Let's break down the problem into its parts. The image provides two tasks, which involve calculations related to tension members and block shear strength.

Problem 1:

A PL 3/8 × 7 tension member is connected with three 1-inch-diameter bolts. The steel has:

• Yield Strength, $F_y = 36 \, \text{ksi}$
• Ultimate Strength, $F_u = 58 \, \text{ksi}$
• $A_e = A_g$, which indicates the effective area is equal to the gross area.

You are required to compute:

1. Design strength for LRFD (Load and Resistance Factor Design) considering tensile yielding.
2. Allowable strength for ASD (Allowable Stress Design) considering tensile yielding.
3. Design strength for LRFD considering tensile rupture.
4. Allowable strength for ASD considering tensile rupture.

Key formulas for LRFD and ASD:

1. LRFD - Tensile Yielding: $\phi P_n = \phi A_g F_y$ where $\phi$ (resistance factor) for yielding is 0.9.

2. ASD - Tensile Yielding: $P_{allow} = \frac{P_n}{\Omega}$ where $\Omega$ (safety factor) is 1.67 for yielding.

3. LRFD - Tensile Rupture: $\phi P_n = \phi A_e F_u$ where $\phi$ for rupture is 0.75.

4. ASD - Tensile Rupture: $P_{allow} = \frac{A_e F_u}{\Omega}$ where $\Omega = 2.0$ for rupture.

Problem 2:

You are required to compute the nominal block shear strength of the tension member in Figure P3.5-1. The given values are:

• $F_u = 400 \, \text{MPa}$ (Ultimate Strength)
• $F_y = 276 \, \text{MPa}$ (Yield Strength)
• Bolts are 7/8-inch diameter.

Key formulas for block shear strength:

1. Nominal block shear strength, $P_{bs}$: $P_{bs} = 0.6 A_{nv} F_u + A_{nt} F_u$ where:
   • $A_{nv}$ is the net area subject to shear.
   • $A_{nt}$ is the net area subject to tension.

I will now calculate the required values using these formulas.Here are the results for Problem 1:

1. Design strength for LRFD considering tensile yielding:
   $P_n = 85.05 \, \text{kips}$

2. Allowable strength for ASD considering tensile yielding:
   $P_n = 56.59 \, \text{kips}$

3. Design strength for LRFD considering tensile rupture:
   $P_n = 114.19 \, \text{kips}$

4. Allowable strength for ASD considering tensile rupture:
   $P_n = 76.13 \, \text{kips}$

Next, I will calculate the nominal block shear strength for Problem 2 using the provided data.The nominal block shear strength for the tension member in Problem 2 is:

$P_{bs} = 58.38 \, \text{kips}$

Let me know if you'd like further explanations or have any additional questions.

Here are five related questions to expand on the current topics:

1. How would the calculations change if the plate's effective area was reduced due to bolt holes?
2. What are the differences between LRFD and ASD approaches in structural design?
3. How does increasing the number of bolts affect the tensile rupture strength?
4. What safety factors are used in other design codes for steel structures?
5. How do the dimensions of the bolts influence block shear strength?

Tip: When calculating block shear strength, it's crucial to account for both tension and shear areas to ensure proper analysis of failure modes.