Calculate the biaxial stresses σ1 and σ2 for the biaxial stress case, where ε1 = .0020 and ε2 = –.0010 are determined experimentally on an aluminum member of elastic constants, E = 71 GPa and v = 0.35. Also, determine the value for the maximum shear stress.

Answers

Answer 1

Answer:

i) σ1 = 133.5 MPa

  σ2 = -2427 MPa

ii) 78.89 MPa

Explanation:

Given data:

ε1 = 0.0020 and ε2 = –0.0010

E = 71 GPa

v = 0.35

i) Determine the biaxial stresses  σ1 and σ2 using the relations below

ε1 = σ1 / E - v (σ2 / E)   -----( 1 )

ε2 = σ2 / E - v (σ1 / E)  -------( 2 )

resolving equations 1 and 2

σ1 = E / 1 - v^2 {  ε1 + vε2 } ---- ( 3 )

σ2 = E / 1 - v^2 {  ε2 + vε1 } ----- ( 4 )

input the given data into equation 3 and equation 4

σ1 = 133.5 MPa

σ2 = -2427 MPa

ii) Calculate the value of the maximum shear stress ( Zmax )

Zmax = ( σ1 - σ2 ) / 2

         = 133.5 - ( - 2427 ) / 2

         = 78.89 MPa


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Answer:

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Answers

Answer:

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Attached below is the remaining answers

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Answers

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A brittle material is subjected to a tensile stress of 1.65 MPa. If the specific surface energy and modulus of elasticity for this material are 0.60 J/m2 and 2.0 GPa, respectively. What is the maximum length of a surface flaw that is possible without fracture

Answers

Answer:

The maximum length of a surface flaw that is possible without fracture is

[tex]2.806 \times 10^{-4} m[/tex]

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The maximum length of a surface flaw that is possible without fracture is

[tex]2.806 \times 10^{-4} m[/tex]

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