The problem requires determining the tensile steel area for a T-beam subjected to specified dead and live loads on a simple span. This is a common design task in reinforced concrete structures.

Step 1: Determine the effective flange width (b_f).

According to ACI 318, for an interior T-beam, the effective flange width (b_f) is the smallest of the following:

• One-fourth of the span length: \(L/4 = 22 \text{ ft} \times 12 \text{ in/ft} / 4 = 66 \text{ in}\)
• \(b_w + 16 h_f = 11 \text{ in} + 16 \times 4.5 \text{ in} = 11 + 72 = 83 \text{ in}\)
• Center-to-center spacing of beams: \(9 \text{ ft} \times 12 \text{ in/ft} = 108 \text{ in}\)

Therefore, the effective flange width, \(b_f = 66 \text{ in}\).

Step 2: Calculate the Dead Load (DL) per unit length.

Assume normal weight concrete with a unit weight of 150 pcf.

Slab weight (per unit length of beam) = \(h_f \times \text{spacing} \times \text{unit weight of concrete}\)

\(= (4.5 \text{ in} / 12 \text{ in/ft}) \times 9 \text{ ft} \times 150 \text{ pcf} = 0.375 \text{ ft} \times 9 \text{ ft} \times 150 \text{ pcf} = 506.25 \text{ plf}\)

Assume a total beam height \(h = d + \text{cover} = 20 \text{ in} + 2.5 \text{ in (assumed cover)} = 22.5 \text{ in}\).

Web (stem) weight below flange = \((h - h_f) \times b_w \times \text{unit weight of concrete}\)

\(= ((22.5 - 4.5) \text{ in} / 12 \text{ in/ft}) \times (11 \text{ in} / 12 \text{ in/ft}) \times 150 \text{ pcf}\)

\(= (18 \text{ in} / 12 \text{ in/ft}) \times (11 \text{ in} / 12 \text{ in/ft}) \times 150 \text{ pcf} = 1.5 \text{ ft} \times 0.9167 \text{ ft} \times 150 \text{ pcf} \approx 206.25 \text{ plf}\)

Total Dead Load, \(DL = 506.25 \text{ plf} + 206.25 \text{ plf} = 712.5 \text{ plf}\)

Step 3: Calculate the Live Load (LL) per unit length.

\(LL = \text{Live Load per unit area} \times \text{spacing}\)

\(= 200 \text{ psf} \times 9 \text{ ft} = 1800 \text{ plf}\)

Step 4: Calculate the Factored Distributed Load (w_u).

Using ACI load factors: \(w_u = 1.2 DL + 1.6 LL\)

\(= 1.2 \times 712.5 \text{ plf} + 1.6 \times 1800 \text{ plf}\)

\(= 855 \text{ plf} + 2880 \text{ plf} = 3735 \text{ plf}\)

Step 5: Calculate the Factored Bending Moment (M_u).

For a simple span beam, \(M_u = w_u L^2 / 8\)

\(= 3735 \text{ plf} \times (22 \text{ ft})^2 / 8\)

\(= 3735 \times 484 / 8 = 225915 \text{ lb-ft}\)

Convert to lb-in: \(M_u = 225915 \text{ lb-ft} \times 12 \text{ in/ft} = 2710980 \text{ lb-in}\)

Step 6: Determine if the Neutral Axis (NA) is within the flange.

To check if the NA is in the flange, calculate the moment capacity if \(a = h_f\).

The compression force in the flange, \(C_f = 0.85 f_c' b_f h_f\)

\(= 0.85 \times 4000 \text{ psi} \times 66 \text{ in} \times 4.5 \text{ in} = 1009800 \text{ lb}\)

The nominal moment capacity of the flange, \(M_{nf} = C_f (d - h_f/2)\)

\(= 1009800 \text{ lb} \times (20 \text{ in} - 4.5 \text{ in}/2) = 1009800 \times (20 - 2.25) \text{ in} = 1009800 \times 17.75 \text{ in} = 17923950 \text{ lb-in}\)

The design strength of the flange (assuming \(\phi = 0.9\) for tension-controlled behavior if NA is in flange and steel yields):

\(\phi M_{nf} = 0.9 \times 17923950 \text{ lb-in} = 16131555 \text{ lb-in}\)

Since the required factored moment \(M_u = 2710980 \text{ lb-in}\) is less than \(\phi M_{nf} = 16131555 \text{ lb-in}\), the neutral axis lies within the flange. Therefore, the T-beam can be designed as a rectangular beam with a width equal to the effective flange width, \(b_f = 66 \text{ in}\).

