### Problem Statement:Compound axial member ABC has a uniform diameter of \( d = 1.45 \, \text{in} \). Segment (1) is an aluminum \([E_1 = 10{,}000 \, \text{ksi}]\) alloy rod with a length of \( L_1 = 85 \, \text{in} \). Segment (2) is a copper \([E_2 = 17{,}000 \, \text{ksi}]\) alloy rod with a length of \( L_2 = 121 \, \text{in} \). When an axial force \( P \) is applied, a strain gage attached to copper segment (2) measures a normal strain of \( \varepsilon_2 = 1800 \, \mu\text{in.} / \text{in.} \) in the longitudinal direction. What is the total elongation of member ABC?### Diagram Explanation:The diagram provided shows a compound member ABC, composed of two segments:- Segment (1) is the aluminum rod, labeled \( L_1 \), starting from point \( A \) and ending at point \( B \).- Segment (2) is the copper rod, labeled \( L_2 \), starting from point \( B \) and ending at point \( C \).Both segments have the same diameter \( d \) and are connected in series. An axial force \( P \) is applied, causing the entire member ABC to elongate.### Required Calculation:To find the total elongation \( \delta_{ABC} \):\[ \delta_{ABC} = \text{Total elongation of member ABC} \]The problem also provides buttons for additional study resources like eTextbook and Media, and a GO Tutorial.Additionally, there are options to save your progress and to submit the final answer, with a note indicating `Attempts: 0 of 5 used`.### Elongation Calculation:The total elongation \( \delta_{ABC} \) can be broken down as the sum of the elongations of each segment due to the applied force \( P \).#### Elongation of Segment (1) - Aluminum:\[ \delta_1 = \frac{P \cdot L_1}{A \cdot E_1} \]#### Elongation of Segment (2) - Copper:Given that the

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Compound axial member ABC has a uniform diameter of d = 1.45 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy rod with a ength of L₁= 85 in. Segment (2) is a copper [E2 = 17,000 ksi] alloy rod with a length of L₂ = 121 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 1800 pin./in. in the longitudinal direction. What is the total elongation of member ABC? LA Aluminum B L2 Copper CP

Compound axial member ABC has a uniform diameter of d = 1.45 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy rod with a ength of L₁= 85 in. Segment (2) is a copper [E2 = 17,000 ksi] alloy rod with a length of L₂ = 121 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 1800 pin./in. in the longitudinal direction. What is the total elongation of member ABC? LA Aluminum B L2 Copper CP

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Thermal Properties of Materials

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Question

### Problem Statement:
Compound axial member ABC has a uniform diameter of \( d = 1.45 \, \text{in} \). Segment (1) is an aluminum \([E_1 = 10{,}000 \, \text{ksi}]\) alloy rod with a length of \( L_1 = 85 \, \text{in} \). Segment (2) is a copper \([E_2 = 17{,}000 \, \text{ksi}]\) alloy rod with a length of \( L_2 = 121 \, \text{in} \). When an axial force \( P \) is applied, a strain gage attached to copper segment (2) measures a normal strain of \( \varepsilon_2 = 1800 \, \mu\text{in.} / \text{in.} \) in the longitudinal direction. What is the total elongation of member ABC?

### Diagram Explanation:
The diagram provided shows a compound member ABC, composed of two segments:

- Segment (1) is the aluminum rod, labeled \( L_1 \), starting from point \( A \) and ending at point \( B \).
- Segment (2) is the copper rod, labeled \( L_2 \), starting from point \( B \) and ending at point \( C \).

Both segments have the same diameter \( d \) and are connected in series. An axial force \( P \) is applied, causing the entire member ABC to elongate.

### Required Calculation:
To find the total elongation \( \delta_{ABC} \):

\[ \delta_{ABC} = \text{Total elongation of member ABC} \]

The problem also provides buttons for additional study resources like eTextbook and Media, and a GO Tutorial.

Additionally, there are options to save your progress and to submit the final answer, with a note indicating `Attempts: 0 of 5 used`.

### Elongation Calculation:
The total elongation \( \delta_{ABC} \) can be broken down as the sum of the elongations of each segment due to the applied force \( P \).

#### Elongation of Segment (1) - Aluminum:
\[ \delta_1 = \frac{P \cdot L_1}{A \cdot E_1} \]

#### Elongation of Segment (2) - Copper:
Given that the

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