(i) N-A depth

(ii) compressive force

(iii) Tensile force

(iv) ultimate moment

(v) safe concentrated load at mid span over an effective span of 6m.

Solution:

$ \mathrm{d}=500 \mathrm{~mm}, \mathrm{~b}=250 \mathrm{~mm} \\ $

$ \mathrm{A}_{\mathrm{st}}=3 \mathrm{X} \frac{\pi}{4} \mathrm{X} 20^{2}=942.48 \mathrm{~mm}^{2} \\ $

$ \mathrm{f}_{\mathrm{ck}}=20 \mathrm{Mpa} \mathrm{f}_{\mathrm{y}}=415 \mathrm{Mpa} \\ $

Step - 1:

$$ \begin{aligned} \frac{x_{u}}{d} &=2.41 \times \frac{f y}{f_{c k}} \times \frac{A_{s t}}{b d} \\\\ &=2,41 \times \frac{415}{20} \times \frac{942.48}{250 \times 250} \times 500 \\\\ \mathrm{x}_{\mathrm{u}} &=188.52 \mathrm{~mm} \\ \end{aligned} $$

Step - 2:

For Fe, $415 \frac{x_{u l i m}}{d}=0.48 ; \mathrm{x}_{\mathrm{ulim}}=0.48 \mathrm{X} 500$ $=240 \mathrm{~mm}$

$\therefore \mathrm{x}_{\mathrm{u}}\lt\mathrm{x}_{\mathrm{ulim}}$, the section is under reinforced.

$$ \mathrm{C}_{\mathrm{u}}=0.36 \mathrm{f}_{\mathrm{ck}} \mathrm{X}_{\mathrm{u}} \mathrm{b}=0.36 \\ $$

$$ \mathrm{X} 20 \mathrm{X} 188.52 \mathrm{X} 250 / 10^{3}=339.34 \mathrm{kN} \text {. } \\ $$

Step - 3:

MR for under reinforced section is,

$$ \begin{aligned} \mathrm{M}_{\mathrm{u}} &=\mathrm{MR}=0.87 \mathrm{f}_{\mathrm{y}} \mathrm{A}_{\mathrm{st}}\left(\mathrm{d}-0.42 \mathrm{x}_{\mathrm{u}}\right) \\\\ &=\frac{0.87 \times 415 \times 942.48(500-0.42 \times 188.52)}{10^{6}} \\\\ &=143.1 \mathrm{kN}-\mathrm{m} \\ \end{aligned} $$

Step - 4:

$$ \mathbf{M}_{\mathrm{u}}=\frac{w_{u} \times L}{4}=\frac{w_{u} \times 6}{4} \\ $$

Equating factored moment to MR,

$$ \begin{aligned} &\frac{W_{u} \times 6}{4}=143.1 \\\\ &\mathrm{~W}_{\mathrm{u}}=95.5 \mathrm{KN} \\ \end{aligned} $$

Safe load, W $=\frac{W_{u}}{1.5}$ load factor/factor of safety.

$$ =63.67 \mathrm{kN} $$