is other important tasks. You may take a factor of safety of 2 for this purpose. These will require the determination of face width F and the module m for each pair.

Shafts

You must also be able to draw the shear force and bending moment diagrams, as well as drawing the free-body diagrams and calculating the reactions at the supports. Since the shafts must be able to accommodate all the elements within the range of the supports, you must assign proper length dimensions for each of the shafts, considering the arrangement of the elements and allowing space for the width of each bearing.

From the bending moment diagrams obtained in task # 4, and after calculating the torque transmitted by each one of the shafts, you will select materials for the shafts and design the shafts. A factor of safety of 2.0 seems to be appropriate. Note that the diameter of the shaft will not be constant. There will be some keyways and/or shoulders on each of the shaft creating stress concentration, which need to be considered.

Bearings

Based on the given data and the reactions found, you will select bearings for each of the supports. In this selection, the design life, the design speed, and reliability requirements should all be considered. The life of the bearings should be at least 108 cycles for this gearbox.

Keys, Enclosure and Assembly

You have to determine the size of the keys and keyways for the application, which also requires selection of materials for the intended application.

A scaled drawing of the entire system will have to show all elements of the gearbox including its housing.

Explanation / Answer

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

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

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

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

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

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