GATE Papers >> Mechanical >> 2016 >> Question No 60

Question No. 60 Mechanical | GATE 2016

The tool life equation for HSS tool is $VT^{0.14}f^{0.7}d^{0.4}$ = Constant. The tool life (T) of 30 min is obtained using the following cutting conditions:
$V=45\mathrm m/\min,f=0.35\mathrm{mm},d=2.0\mathrm{mm}$
If speed $ (\mathrm V) $, feed $ (\mathrm f) $ and depth of cut $ (\mathrm d) $ are increased individually by 25%, the tool life (in min) is

Answer : (B) 1.06

Solution of Question No 60 of GATE 2016 Mechanical Paper

$ \begin{array}{l}\mathrm T=30\;\min\;\;\;\;{\mathrm V}_1=45\;\mathrm m/\min\\{\mathrm f}_1=0.35\;\mathrm{mm}\;\;\;\;{\mathrm d}_1=2\;\mathrm{mm}\\{\mathrm V}_2=1.25\;{\mathrm V}_1\;\;\;\;{\mathrm f}_2=1.25\;{\mathrm f}_1\;\;\;\;{\mathrm d}_2=1.25\;\dot{{\mathrm d}_1}\end{array} $

From tool life equation 

$ \begin{array}{l}{\mathrm V}_1{\mathrm T}_1^{0.14}{\mathrm f}_1^{0.7}{\mathrm d}_1^{0.4}={\mathrm V}_2{\mathrm T}_2^{0.14}{\mathrm f}_2^{0.7}{\mathrm d}_2^{0.4}\\{\mathrm V}_1\left(30\right)^{0.14}{\mathrm f}_1^{0.7}{\mathrm d}_1^{{}^{0.4}}=1.25{\mathrm V}_1{\mathrm T}_2^{0.14}\left(1.25{\mathrm f}_1\right)^{0.7}\left(1.25{\mathrm d}_1\right)^{0.4}\\\left(30\right)^{0.14}=1.25\times{\mathrm T}_2^{0.14}\left(1.25\right)^{0.7}\times\left(1.25\right)^{0.4}\\{\mathrm T}_2=1.06\;\min\end{array} $

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