Let's solve the problem step by step: ### Given Data - **Cross section:** 18" wide, 30" deep (with tension steel at 27.5" from the top) - \( f'_c = 5000 \) psi (concrete compressive strength) - \( f_s = 60 \) ksi (steel yield strength) = 60,000 psi - \( A_s = 6 \times \text{area of #8 bar} \) - Area of #8 bar: \( A_{#8} = 0.79 \) in² (standard value) - So, \( A_s = 6 \times 0.79 = 4.74 \) in² --- ### **a. Cracked Transformed Moment of Inertia** #### **Step 1: Modulus Ratio** \[ n = \frac{E_s}{E_c} \] - \( E_s = 29,000,000 \) psi (steel) - \( E_c \approx 57,000 \sqrt{f'_c} = 57,000 \times \sqrt{5000} = 4,031,129 \) psi \[ n = \frac{29,000,000}{4,031,129} \approx 7.2 \] #### **Step 2: Locate the Neutral Axis (\(x\))** Let \( x \) be the depth of the neutral axis from the top. The equivalent area of steel is \( n A_s \). Sum of moments about NA: \[ 18x \cdot \frac{x}{2} + nA_s \cdot (d - x) = 0 \] But for equilibrium: \[ 18x = nA_s \cdot \frac{d-x}{x} \] But let's set up the centroid equation: Let: - \( b = 18 \) in - \( d = 27.5 \) in Sum of compressive force = sum of tensile force: \[ C = T \implies 0.85 f'_c b x = nA_s (d-x)/x \] But for the transformed section, the centroid is found by: \[ b x = nA_s \] But for the centroid (\( \bar{y} \)), set up the equivalent area: \[ A_c \cdot y_c + nA_s \cdot y_s = (A_c + nA_s) \cdot \bar{y} \] Where: - \( A_c = b x \) - \( y_c = x/2 \) - \( y_s = d \) - \( nA_s \) is at \( d \) from the top. Set sum of moments = 0 about NA: \[ b x \cdot \frac{x}{2} = nA_s (d - x) \] \[ 18x \cdot \frac{x}{2} = 7.2 \cdot 4.74 \cdot (27.5 - x) \] \[ 9x^2 = 34.13(27.5 - x) \] \[ 9x^2 + 34.13x - 938.58 = 0 \] Solve the quadratic: \[ x = \frac{-34.13 \pm \sqrt{34.13^2 + 4 \cdot 9 \cdot 938.58}}{2 \cdot 9} \] \[ x = \frac{-34.13 \pm \sqrt{1165 + 33788}}{18} \] \[ x = \frac{-34.13 \pm \sqrt{34953}}{18} \] \[ \sqrt{34953} \approx 187 \] \[ x = \frac{-34.13 + 187}{18} \approx \frac{152.87}{18} \approx 8.49\, \text{in} \] #### **Step 3: Transformed Moment of Inertia** \[ I_{\text{cr}} = \frac{1}{12} b x^3 + nA_s (d-x)^2 \] \[ I_{\text{cr}} = \frac{1}{12} \cdot 18 \cdot (8.49)^3 + 7.2 \cdot 4.74 \cdot (27.5 - 8.49)^2 \] First term: \[ (8.49)^3 = 611.65 \] \[ \frac{1}{12} \cdot 18 \cdot 611.65 = 1.5 \cdot 611.65 = 917.48 \] Second term: \[ (27.5 - 8.49) = 19.01 \] \[ (19.01)^2 = 361.39 \] \[ 7.2 \cdot 4.74 = 34.13 \] \[ 34.13 \cdot 361.39 = 12,337.17 \] So, \[ I_{\text{cr}} = 917.48 + 12,337.17 = \boxed{13,254.65\, \text{in}^4} \] --- ### **b. Ratio of Cracked to Uncracked Moment of Inertia** Uncracked (gross) moment of inertia: \[ I_g = \frac{1}{12} b h^3 = \frac{1}{12} \cdot 18 \cdot 30^3 = 1.5 \cdot 27,000 = 40,500 \, \text{in}^4 \] \[ \text{Ratio} = \frac{I_{\text{cr}}}{I_g} = \frac{13,254.65}{40,500} \approx \boxed{0.327} \] --- ### **c. Moment for Max Concrete Stress = \(0.45f'_c\)** The extreme fiber stress in concrete: \[ f_c = \frac{M y}{I_g} \] where \( y = 15 \) in (top fiber). Set \( f_c = 0.45 f'_c = 0.45 \times 