Machine Design Miscellaneous
 A planestrain compression (forging) of a block is shown in the figure. The strain in the zdirection is zero. The yield strength (S_{y}) in uniaxial tension/compression of the material of the block is 300 MPa and its follows the Tresca (maximum shear stress) criterion. Assume that the entire block has started yielding. At a point where σ_{x} = 40 MPa (compressive) and τ_{xy} = 0, the stress component σ_{y} is

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σ_{x}, σ_{y} & σ_{z} are principal stresses.
Under plane strain condition, ε_{z} = 0⇒ σ_{z}  μσ_{y}  μσ_{x} = 0 E E E
⇒ σ_{z} = (σ_{x} + σ_{y}) = μ (σ_{x} + σ_{y})
but σ_{x} = 40
⇒ σ_{z} = μ(40 + σ_{y})
From Tresca theory,
= 300
But μ is not given
Hence if we neglect σ_{z} then
max [40 + σ_{y}, σ_{y}, +40] = 300
⇒ 40 + σ_{y} = 300
40 + σ_{y} = 300
⇒ σ_{y} = 260 MPaCorrect Option: B
σ_{x}, σ_{y} & σ_{z} are principal stresses.
Under plane strain condition, ε_{z} = 0⇒ σ_{z}  μσ_{y}  μσ_{x} = 0 E E E
⇒ σ_{z} = (σ_{x} + σ_{y}) = μ (σ_{x} + σ_{y})
but σ_{x} = 40
⇒ σ_{z} = μ(40 + σ_{y})
From Tresca theory,
= 300
But μ is not given
Hence if we neglect σ_{z} then
max [40 + σ_{y}, σ_{y}, +40] = 300
⇒ 40 + σ_{y} = 300
40 + σ_{y} = 300
⇒ σ_{y} = 260 MPa
 A machine element is subjected to the following biaxial state of stress; σ_{x} = 80 MPa; σ_{y} = 20 MPa; τ_{x} = 40 MPa. If the shear strength of the material is 100 MPa, the factor of safety as per Tresca's maximum shear stress theory is

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= 50 + √50² = 100
σ_{2} = 0τ = σ_{1}  σ_{2} = 50 7 FOS = 100 = 2 50
Correct Option: B
= 50 + √50² = 100
σ_{2} = 0τ = σ_{1}  σ_{2} = 50 7 FOS = 100 = 2 50
 The uniaxial yield stress of a material is 300 MPa. According to Von Mises criterion, the shear yield stress (in MPa) of the material is _______.

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If there is uniaxial loading yield stress is σ_{y}
As per Von Mises failure theory
σ²_{y} = σ²_{1} + σ_{1}σ_{2} + σ²_{2}
Under pure shear stress loading
σ_{1} = – σ_{2} = τ
then σ²_{y} = τ² + τ² + τ² = 3τ²⇒ τ = σ_{y} √3 Hence τ = σ_{y} = 0.5577σ_{y} = 173.28 √3
Where τ is shear yield stressCorrect Option: A
If there is uniaxial loading yield stress is σ_{y}
As per Von Mises failure theory
σ²_{y} = σ²_{1} + σ_{1}σ_{2} + σ²_{2}
Under pure shear stress loading
σ_{1} = – σ_{2} = τ
then σ²_{y} = τ² + τ² + τ² = 3τ²⇒ τ = σ_{y} √3 Hence τ = σ_{y} = 0.5577σ_{y} = 173.28 √3
Where τ is shear yield stress
 A shaft is subjected to pure torsional moment. The maximum shear stress developed in the shaft is 100 MPa. The yield and ultimate strengths of the shaft material in tension are 300 MPa and 450 MPa, respectively. The factor of safety using maximum distortion energy (VonMises) theory is _______ .

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σ_{1} = 100 MPa
σ_{2} = 100 MPa
As per Vonmises yield theory.
n² [(σ_{1}  σ_{2})² + (σ_{2}  σ_{3}²) + (σ_{1}  σ_{3}²)]
≤ 2σ_{2}²
2n²[σ_{1}² + σ_{2}²  σ_{1}σ_{2}] ≤ 2σ_{y}²
⇒ n²[(100)² + (100)² + (100)²]≤ (300)²
∴ n = 1.732Correct Option: A
σ_{1} = 100 MPa
σ_{2} = 100 MPa
As per Vonmises yield theory.
n² [(σ_{1}  σ_{2})² + (σ_{2}  σ_{3}²) + (σ_{1}  σ_{3}²)]
≤ 2σ_{2}²
2n²[σ_{1}² + σ_{2}²  σ_{1}σ_{2}] ≤ 2σ_{y}²
⇒ n²[(100)² + (100)² + (100)²]≤ (300)²
∴ n = 1.732
 Consider the two states of stress as shown in configurations I and II in the figure below. From the standpoint of distortion energy (vonMises) criterion, which one of the following statements is true?

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Both yields simultaneously.
Correct Option: C
Both yields simultaneously.