Let's break down and solve the problem step by step: --- ## **Problem Restatement** Two wooden members are joined by plywood splice plates that are fully glued on their outer surfaces (see figure). - Width of plate, \( b = 140 \) mm - Shear strength of glue, \( \tau_{allow} = 2.5 \) MPa - The force applied, \( P = 18 \) kN - **Find:** The length \( L \) (in mm) required for the splice plates so that the average shear stress in the glue does not exceed the allowable value. --- ## **Step 1: Find the Total Glued Area** There are two splice plates (one on each side). Each plate has a glued area: \[ A_{glue, \text{one side}} = \text{width} \times \text{length} = b \times L \] Total glued area (both sides): \[ A_{glue, \text{total}} = 2 \times b \times L \] --- ## **Step 2: Shear Stress Equation** Average shear stress in the glue: \[ \tau_{avg} = \frac{P}{A_{glue, \text{total}}} \] Set \(\tau_{avg}\) equal to the allowable shear stress: \[ \frac{P}{2bL} = \tau_{allow} \] --- ## **Step 3: Substitute Values** Given: - \(P = 18\) kN = **18,000 N** - \(b = 140\) mm - \(\tau_{allow} = 2.5\) MPa = **2.5 N/mm²** Substitute into the equation: \[ \frac{18,000}{2 \times 140 \times L} = 2.5 \] --- ## **Step 4: Solve for \(L\)** \[ 2 \times 140 \times L \times 2.5 = 18,000 \] \[ 2 \times 140 \times 2.5 \times L = 18,000 \] \[ 700 \times L = 18,000 \] \[ L = \frac{18,000}{700} \] \[ L = 25.71 \text{ mm} \] --- ## **Step 5: Final Answer** \[ \boxed{L = 25.7 \ \text{mm}} \] (Rounded to one decimal place as appropriate) --- ### **Summary Table** | Given | Value | |--------------|------------| | Force, \(P\) | 18,000 N | | Width, \(b\) | 140 mm | | Shear Strength, \(\tau_{allow}\) | 2.5 MPa | | **Required Length, \(L\)** | **25.7 mm** | --- **Final Answer:** \[ \boxed{L = 25.7 \ \text{mm}} \] If you need the answer to more decimal places, use your calculator, but this is the clear methodology.