Let's solve each part step-by-step, using the provided data and the formulas for concrete shear strength and tie spacing. ### **Given Data** - Column: 600 mm x 400 mm - Reinforcement: 8-32 mm dia bars - Tie diameter: 12 mm - Clear concrete cover: 40 mm - Concrete strength: \( f'_c = 28 \) MPa - Yield strength of steel: \( f_y = 415 \) MPa - Use \( \lambda = 1.0 \) (normal weight concrete) - Depth to centroid of tension steel: \( d = 532 \) mm (see diagram) - Forces: - **Positive x-direction:** \( M_u = 420 \) kN·m, \( V_u = 370 \) kN, \( N_u = 1320 \) kN - **Negative x-direction:** \( M_u = 420 \) kN·m, \( V_u = 370 \) kN, \( N_u = 450 \) kN --- ## **Step 1: Calculate the Concrete Shear Strength (\( V_{c} \))** The ACI simplified formula for concrete shear strength in columns is: \[ V_{c} = 0.17 \lambda \sqrt{f'_c} b_w d \] Where: - \( \lambda = 1.0 \) - \( f'_c = 28 \) MPa - \( b_w = \) width of web (400 mm) - \( d = 532 \) mm First, let's calculate \( \sqrt{f'_c} \): \[ \sqrt{28} = 5.2915 \] Now substitute values: \[ V_{c} = 0.17 \times 1.0 \times 5.2915 \times 400 \times 532 \] Calculate step-by-step: - \( 0.17 \times 5.2915 = 0.899555 \) - \( 0.899555 \times 400 = 359.822 \) - \( 359.822 \times 532 = 191,465 \) N \( = 191.47 \) kN ### **But we must consider axial load effect (ACI 318 simplified):** \[ V_{c} = 0.17 \lambda \sqrt{f'_c} b_w d \left[1 + \frac{N_u}{14 b_w d}\right] \] Where \( N_u \) is in Newtons (convert kN to N: multiply by \( 10^3 \)). Let’s do for both cases: --- ### **Positive x-direction (\( N_u = 1320 \) kN):** \[ \frac{N_u}{14 b_w d} = \frac{1,320,000}{14 \times 400 \times 532} \] \[ = \frac{1,320,000}{2,982,400} = 0.443 \] So: \[ V_{c} = 191.47 \times (1 + 0.443) = 191.47 \times 1.443 = 276.34 \text{ kN} \] Closest answer: **c. 271.14 kN** --- ### **Negative x-direction (\( N_u = 450 \) kN):** \[ \frac{N_u}{14 b_w d} = \frac{450,000}{2,982,400} = 0.151 \] So: \[ V_{c} = 191.47 \times (1 + 0.151) = 191.47 \times 1.151 = 220.34 \text{ kN} \] Closest answer: **a. 217.06 kN** --- ## **Step 2: Determine the Spacing of 12 mm dia. ties** Per ACI, maximum tie spacing: - Least of: \( 16 \times \) smallest longitudinal bar diameter, \( 48 \times \) tie diameter, or the least column dimension Calculate each: - \( 16 \times 32 = 512 \) mm - \( 48 \times 12 = 576 \) mm - Least column dimension = 400 mm Maximum spacing = **400 mm**, but per code, further limit to \( 0.75d \): - \( 0.75 \times 532 = 399 \) mm So, the provided options are less than this. The closest standard value (for seismic or high load) is often **250 mm**. Correct answer: **b. 250 mm** --- ## **Final Answers** **8. Concrete shear strength for positive x-direction:** **Answer: c. 271.14 kN** **9. Concrete shear strength for negative x-direction:** **Answer: a. 217.06 kN** **10. Spacing of 12 mm dia. ties:** **Answer: b. 250 mm** --- **Summary Table:** | Q# | Final Answer | |----|--------------| | 8 | c. 271.14 kN | | 9 | a. 217.06 kN | | 10 | b. 250 mm |

Let's analyze the problem step-by-step, providing detailed calculations for each part, based on the provided data and relevant codes (such as ACI 318). The problem involves a reinforced concrete column subjected to axial load, bending moment, and shear forces, with reinforcement details given. --- ## **Given Data Recap** - **Column dimensions:** 600 mm x 400 mm - **Reinforcement:** 8 bars of 32 mm diameter - **Tie diameter:** 12 mm - **Concrete cover:** 40 mm - **Concrete strength:** \( f'_c = 28\, \text{MPa} \) - **Steel yield strength:** \( f_y = 415\, \text{MPa} \) - **Type of concrete:** Normal weight (\( \lambda = 1.0 \)) - **Reinforcement centroid depth:** approximately \( d = 532\, \text{mm} \) (from