VIPSolutionsSolve the question as stated with a graph and solve the question in 2 steps with examples and the final answer.05) Actinic br, whichis made to porns AB sd BC, bfckd io ing engoeot shown n Fifer 0.5 Gv hte whole bt rad of he same ea (Mod of Elmicy 205 GPa ad he Pom 036) (0 Fiod he il xinsin ccmeesio of he br (mart) {6 Decmie te changes i ton a coms sci. (6 packs) ooops 48 sd prin BC a rl Fewease ©) The cymbal ABCD wth 100mm amc avd ppt st Dd pr 0h oe oe he in 5 Fiewe 0s 0 De bk rsd et gm 6 (Sm) 0 llcne. be’ pol moet of inn 1) for ‘cylindrical bar. ool 0 CAA ot rn dpe ef od (mc 1
Question:
Solve the question as stated with a graph and solve the question in 2 steps with examples and the final answer.
05) Actinic br, whichis made to porns AB sd BC, bfckd io
ing engoeot shown n Fifer 0.5 Gv hte whole bt
rad of he same ea (Mod of Elmicy 205 GPa ad he
Pom 036)
(0 Fiod he il xinsin ccmeesio of he br (mart)
{6 Decmie te changes i ton a coms sci. (6 packs)
ooops 48 sd prin BC
a rl
Fewease
©) The cymbal ABCD wth 100mm amc avd ppt st Dd
pr 0h oe oe he in 5
Fiewe 0s
0 De bk rsd et gm 6 (Sm)
0 llcne. be’ pol moet of inn 1) for
‘cylindrical bar. ool
0 CAA ot rn dpe ef od (mc
1
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Asked by: Rahul
Created at: 2026-01-02 19:44:13
Asked by: Rahul
Created at: 2026-01-02 19:44:13
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Let's solve **Q.5(a)** step by step.
---
# Q.5(a) Solution
### **Given Data**
- **Material:** Same for both segments
- Modulus of Elasticity, \( E = 205 \) GPa \( = 205 \times 10^9 \) Pa
- Poisson's Ratio, \( \nu = .35 \)
- **AB:**
- Length, \( L_{AB} = 1.2 \) m
- Diameter, \( d_{AB} = 60 \) mm \( = .06 \) m
- Force, \( F_{AB} = 100 \) kN \( = 100,000 \) N
- **BC:**
- Length, \( L_{BC} = .8 \) m
- Diameter, \( d_{BC} = 80 \) mm \( = .08 \) m
- Force, \( F_{BC} = 250 \) kN \( = 250,000 \) N
---
## **Step 1: Find Total Extension/Compression of the Bar**
### **Axial Deformation Formula**
\[
\delta = \frac{FL}{AE}
\]
Where:
- \( F \) = Force (N)
- \( L \) = Length (m)
- \( A \) = Cross-sectional area (\( m^2 \))
- \( E \) = Young’s modulus (Pa)
### **Calculate Areas**
\[
A_{AB} = \frac{\pi}{4} d_{AB}^2 = \frac{\pi}{4}(.06)^2 = .002827 \, m^2
\]
\[
A_{BC} = \frac{\pi}{4} d_{BC}^2 = \frac{\pi}{4}(.08)^2 = .005027 \, m^2
\]
### **Calculate Deformations**
#### **For AB:**
\[
\delta_{AB} = \frac{F_{AB} L_{AB}}{A_{AB} E}
= \frac{100{,}000 \times 1.2}{.002827 \times 205 \times 10^9}
= \frac{120,000}{579,535,000}
= .000207 \text{ m} = .207 \text{ mm}
\]
#### **For BC:**
\[
\delta_{BC} = \frac{F_{BC} L_{BC}}{A_{BC} E}
= \frac{250,000 \times .8}{.005027 \times 205 \times 10^9}
= \frac{200,000}{1,030,535,000}
= .000194 \text{ m} = .194 \text{ mm}
\]
#### **Total Extension**
\[
\delta_{total} = \delta_{AB} + \delta_{BC} = .000207 + .000194 = .000401 \text{ m} = .401 \text{ mm}
\]
---
## **Step 2: Determine Change in Dimensions of the Cross Sections**
### **Lateral/Transverse Strain**
\[
\epsilon_{lat} = -\nu \epsilon_{long}
\]
Where \( \epsilon_{long} = \frac{\delta}{L} \)
#### **For AB:**
\[
\epsilon_{long,AB} = \frac{\delta_{AB}}{L_{AB}} = \frac{.000207}{1.2} = .000172
\]
\[
\epsilon_{lat,AB} = -.35 \times .000172 = -.000060
\]
\[
\Delta d_{AB} = \epsilon_{lat,AB} \times d_{AB} = -.000060 \times .06 = -.0000036 \text{ m} = -.0036 \text{ mm}
\]
#### **For BC:**
\[
\epsilon_{long,BC} = \frac{.000194}{.8} = .0002425
\]
\[
\epsilon_{lat,BC} = -.35 \times .0002425 = -.0000849
\]
\[
\Delta d_{BC} = \epsilon_{lat,BC} \times d_{BC} = -.0000849 \times .08 = -.00000679 \text{ m} = -.0068 \text{ mm}
\]
---
## **Summary Table**
| Section | Axial Deformation (mm) | Lateral Change (mm) |
|---------|------------------------|---------------------|
| AB | .207 | -.0036 |
| BC | .194 | -.0068 |
- **Total Extension:** **.401 mm**
- **Change in diameter for AB:** **-.0036 mm**
- **Change in diameter for BC:** **-.0068 mm**
---
## **Graph: Axial Deformation Along the Bar**
```plaintext
| *
| *
| *
| * *
| * *
| * *
|___________*___*___________
A B C
```
- Slope is higher in AB (smaller area), lower in BC (larger area).
---
# **Final Answers**
**Step 1:**
- **Total extension/compression:**
\(\boxed{.401 \text{ mm}}\)
**Step 2:**
- **Change in diameter for AB:** \(\boxed{-.0036 \text{ mm}}\)
- **Change in diameter for BC:** \(\boxed{-.0068 \text{ mm}}\)
---
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