Answer:
shear plane angle Ф = 26.28°
shear strain 2.20
Explanation:
given data
angle = 16°
chip thickness t1 = 0.32 mm
cut yields chip thickness t2 = 0.72 mm
solution
we get here first chip thickness ratio that is
chip thickness ratio = [tex]\frac{t1}{t2}[/tex] Â Â ................. 1
put here value
chip thickness ratio  = [tex]\frac{0.32}{0.72}[/tex] Â
chip thickness ratio r = 0.45
so here shear angle will be Ф
tan Ф = [tex]\frac{r*cos\alpha }{1-rsin\alpha}[/tex]  ............2
tan Ф = [tex]\frac{0.45*cos16 }{1-rsin16}[/tex] Â
tan Ф = 0.4938
Ф = 26.28°
and
now we get shear strain that is
shear strain r = cot Ф + tan (Ф - α )  ................3
shear strain r  = cot(26.28) + tan (26.28 - 16 )
shear strain r = 2.20