Let's solve the design problem step by step for the **column CD** in the given frame. --- ## **Step 1: Analyze Loads on Column CD** Column CD is subjected to: - Vertical load from node C: **200 kN** - Horizontal load at mid-height: **80 kN** - Column height: **3 m** Column is **fixed at C** (monolithic), **free at D**. --- ## **Step 2: Calculate Forces and Moments on Column CD** ### **A. Axial Force in CD** The direct vertical load at C is **200 kN** (downward). Assume self-weight is negligible for preliminary design. ### **B. Moment Due to Horizontal Load (80 kN)** - The 80 kN force acts **3 m above D** (mid-height). - This produces a moment at C (fixed support): \[ M_{C} = 80\, \text{kN} \times 3\, \text{m} = 240\, \text{kN}\cdot\text{m} \] ### **C. Shear in Column** - Shear at C due to 80 kN = **80 kN** (horizontal). - Shear at D = **0** (free end). --- ## **Step 3: Design the Cross-Section** ### **A. Load Combinations** The column is subjected to: - **Axial load, \( N = 200\,\text{kN} \)** - **Moment, \( M = 240\,\text{kN}\cdot\text{m} \)** - **Shear, \( V = 80\,\text{kN} \)** ### **B. Assume Square Cross-Section** Let side = \( b \) (in mm) ### **C. Check for Short/Slender Column** Effective length, \( l = 3\,\text{m} \) If \( l/b < 12 \), short column. Assume \( b = 300\,\text{mm} \) (trial), \( l/b = 3000/300 = 10 < 12 \) ⇒ **Short column** --- ## **Step 4: Calculate Required Area of Steel** ### **A. Axial + Moment Design (Interaction)** Use **Limit State Method** (per codes like Eurocode 2 or ACI): #### **Factored Loads** Assume partial safety factors: - Load factor = 1.5 \[ N_u = 1.5 \times 200 = 300\,\text{kN} \] \[ M_u = 1.5 \times 240 = 360\,\text{kN}\cdot\text{m} \] #### **Section Capacity** - Assume \( f_{cu} = 25\,\text{MPa} \), \( f_y = 415\,\text{MPa} \) - Use typical cover = 40 mm, effective depth \( d = 300-40 = 260\,\text{mm} \) #### **Interaction Check** The design is for combined axial load and bending: \[ \frac{N_u}{0.45f_{ck}A_g + 0.75f_yA_{sc}} + \frac{M_u}{0.15f_{ck}A_g d} \leq 1 \] Where: - \( A_g = b^2 \) - \( A_{sc} = \) area of steel (to be found) #### **Try with \( b = 300\,\text{mm} \), \( d = 260\,\text{mm} \)** \[ A_g = 300 \times 300 = 90,000\,\text{mm}^2 \] \[ 0.45f_{ck}A_g = 0.45 \times 25 \times 90,000 = 1,012,500\,\text{N} \] \[ 0.15f_{ck}A_gd = 0.15 \times 25 \times 90,000 \times 0.26 = 87,750\,\text{Nm} = 87.75\,\text{kN}\cdot\text{m} \] \[ \frac{N_u}{1,012,500} + \frac{M_u}{87,750} = \frac{300,000}{1,012,500} + \frac{360,000}{87,750} = 0.296 + 4.1 = 4.4 > 1 \] So, need a **larger section**. --- ### **Try \( b = 400\,\text{mm} \), \( d = 360\,\text{mm} \)** \[ A_g = 400 \times 400 = 160,000\,\text{mm}^2 \] \[ 0.45f_{ck}A_g = 0.45 \times 25 \times 160,000 = 1,800,000\,\text{N} \] \[ 0.15f_{ck}A_gd = 0.15 \times 25 \times 160,000 \times 0.36 = 216,000\,\text{Nm} \] \[ \frac{N_u}{1,800,000} + \frac{M_u}{216,000} = \frac{300,000}{1,800,000} + \frac{360,000}{216,000} = 0.167 + 1.67 = 1.84 > 1 \] Still, it's **overstressed** for the moment. --- ### **Try \( b = 500\,\text{mm} \), \( d = 460\,\text{mm} \)** \[ A_g = 500 \times 500 = 250,000\,\text{mm}^2 \] \[ 0.45f_{ck}A_g = 0.45 \times 25 \times 250,000 = 2,812,500\,\text{N} \] \[ 0.15f_{ck}A_gd = 0.15 \times 25 \times 250,000 \times 0.46 = 431,250\,\text{Nm} \] \[ \frac{N_u}{2,812,500} + \frac{M_u}{431,250} = \frac{300,000}{2,812,500} + \frac{360,000}{431,250} = 0.107 + 0.835 = 0.942 < 1 \] **Now, OK!** --- ## **Step 5: Area of Steel** ### **A. Minimum Steel (per code, say 0.8% of gross area)** \[ A_{sc, min} = 0.8\% \times 250,000 = 0.008 \times 250,000 = 2,000\,\text{mm}^2 \] ### **B. Provide Main Reinforcement** - Use 6 bars of 25 mm diameter (each = 491 mm²): \( 6 \times 491 = 2,946\,\text{mm}^2 \) - Distribute bars equally on four faces. ### **C. Shear Reinforcement (Stirrups)** - Shear force = 80 kN. - Provide 8 mm diameter stirrups at 200 mm c/c (check per code for actual requirements). --- ## **Step 6: Final Answer** ### **Column CD Design:** - **Cross-section:** \( 500\,\text{mm} \times 500\,\text{mm} \) (square) - **Main reinforcement:** 6 bars of 25 mm diameter (\( \approx 2,946\,\text{mm}^2 \)), distributed equally - **Stirrups:** 8 mm @ 200 mm c/c (minimum as per code) - **Concrete grade:** \( f_{cu} = 25\,\text{MPa} \) - **Steel grade:** \( f_y = 415\,\text{MPa} \) - **Effective cover:** 40 mm --- ### **Summary Table** | Parameter | Value | |------------------|-----------------------| | Column size | 500 mm × 500 mm | | Main bars | 6 × 25 mm diameter | | Area of steel | ≈ 2,946 mm² | | Stirrups | 8 mm @ 200 mm c/c | | Concrete grade | 25 MPa | | Steel grade | 415 MPa | | Effective cover | 40 mm | --- **If you want more detailed reinforcement arrangement or code references, let me know!**