Information and Pool Etabs Manuals English E-tn-cfd-bs-8110!97!007

7/28/2019 Information and Pool Etabs Manuals English E-tn-cfd-bs-8110!97!007

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COMPUTERS AND STRUCTURES, INC., BERKELEY, CALIFORNIA AUGUST 2002

CONCRETE FRAME DESIGN BS 8110-97

Technical Note

Beam Design

This Technical Note describes how this program completes beam design when

the BS 8110-97 code is selected. The program calculates and reports the re-

quired areas of steel for flexure and shear based upon the beam moments,

shears, load combination factors and other criteria described herein.

Overview

In the design of concrete beams, the program calculates and reports the re-

quired areas of steel for flexure and shear based upon the beam moments,

and shears, load combination factors, and other criteria described herein. The

reinforcement requirements are calculated at a user-defined number of check

stations along the beam span.

All beams are designed for major direction flexure and shear only.

Effects resulting from axial forces, minor direction bending, and tor-

sion that may exist in the beams must be investigated independently

by the user.

The beam design procedure involves the following steps:

Design beam flexural reinforcement

Design beam shear reinforcement

Design Beam Flexural Reinforcement

The beam top and bottom flexural steel areas are designed at a user-defined

number of check stations along the beam span. The following steps are in-volved in designing the flexural reinforcement for the major moment for a

particular beam at a particular section:

Determine the maximum factored moments

Determine the reinforcing steel


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Concrete Frame Design BS 8110-97 Beam Design

Design Beam Flexural Reinforcement Page 2 of 10

Determine Factored Moments

In the design of flexural reinforcement of concrete frame beams, the factored

moments for each load combination at a particular beam station are obtained

by factoring the corresponding moments for different load cases with the

corresponding load factors.

The beam section is then designed for the maximum positive and maximum

negative factored moments obtained from all of the load combinations at that

section.

Negative beam moments produce top steel. In such cases, the beam is al-

ways designed as a rectangular section. Positive beam moments produce

bottom steel. In such cases, the beam may be designed as a rectangular sec-

tion, or T-Beam effects may be included.

Determine Required Flexural Reinforcement

In the flexural reinforcement design process, the program calculates both the

tension and compression reinforcement. Compression reinforcement is added

when the applied design moment exceeds the maximum moment capacity of

a singly reinforced section. The user has the option of avoiding the compres-

sion reinforcement by increasing the effective depth, the width, or the grade

of concrete.

The design procedure is based on the simplified rectangular stress block as

shown in Figure 1 (BS 3.4.4.1). It is assumed that moment redistribution in

the member does not exceed 10% (i.e., b 0.9) (BS 3.4.4.4). The code alsoplaces a limitation on the neutral axis depth, x/d 0.5, to safeguard againstnon-ductile failures (BS 3.4.4.4). In addition, the area of compression rein-

forcement is calculated assuming that the neutral axis depth remains at the

maximum permitted value.

The design procedure used by the program for both rectangular and flanged

sections (L- and T-beams) is summarized in the next section. It is assumed

that the design ultimate axial force does not exceed 0.1fcuAg (BS 3.4.4.1);hence, all the beams are designed for major direction flexure and shear only.

Design of a Rectangular beam

For rectangular beams, the moment capacity as a singly reinforced beam,

Msingle, is obtained first for a section. The reinforcing steel area is determined


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based on whether Mis greater than, less than, or equal to Msingle. See Figure

1.

Figure 1: Design of a Rectangular Beam Section

Calculate the ultimate moment of resistance of the section as a singly re-

inforced beam.

Msingle = K'fcubd2, where (BS 3.4.4.4)

K'= 0.156.

IfMMsingle, no compression reinforcement is required. The area of ten-sion reinforcement,As, is obtained from

As =

z)f95.0(

M

y

, where (BS 3.4.4.4)

z= d

+9.0

K25.05.0 0.95d, and (BS 3.4.4.4)

0.67fcu/c


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K=2

cubdf

M. (BS 3.4.4.4)

This is the top steel if the section is under negative moment and the bot-

tom steel if the section is under positive moment.

