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Analysis of Orthogonal Machining

In an orthogo...

Question

In an orthogonal machining operation, the cutting and thrust forces are 600 N and 173 N respectively. If the value of shear angle is $60^{\circ}$ , what is the approximate value of shear force?

100 N

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150 N

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200 N

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300 N

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Solution

The correct option is B 150 N
$F_{s} = F_{c} cos ϕ - F_{t} sin ϕ$

$= 600 cos 60 - 173 sin 60$

$= 600 \times \frac{1}{2} - 100 \sqrt{3} \times \frac{\sqrt{3}}{2} [∵ 173 \approx 100 \sqrt{3}]$

$= 300 - 150$

$\Rightarrow F_{s} = 150 N$

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Linked Data:
During orthogonal machining of a mild steel specimen with a cutting tool of zero rake angle, the following data is obtained:
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In an orthogonal machining operation, the cutting and thrust forces are 600 N and 173 N respectively. If the value of shear angle is 60∘, what is the approximate value of shear force?

The correct option is B 150 NFs=Fccosϕ−Ftsinϕ =600cos60−173sin60 =600×12−100√3×√32[∵173≈100√3] =300−150 ⇒Fs=150 N

In an orthogonal machining operation, the cutting and thrust forces are 600 N and 173 N respectively. If the value of shear angle is $60^{\circ}$ , what is the approximate value of shear force?

The correct option is B 150 N
$F_{s} = F_{c} cos ϕ - F_{t} sin ϕ$

$= 600 cos 60 - 173 sin 60$

$= 600 \times \frac{1}{2} - 100 \sqrt{3} \times \frac{\sqrt{3}}{2} [∵ 173 \approx 100 \sqrt{3}]$

$= 300 - 150$

$\Rightarrow F_{s} = 150 N$