# 1: Introduction This problem involves the analysis of stresses in a cantilever bar subjected to external forces, specifically for a ductile material. The bar is loaded with a force \( F_y = 250.0 \) lbf, while \( F_z = F_x = 0 \) lbf. The lengths of rods DC and AB are 14.00 in. and 11.00 in., respectively. The bars have specified diameters. The goal is to determine the **principal stresses** (\( \sigma_1 \), \( \sigma_2 \)) and the **maximum shear stress** (\( \tau_{\text{max}} \)) at the critical stress element, typically at the fixed support where stresses are highest due to the applied force and resulting moments. ### Explanation Understanding the stress distribution in structural members is crucial for predicting failure and ensuring safety. Principal stresses represent the maximum and minimum normal stresses at a point, while the maximum shear stress is important for ductile materials, as it often governs yielding. The combination of direct (normal) and bending stresses, along with any shear, must be analyzed using established formulas from mechanics of materials. This introduction sets the stage for methodically evaluating these stresses by integrating knowledge of static equilibrium, mechanics of materials, and Mohr’s Circle for stress transformation. --- # 2: Presentation of Relevant Formulas Required To Solve The Question ### **a. Direct Normal Stress (Axial)** \[ \sigma = \frac{F}{A} \] where \( F \) is the axial load and \( A \) is the cross-sectional area. ### **b. Bending Stress (Due to Moment)** \[ \sigma_b = \frac{M c}{I} \] where \( M \) is the moment at the point of interest, \( c \) is the distance from the neutral axis (outer fiber), and \( I \) is the moment of inertia. ### **c. Shear Stress (Transverse)** \[ \tau = \frac{V}{A} \] where \( V \) is the shear force at the section and \( A \) is the cross-sectional area. ### **d. Principal Stresses** \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] ### **e. Maximum Shear Stress** \[ \tau_{\text{max}} = \frac{|\sigma_1 - \sigma_2|}{2} \] ### Explanation The above formulas are essential for determining the state of stress at a point subjected to combined loading. The direct stress formula calculates the normal stress from axial loading. The bending stress formula determines the maximum normal stress due to the moment induced by the applied force. Shear stress is included if there is a transverse force. The principal stress equations provide the extreme values of normal stress using the computed normal and shear stresses. The maximum shear stress formula quantifies the largest shear force that the material experiences, which is critical for ductile failure analysis. The use of these formulas ensures a complete and accurate assessment of the stresses in the cantilever bar. --- # 3: Detailed Step-by-Step Solution ## **Step 1: Calculation of Geometric Properties** ### **a. Area of Rod at Fixed End (Diameter = 2 in.)** \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (2)^2 = \pi \text{ in}^2 \] ### **b. Moment of Inertia (for a solid circle, diameter = 2 in.)** \[ I = \frac{\pi}{64} d^4 = \frac{\pi}{64} (2)^4 = \frac{\pi}{64} \cdot 16 = \frac{\pi}{4} \text{ in}^4 \] ### Explanation The cross-sectional area and moment of inertia are required to compute the direct, bending, and shear stresses. The area is used for normal and shear calculations, while the moment of inertia is essential for finding bending stresses. The diameter at the fixed end is chosen because this is typically the critical location for maximum stress. --- ## **Step 2: Calculation of Forces and Moments at the Fixed End** ### **a. Axial Force** Given \( F_x = 0 \), so no axial stress. ### **b. Shear Force** Given \( F_y = 250.0 \) lbf acts at the free end. ### **c. Bending Moment at the Wall** \[ M = F_y \cdot L \] Where \( L = 12 \) in (distance from wall to point of force application). \[ M = 250.0 \text{ lbf} \times 12 \text{ in} = 3000.0 \text{ lbf} \cdot \text{in} \] ### Explanation The forces and moments acting at the wall are determined based on the loading and geometry. The moment is a result of the perpendicular force acting at the distance from the fixed support. The maximum moment and thus maximum bending stress occur at the wall. --- ## **Step 3: Calculation of Stresses at the Critical Section** ### **a. Direct Normal Stress** \[ \sigma = \frac{F}{A} = \frac{0}{\pi} = 0 \text{ ksi} \] ### **b. Bending Stress** \[ c = \frac{d}{2} = 1 \text{ in} \] \[ \sigma_b = \frac{M c}{I} = \frac{3000.0 \times 1}{\pi/4} = \frac{3000.0 \times 4}{\pi} = \frac{12000}{\pi} \text{ psi} \approx 3819.718 \text{ psi} = 3.820 \text{ ksi} \] ### **c. Shear Stress** \[ \tau = \frac{V}{A} = \frac{250.0}{\pi} \text{ psi} \approx 79.577 \text{ psi} = 0.080 \text{ ksi} \] ### Explanation The direct normal stress is zero due to the absence of a force in the x-direction. The bending stress is calculated using the moment at the wall, distance to the outer fiber, and the moment of inertia. Shear stress is calculated as the ratio of the vertical force to the area. These values provide the components needed for principal stress analysis. --- ## **Step 4: Principal Stresses Calculation** Given: - \( \sigma_x = \sigma_b = 3.820 \) ksi (tension at the top fiber) - \( \sigma_y = 0 \) ksi (no stress in y-direction) - \( \tau_{xy} = 0.080 \) ksi \[ \sigma_{1,2} = \frac{3.820 + 0}{2} \pm \sqrt{ \left( \frac{3.820 - 0}{2} \right)^2 + (0.080)^2 } \] \[ = 1.910 \pm \sqrt{ (1.910)^2 + 0.0064 } \] \[ = 1.910 \pm \sqrt{3.6481 + 0.0064} \] \[ = 1.910 \pm \sqrt{3.6545} \] \[ = 1.910 \pm 1.911 \] - \( \sigma_1 = 1.910 + 1.911 = 3.821 \) ksi - \( \sigma_2 = 1.910 - 1.911 = -0.001 \) ksi ### Explanation Principal stresses are determined using the combined normal and shear stresses. The maximum principal stress is closely aligned with the calculated bending stress, and the minimum is very close to zero (slightly negative due to the presence of a small shear component). This step ensures the true state of stress at the critical location is captured. --- ## **Step 5: Maximum Shear Stress Calculation** \[ \tau_{\text{max}} = \frac{|\sigma_1 - \sigma_2|}{2} = \frac{3.821 - (-0.001)}{2} = \frac{3.822}{2} = 1.911 \text{ ksi} \] ### Explanation The maximum shear stress is calculated from the difference between the principal stresses, providing a direct measure of the shear experienced by the material. This value is essential in ductile failure criteria such as the maximum shear stress (Tresca) theory. --- # Conclusion - The **principal stress \( \sigma_1 \)** is determined to be **3.821 ksi**. - The **principal stress \( \sigma_2 \)** is determined to be **-0.001 ksi**. - The **maximum shear stress \( \tau_{\text{max}} \)** is determined to be **1.911 ksi**. These results comprehensively address the principal and shear stresses at the critical point of the cantilever bar, ensuring safe and reliable design according to established mechanical principles.