Structural analysis & design of troughed floors

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STRUCTURAL ANALYSIS AND DESIGN OF TROUGHED FLOORS
ACCORDING TO EUROCODE
Omotoriogun Victor Femi
Copyright ©Structures centre, 2020. All rights reserved
structurescentre@gmail.com
Abstract
This article presents the analysis and design of troughed floors. These are slabs that combines the
advantages of ribbed floors with those of a slab with level soffits (flat slab). They provide one of the
most efficient solution for long-span floors, typically up to 12m and are also very effective where
good surface finish is desired. The levelled soffits present the floor with the advantage of exposing
the slab profile which may be used for passive cooling or for aesthetics. They are composed of very
shallow beams spanning between columns with typically equal depth narrow ribs spanning in the
orthogonal direction. A thin slab layer known as the topping completes the system. Overall, they
tend to be dipper than solid or flat slabs but are considerably lighter in weight than an equivalent
flat slab, thus the extent of their foundations is greatly reduced. This paper therefore, aims to
introduce the concept, analysis and design of troughed floors to the reader. Also, a practical worked-
example using the Eurocode has been included in this article to aid the reader in further
understanding the design of this kind of floors.

2
1.0 Introduction
The flexural design of a reinforced concrete element neglects the weight of concrete below its neutral
axis. It assumes that the tensile stress in the concrete element will be fully resisted by the
reinforcement alone. This is because the tensile strength of concrete is negligible and hence, not
considered effective in resisting tensile stresses. It therefore follows that the volume of concrete in
the tensile zone is almost useless except for resisting shear stresses. Thus, by introducing voids to
the soffits of slabs or replacing them with lighter materials, the self-weight of a slab is reduced and
the efficiency of the concrete section in bending is increased. This is the concept behind the ribbed
floor system.
The voids introduced into the slab gives a T section as shown in (figure 1). This consist of closely
spaced concrete elements known as Ribs. At the supports, the slab is either made solid or the ribs
made to terminate into a supporting beam. This is done to enhance the resistance of the section
against shear forces which tends to be critical at the supports. Between the ribs is a thin layer of
concrete connecting the ribs termed as the Topping.
Figure 1: Section Through a Ribbed Slab
The ribbed floor system is applicable in essentially long span floors of up to 12m with relatively
light-moderate loading. They provide a lighter and stiffer slab than an equivalent flat or solid slab,
reducing the extent of foundations. They are also very useful where slab vibration is very critical
such as laboratories and hospitals and generally give good surface finishes, this allows for the use of
the exposed soffits in the finished building. There are several types of ribbed floors, however of
interest are the troughed floors.

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A troughed slab is almost no different from the ribbed slab except that it combines the advantages
of a ribbed slab with those of level soffits (flat slabs). They provide floors with longer span than one-
way solid or flat slabs. The floor soffits are levelled (figure 2) and the provision of holes for building
services causes little or no problem in the ribbed areas. To ensure a levelled soffit, the depth of the
supporting beams is kept equal to that of the ribs. As a result, the depth of the section depends on the
width of the beam used, which tends to be very wide and heavily reinforced, so as to control
deflection. Preferably a trough slab should be oriented such that the ribs span the longest distance
while the shallow supporting beam span the shorter distance.
2.0 Sizing the Elements
Sizing a troughed floor can be sometimes very tricky. This is because the size of each component of
the floor is inter-dependent. Thus, the size of the ribs as well as that of the supporting beams must
be decided upon at the same time in-order to ensure a levelled soffit. Generally, the depth of the
section is more critical and it is almost always governed by deflection requirements, except in certain
instances where the applied actions might significantly influence the sizing.
Guidance on sizing a troughed floor can be obtained from Economic Concrete Frame Elements to
Eurocode 2 by The Concrete Centre. There is a chart within this reference text from which the size
Figure 2: Troughed floor (https://www.formscaff.com/project/toyota-factory-trough -slabs)

