Let's break down and solve the problem **step-by-step**. --- ## **Problem Restatement** Two **10XX steel plates** (E = 200 GPa), each **10 mm thick** and **25 mm wide**, are separated by a **.13 mm thick gasket** and held by a **single M12 bolt** (AISI 103 steel). The **joint is tightened** with a **preload of 10 kN**. **Question:** How much **external load** can be applied **before the joint starts to separate** (i.e., the clamp load is zero)? --- ## **Step-by-Step Solution** ### **1. Understand the Joint Separation Condition** The joint will **start to separate** when the **external tensile force** equals the initial **clamp (preload) force**. But, since the **gasket compresses** and the bolt stretches, **not all external load** is transmitted to the bolt. We need to calculate: - **Bolt stiffness (\( k_b \))** - **Joint stiffness (\( k_j \))** - **Load fraction taken by the bolt (\( C \))** --- ### **2. Calculate Bolt and Joint Stiffness** #### **Bolt Stiffness (\( k_b \))** \[ k_b = \frac{A_b E_b}{L_b} \] - \( A_b \): Tensile stress area of M12 bolt (AISI 103 steel) - \( E_b \): Modulus of elasticity of bolt (200 GPa) - \( L_b \): Length of bolt under tension (across plates + gasket = 10 mm + .13 mm + 10 mm = 20.13 mm) ##### **Find \( A_b \):** For M12 bolt, stress area (\( A_b \)) ≈ **84.3 mm²**. \[ A_b = 84.3\, \text{mm}^2 \] \[ E_b = 200\, \text{GPa} = 200 \times 10^3\, \text{N/mm}^2 \] \[ L_b = 10 + .13 + 10 = 20.13\, \text{mm} \] \[ k_b = \frac{84.3 \times 200 \times 10^3}{20.13} = \frac{16,860,000}{20.13} \approx 837,900\, \text{N/mm} \] --- #### **Joint Stiffness (\( k_j \))** The joint (steel + gasket) is compressed. The **gasket** is much more compressible than steel, so its stiffness is dominant. \[ k_j = \frac{A_j E_j}{L_j} \] - \( A_j \): Area under load = width × thickness = \( 25\, \text{mm} \times 10\, \text{mm} = 250\, \text{mm}^2 \) - \( E_j \): For the gasket (not given, but for separation calculation, assume gasket is much softer, so it controls separation) - \( L_j \): Thickness of gasket = **.13 mm** But since the problem is about the joint **starting to separate** (when the plate just lifts off), we focus on the **initial clamp**. --- ### **3. Load Distribution Coefficient \( C \)** The **fraction of the external load** taken by the bolt: \[ C = \frac{k_b}{k_b + k_j} \] But at **separation**, the **joint stiffness goes to zero** (since plates lose contact). So, the **entire external force** is taken by the bolt at that point. --- ### **4. Maximum External Load Before Separation** The joint starts to separate when the **external load (\( F_{ext} \))** is such that: \[ \text{Initial preload} + C \cdot F_{ext} = \text{Initial preload} \] \[ C \cdot F_{ext} = \] But, in reality, the **joint starts to separate when the compressive force in the joint reaches zero**. \[ F_{ext,max} = \frac{\text{Initial preload}}{1 - C} \] But with the **gasket being very soft** (\( k_j \ll k_b \)), \( C \to 1 \) and \( F_{ext,max} \to \text{preload} \). So, **maximum external load before separation ≈ initial preload**. --- ### **5. Final Answer** \[ \boxed{F_{ext,\ max} \approx 10\ \text{kN}} \] --- ## **Summary Table** | Parameter | Value | |--------------------|-----------------| | Preload | \( 10\, \text{kN} \) | | Maximum ext. load | \( 10\, \text{kN} \) (approx) | | Answer | \( 10,000\, \text{N} \) | --- ### **Final Answer** \[ \boxed{10,\!000\ \text{N}} \] --- If you have questions about any step, please ask!