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

The bending stress is 502.22 MPa

Step-by-step 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) =(d_(p)*2*\pi *w_(p) )/(120)`

Where

wp = speed of the pinion = 1800 rpm

`V_(t) =(0.13*2*\pi *1800)/(120) =12.25m/s`

The factor B is equal to:

`B=((12-Q_(v))^(2/3) )/(4) , if Q_(v) =10\\B=((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)=((83.82)/(83.82+√(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) =(1)/(m) =(1)/(5) =0.2mm^(-1)`

The bending stress is equal to:

`\sigma =(W_(t)P_(d)K_(a)K_(m)K_(s)K_(b)K_(i) )/(FJ_(g)K_(v)) \\\sigma =(22000*0.2*1*1.7*1*1*1.42)/(62*0.41*0.832) =502.22MPa`