Step 7: Calculate the Required Tensile Steel Area (A_s).

For a rectangular beam, the nominal moment capacity \(M_n = A_s f_y (d - a/2)\) where \(a = A_s f_y / (0.85 f_c' b_f)\).

The design moment capacity is \(M_u = \phi M_n = \phi A_s f_y (d - a/2)\).

Substitute \(a\): \(M_u = \phi A_s f_y (d - \frac{A_s f_y}{2 \times 0.85 f_c' b_f})\)

Rearrange into a quadratic equation in terms of \(A_s\):

\(\frac{\phi f_y^2}{1.7 f_c' b_f} A_s^2 - (\phi f_y d) A_s + M_u = 0\)

Given: \(\phi = 0.9\) (assumed tension-controlled), \(f_c' = 4000 \text{ psi}\), \(f_y = 50000 \text{ psi}\), \(b_f = 66 \text{ in}\), \(d = 20 \text{ in}\), \(M_u = 2710980 \text{ lb-in}\).

Let \(K_1 = \frac{\phi f_y^2}{1.7 f_c' b_f} = \frac{0.9 \times (50000 \text{ psi})^2}{1.7 \times 4000 \text{ psi} \times 66 \text{ in}} = \frac{0.9 \times 2.5 \times 10^9}{448800} \approx 5013.368\)

Let \(K_2 = \phi f_y d = 0.9 \times 50000 \text{ psi} \times 20 \text{ in} = 900000\)

The equation becomes: \(5013.368 A_s^2 - 900000 A_s + 2710980 = 0\)

Using the quadratic formula \(A_s = \frac{-(-K_2) \pm \sqrt{(-K_2)^2 - 4 K_1 M_u}}{2 K_1}\):

\(A_s = \frac{900000 \pm \sqrt{(900000)^2 - 4 \times 5013.368 \times 2710980}}{2 \times 5013.368}\)

\(A_s = \frac{900000 \pm \sqrt{8.1 \times 10^{11} - 5.4390 \times 10^{10}}}{10026.736}\)

\(A_s = \frac{900000 \pm \sqrt{7.5561 \times 10^{11}}}{10026.736}\)

\(A_s = \frac{900000 \pm 869258.3}{10026.736}\)

Taking the smaller positive root for \(A_s\):

\(A_s = \frac{900000 - 869258.3}{10026.736} = \frac{30741.7}{10026.736} \approx 3.066 \text{ in}^2\)

Step 8: Check minimum and maximum steel area requirements.

Minimum steel area (\(A_{s,min}\)) for flexural members (ACI 318-19, 9.6.1.2):

\(\rho_{min} = \text{max} \left(\frac{3\sqrt{f_c'}}{f_y}, \frac{200}{f_y}\right)\)

\(\frac{3\sqrt{4000}}{50000} = \frac{3 \times 63.246}{50000} \approx 0.003795\)

\(\frac{200}{50000} = 0.004\)

So, \(\rho_{min} = 0.004\).

For T-beams with flange in compression, \(A_{s,min} = \rho_{min} b_w d = 0.004 \times 11 \text{ in} \times 20 \text{ in} = 0.88 \text{ in}^2\).

Since \(A_s = 3.066 \text{ in}^2 > A_{s,min} = 0.88 \text{ in}^2\), the minimum steel requirement is satisfied.

Maximum steel area for tension-controlled section (\(\epsilon_t \ge 0.005\)):

The depth of equivalent rectangular stress block \(a = \frac{A_s f_y}{0.85 f_c' b_f} = \frac{3.066 \times 50000}{0.85 \times 4000 \times 66} = \frac{153300}{224400} \approx 0.683 \text{ in}\)

The depth of neutral axis \(c = a / \beta_1 = 0.683 \text{ in} / 0.85 \approx 0.804 \text{ in}\)

Tensile strain in steel, \(\epsilon_t = \epsilon_u \frac{d-c}{c} = 0.003 \frac{20 - 0.804}{0.804} = 0.003 \times 23.88 \approx 0.0716\)

Since \(\epsilon_t = 0.0716 \ge 0.005\), the section is tension-controlled, and the assumption of \(\phi = 0.9\) is valid. The calculated steel area is within acceptable limits.

The required tensile steel area is \(3.066 \text{ in}^2\).