Beam Design
RC BEAM DESIGN
Based on BS 8110
Partial Safety Factor γm
Cl. 2.4.4.1
Design strength 
Given in Table 2.2
Characteri stic strength
m
Durability
Cl. 2.2.4 and 3.1.5
Exposure conditions (Integrity of rc concrete
– ability to prevent corrosion) : moderate,
severe, very severe, most severe and
abrasive (Table 3.2)
Protection a agaist corrosion of steel (Table
3.3)
Fire resistance requirement (Table 3.4)
Nominal cover
Cl. 3.3.1.1
Bar Size (Cl. 3.3.1.2)
 Single Bars: nominal cover ≥ main bar diameter, d1
 Paired bars: nominal cover ≥ √2 d1
 Bundles bars: 2 √(Aequivalent/π)
Example:
Using the data given, determine:
(i) the nominal cover required to the underside of
the beam, and
(ii) the minimum width of beam required.
Data:
Exposure condition mild
Characteristic strength of concrete (fcu) 40 N/mm2
Nominal maximum aggregate size (hagg) 20 mm
Diameter of main tension steel 25 mm
Diameter of shear links 8 mm
Minimum required fire resistance 1.5 hours
SOLUTION
 Clause 3.3.1.2 Nominal cover ≥ (main bar diameter − link
diameter)
≥ (25 − 8) = 17 mm
 Clause 3.3.1.3 Nominal cover ≥ nominal maximum
aggregate size > 20 mm
 Clause 3.3.3 Exposure condition is mild
 Grade of concrete is C40
 Table 3.3 Nominal cover ≥ 20 mm*
 Clause 3.3.6 Minimum fire resistance = 1.5 hr
 The beam is simply supported
 Table 3.4 Nominal cover ≥ 20 mm
The required nominal cover = 20 mm
Flexural Strength of Sections
The flexural strength (i.e. the ultimate
moment of resistance of a cross-section)
is determined assuming the following
conditions as given in Clause 3.4.4.1,
BS8110:Part 1
Concrete Capacity
The maximum compressive force which can be
resisted by the concrete corresponds to the
maximum depth permitted for the neutral axis,
as shown in Figure 5.28 (i.e. x = d/2).
Concrete Capacity
 Consider the moment of the compressive force about the
line of action of Ft :
 Mult,concrete = (Fc × z)
where Fc = compressive force = (stress × area)
= [0.45fcu × (b × 0.9x )]
(for the maximum concrete force x = d/2)
= [0.45fcu × (b × 0.45d)] = 0.2bdfcu
z = lever arm
= [d − (0.5 × 0.45d)] = 0.775d
{Note: In general z = [d − (0.5 × 0.9x)] }
Mult,concrete = (0.2bdfcu )×(0.775d) =0.156bd2fcu
 This equation can be rewritten as 0.156 =M/bd2fcu
Steel Capacity
 Mult,steel = (Fs × z)
 where Fs = tensile force = (stress × area)
= (0.95fy × As)
z = lever arm
 Mult,steel = 0.95fyAs z
 As = M/0.95fyz
 Consider z, the lever arm:
 z = [d − (0.5 × 0.9x)] ,
 0.9x = 2(d − z)
 Mult,concrete = [0.45fcu × (b × 0.9x)] × z (substitute for
0.9x)
= 0.9fcu b (d − z) z
Steel Capacity
Mult,concrete = K bd 2 fcu
K bd 2 fcu = 0.9fcu b (d − z) z
Kd 2 = 0.9fcu dz − 0.9z 2
z2 − dz + Kd2/0.9 = 0
The solution of this quadratic equation (ax2 + bx +
c = 0) gives an expression which can be used to
determine the lever arm, z.
Summary
*The smaller of these two values being the critical case
When K ≤ K′ for a singly-reinforced section the maximum
moment permitted, based on the concrete strength, is equal to
0.156 bd2fcu
When K > K′ a section requires compression reinforcement.
Area of steel reinforcement, As
Areas/metre width for various pitches
of bars
Example 1:
A rectangular beam section is shown
in Figure 5.29. Using the data given,
determine the maximum ultimate
moment which can be applied to the
section assuming it to be singly
reinforced.
Data:
 Characteristic strength of concrete (fcu)
40 N/mm2
 Characteristic strength of steel (fy)
460 N/mm2
Solution:
 Strength based on concrete:
Mult = 0.156bd2fcu
= (0.156 × 250 × 4202 × 40)/106
= 275.2 kNm
 Strength based on steel:
As = 1260 mm2
z = 0.775 d = 0.775 x 420 = 325.5
Mult = 0.95fyAs z
= (0.95 × 460 × 1260 × 325.5)/106
= 179.5 kNm
 The maximum design moment which can be applied is:
Mult = 179.5 kNm
 The cross-section of a simply
Example 2:
supported rectangular beam
is shown in Figure 5.30. Using
the data given, determine the Data:
maximum ultimate moment
 Characteristic strength of
which can be applied to the
concrete (fcu) 30 N/mm2
section assuming it to be
 Characteristic strength of steel
singly-reinforced.
(fy) 460 N/mm2
 Nominal maximum aggregate
size (hagg) 20 mm
 Diameter of main tension steel
32 mm
 Diameter of shear links 8 mm
 Exposure condition mild
 Minimum required fire
resistance 1.0 hour
Solution:
Clause 3.3.1 Nominal cover to all steel
Clause 3.3.1.2 ≥ bar size
cover ≥ (32 − 8) = 24 mm
Clause 3.3.1.3 ≥ Nominal maximum aggregate size
cover ≥ 20 mm
Clause 3.3.3 Exposure condition: mild
Table 3.3 and fcu = 30 N/mm2
cover ≥ 25 mm
Clause 3.3.6 Min. fire resistance: 1 hr
beam is simply supported
Table 3.4 cover ≥ 20 mm
 The required nominal cover to all steel = 25 mm
Solution
 Figure 3.2 Minimum width b for 1 hour fire resistance = 200
mm  adequate
 Effective depth
d = (h − cover − link diameter − bar diameter/2)
= (475 − 25 − 8 − 16) = 426 mm
 Strength based on concrete:
Mult = 0.156bd2fcu = (0.156 × 250 × 4262 × 30)/106 = 212.3 kNm
 Strength based on steel:
As = 2410 mm2
z = 0.775 d
Mult = 0.95fyAs z = (0.95 × 460 × 2410 × 0.775 × 426)/106
= 349.9 kNm
 The maximum design moment which can be applied is: Mult =
212.3 kNm
Example 3:
The cross-section of a simply supported rectangular
beam is shown in Figure 5.31. Using the data given,
and assuming the section to be singly-reinforced,
determine the area of tension reinforcement
required to resist an applied ultimate bending
moment of 150 kNm.
Data:







