(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 \\\\ …