In 2001, a land-development company launched an engineering project on land adjacent to a protected wetlands area
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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 = ................. 1
put here value
chip thickness ratio =
chip thickness ratio r = 0.45
so here shear angle will be Ф
tan Ф = ............2
tan Ф =
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