# Step-by-Step Solution Let's break down the problem and solve for the **maximum average normal stress** in the vertical links (at points **B** and **D**). --- ## **Given Data** - **Number of links:** 4 (vertical links) - **Cross section of each link:** \(8\, \text{mm} \times 36\, \text{mm}\) - **Load \(P\):** \(22\, \text{kN} = 22,000\, \text{N}\) - **Points of interest:** Links at **B** and **D** --- ## **Step 1: Find the Force in Each Link** The total load \(P\) is **shared equally among the 4 links** (since the system is symmetric and all links have the same properties): \[ \text{Force per link} = \frac{P}{4} = \frac{22,000\, \text{N}}{4} = 5,500\, \text{N} \] --- ## **Step 2: Calculate the Cross-sectional Area of Each Link** \[ A = \text{width} \times \text{thickness} = 36\, \text{mm} \times 8\, \text{mm} = 288\, \text{mm}^2 \] --- ## **Step 3: Calculate the Average Normal Stress** Average normal stress in a link is: \[ \sigma = \frac{F}{A} \] where: - \(F\) = force in each link = 5,500 N - \(A\) = cross-sectional area = 288 mm² **Convert area to m² for stress in MPa:** \[ 288\, \text{mm}^2 = 288 \times 10^{-6}\, \text{m}^2 \] \[ \sigma = \frac{5,500}{288 \times 10^{-6}} \] \[ \sigma = \frac{5,500}{.000288} \] \[ \sigma = 19,097,222\, \text{N/m}^2 = 19.1\, \text{MPa} \] --- ## **Final Answer** > **The maximum value of the average normal stress in the links connecting points B and D is:** > > \[ > \boxed{19.1\, \text{MPa}} > \] --- ### **Summary Table** | Quantity | Value | |------------------------|------------------| | Load per link (\(F\)) | 5,500 N | | Area (\(A\)) | 288 mm² | | Normal Stress (\(\sigma\)) | 19.1 MPa | --- If you have more questions about this type of problem, let me know!