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of the ribs and the supporting beams can be determined based on span and the magnitude of applied
variable actions.
However as, an initial step, the depth of the section may be determined conservatively based on it
span/depth ratio from equation (1).
[ depth =
span
20−25
] (1)
Typically, the width of the supporting beam is very wide and may vary between 900-2400mm. The
width of ribs is governed by cover and bar spacing. The spacing between the ribs is typically 600,
750, 900 & 1200mm depending on the span with an average rib width of 150mm.
Having determined the sizes of the components of the floor, the average self-weight of the slab may
be determined from expression (2). This is based on the expression given in the Concrete Centre
Publication: Concrete Building Scheme Designers Manual (2009).
Figure 3 : Terms for Estimating the self-weight of Troughed Slabs
self weight =
25
𝑦
{𝑟 [
𝑠
𝑐
( 𝑤 + 𝑘) + 𝑡] + ℎ( 𝑢 + 𝑣)} (2)
3.0 Structural Analysis
Troughed floors are essentially one-way slab systems; therefore, their analysis is carried out in the
same fashion as one-way solid slabs. If the slab consists of multiple span, coefficients for continuous

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one-way solid slabs may be used to obtain the design moment and shear forces provided the
conditions set-out for their use are satisfied. To use the coefficient in Table 1 in analysing a troughed
slab, the following conditions must be satisfied, these are:
a. The area of each bay must exceed 30m2
b. The ratio of the characteristic variable actions to permanent action must not exceed 1.25
c. The characteristic variable actions must not exceed 5kN/m2
d. The spans must be approximately equal (generally, within 85% of the maximum span).
Table 1: Bending Moment and Shear Force for Continuous One-way Slab
The band-beams may also be analysed using simple coefficients provided that the beam is uniformly
loaded, the geometry of the beam spans is within 85% of each other and the variable action on each
span is equal or similar. Where these conditions are satisfied, the coefficients giving in Table 2 may
also be used to determine the bending moment and shear forces in the beams.
Table 2: Bending Moment and Shear Force Coefficients for Continuous Beams
Alternatively, and where the conditions required for using coefficients has not been met, the slab and
the band-beams may be analysed using any chosen method of elastic analysis such as, moment
distribution, three moment equations, slope-deflection equations or even matrix method by assuming

6
linear supports or through the analysis of a sub-frame. If elastic analysis is used to obtain the design
moment and shear force, the slab and beam must be analysed for the most unfavourable arrangement.
This is easily satisfied by considering three load cases.
• All span loaded with the maximum design action. (One load case).
• Alternate spans loaded with the maximum design action while the other spans are loaded
with the maximum permanent actions. (Two load cases)
However, where the analysis is carried out for the single load-case of all span loaded with the
maximum design actions then the resulting support moment must be redistributed by 20% into the
spans except at the supports of cantilevers.
4.0 Structural Design
The critical areas to design for include flexure and shear, this is carried out in accordance with section
6.1 and 6.2 of BS-EN 1992-1-1-Eurocode 2: Design of Concrete Structures – Part 1-1: General
Rules for Building.
Ribs behave like beams and therefore the rules governing the design of beams are observed. At the
supports the ribs are solid and thus designed as a rectangular beam while in the spans the section is
flanged and designed as a flanged beam. However, where the neutral axis is within the flange of the
ribs, the section can also be designed as a rectangular beam at midspan.
Deflection is usually verified by comparing the ratio of the actual span-effective depth ratios with
that of the permissible span- effective depth ratios. Detailing checks are also carried out to ensure
serviceability and ease in constructing the floor. This includes checking for the minimum and
maximum area of steel, minimum and maximum bar spacings and measures for crack control. All of
this can be obtained from section 7 of BS-EN 1992-1-1-Eurocode 2: Design of Concrete Structures
– Part 1-1: General Rules for Building.
Light reinforcement or fabric mesh based on the minimum area of steel required is provided in the
toppings to avoid cracking. However, where the spacing between the ribs exceeds 1200mm, the
toppings must be designed for moment and shear by assuming one-way action between the ribs.

