Answer:
The percentage of the total energy dissipated into shear plane is 89.46%.
Explanation:
Given that,
Rake angle = 10°
Thickness ratio= 0.5
Cutting Force = 400 N
Thrust force = 200 N
Speed =3 m/s
Suppose the shear force is 345.21 N.
We need to calculate the shear plane angle
Using formula shear angle
[tex]\tan\phi=\dfrac{r\cos\alpha}{1-r\sin\alpha}[/tex]
Put the value in to the formula
[tex]\tan\phi=\dfrac{0.5\cos10}{1-0.5\sin10}[/tex]
[tex]\tan\phi=0.539[/tex]
[tex]\phi=\tan^{-1}(0.539)[/tex]
[tex]\phi=28.32^{\circ}[/tex]
We need to calculate the shear velocity
Using formula of shear velocity
[tex]v_{2}=\dfrac{v\cos\alpha}{\cos(\phi-\alpha)}[/tex]
Put the value into the formula
[tex]v_{2}=\dfrac{3\times\cos10}{\cos(28.32-10)}[/tex]
[tex]v_{2}=3.11\ m/s[/tex]
We need to calculate the percentage of the total energy dissipated into shear plane
Using formula of energy dissipated
[tex]\%d=\dfrac{P_{s}}{P}\times100[/tex]
[tex]\%d=\dfrac{F_{s}\times v_{c}}{F_{c}\times v}\times100[/tex]
Put the value into the formula
[tex]\%d=\dfrac{345.21\times3.11}{400\times3}\times100[/tex]
[tex]d=89.46\%[/tex]
Hence, The percentage of the total energy dissipated into shear plane is 89.46%.
