Machine Design Miscellaneous

Machine Design Miscellaneous

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Machine Design

Machine Design Miscellaneous
  1. A bolt of major diameter 12 mm is required to clamp two steel plates. Cross sectional area of the threaded portion of the bolt is 84.3 mm2. The length of the threaded portion in grip is 30 mm, while the length of the unthreaded portion in grip is 8 mm. Young's modulus of material is 200 GPa. The effective stiffness (in MN/m) of the bolt in the clamped zone is _____.








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    d1 = 12 mm;
    l1 = 8 mm
    A2 = 84.33 mm²

    A1 =
    π
    		(d1)²
    4

    l2 = 30 mm
    K1 =
    A1E1
    l1

    K2 =
    A2E2
    l2

    =
    84.33 × 200
    		 = 562.2
    30

    =
    (π/4)(12)² × 200
    		 = 2827.4
    8

    1
    		 =
    1
    		 +
    1
    KK1K2

    =
    π
    		 × 3000 + 1500 × π
    8

    =
    1
    		 +
    1
    2827.4562.2

    ⇒  K = 468.9 MN/m

    Correct Option: A

    d1 = 12 mm;
    l1 = 8 mm
    A2 = 84.33 mm²

    A1 =
    π
    		(d1)²
    4

    l2 = 30 mm
    K1 =
    A1E1
    l1

    K2 =
    A2E2
    l2

    =
    84.33 × 200
    		 = 562.2
    30

    =
    (π/4)(12)² × 200
    		 = 2827.4
    8

    1
    		 =
    1
    		 +
    1
    KK1K2

    =
    π
    		 × 3000 + 1500 × π
    8

    =
    1
    		 +
    1
    2827.4562.2

    ⇒  K = 468.9 MN/m

  1. The primary and secondary shear loads on bolt P, respectively, are








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    Let F1 and F2 resist the 4 kN eccentric load. The primary shear load is due to a direct shear stress of 4kN, which is equally divided amongst the two bolts, i.e. 2 k N,
    ∴  Fprimary on P = 2 kN

    The secondar y load is t o r esist t he moment produced by the eccentric load 4 kN about the C. G. of the joint
    Moment   M = 4 kN × 2000 mm
          = 8000 Nm
    Now this moment must be balanced by a moment pr oduced by equal and opposite forces i .e. secondary loads on the bolts
    i .e.     F × 400 = 800 Nm
    ⇒  F = 20 kN

    Correct Option: A

    Let F1 and F2 resist the 4 kN eccentric load. The primary shear load is due to a direct shear stress of 4kN, which is equally divided amongst the two bolts, i.e. 2 k N,
    ∴  Fprimary on P = 2 kN

    The secondar y load is t o r esist t he moment produced by the eccentric load 4 kN about the C. G. of the joint
    Moment   M = 4 kN × 2000 mm
          = 8000 Nm
    Now this moment must be balanced by a moment pr oduced by equal and opposite forces i .e. secondary loads on the bolts
    i .e.     F × 400 = 800 Nm
    ⇒  F = 20 kN

  1. The resultant shear stress on bolt P is closest to:








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    Resultant load on   P = Fsec – Fprim
    = 20 – 2 = 18 kN

    ∴    Stress =
    18 × 10³
    		 = 159 MPa
    (π/4)(12)²

    Correct Option: B

    Resultant load on   P = Fsec – Fprim
    = 20 – 2 = 18 kN

    ∴    Stress =
    18 × 10³
    		 = 159 MPa
    (π/4)(12)²

  1. For the three bolt system shown in the figure, the bolt material has shear yield strength of 200 MPa. For a factor of safety of 2, the minimum metric specification required for the bolt is








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    σy = 200 MPa
    P = 19 × 10³N

    σ =
    P
    		 ⇒ d = 10.9 mm
    (π/4)(d1)²

    M10

    Correct Option: B

    σy = 200 MPa
    P = 19 × 10³N

    σ =
    P
    		 ⇒ d = 10.9 mm
    (π/4)(d1)²

    M10

  1. The yield strength of a steel shaft is twice its endurance limit. Which of the following torque fluctuations represent the most critical situation according to Soderberg criterion?








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    τa
    		 +
    τm
    		 =
    1
    SeSyFos

    τa
    		 +
    τm
    		 =
    Se
    12Fos

    τa is highest in (a)

    Correct Option: A

    τa
    		 +
    τm
    		 =
    1
    SeSyFos

    τa
    		 +
    τm
    		 =
    Se
    12Fos

    τa is highest in (a)