IfM> Msingle, the area of compression reinforcement,'sA , is given by

'sA = ( ) )'dd(f67.0f

MM

ccu's

single

, (BS 3.4.4.4)

where d'is the depth of the compression steel from the concrete compres-

sion face, and

'

sf = 700

d

d'2

1 0.95 fy. (BS 3.4.4.4)

This is the bottom steel if the section is under negative moment. From

equilibrium, the area of tension reinforcement is calculated as:

sA = ( ) )'dd(f95.0MM

z)f95.0(

M

y

single

y

single

+ , where (BS 3.4.4.4)

z=

+ 9.0

'

25.05.0

K

d = 0.776887

d. (BS 3.4.4.4)

As is to be placed at the bottom of the beam andAs' at the top for positive

bending and vice versa for negative bending.

Design as a T-Beam

(i) Flanged beam under negative moment

The contribution of the flange to the strength of the beam is ignored. The de-

sign procedure is therefore identical to the one used for rectangular beams,except that in the corresponding equations, b is replaced by bw. See Figure 2.

(ii) Flanged beam under positive moment

With the flange in compression, the program analyzes the section by consid-

ering alternative locations of the neutral axis. Initially the neutral axis is as-


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sumed to be located in the flange. On the basis of this assumption, the pro-

gram calculates the depth of the neutral axis. If the stress block does not

extend beyond the flange thickness, the section is designed as a rectangular

beam of width bf. If the stress block extends beyond the flange width, the

contribution of the web to the flexural strength of the beam is taken into

account. See Figure 2.

Figure 2: Design of a T-Beam Section

The T-beam requires only tension reinforcement when the moment is posi-

tive, the flange is in compression, the moment is less than f fcu bd2 and the

flange depth is less than 0.45d. In those conditions, the tension reinforcing

steel area of the T-beam is calculated as follows (BS 3.4.4.5):

( )( )fy

fwcus h5.0df95.0

hd45.0dbf1.0MA

+= (BS 3.4.4.5)

where,

b

b15.0

d2

h1

b

b1

d

h45.0 wf

f

wff +

= (BS 3.4.4.5)

0.67fcu/c


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If the above conditions are not met, the T-beam is designed using the general

principle of the BS 8110 code (BS 3.4.4.4, BS 3.4.4.5), which is as follows:

Assuming that the neutral axis lies in the flange, the normalized moment is

computed as

K=2dbf

M

fcu

. (BS 3.4.4.4)

Then the moment arm is computed as

z=

+9.0

K25.05.0d 0.95d, (BS 3.4.4.4)

the depth of neutral axis is computed as

x=45.0

1(dz), and (BS 3.4.4.4)

the depth of compression block is given by

a = 0.9x. (BS 3.4.4.4)

Ifahf, the subsequent calculations forAs are exactly the same as previ-

ously defined for the rectangular section design. However, in that case thewidth of the compression block is taken to be equal to the width of the

compression flange, bf for design. Compression reinforcement is required

ifK> K'.

Ifa > hf, the subsequent calculations for As are performed in two parts.

The first part is for balancing the compressive force from the flange, Cf,

and the second part is for balancing the compressive force from the web,

Cw, as shown in Figure 2.

In this case, the ultimate resistance moment of the flange is given by

Mf= 0.67 (fcu/c)(bfbw)hf(d 0.5hf), (BS 3.4.4.1)

the balance of moment taken by the web is computed as

Mw= MMf, and


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the normalized moment resisted by the web is given by

Kw= 2dbf

M

wcu

w . (BS 3.4.4.1)

IfKwK', the beam is designed as a singly reinforced concrete beam.The area of steel is calculated as the sum of two parts: one to balance

compression in the flange and one to balance compression in the web.

As =zf95.0

M

)h5.0d(f95.0

M

y

w

fy

f +

, where (BS 3.4.4.1)

z=

+9.0

K25.05.0d w 0.95d. (BS 3.4.4.1)

If Kw > K', compression reinforcement is required. The compression

reinforcing steel area is calculated using the following procedure:

The ultimate moment of resistance of the web only is given by

Muw= K' fcu bwd2. (BS 3.4.4.4)

The compression reinforcement is required to resist a moment of mag-

nitude MwMlw. The compression reinforcement is computed as

( ) )'dd(/f67.0fMM

Accu

's

uww's

=

, (BS 3.4.4.1)

where d'is the depth of the compression steel from the concrete com-

pression face, and

'sf = 700

d

d'21 0.95fy. (BS 3.4.4.1)

