Ernst and Merchant Theory MCQ Quiz - Objective Question with Answer for Ernst and Merchant Theory - Download Free PDF

Last updated on Jun 15, 2025

Latest Ernst and Merchant Theory MCQ Objective Questions

Ernst and Merchant Theory Question 1:

Which of the following relation is correct for Ernst-Merchant Theory? [where ϕ = Shear angle, α = Rake angle, β = Friction angle]

  1. ϕ=π212(αβ)
  2. ϕ=π4+12(αβ)
  3. ϕ=π2+12(αβ)
  4. ϕ=π412(αβ)

Answer (Detailed Solution Below)

Option 2 : ϕ=π4+12(αβ)

Ernst and Merchant Theory Question 1 Detailed Solution

Concept:

From the merchant first analysis

2ϕ+βα=π2

Where, ϕ = shear angle, β = friction angle, α = cutting rake angle

26 June 1

There are several theories for the relationship between α (rake angle), β (friction angle), and ϕ (shear angle).

Ernst – Merchant theory:

ϕ=π4β2+α2

Stabler theory (Modified Ernst – Merchant theory):

ϕ=π4β+α2

Lee and Shaffer theory:

ϕ=π4β+α

Ernst and Merchant Theory Question 2:

The relationship between the shear angle ϕ, friction angle β, and cutting rake angle α, and the machining constant C for the work material is

  1. 2ϕ + β - α = C
  2. 2α + β - ϕ = C
  3. 2β + α - ϕ = C
  4. 2ϕ + α + β = C

Answer (Detailed Solution Below)

Option 1 : 2ϕ + β - α = C

Ernst and Merchant Theory Question 2 Detailed Solution

Explanation:

The relationship between the shear angle (φ), friction angle (β), cutting rake angle (α) and the machining constant (c) for the work material is given by,

Merchant Theory,

  • 2φ + β - α = Cm (for minimum energy consumption)
  • where, φ = shear angle in degree
  • β = friction angle in degree
  • α = rake angle in degree
  • and Cm = machining constant depends on work material.
  • In absence of any machining constant Cm the angle relationship is given by,

Ernst and Merchant Theory:

  • 2φ + β - α = π/2 = 90°.

Important Points

Various other important relationship between the shear angle (φ), friction angle (β), cutting rake angle (α) and the machining constant (c) for the work material is given by,

  • Lee and Shaffer theory:
    • φ + β - α = π/4 = 45°.
  • Stabler theory: (modified Ernst and Merchant theory)
    • φ + β - α/2 = π/4 = 45°.

Ernst and Merchant Theory Question 3:

During a turning operation of a specific work material having a shear strength of 220 MPa under orthogonal cutting condition, the process parameters are

Feed

0.2 mm/rev

Depth of cut

1 mm

Rake angle

-5°


Given chip thickness ratio as 0.5, friction angle as 49.2° and shear angle as 25.4°, the feed force (in N) is ________

Answer (Detailed Solution Below) 450 - 470

Ernst and Merchant Theory Question 3 Detailed Solution

Concept:

Shear Force, 

Fs=t0bsinϕ×τs

Feed force,

Ft=Fssin(βα)cos(ϕ+βα)

Calculation:

Given:

Rake angle, α = - 5°, friction angle, β = 49.2°, shear angle, ϕ = 25.4°, chip thickness ratio, r = 0.5, shear strength, τs = 200 MPa, 

feed, f = uncut chip thickness, t0 = 0.2 mm/rev, Depth of cut, d = width of chip, b = 1 mm

Cross section area of uncut chip = t0 × b = f × d = 0.2 × 1 = 0.2 = A0

Now,

Fs=t0bsinϕ×τs

Fs=0.2×1sin25.4×220=102.58N

Ft=Fssin(βα)cos(ϕ+βα)=102.58×sin(49.2(5))cos(79.6)

∴ Ft = 460.88 N

Top Ernst and Merchant Theory MCQ Objective Questions

Which of the following relation is correct for Ernst-Merchant Theory? [where ϕ = Shear angle, α = Rake angle, β = Friction angle]

  1. ϕ=π212(αβ)
  2. ϕ=π4+12(αβ)
  3. ϕ=π2+12(αβ)
  4. ϕ=π412(αβ)

Answer (Detailed Solution Below)

Option 2 : ϕ=π4+12(αβ)

Ernst and Merchant Theory Question 4 Detailed Solution

Download Solution PDF

Concept:

From the merchant first analysis

2ϕ+βα=π2

Where, ϕ = shear angle, β = friction angle, α = cutting rake angle

26 June 1

There are several theories for the relationship between α (rake angle), β (friction angle), and ϕ (shear angle).

Ernst – Merchant theory:

ϕ=π4β2+α2

Stabler theory (Modified Ernst – Merchant theory):

ϕ=π4β+α2

Lee and Shaffer theory:

ϕ=π4β+α

During a turning operation of a specific work material having a shear strength of 220 MPa under orthogonal cutting condition, the process parameters are

Feed

0.2 mm/rev

Depth of cut

1 mm

Rake angle

-5°


Given chip thickness ratio as 0.5, friction angle as 49.2° and shear angle as 25.4°, the feed force (in N) is ________

Answer (Detailed Solution Below) 450 - 470

Ernst and Merchant Theory Question 5 Detailed Solution

Download Solution PDF

Concept:

Shear Force, 

Fs=t0bsinϕ×τs

Feed force,

Ft=Fssin(βα)cos(ϕ+βα)

