**Tag:** Mechanics, Strength of Materials **Topic:** Bolt Stretch Calculation --- ### **Step 1: Given Data & Conceptual Introduction** **Given Data:** - Bolt: 3/4-16 UNF steel bolt - Clamps two steel plates, each 2 in (50 mm) thick - 0.75 in (19 mm) threaded section remains under nut - Stress in bolt body: 40,000 lbf/in² (280 MPa) - Modulus of elasticity (E): 2.9 × 10⁷ lbf/in² (200 GPa) - Nut has six threads **Conceptual Introduction:** When a bolt is tightened to clamp two plates, it elongates (stretches) due to the applied tensile force. The amount of stretch can be found using the formula for elongation under axial load: \[ \delta = \frac{PL}{AE} \] Where \( P \) is the force, \( L \) is the length, \( A \) is the cross-sectional area, and \( E \) is the modulus of elasticity. **Explanation Block:** The stretch of the bolt is calculated by considering the total length of the bolt under stress, its cross-sectional area, and material properties. The formula relates the applied force, geometry, and elasticity to the elongation. --- ### **Step 2: Step-by-step Calculation (Find Required Length and Area)** **Step 2.1:** **Find the total length of the bolt under stress** - Total plate thickness = \(2 + 2 = 4\) in - Threaded section under nut = \(0.75\) in - Total bolt length under stress = \(4 + 0.75 = 4.75\) in **Step 2.2:** **Find cross-sectional area of the bolt (root area for 3/4-16 UNF)** - For a 3/4-16 UNF bolt, the tensile (root) area \(A \approx 0.373\) in² **Step 2.3:** **Find force in bolt** - Stress, \(\sigma = 40,000\) lbf/in² - Force, \(P = \sigma \times A = 40,000 \times 0.373 = 14,920\) lbf **Explanation Block:** We first determine the total length of the bolt stretched (sum of plate thickness and the threaded section), the root area for stress calculation, and then use given stress to find the tensile force in the bolt. --- ### **Step 3: Step-by-step Calculation (Calculate Stretch)** \[ \delta = \frac{PL}{AE} \] - \(P = 14,920\) lbf - \(L = 4.75\) in - \(A = 0.373\) in² - \(E = 2.9 \times 10^7\) lbf/in² \[ \delta = \frac{14,920 \times 4.75}{0.373 \times 2.9 \times 10^7} \] \[ \delta = \frac{70,370}{10,817,000} \] \[ \delta \approx 0.0065 \text{ in} \] **Explanation Block:** Plugging the determined values into the elongation formula, we solve for bolt stretch by direct substitution and careful arithmetic, ensuring unit consistency. --- ### **Final Answer:** **The bolt stretches approximately \(0.0065\) inches.**