Let's solve the problem step by step: --- ## **Step 1: Find the Support Reactions** The arm is subjected to an **850 N** downward force at its end. To find the stresses at section \( a-a \) (point A), we need internal **normal force**, **shear force**, and **bending moment** at this section. ### **A. Calculate the Moment at Section \( a-a \)** The distance from the wall to the force is: \[ .15\, \text{m} + .15\, \text{m} + .35\, \text{m} = .65\, \text{m} \] So, the **bending moment** at section \( a-a \) (just left of the load): \[ M_a = 850\,\text{N} \times .35\,\text{m} = 297.5\,\text{Nm} \] ### **B. Shear Force at Section \( a-a \)** The vertical **shear force** at section \( a-a \) is equal to the applied load: \[ V_a = 850\, \text{N} \] ### **C. Axial Force** Since the load is vertical and the arm is horizontal, there is **no axial force** at section \( a-a \). --- ## **Step 2: Find Stresses at Point A** ### **A. Section Properties (I-beam)** Given: - Flange width, \( b = 20\, \text{mm} \) - Flange thickness, \( t_f = 7.5\, \text{mm} \) - Web thickness, \( t_w = 7.5\, \text{mm} \) - Depth, \( h = 50\, \text{mm} \) #### **Moment of Inertia (\( I \))** For I-section: \[ I = 2 \left[ \frac{b t_f^3}{12} + b t_f \left(\frac{h}{2} - \frac{t_f}{2}\right)^2 \right] + \frac{t_w (h-2t_f)^3}{12} \] Plug in values: - \( b = 20\, \text{mm} \) - \( t_f = 7.5\, \text{mm} \) - \( t_w = 7.5\, \text{mm} \) - \( h = 50\, \text{mm} \) Calculate the **neutral axis** is at 25 mm from the bottom. #### **Simplified Calculation (approximate):** For now, let's use the web for shear and the furthest fiber for bending: - Distance from neutral axis to extreme fiber, \( c = 25\, \text{mm} = .025\, \text{m} \) --- ### **B. Bending Stress (\( \sigma \))** \[ \sigma = \frac{M_a c}{I} \] --- ### **C. Shear Stress (\( \tau \))** For maximum shear in web: \[ \tau = \frac{V_a Q}{I b} \] Where \( Q \) is the first moment of area about the neutral axis for the area above or below the point of interest. --- ## **Step 3: Calculate \( I \) and \( Q \)** Let's use **approximate** values: ### **A. Moment of Inertia (\( I \))** For the given I-section, 1. **Area of each flange**: \( = 20 \times 7.5 = 150 \, \text{mm}^2 \) 2. **Area of web**: \( = 7.5 \times (50 - 2 \times 7.5) = 7.5 \times 35 = 262.5 \, \text{mm}^2 \) Moment of inertia for full section (about centroid): \[ I \approx \frac{1}{12} b h^3 - \frac{1}{12} (b-t_w)(h-2t_f)^3 \] \[ I \approx \frac{1}{12} (20)(50^3) - \frac{1}{12}(12.5)(35^3) \] \[ I \approx \frac{1}{12}(20)(125000) - \frac{1}{12}(12.5)(42875) \] \[ I \approx \frac{2,500,000}{12} - \frac{535,937.5}{12} \] \[ I \approx 208,333 - 44,661.5 \approx 163,671\, \text{mm}^4 \] --- ### **B. Calculate Bending Stress at A** \[ \sigma = \frac{M_a c}{I} \] \[ M_a = 297.5\, \text{Nm} = 297,500\, \text{Nmm} \] \[ c = 25\, \text{mm} \] \[ I = 163,671\, \text{mm}^4 \] \[ \sigma = \frac{297,500 \times 25}{163,671} = \frac{7,437,500}{163,671} \approx 45.44\, \text{MPa} \] --- ### **C. Shear Stress at A (max in web center)** The shear is max at the web center: \[ \tau_{max} = \frac{V_a}{A_{web}} \] \[ A_{web} = t_w \times (h-2t_f) = 7.5 \times 35 = 262.5\, \text{mm}^2 \] \[ \tau_{max} = \frac{850}{262.5} = 3.24\, \text{N/mm}^2 = 3.24\, \text{MPa} \] --- ## **Step 4: Principal Stresses and Maximum In-Plane Shear Stress** Given: - \( \sigma_x = 45.44\, \text{MPa} \) (tension or compression, depending on location) - \( \tau_{xy} = 3.24\, \text{MPa} \) Principal stresses: \[ \sigma_{1,2} = \frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2} \] \[ \sigma_{1,2} = \frac{45.44}{2} \pm \sqrt{\left(\frac{45.44}{2}\right)^2 + (3.24)^2} \] \[ = 22.72 \pm \sqrt{516.1 + 10.5} \] \[ = 22.72 \pm \sqrt{526.6} \] \[ = 22.72 \pm 22.95 \] So, - \( \sigma_1 = 45.67\, \text{MPa} \) - \( \sigma_2 = -.23\, \text{MPa} \) Maximum in-plane shear stress: \[ \tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{45.67 - (-.23)}{2} = 22.95\, \text{MPa} \] --- ## **Step 5: Orientation** The orientation of principal planes (\( \theta_p \)) is found by: \[ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x} \] \[ \tan 2\theta_p = \frac{2 \times 3.24}{45.44} = \frac{6.48}{45.44} = .143 \] \[ 2\theta_p = \arctan(.143) \approx 8.2^\circ \] \[ \theta_p = 4.1^\circ \] --- ## **Summary Table** | Value | Result | |------------------------------------|---------------| | Bending Stress (\( \sigma_x \)) | 45.44 MPa | | Shear Stress (\( \tau_{xy} \)) | 3.24 MPa | | Principal Stresses (\( \sigma_1 \))| 45.67 MPa | | Principal Stresses (\( \sigma_2 \))| -.23 MPa | | Max In-Plane Shear (\( \tau_{max} \))| 22.95 MPa| | Principal Plane Angle (\( \theta_p \))| 4.1° | --- ### **Final Answers** **Principal stresses at A:** - \( \sigma_1 = 45.67 \) MPa - \( \sigma_2 = -.23 \) MPa **Maximum in-plane shear stress:** - \( \tau_{max} = 22.95 \) MPa **Orientation of principal planes:** - \( \theta_p = 4.1^\circ \) --- If you need a stress element diagram or further breakdown, let me know!