A 12 mm diameter rod has an axial strain of 0.0009. If the metal rod has a Poisson's Ratio of 0.25, what is the transverse deformation in the rod (change in diameter) in mm?
A 12 mm diameter rod has an axial strain of 0.0009. If the metal rod has a Poisson's Ratio of 0.25, what is the transverse deformation in the rod (change in diameter) in mm?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![**Question:**
A 12 mm diameter rod has an axial strain of 0.0009. If the metal rod has a Poisson's Ratio of 0.25, what is the transverse deformation in the rod (change in diameter) in mm?
**Explanation:**
To find the transverse deformation, we can use the formula for Poisson's effect, which relates the lateral strain to the axial strain through Poisson's Ratio (ν):
\[ \text{Lateral Strain} = -\nu \times \text{Axial Strain} \]
Given:
- Axial Strain = 0.0009
- Poisson’s Ratio (ν) = 0.25
The lateral strain will be:
\[ \text{Lateral Strain} = -0.25 \times 0.0009 = -0.000225 \]
The transverse deformation (change in diameter) is calculated by multiplying the lateral strain by the original diameter:
\[ \text{Change in Diameter} = 12 \, \text{mm} \times -0.000225 = -0.0027 \, \text{mm} \]
Thus, the diameter decreases by 0.0027 mm due to the axial strain.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1815343-0e1e-45bc-b31a-0935809f9211%2F6b8181c8-fb1a-4d49-b3d9-8ec144e4f981%2Fzn59zd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
A 12 mm diameter rod has an axial strain of 0.0009. If the metal rod has a Poisson's Ratio of 0.25, what is the transverse deformation in the rod (change in diameter) in mm?
**Explanation:**
To find the transverse deformation, we can use the formula for Poisson's effect, which relates the lateral strain to the axial strain through Poisson's Ratio (ν):
\[ \text{Lateral Strain} = -\nu \times \text{Axial Strain} \]
Given:
- Axial Strain = 0.0009
- Poisson’s Ratio (ν) = 0.25
The lateral strain will be:
\[ \text{Lateral Strain} = -0.25 \times 0.0009 = -0.000225 \]
The transverse deformation (change in diameter) is calculated by multiplying the lateral strain by the original diameter:
\[ \text{Change in Diameter} = 12 \, \text{mm} \times -0.000225 = -0.0027 \, \text{mm} \]
Thus, the diameter decreases by 0.0027 mm due to the axial strain.
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