1. **Problem Statement:** Calculate the allowable load on a simply supported precast plain concrete beam with span $L=3$ m, given the cracking moment $M_{cr}=27$ kN-m. 2. **Given Data:** - Concrete compressive strength $f'_c=17$ MPa - Concrete unit weight $\gamma=17$ kN/m³ - Dimensions: $a=250$ mm, $b=350$ mm, $d=200$ mm, $e=600$ mm, $c=100$ mm 3. **Step 1: Understand the beam and loading** - The beam is simply supported with span $L=3$ m. - The cracking moment $M_{cr}=27$ kN-m is the moment at which the beam starts to crack. - The allowable load $w$ (uniformly distributed load in kN/m) can be found from the bending moment formula for simply supported beams: $$M = \frac{wL^2}{8}$$ 4. **Step 2: Calculate allowable load $w$ from cracking moment** Rearranging the formula: $$w = \frac{8M_{cr}}{L^2}$$ Substitute values (convert $L=3$ m to meters, $M_{cr}=27$ kN-m): $$w = \frac{8 \times 27}{3^2} = \frac{216}{9} = 24 \text{ kN/m}$$ 5. **Step 3: Determine cracking moment for positive and negative bending** - The cracking moment is given as 27 kN-m, which applies to the beam section. - For positive bending (sagging), the moment is $M_{cr} = 27$ kN-m. - For negative bending (hogging), the cracking moment depends on the section properties but is typically similar if the section is symmetric. Since the beam is plain concrete and the problem does not specify different moments, we assume: $$M_{cr}^{+} = M_{cr}^{-} = 27 \text{ kN-m}$$ **Final answers:** - Allowable load $w = 24$ kN/m - Cracking moments $M_{cr}^{+} = 27$ kN-m, $M_{cr}^{-} = 27$ kN-m