Past Year Questions: Sheet Metal Operations

# Past Year Questions: Sheet Metal Operations

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## 40 Questions MCQ Test | Past Year Questions: Sheet Metal Operations

Past Year Questions: Sheet Metal Operations for Mechanical Engineering 2023 is part of Mechanical Engineering preparation. The Past Year Questions: Sheet Metal Operations questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Past Year Questions: Sheet Metal Operations MCQs are made for Mechanical Engineering 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Past Year Questions: Sheet Metal Operations below.
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Past Year Questions: Sheet Metal Operations - Question 1

### The thickness of the blank needed to produce, by power spinning a missile cone of thickness 1.5 mm and half cone angle 30° [ME 1992]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 1

Thickness of blank

Past Year Questions: Sheet Metal Operations - Question 2

### A 50 mm diameter disc is to be punched out from a carbon steel sheet 1.0 mm thick. The diameter of the die should be (Take clearance as 3% of sheet thickness) _______. [ME 1996]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 2

Clearness should be given on punch in Blanking operation. Die should be exact.

Past Year Questions: Sheet Metal Operations - Question 3

### The cutting force in punching and blanking operations mainly depends on [ME 2001]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 3

Cutting force = t × A
t = Shear strength

Past Year Questions: Sheet Metal Operations - Question 4

In blanking operation the clearance is provided on

[ME 2002]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 4

Blanking operation cleanness is provided on.
Punch Die will be exact.

Past Year Questions: Sheet Metal Operations - Question 5

A metal disc of 20 mm diameter is to be punched from a sheet of 2 mm thickness. The punch and the dia clearance is 3%. The required punch diameter is

[ME 2003]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 5

Punch diameter

Past Year Questions: Sheet Metal Operations - Question 6

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

[ME 2004]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 6

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

= 22.61 kN

Past Year Questions: Sheet Metal Operations - Question 7

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

[ME 2011]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 7

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

Past Year Questions: Sheet Metal Operations - Question 8

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

[ME 2012]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 8

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

Past Year Questions: Sheet Metal Operations - Question 9

A rectangular hole of size 100 mm × 50 mm is to be made on a 5 mm thick sheet of steel having ultimate tensile strength and shear strength of 500 MPa and 300 MPa, respectively. The hole is made by punching process. Neglecting the effect of clearance, the punching force (in kN) is

[ME 2014,Set-2]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 9

Perimeter = (50 + 100) × 2 = 300
Thickness = 5
so shown Area = 300 × 5 = 1500 mm2.
Punching force = τ ×As.
= 300 × 1500 = 450 KN

Past Year Questions: Sheet Metal Operations - Question 10

Internal gears are manufactured by

[ME 2014,Set-2]

Past Year Questions: Sheet Metal Operations - Question 11

Calculate the smallest punch diameter that can be designed for piercing sheet metal strip with the following data. Crushing strength of the punch material is 1500 MPa. Thickness of the sheet is 2 mm, factor of safety is 3, shear strength of the sheet material 500 MPa is

[PI 1990]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 11

Past Year Questions: Sheet Metal Operations - Question 12

In blanking operation, the best way to improve the smoothness and secureness of the edges is to

[PI 1994]

Past Year Questions: Sheet Metal Operations - Question 13

For 50% penetration of work material, a punch with single shear equal to thickness will

[PI 2001]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 13

Past Year Questions: Sheet Metal Operations - Question 14

Which one is not a method of reducing cutting forces to prevent the overloading of press?

[PI 2003]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 14

Shear decrease the maximum force but Energy Requirement remains the same.

Past Year Questions: Sheet Metal Operations - Question 15

A hydraulic press is used to produce circular blanks of 10 mm diameter from a sheet of 2 mm thickness. If the shear strength of the sheet material is 400 N/mm2, the force required for producing a circular blank is

[PI 2003]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 15

Punching force = πDtτ
= π × 2 × 10 × 400 = 25.13 KN

Past Year Questions: Sheet Metal Operations - Question 16

Consider the following statements related piercing and blanking
1. Shear on the punch reduces the maximum cutting force
2. Shear increases the capacity of the press needed
3. Shear increases the life of the punch
4. Total energy needed to make the cut remains unaltered due to provision of shear.

Which of the above statements are correct?

[PI 2003]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 16

Over all Energy requirement Unaltered & Maximum Force decreased.

Past Year Questions: Sheet Metal Operations - Question 17

A cylindrical cup of 48.5 mm diameter and 52 mm height has a corner radius of 1.5 mm.Cold rolled steel sheet of thickness 1.5 mm is used to produce the cup. Assume trim allowance as 2 mm per 25 mm of cup diameter.What is the blank size in mm?

[PI 2004]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 17

Past Year Questions: Sheet Metal Operations - Question 18

A hole of 40 mm diameter is pierced in a steel sheet of 4 mm thickness without shear on the tool. Shear strength of steel is 400 N/mm2 and penetration is 25%. What is the expected percentage load reduction if a shear of 1 mm is provided on the punch?

