Improved Crack Width Calculation Method To BS 8007 For Combined Flexure and Direct Tension 2005

 

technical note:

bs 8007

Improved crack width calculation method to BS 8007 for combin combined ed flexure flexure and and direct direct tension tension BS 80071 includes recommendations for the calculation of  design crack widths for sections under flexure and for 

concrete ∆ε s2 strain reduction in reinforcement at face 2 due to tension tension stiffening of  concrete

sections under direct does the not combined provide forces. In recommendations fortension. sectionsItunder a previous technical note2 Erhard Kruger set out a method for  calculating crack widths under combined loads. Now with Robin Atkinson he proposes an improvement on the method.

ρ 1

I

n a previous Technical Note by one of  the authors2 it was shown that the separate equations for flexure and combined tension are based on similar similar premises.A  premises. A  method was proposed to proportion the tensile stiffening force to the two layers of  reinforcement by considering horizontal and moment equilibrium of the stiffening  forces.This forces. This method results in the ‘neutral axis’ of the stiffening strain diagram not coinciding with the neutral axis of the section under the applied forces. Some literature literature suggests, suggests, however however,, that the stiffening strains should emanate from the neutral axis position in all cases.This approach gives a seamless consistency throughout the whole range of possible combinations of moment and tensile force. The authors set out an improved method for proportioning the tensile stiffening  force to the two layers of reinforcement for certain cases in order to achieve achieve this.It this. It also provides revised equations for the case where the neutral axis is between face 2 and its adjacent adjacent reinforcement. reinforcement.

 f ´s2  f stif1  f stif2  f y  F stif 

concrete  F stif1 portion of stiffening tensile force acting at level level of steel at face 1  F stif2 portion of stiffening tensile force acting at level level of steel at face 2 h overall depth of section

a2 acr  As1  As2 b cmin c1 c2  e  Ec

 Es  f c  f cu  f s1  f s2  f ´s1

distance from face face 1 to centroid of  reinforcement at face face 1 distance from face face 2 to centroid of  reinforcement at face face 2 distance from point considered to surface of the nearest longitudinal bar area of reinforcement reinforcement at face 1 area of reinforcement reinforcement at face 2 width of section considered (normally 1m) minimum cover to tension steel minimum cover to reinforcement at face fa ce 1 minimum cover to reinforcement at face fa ce 2 eccentricity =  M  T  modulus of elasticity of concrete (1 / 2 the instantaneous value when used to determine αe) modulus of elasticity of reinforcement compressive stress in concrete 28 day characteristic (cube) strength of concrete stress in reinforcement reinforcement at face 1 stress in reinforcement reinforcement at face 2 stiffening stress in reinforcement at

18|The

< <

a2 h - a1

F F

a constant =

k2

a constant =

 K 

a constant for a particular section under a certain configuration of  moment and axial tension

J   h Ke+ 2 =K K e -  h 2 L

 h - a2 a1

N

- a2 O

O

+ a1 O

P

 M  applied moment at section considered n1 ratio  x h T  applied axial tension at section considered w design surface crack width w1 design surface crack crack width at at face 1 w2 design surface crack width at face 2  x distance to the neutral axis from face fa ce 2  xstif  apparent neutral axis depth of stiffening strain strain from face 2

b

l

α e

modular ratio =  E s

ε 11 11

strain at face face 1 ignoring stiffening  stiffening  effect of concrete strain at face face 2 ignoring stiffening  stiffening  effect of concrete

ε 12 ε m

 E c

average strain at level where cracking is being considered ε s1 strain in reinforcement reinforcement at face face 1 ε s2 strain in reinforcement reinforcement at face face 2 ∆ε s1 strain reduction in reinforcement at face 1 due to tension stiffening of 

Structural Engineer – 17 May 2005

d n =

 A s1 bh

ratio of reinforcement reinforcement at face 2

ρ 2

d

=

 A s2 bh

n

 Note: Gen  Note: Genera erally lly sub subscr script iptss 1 and 2 refe referr to to  faces  fac es 1 and and 2 of of the the sec sectio tion n resp respect ective ively ly

Introduction The author2 showed that the separate equations for flexure and direct tension are based on similar premises, premises, and that eqn (1) of BS 8007 8007:: Appe Appendix ndix B, B, i.e. w=

3acr f m  acr - c min 1+2 h-x

d

...(1)

n

can be used both for flexure and for direct tension.He tension. He then indicated that that it can therefore be assumed that eqn (1) will also apply to the case of combined flexure and direct tension.

Combined flexure and direct tension

k1

Notation a1

face 1 face stiffening stress in reinforcement at face fa ce 2 stiffening tensile stress in concrete at face fa ce 1 stiffening tensile stress in concrete at face fa ce 2 characteristic characterist ic strength of reinforcement total stiffening tensile force in

