A standard steel specimen having a diameter of 0.505 in. and a 2.00-in. gage length is used in a tension test. At what load P will the extensometer read 0.002 in. deformation? Assume a proportional limit of 34,000 psi.



To find the load \( P \) that will result in an elongation (deformation) of 0.002 in. in the steel specimen, we can use **Hooke's Law**. Since the deformation is within the elastic range, the stress is proportional to strain. 

Given dataTo find the load \( P \) that will result in an elongation (deformation) of 0.002 in. in the steel specimen, we can use **Hooke's Law**. Since the deformation is within the elastic range, the stress is proportional to strain. 

Given 
- Diameter of the specimen: \( d = 0.505 \, \text{in.} \)
- Gage length: \( L_0 = 2.00 \, \text{in.} \)
- Deformation (elongation): \( \Delta L = 0.002 \, \text{in.} \)
- Proportional limit (yield stress): \( \sigma = 34,000 \, \text{psi} \)

### Steps to solve:
1. **Determine the strain**:
   \[
   \varepsilon = \frac{\Delta L}{L_0} = \frac{0.002 \, \text{in.}}{2.00 \, \text{in.}} = 0.001
   \]
   Strain is dimensionless.

2. **Using Hooke's Law**:
   Stress \( \sigma \) is related to strain \( \varepsilon \) through the proportional limit (in the elastic range):
   \[
   \sigma = E \varepsilon
   \]
   where \( E \) is the modulus of elasticity. However, since we know the stress is at the proportional limit of 34,000 psi, this is the stress causing the deformation.

3. **Calculate the cross-sectional area of the specimen**:
   The cross-sectional area \( A \) of the specimen is the area of a circle:
   \[
   A = \frac{\pi d^2}{4} = \frac{\pi (0.505 \, \text{in.})^2}{4} = 0.2003 \, \text{in}^2
   \]

4. **Calculate the load**:
   Stress \( \sigma \) is related to load \( P \) and cross-sectional area \( A \) by:
   \[
   \sigma = \frac{P}{A}
   \]
   Rearranging for \( P \):
   \[
   P = \sigma \times A = 34,000 \, \text{psi} \times 0.2003 \, \text{in}^2 = 6,810.2 \, \text{lbs}
   \]

Thus, the load \( P \) required to cause a deformation of 0.002 in. is approximately **6,810 lbs**.

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Would you like more details on any part of the solution? 

Here are some related questions:
1. How is the modulus of elasticity \( E \) used in calculating deformation?
2. What happens to the material after it surpasses the proportional limit?
3. How do different diameters of specimens affect the load required for a given deformation?
4. Why is it important to know the gage length in tension tests?
5. What are the consequences of assuming an ideal elastic behavior in practical applications?

**Tip**: Always check that units are consistent when applying formulas, especially when dealing with stress and strain in different systems of measurement.

A standard steel specimen having a diameter of 0.505 in. and a 2.00-in. gage length is used in a tension test. At what load P will the extensometer read 0.002 in. deformation? Assume a proportional limit of 34,000 psi.

Find the load P required to cause a deformation of 0.002 in. in a standard steel specimen with a diameter of 0.505 in. and a gage length of 2.00 in. using Hooke's Law and a proportional limit of 34,000 psi. This detailed solution includes key concepts such as stress-strain relationships and the use of cross-sectional area calculations.

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