Let's solve the problem step by step: --- ## **Given Data** - **Beam Section**: T-beam (see figure) - **Concrete Strength, \( f'_c \) = 400 psi** - **Steel Yield Strength, \( f_y \) = 60,000 psi** - **Top Flange Width, \( b_f \) = 36 in** - **Flange Thickness, \( t_f \) = 6 in** - **Web Width, \( b_w \) = 12 in** - **Overall Depth, \( h \) = 21.5 in** - **Distance to centroid of steel, \( d \) = 21.5 in** - **Number and Size of Bars**: 6 No. 7 bars (No. 7 bar area \( = .6 \) in² per bar) - **Total Area of Steel, \( A_s \) = \( 6 \times .6 = 3.6 \) in²** --- ## **Step 1: Calculate the Effective Depth (\( d \))** The centroid of the tension reinforcement is at 21.5 in from the top. --- ## **Step 2: Check If Compression Block is in Flange or Web** \[ a = \frac{A_s f_y}{.85 f'_c b} \] But first, assume the neutral axis is in the flange. The maximum depth of compression block that can fit in the flange is: \[ a_{flange} = t_f = 6 \text{ in} \] Calculate \( a \): \[ a = \frac{A_s f_y}{.85 f'_c b_f} \] \[ a = \frac{3.6 \times 60,000}{.85 \times 4,000 \times 36} \] \[ a = \frac{216,000}{122,400} \] \[ a = 1.77 \text{ in} < 6 \text{ in} \] So the compression block is **entirely in the flange**. --- ## **Step 3: Calculate \( \phi M_n \) (Negative Moment Capacity)** - \( \beta_1 \) for \( f'_c = 400 \) psi: \( \beta_1 = .85 \) (ACI Table) ### **a. Calculate \( c \) (neutral axis depth):** \[ a = \beta_1 c \implies c = \frac{a}{\beta_1} = \frac{1.77}{.85} = 2.08 \text{ in} \] ### **b. Calculate Nominal Moment Strength \( M_n \):** \[ M_n = A_s f_y \left[d - a/2\right] \] \[ M_n = 3.6 \times 60,000 \left[21.5 - \frac{1.77}{2}\right] \] \[ = 216,000 \left[21.5 - .885\right] \] \[ = 216,000 \times 20.615 \] \[ = 4,456,840 \text{ in-lb} \] Convert to ft-lb (divide by 12): \[ M_n = \frac{4,456,840}{12} = 371,403 \text{ ft-lb} \] Or to ft-kip (divide by 12,000): \[ M_n = \frac{4,456,840}{12,000} = 371.4 \text{ ft-kip} \] ### **c. Apply Strength Reduction Factor (\( \phi \)):** For flexure, \( \phi = .9 \): \[ \phi M_n = .9 \times 371.4 = 334.3 \text{ ft-kip} \] --- ## **Step 4: Check \( A_{s, min} \) (Minimum Steel Area)** \[ A_{s, min} = \text{max} \left[ 3 \sqrt{f'_c} b_w d / f_y, \; 200 b_w d / f_y \right] \] - \( b_w = 12 \) in - \( d = 21.5 \) in - \( f'_c = 400 \) psi - \( f_y = 60,000 \) psi Calculate both: ### **a.** \[ 3 \sqrt{400} \times 12 \times 21.5 / 60,000 \] \[ = 3 \times 63.25 \times 12 \times 21.5 / 60,000 \] \[ = 189.75 \times 12 \times 21.5 / 60,000 \] \[ = 2,277 \times 21.5 / 60,000 \] \[ = 48,955.5 / 60,000 = .816 \text{ in}^2 \] ### **b.** \[ 200 \times 12 \times 21.5 / 60,000 \] \[ = 2,400 \times 21.5 / 60,000 \] \[ = 51,600 / 60,000 = .86 \text{ in}^2 \] So, use the maximum: \[ A_{s, min} = .86 \text{ in}^2 \] Available \( A_s = 3.6 \) in² \( > .86 \) in² → **OK** --- ## **Summary of Answers** ### **1. Negative-Moment Capacity** \[ \boxed{\phi M_n = 334.3 \text{ ft-kip}} \] ### **2. Minimum Steel Area Required** \[ \boxed{A_{s, min} = .86 \text{ in}^2} \] ### **3. Provided Steel Area** \[ A_s = 3.6 \text{ in}^2 \text{ (OK, greater than } A_{s, min}) \] --- ### **Explanation** - The compression block is fully within the flange (since \( a < t_f \)). - The negative moment capacity is computed using the flexural strength formula for singly-reinforced sections. - Minimum steel area is checked using ACI requirements, and the section is adequately reinforced. --- **If you need a detailed sketch or have further sub-questions, let me know!**