1. **Problem Statement:** Calculate and compare the principal tensile stress at the centroidal axis and at the junction of the web with the lower flange of a prestressed concrete beam with an unsymmetrical I-section. Given: - Fibre stress at top edge = 12 N/mm² (compression) - Stress linearly reduces to zero at bottom - Bottom flange width = 1200 mm - Bottom flange thickness = 900 mm - Web depth = 1000 mm - Web thickness = 600 mm - Total vertical shear load = 2350 kN 2. **Step 1: Understand the stress distribution and geometry** The stress varies linearly from 12 N/mm² compression at the top to 0 at the bottom. This implies a linear stress gradient through the depth. 3. **Step 2: Calculate the centroid of the section** Calculate the area and centroid of each part: - Bottom flange area $A_{bf} = 1200 \times 900 = 1,080,000$ mm² - Web area $A_w = 600 \times 1000 = 600,000$ mm² Assuming the top flange is negligible or not given, centroid $y_c$ from bottom is: $$y_c = \frac{A_{bf} \times \frac{900}{2} + A_w \times (900 + \frac{1000}{2})}{A_{bf} + A_w} = \frac{1,080,000 \times 450 + 600,000 \times 1400}{1,680,000}$$ Calculate numerator: $$1,080,000 \times 450 = 486,000,000$$ $$600,000 \times 1400 = 840,000,000$$ Sum = 1,326,000,000 Divide by total area: $$y_c = \frac{1,326,000,000}{1,680,000} = 789.29 \text{ mm}$$ 4. **Step 3: Calculate moment of inertia $I$ about centroidal axis** Use parallel axis theorem: - $I_{bf} = \frac{1}{12} b h^3 = \frac{1}{12} \times 1200 \times 900^3 = \frac{1}{12} \times 1200 \times 729,000,000 = 72,900,000,000$ mm$^4$ - Distance from bottom flange centroid to neutral axis $d_{bf} = 789.29 - 450 = 339.29$ mm - $I_{bf,total} = I_{bf} + A_{bf} d_{bf}^2 = 72,900,000,000 + 1,080,000 \times 339.29^2$ Calculate: $$1,080,000 \times 115,096 = 124,303,680,000$$ Sum: $$I_{bf,total} = 72,900,000,000 + 124,303,680,000 = 197,203,680,000$$ mm$^4$ - $I_w = \frac{1}{12} b h^3 = \frac{1}{12} \times 600 \times 1000^3 = 50,000,000,000$ mm$^4$ - Distance from web centroid to neutral axis $d_w = 1400 - 789.29 = 610.71$ mm - $I_{w,total} = I_w + A_w d_w^2 = 50,000,000,000 + 600,000 \times 610.71^2$ Calculate: $$600,000 \times 373,000 = 223,800,000,000$$ Sum: $$I_{w,total} = 50,000,000,000 + 223,800,000,000 = 273,800,000,000$$ mm$^4$ - Total moment of inertia: $$I = I_{bf,total} + I_{w,total} = 197,203,680,000 + 273,800,000,000 = 471,003,680,000$$ mm$^4$ 5. **Step 4: Calculate bending moment $M$ from fibre stress** Maximum compressive stress at top fiber $\sigma_{top} = 12$ N/mm² Distance from centroid to top fiber $y_{top} = 789.29$ mm Using flexure formula: $$\sigma = \frac{M y}{I} \Rightarrow M = \frac{\sigma I}{y} = \frac{12 \times 471,003,680,000}{789.29}$$ Calculate: $$M = 7,161,000,000 \text{ Nmm} = 7,161 \text{ kNm}$$ 6. **Step 5: Calculate shear stress $\tau$ at centroidal axis and web-flange junction** Shear force $V = 2350$ kN = 2,350,000 N Shear stress formula: $$\tau = \frac{V Q}{I t}$$ Where: - $Q$ = first moment of area above the point where shear is calculated - $t$ = thickness at the point At centroidal axis: - Area above centroid $A' = A_{bf} + \text{part of web above centroid}$ Web above centroid depth = $1400 - 789.29 = 610.71$ mm Web area above centroid $= 600 \times 610.71 = 366,426$ mm² Total $A' = 1,080,000 + 366,426 = 1,446,426$ mm² Centroid of area above centroid: - Bottom flange centroid at 450 mm - Web centroid above centroid at $789.29 + \frac{610.71}{2} = 1,094.65$ mm Calculate $Q$: $$Q = A_{bf} (y_c - y_{bf}) + A_{web} (y_c - y_{web})$$ $$= 1,080,000 (789.29 - 450) + 366,426 (789.29 - 1,094.65)$$ Calculate: $$1,080,000 \times 339.29 = 366,778,800$$ $$366,426 \times (-305.36) = -111,900,000$$ Sum: $$Q = 366,778,800 - 111,900,000 = 254,878,800$$ mm$^3$ Thickness at centroidal axis $t = 600$ mm (web thickness) Calculate shear stress: $$\tau = \frac{2,350,000 \times 254,878,800}{471,003,680,000 \times 600} = \frac{599,833,680,000,000}{282,602,208,000,000} = 2.12 \text{ N/mm}^2$$ At junction of web and lower flange: - Area above junction = bottom flange area $A_{bf} = 1,080,000$ mm² - Centroid of bottom flange = 450 mm - Distance from neutral axis to bottom flange centroid = $789.29 - 450 = 339.29$ mm Calculate $Q$: $$Q = A_{bf} \times (y_c - y_{bf}) = 1,080,000 \times 339.29 = 366,778,800$$ mm$^3$ Thickness at junction $t = 1200$ mm (bottom flange width) Calculate shear stress: $$\tau = \frac{2,350,000 \times 366,778,800}{471,003,680,000 \times 1200} = \frac{861,927,180,000,000}{565,204,416,000,000} = 1.52 \text{ N/mm}^2$$ 7. **Step 6: Calculate principal tensile stress** Principal tensile stress $\sigma_1$ is given by: $$\sigma_1 = \frac{\sigma_x + \sigma_y}{2} + \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau^2}$$ Assuming $\sigma_y = 0$ (no vertical normal stress), and tensile stress is negative compression: At centroidal axis: $$\sigma_x = -\text{bending stress at centroid}$$ Calculate bending stress at centroid: $$\sigma_x = \frac{M y}{I} = \frac{7,161,000,000 \times 0}{471,003,680,000} = 0$$ Actually, at centroid $y=0$, so bending stress is zero. So principal tensile stress: $$\sigma_1 = \frac{0 + 0}{2} + \sqrt{\left(\frac{0 - 0}{2}\right)^2 + 2.12^2} = 2.12 \text{ N/mm}^2$$ At junction of web and lower flange: Calculate bending stress at junction: Distance from neutral axis $y = 789.29 - 900 = -110.71$ mm (negative means tension side) $$\sigma_x = \frac{7,161,000,000 \times (-110.71)}{471,003,680,000} = -1.68 \text{ N/mm}^2$$ Calculate principal tensile stress: $$\sigma_1 = \frac{-1.68 + 0}{2} + \sqrt{\left(\frac{-1.68 - 0}{2}\right)^2 + 1.52^2} = -0.84 + \sqrt{0.7056 + 2.3104} = -0.84 + 1.79 = 0.95 \text{ N/mm}^2$$ 8. **Final answer:** - Principal tensile stress at centroidal axis = 2.12 N/mm² - Principal tensile stress at junction of web and lower flange = 0.95 N/mm² These results show higher principal tensile stress at the centroidal axis due to shear, while bending stress reduces tensile stress at the junction.