Practical Design to Eurocode 2 - Concrete Centre

1 Practical Design to Eurocode 2 09/11/16. Practical Design to Eurocode 2. The webinar will start at Course Outline Lecture Date Speaker Title 1 21 Sep Charles Goodchild Introduction, Background and Codes 2 28 Sep Charles Goodchild EC2 Background, Materials, Cover and effective spans 3 5 Oct Paul Gregory Bending and Shear in Beams 4 12 Oct Charles Goodchild Analysis 5 19 Oct Paul Gregory Slabs and Flat Slabs 6 26 Oct Charles Goodchild Deflection and Crack Control 7 2 Nov Paul Gregory Detailing 8 9 Nov Jenny Burridge Columns 9 16 Nov Jenny Burridge Fire 10 23 Nov Jenny Burridge Foundations Week 8 1. Practical Design to Eurocode 2 09/11/16. Columns Lecture 8. 9th November 2016. Model Answers Lecture 7 Exercise: Lap length for column longitudinal bars Week 8 2.

2 Practical Design to Eurocode 2 09/11/16. Column lap length exercise Design information H25's C40/50 Concrete 400 mm square column Lap 45mm nominal cover to main bars Longitudinal bars are in compression Maximum ultimate stress in the bars is 390 MPa Exercise: Calculate the minimum lap length using EC2 equation : H32's Column lap length exercise Procedure Determine the ultimate bond stress, fbd EC2 Equ. Determine the basic anchorage length, lb,req EC2 Equ. Determine the Design anchorage length, lbd EC2 Equ. Determine the lap length, l0 = anchorage length x 6. Week 8 3. Practical Design to Eurocode 2 09/11/16. Solution - Column lap length Determine the ultimate bond stress, fbd fbd = 1 2 fctd EC2 Equ.

3 1 = Good' bond conditions 2 = bar size 32.. fctd = ct fctk,0,05/ c EC2 cl (2), Equ ct = = c fctk,0,05 = x fck2/3 EC2 Table = x 402/3. = MPa fctd = ct fctk,0,05/ = = c fbd = x = MPa Solution - Column lap length Determine the basic anchorage length, lb,req lb,req = ( /4) ( sd/fbd) EC2 Equ Max ultimate stress in the bar, sd = 390 MPa. lb,req = ( /4) ( 390 ). = . For Concrete class C40/50. Week 8 4. Practical Design to Eurocode 2 09/11/16. Solution - Column lap length Determine the Design anchorage length, lbd lbd = 1 2 3 4 5 lb,req lb,min Equ. lbd = 1 2 3 4 5 ( ) For Concrete class C40/50. For bars in compression 1 = 2 = 3 = 4 = 5 = Hence lbd = . Solution - Column lap length Determine the lap length, l0 = anchorage length x 6.

4 All the bars are being lapped at the same section, 6 = A lap length is based on the smallest bar in the lap, 25mm Hence, l0 = lbd x 6. l0 = x l0 = = x 25. l0 = 993 mm Week 8 5. Practical Design to Eurocode 2 09/11/16. Columns The Column Design Process Determine the actions on the column Allow for imperfections in the column Determine slenderness, , via effective length lo Determine slenderness limit, lim Yes No Is lim? Column is not slender, Column is slender MEd = Max (M02, NEde0) MEd = Max(M02;M0e+M2;M01+ ;NEde0). Calculate As ( using column chart). Check detailing requirements Week 8 6. Practical Design to Eurocode 2 09/11/16. Actions Actions on the columns are determined using Actions one of the analysis methods we looked at for flexural Design .

5 Imperfections From the analysis obtain the following Slenderness, . data: Effective length, l0. Ultimate axial load, NEd Ultimate moment at the top of Slenderness limit, lim the column, Mtop Yes Slen- Ultimate moment at the bottom Is lim? der of the column, Mbottom No Design Moments, MEd Allow for imperfections .. Calculate As Detailing Geometric Imperfections: Cl. Deviations in cross-section dimensions are normally Actions taken into account in the material factors and should not be included in structural analysis Imperfections Imperfections need not be considered for SLS Slenderness, . But out-of-plumb needs to be considered and is represented by an inclination, i Effective length, l0. i = 0 h m Slenderness limit, lim where 0 = 1/200.

