VIPSolutionsGive step-by-step solution with explanation and final answer:PROBLEM 024,5.6,7 Forthe next questions, please use this information to solve problems. From the tensile test of bronze specimen with gage length of 115 mm and diameter of 20 im the following mechanics properties ae know: £=110000 MP v=034 Question. 03pts Knowing: Sy=samps Su= 190 MP Wht isthe maximum ail fore in Newton that may be applied to the specimen without plastic deformation? Question 5 03pts Knowing: Sy=s0mps Sue 234 MP Wht is the axial elongation (AL in milimeters) that occurred without plastic deformation? Express the answer to three significative figures. Question 6 o02pts Knowing: Sy=semps Su=283P Wht is the axa force (i Newton) that may be applied othe specimen to break Question? 03pts Knowing: Sy=simps Su= 309 MPa Wht is the transverse strin m5) just befor the pli deformstion? Express the answer to three significative figures and absolute value.
Question:
Give step-by-step solution with explanation and final answer:
PROBLEM 024,5.6,7
Forthe next questions, please use this information to solve problems.
From the tensile test of bronze specimen with gage length of 115 mm and diameter of 20
im the following mechanics properties ae know:
£=110000 MP
v=034
Question. 03pts
Knowing:
Sy=samps
Su= 190 MP
Wht isthe maximum ail fore in Newton that may be applied to the specimen without
plastic deformation?
Question 5 03pts
Knowing:
Sy=s0mps
Sue 234 MP
Wht is the axial elongation (AL in milimeters) that occurred without plastic deformation?
Express the answer to three significative figures.
Question 6 o02pts
Knowing:
Sy=semps
Su=283P
Wht is the axa force (i Newton) that may be applied othe specimen to break
Question? 03pts
Knowing:
Sy=simps
Su= 309 MPa
Wht is the transverse strin m5) just befor the pli deformstion? Express the answer
to three significative figures and absolute value.
This is AI generated Response.
Asked by: Rahul
Created at: 2026-01-19 20:11:41
Asked by: Rahul
Created at: 2026-01-19 20:11:41
Final Answer:
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Let's solve each question step by step using the provided data:
- **Gage length, \( L_ = 115\, \text{mm} \)**
- **Diameter, \( d = 20\, \text{mm} \)**
- **Young’s modulus, \( E = 110{,}000\, \text{MPa} \)**
- **Poisson's ratio, \( \nu = .34 \)**
### **Step 1: Calculate Cross-Sectional Area**
\[
A = \frac{\pi d^2}{4} = \frac{\pi (20)^2}{4} = \frac{\pi \times 400}{4} = 100\pi \approx 314.16\, \text{mm}^2
\]
---
## **Question 4**
**Given:** \( S_y = 53\, \text{MPa} \)
**Asked:** Maximum axial force (N) without plastic deformation.
### **Solution:**
\[
F_{\text{max}} = S_y \times A = 53\, \text{MPa} \times 314.16\, \text{mm}^2
\]
\[
= 53 \times 314.16 = 16,650.48\, \text{N}
\]
**Final Answer:**
\[
\boxed{16,700\, \text{N}}
\]
(Rounded to three significant figures)
---
## **Question 5**
**Given:** \( S_y = 60\, \text{MPa} \), \( L_ = 115\, \text{mm} \), \( E = 110{,}000\, \text{MPa} \)
**Asked:** Axial elongation (\( \Delta l \)) in mm, without plastic deformation.
### **Solution:**
Elastic strain at yield:
\[
\epsilon_y = \frac{S_y}{E} = \frac{60}{110,000} = .0005455
\]
Elongation:
\[
\Delta l = \epsilon_y \times L_ = .0005455 \times 115 = .0627\, \text{mm}
\]
**Final Answer:**
\[
\boxed{.0627\, \text{mm}}
\]
---
## **Question 6**
**Given:** \( S_u = 283\, \text{MPa} \)
**Asked:** Axial force (N) to break the specimen.
### **Solution:**
\[
F_{\text{break}} = S_u \times A = 283 \times 314.16 = 88,904.28\, \text{N}
\]
**Final Answer:**
\[
\boxed{88,900\, \text{N}}
\]
(Rounded to three significant figures)
