Sure! Let's **redo the calculation** with your provided data: - \( P = V = 125\, \text{kN} = 125,000\, \text{N} \) - \( b = 125\, \text{mm} \) - \( d = h = 200\, \text{mm} \) - \( t = 9\, \text{mm} \) - \( y_k = 50\, \text{mm} \) (below centroid) We will **follow the same steps** as in your image: --- ### **Step 1: Moment of Inertia about z-axis (\(I_z\))** \[ I_z = \frac{b h^3}{12} - \frac{(b-2t)(h-2t)^3}{12} \] \[ I_z = \frac{125 \times 200^3}{12} - \frac{(125-2\times9) \times (200-2\times9)^3}{12} \] \[ = \frac{125 \times 8,000,000}{12} - \frac{107 \times 146,689}{12} \] \[ = \frac{1,000,000,000}{12} - \frac{107 \times 146,689^3}{12} \] Calculate: - \( 200^3 = 8,000,000 \) - \( b-2t = 125-18 = 107 \) - \( h-2t = 200-18 = 182 \) - \( 182^3 = 6,029,768 \) \[ I_z = \frac{125 \times 8,000,000}{12} - \frac{107 \times 6,029,768}{12} \] \[ = \frac{1,000,000,000}{12} - \frac{645,184,176}{12} \] \[ = 83,333,333.33 - 53,765,348 \] \[ = 29,567,985 \,\text{mm}^4 \] --- ### **Step 2: Shear at point K (\( y_k = 50 \) mm below centroid)** #### **First, locate centroid:** - Centroid from base = \( h/2 = 200/2 = 100\,\text{mm} \) - \( y_k \) is 50 mm below centroid, so from base: \( 100 + 50 = 150\,\text{mm} \) #### **Calculate \( Q_k \):** \[ Q_k = \left[ b t \left( \frac{h}{2} - \frac{t}{2} \right) \right] + \left[ 2 t \left( \frac{h}{2} - t - y_k \right) \left( y_k + \frac{\frac{h}{2} - t - y_k}{2} \right) \right] \] Let's break it down: - \( b = 125 \) - \( t = 9 \) - \( h = 200 \) - \( y_k = 50 \) First term: \[ 125 \times 9 \times (100 - 4.5) = 1,125 \times 95.5 = 107,437.5 \] Second term: - \( \frac{h}{2} - t - y_k = 100 - 9 - 50 = 41 \) - The centroid of this area from the neutral axis is \( 50 + 41/2 = 70.5 \) So, \[ 2 \times 9 \times 41 \times 70.5 = 18 \times 41 \times 70.5 = 738 \times 70.5 = 51,969 \] So, \[ Q_k = 107,437.5 + 51,969 = 159,406.5\, \text{mm}^3 \] --- ### **Step 3: Shear Stress at Point K** \[ \tau_x = \frac{V Q_k}{I_z t} \] \[ = \frac{125,000 \times 159,406.5}{29,567,985 \times 9} \] \[ = \frac{19,925,812,500}{266,111,865} \] \[ = 74.89\,\text{N/mm}^2 \] --- ### **Step 4: Maximum Horizontal Shear Stress** For max Q (at neutral axis): - Use \( y_k = 0 \) - \( Q_{max} = 125 \times 9 \times (100 - 4.5) + 2 \times 9 \times 91 \times (45.5) \) - where \( 91 = 100 - 9 \), and centroid: \( 45.5 = 91/2 \) \[ Q_{max} = 1,125 \times 95.5 + 18 \times 91 \times 45.5 \] \[ = 107,437.5 + 81,969 = 189,406.5\,\text{mm}^3 \] \[ \tau_{max} = \frac{125,000 \times 189,406.5}{29,567,985 \times 9} \] \[ = \frac{23,675,812,500}{266,111,865} \] \[ = 88.98\,\text{N/mm}^2 \] --- ## **Final Answers** ### (a) Shear stress at point K: \[ \boxed{74.89\,\text{N/mm}^2} \] ### (b) Maximum horizontal shear stress: \[ \boxed{88.98\,\text{N/mm}^2} \] --- If you'd like the step-by-step breakdown or a check on any specific value, let me know!