Absolutely, let’s break this down step by step! --- # **Step-by-Step Solution** ## **Step 1: Understanding the Problem** - **What’s given?** - 9 rivets connect two plates. - A 15 kN load is applied as shown. - Ultimate shear strength of a rivet = **350 N/mm²**. - Need to find the **minimum allowable rivet diameter**. --- ## **Step 2: Analyze the Rivet Group** ### **Load Distribution** - The load is **eccentric** (not applied through the centroid of the rivet group). - Each rivet is subjected to: 1. **Direct shear** (from the force itself). 2. **Secondary shear** (from the moment caused by the offset). --- ## **Step 3: Calculate Forces on Rivets** ### **A. Find the Centroid of the Rivet Group** The rivets are in a \(3 \times 3\) square: - Horizontal and vertical spacing = 25 mm. Centroid is at the **center** of the square. ### **B. Direct Shear (Primary Shear)** Total shear from load = 15 kN = **15,000 N** - **Direct shear per rivet**: \[ F_{direct} = \frac{15,000\ \text{N}}{9} = 1666.67\ \text{N} \] ### **C. Secondary Shear Due to Moment** The load is applied **50 mm** from the centroid: - **Moment**: \( M = F \times d = 15,000 \times 50 = 750,000\ \text{N-mm} \) #### **Polar Moment of Area (\( \sum r^2 \))** Each rivet’s distance from centroid (\( r \)): - The rivet group is symmetric, so calculate \( r \) for each rivet and sum \( r^2 \). Let’s number the rivets as in the figure: - The centroid is at (,). Each rivet's coordinates (relative to centroid): - x = -25, , 25 mm - y = -25, , 25 mm So, for rivet at (x, y): \[ r^2 = x^2 + y^2 \] Sum for all 9 rivets: | Rivet | x (mm) | y (mm) | \( r^2 \) (mm²) | |-------|--------|--------|----------------| | 1 | -25 | 25 | \( 125 \) | | 2 | | 25 | \( 625 \) | | 3 | 25 | 25 | \( 125 \) | | 4 | 25 | | \( 625 \) | | 5 | | | \( \) | | 6 | -25 | | \( 625 \) | | 7 | -25 | -25 | \( 125 \) | | 8 | | -25 | \( 625 \) | | 9 | 25 | -25 | \( 125 \) | Sum: \( 4 \times 125 + 4 \times 625 + = 500 + 250 = 750\ \text{mm}^2 \) --- #### **Shear per Rivet Due to Moment** Shear on each rivet by moment is: \[ F_{moment, i} = \frac{M \cdot r_i}{\sum r^2} \] For rivet farthest from centroid (\( r = \sqrt{125} = 35.36 \) mm): \[ F_{moment, max} = \frac{750,000 \times 35.36}{750} = 3536\ \text{N} \] --- ### **D. Resultant Force on Most Stressed Rivet** The worst-case rivet will have direct + moment shear (vector sum): \[ F_{resultant} = \sqrt{F_{direct}^2 + F_{moment, max}^2} \] \[ F_{resultant} = \sqrt{1666.67^2 + 3536^2} = \sqrt{2,777,778 + 12,504,896} \approx \sqrt{15,282,674} \approx 391\ \text{N} \] --- ## **Step 4: Shear Strength of Rivet** - **Ultimate shear strength**: \( 350\ \text{N/mm}^2 \) - Shear area of rivet: \( A = \frac{\pi d^2}{4} \) Set shear strength = resultant force: \[ \text{Shear Strength} = 350 \times \frac{\pi d^2}{4} \] Set equal to force: \[ 350 \times \frac{\pi d^2}{4} = 391 \] \[ \frac{\pi d^2}{4} = \frac{391}{350} \] \[ d^2 = \frac{391 \times 4}{350 \pi} \] \[ d^2 = \frac{15,640}{1099.56} = 14.22 \] \[ d = \sqrt{14.22} \approx 3.77\ \text{mm} \] --- ## **Step 5: Final Answer** **Minimum allowable rivet diameter:** \[ \boxed{3.8\ \text{mm}} \] (Round up to the nearest standard value, typically 4 mm.) --- # **Summary Table** | Quantity | Value | |----------------------------|------------------| | Direct shear per rivet | 1667 N | | Moment shear (max rivet) | 3536 N | | Resultant shear (max) | 391 N | | Shear strength | 350 N/mm² | | Minimum diameter | 3.8 mm | --- ## **Key Concepts Used** - Eccentric loading on a group of fasteners - Direct and moment-induced shear - Resultant force calculation - Shear area formula for a circle Let me know if you want to see a diagram or have any step explained in more detail!