Thiết kế cấu kiện theo eurocode 2

Thể theo nguyện vọng của các đồng nghiệp là Giảng viên, Kỹ sư tại Việt Nam, người viết sách này hy vọng có thể đóng góp được một phần nhỏ trong việc phổ biến bộ tiêu chuẩn mới trên thế giới đến với cộng đồng Xây Dựng Việt Nam. Trong cuốn sách này, bạn đọc sẽ thấy có một số điểm khá tương đồng với những quy chuẩn đã được áp dụng ở Việt Nam, nhưng ở một số trường hợp, nếu quan sát kĩ thì giá trị sử dụng để tính toán sẽ rất khác vì EUROCODES được biên soạn dựa trên những lý thuyết tính toán mới hơn những lý thuyết trước đây. Nhìn chung, thiết kế theo EUROCODE 2 giúp người sử dụng có thể hiểu rõ hơn bản chất của vấn đề hơn so với TCVN hiện hành và cũng dễ áp dụng hơn vì sử dụng các hệ số kinh nghiệm ít hơn và ít phải tra bảng (ví dụ việc xác định cường độ tính toán của bêtông hay của cốt thép). Bộ EUROCODES hiện tại gồm 10 phần khác nhau: từ EUROCODE 0 đến EUROCODE 9. Trong mỗi EUROCODE, ngoài những phần chung thì trong một số trường hợp, EUROCODE cho phép mỗi quốc gia lựa chọn cho mình một số hệ số riêng tùy theo hoàn cảnh, môi trường hay gặp tại quốc gia đó (ví dụ: bề dày lớp bêtông bảo bệ ; gia tốc tính toán tiêu chuẩn tải trọng động đất;…). Những hệ số này được ghi trong “Phụ lục quốc gia” của từng nước. Cuốn sách này sẽ đề cập đến EUROCODE 0 (cơ sở lý thuyết tính toán), EUROCODE 1 (tải trọng) và chủ yếu là EUROCODE 2 (bêtông cốt thép). EUROCODE 8 dùng cho thiết kế kháng chấn cũng sẽ được đề cập, với những phần liên quan đến kết cấu bêtông cốt thép.

Practical Design to Eurocode Beams • Bending / Flexure • Shear • Detailing Anchorage & Laps Members & particular rules Beams Flexure Section Design: Bending • Flexural design is generally the same as BS8110 in principle • Modified for high strength concrete • EC2 presents the principles only • Design manuals will provide the standard solutions for basic design cases Rectangular Concrete Stress Block (3.1.7, Figure 3.5) cu3 Ac fcd Fc x x d As Fs s fck  50 MPa 50 < fck  90 MPa  0.8 = 0.8 – (fck – 50)/400  1.0 = 1,0 – (fck – 50)/200 fcd = cc fck /c = 0.85 fck /1.5 Analysis of a singly reinforced beam Cl 3.1.7 EN 1992-1-1 Design equations can be derived as follows: b M For grades of concrete up to C50/60, εcu= 0.0035,  = and  = 0.8 fcd = 0.85fck/1.5, fyd = fyk/1.15 = 0.87 fyk Fc = (0.85 fck / 1.5) b (0.8 x) = 0.453 fck b x Fst = 0.87As fyk For horizontal equilibrium Fc= Fst 0.453 fck b x = 0.87As fyk Guess As Solve for x z = d - 0.4 x M = Fc z Maximum neutral axis depth b M   Redistributed Bending Moment Elastic Bending Moment   k1  k x u d xu = Neutral axis depth after redistribution EC2 NA gives k1 = 0.4 and k2 = 1.0 K  M bd f ck Value of K for maximum value of M with no compression steel and when x is at its maximum value If K > K’ Compression steel required EC2 Equ 5.10a   0.4  x u d xu   - 0.4 d Analysis of a Singly Reinforced Beam3.1.7 EN 1992-1-1 Design equations can be derived as follows: b M For grades of concrete up to C50/60, εcu= 0.0035,  = and  = 0.8 fcd = 0.85fck/1.5, fyd = fyk/1.15 Fc = (0.85 fck / 1.5) b (0.8 