4. A load P is applied concentrically to the bar shown. If the allowable normal stress is 76 ksi, determine the maximum allowable load P. P 3.5 in. 0.5 in. dia. hole 0.4 in. rad. 2.5 in. P

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### Determining the Maximum Allowable Load on a Bar with Applied Load P

#### Problem Statement:
A load \( P \) is applied concentrically to the bar shown. If the allowable normal stress is 76 ksi, determine the maximum allowable load \( P \).

#### Diagram Description:
The provided diagram displays a vertically oriented bar that is subject to a load \( P \) applied concentrically on both ends, represented by red arrows pointing towards the bar. 

Key dimensions and features highlighted in the diagram include:
- The bar has an overall length, which includes specific sections of 3.5 inches and 2.5 inches.
- There is a reduction area in the bar with dimensions involving a radius of 0.4 inches at a specific spot.
- A circular hole with a diameter of 0.5 inches is present in the section labeled 3.5 inches.

The geometry involves a middle section curving inward (with the specified radius), creating a region of decreased cross-sectional area where the load is concentrated.

#### Calculations:
To determine the maximum allowable load \( P \), follow these steps:
1. **Identify the Critical Cross-Sectional Area**:
   - The critical area is typically where the nominal stress will likely be the highest, often at the location where the bar has the smallest cross-section, which in this case is around the 0.5-inch diameter hole.
   
2. **Calculate the Cross-Sectional Area**:
   - **Width Before and After Hole**: Assume the bar has a consistent width 'w'.
   - **Net Cross-Sectional Area** (area after accounting for the hole): 
     \[
     A = w \times t - (\text{Area of the hole}) \ =\ w \times t - \left( \frac{\pi \times (d/2)^2}{4} \right)
     \]
     where \( w \) is the width of the bar, \( t \) is the thickness, and \( d \) is the diameter of the hole (0.5 inches).

3. **Allowable Stress**:
   - The normal stress \( \sigma \) is \( 76 \) ksi. Using the stress formula:
     \[
     \sigma = \frac{P}{A}
     \]
     where \( P \) is the load and \( A \) is the cross-sectional area.

4. **Determine Maximum Load \( P
Transcribed Image Text:### Determining the Maximum Allowable Load on a Bar with Applied Load P #### Problem Statement: A load \( P \) is applied concentrically to the bar shown. If the allowable normal stress is 76 ksi, determine the maximum allowable load \( P \). #### Diagram Description: The provided diagram displays a vertically oriented bar that is subject to a load \( P \) applied concentrically on both ends, represented by red arrows pointing towards the bar. Key dimensions and features highlighted in the diagram include: - The bar has an overall length, which includes specific sections of 3.5 inches and 2.5 inches. - There is a reduction area in the bar with dimensions involving a radius of 0.4 inches at a specific spot. - A circular hole with a diameter of 0.5 inches is present in the section labeled 3.5 inches. The geometry involves a middle section curving inward (with the specified radius), creating a region of decreased cross-sectional area where the load is concentrated. #### Calculations: To determine the maximum allowable load \( P \), follow these steps: 1. **Identify the Critical Cross-Sectional Area**: - The critical area is typically where the nominal stress will likely be the highest, often at the location where the bar has the smallest cross-section, which in this case is around the 0.5-inch diameter hole. 2. **Calculate the Cross-Sectional Area**: - **Width Before and After Hole**: Assume the bar has a consistent width 'w'. - **Net Cross-Sectional Area** (area after accounting for the hole): \[ A = w \times t - (\text{Area of the hole}) \ =\ w \times t - \left( \frac{\pi \times (d/2)^2}{4} \right) \] where \( w \) is the width of the bar, \( t \) is the thickness, and \( d \) is the diameter of the hole (0.5 inches). 3. **Allowable Stress**: - The normal stress \( \sigma \) is \( 76 \) ksi. Using the stress formula: \[ \sigma = \frac{P}{A} \] where \( P \) is the load and \( A \) is the cross-sectional area. 4. **Determine Maximum Load \( P
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