---
## **Question 7**
**Given:** \( S_y = 51\, \text{MPa} \), \( E = 110,000\, \text{MPa} \), \( \nu = .34 \)
**Asked:** Transverse strain (%) just before plastic deformation.
### **Solution:**
Axial strain at yield:
\[
\epsilon_{\text{axial}} = \frac{S_y}{E} = \frac{51}{110,000} = .0004636
\]
Transverse strain (absolute, using Poisson's ratio):
\[
\epsilon_{\text{trans}} = |\nu \cdot \epsilon_{\text{axial}}| = .34 \times .0004636 = .0001576
\]
Convert to percentage:
\[
\epsilon_{\text{trans}}\, (\%) = .0001576 \times 100 = .0158\%
\]
**Final Answer:**
\[
\boxed{.0158\%}
\]
---
# **Summary Table**
| Question | Final Answer |
|----------|------------------------------|
| 4 | \( 16,700\, \text{N} \) |
| 5 | \( .0627\, \text{mm} \) |
| 6 | \( 88,900\, \text{N} \) |
| 7 | \( .0158\% \) (absolute) |
Question:
Please solved all parts
This is AI generated Response.
Asked by: Rahul
Created at: 2026-01-19 20:12:38
Asked by: Rahul
Created at: 2026-01-19 20:12:38
Final Answer:
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# **Solution to Problems Based on Tensile Test Data**
## **Given Data**
- **Gage length, \( L = 115\, \text{mm} \)**
- **Diameter, \( d = 20\, \text{mm} \)**
- **Young’s modulus, \( E = 110{,}000\, \text{MPa} \)**
- **Poisson's ratio, \( \nu = 0.34 \)**
### **Step 1: Calculate Cross-Sectional Area**
\[
A = \frac{\pi d^2}{4} = \frac{\pi (20\, \text{mm})^2}{4} = \frac{\pi \times 400}{4} \approx 314.16\, \text{mm}^2
\]
---
## **Question 4**
**Given:** \( S_y = 53\, \text{MPa} \)
**Asked:** Maximum axial force (N) without plastic deformation.
### **Solution:**
\[
F_{\text{max}} = S_y \times A = 53\, \text{MPa} \times 314.16\, \text{mm}^2
\]
\[
F_{\text{max}} = 53 \times 314.16 \approx 16,650.48\, \text{N}
\]
**Final Answer:**
\[
\boxed{16,700\, \text{N}}
\]
(Rounded to three significant figures)
---
## **Question 5**
**Given:** \( S_y = 60\, \text{MPa} \)
**Asked:** Axial elongation (\( \Delta l \)) in mm, without plastic deformation.
### **Solution:**
Elastic strain at yield:
\[
\epsilon_y = \frac{S_y}{E} = \frac{60\, \text{MPa}}{110,000\, \text{MPa}} \approx 0.0005455
\]
Elongation:
\[
\Delta l = \epsilon_y \times L = 0.0005455 \times 115\, \text{mm} \approx 0.0627\, \text{mm}
\]
**Final Answer:**
\[
\boxed{0.0627\, \text{mm}}
\]
---
## **Question 6**
**Given:** \( S_u = 283\, \text{MPa} \)
**Asked:** Axial force (N) to break the specimen.
### **Solution:**
\[
F_{\text{break}} = S_u \times A = 283\, \text{MPa} \times 314.16\, \text{mm}^2
\]
\[
F_{\text{break}} = 283 \times 314.16 \approx 88,904.28\, \text{N}
\]
**Final Answer:**
\[
\boxed{88,900\, \text{N}}
\]
(Rounded to three significant figures)
---
## **Question 7**
**Given:** \( S_y = 51\, \text{MPa} \)
**Asked:** Transverse strain (%) just before plastic deformation.
### **Solution:**
Axial strain at yield:
\[
\epsilon_{\text{axial}} = \frac{S_y}{E} = \frac{51\, \text{MPa}}{110,000\, \text{MPa}} \approx 0.0004636
\]
Transverse strain (absolute, using Poisson's ratio):
\[
\epsilon_{\text{trans}} = |\nu \cdot \epsilon_{\text{axial}}| = 0.34 \times 0.0004636 \approx 0.0001576
\]
Convert to percentage:
\[
\epsilon_{\text{trans}}\, (\%) = 0.0001576 \times 100 \approx 0.0158\%
\]
**Final Answer:**
\[
\boxed{0.0158\%}
\]
---
# **Summary Table**
| Question | Final Answer |
|----------|------------------------------|
| 4 | \( 16,700\, \text{N} \) |
| 5 | \( 0.0627\, \text{mm} \) |
| 6 | \( 88,900\, \text{N} \) |
| 7 | \( 0.0158\% \) |
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