Let's break down the solution for the **T-beam section** as shown: ## **Given Data** - **Concrete compressive strength,** \( f'_c = 300 \) psi - **Number of rebars:** 8 bars of No. 9 (\( A_s = 8 \times 1.00 = 8.00 \text{ in}^2 \)) - (Area of No. 9 rebar = 1.00 in²) - **Yield strength of rebar,** \( f_y = 60,000 \) psi (Grade 60) - **Dimensions:** - Flange width, \( b_f = 34 \) in (16 + 18) - Flange thickness, \( h_f = 6 \) in - Web width, \( b_w = 16 \) in - Total height, \( h = 48 \) in - Distance from top to centroid of rebars, \( d = 48 - 3 - .5 \times 1.128 \approx 44.44 \) in (diameter of #9 bar ≈ 1.128 in) --- ## **Step 1: Calculate Depth to Steel (\(d\))** - **Effective depth (\( d \)):** \[ d = 48 - 3 - .5 \times 1.128 \approx 44.44 \text{ in} \] --- ## **Step 2: Check Flange or Web Controls** - **Neutral axis depth (\( c \)) will be checked later**. Assume flange is effective (i.e., neutral axis in flange). --- ## **Step 3: Calculate the Ultimate Strength of Concrete** ### **Assume Neutral Axis in Flange** - **Maximum compressive force in concrete:** \[ C = .85 f'_c b_f a \] where \( a = \beta_1 c \), \( \beta_1 = .85 \) for \( f'_c \leq 400 \) psi. - **Steel tension:** \[ T = A_s f_y \] \[ T = 8 \times 1.00 \times 60,000 = 480,000 \text{ lbs} \] ### **Equilibrium: \( T = C \)** \[ T = .85 f'_c b_f a \] \[ 480,000 = .85 \times 300 \times 34 \times a \] \[ 480,000 = 86,700 a \] \[ a = \frac{480,000}{86,700} = 5.54 \text{ in} \] - **Check if neutral axis is in flange:** - \( a = 5.54 \) in \( < 6 \) in (flange thickness) → **OK!** --- ## **Step 4: Calculate Neutral Axis Depth (\( c \))** \[ a = \beta_1 c = .85 c \] \[ c = \frac{a}{.85} = \frac{5.54}{.85} = 6.52 \text{ in} \] --- ## **Step 5: Nominal Moment Capacity (\( M_n \))** - **Distance from compression resultant to tension resultant:** \[ \text{Lever arm} = d - \frac{a}{2} = 44.44 - \frac{5.54}{2} = 44.44 - 2.77 = 41.67 \text{ in} \] - **Nominal moment:** \[ M_n = T \times (\text{lever arm}) \] \[ M_n = 480,000 \times 41.67 = 20,001,600 \text{ in-lb} \] \[ M_n = \frac{20,001,600}{12} = 1,666,800 \text{ ft-lb} \] \[ M_n = \frac{1,666,800}{12} = 138,900 \text{ ft-kip} \] - **Factored moment (\( \phi M_n \)), \( \phi = .9 \) for flexure** \[ \phi M_n = .9 \times 1,666,800 = 1,500,120 \text{ ft-lb} \] \[ \phi M_n = \frac{1,500,120}{12} = 125,010 \text{ ft-kip} \] --- ## **Step 6: Limiting Curvature (\( \phi \))** - **Strain in steel at ultimate (\( \epsilon_{su} \)):** - For tension-controlled section, \( \epsilon_{su} \geq .005 \) - **Neutral axis depth \( c \):** \[ \epsilon_{su} = \frac{d-c}{c} \times .003 \] \[ \epsilon_{su} = \frac{44.44 - 6.52}{6.52} \times .003 = \frac{37.92}{6.52} \times .003 \approx .0174 \] - **Curvature (\( \phi \)):** \[ \phi = \frac{\epsilon_{cu}}{c} = \frac{.003}{6.52} = .000460 \text{ in}^{-1} \] --- ## **Summary of Answers** ### **1. Moment Capacity (\( \phi M_n \)):** \[ \boxed{1,500,120 \text{ in-lb} \ (\approx 125 \text{ ft-kip})} \] ### **2. Limiting Curvature (\( \phi \)):** \[ \boxed{.000460 \text{ in}^{-1}} \] --- ## **Assumptions/Notes** - All rebar yields; section is tension-controlled. - Compression block entirely within the flange. - Ignore self-weight for this calculation. - Units are consistent. If you need a more detailed check for flange/web control or interaction with code provisions, let me know!