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5.0 Worked Example
A troughed slab is required for a shopping mall building. The layout of the floor is shown in figure
4 below. Preliminary sizing of the structural components has been carried out according to section
2.0 and is indicated on this figure. Design the troughed slab and the band beam along gridline B
completely using concrete of class C30/37 and steel bars having a yield strength of 460Mpa.
Figure 4: Troughed Slab Layout
Figure 5: Section through Ribs

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5.1 Troughed Slab
5.1.1 Actions
Permanent Actions:
i. self weight =
25
y
{r [
s
c
(w + k) + t] + h(u + v)}
=
25
9.5
{8.0 [
0.35
0.75
(0.15 + 0.062) + 0.10] + 0.45(0.75 + 0.75)}=6.0kN/m2
ii. finishes and services say =1.5kN/m2
Total Permanent actions gk = 6.0+1.5 =7.5kN/m2
Variable Actions:
a. Floor imposed load (shopping malls) = 4.0kN/m2
b. Demountable partitions say = 1.0kN/m`2
Total Variable actions qk = 4.0+1.0 = 5.0kN/m2
By inspection the permanent actions are less than 4.5 times the variable actions, therefore equation
6.13b of BS EN 1990 will give the most unfavourable results.
Design Load = 1.35𝜉𝑔 𝑘 + 1.5𝑞 𝑘= (1.35 × 0.925 × 7.5) + (1.5 × 5) = 𝟏𝟔. 𝟖𝟕𝒌𝑵/𝒎 𝟐
Design Permanent Load 1.35𝜉𝑔 𝑘 = 1.35 × 0.925 × 7.5 = 𝟗. 𝟑𝟕𝒌𝑵/𝒎 𝟐
5.1.2 Analysis
Figure 6: Analysis of the Troughed Slab

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The analysis of the troughed slab is carried out in accordance with section 3.0 using the coefficients
for continuous one-way slab if the conditions listed are satisfied.
• Area of each bay = (9.5 × 27) = 256.5𝑚2
> 30𝑚2
• The ratio of the characteristic imposed loads to dead load =
𝑞 𝑘
𝑔 𝑘
=
5
7.5
= 0.66 < 1.25
• The characteristic imposed load = 5kN/m2
• The spans are equal i.e. 9.5m
Since the conditions are satisfied Table 1.0 is applicable.
𝐹 = 𝑛 𝑠 𝐿 = 16.87 × 9.5 = 160.27𝑘𝑁
Table 3: Bending Moment and Shear Force in Trough Slab
End Supports A & C End Spans A-B & B-C Interior Support B
Moment (kN.m) 0.04FL = 60.90 0.075FL = 114.2 0.086FL = 130.94
Shear (kN) 0.45F = 72.12 − 0.6F = 96.16
5.1.3 Flexural Design
End Supports A & C
MEd = 60.9 × 0.75 = 45.68kN. m/rib
Assuming cover to reinforcement of 25mm, 12mm bars, 8mm links
d = h − (cnom + links +
∅
2
) = 450 − (25 + 8 +
12
2
) = 411mm; b = 750mm
k =
MEd
bd2fck
=
45.68 × 106
750 × 4112 × 30
= 0.012
z = d[0.5 + √0.25 − 0.882k] ≤ 0.95d
=d[0.5 + √0.25 − 0.882(0.015)] ≤ 0.95d
=0.95d = 0.95 × 411 = 390.45mm
As =
MEd
0.87fykz
=
45.68 × 106
0.87 × 460 × 390.45
= 292.34mm2
/rib
Try 3T12mm bars per rib (As, prov = 336mm2
).

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End Spans A-B & B-C
MEd = 114.2 × 0.75 = 85.65kN. m/rib
∅
2
) = 450 − (25 + 8 +
20
2
) = 407mm; b = 750mm
k =
MEd
bd2fck
=
85.65 × 106
750 × 4072 × 30
= 0.023
z = d[0.5 + √0.25 − 0.882k] ≤ 0.95d
=d[0.5 + √0.25 − 0.882(0.021)] ≤ 0.95d
=0.95d = 0.95 × 407 = 386.65mm
We have to verify the position of the neutral axis:
x = 2.5(d − z) = 2.5(407 − 386.65) = 50.88mm
Therefore x < hf = 50.88 < 100 (neutral axis is within the flange)
Hence, we can design as a rectangular section.
As =
MEd
0.87fykz
=
85.65 × 106
0.87 × 460 × 386.65
= 553.52mm2
/rib
/rib).
Interior Support B
𝑀 𝐸𝑑 = 130.94 × 0.75 = 98.21𝑘𝑁. 𝑚/𝑟𝑖𝑏
∅
2
) = 450 − (25 + 8 +
16
2
) = 409mm; b = 750mm
k =
MEd
bd2fck
=
98.21 × 106
750 × 4072 × 30
= 0.026
z = d[0.5 + √0.25 − 0.882k] ≤ 0.95d
=𝑑[0.5 + √0.25 − 0.882(0.032)] ≤ 0.95𝑑
=0.95𝑑 = 0.95 × 409 = 388.55𝑚𝑚
As =
MEd
0.87fykz
=
98.21 × 106
0.87 × 460 × 388.55
= 631.58mm2
/rib

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Try 2T 16mm + 3T12mm bars (As, prov = 741mm2
/rib). To be spread across the effective width beff
beff = bw + beff,1 + beff,2 ≤ b
beff,1 = beff,2 = 0.1lo = 0.1 × 0.15(l1 + l2) = 0.1 × 0.15(9.5 + 9.5) = 285mm
beff = 150 + 285 + 285 = 720mm < b(750mm)
Therefore, this reinforcement will be spread across a width of 720mm.
5.1.4 Shear Design
The critical section for shear design is taken at a distance d from the supports and the width of the
rib is taken at middle of the longitudinal bars. By inspection the shear force is more critical at the
interior support B, therefore this can be used to size the shear reinforcement in the ribs.
𝑉𝐸𝑑 = (96.16 − 0.407 × 16.87) × 0.75 = 66.97𝑘𝑁
𝑉𝑅𝑑,𝑐 = (
0.18
𝛾𝑐
) 𝑘(100𝜌1 𝑓𝑐𝑘)
1
3 𝑏 𝑤 𝑑 ≥ 0.035𝑘
3
2√𝑓𝑐𝑘 𝑏 𝑤 𝑑
𝑘 = 1 + √
200
𝑑
= 1 + √
200
407
= 1.7 < 2
𝜌1 =
𝐴 𝑠
𝑏 𝑤 𝑑
As = 402mm2
assuming only 2T16mm bar can be used in resisting shear stresses
𝑏 𝑤 = 150 + 2 × (25 + 8 + 16/2)𝑡𝑎𝑛10 = 164.46𝑚𝑚
𝜌1 =
402
164.46 × 409
= 0.0060
0.18
1.5
) × 1.7 × (100 × 0.0060 × 30)
1
3 ∙ 164.46 × 409
≥ 0.035 × 1.7
3
2 × √30 × 164.46 × 409
=35.96kN
Since 𝑉𝐸𝑑 > 𝑉𝑅𝑑,𝑐 (66.97 > 33.46𝑘𝑁) therefore shear reinforcement is required.
𝜃 = 0.5𝑠𝑖𝑛−1
(
5.56𝑉𝐸𝑑
𝑏 𝑤 𝑑(1 −
𝑓𝑐𝑘
250
)𝑓𝑐𝑘
) = 0.5𝑠𝑖𝑛−1
(
5.56 × 66.97 × 103
164.46 × 409 (1 −
30
250
) 30
) = 6.05°
cot 𝜃 = cot 6.05 = 9.44 > 2.5 Hence take cot 𝜃 = 2.5

12
Asv
Sv
≥
VEd
zcotθfywd
where z = 0.9d = 0.9 × 409 = 368.1mm
Asv
Sv
≥
66.97 × 103
368.1 × 2.5 × 460
= 0.16
max spacing = 0.75d = 0.75 × 409 = 306.75mm
Use T8 @ 300mm centres (0.34).
5.1.5 Deflection Requirement
[
𝐿
𝑑
]
𝑙𝑖𝑚𝑖𝑡
= 𝑁 × 𝐾 × 𝐹1 × 𝐹2 × 𝐹3
𝜌 =
𝐴 𝑠,𝑟𝑒𝑞
𝐴 𝑐
=
𝑐ℎ + 𝑏 𝑤(𝑑 − 𝑡)
=
553.52
(750 × 100) + 164.46(407 − 100)
= 0.44%
ρo = 10−3
√fck = 10−3
× √30 = 0.547% since ρ < ρo
N = [11 +
1.5√fckρo
ρ
+ 3.2√fck (
ρo
ρ
− 1)
3
2
] = [11 +
1.5√30 × 0.547
0.44
+ 3.2√30 (
0.547
0.44
− 1)
3
2
]
= 23.31
F1 = 0.82
K = 1.3 (end spans)
F2 =
7.0
l
=
7.0
9.5
= 0.737
F3 =
310
σs
≤ 1.5
σs =
fy
γs
[
gk + φqk
ns
] (
As,req
As,prov
) ∙
1
δ
=
460
1.15
×
7.5 + 0.6(5)
16.87
×
553.52
628
= 219.4Mpa
F3 =
310
219.4
= 1.41
[
L
d
]
limit
= 23.31 × 1.3 × 0.82 × 0.737 × 1.41 = 25.82
[
L
d
]
actual
=
span
effective depth
=
9500
407
= 23.34
Since actual span-effective depth ratio is less than the limiting span-effective depth ratio, we can
conclude that deflection verification is complete.

13
5.1.6 Detailing Checks.
The minimum area of steel required in ribs:
As,min = 0.26
fctm
fyk
𝑏 𝑤d ≥ 0.0013bd
fctm = 0.30fck
2
3
= 0.3 × 30
2
3 = 2.9Mpa
As,min = 0.26 ×
2.9
460
× 164.46 × 409 ≥ 0.0013 × 164.46 × 409
= 110.25mm2
. By observation it is not critical.
Minimum area of steel required in toppings:
As,min = 0.26
fctm
fyk
𝑏𝑡t ≥ 0.0013bt
As,min = 0.26 ×
2.9
460
× 103
× 100 ≥ 0.0013 × 103
× 100
= 163.91mm2
/𝑚.
Use A142 Mesh in Topping.
Figure 7 summarizes the reinforcement required in the troughed slab. By combining this figure with
the floor layout shown in figure 4, the slab can be detailed to best practice after satisfying other
detailing requirements such as deciding the curtailment points and the required anchorage and
lapping lengths.
Figure 7: Summary of Reinforcement in Trough Slab

14
5.2 Band Beam on Gridline B
5.2.1 Actions on Beam
Permanent Actions:
a. load transferred from troughed slab to beam = 1.1 × (
9.5
2
+
9.5
2
) × 7.5 = 78.375kN/m
b. self-weight of band-beam = (0.45 × 25 − 6.0) × 1.5 = 7.875kN/m
Permanent Actions Gk = 78.375 + 7.875 = 86.25kN/m
Variable Actions:
load transferred from troughed slab to beam = 1.1 × (
9.5
2
+
9.5
2
) × 5.0 = 52.25kN/m
Variable Actions Qk = 52.25kN/m
By inspection the permanent actions are less than 4.5 times the variable actions, therefore equation
6.13b of BS EN 1990 will give the most unfavourable results.
Design Load = 1.35𝜉𝐺 𝑘 + 1.5𝑄 𝑘= (1.35 × 0.925 × 86.25) + (1.5 × 52.25) = 𝟏𝟖𝟔. 𝟏𝒌𝑵/𝒎
Design Permanent Load 1.35𝜉𝐺 𝑘 = 1.35 × 0.925 × 86.25 = 𝟏𝟎𝟕. 𝟕𝟎𝒌𝑵/𝒎
5.2.2 Analysis of Beam
The simplified coefficients for carrying out analysis cannot be applied because the spans are largely
unequal (i.e. outside the range of 85% of the largest span). Hence analysis of a sub-frame must be
carried out to obtain the design moments and shear forces at the critical sections. Only the bending
moment and shear force envelope is presented in this article. The reader is expected to have a basic
knowledge on analysis of subframes. However, guidance can be obtained from: How to Analyse
Element in Frames.
The load cases considered are:
• All spans loaded with the maximum design loads
• Alternate spans loaded with the maximum design load while the other spans are loaded with
the design permanent actions

