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The principal stresses at a point in an elastic material are 60 N/mm2 tensile, 20 N/mm2 tensile and 50 N/mm2 compressive. If the material properties are: μ = 0.35 and E = 105 N/mm2 then the volumetric strain of the material is
  • a)
    9 x 10-5
  • b)
    3 x 10-4
  • c)
    10.5 x 10-5
  • d)
    21 x 10-5
Correct answer is option 'A'. Can you explain this answer?
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The principal stresses at a point in an elastic material are 60 N/mm2 ...
Volumetric strain,
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The principal stresses at a point in an elastic material are 60 N/mm2 ...
Young's modulus E = 200 GPa, Poisson's ratio ν = 0.3, and yield stress σy = 250 N/mm2, determine:

a) The maximum shear stress at the point.

b) The factor of safety against yielding in tension.

c) The factor of safety against yielding in compression.

a) The maximum shear stress at the point can be found using the formula:

τmax = (σ1 - σ3) / 2

where σ1 and σ3 are the maximum and minimum principal stresses, respectively. Substituting the given values, we get:

τmax = (60 - (-50)) / 2 = 55 N/mm2

Therefore, the maximum shear stress at the point is 55 N/mm2.

b) The factor of safety against yielding in tension can be found using the formula:

FS = σy / σ1

where σy is the yield stress and σ1 is the maximum principal stress. Substituting the given values, we get:

FS = 250 / 60 = 4.17

Therefore, the factor of safety against yielding in tension is 4.17.

c) The factor of safety against yielding in compression can be found using the formula:

FS = σy / |σ3|

where |σ3| is the absolute value of the minimum principal stress. Substituting the given values, we get:

FS = 250 / 50 = 5

Therefore, the factor of safety against yielding in compression is 5.
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The principal stresses at a point in an elastic material are 60 N/mm2 tensile, 20 N/mm2 tensile and 50 N/mm2 compressive. If the material properties are: μ= 0.35 and E = 105 N/mm2 then the volumetric strain of the material isa)9 x 10-5b)3 x 10-4c)10.5 x 10-5d)21 x 10-5Correct answer is option 'A'. Can you explain this answer?
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