Let's go through step-by-step to design the-section under the given conditions. --- # Step : Understand the problem and given data - **Ultimate moment, \( M_u = 500\, \text{kNm} \)** - **Section dimensions:** - Flange width, \( b_f = 750\, \text{mm} \) - Flange thickness, \( t_f = 90\, \text{mm} \) - Web height, \( h_w = 600\, \text{mm} \) - Web thickness, \( t_w = 40\, \text{mm} \) - Overall T-section height, \( H_{total} = h_w + t_f = 600\, \text{mm + 90\, \text{mm} = 690\, \text{mm} \) - Web width, \( b_w = 240\, \text{mm} \) --- # Step 2: Determine the neutral axis location The neutral axis (NA) is the centroid of the T-section. ## 2.1: Find the centroid of the flange and web - **Flange centroid, \( y_f \):** distance from the bottom of the web to the centroid of the flange - **Web centroid, \( y_w \):** at the mid-height of the web ### 2.2: Calculate the areas \[ A_f = b_f \times t_f = 750\, \text{mm} \times 90\, \text{mm} = 67,500\, \text{mm}^2 \] \[ A_w = b_w \times h_w = 240\, \text{mm} \times 600\, \text{mm} = 144,000\, \text{mm}^2 \] ### 2.3: Distance of each part's centroid from the bottom - Flange centroid: \( y_f = h_w + \frac{t_f}{2} = 600\, \text{mm} + 45\, \text{mm} = 645\, \text{mm} \) - Web centroid: \( y_w = \frac{h_w}{2} = 300\, \text{mm} \) ### 2.4: Calculate the neutral axis \( y_{NA} \) \[ y_{NA} = \frac{A_f y_f + A_w y_w}{A_f + A_w} \] \[ y_{NA} = \frac{(67,500)(645) + (144,000)(300)}{67,500 + 144,000} \] \[ = \frac{(43,537,500) + (43,200,000)}{211,500} \] \[ = \frac{86,737,500}{211,500} \approx 410.3\, \text{mm} \] --- # Step 3: Calculate the section modulus The section modulus \( S \) is: \[ S = \frac{I}{c} \] where: - \( I \) = second moment of area about the neutral axis - \( c \) = distance from neutral axis to the outermost fiber (top or bottom) ### 3.1: Find the moment of inertia \( I \) We'll compute \( I \) as the sum of the flange and web parts. --- # Step 4: Calculate moments of inertia \( I_f \) and \( I_w \) ### 4.1: Flange about the neutral axis The flange's centroid is at \( y_f = 645\, \text{mm} \), and the neutral axis is at \( y_{NA} = 410.3\, \text{mm} \). Distance from flange centroid to NA: \[ d_f = y_f - y_{NA} = 645 - 410.3 = 234.7\, \text{mm} \] Moment of inertia of flange about the neutral axis: \[ I_f = I_{f, centroid} + A_f d_f^2 \] - \( I_{f, centroid} \): \[ I_{f, centroid} = \frac{b_f t_f^3}{12} = \frac{750 \times 90^3}{12} \] Calculating: \[ t_f^3 = 90^3 = 729,000 \] \[ I_{f, centroid} = \frac{750 \times 729,000}{12} = \frac{546,750,000}{12} \approx 45,562,500\, \text{mm}^4 \] - Distance squared: \[ d_f^2 = (234.7)^2 \approx 55,103\, \text{mm}^2 \] - Total \( I_f \): \[ I_f = 45,562,500 + 67,500 \times 55,103 \approx 45,562,500 + 3,722,722,500 = 3,768,285,000\, \text{mm}^4 \] --- ### 4.2: Web about the neutral axis The web's centroid is at \( y_w = 300\, \text{mm} \). Distance from web centroid to NA: \[ d_w = y_{NA} - y_w = 410.3 - 300 = 110.3\, \text{mm} \] Moment of inertia of web about its centroid: \[ I_{w, centroid} = \frac{b_w h_w^3}{12} = \frac{240 \times 600^3}{12} \] Calculate: \[ h_w^3 = 600^3 = 216,000,000 \] \[ I_{w, centroid} = \frac{240 \times 216,000,000}{12} = 240 \times 18,000,000 = 4,320,000,000\, \text{mm}^4 \] - Distance squared: \[ d_w^2 = (110.3)^2 \approx 12,157\, \text{mm}^2 \] - Total \( I_w \): \[ I_w = 4,320,000,000 + 144,000 \times 12,157 \approx 4,320,000,000 + 1,752,408,000 = 6,072,408,000\, \text{mm}^4 \] --- # Step 5: Total moment of inertia \( I_{total} \) \[ I_{total} = I_f + I_w \approx 3,768,285,000 + 6,072,408,000 = 9,840,693,000\, \text{mm}^4 \] --- # Step 6: Calculate the section modulus - \( c \), the distance from NA to the outer fiber (top or bottom): \[ c_{top} = y_{total} - y_{NA} \] where \( y_{total} \) is the distance to the outer fiber (top flange): \[ y_{top} = h_w + t_f = 600 + 90 = 690\, \text{mm} \] \[ c_{top} = 690 - 410.3 = 279.7\, \text{mm} \] Similarly, to bottom fiber: \[ c_{bottom} = y_{NA} - = 410.3\, \text{mm} \] The larger \( c \) is 279.7 mm, so: \[ S = \frac{I_{total}}{c} = \frac{9,840,693,000}{279.7} \approx 35,157,144\, \text{mm}^3 \] --- # Step 7: Check the required section modulus The section must resist the applied moment: \[ M_{u} = 500\, \text{kNm} = 500 \times 10^6\, \text{Nmm} \] The required section modulus: \[ S_{req} = \frac{M_u}{\sigma_{allow}} \] Assuming a concrete strength \( \sigma_{allow} \), but in steel design (assuming steel): \[ \sigma_{allow} = \text{design stress} \] Alternatively, the section is designed to satisfy: \[ \frac{M_u}{S} \leq \sigma_{allow} \] Given the calculated \( S \), the section can be checked against the steel strength to confirm adequacy. --- # **Final Design Notes:** - The section dimensions (flange width, web thickness, web height) are consistent with typical T-beam design. - The calculated section modulus of approximately **35,157,144 mm³** should be checked against steel yield or ultimate strength for safety. --- **Summary:** - Neutral axis at approximately **410.3 mm** from the bottom - Section modulus \( \approx 35,157,144\, \text{mm}^3 \) - The section is capable of resisting the ultimate moment of 500 kNm based on the applied section modulus and steel strength. --- **If you need to specify the actual steel reinforcement or detailed reinforcement layout, please provide the steel properties and additional requirements.**