III. RESULTADOS
4.2. Recomendaciones
6.1.1. General (additional sub-section)
This section of the guide deals with the design at ultimate limit state of members subject to bending with or without axial force. It is split into the following additional sub-sections for convenience:
. Reinforced concrete beams Section 6.1.2
. Prestressed concrete beams Section 6.1.3
. Reinforced concrete columns Section 6.1.4
. Brittle failure of members with prestress Section 6.1.5
6.1.2. Reinforced concrete beams (additional sub-section) 6.1.2.1. Assumptions
2-1-1/clause 6.1(2)Pmakes standard assumptions for the calculation of ultimate moments of resistance as follows:
(1) Plane sections remain plane.
(2) Strain in bonded reinforcement, whether in tension or compression, is the same as the strain in the concrete at the same level.
(3) Tensile strength of the concrete is ignored.
(4) The stresses in the concrete in compression are given by the design stress–strain relation-ships discussed in section 3.1.7.
(5) The stresses in the reinforcing steel are given by the design stress–strain relationships dis-cussed in section 3.2.7.
(6) The initial strain in prestressing is taken into account.
2-1-1/clause 6.1(2)P
Assumption (1), relating to linear strains, is only appropriate for ‘beam-like’ behaviour and does not apply to deep beams (see definition in 2-1-1/clause 5.3.1) or to local load introduction, such as in prestressed end blocks or in the vicinity of bearings. In these situations the stresses and strains vary in a complex manner and 2-1-1/clause 6.1(1)P notes that strut-and-tie analysis is more appropriate, as discussed in section 6.5.
Local bond slip also means that the strains in reinforcement will not always exactly match those in the surrounding concrete, but the assumption of equal strains in (2) above is adequate for design.
2-1-1/clause 6.1(3)Pdescribes the stress–strain curves to be used for concrete, reinforce-ment and prestressing steel. The design stress–strain curves for reinforcereinforce-ment include one with an inclined branch (representing strain hardening) with a limit on the ultimate strain.
While its use can give rise to a small saving in reinforcement in under-reinforced beams, the calculations involved are more time-consuming and are not suited to hand calculations.
Computer software can be used to automate the process. For the purposes of developing design equations and worked examples, the rest of this chapter considers only the reinforce-ment stress–strain curve with a horizontal top branch and no strain limit. The same princi-ples, however, apply to the use of the inclined branch. The curves for prestressing also include one with an inclined branch and its use is illustrated in Worked example 6.1-5 to illus-trate the improvement over the use of the curve with the horizontal plateau. Whereas use of the curve with horizontal plateau for reinforcement leads to reinforced concrete bending resistances similar to those from BS 5400 Part 4,9 use of the curve with horizontal plateau for prestressing leads to a lower bending resistance.
For concrete, EC2 allows three different stress–strain relationships, as discussed in section 3.1 of this guide and illustrated in Fig. 3.1-3. As was shown in section 3.1.7, the differences between the three alternatives are very small and, in fact, the simple rectangular stress block generally gives the greatest moment of resistance. This is unlike both BS 5400 Part 4 and Model Code 90,6 where the peak stress used in the rectangular block was lower than that in the parabola-rectangle block. (Model Code 90 reduces the peak allowable stress in the rectangular block by a factor equivalent to 0 as used in 2-1-1/clause 6.5.) The design equations developed in the following sections apply to all three concrete stress blocks but it is simplest and most economic to use the simpler rectangular stress block, as illustrated in Worked example 6.1-1. The worked examples, however, generally use the parabolic-rectangular stress block as it is more general.
6.1.2.2. Strain compatibility
The ultimate moment resistance of a section can be determined by using the strain compat-ibility method, by either algebraic or iterative approaches. An iterative approach is possible using the following steps:
(1) Guess a neutral axis depth and calculate the strains in the tension and compression reinforcement by assuming a linear strain distribution and a strain of "cu2 (or "cu3 if not using the parabolic-rectangular stress–strain idealization) at the extreme fibre of the concrete in compression.
(2) Calculate from the stress–strain idealizations the steel stresses appropriate to the calculated steel strains.
(3) Calculate from the stress–strain idealizations the concrete stresses appropriate to the strains associated with the assumed neutral axis depth.
