This document provides design calculations for structural elements of a concrete car park structure according to BS-8110, including: 1. A one-way spanning roof slab with a span of 2.8m, designed as simply supported with 10mm main reinforcement bars at 300mm spacing and 8mm secondary bars. 2. A load distribution beam D and non-load bearing beam E, with calculations provided for beam D's dead and imposed loads. 3. Requirements include individual work submission by January 2nd, 2016 and assumptions to be clearly stated.

1. Question:-
A concrete car park structure is shown in the figure you are requested to design the
following elements using BS-8110
 Slab
 Beam denoted by D and E
 A pad footing
 A column
DATA’S:-
1) Finishes of 1 KN/m2
2) Imposed load on slab
Of = [1 +
𝐵
10
] KN/m2
3) Grade of concrete 25 N/mm2
4) Assume self-weight of block
work is 20 KN/m3
5) Bearing pressure of ground
soil 150 KN/m2
6) Cover for slab is 20 mm
7) Cover for beam is 25 mm
8) Cover for footing is 50 mm
WHERE:-
 “A” is a first digit of your
Registration number.
 “B” is a last digit of your
Registration number.
REQUREMENTS:-
 Write your assumption clearly.
 All work are to be under taken individually.
 Submission date is 02/01/2016.
 Late submission will not be accepted.
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2. P.NIROJAN–JAF/CE/2014/F/0041–CIVILENGINEERING–2NDYEAR–2NDSEMISTER–S.L.I.A.T.EJAFFNACAMPAS–PAGE:2
Calculation:-
[My registration number is 41, First digit number (A) = 4,and second digit number (B) = 1]
1) Imposed load on slab = [1 +
𝐵
10
] KN/m2
Solution: - Imposed load on slab (Qk) is = [1 +
1
10
] KN/m2
= [1 + 0.1] KN/m2
= 1.1 KN/m2
2) Car Park Width is = [2.4 +
𝐴
10
] m
Solution: - Car Park Width is = [2.4 +
4
10
] m
= [2.4 + 0.4] m
= 2.8 m
[This design calculation Assume the finishes load only consider the slab]
 Symbols:-
𝑓𝑦 – Characteristic strength of reinforcement
𝑓𝑦𝑣 – Characteristic strength of links
𝑓𝑐𝑢 – Characteristic strength of concrete
𝛾 𝑚 – Partial safety factor for strength of materials
𝑑 – Effective depth of the tension reinforcement
𝐴 𝑆 – Area of the tension reinforcement
𝐴 𝑆𝑉 – Total cross-section of links at the neutral axis
𝑏 – Width or Effective width of the section
𝐴 𝑆𝐶 – Area of vertical reinforcement
𝐺 𝑘 – Characteristic dead load
𝑄 𝑘 – Characteristic Imposed load
𝑙 𝑥 – Length of shorter side
𝑙 𝑦 – Length of longer side
𝑙 – Span of the beam
𝐻 – High yields steel
𝑀 – Design ultimate moment
𝑉 – Design shear force due to ultimate loads
 𝑐 – Design concrete shear stress
 – Design shear stress at a cross-section
𝑁 – Design axial force
𝑆 𝑉 – Spacing of links along the member
𝑐 – Column width
𝑙 𝐶 – Half the spacing between column centers
𝐴 𝐶 – Area of the concrete section
𝑓𝑆 – Estimated design service stress in the
tension reinforcement
𝑙 𝑂 – Clear height between end restraints
ℎ – Overall depth of the cross-section of a R/F

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REINFORCEMENT
SLAB
BS 8110 : 1997
PART – 1, SECTION 3 , CLAUSE 3.5[Solid slabs supported by
beams or walls]

4. h
d
c
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 Design of a reinforced roof slab ( BS 8110 )
A reinforced concrete roof subject to an imposed load of 1.1 KNm−2 spans between
walls as shown below. Design the roof for 𝑓𝑐𝑢= 25 Nmm-2 and cover for slab is 20mm.
[Assuming the following material strength 𝑓𝑦 = 460 Nmm-2 and Diameter of main steel
(Φ) = 10 mm and Assume this slab is simply supported but normal condition is fixed
support]
a) ANALYSIS
The car park area is 2800 mm x 6000 mm, so 𝑙 𝑥 = 2800 mm and 𝑙 𝑦 = 6000 mm
𝑙 𝑦
𝑙 𝑥
=
6000
2800
= 2.143 > 2, Hence this slab is one-way spanning slab.
b) DEPTH OF SLAB
[Assume the modification factor is 1.5]
Minimum effective depth, dmin =
𝑆𝑝𝑎𝑛
𝐵𝑎𝑠𝑖𝑐 𝑟𝑎𝑡𝑖𝑜 ×𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
=
2800
20 ×1.5
= 93.333 mm
Hence, assume effective depth of slab (d) = 95 mm.
Overall depth of slab (h) = d + /2 + c
= 95 + 10/2 + 20
= 120 mm
c) LOADING
 Dead
Self-weight of slab = 0.120m × 24 kNm−3 = 2.880 KNm−2
Finishes = 1 KNm-2
Total Dead load (Gk) = 2.880 + 1 = 3.880 KNm-2
225 mm
2800 mm

(Basic ratio from table
3.9,BS-8110)

