design charts for corbels

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Paper: Al-Shawi

Paper

Design charts for corbelsF.A. N. Al-Shawi,BEng, PhDSheffield Hallam University

SynopsisThe designof corbels using the strut-and-tie model entails atrial-and-error procedure in the determination of tensionreinforcement. This is a laborious method and mightalso lead tounacceptable depth ofthe compression block or the vertical shearcapacity being exceeded. This means that the whole procedureneeds tobe repeated until all design criteria are satisfied.

In thispaper; design chartsbased on BS 8110and EC2recommendations are presented. These charts take intoconsideration all constraints. Therefore, they can be used as adesign aid.

Notationis the area of tension reinforcement

is the area of concrete cross-section

are the distances from the line of action of the load to the root of

the corbel in BS 8 l0and EC2, respectively

is the width of corbel

is the compressive force developed by the concrete (strut force)

is the effective depth at the root of the corbel

is the total tensile force to be developed by the tension

reinforcement

is the vertical force acting on the corbel (EC2)

is the characteristic compressive strength of concrete cylinders

is the characteristic compressive strength of concrete cubes(BS8110)

are the characteristic strengths of steel reinforcement inBS8 I Oand EC2 respectively

are the depths at the root of corbel in BS 8110and EC2,

respectively

are the horizontal forces acting on the corbel in BS 8110and EC2,

respectively

is the non-dimensional factor,

=V/bdfc,,(BS 8110)

is the depth factor for shear resistance

is the design axial force (compression positive)

is the ratio of the depth of the compression zone to the effective

depth of the corbel=x/d

is the non-dimensional factor,

=A,f,/bdf,.,,(BS8110)

is the. tensile force in the strut-and-tie model

is the vertical force acting on the corbel (BS 8 I O)

is the design shear resistance without shear reinforcement

is the maximum design shear force that can be carried without

crushing

isthe design shear stress atacross-section (BS8110)

is the design concrete shear stress(BS 8110)

is the depth of the compression zone at the root of corbel

=a,/d (BS 8 10)

=a,/d (EC2)

is the shear enhancement factor

is the partial safety factor for concrete

is the partial safety factor for steel reinforcement

is the angle between the direction of the strut compressive forceand the horizontal

is the coefficient of friction between the contact surfaces at the

support

is the efficiency factor used in the assessment of shear strength

is the average stress in concrete due to axial force

is the basic design shear strength of members without shear

reinforcement

=F,/bdf,k (EC2)

=A,fJ bdf,.k (EC2)

Charts based onBS 8110’The recommendations, as laid out in clause 5.2.7, areas follows:

(a) The distancea,.between the line of the reaction to the supported loadV

and the rootof the corbel is less than d (the effective depth of the root

of the corbel); and

(b) the depth at the outer edge of the contact area of the supported load is

not less than one-half of the depth at the root of the corbel.

The design simplifying assumptions are that theconcrete and reinforcement

may be assumed to act aselements of a simple strut-and-tie system. Note

that other systems are also available?.

Referring to Fig l(a)

V=C sin8T = C C O S ~

where

(see clause 3.4.4.4 for the stress block)

and

1-0.45n: tan8=-hence,

(1-0.45n)

sin =JW

....(1)

....(2)

....(3)

....(4)

....(5)

and,

=J....(6)

substitution of eqns. (3), (S) and (6) in eqn. ( 1 ) (noting that ‘yc= l .SO, as

given in clause 2.4.4) gives

v=-) 1,.9xba (1-0.45n)1 ....(7)a2+1-0.45n)’I

dividing by bdf,.,,(noting thatx =nd)and introducing the parameter

K =V&&.,,, eqn. (7)becomes

(0.202K+0.182a)n2-(0.9K+0.405a)n+K(l +a2)=0 .... 8)

solving the above quadratic equation gives

(0.9K +0.40%~) ,/(0.9K +0.405~~)~4(0.202K+0.182a)K(1+a’)n=

2(0.202K+0.182~~) ....(9)

Having foundn, the tensile forceTcan be evauated from (2) as

T=(0.y)0.9xba2

a2+(1-0.45n)’J l O )

If the horizontal frictional forceH (=pV ) s added to the above force, then

the total tensile force is given by:

