= Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in the figure. One load of P 35 kips is applied to the assembly at A, two loads of Q = 25 kips are applied at B, and two loads of R = 34 kips are applied at C. The normal stress magnitude in aluminum rod (1) must be limited to 16 ksi. The normal stress magnitude in steel rod (2) must be limited to 35 ksi. The normal stress magnitude in brass rod (3) must be limited to 21 ksi. Determine the minimum diameter required for each of the three rods. P d₁ d2 d3 = = = A IN i i i IN 1 1 1 B Q in. in. in. R (3) R D

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**Axial Loads and Stress Analysis of Cylindrical Rods**

Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in the figure. It is given that:

- One load of \( P = 35 \) kips is applied to the assembly at point \( A \).
- Two loads of \( Q = 25 \) kips are applied at point \( B \).
- Two loads of \( R = 34 \) kips are applied at point \( C \).

The constraints on the rods are as follows:

- The normal stress magnitude in the aluminum rod (1) must be limited to \( 16 \) ksi.
- The normal stress magnitude in the steel rod (2) must be limited to \( 35 \) ksi.
- The normal stress magnitude in the brass rod (3) must be limited to \( 21 \) ksi.

**Objective:** 
Determine the minimum diameter required for each of the three rods.

### Diagram Explanation:

- The figure depicts a series of three cylindrical rods connected end-to-end, each subjected to different axial loads.
- From left to right, the rods are connected between points \( A \), \( B \), \( C \), and \( D \).
- Rod \( (1) \) experiences the load \( P \) applied at \( A \).
- Rod \( (2) \) experiences two loads \( Q \) applied at \( B \).
- Rod \( (3) \) experiences two loads \( R \) applied at \( C \).

### Calculation Input Fields:

The inputs for determining the minimum required diameters for each rod are depicted as follows:

\[ d_1 = \boxed{\ \ \ \ } \text{ in.} \]
\[ d_2 = \boxed{\ \ \ \ } \text{ in.} \]
\[ d_3 = \boxed{\ \ \ \ } \text{ in.} \]

Where:
- \( d_1 \) is the diameter of the aluminum rod (1).
- \( d_2 \) is the diameter of the steel rod (2).
- \( d_3 \) is the diameter of the brass rod (3).

To find the minimum diameters, apply the fundamental stress formula:

\[ \sigma = \frac{F}{A} \]

Where:
- \( \sigma \) is the stress.
- \( F \) is the force (load).
Transcribed Image Text:**Axial Loads and Stress Analysis of Cylindrical Rods** Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in the figure. It is given that: - One load of \( P = 35 \) kips is applied to the assembly at point \( A \). - Two loads of \( Q = 25 \) kips are applied at point \( B \). - Two loads of \( R = 34 \) kips are applied at point \( C \). The constraints on the rods are as follows: - The normal stress magnitude in the aluminum rod (1) must be limited to \( 16 \) ksi. - The normal stress magnitude in the steel rod (2) must be limited to \( 35 \) ksi. - The normal stress magnitude in the brass rod (3) must be limited to \( 21 \) ksi. **Objective:** Determine the minimum diameter required for each of the three rods. ### Diagram Explanation: - The figure depicts a series of three cylindrical rods connected end-to-end, each subjected to different axial loads. - From left to right, the rods are connected between points \( A \), \( B \), \( C \), and \( D \). - Rod \( (1) \) experiences the load \( P \) applied at \( A \). - Rod \( (2) \) experiences two loads \( Q \) applied at \( B \). - Rod \( (3) \) experiences two loads \( R \) applied at \( C \). ### Calculation Input Fields: The inputs for determining the minimum required diameters for each rod are depicted as follows: \[ d_1 = \boxed{\ \ \ \ } \text{ in.} \] \[ d_2 = \boxed{\ \ \ \ } \text{ in.} \] \[ d_3 = \boxed{\ \ \ \ } \text{ in.} \] Where: - \( d_1 \) is the diameter of the aluminum rod (1). - \( d_2 \) is the diameter of the steel rod (2). - \( d_3 \) is the diameter of the brass rod (3). To find the minimum diameters, apply the fundamental stress formula: \[ \sigma = \frac{F}{A} \] Where: - \( \sigma \) is the stress. - \( F \) is the force (load).
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