1. **Problem Statement:** Calculate the neutral axis depth $x$, moment of inertia $I$, and safe moment capacity $M_c$ for a reinforced concrete section with width $b=300$ mm, effective depth $d=520$ mm, and now with $n=5$ reinforcement bars each having area $A_s = 6 \pi (28)^2 = 3695$ mm$^2$. 2. **Given Data:** - Width $b=300$ mm - Effective depth $d=520$ mm - Number of bars $n=5$ - Area of one bar $A_s=3695$ mm$^2$ - Total steel area $n A_s = 5 \times 3695 = 18475$ mm$^2$ 3. **Set up the equilibrium equation for neutral axis depth $x$:** The force equilibrium equation is: $$300 \times \frac{x}{2} = n A_s (520 - x)$$ Substitute values: $$300 \times \frac{x}{2} = 18475 (520 - x)$$ Simplify: $$150 x = 18475 (520 - x)$$ $$150 x = 18475 \times 520 - 18475 x$$ Bring all terms to one side: $$150 x + 18475 x = 18475 \times 520$$ $$18625 x = 9607000$$ Solve for $x$: $$x = \frac{9607000}{18625} \approx 515.9 \text{ mm}$$ 4. **Calculate moment of inertia $I$:** $$I = \frac{b x^3}{3} + n A_s (d - x)^2$$ Calculate each term: $$\frac{300 \times (515.9)^3}{3} = 300 \times \frac{137,352,000}{3} = 300 \times 45,784,000 = 13,735,200,000$$ $$n A_s (d - x)^2 = 18475 \times (520 - 515.9)^2 = 18475 \times (4.1)^2 = 18475 \times 16.81 = 310,700$$ Sum: $$I = 13,735,200,000 + 310,700 \approx 13,735,510,700 \text{ mm}^4$$ 5. **Calculate safe moment capacity $M_c$:** Given concrete stress $f_c = 12.15$ MPa (assumed from original problem), $$M_c = \frac{f_c I}{x} = \frac{12.15 \times 13,735,510,700}{515.9}$$ Calculate numerator: $$12.15 \times 13,735,510,700 = 166,847,000,000$$ Divide: $$M_c = \frac{166,847,000,000}{515.9} \approx 323,300,000 \text{ N.mm} = 323.3 \text{ kN.m}$$ 6. **Conclusion:** With 5 reinforcement bars, the neutral axis depth $x$ is approximately 515.9 mm, the moment of inertia $I$ is approximately $1.37 \times 10^{10}$ mm$^4$, and the safe moment capacity $M_c$ is approximately 323.3 kN.m. This shows an increase in moment capacity compared to the original 9 bars case due to the change in steel area and neutral axis depth.