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18|The Structural Engineer 17 May 2005

technical note: bs 8007

I

n a previous Technical Note by one of

the authors2 it was shown that the sepa-

rate 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.

Notationa1 distance from face 1 to centroid of

reinforcement at face 1

a2 distance from face 2 to centroid of

reinforcement at face 2

acr distance from point considered to

surface of the nearest longitudinal barAs1 area of reinforcement at face 1

As2 area of reinforcement at face 2

b width of section considered (normally

1m)

cmin minimum cover to tension steel

c1 minimum cover to reinforcement at

face 1

c2 minimum cover to reinforcement at

face 2

e eccentricityT

M=

Ec modulus of elasticity of concrete (1/2

the instantaneous value when used

to determine e)

Es modulus of elasticity of reinforcement

fc compressive stress in concretefcu 28 day characteristic (cube) strength

of concrete

fs1 stress in reinforcement at face1

fs2 stress in reinforcement at face2

fs1 stiffening stress in reinforcement at

face 1

fs2 stiffening stress in reinforcement at

face 2

fstif1 stiffening tensile stress in concrete at

face 1

fstif2 stiffening tensile stress in concrete at

face 2

fy characteristic strength of reinforce-

ment

Fstif total stiffening tensile force in

concrete

Fstif1 portion of stiffening tensile force

acting at level of steel at face 1

Fstif2 portion of stiffening tensile force

acting at level of steel at face 2

h overall depth of section

k1 a constant h a

a1

2

= -< Fk2 a constant a

h a1

2=

-< FK a constant for a particular section

under a certain configuration of

moment and axial tension

eh

a

eh

a

2

2

1

2

=- +

+ -J

L

KKKK

N

P

OOOO

M applied moment at section considered

n1 ratio hx

T applied axial tension at section

consideredw design surface crack width

w1 design surface crack width at face 1

w2 design surface crack width at face 2

x distance to the neutral axis from

face 2

xstif apparent neutral axis depth of stiff-

ening strain from face 2

e modular ratioE

Es

c=b l

11 strain at face 1 ignoring stiffening

effect of concrete

12 strain at face 2 ignoring stiffening

effect of concrete

m average strain at level where crack-ing is being considered

s1 strain in reinforcement at face 1

s2 strain in reinforcement at face 2

s1 strain reduction in reinforcement at

face 1 due to tension stiffening of

concrete

s2 strain reduction in reinforcement at

face 2 due to tension stiffening of

concrete

1 ratio of reinforcement at face1

bhAs1

=d n2 ratio of reinforcement at face2

bhAs2

=d nNote:Generally subscripts 1 and 2 refer to

faces 1 and 2 of the section respectively

IntroductionThe author2 showed that the separateequations for flexure and direct tension

are based on similar premises,and that

eqn (1) of BS8007:Appendix B, i.e.

w

h xa c

a

1 2

3

mincr

cr m=

+-

-

f

d n...(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 tensionFor 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 Fig1.The author2 proposed a

method for proportioning the total stiffen-

ing force to the two layers of reinforcement

by considering horizontal and moment

equilibrium of the stiffening forces,Fstif1andFstif2.

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. xstifx. 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 consis-

tent 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.Sayfs1andfs2 are the tensile stiffening stresses

in the two layers of reinforcement.With

Improved crack width calculation method toBS 8007 for combined flexure and direct tension

BS 80071

includes recommendations for the calculation ofdesign crack widths for sections under flexure and forsections under direct tension. It does not providerecommendations for sections under the combined forces. Ina previous technical note2 Erhard Kruger set out a method forcalculating crack widths under combined loads. Now withRobin Atkinson he proposes an improvement on the method.

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17 May 2005 The Structural Engineer|19

technical note: bs 8007

the neutral axis position at x 0,the

maximum stiffening tensile stress in the

concrete is:

.

.

/

/

f for w

and f for w

N mm mm

N mm mm

23

0 2

1 0 1

stif

stif

12

12

= =

= =

...(2)

Sincex is negative,it follows from the

figure that:

f fh x

xstif stif 2 1=-

-__ ii ...(3)The total stiffening force is:

Ff f

bh2

stifst if stif 1 2

=+e o ...(4)

The tensile stiffening forces in the two

reinforcement layers are:

F f A

F f A

stif s s

stif s s

1 1 1

2 2 2

=

=

l

l...(5)

From horizontal equilibrium:

F f A f Astif s s s s1 1 2 2= +l l ...(6)

Sincex is negative,it follows from the

figure that:

h x a

fa x

fs s

1

1

2

2

- -=

-

l l...(7)

Therefore

fh x a

f a xs

s

21

1 2

=- -

-l

l _ i...(8)

Substituting equation (8) in equation (6)

and re-arranging, it follows that:

fA

h x aa x

A

Fs

s s

stif1

11

22

=+

- --

l ...(9)

Similarly:

f

a xh x a

A A

Fs

s s

stif

2

11 2

2 =

-- -

+

l ...(10)

But, from the diagram:

a xh x a

f

f

s

s

2

1

2

1

-- -

= ...(11)

Substituting equation (11) in equations

(9) and (10):

ff A f A

fFs

s s s s

sstif1

1 1 2 2

1=

+l ...(12)

ff A f A

fFs

s s s s

sstif2

1 1 2 2

2=

+l ...(13)

Fstifcan be determined from equations

(2) to (4).

When the neutral axis position is atx

h (Fig 3):

.

.

/

/

f w

f w

N mm for mm

and N mm for mm

23

0 2

1 0 1

stif

stif

22

22

= =

= =

...(14)

As previously, equations (12) and (13)

can be used to determine the tensile stiff-

ening stress in the two layers of reinforce-

ment.

Determining the average strain: By

dividing both sides of equations (12) and

(13) by Youngs modulus for steel,Es, the

effective reduction in strain in the rein-forcement due to the stiffening effect of

concrete is:

f A f AFs

s s s s

sstif1

1 1 2 2

1=

+f fD ...(15)

f A f A Fs s s s ss

stif1 1 2 22

2

= +f

f

D ...(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 < a2wherefs2 0 (i.e.compression)

x a2wherefs2 > 0 (i.e. tension)

Whenx a2:

The equations previously given by the

author2 apply to this case.The positionx of

the neutral axis is given by equation 17

(originally eqn 34)2 : (see panel 1)

where nhx

1 = andeT

M= .The concrete

stress,fc and steel stresses,fs1 andfs2, can

be determined from equations (35) to (37)

in Kruger2.

When:x < a2:

By settingnh

a1

2= in equation (17), the

value of the eccentricity,e, can be deter-

mined for whichx a2 i.e.(See Eq 18,

panel 1).

For this casefs1 andfs2 are both tensile.

If these are defined as positive,andfc 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 nhx

1 = andeT

M= .

The concrete stress,fc, can be deter-

mined 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 equa-tions (17) and (19) are cubic,and therefore

the solution ofn1 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

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20|The Structural Engineer 17 May 2005

technical note: bs 8007

.

.

,

F f h x b

b h xw

b h xw

for mm and

for mm

21

30 2

20 1

stif stif1= -

=-

=

=-

=

___

iii

...(21)

The values offs1 andfs2 can again be

determined from equations (12) and (13)

respectively.

When a2 x < hAs previously explained by the author2

and from Fig 5, it is clear that the

complete stiffening force acts on the rein-

forcement in face 1 as in the case of flexure

only and equation (43) from Kruger2 still

applies.

Determining the average strain: When

0