technical report: crack widths
Crack width calculation to BS 8007 for combined flexure and direct tension H. G. Kruger derives and sets out the equations for determining surface crack widths for sections under combined flexure and direct tension, which, according to the author, are not covered in the BS 8007 standard
H. G. (Erhard) Kruger, is a specialist engineer associate in the structural division of BKS (Pty) Ltd.
ε12 ε2 ε21 ε22 εm
B
S 8007 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. This Technical Report examines the procedures given in the code, and shows the separate equations for flexure and direct tension to be based on similar premises. The situation where tension exists across the whole of the section is examined, and the limiting values of the depth of the neutral axis are calculated. Equations are developed for surface strains and the stiffening effect of the concrete. Similar equations are developed for the second case where some compression is present on one face of the section. It is shown how the design crack widths for these cases can be determined. Equations are developed with variables allowing for different reinforcement ratios and concrete cover at each face.
Es fc fs fs1 fs2 fstif fstif1 fstif2 Fstif Fstif1
Fstif2
h k1
k2 K
Notation a’
d
distance from the compression face to the point at which crack width is being calculated distance from face 1 to centroid of the reinforcement at face 1 distance from face 2 to centroid of the reinforcement at face 2 distance from point considered to surface of the nearest longitudinal bar area of tension reinforcement area of reinforcement at face 1 area of reinforcement at face 2 width of section at the centroid of the tension steel 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 effective depth
e
eccentricity K=
Ec
modulus of elasticity of concrete (1/2 the instantaneous value when used to determine αe)
a1 a2 acr
As As1 As2 bt b cmin c1 c2
J
L
M T
N O P
modulus of elasticity of reinforcement compressive stress in concrete stress in reinforcement stress in reinforcement at face 1 stress in reinforcement at face 2 stiffening tensile stress in concrete stiffening tensile stress in concrete at face 1 stiffening tensile stress in concrete at face 2 total stiffening tensile force in concrete portion of stiffening tensile force acting at the level of the steel at face 1 portion of stiffening tensile force acting at the level of the steel at face 2 overall depth of section
εm1 εm2 εsm εsm1 εs εs1 εs2 ∆εs ∆εs1 ∆εs2 ρ1
a2 a constant < = F h - a1 a constant < =
strain at face 2 ignoring stiffening effect of concrete strain due to stiffening effect of concrete between cracks strain due to stiffening effect of concrete between cracks at face 1 strain due to stiffening effect of concrete between cracks at face 2 average strain at level where cracking is being considered average strain at face 1 average strain at face 2 average strain in tension reinforcement average strain in tension reinforcement at face 1 strain in tension reinforcement strain in reinforcement at face 1 strain in reinforcement at face 2 strain reduction in tension reinforcement due to tension stiffening of concrete strain reduction in reinforcement at face 1 due to tension stiffening of concrete strain reduction in reinforcement at face 2 due to tension stiffening of concrete ratio of reinforcement at face 1 e=
ρ2
h - a2 F a1
V R S e+ h -a W W S = 2h SS e - 2 + a WW X at the section Tapplied moment
s1
ratio of reinforcement at face 2 e=
a constant for a particular section under a certain configuration of moment and direct tension
A o bh
A s2 o bh
Note: Generally subscripts 1 and 2 refer to faces 1 and 2 of the section respectively.
2
1
M
considered x h
n1
ratio
T
applied direct tension at the 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 distance to the centroid of the concrete stiffening force from face 2
w w1 w2 x x’
αe ε1 ε11
modular ratio b =
Es Ec
l
strain at level considered ignoring the stiffening effect of concrete strain at face 1 ignoring stiffening effect of concrete
18|The Structural Engineer – 17 September 2002
Introduction The British Code of Practice, BS 80071 includes recommendations for the calculation of design surface crack widths for sections under flexure and for sections under direct tension. However, the code does not provide recommendations for sections under combined flexure and direct tension. Furthermore, little guidance is given in the literature. Neither Anchor2 nor Batty3 provides a rational approach to allow for the effect of tension stiffening for the case of combined flexure and direct tension. Since combined flexure and direct tension often exist in structural elements of certain water-retaining structures (e.g. in the horizontal direction of walls of rectangular or square tanks), a need for calculating design crack widths for this case exists. The Technical Report proposes a method for calculating the design crack widths, w1 and w2, at the two faces of a section under these loadings.
technical report: crack widths
Fig 1. Stiffening effect of concrete in flexure
The effective strain reduction at any level in the section, therefore, is: f2 =
2bt h 3Es As
....(6)
Similarly, it can be shown that the value of the tensile stress at the tension face for a design crack width of 0.1mm, is taken as 1MPa.
Crack width formulae The design crack width defined by eqn (1) of BS 8007: Appendix B, is: w=
BS 8007 approach The procedure for the calculation of design crack widths given in Appendix B of BS 8007 can be summarised as follows: • calculate the average strain in the section at the level where cracking is being considered allowing for the stiffening effect of the uncracked concrete between cracks • calculate the design crack width using this value of the strain.
Df s =
f sm = f s -
Fstif =
bt (h - x) 3
....(1)
The effective strain reduction in the steel is:
bt (h - x) 3Es As
=
w = 3acr f m
ad - xk
ad - xk
=f1 - f 2 where f2 =
-
3E s A s a d - x k
2 bt h 3
....(8)
Combined flexure and direct tension ...(4)
The procedure for calculating the design crack width for sections under combined flexure and tension, can be summarised as follows:
bt a h - x k a al- x k
• determine the position of the neutral axis in the cracked section, x • determine the strain due to tension stiffening of concrete between cracks, ε2 • determine the average strain at the level where cracking is considered, εm • determine the crack width according to eqn (7).
3Es As a d - x k
Similarly, by considering the cracked concrete section in direct tension as shown in Fig 2, Eqn (5) of BS 8007: Appendix B for the stiffening effect of concrete in direct tension for a design crack width of 0.2mm, can be derived as follows: The stiffening force of the concrete in tension is: Fstif =
....(7)
Since this is the same as eqn (4) of BS 8007: Appendix B, it can, therefore, be assumed that eqn (7) will also apply to the case of combined flexure and direct tension.
(al- x) f sm
(al- x) f s
acr - cmin m h-x
....(3)
The average value of the strain at a distance al from the compression face, where the crack width is to be calculated, is:
b t a h - x k a al- x k
1 + 2c
In a section under direct tension, the value of the depth of the neutral axis is x = –∞ or x = ∞. By substituting these values in eqn (7) the design crack width for a section under direct tension is:
....(2)
The average value of the steel strain, therefore, is:
fm = The average steel strain may, as an approximation, be determined by calculating the steel stress on the basis of a cracked section, and reducing this by the tensile force due to tension stiffening in the concrete (BS 8110: Part 24 Clause 3.8.3, Assessment of crack widths). By considering the cracked concrete section in flexure as shown in Fig 1 (See BS 8110: Part 2 Fig 3.1), Eqn (2) of BS 8007: Appendix B, which defines the stiffening effect of the concrete for a design crack width of 0.2mm, can be derived as follows5. The value of the tensile stress at the tension face between cracks is assumed as 2/3MPa. The stiffening force of the concrete in tension, therefore, is:
bt (h - x) 3Es As
3acr f m
Two cases can be considered: • Case 1: Complete section in tension • Case 2: Section partially in compression.
