2. 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., βb ≥ 0.9) (BS 3.4.4.4). The code also
places a limitation on the neutral axis depth, x/d ≤ 0.5, to safeguard against
non-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 0.1fcu Ag (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
3. Concrete Frame Design BS 8110-97 Beam Design
Design Beam Flexural Reinforcement Page 3 of 10
based on whether M is 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.
If M ≤ Msingle, 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
K
25.05.0 ≤ 0.95d, and (BS 3.4.4.4)
0.67fcu/γc
4. Concrete Frame Design BS 8110-97 Beam Design
Design Beam Flexural Reinforcement Page 4 of 10
K = 2
cu bdf
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.
• If M > Msingle, the area of compression reinforcement, '
sA , is given by
'
sA =
( ) )'dd(f67.0f
MM
ccu
'
s
single
−−
−
γ
, (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(f95.0
MM
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 and As
'
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-
5. Concrete Frame Design BS 8110-97 Beam Design
Design Beam Flexural Reinforcement Page 5 of 10
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 βf 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
fwcu
s
h5.0df95.0
hd45.0dbf1.0M
A
−
−+
= (BS 3.4.4.5)
where,
b
b
15.0
d2
h
1
b
b
1
d
h
45.0 wf
f
wf
f +
−
−=β (BS 3.4.4.5)
0.67fcu/γc
6. Concrete Frame Design BS 8110-97 Beam Design
Design Beam Flexural Reinforcement Page 6 of 10
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 = 2
dbf
M
fcu
. (BS 3.4.4.4)
Then the moment arm is computed as
z =
−+
9.0
K
25.05.0d ≤ 0.95d, (BS 3.4.4.4)
the depth of neutral axis is computed as
x =
45.0
1
(d − z), and (BS 3.4.4.4)
the depth of compression block is given by
a = 0.9x. (BS 3.4.4.4)
If a ≤ hf, the subsequent calculations for As are exactly the same as previ-
ously defined for the rectangular section design. However, in that case the
width of the compression block is taken to be equal to the width of the
compression flange, bf for design. Compression reinforcement is required
if K > K'.
If a > 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/γc) (bf − bw)hf (d − 0.5hf), (BS 3.4.4.1)
the balance of moment taken by the web is computed as
Mw = M − Mf, and
7. Concrete Frame Design BS 8110-97 Beam Design
Design Beam Flexural Reinforcement Page 7 of 10
the normalized moment resisted by the web is given by
Kw = 2
dbf
M
wcu
w
. (BS 3.4.4.1)
− If Kw ≤ K', 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
K
25.05.0d 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 bw d2
. (BS 3.4.4.4)
The compression reinforcement is required to resist a moment of mag-
nitude Mw Mlw. The compression reinforcement is computed as
( ) )'dd(/f67.0f
MM
A
ccu
'
s
uww'
s
−−
−
=
γ
, (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'2
1 ≤ 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)
8. Concrete Frame Design BS 8110-97 Beam Design
Design Beam Flexural Reinforcement Page 8 of 10
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 percentage
Section Situation
Definition of
percentage fy= 250 MPa fy = 460 MPa
Rectangular 100
bh
As
0.24 0.13
f
w
b
b
< 0.4 100
hb
A
w
s
0.32 0.18
T-Beam with web in
tension
f
w
b
b
≥ 0.4 100
hb
A
w
s
0.24 0.13
T-Beam with web in
compression
100
hb
A
w
s
0.48 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 Situation
Definition of
percentage
Minimum
percentage
Rectangular 100
bh
As
'
0.20
Web in tension 100
ff
s
hb
A
'
0.40
T-Beam
Web in compression 100
hb
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
9. Concrete Frame Design BS 8110-97 Beam Design
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 major
and 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)
v ≤ N/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 = bw d, 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 is
specified in the concrete material properties.
If v exceeds 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 = RLW
4
1
3
1
s
m
21
d
400
bd
A100kk79.0
γ
, (BS 3.4.5.4, Table 3.8)
10. Concrete Frame Design BS 8110-97 Beam Design
Design Beam Shear Reinforcement Page 10 of 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)
γm = 1.25, and (BS 2.4.4.1)
As is the area of tensile steel.
However, the following limitations also apply:
0.15 ≤
bd
As100
≤ 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)
If v ≤ vc +0.4, provide minimum links defined by
yvv
sv
f95.0
b4.0
s
A
≥ , (BS 3.4.5.3)
else if vc +0.4 < v < vmax, provide links given by
yv
c
v
sv
f95.0
b)vv(
s
A −
≥ , (BS 3.4.5.3)
else if v ≥ vmax,
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). If fyv is
defined as greater than 460 MPa, the program designs shear reinforcing
assuming that fyv equals 460 MPa.
