solve with proper explanationThe table shows the results: Elongation in in. and Load in Ib., of the tension test for a new material under development. The gauge length is 2
in. and a diameter of 1/4 in. The final diameter is 0.210 in. Determine the modulus of resilience from your response in (ft-Ib)/in.*
i
in.
[0000 | 0 |
[0008 | 400 |
0016 | 800 |
0018 [900 |
[002 | 1100 |
[0027 | 1200 |
0075 | 1200 |
[0300 [ 1720
| _0400 | 1820 |
[0600 | 1950 |
|__0800 | 1900 |
| 1000 | 1800 |
Question:
solve with proper explanationThe table shows the results: Elongation in in. and Load in Ib., of the tension test for a new material under development. The gauge length is 2
in. and a diameter of 1/4 in. The final diameter is 0.210 in. Determine the modulus of resilience from your response in (ft-Ib)/in.*
i
in.
[0000 | 0 |
[0008 | 400 |
0016 | 800 |
0018 [900 |
[002 | 1100 |
[0027 | 1200 |
0075 | 1200 |
[0300 [ 1720
| _0400 | 1820 |
[0600 | 1950 |
|__0800 | 1900 |
| 1000 | 1800 |
This is AI generated Response.
Asked by: manish kumar
Created at: 2025-08-15 09:36:00
Final Answer:
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Let's solve the problem step by step:
### **Given:**
- Gauge length (\(L\)) = 2 in.
- Initial diameter (\(d_0\)) = 0.25 in.
- Final diameter (\(d_f\)) = 0.210 in.
- Table provides Elongation (in) and Load (lb).
### **Required:**
**Find the modulus of resilience** in \((\text{ft-lb})/\text{in}^3\).
---
## **Step 1: Calculate Stress and Strain**
**Stress (\(\sigma\)):**
\[
\sigma = \frac{\text{Load}}{\text{Area}}
\]
Area (\(A\)) = \(\frac{\pi d_0^2}{4} = \frac{\pi (0.25)^2}{4} = 0.0491~\text{in}^2\)
**Strain (\(\epsilon\)):**
\[
\epsilon = \frac{\text{Elongation}}{\text{Gauge Length}}
\]
---
## **Step 2: Find Yield Point**
Modulus of resilience = **area under the stress-strain curve up to the elastic limit (yield point)**.
From the table, the load and elongation increase linearly up to 0.022 in. elongation and 1100 lb load. After that, the load increases non-linearly, indicating the yield point is around here.
So, use:
- Load = 1100 lb
- Elongation = 0.022 in
---
### **Stress at Yield:**
\[
\sigma_{y} = \frac{1100}{0.0491} = 22400~\text{psi}
\]
### **Strain at Yield:**
\[
\epsilon_{y} = \frac{0.022}{2} = 0.011
\]
---
## **Step 3: Modulus of Elasticity (E)**
\[
E = \frac{\sigma_{y}}{\epsilon_{y}} = \frac{22400}{0.011} = 2,036,364~\text{psi}
\]
---
### **Modulus of Resilience Formula**
\[
\text{Modulus of Resilience} = \frac{\sigma_{y}^2}{2E}
\]
\[
= \frac{(22400)^2}{2 \times 2,036,364}
\]
\[
= \frac{501,760,000}{4,072,728}
\]
\[
= 123.2~\text{psi}
\]
---
### **Convert psi to (ft-lb)/in³**
1 psi = 1 lb/in²
1 (ft-lb)/in³ = 12 lb/in² (since 1 ft = 12 in)
So,
\[
\text{Modulus of Resilience} = \frac{123.2}{12} = 10.27~(\text{ft-lb})/\text{in}^3
\]
---
## **Final Answer**
### **The modulus of resilience is:**
\[
\boxed{10.3~(\text{ft-lb})/\text{in}^3}
\]
(to 1 decimal place)
---
#### **Summary of Steps:**
1. Find area, stress, strain at yield.
2. Calculate modulus of elasticity.
3. Use modulus of resilience formula.
4. Convert units as required.
If you need the calculations in more detail (with more data points), let me know!
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