Characteristic strength of concrete (fcu) 40 N/mm2
Characteristic strength of steel (fy) 460 N/mm2
Nominal maximum aggregate size (hagg) 20 mm
Diameter of main tension steel Assume 25 mm
Diameter of shear links 10 mm
Exposure condition moderate
Minimum required fire resistance 2.0 hours
Solution:
• Clause 3.3.1 Nominal cover to all steel
• Clause 3.3.1.2 ≥ bar size
cover = (25 − 10) = 15 mm
• Clause 3.3.1.3 ≥ Nominal maximum
• aggregate size
cover = 20 mm
• Clause 3.3.3 Exposure condition: moderate
• Table 3.3 fcu = 40 N/mm2
cover ≥ 30 mm
• Clause 3.3.6 Min. fire resistance: 2.0 hr
• beam is simply supported
• Table 3.4 cover ≥ 40 mm
The required nominal cover to the main steel = 40 mm
• Figure 3.2 Minimum width b for 2 hours, fire resistance =
200 mm  adequate
Figure 5.31
Solution:
 Effective depth d = (h − cover − link diameter − bar
diameter/2)
= (450 − 40 − 10 − 13) = 387 mm
 Clause 3.4.4.4
 Check that the section is singly-reinforced:
 K = M/bd2fcu = 0.125 < 0.156
 Since K < K ′ the section is singly-reinforced

K 
z  d 0.5  0.25 
  0.83d  0.95d
0
.
9


 As = M/0.95fyz = [(150 × 106) / (0.95 × 460 × 0.83 × 387)]
= 1068 mm2
 Adopt 4/ 20 mm diameter HYS bars providing 1260
mm2.
Shear Strength of Sections
The equation in BS 8110 (Cl 3.4.5.2)
V
v
bv d
Where




