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3 years ago

8

A pencil sharpener has a handle with a radius of 1.5 cm. The sharpening mechanism has a radius of 0.5 cm. What is the IMA of the

pencil sharpener?

1 answer:

abruzzese [7]3 years ago

8 0

1.5 / 0.5 = 3 I believe this is the right answer

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How many protons are present in an atom

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3 years ago

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 m

Rom4ik [11]

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

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coldgirl [10]

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