Calculation:

Given:

Rake angle, α = - 5°, friction angle, β = 49.2°, shear angle, ϕ = 25.4°, chip thickness ratio, r = 0.5, shear strength, τs = 200 MPa, 

feed, f = uncut chip thickness, t0 = 0.2 mm/rev, Depth of cut, d = width of chip, b = 1 mm

Cross section area of uncut chip = t0 × b = f × d = 0.2 × 1 = 0.2 = A0

Now,

Fs=t0bsinϕ×τs

Fs=0.2×1sin25.4×220=102.58N

Ft=Fssin(βα)cos(ϕ+βα)=102.58×sin(49.2(5))cos(79.6)

∴ Ft = 460.88 N

Ernst and Merchant Theory Question 6:

The relationship between the shear angle ϕ, friction angle β, and cutting rake angle α, and the machining constant C for the work material is

  1. 2ϕ + β - α = C
  2. 2α + β - ϕ = C
  3. 2β + α - ϕ = C
  4. 2ϕ + α + β = C

Answer (Detailed Solution Below)

Option 1 : 2ϕ + β - α = C

Ernst and Merchant Theory Question 6 Detailed Solution

Explanation:

The relationship between the shear angle (φ), friction angle (β), cutting rake angle (α) and the machining constant (c) for the work material is given by,

Merchant Theory,

  • 2φ + β - α = Cm (for minimum energy consumption)
  • where, φ = shear angle in degree
  • β = friction angle in degree
  • α = rake angle in degree
  • and Cm = machining constant depends on work material.
  • In absence of any machining constant Cm the angle relationship is given by,

Ernst and Merchant Theory:

  • 2φ + β - α = π/2 = 90°.

Important Points

Various other important relationship between the shear angle (φ), friction angle (β), cutting rake angle (α) and the machining constant (c) for the work material is given by,

  • Lee and Shaffer theory:
    • φ + β - α = π/4 = 45°.
  • Stabler theory: (modified Ernst and Merchant theory)
    • φ + β - α/2 = π/4 = 45°.

Ernst and Merchant Theory Question 7:

Which of the following relation is correct for Ernst-Merchant Theory? [where ϕ = Shear angle, α = Rake angle, β = Friction angle]

  1. ϕ=π212(αβ)
  2. ϕ=π4+12(αβ)
  3. ϕ=π2+12(αβ)
  4. ϕ=π412(αβ)

Answer (Detailed Solution Below)

Option 2 : ϕ=π4+12(αβ)

Ernst and Merchant Theory Question 7 Detailed Solution

Concept:

From the merchant first analysis

2ϕ+βα=π2

Where, ϕ = shear angle, β = friction angle, α = cutting rake angle

26 June 1

There are several theories for the relationship between α (rake angle), β (friction angle), and ϕ (shear angle).

Ernst – Merchant theory:

ϕ=π4β2+α2

Stabler theory (Modified Ernst – Merchant theory):

ϕ=π4β+α2

Lee and Shaffer theory:

ϕ=π4β+α

Ernst and Merchant Theory Question 8:

During a turning operation of a specific work material having a shear strength of 220 MPa under orthogonal cutting condition, the process parameters are

Feed

0.2 mm/rev

Depth of cut

1 mm

Rake angle

-5°


Given chip thickness ratio as 0.5, friction angle as 49.2° and shear angle as 25.4°, the feed force (in N) is ________

Answer (Detailed Solution Below) 450 - 470

Ernst and Merchant Theory Question 8 Detailed Solution

Concept:

Shear Force, 

Fs=t0bsinϕ×τs

Feed force,

Ft=Fssin(βα)cos(ϕ+βα)

Calculation:

Given:

Rake angle, α = - 5°, friction angle, β = 49.2°, shear angle, ϕ = 25.4°, chip thickness ratio, r = 0.5, shear strength, τs = 200 MPa, 

feed, f = uncut chip thickness, t0 = 0.2 mm/rev, Depth of cut, d = width of chip, b = 1 mm

Cross section area of uncut chip = t0 × b = f × d = 0.2 × 1 = 0.2 = A0

Now,

Fs=t0bsinϕ×τs

Fs=0.2×1sin25.4×220=102.58N

Ft=Fssin(βα)cos(ϕ+βα)=102.58×sin(49.2(5))cos(79.6)

∴ Ft = 460.88 N

Ernst and Merchant Theory Question 9:

A component having tensile strength 320 MPa and shear strength 200 MPa is cut in a turning operation at 3 m/s. Rake angle of the tool is 7° and the feed and depth of cut are 0.3 mm/rev and 3 mm respectively. Using orthogonal model and resulting chip ratio as 0.44, then value of feed force is _______ N

Answer (Detailed Solution Below) 660 - 670

Ernst and Merchant Theory Question 9 Detailed Solution

Concept:

tanϕ=rcosα1rsinα

β = 90° + α – 2ϕ (merchant equation)

Fs=t0bsinϕ×τs

Ft=Fssin(βα)cos(ϕ+βα)

Calculation:

Given:

r = 0.44, α = 7°, τs = 200 MPa, t0 = 0.3 mm/rev, b = 3 mm

ϕ=tan1[0.44cos710.44sin7]

ϕ = 24.77°

β = 90° + 7 – 2 × 24.77

β = 47.46°

Now,

Fs=0.3×3sin24.77×200=429.62N

Ft=Fssin(βα)cos(ϕ+βα)=429.62×sin(47.467)cos(65.23)

Ft = 667.30 N

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