Design Charts for Corbels

Paper: Al-Shawi

Paper

Design charts for corbels F. A. N. Al-Shawi, BEng, PhD Sheffield Hallam University

Synopsis The design of corbels using the strut-and-tie model entails a trial-and-error procedure in the determination of tension reinforcement. This is a laborious method and mightalso lead to unacceptable depth ofthe compression block or the vertical shear capacity being exceeded. This means that the whole procedure needs to be repeated until all design criteria are satisfied. In thispaper; design chartsbased on BS 8110 and EC2 recommendations are presented. These charts take into consideration all constraints. Therefore, they can be used as a design aid. Notation is 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 l l0 and 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 (BS 81 10) are the characteristic strengths of steel reinforcement in BS 8 I I O and EC2 respectively are the depths at the root of corbel in BS 8110 and EC2, respectively are the horizontal forces acting on the corbel in BS 8 1 10 and EC2, respectively is the non-dimensional factor, = V/bdfc,,(BS 8 1 10) = F,/bdf,k (EC2) 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,.,, (BS 8 1 1 0) = A,fJbdf,.k (EC2) is the. tensile force in the strut-and-tie model is the vertical force acting on the corbel (BS 8 l IO) is the design shear resistance without shear reinforcement is the maximum design shear force that can be carried without crushing is the design shear stress at a cross-section (BS 8 110) is the design concrete shear stress (BS 81 10) is the depth of the compression zone at the root of corbel = a,/d (BS 8 l 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 force and 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

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Charts basedon BS 8110’ The recommendations, as laid out in clause 5.2.7, are as follows: (a) The distance a,.between the line of the reaction to the supported load V and the root of 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 as elements of a simple strut-and-tie system. Note that other systems are also available?. Referring to Fig l(a)

....(1) ....(2)

V= C sin 8 T= CCOS~

where

....(3 ) (see clause 3.4.4.4 for the stress block) and

1 - 0.45n :. tan 8 = -

....(4)

a

hence, (1 - 0.45n) sin =

....(5)

JW

and,

=

....(6)

-*J

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,

0.9 x b a (1 - 0.45n) a 2+ (1 - 0.45n)’I

1

....(7)

dividing by bdf,.,, (noting that x = 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 n=

+ 0 . 4 0 5 ~-~4(0.202K )~ + 0.182a)K(1+ a’) 2(0.202K + 0 . 1 8 2 ~ ~ ) ....(9)

Having found n, the tensile forceT can be evaluated from (2) as

(

T=( 0.y) 0.9 x b a 2

a 2+ (1 - 0.45n)’

JlO)

If the horizontal frictional force H ( = p V ) is added to the above force, then the total tensile force is given by:

The Structural EngineerNolume 74/No 13/2 July 1996

Paper: Al-Shawi where 2 2d p=-=f l c

....(21:

a

a"

L.,,should not be taken as greater than 40N/mm2. 1OOA,y

bd

should not be taken asgreater than 3.0

400 should not be taken as less than 1 .O d (4

Equating eqns. (19) and (20) and noting that V/bd = Kf,.,, 100A,/bd = 100~C,.,/f,, and taking the depth factor 400/d here as 1.O (safe), in order not to involve the dimensional quantity d, will give the following equation:

(b)

Fig 1. Strut-and-tie system ( a ) BS 8110 model; (b)EC2 model r=

....(11)

FT=T+pV substituting equation(10) into (1 l ) gives 0.402fC, nd. b a 2 a 2 + (1 - 0.45)2 J + P v

....(12)

0. l 24K3a3f 2CLf f,.

....(22)

f *cu

(NOTE: p,,should not be taken as greater than 40N/mm2 but.f;., can take higher values, and the limit on r is such that 100A,/bd is not more than 3.0.) Solving eqns. (15) and (22) simultaneously for the two unknowns K and r will give the co-ordinates of the point ofintersection, i.e. the limit of applicability of eqn.( 15).

Charts basedon ECZ3

the area of tensile reinforcementis given by

The main recommendations, as laid out in clause 2.5.3.7.2, are as follows: FT A,7 = -

....(13)

(f,/Y,)

substituting eqn. (1 2) into eqn. ( I 3) (noting that 7, = 1.15, as given in clause 2.4.4) gives 0.462&.,., b d n a2

A, =

[a2+ (1 - 0.45n)']

+- 1.l5pV

....(14)

fv

Dividing by bd&'f,, (noting that K = Vhdf,.,,),and introducing the parameter r (= Af,/bdf,.,,), eqn. ( 14) becomes (0.462 n.a

r=[

a 2+ ( I

)

]+1.15pK

....(15)

- 0.45n)'

So, for any value of K and a, n can be evaluated from eqn. (9) and substituted (together with K , a and y) in eqn. (1 5 ) to give r.

Constraints For any given value of a, eqn. (15 ) gives the value of r for different values of K, which is a measure of the magnitude of the applied shear force V relative to the corbel dimensions and the characteristic concrete strength (bdfc,,).Obviously, there are limits on the value of K beyond which the K-r curve cannot be used. These limits are: (1)The ratio of the depth of thecompression zone to the effective depth (dd) must not exceed 0.50 (clause 3.4.4.4). (2)The magnitude of 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 FT2

....( 16)

1/2V+ pV

....( 17)

(3) The vertical design shear stress v should not exceed the lesser of 0 . 8 0 a and 5N/mm2. It should also not exceed the design concrete shear stress v, (clause 3.4.5). This implies that this limit is given by:

v = v,

....( 18)

where

(i) The depth of the equivalent rectangular stress block is 0 . 8 instead ~ of 0 . 9 ~ and , the uniform compressive stress is 0.85jJx. 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 friction p of 0.20. Therefore, the equations for n ( = dd)and r (= A,f,/bdf,,,) are: n = (0.8K+0.453a)-

,(

1)

(0.8K+0.435a)2 -4(0.16K+0.181a)K(l+a2 2(0.16K+O.l81a)

....(23)

and r=

0.521 n a 2 +0.23K + (1 - 0.4n)2

....(24)

(Note that x, = 1.50 and "/S = 1.15 as given in clause 2.3.3.2, i.e. the same as in BS 8 1 10.)

Constraints The limits of applicability of the K-r curve, as given by eqn. (24), are:

(dd5 )0.45 forf,.k5 35N/mm2

(dd) 50.35 for& > 35N/mm2 (2) Clause 4.3.2.3 states that the applied vertical force FVshould not exceed the maximum design shear force which is given by: VRd2 = 0.45 V

k b d

....(25)

Y C

v=- V bd

....(19:

and

where v = 0.7 -

1

v, = 0.79

The equations for the determination of tension reinforcement are similar to those of BS 8 1 10 except as regards the following:

(1) The ratio of the compression zone to the effective depth is restricted by clause 2.5.3.4.2 as follows:

The aboveequation can be simplified and rearranged to give: r 2 1.15K(0.5+p)

(a) Corbels with 0.4hCI a, Ih, (see Fig 1(b)) may bedesigned using a simple strut-and-tie model. (b) Unless special provision is made to limit horizontal forces on the support, or other justification is given, the corbel should be designed for the vertical force F, and a horizontal force H, > 0.2Fv acting at the bearing area.

(%)'p

1

fck but not less than 0.5 200

....(26)

I

( 100As 7 3 (d) 400 ) 4

The Structural Engineer/Volume 74/No 13/2 July 1996

....(20:

In addition, Fv should not exceed the design shear resistance V,,,. This implies that the limit is given by:

223

Paper: Al-Shawi Results

K 0.08

1-

0.07

Fig 2 is a chart for p = 0.00 and 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 example 2). This is on the safe side because the increase in shear resistance due to the additional tension reinforcement is not taken into account. Some typical values of p are given in ref. 4. Fig 4 is the chart based on EC2 recommendations and Fig 3 is the equivalent chart based on BS 8 1 10 with the value of p = 0.20 for comparison.

BS 8110, p=O.O

6 0.06 0.05 0.04 0.03

Conclusions Vertical shear constraint lines - - - - - - - f = 250 N/rnrn*

0.02 0.01

_-_.__

I

I

0 0

0.020.01

f = 460 N/rnrn2 1

I

r

0.03 0.060.050.04

Fig 2 BS 8110 chart with p = 0.00

....(27)

....(28)

p=-=-5d a,

a

(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 l 10 for p = 0.00 can be usedfor other values of p, with the provision of additional tensile reinforcement to resist the horizontal frictional force. (3) There is good agreement between the results obtained from BS 8 1 10and EC2 charts as can be seen from examples 3 and 4 in Appendix A.

References BS 81 10 Structural use ofconcrete: Part l , London, British Standards Institution, I985 Somerville, G: ‘The behaviour and design of reinforced concrete corbels’, Cement & Concrete Association, Technical Report 472, August 1972 Eurocode 2, ‘Design of concrete structures, Part 1, General rules and rules for buildings’, DD ENV 1992-1-1: 1992, British Standards Institution. Structuraljoints in precast concrete,London, Institution of Structural Engineers, August 1978 Handbook to BS 8110: 1985 Structural use of concrete, Viewpoint Publication, I987

1.

2.

with 1 .O I p 55.0 (clause4.3.2.2)

....(29)

3.

(see also clause 3.1.2.3)

....(30)

4.

....(31)

5.

2 Z R d = 0.035 f ,:

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

( k is taken, here, as I .O (safe) in order not to involve the dimensional quantity d)

Appendix A. Examples Calculate the steel reinforcement requirements for the following cases:

p = - -As - rfck but not more than 0.02 fyk

Example I : BS 8110 chart

and

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

A., = 35N/mm2,f, = 250N/mm2 Solution : a

250 500

a = 2 = -= 0.50 < 1.0 ok d

Equating eqns. (27) and (28), simplifying and rearranging gives the folIowing equation: .(32: Solving eqns. (24) and (32) simultaneously for the two unknowns K and r will give the co-ordinates of the point of intersection, i.e. the limit ofapplicability of eqn. (24).

K -

-

c

K r

0.08 0.07

Enter Fig 2 (for BS 8 I IO, p = 0.00) with the value of K = 0.043 and project horizontally to intersect with the line of 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).

0.4

H/= 0.45

BS 81 10, 1.1 = 0.2

0.08

0.07

EC2, equivalent p = 0.2

= 0.5

0.06 -

0.06

0.05 -

0.05

0.04 -

0.04

0.03 -

0.03

0.02 .-

0.02 Vertical shear constraint lines

0.01 n “0

I

0.01

0.02

I

0.03

Fig 3. BS 8110 chart with p = 0.20

2 24

f = 460

N/rnrn2

1

1

l

1

0.04

0.05

0.06

r

0.01 0

0

0.01

0.02

0.03

0.04

0.05

0.06

r

Fig 4. EC2 chart with an equivalent p = 0.20

The Structural EngineedVolurne 74/No 13/2 July 1996

Paper: Al-Shawi Shear reinforcement:

r b d f,, - 0.028 x 400 x 500 x 35 A, --

A,, = 0.50 x 603 = 302mm2

250

fv

Use 2T10 links (four legs, A = 3 12mm2 > 302mm2 ok) at a spacing of llOmm c/c.

= 784mm2

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

100

=

:.

Example 3: EC2 chart

0.24 x 400 x 550 = 528mm2 < 784mm20k 100

use 4R 16 (A = 804mm2 784mm2 ok)

F" = 215kN, a, = 225mm, b = 350mm, h, = 450mm, d = 405mm, fck = 30N/mm2,fq.k = 460N/mm2 Solution:

shear reinforcement: 450

h,

A,, = 0.50 A, (clause 5.2.7.2.3) = 0.50 x 804 = 402mm2

225 405

a = 5 = -= 0.56 (interpolate betweena = 0.50 and a = 0.60)

use 3R 10links (six legs, A = 468mm2 > 402mm2 ok) over a depth of twothirds of the effective depth, i.e. 334mm. This will require a spacing of 115mm c/c.

d

Typical detail of reinforcement is shown in Fig 5(a)5. From Fig 4, r is read as 0.047 Example 2: BS 8110 chart A,

V = 150kN, a , = 200mm, b = 300mm, h = 350mm, d = 3 lOmm, p = 0.35, f,,= 30N/mm2,f,, = 460N/mm2

r b dfck

x--

-

0.047 x 350 X 405 x 30

( A , )m .m =- oh

Solution:

= 435mm2

460

fyk

but not less than 0.0015bd (clause 5.4.2)

fyk

Consider p = 0.00 first and find the reinforcement required.

forfyk = 460N/mm2, (As)min is given by 0.0015 bd

200 a=-=0.65 435mm2 ok)

a = 0.6 and a = 0.7.

Shear reinforcement:

K=

150 x lo3 = 0.054 300x310~30

A,,=0.4AS (clause 5.4.4) A,, = 0.4 x 452 = 181mm2

From Fig 2, r is read as 0.046 A, =

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

0.046 X 300 X 3 10 X 30 = 279mm2 460

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

determine the reinforcement required to resist the horizontal frictional force, H , as follows:

H = pV = 0.35 X 150 =52.5kN

Example 4: repeat example3 using BS 8110 chart Since the equivalent coefficient of friction p is 0.20, Fig 3 is used in this case. Note that the equivalent concrete cube strength to fck = 30N/mm2 is f,,= 37N/mm2 (Table 3.1 ofEC2).

225 405

a = 5= -= 0.56 < 1.0 ok d

:.

total tension steel reinforcement = A, = (A,) additional

(interpolate between a = 0.50 and a = 0.60)

= 279 + 132 = 411mm2

(A,)min =

:.

0100 .13x 300 X 310 = 121mm2 41 Imm20k

*",/-c;

From Fig 3, r is read as 0.038

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

Main tension reinforcement looped horizontally

Main tension reinforcement cross bar

A, =

0.038 X 350 X 405 X 37 = 433mm2 460 0.13 350 X 450 = 205mm2 < 433mm20k 100

(As)min = - X

.:

use 4T12 (A = 452mm2 > 433mm2 ok)

Shear reinforcement: A,,= 0.5A, = 0.5 x 452 = 226mm2

therefore use 2T10 links (4 legs, A = 3 12mm2> 226mm2 ok) at 140mm c/c.

Bars provided to anchor shear reinforcement (a)

(W

Fig 5. Typical detailing (a) UKpractice; (b)EC2 recommendation

The Structural Engineer/Volume 74/No 13/2 July 1996

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