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 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.
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.
h
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 concrete portion of stiffening tensile force acting at level level of steel at face 1 portion of stiffening tensile force acting at level level of steel at face 2 overall depth of section
k1
a constant =
k2
h - a2 a constant = a1
f ´s2 f stif1 f stif2 f y F stif F stif1 F stif2
K
a1 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
N
- a2 O
O
+ a1 O
P
M applied moment at section considered x n1 ratio h T applied axial tension at section considered w design surface crack width crack width at at face 1 w1 design surface crack w2 design surface crack width at face 2 distance to the neutral axis from x face fa ce 2 xstif apparent neutral axis depth of stiffening strain strain from face 2 α e
b
modular ratio = E s
E c
l
strain at face face 1 ignoring stiffening stiffening effect of concrete ε 12 strain at face face 2 ignoring stiffening stiffening effect of concrete ε m 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 ε 11
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
n
...(1)
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
a constant for a particular section under a certain configuration of moment and axial tension
J h Ke+ 2 =K Ke- h 2 L
Notation
< <
concrete ∆ε s2 strain reduction in reinforcement at face 2 due to tension tension stiffening of concrete ρ 1 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: 2 Equations (9) to (15) in Kruger 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. i.e. x However er,, the xstif ≠ x x.. Howev 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. Prop Pr opor 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
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 is negative,it follows from the figure that:
_ i - x
_ i
f stif2 = f stif1
...(3)
h-x
The total stiffening force is:
e
F stif =
f st if1 + f stif2 2
o
...(4)
bh
Fig 2. (below) Stiffening effect of concrete; x≤0
The tensile stiffening forces in the two reinforcement layers are: Fstif1 = f
l
Fstif2 = f
l
s1 s2
As 1
...(5)
As 2
From horizontal equilibrium: Fstif = f
l
s1
A s1 + f
l
s2
...(6)
As 2
Since x is negative,it follows from the figure that: l
l
f s1 f s2 = a2 - x h - x - a1
Therefore l
f s2 =
f
l
s1
_
a2 - x
h - x - a1
...(7)
i
...(8)
Substituting equation (8) in equation (6) and re-arranging, it follows that: l
f s1 =
A s1 +
F stif a2 - x As2 h - x - a1
...(9)
Fig 3. (above) Stiffening effect of concrete; x≥h
Similarly: F stif h - x - a1 a2 - x As1 + As2
l
f s2 =
...(10)
But, from the diagram: f s1 h - x - a1 a2 - x = f s2
...(11)
Substituting equation (11) in equations (9) and (10): l
f s1 =
f s1 F stif f s1 A s1 + f s2 As2
l
f s2 =
f s2 F stif f s1 As1 + f s2 As2
...(12)
...(13)
F stif can be determined from equations (2) to (4). When the neutral axis position is at x ≥ h (Fig 3): f stif 2 = 2
3
N/mm
2
for w = 0.2 mm
...(14)
Df s2 =
f s2 f s1 As1 + f s2 As2
F stif
...(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 Determining the neutral axis position: Strictly speaking, two cases have to be considered, i.e. x < a2 where f s2 ≤ 0 (i.e. compression) x ≥ a2 where f s2 > 0 (i.e. tension) When x ≥ a2: The equations previously given by the author2 apply to this case. The position x of
the neutral axis is given by equation 17 (originally eqn 34)2 : (see panel 1) where n1 = x and e = M . The concrete h T stress, f c and steel stresses, f s1 and f s2, can be determined from equations (35) to (37) in Kruger2. When: x < a2: a By setting n1 = 2 in equation (17), the h value of the eccentricity, e, can be determined for which x ≤ a2 i.e.(See Eq 18, panel 1). For this case f s1 and f s2 are both tensile. If these are defined as positive, and f c as negative,and by considering horizontal and moment equilibrium, the position of the neutral axis, x can be determined from the equation (19) (see panel 1): where n1 = x and e = M . h T The concrete stress, f c, 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. It should be noted that both equations (17) and (19) are cubic, and 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. Proportioning the stiffening force: As shown in Fig 4 and Fig 5, the maximum stiffening tensile stress in the concrete is again given by Eq (2). 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. For the improved method, consider a section with width, b, as shown in Fig 4. From the figure it follows that the total stiffening force is:
2
and f stif 2 = 1N/mm for w = 0.1 mm
As previously, equations (12) and (13) can be used to determine the tensile stiffening stress in the two layers of reinforcement. Determining the average strain: By dividing both sides of equations (12) and (13) by Young’s modulus for steel, Es, the effective reduction in strain in the reinforcement due to the stiffening effect of concrete is:
Df s1 =
f s1 f s1 A s1 + f s2 As2
F stif
...(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
M [kNm] 0 1 5 10 87.8 87.8 87.8 87.8
x [mm] –∞ –10712.1 – 2002.4 –913.7 83.0 84.6 85.0 85.1
w1 [mm] 0.200 0.204 0.217 0.234 0.208 0.202 0.200 0.200
w2 [mm] 0.200 0.197 0.185 0.169 – – – –
Previous method w1 [mm] 0.200 0.204 0.220 0.239 0.208 0.202 0.200 0.200
w2 [mm] 0.200 0.197 0.182 0.164 – – – –
17 May 2005 – The Structural Engineer|19
technical note:
bs 8007
determined by the previous (Kruger) method and the proposed improved method are given in Table 1. 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. As for the previous method,it shows a smooth transition in the values of design surface crack widths from direct tension to combined direct tension and flexure, and from flexure to combined flexure and direct tension.