....(5)
Fig 2. Section in direct tension
Case 1: Complete section in tension Determining the neutral axis depth Consider the cracked concrete section with a width, b, as shown in Fig 3. The position of the neutral axis x, is defined as negative when it is above face 2 of the section, where face 2 is defined as the side of the section under compression when the section is subjected to a moment in an anti-clockwise direction. The complete section would be in tension when x ≤ 0 or x ≥ h. By considering horizontal and moment equilibrium, and keeping in mind that x is negative as shown in Fig 3(c), the position of the neutral axis, x, is2: (See panel at end)
....(9)
17 September 2002 – The Structural Engineer|19
technical report: crack widths
From equation 9 the following equations can be derived for the limits of x as indicated: For x # 0:
and K=
t1 t2
$ - k1 K where k1 =
....(12)
t1 For x ≥ h: t 2 # - k 2 K h - a2 a1
Fig 4. (Below) Relationship between x and ρ1/ρ2 Fig 5. (Bottom) Stiffening effect of concrete, complete section in tension
M + T _ 0.5h - a2 i bht 1 _ h - a1 - a2 i
....(14)
f s2 = t12 d T - t 1 f s1 n bh
....(15)
f s1 =
Consider for example a concrete section, with b = 1000mm, h = 400mm, a1 = 50mm, a2 = 60mm, M = 10kNm and T = 400kN. The relationship between
t1 the depth of the neutral axis, and t 2 , is shown in Fig 4. Determining the stiffening force Consider a section with width, b, as shown in Fig 5(a), with the neutral axis position at x ≤ 0. The maximum stiffening tensile stress in the concrete is: f stif1 = 2 3 MPa for w = 0.2mm, and f stif1 =1MPa for w = 0.lmm
f stif2 = f stif1
e ≠ (h/2) – a1 t1 When - k1 K # t 2 < - K , –∞ < x ≤ 0 h≤x<∞
20|The Structural Engineer – 17 September 2002
....(16)
Since x is negative, it follows from the figure that:
....(13)
When the eccentricity of the tension force coincides with the centroid of the reinforcement at face 1, ie when the denominator of eq 11 equals 0, it can be shown that the section is partially in compression. The equations in the following section should then be used to determine x. In summary, provided that:
When - K < t 1 # - k 2 K , t2
x = –∞ or x = ∞
When none of the above applies, 0 < x < h. The equations for steel stresses are:
a2 h - a1
....(11)
+ a1 m
t1 For x = ∞ or x = –∞: t 2 = - K
where k 2 =
t1 When t 2 = - K ,
....(10)
c e + 2h - a2 m c e - 2h
Fig 3. Case 1 – Complete section in tension
_- xi
_h - xi
....(17)
The total stiffening force is: Fstif = e
fstif1 + fstif2 o bh 2
....(18)
The centroid of the stiffening force is at: x l=
h _ 2f stif1 + f stif2 i 3 _ f stif1 + f stif2 i
....(19)
Taking moments about the steel in face 2 yields: Fstif1 = Fstif d
x l- a2 n h - a1 - a2
From horizontal equilibrium:
....(20)
technical report: crack widths
Fig 6. Case 2 – Section partially in compression
and moment equilibrium, the position of the neutral axis x, can be found from the following equation2 by using trial and error or computer methods: (See panel at end)
....(34)
where n1 = x/h and e = M/T. The concrete stress, fc, can be determined from eqn (35) (See panel at end)
....(35)
The equations for steel stresses, are:
....(21)
Fstif2 = Fstif - Fstif1
f 12 = f s2 When the neutral axis position is at x ≥ h (Fig 5(b)): f stif2 = 2 3 MPa f stif2 =1MPa
for w = 0.2mm, and ....(22) for w = 0.1mm
h f stif1 = f stif2 x x
Determining the average strain The effective reduction in strain in the reinforcement due to the stiffening effect of the concrete is: Fstif1 , and E s A s1 Fstif2 Df s2 = E s A s2
....(24) ....(25)
The effective reduction in strain at the faces of the section is: f 21 = Df s1 -
f 22 = Df s2 +
a1 _ Df s2 - Df s1 i
_ h - a1 - a2 i
a2 _ Df s2 - Df s1 i
_ h - a1 - a2 i
....(26)
....(27)
The strain in the concrete at the faces, ignoring the stiffening effect, is: f 11 = f s1 +
a1 _ f s1 - f s2 i
_ h - a1 - a2 i
....(29)
Therefore, according to BS 8007 Clause B3, the average strains are as follows: At face l: εm1 = ε11– ε21 At face 2: εm2 = ε12– ε22
....(28)
....(36)
f s2 = a e x -x a2 f c
....(37)
Determining the stiffening force As shown in Fig 7, the maximum stiffening tensile stress in the concrete is:
....(30) ....(31)
....(23)
As previously, the total stiffening force is as given in eqn (18), the centroid of the stiffening force as in eqn (19), and the values of Fstif1 and Fstif2 as in equations (20) and (21) respectively.
Df s1 =
a2 _ f s1 - f s2 i
_ h - a1 - a2 i
f s1 = - a e h - xx - a1 f c
Determining the design surface crack width
f stif1 = 2 MPa 3
for w = 0.2mm, and
f stif1 =1MPa
for w = 0.lmm ....(38)
When 0 < x
3a cr f m ....(32) 2 _ a cr - c min i 1+ _h - xi When x ≥ h, this equation has to be modified by replacing (h – x) in the denominator by x. Therefore: When x # 0: w =
Fstif = 1 fstif1 _ h - x i b 2
....(39)
The centroid of the stiffening force is at: x l= x + 2h 3
3a cr f m when x $ h: w = ....(33) 2 _ a cr - c min i 1+ x
....(40)
Taking moments about the reinforcement in face 2 yields:
By substituting the values for εm1 or εm2 in these eqns, the values of the design crack widths, w1 and w2 at faces 1 and 2 respectively, can be calculated. It should be noted that when A2 << A1, w2 can actually be larger than w1, even if the bending moment is applied in an anti-clockwise direction as shown in Fig 5.
Fstif1 = Fstif d
x l- a2 n h - a1 - a2
....(41)
From horizontal equilibrium: Fstif2 = Fstif - Fstif1
....(42)
When a2 ≤ x < h: The complete stiffening force acts on the reinforcement in face 1 as in the case of flexure only. Therefore,
Case 2: Section partially in compression Determining the neutral axis position Consider a cracked section with a width, b, which is partially under compression, as shown in Fig 6. Define fc and fs2 as negative to indicate compression. By considering horizontal
Fstif1 = 1 f stif1 _ h - x i b 2 =
b _h - xi for w = 0.2mm, and 3
=
b _h - xi for w = 0.1mm 2
....(43)
Determining the average strain When 0
Df s1 =
Fig 7. Stiffening effect of concrete, section partially in compression
....(44) ....(45)
Due to the section being partially in compression, cracks can form only in face 1 and the crack width in face 2 need not be calculated. The effective reduction in strain at face 1 is:
17 September 2002 – The Structural Engineer|21
technical report: crack widths
f21 = Dfs1 +
a1 _ Dfs1 - Dfs2 i
Table 2: Surface crack width examples ....(46)
_ h - a1 - a2 i
The strain in the concrete at face 1, ignoring the stiffening effect, is: f 11 = f s1 +
a1 _ f s1 - f s2 i
....(47)
_ h - a1 - a2 i
Therefore, according to BS 8007 Clause B3, the average strain at face 1 is: f m1 = f 11 - f 21 ....(48) When a2 ≤ x < h: The effective reduction in strain in the reinforcement due to the stiffening effect of the concrete is: Df s1 =
f stif1 b _ h - x i 2E s A s1
....(49)
Parameter Case a1 mm a2 mm As1 As2 M kNm T kN ρ1/ρ2 –k1K –K –k2K x mm w1 mm w2 mm1
1
Example 2
x
x>h 48 46 T16@150mm T12@150mm 10 525 1.78 0.27 1.48 7.85 1289.9 0.200 0.177
[ b = 1000mm; h = 300mm; fy = 460MPa; Es = 200GPa; Ec = 27GPa ] 1 Negative value indicates that face is uncracked
The average value of the steel strain, therefore, is: f sm1 = f s1 -
f stif1 b _ h - x i 2E s A s1
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. 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.
....(50)
The average value of the strain at face 1 is: f m1 =
_ h - x i f sm1
_ h - a1 - x i 2 f stif1 b _ h - x i _ h - x i f s1 = .(51) _ h - a1 - x i 2E s A s1 _ h - a1 - x i
Examples Examples of sections checked for design surface crack widths of 0.2mm are given in Table 2.
= f 11 - f 21 where f 21 =
b _h - xi
2
Acknowledgement
3E s A s1 _ h - a1 - x i
Conclusions
for w = 0.2mm, and =
b _h - xi
3 o
direct tension. 2. The equations for the stiffening effect of concrete in flexure and in direct tension given in BS 8007: Appendix B can be derived from first principles by assuming the maximum stiffening tensile stress of the concrete between cracks as 1MPa and 2/3MPa for design crack widths of 0.1mm and 0.2mm respectively. It can, therefore, be assumed that the same principles apply to the case of combined flexure and direct tension. 3. Two cases exist for a section subjected to combined flexure and direct tension, i.e. • complete section is in tension • section is partially in compression. Equations to determine the neutral axis depth, x, for both these cases have been given. 4. The proposed method enables the designer to check design crack widths for different configurations of reinforcement and concrete cover in the two faces of a section, resulting in a more economical design. 5. Although the method involves somewhat cumbersome mathematics, the equations are easily programmable in spreadsheet programmes like Corel Quattro Pro, Microsoft Excel etc, or in Mathcad.
1. It has been shown that the expression for design crack widths in flexure as given in Equation (1) of BS 8007: Appendix B, also applies to the case of direct tension. It can, therefore, be assumed that it will also apply to the case of combined flexure and
2
2E s A s1 _ h - a1 - x i
for w = 0.1mm The equations for flexure in BS 8007, therefore, applies directly to this case. Determining the design surface crack width In both cases, ie when 0 < x < a2 and a2
Equations panel t1 t2 x=
Transition Consider a section with the following characteristics: b = 1000mm; h = 350mm; c1=c2 =40mm As1: T16 @ 150mm; As2: T16 @ 150mm; fy = 460MPa; fcu = 35MPa. The service moment and service tension capacities of the section for a design surface crack width of 0.2mm, are: Ms = 87.8kNm and Ts = 675kN.
The author thanks Robert Anchor for his assistance in the development of this method, both in discussions and by correspondence which stretched over a number of years. Without his help this Technical Report would not have been se possible.
_ h - a 1i c e t1 t2
ce - 2 h
h 2
+ a1
+ a1
m + a 2 c e + 2h - a 2 m
m + ce + 2 - a 2m
2 n1 d 3 - 2n1 + 6e n + 6t2 _ ae - 1i d n1 - a2 n d1 - 2a2 + 2e n + 6t1 ae d1 - a1 - n1 n d1 - 2a1 - 2e n = 0 h h h h h h h
T = x + a - 1 1 - a2 As2 - a h - 1 - a1 As1 _ e id e d x n bh x x n bh fc bh 2h
Table 1: Design surface crack widths for different configurations of flexure and direct tension
....(34)
....(35)
REFERENCES
T [kN]
M [kNm]
x [mm]
w1 [mm]
w2 [mm]
1.
675 675 675 675 20 5 1 0
0 1 5 10 87.8 87.8 87.8 87.8
–∞ –10712.1 –2002.4 –913.7 83.0 84.6 85.0 85.1
0.200 0.204 0.218 0.236 0.208 0.202 0.200 0.200
0.200 0.196 0.180 0.161 – – – –
2.
22|The Structural Engineer – 17 September 2002
....(9)
h
3. 4. 5.
BS 8007: Design of concrete structures for retaining acqueous liquids, London, British Standards Institution, 1987 Anchor, R. D.: Design of liquid retaining concrete structures, (2nd ed.), London, Edward Arnold, 1992, pp 45 – 52 Batty, I., Westbrook, R.: The design of water-retaining structures, Harlow, Longman Scientific and Technical, 1991, pp 58 – 60 BS 8110 Structural use of concrete: Part 2: Code of practice for special circumstances, London, British Standards Institution, 1985, pp 3/2 – 3/4, 3/6 – 3/7 Rowe, R. E. et al: Handbook to British Standard BS 8110:1985 Structural use of concrete, Sussex, Palladian Publications Ltd, 1987, p 179
verulam
Queries, comments, Q correspondence, and curiosities… Codes of practice I have received a lengthy letter from Mr S. L. Hammond of Whitstable, Kent who has given much thought to the use of partial safety factors in Codes of Practice and the design principles on which the codes are based. For the sake of comprehension Mr Hammond’s letter is published in full. He writes: When I entered the field of structural engineering the new design codes BS 8110 and BS 5950 were coming into use, since then I have used these codes along with the masonry code which is in a similar format. As an engineer I have tried to make sense of these publications and understand the theoretical basis of the design rules. Unfortunately the more closely these documents are examined, the less sense they make from the point of view of basic physical laws and mathematics. All these codes, and the new ‘Eurocodes’, make use of so called ‘partial safety factors’ in order to calculate the design loads acting on the structure, or a particular section of a member, the idea being that the various ‘partial factors’ represent different risks of the calculated load being exceeded in the actual structure. Take the following example of a simply supported beam loaded with a variety of non-uniform imposed and dead loads, these different types of load having different partial factors applied to them. Then calculate the design (factored) moments and shears for the beam, and the actual (non factored) moments and shears in the beam. Unless both types of load are applied in an identical way, the ratio (factor of safety) between actual and design forces will vary from location to location in the beam. In the same way a calculated example will show that the ratio between the calculated design loads and the actual loads acting on a structure will vary from loca-