(4) Calculate the net tensile and compressive forces at the section. If these are not equal, adjust the neutral axis depth and return to step (1).
(5) When the net tensile force is equal to the net compressive force, take moments about a common point in the section to determine the ultimate moment of resistance.
The strain compatibility method described above is tedious for hand analysis, but must be used for non-uniform sections (or at least for sections which are non-uniform in the compres-sion zone). This method is illustrated in Worked example 6.1-4 for a flanged beam. A further difficulty, in this case for flanged beams, stems from the provisions of 2-1-1/clause 6.1(5) and 2-1-1/clause
6.1(1)P
2-1-1/clause 6.1(3)P
2-1-1/clause 6.1(5)
2-1-1/clause 6.1(6). The former requires the mean strain in parts of the section which are fully in compression with approximately concentric loading (such as the flanges of box girders where the neutral axis is in the webs) to be limited to "c2 or "c3 (as appropriate).
‘Approximately concentric’ is defined as e=h < 0:1, which is equivalent to a neutral axis at a depth below top of flange greater than 1.33h for a flange of depth h. This statement is a simplification of the range of limiting strain distributions in 2-1-1/Fig. 6.1, referenced by 2-1-1/clause 6.1(6), which is of general application. (The simplification of 2-1-1/clause 6.1(5) does not, however, simplify the resistance calculation itself .) Where a part has zero compressive strain at one face, the limiting strain can still be taken as "cu2 or "cu3 as appropriate for the stress block used. Where the part has equal compressive strains at both faces, a reduced limit of "c2 or "c3 applies as appropriate, depending on the stress–
strain idealization used.
The reduction in limiting strain for pure compression arises because the real concrete behaviour is such that the peak stress is reached at a strain approximating to "c2 (or "c3Þ and then drops off before the final failure strain is obtained. Thus, for pure compression, peak load is obtained at approximately "c2, but for pure flexure the resistance continues to increase beyond the attainment of this strain. For intermediate cases of strain diagram, the limiting strain needs to be obtained by interpolation between these cases and this can be done by rotating the strain diagram about the fixed pivot point, shown in Fig. 6.1-1.
The same applies for entire sections which are wholly in compression. For a flange of thick-ness h wholly in compression, this means limiting the strain to "c2at a height of h"c2="cu2in the flange. The simplified rule in 2-1-1/clause 6.1(5) limits the strain to "c2at mid-height.
The need for this additional level of complexity for bridges is partly mitigated by the use of the recommended value of cc¼ 0:85 which compensates for the drop-off in strength at increasing strain, as discussed in section 3.1.6. While the theoretical need for this additional complexity can be explained as above, the practical need seems dubious and was not required in BS 5400 Part 4.9The method is illustrated in Worked example 6.1-4. For beams with steel that yields with the usual assumption of concrete limiting stress of "cu2or "cu3, the effect of this modification is typically negligible. Where the steel does not yield, the effect can be a little more significant, but still usually relatively small. Where such calculation is required, it is considerably simpler to perform with the rectangular stress block.
For uniform sections (or at least uniform in the compression zone), it is possible to use the simplified design equations which are developed in the following sections.
6.1.2.3. Singly reinforced beams and slabs
Consider the singly reinforced rectangular beam illustrated in Fig. 6.1-2, with Fs¼ ð fyk=sÞAs and Fc¼ favbx acting at a lever arm x from the compression fibre, where fav is the average stress in the concrete above the neutral axis at the ultimate limit state. fav and are parameters relating to the geometry of the concrete stress block being used. Formulae for these are given in section 3.1.7 along with a tabulation of their values for varying concrete strengths and stress–strain idealizations. The failure strain shown in Fig. 6.1-2 of "cu2is only appropriate for the parabolic-rectangular block and should be replaced by "cu3 for the
2-1-1/clause 6.1(6)
Pivot point
Compressive strains in flange
Actual final strain
limiting cases for strain hεc2/εcu2
εc2 εcu2
h
Fig. 6.1-1. Allowable maximum strain in flanges depending on strain distribution ("c3and "cu3used for bilinear and rectangular stress block)
bilinear and rectangular blocks. The equivalent diagram for the rectangular stress block is shown in Fig. 6.1-3.
Equations are now developed with reference to Fig. 6.1-2. From moment equilibrium (assuming the steel yields), taking moments about the centroid of the compressive force:
M¼ Fsz¼ fyk
s Asz (D6.1-1)
Alternatively, taking moments about the centroid of the tensile force:
M¼ Fcz¼ favbxz (D6.1-2)
For equilibrium:
Fc¼ Fs) favbx¼ fyk
s As which can be rewritten as:
x d ¼ fyk
favs (D6.1-3)
with
¼As
bd (D6.1-4)
From the geometry of the stress–strain diagrams:
z¼ d x (D6.1-5)
Substituting this into equation (D6.1-2) gives:
M¼ favbxðd xÞ ¼ favbx
1 x
d
d
(a) Section (b) Forces/stress blocks (c) Strains βx x
d
b
z
εcu2
εs
fcd
Fc
Fs
As
Fig. 6.1-2. Singly reinforced rectangular beam at failure (using parabolic-rectangular concrete stress block)
(a) Section (b) Forces/stress blocks (c) Strains As
βx λx
z Fc
Fs
b
d
εs
x
εcu3
ηfcd
Fig. 6.1-3. Singly reinforced rectangular beam at failure (using rectangular concrete stress block)
or (D6.1-1) and (D6.1-6) can be used as design equations for the section’s ultimate moment resistance. A check must, however, be made to ensure the strain in the reinforcement is sufficient to cause yielding as assumed.
From the reinforcing steel stress–strain idealization (refer to section 3.2):
"s;yield¼ fyk
sEs
and from the strain diagram in Fig. 6.1-2:
"s¼"cu2 which can be expressed as:
x
d 1
fyk
sEs"cu2þ 1
(D6.1-8)
(Note that "cu2should be replaced by "cu3in equation (D6.1-8) if the rectangular or bilinear concrete stress blocks are used.)
If x=d does not satisfy equation (D6.1-8) then the reinforcement does not yield and the expressions in equations (D6.1-1) and (D6.1-3) are not valid, so neither is the resistance in equation (D6.1-6). In this case there are the following options:
(1) increase the section size to comply with equation (D6.1-8);
(2) add compression reinforcement (see section 6.1.2.4 on doubly reinforced beams) to comply with equation (D6.1-8);
(3) use the strain-compatibility method to establish actual reinforcement force and hence moment resistance;
(4) conservatively take the reinforcement strain from equation (D6.1-7) and the lever arm from equation (D6.1-5) using the depth of compression zone, x, from equation (D6.1-3) (determined assuming the steel yields), so that from equation (D6.1-1), M¼ AsEs"sz.
For the simple rectangular block, the above can be simplified by substituting for fav ¼ fcd and ¼ =2 (from the expressions in section 3.1.7) in equations (D6.1-3) and (D6.1-5) above so that:
It is still necessary to check that equation (D6.1-8) is satisfied (but "cu2should be replaced by "cu3Þ in order to use equation (D6.1-9), as it assumes that the steel yields. The modified equation is:
If the steel does not yield, one of the above options needs to be considered. The last option leads to conservatively taking the moment resistance as:
M¼ EsAs"sz with z¼ d In equation (D6.1-11), "sis determined from equation (D6.1-7) but substituting "cu3for "cu2. The equations above are obviously more suited to analysing given sections rather than designing a section to resist a given moment. For design purposes it is therefore convenient to rearrange the equations. If Kav¼ M=bd2fav is defined and substituted into equation (D6.1-6) then: which can be written as a quadratic equation ðx=dÞ2 ðx=dÞ þ Kav ¼ 0 with solutions:
x
The lower root of equation (D6.1-13) is the relevant one. Again, the ratio of x=d should be checked against the limit in equation (D6.1-8) and the section designed from rearranging equation (D6.1-1) to give:
AsMs
fykz (D6.1-14)
with z¼ d x as before from equation (D6.1-5) and x taken from equation (D6.1-13).
Worked example 6.1-1: Reinforced concrete deck slab
A 250 mm thick deck slab has a maximum applied ULS sagging moment of 150 kNm/m.
Assuming 40 mm cover to the main reinforcement and C35/45 concrete, find the required reinforcement area for adequate ultimate moment resistance using both the parabolic-rectangular and simplified parabolic-rectangular stress blocks. Reinforcement is B500B.
fck¼ 35 MPa. Assuming cc ¼ 0.85 and c¼ 1.5, from section 3.1.7 for the parabolic-rectangular concrete stress block, fav¼ 16.056 MPa and ¼ 0.416.
From equation (D6.1-12):
2 0:416 ¼ 0:262 (and an irrelevant root of 2.14)
Check limit from equation (D6.1-8) to ensure reinforcement is yieldingð fyk¼ 500 MPa, Es¼ 200 GPa and s¼ 1:15Þ:
The calculation is now repeated with the rectangular stress block for comparison:
From section 3.1.7 for the rectangular concrete stress block, fav ¼ 15.867 MPa and
¼ 0.400.
From equation (D6.1-12):
Kav¼ 150 106
1000 2002 15:867¼ 0:236 From equation (D6.1-13):
x
d ¼1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4 0:400 0:236 p
2 0:400 ¼ 0:264
Reinforcement will yield by inspection, thus x¼ 0:264 200 ¼ 52:9 mm and, from equation (D6.1-5), z¼ 200 0:400 52:9 ¼ 178:9 mm.
Thus, from equation (D6.1-14):
As150 106 1:15
500 178:9 ¼ 1929 mm2=m
which is slightly less reinforcement than required for the parabolic-rectangular block.
Equation (D6.1-9) can be used as a check on the moment resistance with this steel area:
M¼ Asfydd
1 fydAs 2 fcdbd
¼ 1929 500
1:15 200 1 500=1:15 1929 2 1:0 0:85 35
1:5 1000 200 0
@
1 A
¼ 150:0 kNm/m as required
Worked example 6.1-2: Voided reinforced concrete slab
h = 1500 mm
d 1400 mm 1400 mm
40 mm cover to links of (say) 20 mm dia.
1000 mm dia.
As = 25φ @ 150
Fig. 6.1-4. Voided reinforced concrete slab
Consider the voided reinforced concrete slab section shown in Fig. 6.1-4. Assuming C35/
45 concrete, fyk¼ 500 MPa, s¼ 1.15, c¼ 1.5 and cc ¼ 0.85, find the ultimate sagging moment resistance. If the ultimate applied moment is increased to 3000 kNm/m, find the additional reinforcement required to provide sufficient ultimate resistance.
(1) To find the moment resistance, consider the voided slab as a flanged beam of width, b¼ 1400 mm, slab depth above hole = slab depth below hole ¼ 250 mm and effective depth, d¼ h cover link f bar f=2; therefore d ¼ 1500 40 20 12:5 1425 mm:
As¼ 12:521400
150 ¼ 4581:5 mm2
From section 3.1.7 for the parabolic-rectangular concrete stress block using
cc¼ 0.85 and c¼ 1.5, fav ¼ 16.056 MPa and ¼ 0.416.
6.1.2.4. Doubly reinforced rectangular beams
Where the tension zone is very heavily reinforced, for efficiency it can become necessary to add compression reinforcement to reduce the depth of the concrete compression zone and thereby allow the tensile reinforcement to yield. This situation arises where the neutral axis depth exceeds the limit in (D6.1-8). It may also be necessary to analyse sections with known compression reinforcement for their ultimate flexural resistance. Of note is the fact that EC2 uses the same stress–strain relationship for reinforcement in tension and compression, unlike the relationships in BS 5400 Part 4.9
From equation (D6.1-4):
¼ 4581:5
1400 1425¼ 0:002296 From equation (D6.1-3):
x
d ¼ 500
16:056 1:15 0:002296 ¼ 0:0622
Check against limit from equation (D6.1-8) to ensure reinforcement is yielding:
fyk1
sEs"cu2þ 1
¼ 1
500
1:15 200 103 0:0035þ 1
¼ 0:6169 >x
d therefore okay
Thus x¼ 0:0622 1425 ¼ 88:6 mm < 250 mm; therefore neutral axis lies in the flange and the above equations are still valid.
From equation (D6.1-5), z¼ 1425 0:416 88:6 ¼ 1388:1 mm, and from equation (D6.1-1):
MRd¼ 500
1:15 4581:5 1388:1 106¼ 2765 kNm for a 1.4 m width (2) To find required increase in As to resist M¼ 3000 kNm/m:
For 1400 mm wide section, M¼ 1:4 3000 ¼ 4200 kNm.
From equation (D6.1-12):
Kav¼ 4200 106
1400 14252 16:056¼ 0:0920 From equation (D6.1-13):
x
d ¼1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4 0:416 0:0920 p
2 0:416 ¼ 0:0958
This is less than the limit from equation (D6.1-8); therefore the reinforcement will yield.
Thus x¼ 0:0958 1425 ¼ 136:6 mm < 250 mm; therefore neutral axis remains in the flange and, from equation (D6.1-5), z¼ 1425 0:416 136:6 ¼ 1368 mm.
Thus from equation (D6.1-14):
As4200 106 1:15
500 1368 ¼ 7060 mm2ð¼ 5043 mm2=mÞ Thus, additional
As¼7060 4581:5
1:4 ¼2479
1:4 ¼ 1770 mm2=m
i.e. add 20f at 150 mm centres (additional As¼ 2094 mm2/m) or use 32ff at 150 mm centres in place of 25f bars (giving total As¼ 5361 mm2/m).
Consider the doubly reinforced rectangular beam illustrated in Fig. 6.1-5 with associated strain and stress diagrams.
For equilibrium, assuming that all reinforcement yields:
Fcþ Fs0¼ Fs) favbxþ fyk
s A0s¼ fyk
s As therefore:
x¼ fyk
s
ðAs A0sÞ
favb (D6.1-15)
or alternatively:
A0s¼ Asfavbxs
fyk (D6.1-16)
Equation (D6.1-16) can be used to determine the required compression reinforcement to allow the tensile reinforcement to yield, when x=d for As alone exceeds the limit given in equation (D6.1-8). In using equation (D6.1-16), the value of x should first be reduced to comply with the limit in equation (D6.1-8).
Equation (D6.1-15) can be used to analyse sections with known reinforcement. x should be checked against the limit given in equation (D6.1-8), as before, to ensure the reinforcement is yielding. A further check is required to ensure that the compression reinforcement is also yielding:
"0s "s;yield) "cu2
1d0
x
fyk
sEs so that:
x d0 1
ð1 CÞ (D6.1-17)
where:
C¼ fyk
sEs"cu2 (D6.1-18)
If the reinforcement does not yield, then the ultimate resistance must be determined using the strain compatibility method. An example is given in section 6.1.2.5.
dʹ
(a) Section (b) Forces/stress blocks (c) Strains x βx
d
εs
Fc
Fʹs
Aʹs
εʹs
Fs
As
b fcd εcu2
Fig. 6.1-5. Doubly reinforced rectangular beam at failure with parabolic-rectangular stress block
Worked example 6.1-3: Doubly reinforced concrete slab
Consider a 1000 mm wide, class C35/45 concrete slab with effective depth of 275 mm to 40f reinforcement at 150 mm centres. Assuming fyk¼ 500 MPa, s¼ 1.15, c¼ 1.5,
cc¼ 0.85 and a parabolic-rectangular concrete stress distribution, find the area of compression reinforcement (at a depth, d0¼ 50 mm) required to fully utilize the tension reinforcement and determine the ultimate moment resistance.
From section 3.1.7 with fck¼ 35 MPa, fav ¼ 16.056 MPa and ¼ 0.416.
From equation (D6.1-4):
¼ 8377
1000 275¼ 0:0305 (3.05%) From equation (D6.1-3):
x
d ¼ 500
1:15 16:056 0:0305 ¼ 0:825 Check against limit in equation (D6.1-8):
1
500
1:15 200 103 0:0035þ 1
¼ 0:6169
but this is not greater than the actual x=d above; therefore it is necessary to add com-pression reinforcement until x=d ¼ 0:6169.
) xmax¼ 0:6169 275 ¼ 169:6 mm
For this depth of concrete in compression, check yielding of compression reinforcement at depth d0.
From equation (D6.1-18):
C¼ 500
1:15 200 103 0:0035¼ 0:6211
From equation (D6.1-17) for compression reinforcement to yield:
x > d0
ð1 CÞ¼ 50
ð1 0:6211Þ¼ 132 mm and as x¼ 169.6 mm, the steel yields.
