Question

ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 10. The transmitted tangential load is 22 kN. Conditions are such that Km = 1.7. The teeth are standard 20-degree, full-depth. The module is 5 and the face width 62 mm. Determine the bending stress when the mesh is at the highest point of single tooth contact.

Answer 1

Answer:

The bending stress is 502.22 MPa

Explanation:

The diameter of the pinion is equal to:

$d_{p} =mN_{p}$

Where

m = module = 5

Np = number of teeth of pinion = 26

$d_{p} =5*26=130mm$ = 0.13 m

The pitch line velocity is equal to:

$V_{t} =\frac{d_{p}*2*\pi *w_{p} }{120}$

Where

wp = speed of the pinion = 1800 rpm

$V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s$

The factor B is equal to:

$B=\frac{(12-Q_{v})^{2/3} }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396$

The factor A is equal to:

A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82

The dynamic factor is:

$K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832$

The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables

JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

The diametrical pitch is equal to:

$P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}$

The bending stress is equal to:

$\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}} \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa$