Let's break down **Problem 1** step by step: --- ### **Given Data:** - **Board size:** 8 in × 1 in × 6 in (likely means flanges are 8" wide, 1" thick, webs are 6" high, 1" thick) - **Allowable shear per screw:** \( V_{\text{all}} = 250 \) lb - **Maximum transverse shear force:** \( V_{\text{max}} = 1200 \) lb - **Two beam sections are shown:** Section A and Section B --- ## **Step 1: Shear Flow Calculation** The **shear flow** (\( q \)) at the location of screws is: \[ q = \frac{VQ}{I} \] Where: - \( V \) = Transverse shear force (1200 lb) - \( Q \) = First moment of area above (or below) the screw location about the neutral axis - \( I \) = Moment of inertia of the whole section about the neutral axis --- ## **Step 2: Section Properties** Let's analyze both sections: ### **Section 1: Flanges on top and bottom, webs at sides** #### **Dimensions:** - **Flanges:** 8" wide × 1" thick - **Webs:** 6" high × 1" thick #### **Moment of Inertia, \( I \):** - The section is a box, so use the formula for a rectangular tube: \[ I = \frac{1}{12} (b_o h_o^3 - b_i h_i^3) \] Where: - \( b_o = 8 \) in, \( h_o = 8 \) in (outside dimensions) - \( b_i = 6 \) in, \( h_i = 6 \) in (inside dimensions) \[ I = \frac{1}{12} [8 \times 8^3 - 6 \times 6^3] = \frac{1}{12} [8 \times 512 - 6 \times 216] \] \[ 8 \times 512 = 4096 \] \[ 6 \times 216 = 1296 \] \[ I = \frac{1}{12} (4096 - 1296) = \frac{1}{12} (2800) = 233.33 \text{ in}^4 \] --- ## **Step 3: First Moment of Area, \( Q \)** We need \( Q \) for the flange (since screws are holding the flange to the web). - **Flange area:** \( A = 8 \times 1 = 8 \) in² - **Distance from centroid of flange to neutral axis:** The neutral axis is at the mid-height (since the section is symmetric). The distance from the centroid of the top flange to the NA is \( 8/2 - 1/2 = 3.5 \) in. \[ Q = A \cdot \bar{y} = 8 \cdot 3.5 = 28 \text{ in}^3 \] --- ## **Step 4: Shear Flow, \( q \)** \[ q = \frac{VQ}{I} = \frac{1200 \times 28}{233.33} = \frac{33,600}{233.33} \approx 144 \text{ lb/in} \] --- ## **Step 5: Screw Spacing** Each screw can resist 250 lb. \[ \text{Spacing, } s = \frac{\text{Allowable per screw}}{q} = \frac{250}{144} \approx 1.74 \text{ in} \] --- ### **Section 2: Flanges on sides, webs on top and bottom** Repeat above steps for the second section. #### **Dimensions:** - **Flanges:** 6" wide × 1" thick (vertical) - **Webs:** 8" wide × 1" thick (horizontal) #### **Moment of Inertia, \( I \):** \[ I = \frac{1}{12} [6 \times 8^3 - 4 \times 6^3] \] \[ 6 \times 512 = 3072 \] \[ 4 \times 216 = 864 \] \[ I = \frac{1}{12} (3072 - 864) = \frac{1}{12} (2208) = 184 \text{ in}^4 \] #### **First Moment of Area, \( Q \):** - **Flange area:** \( A = 6 \times 1 = 6 \) in² - **Distance from centroid of flange to NA:** \( 8/2 - 1/2 = 3.5 \) in \[ Q = 6 \cdot 3.5 = 21 \text{ in}^3 \] #### **Shear Flow, \( q \):** \[ q = \frac{1200 \times 21}{184} = \frac{25,200}{184} \approx 137 \text{ lb/in} \] #### **Screw Spacing:** \[ s = \frac{250}{137} \approx 1.82 \text{ in} \] --- ## **Step 6: Cost Comparison** - **Section 1:** Screw spacing = **1.74 in** - **Section 2:** Screw spacing = **1.82 in** (larger spacing, so fewer screws needed, thus lower cost) --- ## **Final Answers** ### **(a) Maximum permissible spacing between screws:** - **Section 1:** **1.74 in** - **Section 2:** **1.82 in** ### **(b) Lower cost section:** - **Section 2** will result in a lower screw cost due to greater permissible screw spacing. --- ### **Summary Table:** | Section | \(I\) (in⁴) | \(Q\) (in³) | \(q\) (lb/in) | Max Spacing (in) | Cost | |------------|-------------|-------------|--------------|------------------|----------| | Section 1 | 233.33 | 28 | 144 | 1.74 | Higher | | Section 2 | 184 | 21 | 137 | 1.82 | Lower | --- Let me know if you’d like help with Problem 2!

Certainly! Let's carefully go through **Problem 1** step by step, providing detailed calculations and clear explanations. --- ## **Problem Restatement:** Given a **composite section** (probably a box or built-up section) subjected to a **transverse shear force \(V = 1200\, lb\)**, determine: - The **maximum spacing** of shear screws permitted (assuming each screw can carry **250 lb** of shear load). - Which of the two section configurations results in **lower cost** (based on screw spacing capacity). --- ## **Step 1: Understand the Section and Data** Since the image isn't accessible here, based on typical problems, the section comprises: - Flanges (horizontal or vertical) - Webs (vertical or horizontal) - Dimensions (probably in inches) From the problem text, it appears the section is a *box* or *built-up* I-beam with: - **Section 1:** Flanges of 8" width, 1" thickness; Webs of 6" height, 1" thickness. - **Section 2:** Flanges of 6" width, 1" thickness; Webs of 8" height, 1" thickness. --- ## **Step 2: Calculate Moment of Inertia (\(I\))** ### **Section 1:** **Dimensions:** - Flanges: 8" width, 1" thick - Webs: 6" high, 1" thick **Assumption:** The section is a rectangular hollow box (like a channel or box section). **Method:** Use the *parallel axis theorem* for a rectangular hollow section: \[ I = \frac{1}{12} (b_o h_o^3 - b_i h_i^3) \] Where: - \(b_o\), \(h_o\): Outer dimensions - \(b_i\), \(h_i\): Inner dimensions (cut-out for web openings). Since it's a box, the web is the opening. But since the section is a simple box, better to treat as a rectangular tube: \[ I_{section} = \frac{1}{12} (B H^3 - b h^3) \] **For Section 1:** - **Outer rectangle:** width = 8 in, height = 8 in (assuming the web is between flanges) - **Inner rectangle:** width = 6 in, height = 6 in (web opening) \[ I_{section1} = \frac{1}{12} [8 \times 8^3 - 6 \times 6^3] \] Calculate: \[ 8 \times 512 = 4096 \] \[ 6 \times 216 = 1296 \] \[ I_{section1} = \frac{1}{12} (4096 - 1296) = \frac{2800}{12} \approx 233.33\, \text{in}^4 \] --- ### **Section 2:** **Dimensions:** - Flanges: 6" width, 1" thickness - Webs: 8" high, 1" thickness Similarly: \[ I_{section2} = \frac{1}{12} [6 \times 8^3 - 4 \times 6^3] \] Calculate: \[ 6 \times 512 = 3072 \] \[ 4 \times 216 = 864 \] \[ I_{section2} = \frac{1}{12} (3072 - 864) = \frac{2208}{12} = 184\, \text{in}^4 \] --- ## **Step 3: Find the First Moment of Area (\(Q\))** The shear flow depends on \(Q\), the first moment of area about the neutral axis, at the point of shear. ### **Locate the neutral axis:** For symmetric sections, the neutral axis passes through the centroid. Assume the centroid is at mid-height. ### **Section 1:** - Shear is transferred through the top flange. - Area of the flange: \[ A_{flange} = 8\, in \times 1\, in = 8\, in^2 \] - Distance from centroid of flange to the neutral axis: \[ \text{distance} = \frac{8}{2} - \frac{1}{2} = 4 - 0.5 = 3.5\, in \] - **Calculate \(Q\):** \[ Q = A_{flange} \times \text{distance} = 8 \times 3.5 = 28\, in^3 \] --- ### **Section 2:** - Flange area: \[ A_{flange} = 6\, in \times 1\, in = 6\, in^2 \] - Distance from centroid to shear point: \[ \frac{8}{2} - \frac{1}{2} = 4 - 0.5 = 3.5\, in \] - **Calculate \(Q\):** \[ Q = 6 \times 3.5 = 21\, in^3 \] --- ## **Step 4: Compute Shear Flow \(q\)** \[ q = \frac{V \times Q}{I} \] Given: \[ V = 1200\, lb \] ### **Section 1:** \[ q_1 = \frac{1200 \times 28}{233.33} \approx \frac{33,600}{233.33} \approx 144\, lb/in \] ### **Section 2:** \[ q_2 = \frac{1200 \times 21}{184} \approx \frac{25,200}{184} \approx 137\, lb/in \] --- ## **Step 5: Determine the **Maximum Allowed Spacing** of Screws** Each screw can carry **250 lb** shear. The maximum spacing \(s\) between screws is: \[ s = \frac{\text{Allowable shear per screw}}{\text{Shear flow } q} \] ### **Section 1:** \[ s_1 = \frac{250}{144} \approx 1.74\, in \] ### **Section 2:** \[ s_2 = \frac{250}{137} \approx 1.82\, in \] --- ## **Final Results:** | Section | Shear flow \(q\) (lb/in) | Max screw spacing \(s\) (in) | |---------|---------------------------|------------------------------| | Section 1 | 144 | **1.74 in** | | Section 2 | 137 | **1.82 in** | --- ## **Conclusion:** - **Maximum safe screw spacing:** - **Section 1:** **1.74 inches** - **Section 2:** **1.82 inches** - **Lower cost section:** Since **Section 2** allows a larger screw spacing, requiring fewer screws, **Section 2** will result in **lower cost**. --- ## **Summary:** - **Shear flow calculations** determine how shear is distributed. - The **smaller shear flow** means **less frequent screws**. - **Section 2** has a **lower shear flow** and thus a **larger screw spacing**, making it more economical. --- **Let me know if you'd like further clarifications or assistance!**

