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Mekhanik [1.2K]
3 years ago
9

Pam Jones worked for 41 years at the same company and had positive performance ratings and personnel records. She needed a calcu

lator for work which she purchased with her own money but was not reimbursed because she lost the receipt. Later, a security guard stopped her as she was leaving work and discovered the calculator in her belongings. After a brief internal investigation, she was fired and it was announced through internal notices that she had committed a theft. The employee sued for libel, saying the company used her as an example to prevent other thefts.a) What are the key issues in this case?b) As an HR Director, how would you have handled this case?
Engineering
1 answer:
UNO [17]3 years ago
5 0

Answer:

a) The key issues are the sue for libel and the evidence.

b) I would make a deal with her and implement a security program in the company.

Explanation:

The main issue in this case is that Pam Jones sued the company for libel, and the company remains in a position in which it has to prove that the internal investigation followed the right steps and indeed, the proves reflected that she was guilty and the fact that she got fired was correct.

The important here is exactly that the theft can be proved.

As an HR Director, I would give the correct proves in order to win the case, and if that is impossible, because of the tools and evidence, I would make a deal with her where both parts can be adequate to the problem.

She can´t be working again in the company but she can get financed according to her working years; also I would use this case as a growing opportunity by implementing new security methods that give more confidence between the company and its employees.

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A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

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