Fig 4. (left) Stiffening effect of concrete, section partially in compression; x
Fig 5. (left) Stiffening effect of concrete, section partially in compression; x≥a2
Examples Four examples of sections checked for design surface crack widths of 0.2mm with both the improved and previous methods are given in Table 2.
Comparison with previous method
Table 2: Examples Parameter
Tables 1 and 2 show that the previous method underestimated the value of w2, and overestimated the value of w1.
Example
Case a1 mm a2 mm As1 As2 M kNm T kN fc –k1K –K –k2K x mm w1 (improved) mm w2 (improved) mm 1 w1 (previous) mm w2 (previous) mm 1
1 x<0 48 55 T16@125 mm T10@150 mm 30 400 3.07 1.37 6.30 32.14 –132.7 0.151 0.056 0.188 –0.038
2 x>h 48 46 T16@150 mm T12@150 mm 10 525 1.78 0.27 1.48 7.85 1289.9 0.169 0.223 0.200 0.177
3 a 2
4 0
_ i _ i _ i
1 f stif1 h - x b 2 =
=
b h-x 3
b h-x 2
for w = 0.2 mm, and
for w = 0.1 mm
...(21)
The values of f´ s1 and f´ s2 can again be determined from equations (12) and (13) respectively. When a2 ≤ x < h As previously explained by the author2 and from Fig 5, 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. 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).The values of the average strains can be determined from equations (26) to (31) given previously in Kruger2. When a2 ≤ x < h The method described by the author2 still applies to this case, and equations (49) to (51) given there are equally valid here. 20|The
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. An improved method for proportioning the tensile stiffening force is proposed, which results in the stiffening strains emanating from the neutral axis position. Generally, the previous method underestimated the design surface crack width at face 2, w2, and overestimated the design surface crack width at face 1, w1.This should not have serious consequences on calculations performed with the previous method, but the authors are of the opinion that the method proposed in this report is technically more consistent than the previse ous method.
H. G. Kruger BEng (Civil), BEng (Hons) Struct, AIStructE, PrEng, MSAICE, is a specialist engineer associate of BKS (Pty) Ltd, Pretoria, South Africa and H. R. C.Atkinson, BSc. Civ Eng, MIStruct E, MICE, Chartered Engineer, is a partner of Howes Atkinson Partnership (HAP) Consulting Engineers in Marlow, Buckinghamshire.
(b = 1000mm h = 300mm f y = 460N/mm2 Es = 200kN/mm2 Ec = 27kN/mm2); 1Negative value indicates that face is uncracked; 2Eqn (17) yields a result of 19.6mm
Fstif =
Conclusions
About the authors
Determining the design surface crack width: In both cases,i.e.when 0 < x < a2 and a2 ≤ x < h, the crack width at face 1 can be determined from equation (1).
Transition
REFERENCES
Consider a section with the following characteristics: b=1000mm h=350mm c1=c2=40mm As1: T16 @ 150mm As2: T16 @ 150mm f y = 460N/mm2 f cu = 35N/mm2 The service moment and service tension capacities of the section for a design crack width of 0.2mm, are:
1.
BS 8007 Design of concrete structures for retaining aqueous liquids , London, British Standards Institution, 1987 2. Kruger, H. G.: ‘Crack width calculation to BS 8007 for combined flexure and direct tension’, The Structural Engineer, 80/18, 17 September 2002, p 18-22 3. Crack width equations for combined flexure and tension, Course : BS 8007 – design of concrete liquid-retaining structures , British Cement Association, 51024 4. Tuma, J. J., Walsh, R. A.: Engineering Mathematics Handbook, (4th ed.), New York, McGraw-Hill, 1997, p 7
M s = 87.8kNm and T s= 675kN A comparison between the crack widths
Panel 1 n
2
1
d
6e h
3 - 2n1 +
R S a hS 2 S h T
#
Structural Engineer – 17 May 2005
2
1
d
3 - 2n1 +
T x = + f c bh 2h
2a2 h
n
d ah n
2
3-
id
6
2
+6
d
a1 h
+ 6 t1 a e 1 -
e
1-
2
2
e
1
s2
-
-
n ta d V a a W n d h n WW h X a n h a e n t a d ah h h a A x n bh
a2 h
+ 12 t1 a e 1 -
2
nd
n1 -
d e n t a d n ah n d h a A a d n bh a d xh -6
n
_
+ 6 t 2 a e - 1
2
d nd
e
n
2
e
1-
2a2 2e + h h
2
1-
2
+6
1
e
1-
a1 - n1 h
nd
1 -
2a1 2e h h
n
= 0 ...( 17)
1
...( 18)
a1 h
1-
2
2
2
-1-
+ 1
2
+6
s1
1
e
...( 20)
1-
1
- n1
nd
1-
2a1 2e h h
n
= 0 ...( 19)
