Question

Shear plane angle and shear strain: In an orthogonal cutting operation, the tool has a rake angle = 16°. The chip thickness before the cut = 0.32 mm and the cut yields a deformed chip thickness = 0.72 mm. Calculate (a) the shear plane angle and (b) the shear strain for the operation.

Answer 1

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 = $\frac{t1}{t2}$    ................. 1

put here value

chip thickness ratio  = $\frac{0.32}{0.72}$  

chip thickness ratio r = 0.45

so here shear angle will be Ф

tan Ф = $\frac{r*cos\alpha }{1-rsin\alpha}$   ............2

tan Ф = $\frac{0.45*cos16 }{1-rsin16}$  

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