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In a circular tapering bar of length 120 mm, diameter varies from 20 mm to 45 mm. If a tensile load of 20 kN is applied axially, find the total deflection (in mm) of the bar. (Take E = 85 MPa) (Answer up to the nearest integer)
Correct answer is '39'. Can you explain this answer?
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In a circular tapering bar of length 120 mm, diameter varies from 20 m...
Theory:
The total deflection of a circular tapering bar can be calculated using the formula:
Δ = (F * L) / (π * E * R)
Where:
Δ = Total deflection
F = Applied load
L = Length of the bar
E = Young's modulus of the material
R = Average radius of the bar
Given:
Length of the bar (L) = 120 mm
Diameter at one end (d1) = 20 mm
Radius at one end (r1) = d1/2 = 10 mm
Diameter at the other end (d2) = 45 mm
Radius at the other end (r2) = d2/2 = 22.5 mm
Applied load (F) = 20 kN = 20000 N
Young's modulus (E) = 85 MPa = 85 * 10^6 N/m²
Solution:
1. Calculate the average radius (R) of the bar:
R = (r1 + r2) / 2
R = (10 + 22.5) / 2
R = 16.25 mm
2. Substitute the given values into the formula for total deflection:
Δ = (F * L) / (π * E * R)
Δ = (20000 * 120) / (π * 85 * 10^6 * 16.25)
Δ ≈ 39.46 mm
Therefore, the total deflection of the bar is approximately 39 mm when a tensile load of 20 kN is applied axially.
The total deflection of a circular tapering bar can be calculated using the formula:
Δ = (F * L) / (π * E * R)
Where:
Δ = Total deflection
F = Applied load
L = Length of the bar
E = Young's modulus of the material
R = Average radius of the bar
Given:
Length of the bar (L) = 120 mm
Diameter at one end (d1) = 20 mm
Radius at one end (r1) = d1/2 = 10 mm
Diameter at the other end (d2) = 45 mm
Radius at the other end (r2) = d2/2 = 22.5 mm
Applied load (F) = 20 kN = 20000 N
Young's modulus (E) = 85 MPa = 85 * 10^6 N/m²
Solution:
1. Calculate the average radius (R) of the bar:
R = (r1 + r2) / 2
R = (10 + 22.5) / 2
R = 16.25 mm
2. Substitute the given values into the formula for total deflection:
Δ = (F * L) / (π * E * R)
Δ = (20000 * 120) / (π * 85 * 10^6 * 16.25)
Δ ≈ 39.46 mm
Therefore, the total deflection of the bar is approximately 39 mm when a tensile load of 20 kN is applied axially.
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Community Answer
In a circular tapering bar of length 120 mm, diameter varies from 20 m...
Here, D1 = 20 mm = 0.020 m, D2 = 45 mm = 0.045 m,
P = 20 kN = 20000 N and L = 120 mm = 0.120 m
Total deflection is
= 0.039 m
= 39 mm
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In a circular tapering bar of length 120 mm, diameter varies from 20 mm to 45 mm. If a tensile load of 20 kN is applied axially, find the total deflection (in mm) of the bar. (Take E = 85 MPa) (Answer up to the nearest integer)Correct answer is '39'. Can you explain this answer?
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In a circular tapering bar of length 120 mm, diameter varies from 20 mm to 45 mm. If a tensile load of 20 kN is applied axially, find the total deflection (in mm) of the bar. (Take E = 85 MPa) (Answer up to the nearest integer)Correct answer is '39'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about In a circular tapering bar of length 120 mm, diameter varies from 20 mm to 45 mm. If a tensile load of 20 kN is applied axially, find the total deflection (in mm) of the bar. (Take E = 85 MPa) (Answer up to the nearest integer)Correct answer is '39'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a circular tapering bar of length 120 mm, diameter varies from 20 mm to 45 mm. If a tensile load of 20 kN is applied axially, find the total deflection (in mm) of the bar. (Take E = 85 MPa) (Answer up to the nearest integer)Correct answer is '39'. Can you explain this answer?.
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