A gear drive required to transmit maximum power of 25 kw . the pinion ...
Problem Statement:
A gear drive is required to transmit a maximum power of 25 kW. The pinion runs at 200 rpm and has a velocity ratio of 1.25. The center distance between the shafts is 650 mm. The teeth have a 20 stub involute profile. The static stress for the cast iron gear may be taken as 60 MPa, and the face width as 10 times the module. Find the module, face width, and number of teeth in each gear. Check the design on dynamic and wear load, considering the dynamic factor in the Buckingham equation as 80.
Solution:
Finding the Module:
Power transmitted by the gear drive is given by:
P = (2πNT)/60
Where P = 25 kW, N = 200 rpm, and T is the torque transmitted.
Therefore, T = (P × 60)/(2πN) = (25 × 60)/(2π × 200) = 2.98 kNm
From the velocity ratio, we know that:
N1/N2 = 1/1.25 = 0.8
Where N1 is the speed of the pinion and N2 is the speed of the gear.
Since we know the speed of the pinion, we can find the speed of the gear:
N2 = N1/0.8 = 250 rpm
Now, we can find the pitch line velocity:
V = (πD1N1)/60 = (πD2N2)/60
Where D1 is the pitch diameter of the pinion and D2 is the pitch diameter of the gear.
Since the teeth have a 20 stub involute profile, we can use the following equation to find the pitch diameter:
D = (2Zmcosϕ)/π
Where Z is the number of teeth, m is the module, and ϕ is the pressure angle (which we assume to be 20°).
Substituting these values, we get:
D1 = (2 × 20 × m × cos20°)/π
D2 = (2 × Z2 × m × cos20°)/π
Equating the two equations and solving for m, we get:
m = 3.15 mm
Finding the Face Width:
The face width is given as 10 times the module, so:
B = 10m = 31.5 mm
Finding the Number of Teeth:
Using the pitch diameter equations from before, we can find the number of teeth:
Z1 = (πD1)/m = 126.5
Z2 = (πD2)/m = 400
Checking the Design:
To check the design on dynamic and wear load, we need to use the Buckingham equation:
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