15
Figure 8: Sub-Frame B-B
Figure 9: Bending Moment Envelope
Figure 10: Shear Force Envelope

16
5.2.3 Flexural Design
End Supports 1 (same as 5)
MEd = 225kN. m
∅
2
) = 450 − (25 + 8 +
16
2
) = 409mm; b = 1500mm
k =
MEd
bd2fck
=
225 × 106
1500 × 4092 × 30
= 0.030
z = d[0.5 + √0.25 − 0.882k] ≤ 0.95d
=d[0.5 + √0.25 − 0.882(0.057)] ≤ 0.95d
=0.95d = 0.95 × 409 = 388.55mm
As =
MEd
0.87fykz
=
225 × 106
0.87 × 460 × 388.55
= 1446.97mm2
/rib
).
End Span 1-2 (same as 4-5)
MEd = 434kN. m
∅
2
) = 450 − (25 + 8 +
20
2
) = 407mm
beff,1 = beff,2 = 0.2b + 0.1𝑙 𝑜 ≤ 0.2𝑙 𝑜
𝑏 = (
9500 − 750 − 750
2
) = 4000𝑚𝑚
𝑙 𝑜 = 0.85𝑙 = 0.85 × 6000 = 5100𝑚𝑚
beff,1 = beff,2 = (0.2 × 4000) + (0.1 × 5100) ≤ (0.2 × 5100) = 1020𝑚𝑚
beff = 1500 + 1020 + 1020 = 3540mm ≤ 4050mm
k =
MEd
bd2fck
=
434 × 106
3540 × 4072 × 30
= 0.025
z = d[0.5 + √0.25 − 0.882k] ≤ 0.95d

17
=d[0.5 + √0.25 − 0.882(0.020)] ≤ 0.95d
=0.95d = 0.95 × 407 = 386.65mm
x = 2.5(d − z) = 2.5(407 − 386.65) = 50.88mm
As =
MEd
0.87fykz
=
434 × 106
0.87 × 460 × 386.65
= 2804.75mm2
/rib
).
Interior Support 2 (same as 4)
MEd = 809kN. m
∅
2
) = 450 − (25 + 8 +
25
2
) = 404.5mm; b = 1500mm
k =
MEd
bd2fck
=
809 × 106
1500 × 404.52 × 30
= 0.110 < 0.168 (section is singly reinforced)
z = d[0.5 + √0.25 − 0.882k] ≤ 0.95d
=d[0.5 + √0.25 − 0.882(0.110)] ≤ 0.95d
=0.89d = 0.89 × 404.5 = 360.0mm
As =
MEd
0.87fykz
=
809 × 106
0.87 × 460 × 360.0
= 5615.25mm2
/rib
).
Interior Span 2-3 (same as 3-4)
MEd = 543kN. m
∅
2
) = 450 − (25 + 8 +
20
2
) = 407mm
beff,1 = beff,2 = 0.2b + 0.1𝑙 𝑜 ≤ 0.2𝑙 𝑜

18
𝑏 = (
9500 − 750 − 750
2
) = 4000𝑚𝑚
𝑙 𝑜 = 0.70𝑙 = 0.70 × 7500 = 5250𝑚𝑚
beff,1 = beff,2 = (0.2 × 4000) + (0.1 × 5250) ≤ (0.2 × 5250) = 1050𝑚𝑚
beff = 1500 + 1050 + 1050 = 3600mm ≤ 4050mm
k =
MEd
bd2fck
=
543 × 106
3600 × 4072 × 30
= 0.030
z = d[0.5 + √0.25 − 0.882k] ≤ 0.95d
=d[0.5 + √0.25 − 0.882(0.030)] ≤ 0.95d
=0.95d = 0.95 × 407 = 386.65mm
x = 2.5(d − z) = 2.5(407 − 386.65) = 50.88mm
As =
MEd
0.87fykz
=
543 × 106
0.87 × 460 × 386.65
= 3509.17mm2
/rib
).
Interior Support 3
MEd = 902kN. m
∅
2
) = 450 − (25 + 8 +
25
2
) = 404.5mm; b = 1500mm
k =
MEd
bd2fck
=
902 × 106
1500 × 404.52 × 30
= 0.123 < 0.168 (section is singly reinforced)
z = d[0.5 + √0.25 − 0.882k] ≤ 0.95d
=d[0.5 + √0.25 − 0.882(0.123)] ≤ 0.95d
=0.88d = 0.88 × 404.5 = 355.96mm
As =
MEd
0.87fykz
=
902 × 106
0.87 × 460 × 355.96
= 6331.82mm2
/rib
).

19
5.2.4 Shear Design
The critical section for shear design is taken at a distance d from the supports and by inspection,
shear is more critical at the interior support C, hence can be used to conservatively size the shear
reinforcement for the whole beam.
𝑉𝐸𝑑 = (710 − 0.405 × 186.1) = 634.63𝑘𝑁
0.18
𝛾𝑐
) 𝑘(100𝜌1 𝑓𝑐𝑘)
1
3 𝑏 𝑤 𝑑 ≥ 0.035𝑘
3
2√𝑓𝑐𝑘 𝑏 𝑤 𝑑
𝑘 = 1 + √
200
404.5
= 1 + √
200
404.5
= 1.7 < 2
𝐴 𝑠 = 3768𝑚𝑚2
𝑏 𝑤 = 1500𝑚𝑚
𝜌1 =
𝐴 𝑠
𝑏 𝑤 𝑑
=
3768
1500 × 404.5
= 0.0062
0.18
1.5
) × 1.7 × (100 × 0.0062 × 30)
1
3 ∙ 1500 × 404.5
≥ 0.035 × 1.7
3
2 × √30 × 1500 × 404.5 = 327.95kN
Since 𝑉𝐸𝑑 > 𝑉𝑅𝑑,𝑐 (634.63𝑘𝑁 > 327.95𝑘𝑁) therefore shear reinforcement is required.
𝜃 = 0.5𝑠𝑖𝑛−1
(
5.56𝑉𝐸𝑑
𝑏 𝑤 𝑑(1 −
𝑓𝑐𝑘
250
)𝑓𝑐𝑘
) = 0.5𝑠𝑖𝑛−1
(
5.56 × 634.63 × 103
1500 × 404.5 (1 −
30
250
) 30
) = 6.36°
cot 𝜃 = cot 6.36 = 8.97 > 2.5 Hence take cot 𝜃 = 2.5
Asv
Sv
≥
VEd
zcotθfywd
where z = 0.9d = 0.9 × 404.5 = 364.05mm
Asv
Sv
≥
634.63 × 103
364.05 × 2.5 × 460
= 1.52
max spacing = 0.75d = 0.75 × 404.5 = 303.4mm
𝐴 𝑠𝑣,𝑚𝑖𝑛
𝑆 𝑣
=
0.08√𝑓𝑐𝑘 𝑏 𝑤
𝑓𝑦𝑘
=
0.08 × √30 × 1500
460
= 1.43
Use 6T8 @ 175mm centres (1.72).

20
5.2.5 Deflection Requirement
By inspection, deflection check is not necessary in the band beams, since the ribs with considerably
longer spans have been verified and found to satisfy the basic requirement. However, for the purpose
of illustration we can check the beams for deflection.
End Spans
[
𝐿
𝑑
]
= 𝑁 × 𝐾 × 𝐹1 × 𝐹2 × 𝐹3
𝜌 =
𝐴 𝑐
=
𝑏 𝑤 𝑑 + (𝑏 𝑒𝑓𝑓 − 𝑏 𝑤)ℎ𝑓
=
2804.75
(1500 × 407) + (3540 − 1500)100
= 0.34%
ρo = 10−3
√fck = 10−3
× √30 = 0.547% since ρ < ρo
N = [11 +
1.5√fckρo
ρ
+ 3.2√fck (
ρo
ρ
− 1)
3
2
] = [11 +
1.5√30 × 0.547
0.34
+ 3.2√30 (
0.547
0.34
− 1)
3
2
]
= 32.54
F1 = 0.82
K = 1.3 (end spans)
F2 = 1.0
F3 =
310
σs
≤ 1.5
σs =
fy
γs
[
gk + φqk
ns
] (
As,req
As,prov
) ∙
1
δ
=
460
1.15
×
86.25 + 0.6(52.25)
186.1
×
2804.75
3140
= 225.8Mpa
F3 =
310
225.8
= 1.37
[
L
d
]
limit
= 32.54 × 1.3 × 0.82 × 1.0 × 1.37 = 47.52
[
L
d
]
actual
=
span
effective depth
=
6000
407
= 14.74
Since the limiting span-effective depth ratio is greater than the actual span-effective depth ratio. It
therefore follows that deflection has been satisfied in the end spans.

