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3 years ago

In engineering, economic cost is a decision-making tangible factor. Group of answer choices True False

Engineering

2 answers:

salantis [7]3 years ago

8 0

Economic cost is a rescission making tangible factor true

polet [3.4K]3 years ago

6 0

Answer:

True

Explanation:

Economic cost is a very important factor to consider in decision making in Engineering. It is part of the four essential elements involved in decision making in engineering analysis.

Economic cost is part of the criteria to evaluate alternatives considering the time value of money by estimating a specific measure of worth of estimated commodity cost over a period of time.

Other Factors to consider in Engineering Economics as related to the Economy are:

(1) Time of Occurence of Cash flows

(2) Interest Rates

(3) Cash Flows

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A 10-mm steel drill rod was heat-treated and ground. The measured hardness was found to be 290 Brinell. Estimate the endurance s

grandymaker [24]

Answer:

the endurance strength $S_e$ = 421.24 MPa

Explanation:

From the given information; The objective is to estimate the endurance strength, Se, in MPa .

To do that; let's for see the expression that shows the relationship between the ultimate tensile strength and Brinell hardness number .

It is expressed as:

$200 \leq H_B \leq 450$

$S_{ut} = 3.41 H_B$

where;

$H_B$ = Brinell hardness number

$S_{ut}$ = Ultimate tensile strength

From ;

$S_{ut} = 3.41 H_B$ ; replace 290 for $H_B$ ; we have

$S_{ut} = 3.41 (290)$

$S_{ut} =$ 988.9 MPa

We can see that the derived value for the ultimate tensile strength when the Brinell harness number = 290 is less than 1400 MPa ( i.e it is 988.9 MPa)

So; we can say

$S_{ut} < 1400$

The Endurance limit can be represented by the formula:

$S_e ' = 0.5 S_{ut}$

$S_e ' = 0.5 (988.9)$

$S_e '$ = 494.45 MPa

Using Table 6.2 for parameter for Marin Surface modification factor. The value for a and b are derived; which are :

a = 1.58

b = -0.085

The value of the surface factor can be calculate by using the equation

$k_a = aS^b_{ut}$

$K_a = 1.58 (988.9)^{-0.085$

$K_a = 0.8792$

The formula that is used to determine the value of $k_b$ for the rotating shaft of size factor d = 10 mm is as follows:

$k_b = 1.24d^{-0.107}$

$k_b = 1.24(10)^{-0.107}$

$k_b = 0.969$

Finally; the the endurance strength, Se, in MPa if the rod is used in rotating bending is determined by using the expression;

$S_e =k_ak_b S' _e$

$S_e$ = 0.8792×0.969×494.45

$S_e$ = 421.24 MPa

Thus; the endurance strength $S_e$ = 421.24 MPa

8 0

3 years ago

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

3 0

3 years ago

Single point cutting tool removes material from a rotating work piece to generate a cylinder is called • Facing Tuming • Both 1

madam [21]

Answer:Turning

Explanation: Turning is the process in which the work piece is subjected to machining so that excess part is removed with the help of rotation by turning machine or lathe machine.The cutter tool is used for cutting the excess of the work piece and it is mostly single-pointed so that give accurate removal of the excess of work piece.At times , according to the requirement multi-pointed tool is also used Therefore, the correct option is turning.

6 0

3 years ago

Dear sir i want to ask something about the solution of my question?

Eva8 [605]

No you may not ask the question

3 0

3 years ago

List all the qualities of an engineer?

Dennis_Churaev [7]

Answer:

Math and Computer Skills. A qualified engineer should be good at math, at least through the level of calculus and trigonometry, and understand the importance of following the data when making design decisions.

Organization and Attention to Detail.

Curiosity.

Creativity.

Critical Thinking.

Intuition.

Explanation:

6 0

3 years ago