Lecture_26

Lecture_26 - 7.5 Biaxial Bending Short Columns 7.5.1...

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Unformatted text preview: 7.5 Biaxial Bending Short Columns 7.5.1 Section subjected to an axial load acting at two major eccentricities Consider the column section shown in Fig. 7.5-1 is subjected to a load N acting at eccentricities ex and ey : Mx = Ney and My = Nex y ex N ey x (a) (b) (c) Fig. 7.5-1 (a) Column section subjected to an axial load N acting at eccentricities ex and ey ; (b) bending moments about two axes; (c) section, strains, and stress block at failure 258 (a) (b) Fig. 7.5-2 (a) M-N interactive surface (failure surface) for column section with biaxial bending; (b) M-N interactive diagrams about x-x and y-y axes • Biaxial bending will generally not govern the design for most columns in a building. • Corner columns may have to resist significant bending about both axes, but the axial loads are usually small. • In practice, a simplified method describe in BS 8110 is generally used in the design for biaxial bending 259 7.5.2 BS 8110 design method (BS 8110: Part 1, clause 3.8.4.5) • A simplified design method for biaxial bending of symmetrically- reinforced rectangular sections. • A column subjected to an ultimate load N and moments Mx and My may be designed to withstand an increased design moment about one axis. Fig. 7.5-3 Section with biaxial bending (i) If Mx My ≥ , the effective design moment about x-x axis is h' b' ′ Mx = Mx + β (ii) If h′ My b′ (7.6) Mx My < , the effective design moment about y-y axis is h' b' M′ = My + β y b' Mx h' (7.7) Table 7.5-1 260 Text Example 9.7 (pp.240-241): Design of a column for biaxial bending The column section shown in the figure is to be designed to resist an ultimate axial load of 1200 kN plus moments of Mx = 75 kNm and My = 80 kNm. fcu = 30 N/mm2 and fy = 460 N/mm2. Mx 75 = = 0.268 h' (350 − 70) My b' = 80 = 0.333 (300 − 60) M x / h' < M y / b' Therefore the increased single axial design moment about y-y axis is M′ = My +β y b' ×Mx h' N = 1200 × 10 3 /(300 × 350 × 30) = 0.38 bhf cu From Table 7.5-1, β = 0.55 M ′ = 80 + 0.55 y 240 × 75 = 115.4 kNm 280 N 1200 × 10 3 = = 11.4 bh 350 × 300 M 115.4 × 10 6 = = 3.66 bh 2 350 × 300 2 From the design chart of Fig. 7.4-2(b) 100Asc / bh ≈ 2.6. Therefore required Asc = 2730 mm2. Provide 4T32 (3216 mm2). 261 7.6* Slender Columns 7.6.1 Slenderness ratio ley lex or > 15 for a braced column • h b l ey lex or > 10 for an unbraced column • h b • Slenderness limits: o lo > 60 times the minimum thickness of a column o If one end of an unbraced column is unrestrained (e.g. a cantilever column), 100b 2 lo = ≤ 60b h 7.6.2 Design moment (BS 8110, Part 1, clause 3.8.3) • A slender column should be designed for an ultimate axial load plus a design moment (an additional moment caused by lateral deflection) N Δ ⇒ Madd = NΔ N Fig. 7.6-1 Additional moment * Optional course materials for CIVL 232 262 • Design Moment (only applicable to columns of a retangular or circular section with symmetrical reinforcement) M t = M i + M add M t = M i + Nau or (7.8) (7.9) where Mi – the initial moment in the column Madd – the moment caused by the deflection of the column au – the deflection of the column o The deflection of a rectangular or circular column is au = β a Kh (7.10) where 1 ⎛ le ⎞ βa = ⎜⎟ 2000 ⎝ b′ ⎠ 2 (7.11) with b’ being the smaller dimension of the column section; and K= N uz − N ≤ 1.0 N uz − N bal (7.12) in which Nuz is the ultimate axial load, Nuz = 0.45fcuAc + 0.95fyAsc and Nbal is the axial load at balanced failure and may be taken as Nbal = 0.25fcuAc for symmetrically reinforced rectangular sections. o To calculate K (Eq. 7.12), Asc must be known and hence a trial and error approach is necessary, taking an initial conservative value of K = 1.0. 263 7.6.3 Braced slender columns • Typical bending moment diagrams Fig. 7.6-2 Design moments of braced slender columns o The maximum additional moment Madd occurs near the mid- height. o The initial moment (at the same location as Madd) is taken as M i = 0.4M 1 + 0.6M 2 ≥ 0.4M 2 (7.13) 264 where M1 – the smaller initial end moment due to design loads (usually taken as negative) M2 – the larger initial end moment due to design loads (usually taken as positive) • The maximum design moment Mt will be the greatest of (a) to (d): (a) M2 (b) Mi + Madd (c) M1 + Madd / 2 (d) Nemin with emin > 0.05h or 20 mm Text Example 9.8 (pp.243-245): Design of a slender braced column A braced column of 300 × 450 cross-section resists at the ultimate limit state an axial load of 1700 kN and end moments of 70 kNm and 10 kNm causing double curvature about the minor axis x-x as shown in the following figure. The column’s effective heights are lex = 6.75 m and ley = 8.0 m. fcu = 30 N/mm2 and fy = 460 N/mm2. Example of braced slender column 265 Slenderness ratios l ex 6.75 = = 22.5 > 15 h 0.3 l ey 8.0 = = 17.8 > 15 h 0.45 Therefore the column is slender. As the column is bent in double curvature M1 = −10 kNm and M2 = 70 kNm Mi = 0.4M1 + 0.6M2 = 0.4 × (−10) + 0.6 × 70 = 38 kNm > 0.4M2 = 0.4 × 70 = 28 kNm With K = 1.0 for the initial value, 2 M add 1 ⎛ le ⎞ = Nau = N ⎜ ⎟ Kh 2000 ⎝ b' ⎠ 2 = 1700 1 ⎛ 6750 ⎞ −3 ×⎜ ⎟ × 1.0 × 300 × 10 2000 ⎝ 300 ⎠ = 129 kNm For the first iteration, the design moment is Mt = Mi + Madd = 38 + 129 = 167 kNm 1700 × 10 3 N / bh = = 12.6 450 × 300 167 × 10 6 = 4.12 M / bh = 450 × 300 2 2 266 From the design chart, 100Asc / bh = 3.0 then K = 0.67 This new value for K is used to recalculate Madd and hence Mt for the second iteration. The design chart is again used to determine 100Asc / bh and a new value of K is shown in Table 7.6-1. This iterations are continued until the value of K in columns (1) and (5) of Table 7.6-1 are in reasonable agreement, which in this design occurs after two iterations. 7.6-1 The steel area required is Asc = 2.1bh / 100 = 2.1 × 450 × 300 / 100 = 2835 mm 2 As a check on the final value of K interpolated from the design chart: N bal = 0.25 f cu bd = 0.25 × 30 × 450 × 240 × 10 −3 = 810 kN N uz = 0.45 f cu bd + 0.95 f y Asc = (0.45 × 30 × 450 × 300 + 0.95 × 460 × 2835 )10 −3 = 3061 kN 267 K= N uz − N N uz − N bal = 3061 − 1700 3061 − 810 = 0.6 which agrees with the final value in column 5 of the table. 7.6.4 Unbraced slender columns • The sway of an unbraced structure causes larger additional moments in the columns. • Maximum design moment always occurs at the stiffer end joint of the column. Fig. 7.6-3 Design moments of unbraced slender columns 268 Text 1. Kuang, J.S., Structural Concrete Design, Course Notes for CIVL 232, The Hong Kong University of Science & Technology, 2006, 269 pp. 2. Mosley, W.H., Bungey, J.H., and Hulse, R., Reinforced Concrete Design, 5th edition, MacMillan, London, 1999, 385 pp. References 1. IStructE and ICE, Manual for the design of reinforced concrete building structures, 2nd edn, Institution of Structural Engineers, London, 2002, 93 pp. 2. Kong, F.K., and Evans, R.H., Reinforced and Prestressed Concrete, 3rd edn, Chapman and Hall, London, 1987, 508 pp. 3. MacGinley, T.J., and Choo, B.S., Reinforced Concrete: Design Theory and Examples, 2nd edn, E & F N Spon, London, 1990, 519 pp. 4. Reynolds, C.E., and Steedman, J.C., 10th edn, Reinforced Concrete Designer’s Handbook, E & FN Spon, London, 1988, 436 pp. Design Codes 1. British Standards Institution, Structural Use of Concrete, BS 8110: 1997. 2. Buildings Department of The Hong Kong Special Administrative Region, The Structural Use of Concrete, 2004. 3. Hong Kong Government, Building (Construction) Regulations, Chapter 123, 1992. 269 ...
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