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To solve the given problem, we need to use the following equations and parameters for tensile yielding and tensile rupture.

The given information includes:
- Thickness of the plate: \( t = \frac{3}{8} \) inches
- Length of the plate: 7 inches (though the length isn't directly necessary for tensile strength calculations)
- Steel yield stress: \( F_y = 36 \, \text{ksi} \)
- Steel ultimate stress: \( F_u = 58 \, \text{ksi} \)
- Effective net area \( A_e \) is assumed to be equal to the gross area \( A_g \) (given as \( A_e = A_g \))

### 1. Design Strength for LRFD (Tensile Yielding):
The design strength considering tensile yielding in LRFD is calculated using:

\[
\phi P_n = \phi A_g F_y
\]

Where:
- \( \phi = 0.9 \) (resistance factor for yielding in LRFD)
- \( A_g \) is the gross area, calculated as \( A_g = \text{width} \times \text{thickness} \)
- \( F_y = 36 \, \text{ksi} \)

### 2. Allowable Strength for ASD (Tensile Yielding):
For ASD, the allowable strength considering tensile yielding is:

\[
P_n = \frac{A_g F_y}{\Omega}
\]

Where:
- \( \Omega = 1.67 \) (safety factor for yielding in ASD)
- \( A_g \) is the gross area.

### 3. Design Strength for LRFD (Tensile Rupture):
For tensile rupture under LRFD:

\[
\phi P_n = \phi A_e F_u
\]

Where:
- \( \phi = 0.75 \) (resistance factor for rupture in LRFD)
- \( A_e \) is the effective net area.
- \( F_u = 58 \, \text{ksi} \)

### 4. Allowable Strength for ASD (Tensile Rupture):
For ASD, the allowable strength considering tensile rupture is:

\[
P_n = \frac{A_e F_u}{\Omega}
\]

Where:
- \( \Omega = 2.00 \) (safety factor for rupture in ASD).

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I will now proceed with the calculations, assuming the plate has a width \( w \) of 7 inches and thickness \( t = \frac{3}{8} \).Here are the computed strengths based on the given dataHere are the computed strengths based on the given 

1. **Design strength for LRFD (Tensile Yielding):**  
   \( \phi P_n = 85.05 \, \text{kips} \)

2. **Allowable strength for ASD (Tensile Yielding):**  
   \( P_n = 56.59 \, \text{kips} \)

3. **Design strength for LRFD (Tensile Rupture):**  
   \( \phi P_n = 114.19 \, \text{kips} \)

4. **Allowable strength for ASD (Tensile Rupture):**  
   \( P_n = 76.13 \, \text{kips} \)

Let me know if you'd like more details or explanations.

Here are 5 related questions to explore:
1. How do bolt hole reductions affect net area in tension members?
2. What are the differences between gross and net area in steel design?
3. How do we account for staggered bolt holes in tension member designs?
4. What are the implications of using a higher yield strength steel in tension member designs?
5. How does the resistance factor \( \phi \) vary for different loading conditions in LRFD?

**Tip:** Always double-check the net area calculation when bolt holes or reductions are involved.

A PL 3/8 x 7 tension member is connected with three 1-inch-diameter bolts. The steel is Fy = 36 ksi and Fu = 58 ksi. Assume that Ae = Ag and compute: a) the design strength for LRFD considering tensile yielding, b) the allowable strength for ASD considering tensile yielding, c) the design strength for LRFD considering tensile rupture, d) the design strength for ASD considering tensile rupture.

Learn how to compute the design and allowable strengths of a PL 3/8 x 7 tension member connected with three 1-inch-diameter bolts using LRFD and ASD methods. This problem involves calculating tensile yielding and tensile rupture using the steel yield and ultimate stresses (Fy = 36 ksi, Fu = 58 ksi). Suitable for undergraduate structural engineering students.

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