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4 years ago

What is the effect of the workpiece specific cutting energy on the cutting forces, and why?

Engineering

1 answer:

ella [17]4 years ago

8 0

Explanation:

Specific cutting energy:

It the ratio of power required to cut the material to metal removal rate of material.If we take the force required to cut the material is F and velocity of cutting tool is V then cutting power will be the product of force and the cutting tool velocity.

Power P = F x V

Lets take the metal removal rate =MRR

Then the specific energy will be

$sp=\dfrac{F\times V}{MRR}$

If we consider that metal removal rate and cutting tool velocity is constant then when we increases the cutting force then specific energy will also increase.

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Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th

Nikitich [7]

Answer:

critical stress = 595 MPa

Explanation:

given data

fracture toughness = 74.6 MPa- $\sqrt{m}$

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress = $\frac{k}{f\sqrt{\pi a}}$ ......................1

put here value and we get

critical stress = $\frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}$

critical stress = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

plain stress condition occur in thin body where stress through thickness not vary by the thinner section.

6 0

3 years ago

A sensor produces a signal with amplitude 15 mV. A voltage amplifier must amplify the signal such that the amplitude of the outp

Nady [450]

Answer:

42.50 dB

Explanation:

Determine the minimum voltage gain

amplitude of input signal ( Vi ) = 15 mV

amplitude of output signal ( Vo) = 2 V

Vo = 2 v

therefore ; minimum gain = Vo / Vi = 2 / ( 15 * 10^-3 )

= 133.33

Minimum gain in DB = 20 log ( 133.33 )

= 42.498 ≈ 42.50 dB

8 0

3 years ago

How to find the voltage(B Aab) in series parallel circuit?

Sindrei [870]

Answer:

Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

(1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

(V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.