In an orthogonal machining operation-if U - specific cutting energy- - - friction angle- - shear plane angle- - - Tool rake angle- then the shear stress - will be expressed as-

Shear stress, τ=F_sbt_1sinϕ [where, F_s= shear force] ⇒ F_s=F_ccos(ϕ+β α)cos(β α) [From Merchant's circle diagram] ∴ nbsp; nbsp;τ=F_csinϕcos(ϕ+β α)bt_1c ...

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Analysis of Orthogonal Machining

In an orthogo...

Question

In an orthogonal machining operation,
if U = specific cutting energy, $β =$ friction angle, $ϕ =$ shear plane angle, $α =$ Tool rake angle, then the shear stress $(τ)$ will be expressed as:

\frac{U sin ϕ cos (β - α)}{cos (ϕ + β - α)}

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\frac{U cos α cos (ϕ + β - α)}{cos (β - α)}

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\frac{U sin ϕ cos (ϕ + β - α)}{cos (β - α)}

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\frac{U sin α cos (ϕ - α)}{cos (β - α)}

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Solution

The correct option is C $\frac{U sin ϕ cos (ϕ + β - α)}{cos (β - α)}$
$Shear stress, τ = \frac{F_{s}}{b t_{1}} sin ϕ$

$[where, F_{s} = shear force]$

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

[From Merchant's circle diagram]

$∴ τ = \frac{F_{c} sin ϕ cos (ϕ + β - α)}{b t_{1} cos (β - α)}$

$Now, \frac{F_{c}}{b t_{1}} = specific cutting energy (U)$

$∴ τ = \frac{U sin ϕ cos (ϕ + β - α)}{cos (β - α)}$

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In an orthogonal machining operation, if U = specific cutting energy, β= friction angle, ϕ= shear plane angle, α= Tool rake angle, then the shear stress (τ) will be expressed as:

In an orthogonal machining operation,
if U = specific cutting energy, $β =$ friction angle, $ϕ =$ shear plane angle, $α =$ Tool rake angle, then the shear stress $(τ)$ will be expressed as: