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 2

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.

Analysis of a singly reinforced beam

Cl 3.1.7 EN 1992-1-1

Design equations can be derived as follows:

For grades of concrete up to C50/60, εcu = 0.0035,  = 1 and  = 0.8

M b

Maximum neutral axis depth

M b

Moment Bending

Elastic

Moment Bending

ted Redistribu



   k1  k2 xu d EC2 Equ 5.10a

xu = Neutral axis depth after redistribution

EC2 NA gives k1 = 0.4 and k2 = 1.0

d

ux 0.4  



0.4 -

d

M

K 

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

Analysis of a Singly Reinforced

Beam3.1.7 EN 1992-1-1

Design equations can be derived as follows:

For grades of concrete up to C50/60, εcu = 0.0035,  = 1 and  = 0.8

Now z = d - 0.4 x

& 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

bz f bd

f

bdz f

bd f

M

ck

ck ck

ck ck

0 = (z/d)2 – (z/d) + 0.88235K Solving the quadratic equation:

z/d = [1 + (1 - 3.529K)0.5 ]/2

z = d [ 1 + (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

)]/2 3,715K

(1 d[1

(1 d[1

)]/2 4,152K

(1 d[1

)]/2 4,412K

(1 d[1

Take moments about the centre of the compression force

M = 0.87As fyk z

Rearranging

As = M /(0.87 fyk z)

The required area of reinforcement can now be:

• calculated using these expressions

• obtained from Tables (eg Table 5 of How to beams and ConciseTable 15.5 )

• obtained from graphs (eg from the ‘Green Book’)

However , it is often considered good practice to limit the depth of the

neutral axis to avoid ‘over-reinforcement’ (ie to ensure a ductile failure)

This is not an EC2 requirement and is not accepted by all engineers A

limiting value for K can be calculated (denoted K’) as follows.

Design aids for flexure

Concise: Table 15.5

Beams with Compression

Reinforcement

There is now an extra force Fsc = 0.87As2 fyk

The area of tension reinforcement can now be

considered in two parts, the first part to balance the compressive force in the concrete, the second part is to balance the force in the compression steel The area of reinforcement required is therefore:

As = K’ fck b d 2 /(0.87 fyk z) + As2

where z is calculated using K’ instead of K

A s2 can be calculated by taking moments about the centre of the tension force:

M = K’ fck b d 2 + 0.87 fyk As2 (d - d2)

Rearranging

As2 = (K - K’) fck b d 2 / (0.87 fyk (d - d2))

The following flowchart outlines the design procedure for rectangular beams with concrete classes up to C50/60 and grade 500 reinforcement

Determine K and K’ from:

Note:  =1.0 means no redistribution and  = 0.8 means 20% moment redistribution.

Compression steel needed

M

K  & K' 0 6 0 18 2 0 21

Carry out analysis to determine design moments (M)

It is often recommended in the UK that K’ is limited to 0.168 to ensure ductile failure

1.00 0.208 0.95 0.195 0.90 0.182 0.85 0.168 0.80 0.153 0.75 0.137 0.70 0.120Design Flowchart

Calculate lever arm z from:

* A limit of 0.95d is considered good practice, it is not a requirement of Eurocode 2

d b f

yk

t

ctm min

,

Check max reinforcement provided As,max  0.04Ac (Cl 9.2.1.1) Check min spacing between bars > Øbar > 20 > Agg + 5

Check max spacing between bars

Calculate tension steel required from: A f M z

yd

s 

Flow Chart for Singly-reinforced

Beam

Flow Chart for

Calculate excess moment from: M '  bd2fck  K  K ' 

Calculate compression steel required from:

 2

yd 2

s

'

d d f

M A





Calculate tension steel required from:

Check max reinforcement provided A s,max  0.04A c (Cl 9.2.1.1) Check min spacing between bars > Ø bar > 20 > A gg + 5

2

s yd

2

A z

f

bd f

K

Flexure Worked Example

Initially assume 32 mm  for tension reinforcement with 25 mm nominal cover

to the link (allow 10 mm for link) and 20mm  for compression reinforcement with 25 mm nominal cover to link

Nominal side cover is 35 mm

 provide compression steel

mm 368

168

0 53 3 1

1 2

449

' 53 3 1

1 2

.

0

30 449

300

10

370

2 6 ck

2

K

f bd

M K

 

kNm

K K f bd

M ck

3.65

10)168.0204.0(30449

300

''

6 2

2

s

mm372

45) – (449435

10

x 65.3'

M A

2

6

2 s yd

s

mm 2293

372 364

435

10 )

3 65 370

M M

A

368

Provide 2 H20 for compression steel = 628mm 2 (372 mm 2 req’d)

and 3 H32 tension steel = 2412mm 2 (2296 mm 2 req’d)

By inspection does not exceed maximum area or maximum spacing of reinforcement rules

Check minimum spacing, assuming H10 links

Space between bars = (300 – 35 x 2 - 10 x 2 - 32 x 3)/2

Factors for NA depth (=nd) and lever arm (=zd) for concrete grade  50 MPa

(1)

Factors for NA depth (=nd) and lever arm (=zd) for concrete grade 70 MPa

Shear in Beams

• Shear design is different from BS8110

• Shear strength should be limited to the value for

C50/60

• The shear effects in links and longitudinal steel have to

be considered explicitly

• Resistance of member governed by the crushing of

compression struts - VRd,max

Applied shear force - VEd

Members Requiring Shear

 angle between shear reinforcement and the beam axis

 angle between the concrete compression strut and the beam axis

z inner lever arm In the shear analysis of reinforced concrete

without axial force, the approximate value z = 0,9d may normally

be used

cot

sw s

1 max

cd w

cw b z f V

21.8 <  < 45

Strut Inclination Method

We can use the following expressions from the code to calculate shear

reinforcement for a beam (Assumes shear reinforcement is always

Shear Design: Links

Variable strut method allows a shallower strut angle –

hence activating more links

As strut angle reduces concrete stress increases

Angle = 45° V carried on 3 links Angle = 21.8° V carried on 6 links

shear reinforcement control

concrete strut control

VRd,max = z bw 1 fcd /(cotθ + tanθ) = 0,5 z bwfcd sin 2

Exp (6.9) where   = 0,6(1-fck/250) Exp (6.6N)

Procedure for design with variable strut

1 Determine maximum applied shear force at support, VEd

2 Determine VRd,max with cot  = 2.5

3 If VRd,max > VEd cot  = 2.5, go to step 6 and calculate required shear

reinforcement

4 If VRd,max < VEd calculate required strut angle:

 = 0.5 sin -1[(vEd/(0.20fck(1-fck /250))]

5 If cot is less than 1 re-size element, otherwise

6 Calculate amount of shear reinforcement required

Asw/s = vEd bw/(fywd cot  ) = VEd /(0.78 d fyk cot )

7 Check min shear reinforcement, Asw/s ≥ bw ρw,min and max spacing,

sl,max = 0.75d ρ w,min = (0.08 √fck)/fyk cl 9.2.2

Shear Resistance with Shear

Reinforcement

Shear - Variable strut method

Concise Fig 15.1 a)

Shear - Variable strut method

Concise Fig 15.1 b)

• Where av  2d the applied shear force, VEd, for a point load

(eg, corbel, pile cap etc) may be reduced by a factor av/2d where 0.5  av  2d provided:

d d

− The longitudinal reinforcement is fully anchored at the support.

− Only that shear reinforcement provided within the central 0.75av is included in the resistance.

Short Shear Spans with Direct

Strut Action (6.2.3)

Note: see PD6687-1:2010 Cl 2.14 for more information

‘Shift’ Rule for Shear

Horizontal component of diagonal shear force

= (V/sin) cos = V cot

Applied shear V

• Flexural principles similar

• Shear approach different – should result in less shear reinforcement

• We will look at the SLS and detailing rules later

Shear Example

Design of shear reinforcement using

Eurocode 2

Design Flow Chart for Shear

Find the minimum area of shear reinforcement required to resist the design

shear force using EC2.

250 / 30 - 1 30 x 20 0

96 4 sin

5 0

) 250 /

1 ( 20 0

sin 5 0

1

ck ck

Ed 1





f f

v

43 1 cot 

(try 10mm links and 32mm main steel)

G k = 75 kN/m, Q k = 50 kN/m , assume no redistribution and use

equation 6.10 to calculate ULS loads.

8 m

450

1000

Beam Example 1 – Bending

ULS load per m = (75 x 1.35 + 50 x 1.5) = 176.25

0 30 934

450

10

1410

2 6

M K

3 1

1



2 6

yd

822 x

435

10 x





z f

M A

Provide 5 H32 (4021 mm 2 )

Beam Example 1 – Shear

Workshop Problem

Workshop Problem

Cover = 35 mm to each face

fck = 30MPaCheck the beam in flexure and shear

G k = 10 kN/m, Q k = 6.5 kN/m (Use eq 6.10)

8 m

300

450

0 30 389

M K

3 1

1



2 6

yd

334 x

435

10 x





z f

M A

Provide 3 H25 (1470 mm 2 )

Detailing

UK CARES (Certification - Product & Companies)

1 Reinforcing bar and coil

2 Reinforcing fabric

3 Steel wire for direct use of for further

processing

4 Cut and bent reinforcement

5 Welding and prefabrication of reinforcing

steel

www.ukcares.co.uk

www.uk-bar.org

Identification of bars

Class A

Class B

Class C

Reinforced Concrete Detailing

• Clear horizontal and vertical distance  , (dg +5mm) or 20mm

• For separate horizontal layers the bars in each layer should be

located vertically above each other There should be room to allow access for vibrators and good compaction of concrete.

Section 8 - General Rules

Spacing of bars

• To avoid damage to bar is

Bar dia  16mm Mandrel size 4 x bar diameterBar dia > 16mm Mandrel size 7 x bar diameterThe bar should extend at least 5 diameters beyond a bend

Minimum mandrel size, m

Min Mandrel Dia for bent bars

Minimum mandrel size, m

• To avoid failure of the concrete inside the bend of the bar:

 m,min  Fbt ((1/ab) +1/(2 )) / fcd

Fbt ultimate force in a bar at the start of a bend

ab for a given bar is half the centre-to-centre distance between bars

For a bar adjacent to the face of the member, ab should be taken as the cover plus  /2

Mandrel size need not be checked to avoid concrete failure if :

– anchorage does not require more than 5 past end of bend

– bar is not the closest to edge face and there is a cross bar  inside bend

– mandrel size is at least equal to the recommended minimum value

Min Mandrel Dia for bent bars

Bearing stress inside bends

Anchorage of reinforcement

EC2: Cl 8.4

The design value of the ultimate bond stress, fbd = 2.25 12fctd

where fctd should be limited to C60/75

1 =1 for ‘good’ and 0.7 for ‘poor’ bond conditions

unhatched zone – ‘good’ bond conditions

hatched zone - ‘poor’ bond conditions

• For bent bars lb,rqd should be measured along the

centreline of the bar

EC2 Figure 8.1

Concise Fig 11.1

Table requires values for:

Cd Value depends on cover and bar spacing, see Figure 8.3

K Factor depends on position of confinement reinforcement,

see Figure 8.4

λ = (∑Ast – ∑ Ast,min)/ As Where Ast is area of transverse reinf.

Table 8.2 - C d & K factors

Concise: Figure 11.3

EC2: Figure 8.3

EC2: Figure 8.4

Table 8.2 - Other shapes

Concise: Figure 11.1

EC2: Figure 8.1

Alpha values

EC2: Table 8.2 Concise: 11.4.2

Anchorage of links

Concise: Fig 11.2

EC2: Cl 8.5

EC2: Cl 8.7

l0 = α1 α2 α3 α5 α6 lb,rqd  l0,min

α6 = ( 1/25)0,5 but between 1.0 and 1.5

where 1 is the % of reinforcement lapped within 0.65l0 from the centre of the lap

Percentage of lapped bars

relative to the total

cross-section area

< 25% 33% 50% >50%

Note: Intermediate values may be determined by interpolation.

