Materials Science and Manufacturing Engineering Miscellaneous

Materials Science and Manufacturing Engineering Miscellaneous

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Materials Science and Manufacturing Engineering

Materials Science and Manufacturing Engineering Miscellaneous
  1. Calculate the punch size in mm, for a circular blanking operation for which details are given below. Size of the blank 25 mm Thickness of the sheet 2 mm Radial clearance between punch and die 0.06 mm Die allowance 0.05 mm








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    Diameter of punch = Diameter of Blank – 2 × radial clearance – die allowance = 25 – 2 × 0.06 – 0.05 = 24.83 mm

    Correct Option: A

    Diameter of punch = Diameter of Blank – 2 × radial clearance – die allowance = 25 – 2 × 0.06 – 0.05 = 24.83 mm

  1. The shear strength of a sheet metal is 300 MPa. The blanking force required to produce a blank of 100 mm diameter from a 1.5 mm thick sheet is close to








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    Blanking force = τ × As = 300 × πdt = 300 × π × 100 × 1.5 = 141 kN

    Correct Option: C

    Blanking force = τ × As = 300 × πdt = 300 × π × 100 × 1.5 = 141 kN

Direction: In shear cutting operation, a sheet of 5 mm thickness is cut along a length of 200 mm. The cutting blade is 400 mm long (see figure) and zero-shear (S = 0) is provided on the edge. The ultimate shear strength of the sheet is 100 MPa and penetration to thickness ratio is 0.2. Neglect friction.

  1. A shear of 20 mm {S = 20 mm) is now provided on the blade. Assuming force vs displacement curve to be trapezoidal, the maximum force (in kN) exerted is








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    Length = 200 mm , Blade length = 400 mm , Zero shear (s = 0)
    Ultimate shear strength of sheet = 100 MPa

    Penetration
    		 = 0.2 = k
    Thickness

    ∴ Shear area = (200 + 200)  5 = 2000 mm2
    Now Work done = τ × shear area × k.t.
    Work done = 100 × 2000 × 5 × 0.2 = 200 Joules
    ∴ Maximum force =
    Work done
    (kt + shear)

    Maximum force =
    200
    		 = 9.5 ≈ 10 kN
    (5 × 0.2 + 20)

    Correct Option: D

    Length = 200 mm , Blade length = 400 mm , Zero shear (s = 0)
    Ultimate shear strength of sheet = 100 MPa

    Penetration
    		 = 0.2 = k
    Thickness

    ∴ Shear area = (200 + 200)  5 = 2000 mm2
    Now Work done = τ × shear area × k.t.
    Work done = 100 × 2000 × 5 × 0.2 = 200 Joules
    ∴ Maximum force =
    Work done
    (kt + shear)

    Maximum force =
    200
    		 = 9.5 ≈ 10 kN
    (5 × 0.2 + 20)

  1. Assuming force vs displacement curve to be rectangular, the work done (in J) is








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    Thickness = 5 mm

    Correct Option: A

    Thickness = 5 mm

  1. 10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness. Shear strength of the material is 400 N/mm2 and penetration is 40%. Shear provided on the punch is 2 mm. The blanking force during the operation will be








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    Given, Shear strength, τ = 400 N/mm2, Sheet thickness, t = 3 mm Diameter of hole, d = 10 mm, Shear on punch, t1 = 2 mm
    Penetration of punch on fraction, 2p = 0.4

    Blanking force when shear is applied on the punch = πdt
    tp
    		τ
    t1

    = π × 10 × 3 × 400 ×
    3 × 0.4
    		 = 22.61 kN
    2

    Correct Option: A

    Given, Shear strength, τ = 400 N/mm2, Sheet thickness, t = 3 mm Diameter of hole, d = 10 mm, Shear on punch, t1 = 2 mm
    Penetration of punch on fraction, 2p = 0.4

    Blanking force when shear is applied on the punch = πdt
    tp
    		τ
    t1

    = π × 10 × 3 × 400 ×
    3 × 0.4
    		 = 22.61 kN
    2