# Step-by-Step Solution ## a. Copy the Drawing and Sketch Force Paths **Description of Force Paths:** 1. **Gas Force on Piston:** The total gas force \( F \) is applied vertically down on the top of the piston. 2. **Force Transfer through Piston:** This force is transmitted downwards through the body of the piston. 3. **Force on Piston Pin:** The force then acts on the piston pin (the horizontal cylindrical pin shown) at both sides. 4. **Force into Connecting Rod:** The piston pin transmits the force into the connecting rod, which is connected at the middle of the pin. **Force Path Sketch (Described):** - Draw arrows from the top of the piston downward. - Show the force splitting and transferring horizontally onto the piston pin (on both sides). - Indicate the force being carried through the pin toward the center (where the connecting rod is attached). - Show the force then proceeding downward through the connecting rod. --- ## b. Stresses in the Piston Pin & Equations ### **Simplifying Assumptions:** - The force \( F \) is evenly distributed on both sides of the piston pin. - The pin is simply supported at both ends and loaded at the center by the connecting rod. - The pin is assumed to be uniform and the contact is perfectly rigid. ### **Identification of Stresses:** #### 1. **Shear Stress in the Pin** - **Cause:** The vertical force from the piston is transferred to the pin at both sides; the connecting rod pulls down at the center. - **Shear force at the center:** \[ \tau = \frac{F}{A} \] Where \( A = \pi d^2 / 4 \) is the cross-sectional area of the pin (if loaded in pure shear at the center). #### 2. **Bending Stress in the Pin** - **Cause:** The pin acts as a simply supported beam, loaded at the center by the force \( F \). - **Maximum bending moment at the center:** \[ M_{\text{max}} = \frac{F \cdot l}{4} \] Where \( l = 2a \) (the distance between supports, i.e., the width of the piston). - **Bending stress:** \[ \sigma_{\text{bending}} = \frac{M_{\text{max}} \cdot c}{I} \] Where: - \( c = d/2 \) (distance from neutral axis to outer fiber) - \( I = \frac{\pi d^4}{64} \) (moment of inertia of circular cross-section) Substitute values: \[ \sigma_{\text{bending}} = \frac{M_{\text{max}} \cdot (d/2)}{\frac{\pi d^4}{64}} = \frac{F \cdot l \cdot d}{8 \cdot \frac{\pi d^4}{64}} = \frac{F \cdot l \cdot 64}{8 \pi d^3} = \frac{8 F l}{\pi d^3} \] If \( l = 2a \): \[ \sigma_{\text{bending}} = \frac{8 F (2a)}{\pi d^3} = \frac{16 F a}{\pi d^3} \] --- ### **Summary of Equations:** - **Shear Stress:** \[ \tau = \frac{F}{\pi d^2 / 4} = \frac{4F}{\pi d^2} \] - **Maximum Bending Stress:** \[ \sigma_{\text{bending}} = \frac{16 F a}{\pi d^3} \] --- ### **Assumptions and Their Effects:** - **Uniform Load Distribution:** Real contact may cause stress concentrations; results are idealized. - **Rigid Supports:** Actual supports may deform, causing more complex stress patterns. - **Neglecting Pin Deflection:** If the pin is long or heavily loaded, deflection could increase stresses. --- ## **Final Answer** ### **a. Force Paths:** - Force goes from gas → piston → piston pin (at both sides) → connecting rod. ### **b. Stresses & Equations:** - **Shear stress:** \( \tau = \frac{4F}{\pi d^2} \) - **Bending stress:** \( \sigma_{\text{bending}} = \frac{16 F a}{\pi d^3} \) - **Assumptions:** Uniform load, rigid supports, neglecting deflection. Let me know if you need the force path sketch or more details!