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for an element with 104 mpa, -48 mpa, and the other stresses equal to zero, what is the safety factor according to the criterion of maximum shear stress if the material has a yield strength of 200 mpa?

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Final answer:

The safety factor according to the criterion of maximum shear stress for the given stresses and a yield strength of 200 MPa is approximately 2.63.

Step-by-step explanation:

To calculate the safety factor according to the criterion of maximum shear stress for a material with given stresses, we must first find the maximum shear stress experienced by the material.

The given normal stresses are 104 MPa in tension and -48 MPa in compression. According to the maximum shear stress theory, also known as Tresca's criterion, the maximum shear stress (τmax) is half the difference between the largest and smallest principal stresses.

τmax = (104 MPa - (-48 MPa)) / 2

= (104 + 48) / 2

= 76 MPa.

The yield strength of the material is given as 200 MPa.

The safety factor (n) is the ratio of the yield strength to the maximum shear stress.

n = Yield Strength / τmax = 200 MPa / 76 MPa

≈ 2.63.

Therefore, the safety factor according to the criterion of maximum shear stress is approximately 2.63.

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