Let's break down the solution step by step for each part: --- ## **Given Data** - **Beam width, \( b \) = 18"** - **Overall depth, \( h \) = 30"** - **Distance to centroid of steel, \( d \) = 27.5"** - **Concrete compressive strength, \( f'_c = 5000 \) psi** - **Steel yield strength, \( f_y = f_s = 60,000 \) psi** - **Number of bars = 6, bar size = #8** - **Area of #8 bar, \( A_{bar} = 0.79 \) in²** - **Total steel area, \( A_s = 6 \times 0.79 = 4.74 \) in²** --- ### **a. The cracked transformed moment of inertia (\( I_{cr} \)):** #### **Step 1:** Find the modular ratio \( n \): \[ n = \frac{E_s}{E_c} \] Where: - \( E_s \) (modulus of steel) ≈ 29,000,000 psi - \( E_c \) (modulus of concrete) ≈ \( 57,000 \sqrt{f'_c} \) - \( f'_c = 5000 \) psi \[ E_c = 57,000 \times \sqrt{5000} \approx 4,030,000 \text{ psi} \] \[ n = \frac{29,000,000}{4,030,000} \approx 7.2 \] #### **Step 2:** Assume the NA depth = \( x \) from top, set up equilibrium - Compression block: \( b x \) - Tension steel: \( nA_s \) located \( d \) from top \[ b x = nA_s \] \[ 18x = 7.2 \times 4.74 \] \[ 18x = 34.13 \implies x = 1.90 \text{ in} \] #### **Step 3:** Calculate \( I_{cr} \): \[ I_{cr} = I_{concrete,above\ NA} + nA_s (d - x/2)^2 \] - \( I_{concrete,above\ NA} = \frac{b x^3}{3} \) - \( y_s = d - x/2 \) \[ I_{concrete,above\ NA} = \frac{18 \times (1.90)^3}{3} = \frac{18 \times 6.859}{3} = 41.15 \text{ in}^4 \] \[ y_s = 27.5 - \frac{1.90}{2} = 26.55 \text{ in} \] \[ I_{steel} = nA_s (y_s)^2 = 34.13 \times (26.55)^2 = 34.13 \times 705.6 = 24,097 \text{ in}^4 \] \[ I_{cr} = 41.15 + 24,097 = 24,138 \text{ in}^4 \] --- ### **b. Ratio of cracked to uncracked moment of inertia** Uncracked (gross) \( I_g \): \[ I_g = \frac{b h^3}{12} = \frac{18 \times 30^3}{12} = \frac{18 \times 27,000}{12} = 40,500 \text{ in}^4 \] \[ \text{Ratio} = \frac{I_{cr}}{I_g} = \frac{24,138}{40,500} = 0.60 \] --- ### **c. Moment for max concrete stress = \( 0.45f'_c \)** \[ f_t = \frac{M y}{I_g} \] Here, \( y = h/2 = 15 \) in Set \( f_t = 0.45 \times 5000 = 2250 \) psi: \[ 2250 = \frac{M \times 15}{40,500} \implies M = \frac{2250 \times 40,500}{15} = 6,075,000 \text{ in-lb} = 506 \text{ kip-in} = 42.2 \text{ kip-ft} \] --- ### **d. Steel force at moment in (c):** \[ C = T = \frac{0.85 f'_c b a}{f_y} \] But for elastic, use: \[ T = \frac{M}{jd} \] For elastic, \( j \approx 0.9 \): \[ jd = 0.9 \times 27.5 = 24.75 \text{ in} \] \[ T = \frac{6,075,000}{24.75} = 245,455 \text{ lb} = 245.5 \text{ kip} \] --- ### **e. Yield Moment (\( M_y \))** \[ M_y = A_s f_y (d - a/2) \] Where \( a = \frac{A_s f_y}{0.85 f'_c b} \) \[ a = \frac{4.74 \times 60,000}{0.85 \times 5,000 \times 18} = \frac{284,400}{76,500} = 3.72 \text{ in} \] \[ M_y = 4.74 \times 60,000 \times (27.5 - 3.72/2) = 284,400 \times 25.64 = 7,293,216 \text{ in-lb} = 608 \text{ kip-ft} \] --- ### **f. Nominal Moment (\( M_n \))** Same as \( M_y \) if singly reinforced, under-reinforced: \[ M_n = A_s f_y (d - a/2) = 7,293,216 \text{ in-lb} = 608 \text{ kip-ft} \] --- ### **g. Max compressive strain in concrete at \( M_n \):** For flexure, \( \epsilon_{cu} = 0.003 \) (code value for ultimate) So, at nominal moment, \( \epsilon_{c,max} = 0.003 \). --- ### **h. Tensile strain in steel at \( M_n \):** From similar triangles: \[ \frac{\epsilon_s}{\epsilon_{cu}} = \frac{d - c}{c} \] Where \( c = a/\beta_1 \), with \( \beta_1 = 0.85 \) for \( f'_c \leq 4000 \) psi, but for 5000 psi, \( \beta_1 = 0.85 - 0.05 \times (5000-4000)/1000 = 0.8 \): \[ c = a/0.8 = 3.72/0.8 = 4.65 \text{ in} \] \[ \epsilon_s = 0.003 \times \frac{27.5 - 4.65}{4.65} = 0.003 \times 4.91 = 0.0147 \] --- ## **Summary Table** | Item | Value | |------|-------| | a | \( I_{cr} = 24,138 \) in⁴ | | b | Ratio = 0.60 | | c | \( M = 42.2 \) kip-ft | | d | Steel force = 245.5 kip | | e | \( M_y = 608 \) kip-ft | | f | \( M_n = 608 \) kip-ft | | g | \( \epsilon_{c,max} = 0.003 \) | | h | \( \epsilon_{s} = 0.0147 \) | If you need step-by-step calculations for any part, feel free to ask!