Answer:
r=0.31
Ф=18.03°
Explanation:
Given that
Diameter of bar before cutting = 75 mm
Diameter of bar after cutting = 73 mm
Mean diameter of bar d= (75+73)/2=74 mm
Mean length of uncut chip = πd
Mean length of uncut chip = π x 74 =232.45 mm
So cutting ratio r
[tex]Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}[/tex]
[tex]r=\dfrac{73.5}{232.45}[/tex]
r=0.31
So the cutting ratio is 0.31.
As we know that shear angle given as
[tex]tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }[/tex]
Now by putting the values
[tex]tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }[/tex]
[tex]tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }[/tex]\
Ф=18.03°
So the shear angle is 18.03°.