α1 α2 α3 α5 are as defined for anchorage length

l0,min  max{0.3 α6 lb,rqd; 15; 200}

Arrangement of Laps

EC2: Cl 8.7.3, Fig 8.8

Worked example

Anchorage and lap lengths

Anchorage Worked Example

Calculate the tension anchorage for an H16 bar in the bottom of a slab:

a) Straight bars b) Other shape bars (Fig 8.1 b, c and d)

Concrete strength class is C25/30 Nominal cover is 25mm

Assume maximum design stress in the bar

Basic anchorage length, l b,req

lb.req = (Ø/4) ( σsd/fbd) EC2 Equ 8.3

Max stress in the bar, σsd = fyk/γs = 500/1.15

Design anchorage length, l bd

lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min

lbd = α1 α2 α3 α4 α5 (40.36Ø) For concrete class C25/30

Alpha values

EC2: Table 8.2 Concise: 11.4.2

Table 8.2 - C d & K factors

Concise: Figure 11.3

EC2: Figure 8.3

EC2: Figure 8.4

Design anchorage length, l bd

lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min

lbd = α1 α2 α3 α4 α5 (40.36Ø) For concrete class C25/30

a) Tension anchorage – straight bar

Design anchorage length, l bd

lbd = α1 α2 α3 α4 α5 lb.req ≥ lb,min

lbd = α1 α2 α3 α4 α5 (40.36Ø) For concrete class C25/30

b) Tension anchorage – Other shape bars

Worked example - summary

H16 Bars – Concrete class C25/30 – 25 Nominal cover

Tension anchorage – straight bar lbd = 36.97Ø = 592mm Tension anchorage – Other shape bars lbd = 40.36Ø = 646mm

lbd is measured along the centreline of the bar

Compression anchorage (α1 = α2 = α3 = α4 = α5 = 1.0)

lbd = 40.36Ø = 646mm Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7

Lap length = anchorage length x α6

Anchorage & lap lengths

How to design concrete structures using Eurocode 2

Table 5.25: Typical values of anchorage and lap lengths for slabs

Bond Length in bar diametersconditions f ck /fcu

compression anchorage

length, lbd

Full tension and

compression lap length, l0

Note: The following is assumed:

- bar size is not greater than 32mm If >32 then the anchorage and lap lengths should be increased by a factor (132 - bar size)/100

- normal cover exists

- no confinement by transverse pressure

- no confinement by transverse reinforcement

- not more than 33% of the bars are lapped at one place

Lap lengths provided (for nominal bars, etc.) should not be less than 15 times the bar size

or 200mm, whichever is greater.

Anchorage /lap lengths for slabs

Manual for the design of concrete structures to Eurocode 2

Laps between bars should normally be staggered and not located in regions

of high stress, the arrangement of lapped bars should comply with the

following (see Figure 8.7 on next slide):

• The clear distance between lapped bars should not be greater than 4φ or 50 mm otherwise the lap length should be increased by a length equal to the clear space where it exceeds 4Ø or 50 mm

• The longitudinal distance between two adjacent (staggered?) laps

should not be less than 0,3 times the lap length, l 0 ;

• In case of adjacent laps, the clear distance between adjacent bars should not be less than 2Ø or 20 mm

When the provisions comply with the above, the permissible percentage of lapped bars in tension may be 100% where the bars are all in one layer Where the bars are in several layers the percentage should be reduced to 50%

All bars in compression and secondary (distribution) reinforcement may be lapped in one section

Arrangement of Laps

EC2: Cl 8.7.2 Concise: Cl 11.6

Arrangement of Laps

EC2: Cl 8.7.2, Fig 8.7 Concise: Cl 11.6.2

• 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