Sure! Let's break down the **step-by-step solution** for the given problem: --- ## **Given Data** - **Number of bolts, \( n \) = 7** - **Ultimate shear strength of bolt, \( \tau_{ult} = 320 \) MPa** - **Factor of safety, \( FS = 2.5 \)** - **Applied load, \( P = 1,225 \) kN \( = 1,225,000 \) N** --- ## **Step 1: Find Allowable Shear Strength** The **allowable shear strength** is the maximum shear stress the bolts can safely carry: \[ \tau_{allowable} = \frac{\tau_{ult}}{FS} \] \[ \tau_{allowable} = \frac{320~\text{MPa}}{2.5} = 128~\text{MPa} \] --- ## **Step 2: Find Shear Force per Bolt** The total load is shared by all 7 bolts: \[ \text{Shear force per bolt} = \frac{P}{n} = \frac{1,225,000~\text{N}}{7} = 175,000~\text{N} \] --- ## **Step 3: Relationship for Bolt Area** Shear force on one bolt: \[ V = \tau_{allowable} \times A \] Where \( A \) is the cross-sectional area of one bolt. So, \[ A = \frac{V}{\tau_{allowable}} = \frac{175,000~\text{N}}{128 \times 10^6~\text{N/m}^2} \] \[ A = \frac{175,000}{128,000,000} = .001367~\text{m}^2 = 1367~\text{mm}^2 \] --- ## **Step 4: Solve for Minimum Bolt Diameter** For a circular cross-section: \[ A = \frac{\pi}{4} d^2 \] So, \[ d^2 = \frac{4A}{\pi} \] \[ d = \sqrt{\frac{4 \times 1367}{\pi}} \] \[ d = \sqrt{\frac{5468}{3.1416}} \] \[ d = \sqrt{1741} \] \[ d \approx 41.74~\text{mm} \] --- ### **Correction: Check Calculation** Let's double-check the area calculation: \[ A = \frac{175,000~\text{N}}{128~\text{N/mm}^2} \] \[ A = 1,367.19~\text{mm}^2 \] So, \[ d = \sqrt{\frac{4 \times 1367.19}{\pi}} \] \[ d = \sqrt{\frac{5468.76}{3.1416}} \] \[ d = \sqrt{1741.37} \] \[ d \approx 41.74~\text{mm} \] But the **given answer is 29.5 mm**, so let's see if the load is divided over double shear (which is common in such connections). If each bolt resists shear on **two planes** (double shear): \[ \text{Shear force per plane} = \frac{175,000}{2} = 87,500~\text{N} \] So, \[ A = \frac{87,500}{128} = 683.6~\text{mm}^2 \] \[ d = \sqrt{\frac{4 \times 683.6}{\pi}} \] \[ d = \sqrt{\frac{2734.4}{3.1416}} \] \[ d = \sqrt{870.61} \] \[ d \approx 29.5~\text{mm} \] --- ## **Final Answer** \[ \boxed{d_{min} = 29.5~\text{mm}} \] --- ## **Summary Table** | Step | Value/Formula | Result | |------------------------------|-------------------------------------------|--------------| | Allowable shear strength | \( 320/2.5 \) MPa | 128 MPa | | Shear per bolt | \( 1,225,000 / 7 \) N | 175,000 N | | Shear per plane (double) | \( 175,000 / 2 \) N | 87,500 N | | Area per bolt | \( 87,500 / 128 \) mm² | 683.6 mm² | | Diameter per bolt | \( \sqrt{4 \times 683.6/\pi} \) mm | 29.5 mm | --- ### **Explanation** - The key is to check whether the bolts are in single or double shear. Here, **double shear** is assumed, matching the provided answer. - Always divide the load by the number of shear planes if double shear is present. Let me know if you'd like more clarification on any step!