Respuesta :
Answer:
The bending stress is 502.22 MPa
Explanation:
The diameter of the pinion is equal to:
[tex]d_{p} =mN_{p}[/tex]
Where
m = module = 5
Np = number of teeth of pinion = 26
[tex]d_{p} =5*26=130mm[/tex] = 0.13 m
The pitch line velocity is equal to:
[tex]V_{t} =\frac{d_{p}*2*\pi *w_{p} }{120}[/tex]
Where
wp = speed of the pinion = 1800 rpm
[tex]V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s[/tex]
The factor B is equal to:
[tex]B=\frac{(12-Q_{v})^{2/3} }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396[/tex]
The factor A is equal to:
A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82
The dynamic factor is:
[tex]K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832[/tex]
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:
[tex]P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}[/tex]
The bending stress is equal to:
[tex]\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[/tex]