v = design shear stress
V = design shear force due to ultimate loads
bv=breadth of the section
d =effective depth
v ≤ 0.8 √(fcu) or ≤ 5 N/mm2
Shear Strength of Sections
This is the equation given in Table 3.7 of the
code for the cross-sectional area of designed
links when the design shear stress v at a crosssection is greater than (vc + 0.4). It is recognised
that the truss analogy produces conservative
results and the code specifies that: designed
links are required when:
(vc + 0.4) < v < 0.8 (fcu)1/2 or 5 N/mm2
Shear Strength of Sections
 It is important to provide minimum areas of steel in
concrete to minimize thermal and shrinkage cracking,
etc.
 Minimum links are specified in the code which provide a
shear resistance of 0.4N/mm2 in addition to the design
concrete shear stress vc, and consequently designed
links are required when v > (vc + 0.4) as indicated.
 In Table 3.7: minimum links are required when 0.5vc <
v < (vc + 0.4)
 The cross-sectional area of the links required is given by:
Example 3:
A concrete beam is simply supported over a 5.0
m span as shown in Figure 5.48. Using the data
given, determine suitable shear reinforcement.
Data:
 Characteristic strength of concrete (fcu) 40 N/mm2
 Characteristic strength of mild steel (fyv) 250 N/mm2
 Maximum shear force at the support 40 kN
Solution:
Solution:
Consider the end of the beam at the support
where the shear force is a maximum = 40 kN
Clause 3.4.5.2
V
40 x103
v

 0.89 N / mm2
bv d 200 x 225
0.8 f cu  0.8 40  5.06 N / mm2
 v  0.8 f cu &  5.0 N / mm2
Solution:
Table 3.7 To determine the required
reinforcement, evaluate vc and either adopt
minimum links throughout or use designed links.
As = area of 2/16 mm diameter bars = 402 mm2
100 As 100x 402

 0.893; d  225mm
bv d
200x 225
From Table 3.8
vc  {0.66  [(0.73  0.66) x (0.893  0.75) / 0.25]}  0.7 N / mm2
Since f cu  25 N / mm2
vc  0.73 ( f cu / 25)  0.73 (40 / 25)  0.82 N / mm2
(vc  0.4)  (0.82  0.4)  1.22 N / mm2
Solution:
 Since 0.5vc < v < (vc + 0.4), minimum links are required for the
whole length of the beam.
 The cross-sectional area of the links required is given by
Option 1:
 Assume 2-legged / 6 mm diameter mild steel links.
Asv = 56.6 mm2
0.95 f yv Asv 0.95x 250x56.6
 Table 3.7 The spacing required , sv 

 169mm
0.4bv
0.4 x200
 Clause 3.4.5.5
 The maximum spacing of the links ≤ 0.75d
= (0.75 x 225) = 169 mm
 Adopt 6 mm diameter mild steel links @ 150 mm centres
throughout the length of the beam.
Solution:
Option 2:
Clause 3.4.5.5
The maximum spacing of the links ≤ 0.75d
= (0.75 × 225) = 169 mm
Assume sv = 150 mm.
*Adopt 6 mm diameter mild steel links (56.6 mm2) @ 150 mm
centres throughout the length of the beam as in option 1.
Example 4:
A concrete beam is simply supported over a 7.0
m span as shown in Figure 5.49. Using the data
given, determine suitable shear reinforcement.
Data:




