In turning trail using orthogonal cutting, a chip length of 84mm was obtained for uncut chip length of 200 mm. The cutting conditions were:
$V_{c} = 30$ m/min, uncut chip thickness is 0.5 mm, rake angle is $20^{\circ}$ , width of cut is 2 mm, cutting force is 900 N, and cutting tool is HSS.

The shear stress induced in the layer of work material is

284 MPa

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267 MPa

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239 MPa

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None

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Solution

The correct option is A 284 MPa
Shear stress,

$τ = \frac{F_{s}}{Sheared area}$

$= \frac{F_{s}}{L_{s} \times b} = \frac{F_{s}}{t_{1} \times b} \times sin ϕ$

$= \frac{F_{s}}{cos (β - α)} \times cos (ϕ + β - α)$

$= \frac{900}{cos (40.26 - 20)} \times cos (24.74 + 40.26 - 20)$

$F_{s} = 678.36 N$

$τ = \frac{678.36 sin 24.74}{0.5 \times 2} = 283.89 = 284 MPa$

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Q. In turning trail using orthogonal cutting, a chip length of 84mm was obtained for uncut chip length of 200 mm. The cutting conditions were:

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The correct option is A 284 MPaShear stress, τ=FsSheared area =FsLs×b=Fst1×b×sinϕ =Fscos(β−α)×cos(ϕ+β−α) =900cos(40.26−20)×cos(24.74+40.26−20) Fs=678.36 N τ=678.36sin24.740.5×2=283.89=284 MPa