tion to location in a structure, thus showing that the use of partial safety factors negates the objective of structural analysis which is to calculate accurately the distribution of forces in a structure due to a given set of loads. Investigation of the statistical basis of the ‘partial factors’ also reveals faulty mathematics. We are told that the partial factors are set according to the degree of risk of the particular type of load exceeding the factored load (whether any proper assessment of real loads has ever been carried out we are not told). Thus for the dead load we assume that, say, 99% of the dead loads will be less than or equal to 1.4 times the calculated dead load, and for the live load, 99% of the actual live loads will be less than or equal to 1.6 times the calculated live load. Thus the actual probability that the calculated factored load will be exceeded depends on a combination of probabilities that varies with the proportion of dead to live load (as the factors of safety are not equal). In fact the highest probability of failure occurs when the load is all of one type (dead or live load), and the minimum probability of failure occurs when the two loads are approximately equal. –There is an appreciable difference in the probabilities of the order of a factor of 100, thus showing that the stated objective of providing a consistent probability of the structure not failing is not met. The introduction of further partial factors such as for wind load only serves to complicate the statistics even more. (As an elementary example of the probability of combined events consider the probability of throwing two sixes in succession with a dice, the probability is one in thirty-six as opposed to one in six for a single six. Loads are continuous variables and the probabilities that the load will exceed the total factored load varies with the ratio of the loads, but unless there is only one type of load, the probability of exceedance bears no relationship to the probabilities on which the partial factors are
based.) This covers the general format of the current codes. However there are also specific mistakes in the individual codes. I will try and briefly list some of the worst errors that I have come across in each of the codes: BS 8110: To achieve the stress distribution assumed in the section design for beams, a considerable amount of deflection must occur, sufficient to lead to a rearrangement of forces in the structure, and possible collapse. Certainly the rotations required to develop the design stresses in the beam negate any question of elastic analysis, although the code assumes an elastic distribution of forces in subframe analysis. Column design: as far as I have been able to find out (Kong and Evans) the design of column sections also assumes yield in the steel and failure in the concrete. This involves the structure in further contortions in order that it may achieve the assumptions on which the section design is based. No account is taken of elastic instability or deflection of the column. The same remarks apply to the design of concrete walls. BS 5950: Allowing for the incorrect format of safety factors the beam design appears to be adequate and cover more design situations than previous codes. However, the column design is based upon inelastic buckling, which for open sections such as channels and H section columns is inadequate. Examination of Timoshenko & Gere’s Theory of elastic stability shows that the stability of open section columns is treated in Chapter 5 ‘Torsional buckling’ and that for the commonly used Hsection, failure occurs in a combined torsional and lateral buckling mode. (This is not to be confused with the torsional buckling of beams). No account is taken of this in allowable stresses for columns either in BS 449 or BS 5950. In any case the use of open
sections for columns is inefficient. S.H.S., R.H.S. and C.H.S. sections should be used, with built up or fabricated box section columns for heavier loads. With these type of sections the treatment and formulae given in Chapter 1 of Timoshenko and Gere, ‘Beam columns’ can be applied. As far as existing structures go, the most appropriate form of remedial work to the columns is to weld plates across the toes of the flanges, thus transforming the open section into a closed section and avoiding the problem of torsional instability. BS 5400: Generally the same comments as BS 5950 apply, with the added observation that the ratios between dead and live loads vary more in a bridge structure than building structures, and therefore the actual factors of safety vary more greatly across the structure. BS 5628: Apart from the format of partial safety factors the method given for the design of elements under vertical load is wrong. The capacity of masonry walls and columns is dependent on the elastic properties of the masonry as well as its crushing strength. No account is taken of this. Mr Hammond says in his covering note that his conclusions are based on sound mathematical reasoning and that not only is he alarmed by the inadequacy of the current codes but by the fact that they do not offer an adequate means of assessing existing structures for strength. Is there support for Mr Hammond’s conclusions among the membership? If so please write in.
Promoting the profession Andrew Sandford has sent in this letter after reading the interview with Professor David Blockley which appeared in NCE 8-15 August.
17 September 2002 – The Structural Engineer|23