output_mds/page_24.md

5. Confirm that tension steel area exceeds $A_{s,\min}$. For Eq. (4-11), there is a requirement to use the larger of $3\sqrt{f_c'}$ or 200 psi in the numerator. In this case, $3\sqrt{4000 \text{ psi}} = 190$ psi, so use 200 psi. Thus,

$$A_{s,\min }=\frac{200\text{ psi}}{f_y}{b}_{w}d=\frac{200\text{ psi}}{60,000\text{ psi}}\times 12\text{ in.}\times 17.5\text{ in.}=0.70{\text{ in.}}^{2}A\_\{s,\backslash min\}\; =\; \backslash frac\{200\; \backslash text\{\; psi\}\}\{f\_y\}\; b\_w\; d\; =\; \backslash frac\{200\; \backslash text\{\; psi\}\}\{60,000\; \backslash text\{\; psi\}\}\; \backslash times\; 12\; \backslash text\{\; in.\}\; \backslash times\; 17.5\; \backslash text\{\; in.\}\; =\; 0.70\; \backslash text\{\; in.\}^2$$

$A_s$ exceeds $A_{s,\min}$, so this section satisfies the ACI Code requirement for minimum tension reinforcement.

Example 4-1M Analysis of Singly Reinforced Beams: Tension Steel Yielding—SI Units

Compute the nominal moment strength, $M_n$, of a beam (Fig. 4-19b) with $f_c' = 20$ MPa ($\beta_1 = 0.85$), $f_y = 420$ MPa, $b = 250$ mm, $d = 500$ mm, and three No. 25 bars (Table A-1M) giving $A_s = 3 \times 510 = 1530 \text{ mm}^2$. Note that the difference between the total section depth, $h$, and the effect depth, $d$, is 65 mm, which is a typical value for beam sections designed with metric dimensions.

1. Compute $a$ (assuming the tension steel is yielding).

$$a=\frac{A_s{f}_y}{0.85f_c^{\mathrm{\prime }}b}a\; =\; \backslash frac\{A\_s\; f\_y\}\{0.85\; f\_c\text{'}\; b\}$$

$$=\frac{1530{\text{ mm}}^2\times 420\text{ MPa}}{0.85\times 20\text{ MPa}\times 250\text{ mm}}=151\text{ mm}=\; \backslash frac\{1530\; \backslash text\{\; mm\}^2\; \backslash times\; 420\; \backslash text\{\; MPa\}\}\{0.85\; \backslash times\; 20\; \backslash text\{\; MPa\}\; \backslash times\; 250\; \backslash text\{\; mm\}\}\; =\; 151\; \backslash text\{\; mm\}$$

Therefore, $c = a / \beta_1 = 151 / 0.85 = 178$ mm.

2. Check whether the tension steel is yielding. The yield strain for the reinforcing steel is

$${ϵ}_y=\frac{f_y}{E_s}=\frac{420\text{ MPa}}{200,000\text{ MPa}}=0.0021\backslash epsilon\_y\; =\; \backslash frac\{f\_y\}\{E\_s\}\; =\; \backslash frac\{420\; \backslash text\{\; MPa\}\}\{200,000\; \backslash text\{\; MPa\}\}\; =\; 0.0021$$

From Eq. (4-18),

$${ϵ}_s=\left(\frac{500\text{ mm}-178\text{ mm}}{178\text{ mm}}\right)\times 0.003=0.00543\backslash epsilon\_s\; =\; \backslash left(\; \backslash frac\{500\; \backslash text\{\; mm\}\; -\; 178\; \backslash text\{\; mm\}\}\{178\; \backslash text\{\; mm\}\}\; \backslash right)\; \backslash times\; 0.003\; =\; 0.00543$$

Thus, the steel is yielding as assumed in step 1.

3. Compute the nominal moment strength, $M_n$. From Eq. (4-21), $M_n$ is (where 1 MPa $= 1 \text{ N/mm}^2$)

$$M_n=A_s{f}_y(d-\frac{a}{2})=1530{\text{ mm}}^2\times 420{\text{ N/mm}}^2(500-\frac{151}{2})\text{mm}M\_n\; =\; A\_s\; f\_y\; \backslash left(\; d\; -\; \backslash frac\{a\}\{2\}\; \backslash right)\; =\; 1530\; \backslash text\{\; mm\}^2\; \backslash times\; 420\; \backslash text\{\; N/mm\}^2\; \backslash left(\; 500\; -\; \backslash frac\{151\}\{2\}\; \backslash right)\; \backslash text\{mm\}$$

$$=273\times {10}^6\text{ N-mm}=273\text{ kN-m}=\; 273\; \backslash times\; 10^6\; \backslash text\{\; N-mm\}\; =\; 273\; \backslash text\{\; kN-m\}$$

Therefore, the design or factored moment strength, $\phi M_n$, of this beam is $0.9 \times 273 = 246$ kN-m.

4. Confirm that the tension steel area exceeds $A_{s,\min}$. For the given concrete strength of 20 MPa, the quantity $0.25\sqrt{f_c'} = 1.12$ MPa, which is less than 1.4 MPa. Therefore, the second part of Eq. (4-11M) governs for $A_{s,\min}$ as

$$A_{s,\min }=\frac{1.4b_{w}d}{f_y}=\frac{1.4\text{ MPa}\times 250\text{ mm}\times 500\text{ mm}}{420\text{ MPa}}=417{\text{ mm}}^{2}A\_\{s,\backslash min\}\; =\; \backslash frac\{1.4\; b\_w\; d\}\{f\_y\}\; =\; \backslash frac\{1.4\; \backslash text\{\; MPa\}\; \backslash times\; 250\; \backslash text\{\; mm\}\; \backslash times\; 500\; \backslash text\{\; mm\}\}\{420\; \backslash text\{\; MPa\}\}\; =\; 417\; \backslash text\{\; mm\}^2$$

$A_s$ exceeds $A_{s,\min}$, so this section satisfies the ACI Code requirement for minimum tension reinforcement.