the diagram) - **Forces in positive x-direction:** - \( M_u = 420\, \text{kN·m} \) - \( V_u = 370\, \text{kN} \) - \( N_u = 1320\, \text{kN} \) - **Forces in negative x-direction:** - \( M_u = 420\, \text{kN·m} \) - \( V_u = 370\, \text{kN} \) - \( N_u = 450\, \text{kN} \) --- ## **Part 1: Calculate the Concrete Shear Strength (\( V_c \))** ### **Step 1: Basic formula for \( V_c \) (from ACI 318 simplified expression):** \[ V_c = 0.17 \lambda \sqrt{f'_c} \, b_w \, d \times \left[ 1 + \frac{N_u}{14\,b_w\,d} \right] \] Where: - \( b_w = 400\, \text{mm} \) - \( d = 532\, \text{mm} \) - \( f'_c = 28\, \text{MPa} \) ### **Step 2: Calculate \( \sqrt{f'_c} \):** \[ \sqrt{28} \approx 5.2915 \] ### **Step 3: Calculate the base \( V_c \) (without axial load effect):** \[ V_{c,\,base} = 0.17 \times 1.0 \times 5.2915 \times 400 \times 532 \] Step-by-step: - \( 0.17 \times 5.2915 \approx 0.89955 \) - \( 0.89955 \times 400 = 359.82 \) - \( 359.82 \times 532 \approx 191,465\, \text{N} = 191.47\, \text{kN} \) --- ### **Step 4: Adjust for axial load \( N_u \)** **In the formula, the axial load effect:** \[ V_c = V_{c,\,base} \times \left[ 1 + \frac{N_u}{14\,b_w\,d} \right] \] Convert \( N_u \) to Newtons: - For positive \( N_u = 1320\, \text{kN} \): \[ \frac{N_u}{14\,b_w\,d} = \frac{1320 \times 10^3}{14 \times 400 \times 532} \] Calculate denominator: \[ 14 \times 400 \times 532 = 14 \times 400 = 5600; \quad 5600 \times 532 = 2,982,400 \] Calculate numerator: \[ 1320 \times 10^3 = 1,320,000 \] So: \[ \frac{1,320,000}{2,982,400} \approx 0.443 \] **Final \( V_c \) for positive \( N_u \):** \[ V_c = 191.47 \times (1 + 0.443) = 191.47 \times 1.443 \approx 276.34\, \text{kN} \] **Closest option from multiple choices:** **c. 271.14 kN** --- ### **For negative \( N_u = 450\, \text{kN} \):** Numerator: \[ 450 \times 10^3 = 450,000 \] Denominator remains same: \[ 2,982,400 \] Calculate: \[ \frac{450,000}{2,982,400} \approx 0.151 \] Thus, \[ V_c = 191.47 \times (1 + 0.151) = 191.47 \times 1.151 \approx 220.34\, \text{kN} \] Closest option: **a. 217.06 kN** --- ## **Part 2: Tie Spacing for 12 mm diameter ties** ### **Step 1: Determine maximum permissible spacing** Standard maximum spacing per ACI: - \( 16 \times \text{diameter of longitudinal bars} \) Given: - Longitudinal bars: 8 bars of 32 mm diameter Calculate: \[ 16 \times 32\, \text{mm} = 512\, \text{mm} \] Also, maximum spacing of ties: \[ 48 \times \text{tie diameter} = 48 \times 12 = 576\, \text{mm} \] And **limit by column dimension**: - Column dimension in the direction of loading: 400 mm Additionally, code limits the spacing to \( 0.75d \): \[ 0.75 \times 532\, \text{mm} \approx 399\, \text{mm} \] **So, the smallest of these is approximately 400 mm, but to be safe and considering standard practice, a spacing of 250 mm is typical and acceptable.** ### **Final choice:** **b. 250 mm** --- ## **Summary of Final Answers:** | Part | Final Answer | Explanation | |---------|------------------|--------------| | **8.** Concrete shear strength in positive x-direction | **c. 271.14 kN** | Calculated using \( V_c \) formula with axial load adjustment. | | **9.** Concrete shear strength in negative x-direction | **a. 217.06 kN** | Similar calculation with lower axial load. | | **10.** Tie spacing for 12 mm ties | **b. 250 mm** | Based on code limits and practical spacing considerations. | --- ## **Note:** - These calculations are based on simplified ACI formulas and typical assumptions. - Actual design should verify with detailed code provisions and consider factors like seismic, load combinations, and detailed reinforcement detailing. --- **Hope this detailed step-by-step solution clarifies the approach!**