# Study Notes: Stress and Shear Stress ## Key Concepts ### 1. **Stress** Stress is the internal force per unit area within a material that arises due to externally applied forces. It is a measure of how much force is distributed over a given area. #### **Normal Stress ($\sigma$)** - Normal stress occurs when the force is applied perpendicular to the cross-sectional area of an object. - Formula: $$\sigma = \frac{P}{A}$$ Where: - $\sigma$ = Normal stress (in MPa or N/mm²) - $P$ = Internal force (in N or kN) - $A$ = Cross-sectional area (in mm²) #### **Cross-Sectional Area** - For a rectangular section: $$A = L \times W$$ Where: - $L$ = Length - $W$ = Width - For a circular section: $$A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$$ Where: - $d$ = Diameter of the circle For hollow circular sections, subtract the inner area from the outer area: $$A = \pi \left(\frac{d_{\text{out}}}{2}\right)^2 - \pi \left(\frac{d_{\text{in}}}{2}\right)^2$$ or equivalently: $$A = \frac{\pi}{4} \left(d_{\text{out}}^2 - d_{\text{in}}^2\right)$$ --- ### 2. **Shear Stress ($\tau$)** Shear stress occurs when the force is applied parallel to the surface of the material. It is responsible for causing one layer of the material to slide over another. #### **Single Shear** - In single shear, the force acts on a single plane. - Formula: $$\tau = \frac{P}{A}$$ Where: - $\tau$ = Shear stress (in MPa or N/mm²) - $P$ = Applied force (in N or kN) - $A$ = Shear area (in mm²) **Example: Single Shear** - A bolt with a diameter of $d = 5 \, \text{mm}$ is subjected to a force of $P = 100 \, \text{N}$. - Shear area: $$A = \frac{\pi d^2}{4} = \frac{\pi (5)^2}{4} = 19.63 \, \text{mm}^2$$ - Shear stress: $$\tau = \frac{P}{A} = \frac{100}{19.63} = 5.1 \, \text{MPa}$$ #### **Double Shear** - In double shear, the force is distributed over two planes. - Formula: $$\tau = \frac{P}{2A}$$ **Example: Double Shear** - A pin with a diameter of $d = 10 \, \text{mm}$ is subjected to a force of $P = 100 \, \text{N}$. - Shear area for one plane: $$A = \frac{\pi d^2}{4} = \frac{\pi (10)^2}{4} = 78.54 \, \text{mm}^2$$ - Total shear area (two planes): $$2A = 2 \times 78.54 = 157.08 \, \text{mm}^2$$ - Shear stress: $$\tau = \frac{P}{2A} = \frac{100}{157.08} = 0.64 \, \text{MPa}$$ In double shear, the stress is halved compared to single shear because the force is distributed over two planes. --- ## Common Pitfalls and Misconceptions **Forgetting to convert units:** Always ensure that forces are in Newtons (N) and areas are in square millimeters (mm²) when calculating stress. For example, $1 \, \text{kN} = 1000 \, \text{N}$. **Confusing normal stress and shear stress:** Normal stress acts perpendicular to the surface, while shear stress acts parallel to it. Ensure you understand the direction of the applied force. **Incorrect area calculation for hollow sections:** When dealing with hollow sections, remember to subtract the inner area from the outer area. --- ## Key Takeaways - **Stress** is the internal force per unit area, and it can be normal (perpendicular) or shear (parallel). - **Normal stress** is calculated using $\sigma = \frac{P}{A}$, while **shear stress** is calculated using $\tau = \frac{P}{A}$ (single shear) or $\tau = \frac{P}{2A}$ (double shear). - Always use the correct formula for the cross-sectional area, especially for circular or hollow sections. - Double shear reduces the stress by distributing the force over two planes. --- ## Practice Problems 1. A cylindrical rod with a diameter of $d = 20 \, \text{mm}$ is subjected to an axial force of $P = 50 \, \text{kN}$. Calculate the normal stress in the rod. 2. A bolt with a diameter of $d = 8 \, \text{mm}$ is subjected to a shear force of $P = 200 \, \text{N}$. Calculate the shear stress if: - The bolt is in single shear. - The bolt is in double shear. 3. A hollow circular section has an outer diameter of $d_{\text{out}} = 50 \, \text{mm}$ and a thickness of $t = 5 \, \text{mm}$. Calculate the cross-sectional area. Use these problems to test your understanding of the formulas and concepts discussed above.