222 The Structural EngineerNolume74/No 13/2 July 1996



Paper:Al-Shawi

( 4 (b )

Fi g 1.Strut-and-tiesystem (a)BS 8110model; (b)EC2 model

FT =T +pV

substituting equation 10) nto(1 ) gives

0.402fC,nd. ba2a2+1-0.45)2 J +P v

f l c

....(11)

....(12)

the area of tensile reinforcementsgiven by

FTA,7=- ....(13)( f, Y, )

substituting eqn. (1 2) into eqn. ( I 3) (noting that7,=1.15, as given in clause

2.4.4) gives

0.462&.,.,bdn a2 1. 5pV

[a2+1-0.45n)'] f v

A, = + ....(14)

Dividing by bd&'f,, (noting thatK =Vhdf,.,,),and introducing the parame-

ter r (=Af,/bdf,.,,),eqn. (14) becomes

r =[ (0.462n.a ) ]+1.15pK

a2 (I -0.45n)'....(15)

So, for any value of K anda,n can be evaluated from eqn. (9) and substi-

tuted (together withK ,aandy) in eqn. (1 5) to giver.

ConstraintsFor any given value ofa,eqn. (15) gives the value of r for different values

of K , which is a measure of the magnitude of the applied shear forceV rel-

ative to the corbel dimensions and the characteristic concrete strength

(bdfc,,).Obviously, there are limits on the value of K beyond which theK-r

curve cannot be used. These limits are:

(1)The ratio of the depth of the compression zone to the effective depth( dd)

must not exceed 0.50 (clause 3.4.4.4).(2)The magnitudeof the resistance provided to horizontal force should not

be less than one-half of the design vertical load (clause 5.2.7.2.1(a)), i.e.

T 2 V/2. It follows from eqn. (1 1) that

F T 2 1/2V+pV ....(16)

The above equation can be simplified and rearranged togive:

r 2 1.15K(0.5+p) ....(17)

(3) The vertical design shear stress v should not exceed the lesser of

0 .80aand 5N/mm2. It should also not exceed the design concrete shear

stress v, (clause 3.4.5). This implies that this limit isgiven by:

v =v, ....(18)

where

V

b dv =- ....(19:

and

1 1 I

v, =0.79 (%)'p (7)00As 3( d)00 4 ....(20:

....(21:

where

2d

a" ap= -= -

L.,,should not be taken as greater than 40N/mm2.

1OOA,y

b dshould not be taken as greater than 3.0

400 shouldnot be taken as less than 1Od

Equating eqns. (19) and (20) and noting that V/bd =Kf , . , ,00A,/bd =

100~C,.,/f,,nd taking the depth factor 400/d here as 1O (safe), in order not

to involve the dimensional quantityd,will give the following equation:

0. 24K3a3 CLf f , .r = ....(22)

(NOTE:p, , hould not be taken as greater than 40N/mm2 but.f;., can take

higher values, and the limit onr is such that 100A,/bd s not more than 3.0.)

Solving eqns. (15)and (22) simultaneously for the two unknownsK and rwill give the co-ordinates of the point of ntersection, i.e. the limit of applic-

ability of eqn.( 15).

Charts basedon ECZ3The main recommendations, as laid out in clause 2.5.3.7.2, are as follows:

(a) Corbels with 0.4hC a, I h,(see Fig 1(b)) may be designed using a sim-

ple strut-and-tie model.

(b) Unless special provision is made to limit horizontal forces on the sup-

port, or other ustification is given, the corbel should be designed for the

vertical forceF,and a horizontal forceH, >0.2Fv acting at the bearing

area.

f cu

The equations for the determination of tension reinforcement are similar to

those of BS 8110except as regards the following:

(i) The depth of the equivalent rectangular stress block is 0. 8~nstead of

0.9~,nd the uniform compressive stress is 0.85jJ x. instead of

0.67f,&. (clause 4.2.l .3.3).(ii) The factor 0.20 associated with the horizontal force is equivalent to a

coefficient of frictionp of 0.20.

Therefore, the equations for n (=dd) nd r (=A,f,/bdf,,,) are:

(0.8K+0.453a)- (0.8K+0.435a)2 -4(0.16K+0.181a)K (l+a2n= ,( 1)2(0.16K+O.l81a) ....(23)

and

r =0.521na2 +0.23K

+1-0.4n)2....(24)

(Note thatx,=1.50 and"/S=1.15 as given in clause 2.3.3.2, i.e. the same asinBS 8110.)

ConstraintsThe limits of applicability of theK -r curve, as given by eqn. (24), are:

(1) The ratio of the compression zone to the effective depth is restricted by

clause 2.5.3.4.2 as follows:

(dd) 0.45 forf,.k535N/mm2

(dd)0.35 for& >35N/mm2

(2) Clause 4.3.2.3 states that the applied vertical forceFV hould not exceed

the maximum design shear force which is given by:

VRd2 =0.45V k b d ....( 5)Y C

where

v =0.7- ck but not less than0.5200

....(26)

In addition, Fvshould not exceed the design shear resistance V,,,. This

implies that the limit is given by:

TheStructural Engineer/Volume 74/No 13/2 July 1996 223



Paper: Al-Shawi

ResultsFig 2 is a chart for p=0.00and in accordance with BS8110. It can also be

used ‘safely’ for other values of p provided that the horizontal frictional

force pVis catered for by an additional amount of tension reinforcement (see

example2).This is on the safe side because the increase in shear resistance

due to the additional tension reinforcement is not taken nto account. Some

typical values of p are given in ref. 4.Fig4 is the chart based on EC2 rec-

ommendations and Fig 3 is the equivalent chart based on BS 8110 with the

value of p =0.20 for comparison.

K

0.08

0.07

0.06

0.05

0. 04

0. 03

0.02

0.01

0

1- BS 8110, p=O.O

6

Conclusions(1) The charts presented in this paper can be used for the direct design of

corbels, thus avoiding the lengthy trial-and-error process.

(2) Charts based on BS 8 10forp=0.00can be used or other values of p,with the provision of additional tensile reinforcement to resist the horizon-

tal frictional force.

(3) Thereisgood agreement between the results obtained fromBS 8110and

EC2 charts as can be seen from examples3and4 in Appendix A.

Vertical shear constraint ines

- - - - - - f =250N/rnrn*

_-_ .__ f =460 Nrnrn2

I I I 10 0.01.02 0.03 0.04.05.06 r

Fig 2 BS 8110 chart withp =0.00

References1. BS 81 10Structural use ofconcrete: Part l ,London, British Standards

Institution, I985

2. Somerville, G: ‘The behaviour and design of reinforced concrete cor-

bels’, Cement&Concrete Association, Technical Report 472,August

1972

3. Eurocode 2, ‘Design of concrete structures, Part 1,General rules and

rules for buildings’, DD ENV 1992-1-1: 1992, British Standards

Institution.

4. Structural joints n precast concrete,London, Institution of Structural

Engineers, August 1978

5. Handbook to BS 8110: 1985 Structural use of concrete, Viewpoint

Publication, I987

....(27)

....(28)

p=-=- d with 1 .O I 55.0 (clause.3.2.2) ....(29)a, a

2-ZRd =0.035 ,: (seealsoclause.1.2.3) ....(30)

k=(1.6-d)butnotlessthan.O(dinm) ....(31)

(k s taken, here, as I .O (safe) in order not to involve the dimensional quan-

tity d) Appendix A. ExamplesCalculate the steel reinforcement requirements for the following cases:

p = - - s - fc k but not more than.02

f y k

Example I : BS 8110chart

and V=300kN,a, =250mm,b=400mm, h=550mm,d=500mm,p=0.00,

A., =35N/mm2,f,=250N/mm2

Solution:

a 250

d 500a=2=- 0.50<1.0ok

Equating eqns. (27) and (28), simplifying and rearranging gives the fol-

Iowing equation:

.(32:Enter Fig 2 (for BS 8 I IO, p =0.00)with the value of K =0.043and project

horizontally to intersect with the lineof a=0.50,then project vertically to

read the values of r as 0.028 (note the vertical shear constraint line for

A., =35N/mm2 andf,,=250N/mm2).

Solving eqns. (24) and (32) simultaneously for the two unknownsK and r

will give the co-ordinates of the point of intersection, i.e. the limit ofapplic-ability of eqn. (24).

K -

0.08 -

0.07 -

0.06-

0. 05-

0.04 -

0.03 -

0. 02.-

0.01 -

K r0.4

0.08 cBS 8110, 11 =0. 2 H/=. 45

0.07 EC2, equivalent p =0.2

0.06

0. 05

0.04

0. 03

0. 02

0.01

00 0.01 0. 02 0.03 0. 04 0.05 0.06 r

Fig 4. EC2 chart with an equivalentp =0.20

The Structural EngineedVolurne 74/No 13/2 July1996

=0. 5

Vertical shear constraint lines

f =460 Nrnrn2

n I I 1 1 l 1

“ 0 0.01 0.02 0.03 0. 04 0.05 0.06 r

Fig 3.BS 8110 chart with p =0.20

224



Paper: Al-Shawi

r b df,, - 0.028x400x500x35A,-

f v 250

Shear reinforcement:

A,, =0.50x 603=302mm2

Use 2T10 links (four legs, A =3 12mm2>302mm2ok) at a spacing of

llOmm c/c.=784mm2

(As)min=z b h clause 3.12.5)100

=0.24x 400x550=528mm2<784mm20k

100

: use 4R 16(A =804mm2 784mm2ok)

shear reinforcement:

A, , =0.50A, (clause 5.2.7.2.3)

=0.50x 804 =402mm2

use 3R 10 inks (six legs,A =468mm2>402mm2ok)over a depthof two-

thirds of the effective depth, i.e. 334mm. This will require a spacing of

115mm c/c.

Typical detail of reinforcement is shown in Fig 5(a)5.

Example 2:BS8110chart

V =150kN,a, =200mm,b=300mm, h=350mm, d=3 Omm,p=0.35,

f , ,=30N/mm2,f,,=460N/mm2

Solution:

Considerp =0.00first and find the reinforcement required.

200

310a=-=0.65<l.Ook

sincea is 0.65, then interpolate betweena=0.6 anda=0.7.

K = 150x lo3 =0.054300x310~30

From Fig 2, r is read as 0.046

A, =

0.046X 300X 3 10X 30=279mm2

460

determine the reinforcement required to resist the horizontal frictional force,

H , as follows:

H =pV=0.35X 150 =52.5kN

: total tension steel reinforcement=A, =(A,) addit ional

=279+132=411mm2

(A,)min=0.13x300X 310=121mm2 41 Imm20k100

: use 3T 16(A =603mm2>4 11mm2ok)

*",/-c;ross barMain tension reinforcement

Bars providedtoanchor shearreinforcement

Main tension reinforcementlooped horizontally

Example 3: EC2 chart

F" =215kN, a, =225mm,b=350mm, h, =450mm, d =405mm, ck=

30N/mm2,q.k=460N/mm2

Solution:

h, 450

225

d 405a=5=- 0.56 (interpolate betweena =0.50 anda =0.60)


r b dfck - 0.047 x 350X 405 x 30=435mm2A, x--

f y k 460

(A , ) . =-oh but not ess than 0.0015d (clause 5.4.2)

forfyk=460N/mm2,(A s )m i ns given by 0.0015bd

(A&,, =0,0015 x 350x405=213mm2 435mm2ok

use 4T12 (A =452mm2>435mm2ok)

mmf y k


A,,=0.4AS (clause 5.4.4)

A,, =0.4x452=181mm2

use 2T8 links (4 legs,A =200mm2>18lmm2ok)

Typical detail of reinforcement is shown in Fig 5(b)3

Example4: repeat example3using BS 8110 chartSince the equivalent coefficient of friction p is 0.20, Fig 3 is used in this

case. Note that the equivalent concrete cube strength to ck =30N/mm2 s

f,,=37N/mm2 (Table 3.1 of EC2).

225

d 405a=5 - 0.56<1.0 ok

(interpolate betweena=0.50anda=0.60)


A, =0.038X 350X 405X 37=433mm2

460

(As)min= X 350X 450=205mm2<433mm20k.13

100

: use 4T12 (A =452mm2>433mm2ok)


A,,= 0.5A,=0.5 x 452=226mm2

therefore use 2T10 links (4 legs,A =312mm2>226mm2ok)at 140mm c/c.

(a) (W

Fig 5. Typical detailing (a) UKpractice; (b)EC 2 recommendation

TheStructural Engineer/Volume74/No 13/2 July 1996 225

design charts for corbels

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