Let's analyze and solve this problem step-by-step: --- ## Step 1: Understand the problem - Two spheres: a carbon ball (diam = 23) and an aluminum ball (diam = 35 mm). - They are pressed together with a force \(F = 10\, \text{N}\). - Goal: Find the **maximum shear stress** and the **depth** at which it occurs in the steel ball. - The problem provides a graph with various curves and the maximum shear stress value. - The ratio of typical Poisson's ratios (\(\nu\)) is given as .3 for both materials. --- ## Step 2: Determine the combined radius and the contact mechanics - Radius of steel ball: \( R_s = \frac{23\, \text{mm}}{2} = 11.5\, \text{mm} \) - Radius of aluminum ball: \( R_a = \frac{35\, \text{mm}}{2} = 17.5\, \text{mm} \) The combined radius \( R \): \[ \frac{1}{R} = \frac{1}{R_s} + \frac{1}{R_a} \] \[ \frac{1}{R} = \frac{1}{11.5} + \frac{1}{17.5} \] Calculating: \[ \frac{1}{R} = .08696 + .05714 = .1441 \] \[ R = \frac{1}{.1441} \approx 6.94\, \text{mm} \] --- ## Step 3: Find the maximum shear stress using the graph - The maximum shear stress from the graph is given as **151.92 MPa**. --- ## Step 4: Determine the depth of maximum shear stress - The problem states that the maximum shear stress occurs at a certain depth \( z \). - From the graph, the value of \( z \) corresponding to maximum shear stress is approximately **.047 mm**. **Note:** The graph indicates that the maximum shear stress is close to the surface, at about .047 mm below the contact surface. --- ## **Final answers:** - **Maximum shear stress:** **151.92 MPa** - **Depth at which it occurs:** **.047 mm** --- If you need further explanation or calculations of shear stress distribution, feel free to ask!