Let's solve the **nominal moment strength** (\(\phi M_n\)) for the given reinforced concrete section per ACI 318. We’ll use these steps for both Q1 and Q2: --- ## **Step 1: Section and Reinforcement Details** - **Width, \(b\):** 14 in - **Total depth, \(h\):** in - **Top cover:** 1.5 in - **Tension reinforcement:** 6 - #9 bars (bottom) - **Compression reinforcement:** 2 - #7 bars (top) ### **Bar Areas** - #9 bar: \(A_{bar} = 1.00\, \text{in}^2\) - #7 bar: \(A_{bar} = .60\, \text{in}^2\) ### **Location of Bars** - **Tension bar centroid (\(d\)):** Bottom cover = \(1.5\) in \(d = h - \text{cover} = 30 - 1.5 = 28.5\) in (Assuming bar diameter is small; for exact, subtract half the bar diameter.) - **Compression bar centroid (\(d'\)):** \(d' = 1.5\) in --- ## **Step 2: Calculate Areas of Steel** - **Tension steel (\(A_s\)):** \(6 \times 1.00 = 6.00\) in² - **Compression steel (\(A_s'\)):** \(2 \times .60 = 1.20\) in² --- ## **Step 3: Material Properties** **Q1:** - \(f_y = 60 \text{ ksi}\) - \(f'_c = 4 \text{ ksi}\) **Q2:** - \(f_y = 80 \text{ ksi}\) - \(f'_c = 4 \text{ ksi}\) --- ## **Step 4: Calculate \(\phi M_n\) for Q1** ### **A) Find Neutral Axis Depth (\(c\))** Use equilibrium (compression = tension): \[ .85 f'_c b a + A_s' f_y = A_s f_y \] Where \(a = \beta_1 c\), and \(\beta_1 = .85\) for \(f'_c = 4\) ksi. \[ .85 \times 4 \times 14 \times a + 1.2 \times 60 = 6 \times 60 \] \[ .85 \times 4 \times 14 \times a + 72 = 360 \] \[ .85 \times 4 \times 14 \times a = 288 \] \[ .85 \times 4 \times 14 = 47.6 \] \[ 47.6a = 288 \] \[ a = \frac{288}{47.6} = 6.05\,\text{in} \] Now, \(c = \frac{a}{\beta_1} = \frac{6.05}{.85} = 7.12\,\text{in}\) --- ### **B) Nominal Moment (\(M_n\))** \[ M_n = A_s f_y (d - a/2) + A_s' f_y (d - d') \] But per ACI, the moment is calculated by taking moments about the centroid of compression force: \[ M_n = A_s f_y (d - a/2) + A_s' f_y (d - d') \] (Here, both steel layers use \(f_y\) since both are yielding; typically, compression steel may not yield, but with high ratio, assume yielding.) Plug in values: - \(d = 28.5\) in - \(d' = 1.5\) in - \(A_s = 6.\) in² - \(A_s' = 1.2\) in² - \(f_y = 60\) ksi - \(a = 6.05\) in Calculate: \[ M_n = 6. \times 60 \times (28.5 - 6.05/2) + 1.2 \times 60 \times (28.5 - 1.5) \] \[ = 360 \times (28.5 - 3.025) + 72 \times 27 \] \[ = 360 \times 25.475 + 72 \times 27 \] \[ = 9171 + 1944 \] \[ = 11,\!115\, \text{in-kips} \] Convert to ft-kips: \[ M_n = \frac{11,\!115}{12} = 926.25\, \text{ft-kips} \] ### **C) Apply \(\phi\) (Strength Reduction Factor)** For flexure, \(\phi = .9\): \[ \phi M_n = .9 \times 926.25 = 833.6\, \text{ft-kips} \] --- ## **Step 5: Calculate \(\phi M_n\) for Q2 (\(f_y = 80\,\text{ksi}\))** Repeat the same steps with \(f_y = 80\) ksi: \[ .85 \times 4 \times 14 \times a + 1.2 \times 80 = 6 \times 80 \] \[ 47.6a + 96 = 480 \] \[ 47.6a = 384 \] \[ a = \frac{384}{47.6} = 8.07\,\text{in} \] \[ c = \frac{8.07}{.85} = 9.49\,\text{in} \] \[ M_n = 6. \times 80 \times (28.5 - 8.07/2) + 1.2 \times 80 \times (28.5 - 1.5) \] \[ = 480 \times (28.5 - 4.035) + 96 \times 27 \] \[ = 480 \times 24.465 + 96 \times 27 \] \[ = 11,\!743.2 + 2,\!592 \] \[ = 14,\!335.2\, \text{in-kips} \] \[ M_n = \frac{14,\!335.2}{12} = 1,\!194.6\, \text{ft-kips} \] \[ \phi M_n = .9 \times 1,\!194.6 = 1,\!075.1\, \text{ft-kips} \] --- ## **Step 6: **Final Answers** ### **Q1 Answer:** \[ \boxed{\phi M_n = 834\, \text{ft-kips}} \] ### **Q2 Answer:** \[ \boxed{\phi M_n = 1,\!075\, \text{ft-kips}} \] --- ### **Summary Table** | Question | \(f_y\) (ksi) | \(\phi M_n\) (ft-kips) | |----------|---------------|-----------------------| | Q1 | 60 | 834 | | Q2 | 80 | 1,075 | --- Let me know if you have any questions or need the calculations shown in more detail!