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3 years ago

12

You are performing a machining operation that approximates orthogonal cutting. Given that the chip thickness prior to chip forma

tion is 0.5 inches and the chip thickness after separation is 1.125 inches, calculate the shear plane angle and shear strain. Use a rake angle of 10 degrees. 21. Suppose in the prior problem that the cutting force and thrust force were measured as 1559 N and 1271 N, respectively. The width of the cut is 3.0mm. Using this new information, calculate the shear strength of the material.

1 answer:

Mnenie [13.5K]3 years ago

3 0

Answer:

A)

shear plane angle = 31.98°

shear strain = cot ( 31.98° ) + tan ( 31.98 - 10 )

B) shear strength = 7339.78

Explanation:

<u>a) Determine the shear plane angle and shear strain</u>

Given data :

Chip thickness before chip formation = 0.5 inches

Chip thickness after separation = 1.125 inches

rake angle ( ∝ ) = 10°

<u>shear plane angle </u>: Tan ∅ = $\frac{rcos\alpha }{1-sin\alpha }$ ----- ( 1 )

r = chip thickness ratio = 0.5 / 1.125 = 0.4444

back to equation 1 : Tan ∅ = ( 0.444 ) * cos 10 / 1 - sin 10

Tan ∅ = 0.4444 * 0.9848 / 1 - 0.1736 = 0.5296

hence ∅ = tan^-1 ( 0.5296 ) = 31.98°

<u>shear strain</u> : R = cot ∅ + tan ( ∅ - ∝ ) ---------- ( 1 )

R = cot ( 31.98° ) + tan ( 31.98 - 10 )

<u>B) determine the shear strength of the material</u>

cutting force = 1559 N

thrust force = 1271 N

width of cut ( diameter ) = 3.0 mm

shear strength = c + σ.tan ∅

c = cohesion force = 1271 * 3 = 3813

σ = normal stress = F / A = 1559 / π/4 * ( 0.5 )^2 = 1559 / 0.1963 = 7941.94

hence : shear strength of material = 3813 + 7941.94 * 0.6244 = 7339.78

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