x) = 0.453 fck b x Fst = 0.87As fyk Take moments about the centre of the tension force M = 0.453 fck b x z …1 Section analysis Now z = d - 0.4 x  x = 2.5(d - z) & M = 0.453 fck b 2.5(d - z) z = 1.1333 (fck b z d - fck b z2) Let K = M / (fck b d 2) K can be considered as the normalised bending resistance  fckbdz M fckbz  1.1333  2 fckbd fckbd  fckbd  = 1.1333 [(z/d)2 – (z/d)] + K = (z/d)2 – (z/d) + 0.88235K    = (z/d)2 – (z/d) + 0.88235K Solving the quadratic equation: z/d = [1 + (1 - 3.529K)0.5]/2 z = d [ + (1 - 3.529K)0.5]/2 Rearranging z = d [ 0.5 + (0.25 – K / 1.134)0.5] This compares to BS 8110 z = d [ 0.5 + (0.25 – K / 0.9)0.5] The lever arm for an applied moment is now known Higher Concrete Strengths fck ≤ 50MPa z  d[1  (1 3,529K )]/2 fck = 60MPa z  d[1  (1 3,715K )]/2 fck = 70MPa z  d[1  (1 3,922K )]/2 fck = 80MPa z  d[1  (1 4,152K )]/2 fck = 90MPa z  d[1  (1 4,412K )]/2 Arrangement of Laps EC2: Cl 8.7.2, Fig 8.7 Concise: Cl 11.6.2 Transverse Reinforcement at Laps Concise: Cl 11.6.4 Bars in tension EC2: Cl 8.7.4, Fig 8.9 Rules apply if bar diameter ≥ 20mm • Where the diameter, , of the lapped bars  20 mm, the transverse reinforcement should have a total area, Ast  1,0As of one spliced bar It should be placed perpendicular to the direction of the lapped reinforcement and between that and the surface of the concrete • If more than 50% of the reinforcement is lapped at one point and the distance between adjacent laps at a section is  10  transverse bars should be formed by links or U bars anchored into the body of the section • The transverse reinforcement provided as above should be positioned at the outer sections of the lap as shown below Ast /2 Ast /2 l /3 l /3 150 mm F s Fs Figure 8.9 – bars in tension l0 Transverse Reinforcement at Laps Bars in compression Concise: Cl 11.6.4 EC2: Cl 8.7.4, Fig 8.9 In addition to the rules for bars in tension one bar of the transverse reinforcement should be placed outside each end of the lap length Figure 8.9 – bars in compression SECTION Detailing of members and particular rules Beams (9.2) • As,min = 0,26 (fctm/fyk)btd but  0,0013btd • As,max = 0,04 Ac • Section at supports should be designed for a hogging moment  0,25 max span moment • Any design compression reinforcement () should be held by transverse reinforcement with spacing 15  Beams (9.2) • Tension reinforcement in a flanged beam at supports should be spread over the effective width (see 5.3.2.1) Curtailment (9.2.1.3) (1) Sufficient reinforcement should be provided at all sections to resist the envelope of the acting tensile force, including the effect of inclined cracks in webs and flanges (2) For members with shear reinforcement the additional tensile force, ΔFtd, should be calculated according to 6.2.3 (7) For members without shear reinforcement ΔFtd may be estimated by shifting the moment curve a distance al = d according to 6.2.2 (5) This "shift rule” may also be used as an alternative for members with shear reinforcement, where: al = z (cot θ - cot α)/2 = 0.5 z cot θ for vertical shear links z= lever arm, θ = angle of compression strut al = 1.125 d when cot θ = 2.5 and 0.45 d when cot θ = Curtailment of reinforcement EC2: Cl 9.2.1.3, Fig 9.2 Concise: 12.2.2 Envelope of (M Ed /z +N Ed) lbd lbd Acting tensile force Resisting tensile force lbd al lbd Ftd al Ftd lbd lbd lbd lbd “Shift rule” • For members without shear reinforcement this is satisfied with al = d • For members with shear reinforcement: al = 0.5 z Cot  But it is always conservative to use al = 1.125d Anchorage of Bottom Reinforcement at End Supports (9.2.1.4) Tensile Force Envelope al Shear shift rule Simple support (indirect) Simple support (direct) • As bottom steel at support  0.25 As provided in the span • lbd is required from the line of contact of the support • Transverse pressure may only be taken into account with a ‘direct’ support Simplified Detailing Rules for Beams Concise: Cl 12.2.4 How to….EC2 Detailing section Supporting Reinforcement at ‘Indirect’ Supports Concise: Cl 12.2.8 EC2: Cl 9.2.5 A supporting beam with height h1 B supported beam with height h2 (h1  h2) B  h /3  h /2 Plan view • The supporting reinforcement is in addition to that required for other reasons  h /3 A  h /2 • The supporting links may be placed in a zone beyond the intersection of beams Solid slabs EC2: Cl 9.3 • Curtailment – as beams except for the “Shift” rule al = d may be used • Flexural Reinforcement – and max areas as beam • Secondary transverse steel not less than 20% main reinforcement • Reinforcement at Free Edges Detailing Comparisons Beams EC2 BS 8110 Main Bars in Tension Clause / Values Values As,min 9.2.1.1 (1): 0.0013 bd 0.26 fctm/fykbd  0.0013 bh As,max 9.2.1.1 (3): 0.04 bd 0.04 bh Main Bars in Compression As,min 0.002 bh As,max 9.2.1.1 (3): 0.04 bd 0.04 bh Spacing of Main Bars dg + mm or  or 20mm smin 8.2 (2): Smax Table 7.3N dg + mm or  Table 3.28 Links Asw,min 9.2.2 (5): (0.08 b s fck)/fyk 0.4 b s/0.87 fyv sl,max 9.2.2 (6): 0.75 d 0.75d st,max 9.2.2 (8): 0.75 d  600 mm d or 150 mm from main bar 9.2.1.2 (3) or 15 from main bar Detailing Comparisons Slabs EC2 BS 8110 Main Bars in Tension Clause / Values Values As,min 9.2.1.1 (1): As,max 0.04 bd 0.26 fctm/fykbd  0.0013 bd 0.0013 bh 0.04 bh Secondary Transverse Bars As,min 9.3.1.1 (2): 0.2As for single way slabs 0.002 bh As,max 9.2.1.1 (3): 0.04 bd 0.04 bh Spacing of Bars smin 8.2 (2): dg + mm or  or 20mm dg + mm or  9.3.1.1 (3): main 3h  400 mm Smax secondary: 3.5h  450 mm places of maximum moment: main: 2h  250 mm secondary: 3h  400 mm 3d or 750 mm Detailing Comparisons Punching Shear EC2 BS 8110 Links Clause / Values Values Asw,min 9.4.3 (2): (fck)/fyk Link leg = 0.053 sr st Total = 0.4ud/0.87fyv Sr 9.4.3 (1): 0.75d 0.75d St 9.4.3 (1): Spacing of Links 1.5d within 1st control perim.: 1.5d outside 1st control perim.: 2d Columns Main Bars in Compression As,min 9.5.2 (2): 0.10NEd/fyk  0.002bh 0.004 bh As,max 9.5.2 (3): 0.06 bh 0.04 bh Links Min size 9.5.3 (1) 0.25 or mm 0.25 or mm Scl,tmax 9.5.3 (3): (12min; 0.6 b;240 mm) 12 9.5.3 (6): 150 mm from main bar 150 mm from main bar