The area of tension reinforcement is obtained from equilibrium

As =

++ 'dd

MM

d777.0

M

h5.0d

M

f95.0

1 uwwuw

f

f

y

. (BS 3.4.4.1)


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Determination of the Required Minimum Flexural Reinforcing

The minimum flexural tensile reinforcing steel required for a beam section is

given by the following table, which is taken from BS Table 3.25 (BS 3.12.5.3)

with interpolation for reinforcement of intermediate strength:

Table 1 Minimum Percentage of Tensile Reinforcing

Minimum percentageSection Situation

Definition of

percentage fy= 250 MPa fy = 460 MPa

Rectangular 100bh

As0.24 0.13

f

w

b

b< 0.4 100

hb

A

w

s0.32 0.18

T-Beam with web in

tension

f

w

bb 0.4 100

hbA

w

s0.24 0.13

T-Beam with web in

compression 100

hb

A

w

s0.48 0.26

The minimum flexural compression steel, if it is required, provided in a rec-

tangular or T-beam section is given by the following table, which is taken

from BS Table 3.25 (BS 3.12.5.3) with interpolation for reinforcement of in-

termediate strength:

Table 2 Minimum Percentage of Compression Reinforcing (if required)

Section SituationDefinition of

percentage

Minimum

percentage

Rectangular 100bh

As'

0.20

Web in tension 100ff

s

hb

A'

0.40

T-Beam

Web in compression 100hb

A

w

s

'

0.20

In addition, an upper limit on both tension reinforcement and compression

reinforcement has been imposed to be 0.04 times the gross cross-sectional


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Design Beam Shear Reinforcement Page 9 of 10

area (BS 3.12.6.1). The program reports an overstress when the ratio exceed

4 percent.

Design Beam Shear Reinforcement

The shear reinforcement is designed for each load combination in the majorand minor directions of the column. The following steps are involved in de-

signing the shear reinforcement for a particular beam for a particular load

combination resulting from shear forces in a particular direction (BS 3.4.5):

Calculate the design shear stress and maximum allowable shear stress as

v=cvA

V, where (BS 3.4.5.2)

v 0.8 RLW cuf , (BS 3.4.5.2, BS 3.4.5.12)

vN/mm2 , (BS 3.4.5.2, BS 3.4.5.12)

vmax= min {0.8RLW cuf , 5.0 MPa}, (BS 3.4.5.2, BS 3.4.5.12)

Acv= bwd, and

RLW is a shear strength reduction factor that applies to light-weight

concrete. It is equal to 1 for normal weight concrete. The factor isspecified in the concrete material properties.

Ifvexceeds 0.8RLW cuf or 5 MPa, the program reports an overstress. In

that case, the concrete shear area should be increased.

Note

The program reports an overstress message when the shear stress exceed 0.8RLW cuf

or 5 MPa (BS 3.4.5.2, BS 3.4.5.12).

Calculate the design concrete shear stress from

vc= RLW4

13

1

s

m

21

d

400

bd

A100kk79.0

, (BS 3.4.5.4, Table 3.8)


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D i B Sh R i f t P 10 f 10

where,

k1 is the enhancement factor for support compression,

and is conservatively taken as 1, (BS 3.4.5.8)

k2 =3

1

25

cuf 1, (BS 3.4.5.4, Table 3.8)

m = 1.25, and (BS 2.4.4.1)

As is the area of tensile steel.

However, the following limitations also apply:

0.15 bdAs100

3, (BS 3.4.5.4, Table 3.8)

d

400 1, and (BS 3.4.5.4, Table 3.8)

fcu 40 N/mm2 (for calculation purpose only). (BS 3.4.5.4, Table 3.8)

Ifvvc+0.4, provide minimum links defined by

yvv

sv

f95.0

b4.0

s

A

, (BS 3.4.5.3)

else ifvc+0.4 < v< vmax, provide links given by

yv

c

v

sv

f95.0

b)vv(

s

A , (BS 3.4.5.3)

else ifvvmax,

a failure condition is declared. (BS 3.4.5.2, 3.4.5.12)

In shear design, fyv cannot be greater than 460 MPa (BS 3.4.5.1). Iffyv is

defined as greater than 460 MPa, the program designs shear reinforcing

assuming that fyvequals 460 MPa.

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Information and Pool Etabs Manuals English E-tn-cfd-bs-8110!97!007