Equation (D6.1-16) can therefore be used to find the required area:
A0s¼ 8377 16:056 1000 169:6 1:15
500 ¼ 2113:1 mm2=m
Therefore adopt 25f compression reinforcement at 225 mm centres (2182 mm2/m). The moment of resistance is now found.
From equation (D6.1-15):
x¼ 500
1:15ð8377 2182Þ
16:056 1000¼ 167:8 mm < xmax
limit above for tensile yield. This depth also exceeds the critical depth of 132 mm above for compression reinforcement yield. Therefore, the forces in concrete and steel are as follows:
Fc¼ favbx¼ 16:056 1000 167:8 103¼ 2693:9 kN Fs0¼ fyk
s A0s¼500
1:15 2182 103¼ 948:5 kN Fs¼ fyk
s As¼500
1:15 8377 103¼ 3642:4 kN
ðFcþ Fs0 Fs¼ 2693:9 þ 948:5 3642:4 ¼ 0 kN therefore section balances) Taking moments about top fibre to find moment of resistance:
MRd¼ Fsd Fcx Fs0d0
¼ ð3642:4 275 2693:9 0:416 167:8 948:5 50Þ 103¼ 764:1 kNm
6.1.2.5. Flanged beams
Flanged sections may be treated as a rectangular section using the equations derived above, providing that the neutral axis at the ultimate limit state remains within the flange. If the simplified rectangular concrete stress block is assumed, a flanged section could be considered as rectangular for neutral axis depths up to 1= times the flange thickness, which is 1.25 times the flange thickness for fck <50 MPa. This is because the stress block remains in the flange and 2-1-1/clause 6.1(5) only applies for approximately ‘concentric’ loading which is equiva-lent to a neutral axis at a depth of 1.33h for a flange of depth h, as discussed in 6.1.2.2 above.
Strictly, 2-1-1/clause 6.1(6) could be considered to require adjustment to the limiting concrete strain for a neutral axis depth of 1.25h, but this will generally have minimal effect.
Where the neutral axis lies within the web, the section should be analysed using the strain compatibility method discussed in section 6.1.2.2 above and illustrated in Worked example 6.1-4. A further complexity for flanged beams is that the variable concrete strain limit, dis-cussed in section 6.1.2.2, should theoretically be applied to the flange regions, although some judgement can be applied here to small flange projections – EC2-1-1 specifically refers to box girder flanges which are usually wide. This adds to the iteration necessary and gives greater need for the analysis by computer.
Worked example 6.1-4: Flanged reinforced concrete beam
1100 mm
1180 mm
900 mm 200 mm
1350 mm
As = 12 no. 40φ bars = 15 079.6 mm2
Fig. 6.1-6. Flanged reinforced concrete section
Consider the flanged section shown in Fig. 6.1-6. Assuming C35/45 concrete, fyk¼ 500 MPa, s¼ 1.15, c¼ 1.5 and cc¼ 0.85, find the ultimate sagging moment resistance using the parabolic-rectangular stress block.
By trial and error, guess a depth to neutral axis of 331 mm and consider the concrete stress block and strains illustrated in Fig. 6.1-7.
(a) Compressive forces/stress blocks (b) Strains 200 h1
h2
h3
a1
a2
a3
Fc1
Fc2
Fc3
x
fcd εcu2 = 0.0035
εc2 = 0.0020 σh1
σh2
Fig. 6.1-7. Compressive stresses in flanged reinforced concrete section From geometry of parabolic-rectangular concrete stress block:
h1¼
10:0020 0:0035x
¼3
7 331 ¼ 141:9 mm h2¼ 200 h1¼ 200 141:9 ¼ 58:1 mm
and
(Note that the 0.358 factor above, for calculating the centroid location, must be derived for the specific geometry of the remaining part of the parabola for each neutral axis depth considered in the iteration. This is particularly tedious for hand calculations and therefore using the simplified rectangular stress block is vastly easier – as illustrated at the end of this example.)
The reinforcement strain isð1180 331Þ=331 0:0035 ¼ 0:009 so it will yield.
The reinforcement strain isð1180 331Þ=331 0:0035 ¼ 0:009 so it will yield.