technical note: bs 8007Improved crack width calculation method to BS 8007 for combined flexure and direct tension BS 80071 includes recommendations for the calculation of design crack widths for sections under flexure and for sections under direct tension. It does not provide recommendations for sections under the combined forces. In 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. concrete ∆εs2 strain reduction in reinforcement at face 2 due to tension stiffening of concrete ρ1 ratio of reinforcement at face 1 d= A s1 n bh ρ2 ratio of reinforcement at face 2 d= A s2 n bh 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 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 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 suggests, 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 this. It also provides revised equations for the case where the neutral axis is between face 2 and its adjacent reinforcement. f´s2 fstif1 fstif2 fy Fstif Fstif1 Fstif2 h k1 k2 K face 1 stiffening stress in reinforcement at face 2 stiffening tensile stress in concrete at face 1 stiffening tensile stress in concrete at face 2 characteristic strength of reinforcement total stiffening tensile force in concrete portion of stiffening tensile force acting at level of steel at face 1 portion of stiffening tensile force acting at level of steel at face 2 overall depth of section a constant = < a2 h - a1 F a2 a constant = < h a1 F a constant for a particular section under a certain configuration of moment and axial tension J Ke+ =K KeL h -a N 2O 2 O h +a O 1 2 P Note: Generally subscripts 1 and 2 refer to faces 1 and 2 of the section respectively Introduction The author2 showed that the separate equations for flexure and direct tension are based on similar premises, and that eqn (1) of BS 8007: Appendix B, i.e. w= 3acr f m a - c min 1 + 2 d cr n h-x ...(1) can be used both for flexure and for direct tension. He then indicated 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 For combined flexure and direct tension, two cases; (i) Complete section in tension and; (ii) Section partially in compression, can be considered: Case 1: Complete section in tension Determining the neutral axis depth: Equations (9) to (15) in Kruger2 still apply. Proportioning the stiffening force: Previous method: Consider a section as shown in Fig 1. The 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, Fstif1 and Fstif2. Apportionment according to 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, i.e. xstif ≠ x. However, the lecture notes of the British Cement Association3 contains a figure that seems to suggest that the stiffening strains should emanate from the neutral axis position. It is generally accepted that this is the case when the neutral axis lies within the section, so it would be consistent to adopt the same approach when the neutral axis is beyond the section. Proportioning the stiffening force: Improved method: Consider a section with width, b, as shown in Fig 2. Say f´s1 and f´s2 are the tensile stiffening stresses in the two layers of reinforcement. With Notation a1 a2 acr As1 As2 b cmin c1 c2 e Ec distance from face 1 to centroid of reinforcement at face 1 distance from face 2 to centroid of reinforcement at face 2 distance from point considered to surface of the nearest longitudinal bar area of reinforcement at face 1 area of reinforcement at face 2 width of section considered (normally 1m) minimum cover to tension steel minimum cover to reinforcement at face 1 minimum cover to reinforcement at face 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 at face 1 stress in reinforcement at face 2 stiffening stress in reinforcement at M n1 T w w1 w2 x xstif applied moment at section considered ratio x h applied axial tension at section considered design surface crack width design surface crack width at face 1 design surface crack width at face 2 distance to the neutral axis from face 2 apparent neutral axis depth of stiffening strain from face 2 modular ratio b = E s E l c strain at face 1 ignoring stiffening effect of concrete strain at face 2 ignoring stiffening effect of concrete average strain at level where cracking is being considered strain in reinforcement at face 1 strain in reinforcement at face 2 strain reduction in reinforcement at face 1 due to tension stiffening of αe ε11 ε12 εm εs1 εs2 ∆εs1 Es fc fcu fs1 fs2 f´s1 18|The Structural Engineer – 17 May 2005 217 0.x f s2 .x .197 0.1mm As previously.xi _.200 0.200 0.x .220 0.164 – – – – 17 May 2005 – The Structural Engineer|19 .8 87. Proportioning the stiffening force: As shown in Fig 4 and Fig 5. and therefore the solution of n1 can also be found directly as described by Tuma4 or on the web page: http://mathforum. two cases have to be considered.202 0.e. x≥h Df s2 = f s2 Fstif f s1 As1 + f s2 As2 .org/dr.x .8 –∞ –10712.(4) The tensile stiffening forces in the two reinforcement layers are: Fstif1 = f ls1 As1 Fstif2 = f ls2 As2 From horizontal equilibrium: Fstif = f ls1 As1 + f ls2 As2 . compression) x ≥ a2 where fs2 > 0 (i. the maximum stiffening tensile stress in the concrete is: f stif1 = 2 3 N/mm 2 for w = 0. x≤0 the neutral axis is given by equation 17 (originally eqn 34)2 : (see panel 1) where n1 = x and e = M .0 85.... e.4 –913.x . (right) Previous method for proportioning stiffening effect _h .(16) .a1 s2 . When: x < a2: By setting n1 = a2 in equation (17).234 0..xi . the effective reduction in strain in the reinforcement due to the stiffening effect of concrete is: Df s1 = f s1 Fstif f s1 As1 + f s2 As2 . and fc as negative.(3) Fig 2. Determining the average strain: By dividing both sides of equations (12) and (13) by Young’s modulus for steel.197 0. equations (12) and (13) can be used to determine the tensile stiffening stress in the two layers of reinforcement.0 84. When the neutral axis position is at x ≥ h (Fig 3): f stif2 = 2 3 N/mm 2 for w = 0..(15) Table 1: Comparison between design surface crack widths for the improved and previous methods Improved method T [kN] 675 675 675 675 20 5 1 0 Previous method w1 [mm] 0...208 0..x A h .x i h ..cubi c. When 0 < x < a2: Apportionment according to the previous method suggested by the author2 again results in the ‘neutral axis’ of the stiffening strain diagram not coinciding with the neutral axis of the section.e..204 0.169 – – – – w2 [mm] 0.(9) Fig 3.(7) Substituting equation (8) in equation (6) and re-arranging.... It should be noted that both equations (17) and (19) are cubic.math/faq/faq.x . the position of the neutral axis. For the improved method..x But. x < a2 where fs2 ≤ 0 (i.. The position x of Substituting equation (11) in equations (9) and (10): f ls1 = f s1 Fstif f s1 As1 + f s2 As2 f s2 Fstif f s1 As1 + f s2 As2 .(5) Since x is negative.239 0.equations2.8 87.(14) and f stif2 = 1N/mm 2 for w = 0. as shown in Fig 4.(8) .a1 A + A s1 s2 a2 .8 87.(13) Fstif can be determined from equations (2) to (4). the h value of the eccentricity. can be determined from equations (35) to (37) in Kruger2.. For this case fs1 and fs2 are both tensile.. can be determined for which x ≤ a2 i. i. and by considering horizontal and moment equilibrium.200 M [kNm] 0 1 5 10 87.technical note: bs 8007 the neutral axis position at x ≤ 0.(2) and f stif1 = 1N/mm 2 for w = 0.. Equations (26) to (33) in Kruger2 can still be used to eventually calculate the design crack widths. it follows from the figure that: f ls1 f ls2 = h .. T h The concrete stress.182 0. tension) When x ≥ a2: The equations previously given by the author2 apply to this case.200 w2 [mm] 0. Case 2: Section partially in compression Determining the neutral axis position: Strictly speaking...200 0. consider a section with width. (See Eq 18..6 85. Es.1 – 2002.1 x [mm] w1 [mm] 0.. b. x can be determined from the equation (19) (see panel 1): where n1 = x and e = M . If these are defined as positive.202 0.208 0.200 0. it follows from the figure that: f stif2 = f stif1 Fig 1.(11) These equations differ from those in the previous method (equations (24) and (25)).2mm .. fc and steel stresses.1mm Since x is negative. the maximum stiffening tensile stress in the concrete is again given by Eq (2). can be determined from Eq (20) (see panel 1) The equations for steel stresses are again given by equations (36) and (37) previously presented in Kruger2.(10) Fstif a2 .185 0.e.html...204 0. panel 1). it follows that: f ls1 = As1 + Similarly: Fstif h .(6) . fs1 and fs2.200 0. from the diagram: f ls2 = h ..(12) f ls2 = .7 83. (above) Stiffening effect of concrete.x Therefore f ls2 = f ls1 _ a2 .a1 .200 0. The concrete T h stress. fc.e.2mm .a1 a2 . From the figure it follows that the total stiffening force is: The total stiffening force is: Fstif = e f stif1 + f stif2 o bh 2 . (below) Stiffening effect of concrete...a1 = f s1 a2 .. 2e n = 0 . (b = 1000mm h = 300mm fy = 460N/mm2 Es = 200kN/mm2 Ec = 27kN/mm2). 4. Table 1 indicates how the values of the design crack width at the two faces of the section vary with different configurations of flexure and direct tension.2mm with both the improved and previous methods are given in Table 2. This should not have serious consequences on calculations performed with the previous method.199 – Tables 1 and 2 show that the previous method underestimated the value of w2. H. BS 8007 Design of concrete structures for retaining aqueous liquids.x i b 2 b _h .a2 n d1 .2mm. MIStruct E. Table 2: Examples Parameter 1 Case a1 mm a2 mm As1 As2 M kNm T kN fc –k1K –K –k2K x mm w1 (improved) mm w2 (improved) mm1 w1 (previous) mm w2 (previous) mm1 x<0 48 55 T16@125 mm T10@150 mm 30 400 3. is a specialist engineer associate of BKS (Pty) Ltd.188 –0. 17 September 2002.44 72. 1Negative value indicates that face is uncracked.48 7.. A comparison between the crack widths Panel 1 2 n d 3 .22 -6. BSc. and overestimated the value of w1.a1 As1 n ed ed x x n bh 2 bh f c bh 2h .192 – Example 2 x>h 48 46 T16@150 mm T12@150 mm 10 525 1.78 -0.2a2 + 2e n + 6t1 a e d1 .a2 n d1 . Course : BS 8007 – design of concrete liquid-retaining structures.). x<a2.2a2 n + 6t1 a e d1 . Conclusions The previous method for proportioning the tensile stiffening force proposed by Kruger2 resulted in the ‘neutral axis’ of the stiffening strain diagram not coinciding with the neutral axis of the section under the applied forces. w1.: ‘Crack width calculation to BS 8007 for combined flexure and direct tension’. When a2 ≤ x < h As previously explained by the author2 and from Fig 5. The Structural Engineer. 51024 Tuma.. London.a h . British Cement Association.n1 n d1 . British Standards Institution. i. Generally. As for the previous method. G..223 0. p 7 3. MICE. section partially in compression.62 102.a1 ..e.22 0.038 Comparison with previous method 3 4 0<x<a2 48 46 T16@150 mm T12@150 mm 30 300 1. and overestimated the design surface crack width at face 1. Examples Four examples of sections checked for design surface crack widths of 0.2 0.( 18) 2 . An improved method for proportioning the tensile stiffening force is proposed.175 – 0. BEng (Hons) Struct.056 0. 2Eqn (17) yields a result of 19. Fig 5.a2 n h h h 2 n d 3 .169 0. the crack width at face 1 can be determined from equation (1). (left) Stiffening effect of concrete.2e n = 0 . Walsh.7 0. and = 3 b _h .22 -1. About the authors H.30 32. Pretoria. Kruger BEng (Civil).. is a partner of Howes Atkinson Partnership (HAP) Consulting Engineers in Marlow. and equations (49) to (51) given there are equally valid here.9 0.a1 ..78 0.1i d n1 .199 – 0.: Engineering Mathematics Handbook.a2 As2 .technical note: bs 8007 Fig 4. Transition Consider a section with the following characteristics: b=1000mm h=350mm c1=c2=40mm As1: T16 @ 150mm As2: T16 @ 150mm fy = 460N/mm2 fcu = 35N/mm2 The service moment and service tension capacities of the section for a design crack width of 0. p 18-22 Crack width equations for combined flexure and tension. South Africa and H. (left) Stiffening effect of concrete.6 d a2 n + 12t1 a e d1 .2a1 n W h h h h W S h X e# T .n1 n d1 . are: Ms = 87. w2. 2.1 . the previous method underestimated the design surface crack width at face 2.xi for w = 0.( 17) 1 h h h h h h h V R W S a2 2 h S d n d 3 ..xi for w = 0. J. but the authors are of the opinion that the method proposed in this report is technically more consistent than the previse ous method. A.. (4th ed.. PrEng.2a1 . Determining the design surface crack width: In both cases.200 0. AIStructE.2n1 + 6e n + 6t 2 _ a e . Chartered Engineer.(21) The values of f´s1 and f´s2 can again be determined from equations (12) and (13) respectively. G. J.. McGraw-Hill. R.( 20) 20|The Structural Engineer – 17 May 2005 . C.07 1.151 0.27 1.75 20.( 19) 1 h h h h h h h T = x + a 1 . Buckinghamshire.78 18. R..1mm = 2 . 1987 Kruger. The values of the average strains can be determined from equations (26) to (31) given previously in Kruger2.85 1289.177 a2<x<h 48 46 T16@150 mm T12@150 mm 63 60 1.14 –132. 1997. when 0 < x < a2 and a2 ≤ x < h. which results in the stiffening strains emanating from the neutral axis position. Atkinson. When a2 ≤ x < h The method described by the author2 still applies to this case.2n1 + 6e n + 6t 2 a e d n1 . x≥a2 determined by the previous (Kruger) method and the proposed improved method are given in Table 1.a1 .00 539. and from flexure to combined flexure and direct tension.6mm Fstif = 1 f stif1 _ h . it is clear that the complete stiffening force acts on the reinforcement in face 1 as in the case of flexure only and equation (43) from Kruger2 still applies. New York.2a2 + 2e n + 6t1 a e d1 . Determining the average strain: When 0 < x < a2 The values of the effective reduction in strain in the reinforcement due to the stiffening effect of concrete are given by equations (15) and (16).2a1 . Civ Eng. section partially in compression. 80/18.8kNm and Ts= 675kN REFERENCES 1.a2 n d1 . MSAICE.37 6.a1 . it shows a smooth transition in the values of design surface crack widths from direct tension to combined direct tension and flexure.2mm.