Machine Design Miscellaneous Easy Questions and Answers | Page - 5

Machine Design Miscellaneous

A plane-strain 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

1. 260 MPa (compressive)
1. 260 MPa (tensile)
1. 340 MPa (tensile)
1. 340 MPa (compressive)

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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 MPa

Correct 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 bi-axial 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

1. 1.0
1. 2.0
1. 2.5
1. 3.3

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= 50 + √50² = 100
σ₂ = 0
τ = σ₁ - σ₂ = 50
7

FOS = 100 = 2
50

Correct Option: B

= 50 + √50² = 100
σ₂ = 0
τ = σ₁ - σ₂ = 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 _______.

1. 173.28
1. 172.28
1. 171.28
1. 169.28

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If there is uniaxial loading yield stress is σ_y
As per Von Mises failure theory
σ²_y = σ²₁ + σ₁σ₂ + σ²₂
Under pure shear stress loading
σ₁ = – σ₂ = 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

Correct Option: A

If there is uniaxial loading yield stress is σ_y
As per Von Mises failure theory
σ²_y = σ²₁ + σ₁σ₂ + σ²₂
Under pure shear stress loading
σ₁ = – σ₂ = 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 (Von-Mises) theory is _______ .

1. 1.732
1. 2.732
1. 0.732
1. 1.622

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σ₁ = 100 MPa
σ₂ = 100 MPa
As per Von-mises yield theory.

n² [(σ₁ - σ₂)² + (σ₂ - σ₃²) + (σ₁ - σ₃²)]
≤ 2σ₂²
2n²[σ₁² + σ₂² - σ₁σ₂] ≤ 2σ_y²
⇒ n²[(100)² + (100)² + (100)²]≤ (300)²
∴ n = 1.732

Correct Option: A

σ₁ = 100 MPa
σ₂ = 100 MPa
As per Von-mises yield theory.

n² [(σ₁ - σ₂)² + (σ₂ - σ₃²) + (σ₁ - σ₃²)]
≤ 2σ₂²
2n²[σ₁² + σ₂² - σ₁σ₂] ≤ 2σ_y²
⇒ 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?

1. I yields after II
1. II yields after I
1. Both yield simultaneously
1. Nothing can be said about their relative yielding

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Both yields simultaneously.

Correct Option: C

Both yields simultaneously.