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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?
- a)10%
- b)5%
- c)1%
- d)0.5%
Correct answer is option 'C'. Can you explain this answer?
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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0...
Actual elongation of the bar 
Calculated elongation of the bar
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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0...
Given:
- Length of bar, L
- Diameter at one end, 1.1D
- Diameter at the other end, 0.9D
- Elongation is computed using mean diameter, D
To find: Approximate error in computed elongation
Solution:
Let's assume that the axial pull is applied at the center of the bar, and the bar is in the elastic region of the stress-strain curve.
The elongation, δ, of a bar due to axial pull is given by:
δ = PL / AE
where
- P is the axial load applied
- A is the cross-sectional area of the bar
- E is the Young's modulus of elasticity of the material
The cross-sectional area of the bar can be approximated as the area of a frustum of a cone, given by:
A = π/4 (D^2 + d^2 + Dd)
where
- D is the diameter at one end of the bar
- d is the diameter at the other end of the bar
The mean diameter, Dm, can be calculated as the arithmetic mean of the diameters at the two ends:
Dm = (D + d) / 2
Substituting the values of A and Dm in the elongation formula, we get:
δ = PL / πE (D^2 + d^2 + Dd) / 4 (D + d)
Now, let's calculate the elongation using the mean diameter, D. This means that we are assuming the bar has a constant diameter of D throughout its length. Using this assumption, the elongation can be calculated as:
δ1 = PL / πE D^2
The error in the computed elongation, δe, can be calculated as the difference between the actual elongation and the elongation calculated using the mean diameter:
δe = δ - δ1
Substituting the values of δ and δ1 in the above equation, we get:
δe = PL / πE [ (D^2 + d^2 + Dd) / 4 (D + d) - D^2 ]
Simplifying the above equation, we get:
δe = PL / πE [ (D - d)^2 / 4 (D + d) ]
Now, substituting the given values in the above equation, we get:
δe = PL / πE [ (0.1D)^2 / 4 (1.0D) ]
δe = 0.0025 PL / πE D
Therefore, the approximate error in computed elongation is 0.25% of the actual elongation. Hence, the correct answer is option (c) 1%.
- Length of bar, L
- Diameter at one end, 1.1D
- Diameter at the other end, 0.9D
- Elongation is computed using mean diameter, D
To find: Approximate error in computed elongation
Solution:
Let's assume that the axial pull is applied at the center of the bar, and the bar is in the elastic region of the stress-strain curve.
The elongation, δ, of a bar due to axial pull is given by:
δ = PL / AE
where
- P is the axial load applied
- A is the cross-sectional area of the bar
- E is the Young's modulus of elasticity of the material
The cross-sectional area of the bar can be approximated as the area of a frustum of a cone, given by:
A = π/4 (D^2 + d^2 + Dd)
where
- D is the diameter at one end of the bar
- d is the diameter at the other end of the bar
The mean diameter, Dm, can be calculated as the arithmetic mean of the diameters at the two ends:
Dm = (D + d) / 2
Substituting the values of A and Dm in the elongation formula, we get:
δ = PL / πE (D^2 + d^2 + Dd) / 4 (D + d)
Now, let's calculate the elongation using the mean diameter, D. This means that we are assuming the bar has a constant diameter of D throughout its length. Using this assumption, the elongation can be calculated as:
δ1 = PL / πE D^2
The error in the computed elongation, δe, can be calculated as the difference between the actual elongation and the elongation calculated using the mean diameter:
δe = δ - δ1
Substituting the values of δ and δ1 in the above equation, we get:
δe = PL / πE [ (D^2 + d^2 + Dd) / 4 (D + d) - D^2 ]
Simplifying the above equation, we get:
δe = PL / πE [ (D - d)^2 / 4 (D + d) ]
Now, substituting the given values in the above equation, we get:
δe = PL / πE [ (0.1D)^2 / 4 (1.0D) ]
δe = 0.0025 PL / πE D
Therefore, the approximate error in computed elongation is 0.25% of the actual elongation. Hence, the correct answer is option (c) 1%.
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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0...
Answer is correct use the formula for tapered bar then $=4PL/πEd1*D2 and other is $=4PL/πDsquare*E.
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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. Can you explain this answer?
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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. 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 bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. 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 bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. Can you explain this answer?.
A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. 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 bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. 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 bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. Can you explain this answer?.
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A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation?a)10%b)5%c)1%d)0.5%Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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