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A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100mm2. Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, Where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm.
Correct answer is between '1.70,1.72'. Can you explain this answer?
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Verified Answer
A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and c...
P = 10kN = 10 *103 N, A = 100mm2
Total change in the length of the bar 
Most Upvoted Answer
A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and c...
Given information:
- Length of the bar, L = 1 m
- Cross-sectional area, A = 100 mm^2 = 100 × 10^-6 m^2
- Elastic modulus, E(x) = 100e-x GPa
- Axial tensile load, P = 10 kN = 10 × 10^3 N
Approach:
To find the axial displacement of the free end, we need to calculate the strain at the free end and then multiply it by the length of the bar. The strain can be obtained by integrating the stress along the length of the bar using Hooke's law.
Step 1: Calculate the stress at each point:
The stress at any point x is given by σ(x) = E(x) × ε(x), where ε(x) is the strain at that point.
Step 2: Calculate the strain at each point:
The strain at any point x is given by ε(x) = δ(x) / L, where δ(x) is the displacement at that point.
Step 3: Apply Hooke's law:
Since the bar is made of a linearly elastic material, Hooke's law can be applied to relate stress and strain. For a uniaxial stress condition, Hooke's law is given by σ(x) = E(x) × ε(x).
Step 4: Integrate the stress equation:
To find the displacement at the free end, we need to integrate the stress equation from x = 0 to x = 1.
Step 5: Calculate the displacement at the free end:
Finally, we can calculate the displacement at the free end by multiplying the strain at x = 1 by the length of the bar.
Calculations:
- Given: E(x) = 100e-x GPa, A = 100 × 10^-6 m^2, P = 10 × 10^3 N, L = 1 m
Step 1: Calculate the stress at each point:
σ(x) = E(x) × ε(x)
Step 2: Calculate the strain at each point:
ε(x) = δ(x) / L
Step 3: Apply Hooke's law:
σ(x) = E(x) × ε(x)
Step 4: Integrate the stress equation:
∫[0,1] σ(x) dx = ∫[0,1] E(x) × ε(x) dx
∫[0,1] E(x) × ε(x) dx = ∫[0,1] E(x) × (δ(x) / L) dx
The integral of E(x) can be solved as:
∫[0,1] E(x) dx = 100 ∫[0,1] e-x dx = 100 [-e-x] [0,1] = 100 (1 - e^-1) GPa.m
Similarly, the integral of ε(x) can be solved as:
∫[0,1] ε(x) dx = ∫[0,1] (δ(x) / L) dx = (1 / L) ∫[0,1]
- Length of the bar, L = 1 m
- Cross-sectional area, A = 100 mm^2 = 100 × 10^-6 m^2
- Elastic modulus, E(x) = 100e-x GPa
- Axial tensile load, P = 10 kN = 10 × 10^3 N
Approach:
To find the axial displacement of the free end, we need to calculate the strain at the free end and then multiply it by the length of the bar. The strain can be obtained by integrating the stress along the length of the bar using Hooke's law.
Step 1: Calculate the stress at each point:
The stress at any point x is given by σ(x) = E(x) × ε(x), where ε(x) is the strain at that point.
Step 2: Calculate the strain at each point:
The strain at any point x is given by ε(x) = δ(x) / L, where δ(x) is the displacement at that point.
Step 3: Apply Hooke's law:
Since the bar is made of a linearly elastic material, Hooke's law can be applied to relate stress and strain. For a uniaxial stress condition, Hooke's law is given by σ(x) = E(x) × ε(x).
Step 4: Integrate the stress equation:
To find the displacement at the free end, we need to integrate the stress equation from x = 0 to x = 1.
Step 5: Calculate the displacement at the free end:
Finally, we can calculate the displacement at the free end by multiplying the strain at x = 1 by the length of the bar.
Calculations:
- Given: E(x) = 100e-x GPa, A = 100 × 10^-6 m^2, P = 10 × 10^3 N, L = 1 m
Step 1: Calculate the stress at each point:
σ(x) = E(x) × ε(x)
Step 2: Calculate the strain at each point:
ε(x) = δ(x) / L
Step 3: Apply Hooke's law:
σ(x) = E(x) × ε(x)
Step 4: Integrate the stress equation:
∫[0,1] σ(x) dx = ∫[0,1] E(x) × ε(x) dx
∫[0,1] E(x) × ε(x) dx = ∫[0,1] E(x) × (δ(x) / L) dx
The integral of E(x) can be solved as:
∫[0,1] E(x) dx = 100 ∫[0,1] e-x dx = 100 [-e-x] [0,1] = 100 (1 - e^-1) GPa.m
Similarly, the integral of ε(x) can be solved as:
∫[0,1] ε(x) dx = ∫[0,1] (δ(x) / L) dx = (1 / L) ∫[0,1]
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A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100mm2. Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, Where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm.Correct answer is between '1.70,1.72'. Can you explain this answer?
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A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100mm2. Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, Where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm.Correct answer is between '1.70,1.72'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100mm2. Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, Where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm.Correct answer is between '1.70,1.72'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100mm2. Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, Where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm.Correct answer is between '1.70,1.72'. Can you explain this answer?.
A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100mm2. Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, Where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm.Correct answer is between '1.70,1.72'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100mm2. Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, Where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm.Correct answer is between '1.70,1.72'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and cross-sectional area of 100mm2. Its elastic modulus varies along its length as given by E(x) = 100e-x GPa, Where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm.Correct answer is between '1.70,1.72'. Can you explain this answer?.
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