[PI 2004]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 18

Past Year Questions: Sheet Metal Operations - Question 19

Determine the load required to punch a 20 mm diameter hole in 2 mm thick sheet. The properties of the material of the sheet are: Tensile strength – 580 MPa
Yield strength in tension – 410
MPa Shear strength – 350 MPa
Yield strength in shear – 250 MPa

[PI 2004]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 19

Punching force = πDtτ
= π × 20 × 2 × 350 = 43.98 KN

Past Year Questions: Sheet Metal Operations - Question 20

A ϕ 25 mm hole is pierced in a t = 2.5 mm thick steel sheet having shear strength τ = 350 MPa.If the diametral clearance is given by the expression c = 0.0064 √τ, the die diameter (in mm), punch diameter (in mm) and punch force in (kN) respectively are

[PI 2005]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 20

t = 2.5
τ = 350 MPa
dhole = 25 mm
Punch dia = 25 mm
Clearance = = 0.0064 x 2.5 x
= 0.3mm
Die Dia = Punch Dia +  Clearance
= 25 + 0.3 = 25.3
Punching force = π Dt τ = π × 25 × 2.5 × 350
= 68.72 KN

Past Year Questions: Sheet Metal Operations - Question 21

It is require to punch a hole of diameter 10 mm on a sheet of thickness 3 mm. The shear strength of the work material is 420 MPa. The required punch force is

[PI 2006]

Past Year Questions: Sheet Metal Operations - Question 22

It is required to punch a hole of diameter 10 mm on a sheet of thickness 3 mm. The shear strength of the work material is 420 MPa.The required punch force is

[PI 2006]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 22

Punching force = πDtτ
= π × 10 × 3 × 420 = 39.58 kN

Past Year Questions: Sheet Metal Operations - Question 23

Circular blanks of 35 mm dia are punched from a steel sheet of 2 mm thickness. If the clearance per side between the punch and die is to be kept as 40 microns. The sizes of punch and die should respectively be

[PI 2007]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 23

Dia of Die = 35 mm
Dia of Punch = 35 – 2 × 0.04 = 34.92mm

Past Year Questions: Sheet Metal Operations - Question 24

A blank of 50 mm diameter is to be sheared from a sheet of 2.5 mm thickness. The required radial clearance between the die and the punch is 6% of sheet thickness. The punch and die diameter (mm) for this blanking operation respectively are

[PI 2008]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 24

Total clearance = 2 × 0.15 = 0.3
Die Dia will be exact = 50 mm
Clearance will be provided on punch
= 50–0.30 = 49.70 mm

Past Year Questions: Sheet Metal Operations - Question 25

A disc of 200 mm diameter is blanked from a strip of an aluminum alloy of thickness 3.2 mm.The material shear strength to fracture is 150 MPa.The blanking force (in kN) is

[PI 2009]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 25

Banking force = πDtτ
= π × 200 × 3.2 × 150 = 301.59 KN

Past Year Questions: Sheet Metal Operations - Question 26

Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is

[PI 2013]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 26

Punching force = πDtτ
= π × 10 × 2 × 80
= 5.023 KN

Past Year Questions: Sheet Metal Operations - Question 27

Wrinkling is a common defect found is

[ME 1991]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 27

Deep drawn components having defects of wrinkles due to less blank holding force.

Past Year Questions: Sheet Metal Operations - Question 28

In deep drawing of sheets, the values of limiting drawing ratio depends on

[ME 1994]

Past Year Questions: Sheet Metal Operations - Question 29

Identify the stress strain in flange portion of a partially drawn cylindrical cup when deep drawing without a blank holder

[ME 1999]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 29

Tensile in Radial direction & compressive in circumferential direction.

Past Year Questions: Sheet Metal Operations - Question 30

A shell of 100 mm diameter and 100 mm height with the corner radius of 0.4 mm is to be produced by cup drawing. The required blank diameter is

[ME 2003]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 30

Past Year Questions: Sheet Metal Operations - Question 31

In the deep drawing of cups, blanks show a tendency to wrinkle up around the periphery (flange). The most likely cause and remedy of the phenomenon are, respectively.

[ME 2008]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 31

Due to circumferential compression buckling in circumference, by increasing blank holding force buckling decrease in lateral direction of load applied.

Past Year Questions: Sheet Metal Operations - Question 32

The value of true strain produced in compressing a cylinder to half its original length is

[ME 2016,Set-2]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 32

Past Year Questions: Sheet Metal Operations - Question 33

Consider a linear elastic rectangular thin sheet of metal, subjected to uniform uniaxial tensile stress of 100 MPa along the length direction. Assume plane stress condition in the plane normal to the thickness. The Young’s modulus E = 200 MPa and Poisson’s ratio μ = 0.3 are given. The principal strains in the plane of the sheet are

[ME 2019,Set-2]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 33

Given
E = 200 MPa
σx = 100 MPa
σy = 0
Principal strain in x-direction

Past Year Questions: Sheet Metal Operations - Question 34

In progressive die (for sheet metal), spring loaded stripper plate is used to clamp the stock until

[PI 1991]

Past Year Questions: Sheet Metal Operations - Question 35

A cup of 10 cm height and 5 cm diameter is to be made from a sheet metal of 2 mm thickness.The number of deductions necessary will be

[PI 1991]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 35

first reduction 50%
Next reduction 30%

Past Year Questions: Sheet Metal Operations - Question 36

Wall thickness of drawn cup is controlled by

[PI 1992]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 36

It is clearance between punch & die in subsequent reduction in single stroke.

Past Year Questions: Sheet Metal Operations - Question 37

In a bending operation, if the modulus of elasticity E is increased keeping all other parameters unehanged, the spring back will

[PI 1994]

Past Year Questions: Sheet Metal Operations - Question 38

The function of draw bead in the deep drawing operation is to

[PI 1994]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 38

Draw beads decreases circumferential stress or buckling in flange.

Past Year Questions: Sheet Metal Operations - Question 39

In deep drawing of sheet metal, spring loaded stripper clamps the work until:

[PI 2004]

Past Year Questions: Sheet Metal Operations - Question 40

In a surface broaching process, a broach having 5 mm pitch and cut per teeth 0.05 mm is used for reducing the thickness by 6 mm. If the length of the workpiece is 50 mm and the cutting speed is 0.5 m/s, the time required/pass in second is

[PI 2006]

Detailed Solution for Past Year Questions: Sheet Metal Operations - Question 40

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