--- title: "Fixed-end column 4 m, 1000 kN" subject: "Design of RCC Structure" chapter: 10 key: "semester_7/design_of_rcc_structure:fixed-end-column-4-m-1000" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/fixed-end-column-4-m-1000 site: BCT Engineering --- # Fixed-end column 4 m, 1000 kN Design of RCC Structure · Chapter 10: Design of compression members: Columns - Asked in: 2069 Chaitra, 14 marks - Video: [Design of Fixed ended Column || Example Solved](https://www.youtube.com/watch?v=S1RIOoOqyTM) — Civil Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/fixed-end-column-4-m-1000 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10 ## Question A column of 4 m length with both ends fixed and effectively held in position is subjected to a design axial load of 1000 kN and factored bending moment of 100 kN-m. Design the rectangular column with its longitudinal and transverse reinforcements. ## Short answer Design a 300×350 mm short column with 14-16 mm bars (2.68%) and 8 mm ties at 250 mm c/c for a 4 m fixed–fixed column under 1000 kN axial load and 100 kNm uniaxial moment. ## Answer #### Data and assumptions - Column 300 mm (width $b$) × 350 mm (depth $D$); $P_u = 1000$ kN; $f_{ck} = 25$ MPa, $f_y = 415$ MPa. Clear cover 40 mm (IS 456 cl. 26.4.2.1), 8 mm ties. - Moment about major axis: $M_{ux}$ = 100 kN·m. - Both ends fixed and held in position: $l_e = 0.65L = 2.6$ m. M25 concrete and Fe415 steel are assumed (not stated). The 100 kN·m moment acts about the major axis. Section size is selected by trial. - Sizes are given as $b \times D$ with $D \ge b$; moments $M_{ux}$ act about the major axis (depth $D$) and $M_{uy}$ about the minor axis (depth $b$). #### Slenderness | Axis | $l_e$ (mm) | Dimension (mm) | $l_e/$dim | Type | |---|---|---|---|---| | x-x (about major) | 2600 | 350 | 7.43 | short | | y-y (about minor) | 2600 | 300 | 8.67 | short | Both ratios are below 12, so the column is **short**; no additional moment is required. #### Minimum eccentricity (cl. 25.4) $$ e_{min} = \max\left(\frac{L}{500} + \frac{D}{30},\ 20\right) $$ - About x-x: $e_x = 4000/500 + 350/30 = 19.7$ mm (min. 20 mm); adopt 20.0 mm; $M_{ux,min} = P_u e = 20.0$ kN·m - About y-y: $e_y = 4000/500 + 300/30 = 18.0$ mm (min. 20 mm); adopt 20.0 mm; $M_{uy,min} = 20.0$ kN·m #### Design moments - Design moments: $M_{ux} = 100.0$ kN·m (not less than the minimum-eccentricity values). #### Longitudinal reinforcement $\dfrac{P_u}{f_{ck} b D} = \dfrac{1000000}{25 \times 300 \times 350} = 0.381$ Bars are chosen so that $p$ lies between 0.8% (minimum) and 4% (practical maximum; IS 456 cl. 26.5.3.1 allows 6%). The uniaxial capacities $M_{ux1}$, $M_{uy1}$ at $P_u$ are read from the SP-16 interaction charts for rectangular sections; here they are computed by strain compatibility with the IS 456 stress block and the design stress-strain curve of steel (same basis as the charts), with effective cover $d' = 40 + 8 + \phi/2$. | Trial | $p$ (%) | $d'/D$ | $M_{ux1}$ (kN·m) | $M_{ux}$ required | Ratio | |---|---|---|---|---|---| | 12-16 φ | 2.30 | 0.160 | 96.0 | 100.0 | 1.042 not ok | | 8-20 φ | 2.39 | 0.166 | 99.2 | 100.0 | 1.008 not ok | | 14-16 φ | 2.68 | 0.160 | 103.3 | 100.0 | 0.968 ok | #### Check $M_{ux} = 100.0$ kN·m $\le M_{ux1} = 103.3$ kN·m. Minimum eccentricity about the minor axis: $P_u e_y = 20.0$ kN·m $\le M_{uy1} = 88.5$ kN·m. Both are satisfied; the section is **safe**. Provide **14-16 mm** bars ($A_{sc} = 2815$ mm², $p = 2.68\%$), within 0.8% to 4%. #### Lateral ties - Diameter $\ge \max(\phi_{max}/4,\ 6) = \max(16/4, 6) = 6$ mm → provide 8 mm. - Pitch $\le \min$(least lateral dimension, $16\phi_{min}$, 300 mm) $= \min(300, 256, 300) = 256$ mm → provide 8 mm ties @ 250 mm c/c. #### Detailing ```text CROSS-SECTION 300 x 350 +-------------------------+ | | | o o o o | | | | o o | | | | o o | | | | o o | | | | o o o o | | | +-------------------------+ o = 16 mm bars, ties 8 mm @ 250 c/c ``` **Answer:** 300 × 350 mm column with 14-16 mm longitudinal bars ($p = 2.68\%$) and 8 mm ties @ 250 mm c/c. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A 4 m column fixed at both ends and held in position has an effective length of A. 4.0 m B. 2.6 m (0.65L) C. 2.0 m (0.5L) D. 8.0 m (2L) Answer: B. IS 456 gives le = 0.65L for both ends fixed, so 2.6 m. (Key point: Effective length) 2. With le = 2600 mm and a 300 x 350 section, the column is classified as A. Slender, since le/b = 8.67 > 12 B. Short, since le/D = 7.43 and le/b = 8.67 are both below 12 C. Slender, since L = 4 m is large D. Short, since only le/D is checked Answer: B. Both ratios are below 12, so there is no additional moment. (Key point: Short or slender) 3. The minimum eccentricity formula in this answer uses which length? A. Effective length 2600 mm B. Unsupported length 4000 mm C. Zero D. Twice the depth Answer: B. emin = L/500 + D/30 uses the unsupported length L = 4000 mm, giving 19.7 mm, so 20 mm is adopted. (Key point: Which length in emin) 4. Only Mux = 100 kN-m is applied. What is checked about the minor axis? A. Nothing B. Pu x emin (20 kN-m) must not exceed Muy1 C. Muy must equal Mux D. The column must be redesigned as slender Answer: B. The minimum eccentricity moment of 20 kN-m about the minor axis must be within Muy1 (88.5 kN-m), so it is satisfied. (Key point: Uniaxial design logic) 5. Why are 14-16 mm bars adopted? A. Mux1 = 103.3 kN-m >= 100 kN-m required, while smaller trials gave less B. p is below 0.8% C. The ties become unnecessary D. They are the only bars with 16 mm size Answer: A. The capacity must be at least the applied moment; 12-16 (96.0) and 8-20 (99.2) fall short. 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