# Step-by-Step Solution: Rectangular Reinforced Concrete Beam Design ## **Given Data** - **Factored Moment (Mu):** 300 kip-ft = 300 × 12 = **360 kip-in** - **Concrete compressive strength (fc'):** 400 psi - **Steel yield strength (fy):** 60,000 psi - **Maximum beam width (b):** 10 in - **Maximum overall depth (h):** 22 in - **Assume εt = .004** - **Strength reduction factor (ϕ):** For εt ≥ .005, ϕ = .9 (with εt = .004, ϕ ≈ .85, but let's use ϕ = .9 conservatively) --- ## **Step 1: Convert Mu to Inch-pounds** \[ Mu = 300 \text{ kip-ft} = 300 \times 12 = 360 \text{ kip-in} = 3,600,000 \text{ lb-in} \] --- ## **Step 2: Choose b and h** - Maximum **b = 10 in** - Maximum **h = 22 in** - Assume cover and bar diameter: effective depth \( d = h - \text{cover} - \frac{\text{bar diameter}}{2} \approx 22 - 2.5 = 19.5 \) in --- ## **Step 3: Calculate Required Nominal Moment Capacity (\( M_n \))** \[ \phi M_n \geq M_u \] \[ M_n = \frac{M_u}{\phi} \] \[ M_n = \frac{3,600,000}{.9} = 4,000,000 \text{ lb-in} \] \[ M_n = \frac{4,000,000}{12} = 333,333 \text{ lb-ft} = 333.3 \text{ kip-ft} \] --- ## **Step 4: Assume Section is Singly Reinforced** We use the ACI equation: \[ M_n = A_s f_y \left( d - a/2 \right) \] where \( a = \frac{A_s f_y}{.85 f'_c b} \). --- ## **Step 5: Estimate Depth Using ACI Simplified Formula** A more direct approach: \[ M_n = .9 \cdot .85 f'_c b a (d - a/2) \] But first, let's estimate \( a \): Let’s assume \( a \ll d \Rightarrow M_n \approx .9 \cdot .85 f'_c b d^2 \rho_{bal} \) (for a quick estimate). But let's use the *balanced steel ratio* for εt = .004: \[ \epsilon_t = .004 = \frac{(d-c)}{c} \cdot .003 \Rightarrow c = \frac{.003}{.003 + .004} d = .43 d \] \[ a = \beta_1 c \] where \( \beta_1 = .85 \) for \( f'_c = 400 \) psi \[ a = .85 \times .43 d = .3655 d \] --- ## **Step 6: Solve for Effective Depth (d)** \[ M_n = .85 f'_c b a (d - a/2) \] \[ M_n = .85 f'_c b (a) (d - a/2) \] Plug \( a = .3655d \): \[ M_n = .85 f'_c b (.3655d) (d - .3655d/2) \] \[ = .85 f'_c b (.3655d) (d - .18275d) \] \[ = .85 f'_c b (.3655d) (.81725d) \] \[ = .85 \times 400 \times 10 \times .3655 \times .81725 d^2 \] \[ = 340 \times 10 \times .29894 d^2 \] \[ = 34,000 \times .29894 d^2 \] \[ = 10,164 d^2 \] Set \( M_n = 4,000,000 \) lb-in: \[ 4,000,000 = 10,164 d^2 \] \[ d^2 = \frac{4,000,000}{10,164} = 393.7 \] \[ d = \sqrt{393.7} = 19.85 \text{ in} \] This is slightly more than the assumed maximum \( d \approx 19.5 \) in (from earlier). Let's check using **maximum possible \( d \) for \( h = 22 \) in**: Assume cover + stirrup + half bar diameter ≈ 2.5 in: \[ d = h - 2.5 = 22 - 2.5 = 19.5 \text{ in} \] --- ## **Step 7: Recalculate Actual Capacity with \( d = 19.5 \) in** \[ M_n = 10,164 \times (19.5)^2 = 10,164 \times 380.25 = 3,867,471 \text{ lb-in} \] \[ \phi M_n = .9 \times 3,867,471 = 3,480,724 \text{ lb-in} = 290,060 \text{ ft-lb} = 241.7 \text{ kip-ft} \] This is **less** than required (\( 300 \) kip-ft). **Conclusion:** The section with maximum width and depth is slightly understrength. We must use compression reinforcement (**doubly reinforced section**) or increase beam dimensions. --- ## **Step 8: Doubly Reinforced Design** We need to provide additional steel in compression. ### **a. Calculate the moment that can be resisted by a singly reinforced section (\( M_{u1} \)):** From above, \( \phi M_{n, singly} = 241.7 \) kip-ft. ### **b. Additional moment to be resisted (\( M_{u2} \)):** \[ M_{u2} = M_u - \phi M_{n, singly} = 300 - 241.7 = 58.3 \text{ kip-ft} = 699,600 \text{ lb-in} \] ### **c. Compression steel required:** \[ M_{u2} = \phi A_s' f_y (d - d') \] Assume \( d' = 2.5 \) in. \[ 699,600 = .9 \times A_s' \times 60,000 \times (19.5 - 2.5) \] \[ 699,600 = .9 \times 60,000 \times 17 \times A_s' \] \[ 699,600 = 918,000 \times A_s' \] \[ A_s' = \frac{699,600}{918,000} = .762 \text{ in}^2 \] ### **d. Total Tension steel:** - Tension steel for singly reinforced section: \[ A_{s1} = \frac{.85 f'_c b a}{f_y} \] Find \( a \): \[ a = .3655 \times 19.5 = 7.13 \text{ in} \] \[ A_{s1} = \frac{.85 \times 400 \times 10 \times 7.13}{60,000} = \frac{242,420}{60,000} = 2.914 \text{ in}^2 \] - Tension steel for compression reinforcement: \[ A_{s2} = A_s' = .762 \text{ in}^2 \] - **Total \( A_s = A_{s1} + A_{s2} = 2.914 + .762 = 3.676 \) in²** --- ## **Step 9: Final Selections** - **Width (\( b \)):** 10 in - **Overall depth (\( h \)):** 22 in - **Effective depth (\( d \)):** 19.5 in - **Tension steel (\( A_s \)):** 3.68 in² - **Compression steel (\( A_s' \)):** .76 in² --- ## **Step 10: Bar Selection** - For \( A_s = 3.68 \) in²: Use **#8 bars** (\( A = .79 \) in² each) - Number required: \( 3.68 / .79 = 4.66 \) ≈ **5 #8 bars** - For \( A_s' = .76 \) in²: Use **#6 bar** (\( A = .44 \) in² each) - Number required: \( .76 / .44 = 1.73 \) ≈ **2 #6 bars** --- ## **Summary Table** | Parameter | Value | |---------------------|------------------| | Beam width (\(b\)) | 10 in | | Overall depth (\(h\)) | 22 in | | Effective depth (\(d\)) | 19.5 in | | Tension steel (\(A_s\)) | 3.68 in² (5 #8 bars) | | Comp. steel (\(A_s'\)) | .76 in² (2 #6 bars) | --- ## **Final Answer** ### **Select:** - **Width:** \( \boxed{10\ \text{in}} \) - **Depth:** \( \boxed{22\ \text{in}} \) - **Effective Depth:** \( \boxed{19.5\ \text{in}} \) - **Tension Steel (\( A_s \)):** \( \boxed{3.68\ \text{in}^2} \) (**5 #8 bars**) - **Compression Steel (\( A_s' \)):** \( \boxed{.76\ \text{in}^2} \) (**2 #6 bars**) --- ### **All architectural and strength requirements are satisfied with this design.**