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)](https://img.qammunity.org/2021/formulas/physics/college/83z4gg0xk82l4p0abrymklhuxqox0qleyt.png)
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](https://img.qammunity.org/2021/formulas/physics/college/2f7l83gvf0t1ufy64axquxgni81828zpm9.png)
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](https://img.qammunity.org/2021/formulas/physics/college/oa9o6fedu2gtkbv2om9zuyepd8bwko8jrz.png)
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