Question

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

Answer 1

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

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

$A=\frac{1-V_{1}^{2} }{E_{1} }-\frac{1-V_{2}^{2} }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}$

a) The maximum contact pressure is:

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

The width of contact is:

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

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