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3 years ago

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Tool life testing on a lathe under dry cutting conditions gauge 'n' and 'C' of Taylor tool life equation as 0.12 and 130 m/min.

respectively. When a coolant was used, 'C' increased by 10%. The increased tool life with the use of coolant at a cutting speed of 90 m/min is

1 answer:

Tom [10]3 years ago

7 0

Answer:

So % increment in tool life is equal to 4640 %.

Explanation:

Initially n=0.12 ,V=130 m/min

Finally C increased by 10% , V=90 m/min

Let's take the tool life initial condition is $T_1$ and when C is increased it become $T_2$ .

As we know that tool life equation for tool

$VT^n=C$

At initial condition $130\times (T_1)^{0.12}=C$ ------(1)

At final condition $90\times (T_2)^{0.12}=1.1C$ -----(2)

From above equation

$\dfrac{130\times (T_1)^{0.12}}{90\times (T_2)^{0.12}}=\dfrac{1}{1.1}$

$T_2=47.4T_1$

So increment in tool life = $\dfrac{T_2-T_1}{T_1}$

= $\dfrac{47.4T_1-T_1}{T_1}$

So % increment in tool life is equal to 4640 %.

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Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

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Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

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From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

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$Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

$Elongated \ Length = 491.28 - 470$

$=21.29mm$

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