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Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 mm. A force of 250 N is transmitted between them. (a) Compute the maximum contact pressure and the width of contact. (b) How deep below the surface is the maximum shear stress, and what is its value

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Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Step-by-step explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa


A=(1-V_(1)^(2) )/(E_(1) )-(1-V_(2)^(2) )/(E_(2) ) =(1-0.3^(2) )/(2.07x10^(5) ) *2=8.79x10^(-6)

a) The maximum contact pressure is:


P=0.564*\sqrt{(F*((1)/(r_(1) )+(1)/(r_(2) )) )/(LA) } =0.564*\sqrt{(250*((1)/(10) +(1)/(15) ))/(20*8.79x10^(-6) ) } =274.58MPa

The width of contact is:


b=1.13*\sqrt{(FA)/(L((1)/(r_(1) )+(1)/(r_(2) )) ) } =1.13*\sqrt{(250*8.79x10^(-6) )/(20*((1)/(10) +(1)/(15) )) } =0.029mm\\2*b=0.058mm

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

T = 0.3*P = 0.3 * 274.58 = 82.37 MPa

At a distance of

0.8*b = 0.8*0.029 = 0.023 mm

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