A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be deformed using a tensile load of 18,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.5 x 10-2 mm. Would the 1040 steel be a possible candidate for this application

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Here are the eight stages of the recruitment process and what Garth might expect to occur or carry out at each stage in the explanation part.

What is recruitment?

The process of identifying, attracting, and selecting qualified candidates for a job opening in an organisation is known as recruitment.

Here are the eight stages of the recruitment process, as well as what Garth might expect to happen or do at each stage:

Identifying the Need for the Position: Garth should review the company's staffing needs and determine if a position needs to be filled. Once a decision has been made, he should create a job description and identify the position's requirements.

Garth should create a recruitment plan that includes a timeline for the recruitment process, a list of recruitment sources, and an advertising strategy.

Garth should actively seek qualified candidates through various recruitment channels such as job boards, social media, referrals, and recruiting events.

Screening Candidates: Garth should go over resumes, cover letters, and other application materials to see if candidates meet the job requirements.

Garth should conduct interviews with the most qualified candidates to assess their skills, experience, and fit for the position.

Garth should review all of the information gathered during the recruitment process and choose the best candidate for the position.

Garth should ensure that the new hire has all of the necessary information and resources to succeed in their new role.

Evaluating the Recruitment Process: Garth should go over the recruitment process to see where he can improve.

Thus, these are the stages of recruitment.

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Answer:

the required time for the specimen is  1109.4 min

Explanation:

Given that;

diameter of metal alloy d₀ = 1.7 × 10² mm

Temperature of heat treatment T = 450°C = 450 + 273 = 723 K

Time period of heat treatment t = 250 min

Increased grain diameter 4.5 × 10² mm

grain diameter exponent n = 2.1

First we calculate the time independent constant K

dⁿ - d₀ⁿ = Kt

K = (dⁿ - d₀ⁿ) / t

we substitute

K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250

K = (373032.163378 - 48299.511117) / 250

K = 1298.9306 mm²/min

Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )

dⁿ - d₀ⁿ = Kt

t = (dⁿ - d₀ⁿ) / K

t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306

t = ( 1489328.26061158 - 48299.511117) / 1298.9306

t = 1441028.74949458 / 1298.9306

t = 1109.4 min

Therefore, the required time for the specimen is  1109.4 min

Answer:

[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]

Explanation:

From the information given:

Life requirement = 40 kh = 40 [tex]40 \times 10^{3} \ h[/tex]

Speed (N) = 520 rev/min

Reliability goal [tex](R_D)[/tex] = 0.9

Radial load [tex](F_D)[/tex] = 2600 lbf

To find C10 value by using the formula:

[tex]C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}[/tex]

where;

[tex]x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}[/tex]

[tex]\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}[/tex]

The Weibull parameters include:

[tex]x_o = 0.02[/tex]

[tex](\theta - x_o) = 4.439[/tex]

[tex]b= 1.483[/tex]

∴

Using the above formula:

[tex]C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}[/tex]

[tex]C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}[/tex]

[tex]C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}[/tex]

[tex]C_{10} = 30962.449 \ lbf[/tex]

Recall that:

1 kN = 225 lbf

∴

[tex]C_{10} = \dfrac{30962.449}{225}[/tex]

[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]

Yes! That is true!
When framing a wall, temporary bracing is
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removed before the next level is constructed.