ratio of reinforcement reinforcement at face 1

For combined flexure and direct tension, two cases; (i) Complete section in tension tension and; (ii) Section partially partially in compression, can be considered: Case 1: Complete section in tension  Determ  Dete rmin inin ing g th the e ne neutr utral al axi axis s de dept pth: h: Equations (9) to (15) in Kruger2 still apply apply..  Prop  Pr opor orti tion oning ing th the e st stiff iffen enin ing g for force ce::  Prev  Pr eviou ious s me metho thod: d: Consider a section as shown in in Fig 1. The author author2 proposed a method for proportioning the total stiffening force to the two layers of reinforcement by considering horizontal and moment equilibrium of the stiffening forces, F  forces, F stif1 and F  and  F stif2.  Apport  App ortion ionmen mentt acc accord ording ing to thi thiss method results in the ‘neutral axis’ of the stiffening strain diagram not coinciding  with the neutral axis of the section under the applied forces, i.e.  xstif ≠ x  x.. Howev However er,, the lecture notes of the British Cement  Associ  Ass ociat ation ion3 contains a figure that seems to suggest that the stiffening strains should emanate from the neutral axis position. It is generally accepted that this this is the case when the neutral axis lies within the section,so section, so it would be consistent to adopt the same approach when the neutral axis is beyond the section.  Propor  Prop orti tion oning ing th the e st stiff iffen enin ing g for force ce::  Impr  Im prove oved d me meth thod: od: Consider a section with width, b, as shown shown in in Fig Fig 2. Say Say f´   f´ s1 and f´  and  f´ s2 are the tensile stiffening stresses in the two layers of reinforcement.With reinforcement. With

 

technical note:

the neutral axis position at x ≤ 0,the maximum stiffening tensile stress in the concrete is:  f stif 1 = 2

N/mm

3

2

  for w = 0.2 mm  

and f stif 1 = 1N/mm 2

bs 8007

Fig 1. (right) Previous method for proportioning stiffening effect

...(2)

for w = 0.1 mm  

Since x Since  x is negative,it negative, it follows from the figure that:  f stif2 = f stif1 

_ i - x

...(3)

_ i h-x

The total stiffening force is:    F stif  =

f st ifif1 + f stif2  2

e

o

bh

Fig 2. (below) ...(4)

Stiffening effect of concrete; x≤0

The tensile stiffening forces in the two reinforcement layers are:  Fstif1 = f

l

 Fstif2 = f

l

s1

As 1

s2

the neutral axis is given by equation 17 (originally eqn 34)2 : (see panel panel 1)  x  M  where n1 = and e = . The concrete h T  stress, f  stress,  f c and steel stresses, f  stresses, f s1 and and f   f s2, ca can n be determined from equations (35) to (37) in Kruger2. When: x When:  x < a2: a By setting n1 = 2 in equation (17), the h  value  va lue of th the e ecce eccentr ntrici icity ty, e  e,, can be deterdetermined for which x which x ≤ a2 i.e.(See Eq 18, panel 1). For this case f  case f s1 and and f   f s2 are both tensile. If these are defined as positive, positive, and and f   f c as negative,and negative, and by considering horizontal horizontal and moment equilibrium, equilibrium, the position of  the neutral axis, x axis, x can be determined from

...(5)

As 2

From horizontal equilibrium:  Fstif = f

l

s1

A s1 + f

l

s2

As2

...(6)

Since x Since  x is negative,it negative, it follows from the figure that: l

l

 f  s1  f  s2   = a2 - x h - x - a1

...(7)

Therefore l

 f s1 a2 - x  f  s2 = h - x - a1 l

 _

...(8)

i

Substituting equation (8) in equation (6) and re-arranging, re-arranging, it follows that: l

 f  s1 =

 F stif  a2 - x  A s1 +  A s2 h - x - a1

...(9)

Fig 3. (above) (above) Stiffening effect of  concrete; x≥h

Similarly: l

 f  s2 =

 F stif  h - x - a1 a2 - x  As1 + As2

...(10)

But, from the diagram: diagram:  f s1 h - x - a1 a2 - x =  f s2

...(11)

Substituting equation (11) in equations (9) and (10):  f s1

l

 f  s1 =

 F stif 

...(12)

Df s2 =  

f s2

  F stif   f s1 As1 + f s2 A s2

...(16)

These equations differ from those in the previous method (equations (24) and (25)). Equations (26) to (33) in Kruger2 can still be used to eventually calculate the design crack widths. Case 2: Section partially in compression compression  Determ  Dete rmin ining ing th the e neu neutr tral al axi axis s po posi siti tion on:: Strictly speaking, speaking, two cases have to be considere consi dered, d, i.e.

 f s1 A s1 + f s2 As2 l

 f  s2 =

 f s2  F stif   f s1 As1 + f s2 A s2

...(13)

 F stif  can be determined from equations (2) to (4). When the neutral axis position is at x at x ≥ h (Fig 3):  f stif 2 = 2

2

3

N/mm   for w = 0.2 mm  

...(14)

 x < a2 where f  where  f s2 ≤ 0 (i.e. compression)  x ≥ a2 where f  where  f s2 > 0 (i.e. tens tension) ion) When x ≥ a2: When x The equations previously given by the author2 apply to this case.The case. The position x position x of 

the equation (19) (see panel 1):  x  M  where n1 = and e = . h T  The concrete stress, f  stress, f c, can be deterdetermined from Eq (20) (see panel 1) The equations for steel stresses are again given by equations equations (36) and (37) previously presented in Kruger2. It should be noted that both equations (17) and (19) (19) are cubic, and therefore therefore the solution of n1 can also be found directly as described by Tuma4 or on the web page: http://mathforum.org/dr.math/faq/faq.cubi c.equations2.html.  Prop  Pr opor orti tion oning ing the st stif iffen fenin ing g for force ce:: As shown in Fig Fig 4 and Fig Fig 5, the maximum maximum stiffening tensile stress in the concrete is again given by Eq (2). When 0