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Chapter:
1. Find moment of resistance of a reinforced concrete rectangular beam of width $300 \mathrm{~mm}$ and overall depth $600 \mathrm{~mm}$ if reinforcement is $4-16 \mathrm{~mm}$ dia. bars at a distance of $40 \mathrm{~mm}$ from bottom. M20 grade concrete mix and TOR steel (Fe415 grade) have been used.
Solution:
1. Permissible stresses:
For M20 grade concrete $\sigma_{\text {cbc }}=7 \mathrm{~N} / \mathrm{mm}^{2}, \quad \mathrm{~m}=\frac{280}{3 \sigma_{\mathrm{cbc}}}=\frac{280}{3 \times 7}=13.33 \approx 13$
and for Fe415 grade steel, $\sigma_{s t}=230 \mathrm{~N} / \mathrm{mm}^{2}$
2. Neutral axis $A_{s t}=804 \mathrm{~mm}^{2}, \quad b=300 \mathrm{~mm}, \quad d=600-40=560 \mathrm{~mm}$
$\mathrm{p}=\frac{\mathrm{A}_{\mathrm{st}}}{\mathrm{bd}}=\frac{804}{300 \times 560}=0.004786$, and $\mathrm{mp}=13 \times 0.004786=0.06221$
Coefficient of neutral axis (k) shall be found from the relation $k=-m p+\sqrt{m^{2} p^{2}+2 m p}$
or, $k=-0.06221+\sqrt{(0.06221)^{2}+2 \times 0.06221}=0.296$
3. Lever arm coefficient, $j=1-k / 3=1-0.296 / 3=0.901$
4. Moment of Resistance Moment of resistance with respect to compression concrete $=\frac{1}{2} \sigma_{c b c} k j b d^{2}$
$=\frac{1}{2} \times 7 \times 0.901 \times 0.296 \times 300 \times 560^{2} \quad \mathrm{~N} . \mathrm{mm}$
$=87.82 \mathrm{KN} m$
Moment of resistance of beam section with respect to tensile steel
$=\sigma_{s t} A_{s t} j d$
$=230 \times 804 \times 0.901 \times 560 \quad \mathrm{~N} . \mathrm{mm}$
$=93.30 \mathrm{KN} \cdot \mathrm{m}$
The actual moment of resistance is smaller of the two values. Therefore, moment of resistance of the beam section $=87.82 \mathrm{KNm}$.
2. A simply supported reinforced concrete beam of rectangular cross-section is 300 $\mathrm{mm}$ wide and $450 \mathrm{~mm}$ deep up to centre of tensile reinforcement which consists of $4-20 \mathrm{~mm}$ dia. bars. Determine depth of neutral axis and maximum tensile stress in steel reinforcement when concrete at extreme compression fibreis stressed to $7 \mathrm{~N} / \mathrm{mm}^{2}$. Take $\mathrm{m}=9$.
So l u t i o n : ~
$$\text { Given: } \mathrm{b}=300 \mathrm{~mm}, \mathrm{~d}=450 \mathrm{~mm}, A_{s t}=4 \times \frac{\pi \times 20^{2}}{4}=1257 \mathrm{~mm}^{2}$$
From the first principle, we have
$$\text { b.n. } \frac{n}{2}=m A_{s t}(d-n)$$
or, $300 \times \frac{n^{2}}{2}=9 \times 1257 \times(450-n)$
or, $\quad n^{2}+75.42 n-33939=0$
Solving the quadratic equation, we get
$$n=150.33 \mathrm{~mm}$$
Using similar triangles of triangular stress distribution diagram (Fig. 2.6), we get
$$\frac{\sigma_{c b c}}{\left(\sigma_{s t} / m\right)}=\frac{n}{d-n} \Rightarrow \sigma_{s t}=\frac{m \sigma_{c b c}(d-n)}{n}=\frac{9 \times 7 \times(450-150.33)}{150.33}=125.6 \mathrm{~N} / \mathrm{mm}^{2}(\mathrm{Ans})$$
3. A rectangular beam is $250 \mathrm{~mm}$ wide and $500 \mathrm{~mm}$ deep up to centre of reinforcement. Find reinforcement required if it has to resist a moment of $45 \mathrm{KNm}$. Assume that M20 grade concrete mix and HYSD (Grade Fe415) bars have been used
Solution
1. Neutral axis The position of neutral axis shall be found by calculating $\mathrm{k}$ from
$$k=-m p+\sqrt{m^{2} p^{2}+2 m p}$$
Where,
$$\begin{array}{l} m=\frac{280}{3 \sigma_{c b c}}=\frac{280}{3 \times 7}=13.33 \approx 13 \\p=\frac{A_{s t}}{b d}, \quad \text { or } \quad A_{s t}=p b d=p \times 250 \times 500=125000 p \\\therefore k=-13 p+\sqrt{169 p^{2}+26 p} \text { (i) }\end{array}$$
2. Reinforcement required Moment of resistance of beam section with respect to steel
$$=\sigma_{s t} A_{s t} j d=(125000 p) \times 230 \times\left(1-\frac{k}{3}\right) \times 500=45 \times 10^{6}$$
Solving Eqns (i) and (ii) by trial and error, we get $p=0.003424$
$\therefore$ Reinforcement required $=0.003424 \times 250 \times 500=428 \mathrm{~mm}^{2}$