written 5.2 years ago by | • modified 2.2 years ago |
Mumbai University > Mechanical Engineering > Sem 4 > Production Process II
Marks: 5M
written 5.2 years ago by | • modified 2.2 years ago |
Mumbai University > Mechanical Engineering > Sem 4 > Production Process II
Marks: 5M
written 5.2 years ago by |
therefore $R=\frac{F_{s}}{cos[ ∅+β-α])}$ ………………2
Shear stress; $f_{s}=\frac{F_{s}}{A_{s}}$
therefore $F_{s}= f_{s} \times \frac{t_{1}\times b}{sin∅}$ ………………………..3
Putting the value of $F_{s}$ from eqn. 3 in eqn. 2 and then the value of R in eqn. 1
$F_{c}= f_{s} \times \frac{t_{1}×b}{sin∅} \times \frac{1}{cos[ ∅+β-α])} x cos (β-α)$
Work consumed in cutting; $W_{c}= F_{c} \times V_{c}$
Now, $W_{c}= f_{s} \times \frac{t_{1}×b}{sin∅} \times \frac{1}{cos[ ∅+β-α]} times cos (β-α) x V_{c}$
All except φ are constant and for $W_{c}$ to be minimum the denominator must be maximum which is a function of φ.
$\frac{d}{d∅}$(Denominator)=0
Therefore, $sinφ [-sin (φ+β –α)] +cos (φ+β –α) .cos φ=0$
$∴cos [2∅+β-α]=0$
Therefore, $2 φ+β –α=\frac{π}{2}$
Hence the expression for merchant theory is derived.
Assumptions of merchant theory:
1) The tool is perfectly sharp and there is no contact along the clearance face.
2) The shear surface is plane extending upward from the cutting edge
3) The cutting edge is a straight line, extending perpendicular to the direction of motion and generates a plane surface as the work moves past it.
4) The chip does not flow to either sides.
5) The depth of cut is constant.
6) Width of tool is greater than that of work piece.
7) The work moves relative to the tool with uniform velocity.
8) A continuous chip is produced with no built up edge.