You are performing a machining operation that approximates orthogonal cutting. Given that the chip thickness prior to chip formation is 0.5 inches and the chip thickness after separation is 1.125 inches, calculate the shear plane angle and shear strain. Use a rake angle of 10 degrees. 21. Suppose in the prior problem that the cutting force and thrust force were measured as 1559 N and 1271 N, respectively. The width of the cut is 3.0mm. Using this new information, calculate the shear strength of the material.

Answers

Answer 1

Answer:

A)

shear plane angle = 31.98°

shear strain = cot ( 31.98° ) + tan ( 31.98 - 10 )

B) shear strength = 7339.78

Explanation:

a) Determine the shear plane angle and shear strain

Given data :

Chip thickness before chip formation = 0.5 inches

Chip thickness after separation = 1.125 inches

rake angle ( ∝ ) = 10°  

shear plane angle : Tan ∅ = [tex]\frac{rcos\alpha }{1-sin\alpha }[/tex]    ----- ( 1 )

r = chip thickness ratio = 0.5 / 1.125 = 0.4444

back to equation 1 : Tan ∅ = ( 0.444 ) * cos 10 / 1 - sin 10

Tan ∅ = 0.4444 * 0.9848 / 1 - 0.1736  = 0.5296

hence ∅ = tan^-1 ( 0.5296 ) = 31.98°

shear strain :  R = cot ∅ + tan ( ∅ - ∝ ) ---------- ( 1 )

R = cot ( 31.98° ) + tan ( 31.98 - 10 )

  B) determine the shear strength of the material

cutting force = 1559 N

thrust force = 1271 N

width of cut ( diameter ) = 3.0 mm

shear strength = c + σ.tan ∅

c = cohesion force  = 1271 * 3  = 3813

σ = normal stress  = F / A = 1559 / π/4 * ( 0.5 )^2 = 1559 / 0.1963 = 7941.94

hence : shear strength of material = 3813 + 7941.94 * 0.6244 = 7339.78


Related Questions

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial temperature of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath

Answers

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]

[tex]\rho = 7900 \ kg/m^3[/tex]

[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]

[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]

Here, we can't apply the lumped capacitance method, since Bi > 0.1

[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]

[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]

[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]

However, on a single rod, the energy extracted is:

[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]

Hence, for centerline temperature at 50 °C;

The surface temperature is:

[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]

Assume, X Company Limited (XCL) is one of the leading 4th generation Life Insurance
Companies in Bangladesh. The Company is fully customer focused. This Life insurance company are
experimenting with analysis of consumer profiles (to determine whether a person eats healthy food,
exercises, smokes or drinks too much, has high-risk hobbies, and so on) to estimate life expectancy.
Companies might use the analysis to find populations to market policies to. From the perspective of
privacy, what are some of the key ethical or social issues raised? Evaluate some of them.

Answers

Answer:

The issues related to the privacy are:

1. Informational privacy

2. Discrimination factors

3. Biased grouping on the basis of Data mining

4. Lack of consent

5. Morally wrong

6. Illegal distribution of information risks

7. Possibility of threat to life

Let's look at some major concerns:

1. Informational privacy : The concept of privacy of the personal information is totally nullified when the information is being used for a purpose other than the intended one for which it was given. This unethical use of information even for general purposes is not correct and is a matter of concern. It is more like using the sensitive data of others for personal benefit which is purely objectionable and raises security issues. Sometimes the data is also shared with the potential employers which might have certain impacts we are unaware of.

2. Data mining issues : The process of using a certain information to arrive and understand the trend and outcomes is called data mining. In this case, the consumer's data undergoes grouping and might get placed in the wrong group rather than the actual one. Also, there can be a case of biasing towards the groups which are not be focused on, or are not a part of the intended audience. This leads to the discrimination factors if we see it from a social point of view.

3. Lack of consent : Use of information without the consent or awareness of the consumers raises concern over the business ethics followed by the company. No one deserves the right to misuse information for his personal benefits without any of its information to the consumer. It is morally wrong and againt the work ethics. Moreover, it raises trust issues between the two involved, and hence is socially unacceptable.Explanation:

What regulates the car engines temperature

Answers

Answer:

car’s thermostat is used to regulate the temperature of the engine to help the engine stay cool.

Explanation:

Answer:

Coolant

Explanation:

Compute the volume percent of graphite, VGr, in a 3.9 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.

Answers

Answer:

Vgr = 0.122 = 12.2 vol %

Explanation:

Density of ferrite = 7.9 g/cm^3

Density of graphite = 2.3 g/cm^3

compute the volume percent of graphite

for a 3.9 wt% cast Iron

W∝ =  (100 - 3.9) / ( 100 -0 ) = 0.961

Wgr = ( 3.9 - 0 ) / ( 100 - 0 ) = 0.039

Next convert the weight fraction to volume fraction using the equation attached below

Vgr = 0.122 = 12.2 vol %

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