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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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Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

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Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

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volume = 3³ = 27 ft³

mass =  3.1  x 27

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