21
Interior Spans
. [
𝐿
𝑑
]
= 𝑁 × 𝐾 × 𝐹1 × 𝐹2 × 𝐹3
𝜌 =
𝐴 𝑐
=
𝑏 𝑤 𝑑 + (𝑏 𝑒𝑓𝑓 − 𝑏 𝑤)ℎ𝑓
=
3509.17
(1500 × 407) + (3600 − 1500)100
= 0.43%
ρo = 10−3
√fck = 10−3
× √30 = 0.547% since ρ < ρo
N = [11 +
1.5√fckρo
ρ
+ 3.2√fck (
ρo
ρ
− 1)
3
2
] = [11 +
1.5√30 × 0.547
0.43
+ 3.2√30 (
0.547
0.43
− 1)
3
2
]
= 23.94
F1 = 0.82
K = 1.5 (Interior spans)
F2 =
7.0
𝑙
=
7
7.5
= 0.93
F3 =
310
σs
≤ 1.5
σs =
fy
γs
[
gk + φqk
ns
] (
As,req
As,prov
) ∙
1
δ
=
460
1.15
×
86.25 + 0.6(52.25)
186.1
×
3509.17
3768
= 235.4Mpa
F3 =
310
235.4
= 1.32
[
L
d
]
limit
= 23.94 × 1.5 × 0.82 × 0.93 × 1.32 = 36.15
[
L
d
]
actual
=
span
effective depth
=
7500
407
= 18.43
Since the limiting span-effective depth ratio is greater than the actual span-effective depth ratio. It
therefore follows that deflection has been satisfied in the interior spans.
5.2.6 Detailing Checks.
Minimum Area of Steel
As,min = 0.26
fctm
fyk
𝑏 𝑤d ≥ 0.0013bd

22
fctm = 0.30fck
2
3
= 0.3 × 30
2
3 = 2.9Mpa
Hogging at Supports
As,min = 0.26 ×
2.9
460
× 1500 × 404.5 ≥ 0.0013 × 1500 × 404.5
= 994.54mm2
. By observation it is not critical anywhere at the supports.
Sagging in Spans
As,min = 0.26 ×
2.9
460
× 1500 × 407 ≥ 0.0013 × 1500 × 404.5
= 1000.69mm2
. By observation it is not critical anywhere in the spans.
Finally, before carrying out the detailing of the beam other detailing checks must be carried out. This
include calculating bar curtailments points, anchorage and lap lengths in line with the code
specifications and best practices.
The detailing of the beam can be obtained from figure 11 which summarizes the reinforcement
required in the various section of the beam.
Figure 11: Summary of Band Beam on Gridline B

23
6.0 References
BS EN 1990: Eurocode Basis of Structural Design.
BS EN 1990: UK National Annex to Eurocode: Basis of Structural Design.
BS EN 1991-1: Eurocode 1: Actions on Structures – Part 1-1: General actions – densities, self-
weight, imposed loads for buildings.
BS EN 1991-1: UK National Annex to Eurocode 1: Actions on Structures – Part 1-1– densities, self-
weight, imposed loads for buildings
BS EN 1992-1-1 Eurocode 2: Design of Concrete Structures – Part 1-1: General Rules for Buildings.
BS EN 1992-1-1 UK National Annex to Eurocode 2: Design of Concrete Structures – Part 1-1:
General Rules for Buildings.
Goodchild, C.H., Webster, R.M & Elliot, K.S (2009): Economic Concrete Frame Elements. The
Concrete Centre
Brooker O. (2009): Concrete Building Scheme Designers Manual: The Concrete Centre.
The Institution of Structural Engineers (2006) Manual for the design of concrete building structures
to Eurocode 2 London: The Institution of Structural Engineer.
The Concrete Centre (2009) Worked Examples to Eurocode 2: Volume 1 [Online] Available at:
www.concretecentre.com/ pdf/Worked_Example_Extract_Slabs.pdf (Accessed: February 2013).
The Concrete Centre 2016 - Concrete Framed Buildings.

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