A sample of steel with a diameter of 0.49 in. is tested in tension using an extensometer with a gauge length of 2.0 in.
A load of 9000 lbs is applied. If the modulus of elasticity is 30,000 ksi, what is the stress? Assume a yield stress of 60 ksi.
ksi
What is the elongation (deformation) in the sample at the same point?
in.
If the load were released, how much permanent deformation would be in the sample?
in.

Let's break down the problem step-by-step:

### **1. Calculating the Stress:**

Stress (\(\sigma\)) is calculated using the formula:
\[
\sigma = \frac{F}{A}
\]
Where:
- \(F\) is the force (load) applied = 9000 lbs
- \(A\) is the cross-sectional area of the steel sample

First, we need to calculate the cross-sectional area \(A\) for the sample with a diameter \(d = 0.49\) in.

The cross-sectional area of a circular sample is:
\[
A = \frac{\pi d^2}{4}
\]

Let's calculate this value:

\[
A = \frac{\pi (0.49)^2}{4}
\]

Next, calculate the stress by dividing the force by the area:
\[
\sigma = \frac{9000}{A}
\]

I'll perform these calculations now.The cross-sectional area of the sample is approximately \(0.1886 \, \text{in}^2\). The stress (\(\sigma\)) applied to the sample is approximately \(47.73 \, \text{ksi}\).

### **2. Calculating the Elongation (Deformation):**

The elongation \(\Delta L\) can be found using Hooke's law:
\[
\Delta L = \frac{\sigma L}{E}
\]
Where:
- \(L = 2.0 \, \text{in}\) (gauge length)
- \(\sigma = 47.73 \, \text{ksi}\)
- \(E = 30,000 \, \text{ksi}\) (modulus of elasticity)

Now, let's calculate \(\Delta L\).The elongation (deformation) in the sample at the applied load is approximately \(0.00318 \, \text{in}\).

### **3. Determining the Permanent Deformation:**

Since the applied stress (47.73 ksi) is less than the yield stress (60 ksi), the material remains in the elastic region. This means that once the load is released, the material will return to its original length with **no permanent deformation**.

### **Summary:**
- **Stress:** \(47.73 \, \text{ksi}\)
- **Elongation:** \(0.00318 \, \text{in}\)
- **Permanent Deformation:** \(0 \, \text{in}\)

Do you want more details or have any questions?

### **Further Questions:**
1. How would the stress change if the applied force were doubled?
2. What would be the elongation if the modulus of elasticity was lower?
3. What happens to the sample if the applied load exceeds the yield stress?
4. How is the modulus of elasticity related to material stiffness?
5. What is the difference between elastic and plastic deformation?

### **Tip:** Remember, as long as stress is below the yield stress, materials return to their original shape once the load is removed.

A sample of steel with a diameter of 0.49 in. is tested in tension using an extensometer with a gauge length of 2.0 in. A load of 9000 lbs is applied. If the modulus of elasticity is 30,000 ksi, what is the stress? Assume a yield stress of 60 ksi. What is the elongation (deformation) in the sample at the same point? If the load were released, how much permanent deformation would be in the sample?

Explore the stress, elongation, and permanent deformation in a steel sample with a diameter of 0.49 in. and a gauge length of 2.0 in. under a 9000 lbs load. This step-by-step solution explains how to calculate stress (47.73 ksi), elongation (0.00318 in.), and the absence of permanent deformation when the load is released.

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