# Q2 Solution: Shear Force, Bending Moment, and Maximum Shear Stress ## **Given Data** - **Beam type:** Simply supported, T-section - **Length (L):** 6 m - **Distributed load (w):** 1.5 kN/m - **Point load (P):** 3 kN at end C - **T-section dimensions:** - Flange: 150 mm × 25 mm - Web: 25 mm × 125 mm --- # **Step 1: Draw Shear Force & Bending Moment Diagrams** ### **A. Calculate Support Reactions** **1. Convert all to kN and m:** - \( w = 1.5\, \text{kN/m} \) - \( P = 3\, \text{kN} \) - \( L = 6\, \text{m} \) #### **Sum of Vertical Forces:** Let \( A_y \) and \( C_y \) be the reactions at supports A and C. \[ A_y + C_y = (1.5 \times 6) + 3 = 9 + 3 = 12\, \text{kN} \] #### **Taking Moments about A:** \[ \sum M_A = = (C_y \times 6) - (1.5 \times 6 \times 3) - (3 \times 6) \] \[ C_y \times 6 = 1.5 \times 6 \times 3 + 3 \times 6 \] \[ C_y \times 6 = (1.5 \times 18) + 18 = 27 + 18 = 45 \] \[ C_y = \frac{45}{6} = 7.5\, \text{kN} \] \[ A_y = 12 - 7.5 = 4.5\, \text{kN} \] --- ### **B. Shear Force Diagram (SFD)** - At A: \( V = +4.5 \) kN - Just left of C: \( V = 4.5 - (1.5 \times 6) = 4.5 - 9 = -4.5\, \text{kN} \) - Just right of C: \( V = -4.5 + 7.5 = 3\, \text{kN} \) - At C (after 3 kN point load): \( V = 3 - 3 = \) #### **Shear Force Diagram Sketch** ``` |\ | \ | \ | \ | \ | |______ ``` - Drop by 1.5 kN/m slope from A to C, then +7.5 kN jump at C (reaction), drop 3 kN at point load. --- ### **C. Bending Moment Diagram (BMD)** \[ M(x) = A_y \cdot x - \frac{w x^2}{2} \] - At A: \( M = \) - At C: \( M = \) - At any \( x \) from A ( to 6 m): \[ M(x) = 4.5x - .75x^2 \] - **Maximum moment at location where shear = :** \[ V = 4.5 - 1.5x = \implies x = 3\, \text{m} \] \[ M_{max} = 4.5 \times 3 - .75 \times 9 = 13.5 - 6.75 = 6.75\, \text{kN} \cdot \text{m} \] #### **Bending Moment Diagram Sketch** - Parabolic curve, maximum at center. --- ## **Step 2: Calculate Absolute Maximum Shear Stress** ### **A. Cross Section Properties** #### **Area \( A \):** - Flange: \( 150 \times 25 = 375\, \text{mm}^2 \) - Web: \( 25 \times 125 = 3125\, \text{mm}^2 \) - Total: \( 6875\, \text{mm}^2 = 6.875 \times 10^{-3}\, \text{m}^2 \) #### **Maximum Shear Force \( V_{max} \):** - At support: \( 4.5\, \text{kN} = 450\, \text{N} \) #### **Shear Stress Formula:** \[ \tau_{max} = \frac{V_{max}}{A} = \frac{450}{6875 \times 10^{-6}} = 654\, \text{kPa} = .654\, \text{MPa} \] - For T-sections, shear is maximum at the neutral axis in the web. --- ## **Summary Table** | Step | Description | Value | |------|--------------------------------------|--------------------------| | 1 | Max Shear Force | \( 4.5\, \text{kN} \) | | 2 | Max Bending Moment | \( 6.75\, \text{kN} \cdot \text{m} \) | | 3 | Max Shear Stress | \( .654\, \text{MPa} \) | --- ## **Final Answers** ### **a) Shear force and bending moment diagrams are shown above.** ### **b) The absolute maximum shear stress in the cross section is:** \[ \boxed{.654\, \text{MPa}} \] --- ### **Example References** - If \( V = 4.5\, \text{kN} \) and area is \( 6875\, \text{mm}^2 \), max shear stress is \( .654\, \text{MPa} \). - For a different loading, recalculate \( V_{max} \) and use the same formula. --- **[Alt text: Diagram showing beam with loading, shear force, and bending moment diagrams.]**