Let's break down the solution step by step for both **LRFD** and **ASD**. We'll check the adequacy of the joint for the given loads. --- ## **Step 1: Given Data** - **Section:** L6x3½x½ (Angle) - **Steel:** ASTM A36 (Fy = 36 ksi, Fu = 58 ksi) - **Plate Thickness:** 3/8 in. - **Bolt:** 11/8 in. diameter (1.125 in.), Group A (A325) - **Length:** 9 ft - **Loads:** - Dead Load (DL) = 20 kips - Live Load (LL) = 60 kips - Wind Load (WL) = 20 kips - **Bolt spacing:** 3½ in. (center-to-center); edge distance = 2 in. - **Number of bolts:** 6 (from the figure) --- ## **Step 2: Factored Load (for LRFD) & Service Load (for ASD)** ### **LRFD:** \[ \text{Factored Load} = 1.2 \cdot DL + 1.6 \cdot LL + 0.5 \cdot WL \] \[ = 1.2 \cdot 20 + 1.6 \cdot 60 + 0.5 \cdot 20 = 24 + 96 + 10 = 130 \text{ kips} \] ### **ASD:** \[ \text{Service Load} = DL + LL + WL = 20 + 60 + 20 = 100 \text{ kips} \] --- ## **Step 3: Bolt Strength Check** ### **A325 Bolts (Group A):** - **Diameter:** 1.125 in. - **Shear strength per bolt (AISC Table J3.2):** - **LRFD:** \( \phi R_n = 47 \text{ kips} \) per bolt (for threads included in shear plane, single shear) - **ASD:** \( R_n/\Omega = 31 \text{ kips} \) per bolt **Total bolts:** 6 - **LRFD Total Bolt Strength:** \( 6 \times 47 = 282 \text{ kips} \) - **ASD Total Bolt Strength:** \( 6 \times 31 = 186 \text{ kips} \) --- ## **Step 4: Plate/Angle Net Area Strength Check** ### **Gross Area (AISC Table 1-7):** - L6x3½x½: \( \text{Area} = 4.36 \text{ in}^2 \) ### **Net Area Calculation:** - Bolt hole diameter = 1.125 in. + 1/8 in. = **1.25 in.** - **Net width loss per hole:** 1.25 in. **Critical Leg (6 in. leg):** - Net area = Gross area - (number of bolts in line × hole diameter × thickness) Number of bolts in one line (assuming bolts are staggered along 6 in. leg): 6 Net area loss = 6 × 1.25 × 0.5 = 3.75 in² \[ A_n = 4.36 - 3.75 = 0.61 \text{ in}^2 \] But this seems too small. Let's check if bolts are staggered or in 2 lines (typically, angles have one bolt per gage per line). From the figure, bolts are along the 6 in. leg, one line. So, - Holes in net section: **1** (critical section passes through one bolt at once, not all bolts). - Net area loss = 1 × 1.25 × 0.5 = 0.625 in² \[ A_n = 4.36 - 0.625 = 3.735 \text{ in}^2 \] #### **Strength Calculation** ##### **LRFD:** \[ \phi = 0.75 \] \[ F_u = 58 \text{ ksi} \] \[ \phi P_n = 0.75 \times 58 \times 3.735 = 162.5 \text{ kips} \] ##### **ASD:** \[ \Omega = 2.00 \] \[ P_n/\Omega = \frac{58 \times 3.735}{2.00} = 108.3 \text{ kips} \] --- ## **Step 5: Compare with Applied Load** ### **LRFD:** - **Bolt strength:** 282 kips > 130 kips (OK) - **Net section strength:** 162.5 kips > 130 kips (OK) ### **ASD:** - **Bolt strength:** 186 kips > 100 kips (OK) - **Net section strength:** 108.3 kips > 100 kips (OK) --- ## **Step 6: Final Answers** ### **1. LRFD:** - **Required strength:** 130 kips - **Bolt strength available:** 282 kips - **Net section strength available:** 162.5 kips **Joint is Satisfactory** --- ### **2. ASD:** - **Required strength:** 100 kips - **Bolt strength available:** 186 kips - **Net section strength available:** 108.3 kips **Joint is Satisfactory** --- ## **Summary Table** | Check | LRFD (kips) | ASD (kips) | |---------------------|-------------|------------| | Required Strength | 130 | 100 | | Bolt Strength | 282 | 186 | | Net Section Strength| 162.5 | 108.3 | | **Satisfactory?** | **YES** | **YES** | --- **If you need the block shear check as well, let me know!**