6 Yes h = 2/ l; 2/3 h 1 Slen- Is lim? m = ( (1+1/m)) der where l = the length or height (m) No (see (6)) Design Moments, MEd m = no. of vert. members Calculate As For isolated columns in braced systems, m and h may be taken as Detailing i = 0 = 1/200. Week 8 7. Practical Design to Eurocode 2 09/11/16. Geometric Imperfections Cl. (7) & (9) For isolated members at ULS, the effect of imperfections may be taken into account in two ways: a) as an eccentricity, ei = i l0/2. So for isolated columns in a braced system, ei = l0/400 may be used. b) as a transverse force, Hi Hi = i N for un-braced members Hi = 2 i N for braced members = N/100. Examples of Isolated Members Figure Braced Unbraced Week 8 8.

7 Practical Design to Eurocode 2 09/11/16. Geometric Imperfections: Cl (4). Subject to a minimum eccentricity: e0 = h/30 but 20 mm Design Moments Stocky Columns M01 = [Min{|Mtop|,|Mbottom|} eiNEd] *. M02. M eiNEd M02 = [Max{|Mtop|,|Mbottom|} + eiNEd] *. ei = lo/400. NEd = Design load in the column For a stocky column, Design moment MEd = Max{M02, e0 NEd}. e0 = Max{h/30,20mm}. M01. * Note: M01 and M02 return to their original signs after this calculation has been carried out Week 8 9. Practical Design to Eurocode 2 09/11/16. Moments in Slender Columns Cl. Fig M02. M eiNEd Design Moment, MEd = Max {M02;. M0e + M2;. M0e M01 + ;. NEde0}. M01. Is the Column Slender? Cl. , Actions 2nd order effects Imperfections Second order effects may be ignored if they Slenderness.

8 Are less than 10% of the corresponding first order effects Effective length, l0. Second order effects may be ignored if the Slenderness limit, lim slenderness, is less than lim where lim = 20 A B C/ n Yes Slen- Is lim? der No With biaxial bending the slenderness should be Design Moments, MEd checked separately for each direction and only need be considered in the directions where lim is Calculate As exceeded Detailing Week 8 10. Practical Design to Eurocode 2 09/11/16. Slenderness Actions Slenderness, = l0/i Imperfections l0 =effective length = Slenderness, . (l = actual length) Effective length, l0. i = radius of gyration = (I/A). Slenderness limit, lim hence for a rectangular section = l0 / h Is lim?

9 Yes Slen- for a circular section = 4 l0 / h No der Design Moments, MEd Calculate As 21. Detailing Effective length Figure , Figure , Effective length, l0 = Fl Actions Imperfections Slenderness, .. Effective length, l0.. M Slenderness limit, lim l0 = l l0 = 2l l0 = 0,7l l0 = l / 2 l0 = l l /2 <l0< l l0 > 2l Yes Slen- Is lim? Braced members: der k1 k2 No F = 0,5 1 + 0,45 + k 1 + 0,45 + k Design Moments, MEd 1 2 . Unbraced members: k k k1 k2 . F= max 1 + 10 1 2 ; 1 + 1 + Calculate As k1 + k 2 1 + k1 1 + k 2 . 22. where k = ( / M) (E / l) Detailing Week 8 11. Practical Design to Eurocode 2 09/11/16. Effective length & k factors PD 6687 - Non failing column From PD 6687. The contribution of non failing'.

10 K1 at columns to the joint stiffness End 1 Failing column may be ignored k2 at For beams /M may be taken as End 2 l/2EI (allowing for cracking in Non failing the beams). column k1= [EI`/`l]col / [ 2EI / l]beams1 k2 = [EI`/`l]col / [ 2EI / l]beams2. Assuming that the beams are symmetrical about the column and their sizes are the same in the two storeys shown, then: k1 = k2 = [EI /`l]col / [2 x 2EI / l]beams Effective length & Slenderness Cl. Slenderness = l0/i Actions where Imperfections l0 = Fl Slenderness, . k = relative stiffness Effective length, l0. E Ic F. lc Slenderness limit, lim k= 2E I b . lb Yes Slen- Is lim? der No Design Moments, MEd i = (I/A) Calculate As for a rectangular section = l0 / h for a circular section = 4 l0 / h Detailing 24.