Certainly! Here is a comprehensive, step-by-step detailed solution to **Problem 1**, including all calculations and explanations. --- ## **Problem Restatement:** Given a structural section subjected to a transverse shear force \( V = 1200\, lb \), determine: 1. The **maximum spacing** of shear screws (each can carry 250 lb shear). 2. Which section configuration results in **lower cost** based on screw spacing capacity. --- ## **Step 1: Understand the Sections and Data** From the problem description, there are two sections: - **Section 1:** Flanges of 8" width, 1" thick, Webs of 6" height, 1" thick. - **Section 2:** Flanges of 6" width, 1" thick, Webs of 8" height, 1" thick. Assuming these are built-up box sections, the goal is to find the shear flow and then the maximum spacing of shear screws. --- ## **Step 2: Calculate the Moment of Inertia \( I \) for Each Section** ### **Section 1:** - **Outer dimensions:** - Width = 8 in - Height = 8 in (assuming the section is a rectangle with flanges and web) - **Inner cut-out:** - Width = 6 in - Height = 6 in The **section** is a hollow rectangle (like a box section). The **Moment of Inertia** about the neutral axis (assumed horizontal axis) is: \[ I = \frac{1}{12} (b_o h_o^3 - b_i h_i^3) \] Where: - \(b_o = 8\, in\), \(h_o = 8\, in\) - \(b_i = 6\, in\), \(h_i = 6\, in\) Calculating: \[ I_{section1} = \frac{1}{12} [8 \times 8^3 - 6 \times 6^3] \] Compute: \[ 8 \times 8^3 = 8 \times 512 = 4096 \] \[ 6 \times 6^3 = 6 \times 216 = 1296 \] Thus, \[ I_{section1} = \frac{1}{12} (4096 - 1296) = \frac{2800}{12} \approx 233.33\, in^4 \] --- ### **Section 2:** - **Outer dimensions:** - Width = 6 in - Height = 8 in - **Inner cut-out:** - Width = 4 in - Height = 6 in Calculate: \[ I_{section2} = \frac{1}{12} [6 \times 8^3 - 4 \times 6^3] \] Compute: \[ 6 \times 512 = 3072 \] \[ 4 \times 216 = 864 \] Therefore: \[ I_{section2} = \frac{1}{12} (3072 - 864) = \frac{2208}{12} = 184\, in^4 \] --- ## **Step 3: Calculate the First Moment of Area \( Q \) for Shear Flow** \(Q\) is the first moment of area about the neutral axis at the shear point, typically across the flange or web. ### **Location of Shear:** Assuming shear is transmitted through the web, the shear flow at the web-flange junction is calculated at the centroid of the flange. --- ### **Section 1:** - **Area of top flange:** \[ A_{flange} = 8\, in \times 1\, in = 8\, in^2 \] - **Distance from centroid of flange to neutral axis:** Since the section is symmetric, the neutral axis is at mid-height: \[ \text{Distance} = \frac{\text{section height}}{2} - \frac{\text{flange thickness}}{2} = 4 - 0.5 = 3.5\, in \] - **Calculate \(Q\):** \[ Q = A_{flange} \times \text{distance} = 8 \times 3.5 = 28\, in^3 \] --- ### **Section 2:** - **Area of flange:** \[ A_{flange} = 6\, in \times 1\, in = 6\, in^2 \] - **Distance from centroid to neutral axis:** \[ = 4 - 0.5 = 3.5\, in \] - **Calculate \(Q\):** \[ Q = 6 \times 3.5 = 21\, in^3 \] --- ## **Step 4: Calculate Shear Flow \( q \)** Shear flow: \[ q = \frac{V \times Q}{I} \] Given: \[ V = 1200\, lb \] ### **Section 1:** \[ q_{1} = \frac{1200 \times 28}{233.33} \approx \frac{33,600}{233.33} \approx 144\, lb/in \] ### **Section 2:** \[ q_{2} = \frac{1200 \times 21}{184} \approx \frac{25,200}{184} \approx 137\, lb/in \] --- ## **Step 5: Determine Maximum Spacing of Shear Screws** Each screw can carry **250 lb** shear. Maximum spacing \(s\): \[ s = \frac{\text{per screw shear capacity}}{\text{shear flow } q} \] ### **Section 1:** \[ s_1 = \frac{250}{144} \approx 1.74\, in \] ### **Section 2:** \[ s_2 = \frac{250}{137} \approx 1.82\, in \] --- ## **Final Answers:** | Section | Shear flow \( q \) (lb/in) | Max screw spacing \( s \) (in) | |---------|---------------------------|------------------------------| | **Section 1** | **144** | **1.74 in** | | **Section 2** | **137** | **1.82 in** | --- ## **Conclusion:** - The **screw spacing** **must be less than or equal to** approximately **1.74 inches** for Section 1, and **1.82 inches** for Section 2. - Since **Section 2** allows **larger screw spacing**, it **requires fewer screws** and thus **costs less**. --- **This completes the detailed step-by-step solution.** Let me know if you need further explanation or assistance!