Characteristic strength of concrete (fcu) 40 N/mm2
Characteristic strength of mild steel (fyv) 250 N/mm2
Characteristic dead load (gk) 5.0 kN/m
Characteristic imposed load (qk) 30.0 kN/m
Solution:
Ultimate design load = [(1.4 × 5.0) + (1.6 × 30.0)]
= 55.0 kN/m
Ultimate design shear force at the support V =
(55.0 × 3.5) = 192.5 kN
Solution:
• Table 3.7 : To determine the required
reinforcement evaluate vc and to determine the
requirement for either minimum links or
designed links.
Solution:
 Table 3.7 Since (vc + 0.4) < v < 0.8 √(fcu)
< 5 N/mm2
designed links are required
 The maximum shear force which can be resisted by the
provision of minimum links is given by:
 V = (vc + 0.4)bvd = (1.06 × 240 × 570)/103
= 145 kN
Solution:
 Provide designed links for the first metre from each end
and minimum links elsewhere.
 The cross-sectional area of the designed links required
is given by
 Assuming 8 mm diameter mild steel links,
mm2 and fyv = 250 N/mm2.
Asv = 101
Solution:
 Clause 3.4.5.5
The maximum spacing of the links ≤ 0.75d
= (0.75 × 570) = 428 mm
 Adopt 8 mm diameter mild steel links @ 125 mm centres
for 1.0 m from each end of the beam.
 The cross-sectional area of the minimum links required is
given by Asv =
 Adopt 8 mm diameter mild steel links @ 250 mm centres
throughout the central 5.0 m of the beam.
Deflections (Cl. 3.4.6)
Allowances for these factors to reflect the actual
behaviour are made in the BS 8110 in terms of:
♦ using a basic ‘span/effective depth’ ratio (span/d)
rather than a ‘span/h ’ ratio (Table 3.9),
♦ applying a modifying factor which is dependent on the
steel strength (steel percentage and steel strength
are directly related through the factor K = M/bd2) and
the steel stress (fs) under service conditions
(Table
3.10),
♦ applying a modifying factor which is dependent on the
percentage of compression steel provided
(Table 3.11).
Deflections
The basic ratios given in Table 3.9 relate
to three support conditions:
 cantilevers,
 simply supported beams,
 continuous beams.
*Note: Continuous beams are considered to
be any beam in which at least one end of
the beam is continuous, i.e. this includes
propped cantilevers at the end of a series
of continuous beams.
Example 5:
A rectangular concrete beam 250 mm wide ×
475 mm overall depth is simply supported over a
7.0 m span. Using the data given, check the
suitability of the beam with respect to deflection.
Data:
 Characteristic strength of concrete (fcu) 40 N/mm2
 Characteristic strength of main steel (fy) 460 N/mm2
 Design ultimate bending moment at mid-span (M)
150.0 kNm
 Assume the distance to the centre of the main steel
from the tension face is 50 mm
Solution:
Effective depth d = (475 − 50) = 425 mm
Check that section is singly-reinforced:
Adopt 3/20 mm diameter HYS bars providing 943 mm2
Solution:
Clause 3.4.6 There is no redistribution of
moments  β b = 1.0
Table 3.9 The beam is simply supported:
L/d = 20
Clause 3.4.6.5
M/bd 2 = (K × fcu )
= (0.082 × 40) = 3.32 N/mm2
fs 
2 fyAsrequired
3 As provided
x
1
b
 291.7 N / mm2
Interpolate from Table 3.10  0.92
Solution:
L/d x value from Table 3.10 = L/d x 0.92
Allowable deflection = 20 x 0.92 = 18.4
Actual deflection = L/d = 6000/425 = 14.12
So, the deflection is acceptable.
Minimum Areas of Steel (Cl. 3.12.5)
Minimum areas of steel are required in structural
elements to ensure that any unnecessary
cracking due to thermal/shrinkage effects or
tension induced by accidental loading can be
minimized.
The required minimum steel areas for different
situations, i.e. tension or compression
reinforcement, rectangular beams/slabs, flanged
beams and columns/walls, are given in
Table 3.25
Example 5: Minimum Areas of Steel
Determine the minimum areas of steel required
for the cross-sections shown in Figure 5.57; in
all cases fy = 460 N/mm2.
(a) Rectangular beam
The required area of steel is given by:
100As /Ac = 0.13
where:
 As is the minimum recommended area of
reinforcement,
 Ac is the total area of concrete.
Ac = (550 × 320) = 176 × 103 mm2
As = (0.13×176×103)/100= 229 mm2
(b) Flanged beam (see Cl
3.4.1.5)
where:
 b is the breadth of the section bw is the breadth or
effective breadth of the rib; for a box, T or I section, bw is
 taken as the average breadth of the concrete below the
flange,
 h is the overall depth of the cross-section of a reinforced
member.
bw / b = (300 / 1200) = 0.25 < 0.4
 the required area of steel is given by:
100As /bwh = 0.18
bw h = (300 × 600) = 180 × 103 mm2
As = 0.18×180×103/100 = 324 mm2
Maximum Areas of Steel (Cl. 3.12.6)
The maximum percentages of steel given
in the code are based on the physical
requirements for placing the concrete and
fixing the steel. They are:
In beams (Clause 3.12.6.1):
 As ≤ 4% Ac
 As′ ≤ 4% Ac
Minimum Spacing of Bars (Cl.
3.12.11.1)
Guidance is given in the code for minimum bar
spacing to ensure that members can be
constructed achieving adequate penetration and
compaction of the concrete to enable the
reinforcement to perform as designed.
It is important that reinforcing bars are
surrounded by concrete for two main reasons:
 (i) to develop sufficient bond between the concrete
and the bars such that the required forces are
transferred between the steel and the concrete, and
 (ii) to provide protection to the steel against corrosion,
fire, etc.
Maximum Spacing of Bars (Clause
3.12.11.2)
The requirement to limit the maximum spacing of
reinforcement is to minimize surface cracking.
Using Table 3.28.
Since very small bars when mixed with larger
bars would invalidate the assumptions on which
the Table 3.28 values are based, the code
specifies that any bar in a section with a
diameter < (0.45 × the largest bar in the section)
should be ignored except when considering
those in the side faces of deep beams.
Maximum Spacing of Bars
The minimum size of bar to be used in the
side faces is given in Clause 3.12.5.4 as:
bar diameter ≥
where: sb is the bar spacing (e.g. 250 mm)
b is the breadth of the section ≤ 500 mm