--- title: "Design steps for long column, unbraced frame" subject: "Design of RCC Structure" chapter: 10 key: "semester_7/design_of_rcc_structure:design-steps-for-long-column-unbraced" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-steps-for-long-column-unbraced site: BCT Engineering --- # Design steps for long column, unbraced frame Design of RCC Structure · Chapter 10: Design of compression members: Columns - Asked in: 2065 Kartik (old course), 10 marks - Video: [Level 1- Lecture 4-3 Design of Long Columns - Braced & Unbraced Columns](https://www.youtube.com/watch?v=ugc0mwompQ0) — Ragab Elmasry - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-steps-for-long-column-unbraced - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-10 ## Question Write down design steps for the design of RC long column of unbraced frame. ## Short answer For an unbraced slender column, compute effective lengths from sway-frame charts, check l_e/D > 12, then add additional moment M_a = P_u D/2000 (l_e/D)^2 reduced by k to the larger end moment. Check biaxial interaction using SP-16 uniaxial capacities and provide ties; revise steel if needed. ## Answer A column is **long (slender)** when $l_{ex}/D$ or $l_{ey}/b$ exceeds 12. In an **unbraced frame** (sway frame) the lateral stability depends on the columns themselves, so the effective length is larger than the unsupported length and the additional moments due to slenderness must be added (IS 456 cl. 25.1.2, 39.7). #### Design steps 1. **Data**: size $b \times D$, factored loads $P_u$, end moments $M_{ux1}, M_{ux2}, M_{uy1}, M_{uy2}$, unsupported length $L$, grades of concrete and steel. 2. **Effective length**: for a sway frame, find $l_{ex}$ and $l_{ey}$ from the charts of IS 456 Annex E (Fig. 27 / Fig. 28) using $\beta_1, \beta_2$ (ratio of column stiffness to beam stiffness at both ends), or from Table 28 when end conditions are known. 3. **Slenderness ratio**: $l_{ex}/D$ and $l_{ey}/b$. If either is above 12, the column is slender about that axis; if above 60, redesign (cl. 25.3.1: $l/b \le 60$). 4. **Minimum eccentricity**: $e_{min} = \dfrac{L}{500} + \dfrac{D}{30} \ge 20$ mm for both axes. 5. **Assume a trial steel** percentage (say 1% to 3%) and bar arrangement; $d'/D$ for cover. 6. **Additional moments** due to slenderness for the slender axis: $$ M_{ax} = \frac{P_u D}{2000}\left(\frac{l_{ex}}{D}\right)^2, \qquad M_{ay} = \frac{P_u b}{2000}\left(\frac{l_{ey}}{b}\right)^2 $$ and reduce by $k = \dfrac{P_{uz} - P_u}{P_{uz} - P_b} \le 1$, with $P_{uz} = 0.45f_{ck}A_c + 0.75f_yA_{sc}$. 7. **Total design moments**: in an unbraced column the additional moment is added to the larger end moment, and the end moment is not reduced (for a sway frame there is no reduction for single/double curvature): $M_{ux} = M_{ux,2} + M_{ax}$, $M_{uy} = M_{uy,2} + M_{ay}$, not less than $P_u e_{min}$. 8. **Check the section**: find $M_{ux1}, M_{uy1}$ (uniaxial capacities at $P_u$) from SP-16 charts, and check the biaxial interaction $$ \left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n} + \left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n} \le 1 $$ with $\alpha_n$ from 1.0 ($P_u/P_{uz} \le 0.2$) to 2.0 ($\ge 0.8$) (cl. 39.6). 9. **Revise**: if not satisfied, increase steel (limit 0.8% to 4%, maximum 6%) or the size, and repeat from step 6. 10. **Lateral ties**: diameter $\ge \max(\phi/4, 6)$ mm, pitch $\le \min(b, 16\phi, 300)$ mm; in a seismic zone use ductile detailing (IS 13920). 11. **Draw** the section and elevation, showing bars, ties and laps. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the design steps, what must be done if the slenderness ratio of the column exceeds 60? A. Add the additional moment Ma and continue B. Redesign the column (larger size), since l/b must not exceed 60 C. Reduce Ma by the factor k and continue D. Treat it as a short column Answer: B. IS 456 cl. 25.3.1 limits l/b to 60, so a column beyond this must be redesigned. Between 12 and 60 it is slender and additional moments are added. (Key point: Slenderness limits) 2. How are the total design moments obtained for a long column in an unbraced (sway) frame? A. Mu = 0.6M2 + 0.4M1, then add Ma B. Mu = Ma only C. Additional moment is added to the larger end moment, with no reduction for curvature D. Mu = the smaller end moment + Ma Answer: C. In a sway frame the end moment is not reduced for single or double curvature. Ma is added to the larger end moment, and the result is not less than Pu*emin. (Key point: Moments in unbraced column) 3. What is the purpose of the factor k = (Puz - Pu)/(Puz - Pb) <= 1 applied to the additional moment Ma? A. It reduces Ma, because the axial load reduces the bending needed near Puz, where capacity is governed by squash load B. It increases Ma for heavily loaded columns C. It converts Ma to a minimum eccentricity moment D. It gives the effective length of the column Answer: A. k is a reduction factor (at most 1) applied to the computed additional moment Ma. Puz = 0.45 fck Ac + 0.75 fy Asc is the axial capacity with zero moment. (Key point: Reduction factor k) 4. In the biaxial check (Mux/Mux1)^αn + (Muy/Muy1)^αn <= 1, how does αn vary? A. It is always 1.0 B. It is always 2.0 C. It varies from 0.8 to 1.0 with Pu/Puz D. It varies linearly from 1.0 (Pu/Puz <= 0.2) to 2.0 (Pu/Puz >= 0.8) Answer: D. IS 456 cl. 39.6 gives αn = 1.0 for Pu/Puz <= 0.2 and 2.0 for Pu/Puz >= 0.8, with linear variation between. 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