5000 = 2250 \) psi. \[ 2250 = \frac{M \cdot 15}{40,500} \implies M = \frac{2250 \times 40,500}{15} = 6,075,000 \, \text{in-lb} \] \[ = \frac{6,075,000}{12} = \boxed{506,250\, \text{ft-lb} = 506.3\, \text{kip-ft}} \] --- ### **d. Steel Force at Moment from Part (c)** First, find the strain in steel, assuming section is uncracked (elastic), and use transformed section. Steel force: \[ f_s = n f_c \] \[ f_s = 7.2 \times 2250 = 16,200 \, \text{psi} \] \[ F_s = f_s \cdot A_s = 16,200 \times 4.74 = 76,788\, \text{lb} = \boxed{76.8\, \text{kips}} \] --- ### **e. Yield Moment (\(M_y\))** Steel yields when \( f_s = 60,000 \) psi. Strain compatibility: \[ \epsilon_s = \frac{d - c}{c} \epsilon_{cu} \] But for linear, when \( f_s = 60,000 \), use the moment with the cracked, transformed section: \[ f_s = E_s \epsilon_s \] \[ \epsilon_s = \frac{f_s}{E_s} = \frac{60,000}{29,000,000} = 0.00207 \] Assume strain varies linearly from zero at NA to max at steel. NA is at \( x = 8.49 \) in, steel is at \( d = 27.5 \) in. \[ \epsilon_{top}/\epsilon_s = x/(d-x) \] \[ \epsilon_{top} = \epsilon_s \cdot \frac{x}{d-x} = 0.00207 \times \frac{8.49}{19.01} = 0.00207 \times 0.447 = 0.000925 \] \[ f_c = E_c \cdot \epsilon_{top} = 4,031,129 \cdot 0.000925 = 3,728\, \text{psi} \] Moment at yield: \[ M_y = \frac{f_c I_{\text{cr}}}{y} \] Here, \( y = d - x = 19.01 \) in \[ M_y = \frac{3,728 \times 13,254.65}{19.01} = \frac{49,426,174}{19.01} = 2,600,000\, \text{in-lb} \] \[ = \frac{2,600,000}{12} = 216,667\, \text{ft-lb} = \boxed{216.7\, \text{kip-ft}} \] --- ### **f. Nominal Moment (\(M_n\))** Use ACI 318: \[ M_n = A_s f_y (d - a/2) \] Where \( a = \frac{A_s f_y}{0.85 f'_c b} \) \[ a = \frac{4.74 \times 60,000}{0.85 \times 5,000 \times 18} = \frac{284,400}{76,500} = 3.72\, \text{in} \] \[ M_n = 4.74 \times 60,000 \times (27.5 - 3.72/2) = 284,400 \times (27.5 - 1.86) = 284,400 \times 25.64 = 7,296,816\, \text{in-lb} \] \[ = 608,068\, \text{ft-lb} = \boxed{608.1\, \text{kip-ft}} \] --- ### **g. Max Compressive Strain in Concrete for \(M_n\)** \[ \epsilon_{cu} = 0.003 \] By ACI, maximum strain at nominal moment is **0.003**. --- ### **h. Tensile Strain in Steel for \(M_n\)** Depth to NA at ultimate: \[ c = \frac{a}{\beta_1} \] ACI: For \( f'_c = 5000 \), \( \beta_1 = 0.85 - 0.05 \times ((f'_c - 4000)/1000) = 0.85 - 0.05 \times (1) = 0.8 \) \[ c = \frac{3.72}{0.8} = 4.65\, \text{in} \] Tensile strain in steel: \[ \epsilon_s = \epsilon_{cu} \frac{d-c}{c} = 0.003 \frac{27.5 - 4.65}{4.65} = 0.003 \times 4.935 = 0.0148 \] --- ## **Summary Table** | Part | Answer | |------|--------| | a | \( I_{\text{cr}} = 13,254.65\, \text{in}^4 \) | | b | Ratio = 0.327 | | c | \( M = 506.3\, \text{kip-ft} \) | | d | \( F_s = 76.8\, \text{kips} \) | | e | \( M_y = 216.7\, \text{kip-ft} \) | | f | \( M_n = 608.1\, \text{kip-ft} \) | | g | \( \epsilon_{cu} = 0.003 \) | | h | \( \epsilon_s = 0.0148 \) | If you want detailed calculation for any step or a check on a specific part, let me know!