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A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).
Correct answer is '930'. Can you explain this answer?
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Most Upvoted Answer
A 200 mm wide plate having a thickness of 20 mm is fed through a rolli...
Given Data:
Width of plate (W) = 200 mm
Thickness of plate before rolling (t1) = 20 mm
Thickness of plate after rolling (t2) = 18 mm
Radius of each roll (R) = 300 mm
Rolling speed (v) = 50 rpm
Strength coefficient of work material flow curve (K) = 300 MPa
Strain hardening exponent (n) = 0.2
Coefficient of friction (μ) = 0.1
Roll force in rolling process:
The roll force (Fr) required in the rolling process can be calculated by the following formula:
Fr = (0.5 × K × ln(Ai/Af) × L × W) / (t1 - t2)
where,
Ai = initial cross-sectional area of plate = W × t1
Af = final cross-sectional area of plate = W × t2
L = length of contact between plate and rolls = π × R
Calculation:
Initial cross-sectional area of plate:
Ai = W × t1 = 200 mm × 20 mm = 4000 mm²
Final cross-sectional area of plate:
Af = W × t2 = 200 mm × 18 mm = 3600 mm²
Length of contact between plate and rolls:
L = π × R = 3.14 × 300 mm = 942 mm
Natural logarithm of initial to final thickness ratio:
ln(Ai/Af) = ln(4000/3600) = 0.0986
Substituting the given values in the formula for roll force:
Fr = (0.5 × K × ln(Ai/Af) × L × W) / (t1 - t2)
Fr = (0.5 × 300 MPa × 0.0986 × 942 mm × 200 mm) / (20 mm - 18 mm)
Fr = 930 kN (rounded off to nearest integer)
Therefore, the roll force required in the rolling process is 930 kN.
Width of plate (W) = 200 mm
Thickness of plate before rolling (t1) = 20 mm
Thickness of plate after rolling (t2) = 18 mm
Radius of each roll (R) = 300 mm
Rolling speed (v) = 50 rpm
Strength coefficient of work material flow curve (K) = 300 MPa
Strain hardening exponent (n) = 0.2
Coefficient of friction (μ) = 0.1
Roll force in rolling process:
The roll force (Fr) required in the rolling process can be calculated by the following formula:
Fr = (0.5 × K × ln(Ai/Af) × L × W) / (t1 - t2)
where,
Ai = initial cross-sectional area of plate = W × t1
Af = final cross-sectional area of plate = W × t2
L = length of contact between plate and rolls = π × R
Calculation:
Initial cross-sectional area of plate:
Ai = W × t1 = 200 mm × 20 mm = 4000 mm²
Final cross-sectional area of plate:
Af = W × t2 = 200 mm × 18 mm = 3600 mm²
Length of contact between plate and rolls:
L = π × R = 3.14 × 300 mm = 942 mm
Natural logarithm of initial to final thickness ratio:
ln(Ai/Af) = ln(4000/3600) = 0.0986
Substituting the given values in the formula for roll force:
Fr = (0.5 × K × ln(Ai/Af) × L × W) / (t1 - t2)
Fr = (0.5 × 300 MPa × 0.0986 × 942 mm × 200 mm) / (20 mm - 18 mm)
Fr = 930 kN (rounded off to nearest integer)
Therefore, the roll force required in the rolling process is 930 kN.
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Community Answer
A 200 mm wide plate having a thickness of 20 mm is fed through a rolli...
Given : w = 200 mm , h0 = 20 mm
R = 300 mm , hf = 18 mm
k = 300 MPa , n = 0.2 and μ= 0.1
σ0 = 159.39 MPa
F = 930 kN
R = 300 mm , hf = 18 mm
k = 300 MPa , n = 0.2 and μ= 0.1
σ0 = 159.39 MPa
F = 930 kN
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A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer?
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A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer?.
A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer?.
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A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer?, a detailed solution for A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer? has been provided alongside types of A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be____________ kN (round off to the nearest integer).Correct answer is '930'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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