5. P.NIROJAN–JAF/CE/2014/F/0041–CIVILENGINEERING–2NDYEAR–2NDSEMISTER–S.L.I.A.T.EJAFFNACAMPAS–PAGE:5  Imposed
Total imposed load (Qk) = 1.1 KNm−2
 Ultimate design load
(For 1 m width of slab total ultimate load), W, is
W = (1.4 Gk + 1.6 Qk) x width of slab × span
= (1.4 × 3.880 + 1.6 × 1.1) x 1 × 2.80
= 20.138 KN
d) MOMENT
 Design moment (M) =
𝑊 ×𝐿
8
=
20.138 𝐾𝑁 ×2.80 𝑚
8
= 7.048 KNm
 Ultimate design moment (Mu) = 0.156 𝑓𝑐𝑢bd2
= 0.156 x 25 Nmm-2 x 1 m x 952 mm2
= 35.197 KNm
Since M < Mu, no compression reinforcement is required.
e) MAIN STEEL
 K =
𝑀
𝑓 𝑐𝑢 b𝑑2 =
7.048 × 106
25 ×1000 ×952 = 0.031
 Z = 𝑑 [ 0.5 + √0.25 − 𝐾
0.9⁄ ]
= 95 [ 0.5 + √0.25 − 0.031
0.9⁄ ]
= 95 [ 0.964 ] mm = 91.607 mm > 0.95𝑑 (= 90.25 mm)
Hence Z = 0.95𝑑 (= 90.25 mm)
 As =
𝑀
0.95 × 𝑓𝑦 ×𝑍
=
7.048 × 106
0.95 ×460 ×90.25
= 178.705 mm2/m
 Asmin = 0.13% Ac = 0.0013 x 1000 x 120 = 156 mm2/m < As
Width of slab
(Asmin from table 3.25,BS-8110)

6. For detailing purposes this area of steel has to be transposed into bars of a given
diameter and spacing using steel area tables. Provide 10 mm diameter bars spaced at
300 mm
 H10 @ 300 mm centers (As = 262mm2/m).
f) ACTUAL MODIFICATION FACTOR
 Design services stress, 𝑓𝑠 =
2
3
× 𝑓𝑦 ×
𝐴𝑠 𝑟𝑒𝑞
𝐴𝑠 𝑃𝑟𝑜𝑣
=
2
3
× 460 ×
178.705
262.00
= 209.171 Nmm-2
 Modification factor = 0.55 +
(477 − 𝑓𝑠)
120 × [0.9 +
𝑀
𝑏𝑑2]
= 0.55 +
(477−209.171)
120 × [0.9+
7.048 ×106
1000 ×952 ]
= 0.55 +
267.829
120 × [0.9+ 0.781]
= 0.55 + 1.328
= 1.878 < 2 , This factor satisfy for this case.
 Hence, New dmin =
𝑆𝑝𝑎𝑛
𝐵𝑎𝑠𝑖𝑐 𝑟𝑎𝑡𝑖𝑜 ×𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
=
2800
20 ×1.878
= 74.547 mm < assume d = 95 mm
Therefore take d = 95 mm and provide H10 @ 300 mm Centers as main steel.
g) SECONDARY REINFORCEMENT
This is one way-spanning slab, hence secondary reinforcement Based on minimum steel
area = 156 mm2/m. Assume secondary reinforcement diameter () 8 mm.
Hence from using steel area tables, provide H8 @ 300 mm Centre’s (As = 168 mm2/m).
(Note that the condition 𝛽 𝑏 ≤ 1does not apply hear)
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7. d=95 mm
h) DEFLECTION
 Actual
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
=
2800
95
= 29.474
 Permissible
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
= Basic ratio x modification factor
= 20 x 1.878
= 37.560
Hence Actual
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
< Permissible
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
Hence defection check is satisfy.
i) SHEAR REINFORCEMENT
Since slab is symmetrically loaded
RA = RB =
𝑊
2
=
20.138
2
= 10.069 KN
 Ultimate shear force (V ) = 10.069 KN
And design shear stress, is
 =
𝑉
𝑏 𝑣 𝑑
=
10.069 × 103
1000 × 95
= 0.106 Nmm-2
 Design concrete shear stress, c
c = 0.79 {
100 𝐴𝑠
𝑏 𝑣 𝑑
}
1
3⁄
(
400
𝑑
)
1
4⁄
𝛾 𝑚
c = 0.79 {
100 ×262
1000 × 95
}
1
3⁄
(
400
95
)
1
4⁄
1.25
c =
0.79 ×0.6509 ×1.4325
1.25
= 0.7366 Nmm-2
From Table 3.16 for BS-8110, since  < c, no shear reinforcement is required.
Main steel
Secondary steel
RBRA 2800 mm
W
V
V
(c from table 3.8,BS-8110)
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8. j) REINFORCEMENT DETAILS
k) CHECK SPACING BETWEEN BARS
Maximum spacing between bars should not exceed the lesser of 3d (= 285 mm) or
750 mm. Actual spacing = 300 mm main steel and 300 mm secondary steel. OK.
l) MAXIMUM CRACK WITH
Since the slab depth does not exceed 200 mm, the above spacing between bars will
automatically ensure that the maximum permissible crack width of 0.3 mm will not be
exceeded.
(from clause 3.12.11.2.7,BS-
88110)
(from clause 3.12.11.2.1,BS-
88110)
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9. REINFORCEMENT
BEAM
BS 8110 : 1997
PART – 1 , SECTION 3 , CLAUSE 3.4 [ Beams ]
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10.  Design of a reinforced beam ( BS 8110 )
A reinforced concrete roof slab to an imposed load of 1.1 KNm−2, self-weight of block
work is 20 KN/m3. Design the beam for 𝑓𝑐𝑢= 25 Nmm-2 and cover for beam is 25 mm.
[Assuming the following material strength 𝑓𝑦 = 460 Nmm-2 and 𝑓𝑦𝑣= 250 Nmm-2]
Figure of the one-way spanning slab load distribution as shown above. It is supported
by beams in only 2 sides, so beam-D is load distribution beam and beam-E is non load
distribution beam
Design of a beam - D
D – Beam as shown below:
a) LOADING
 Dead
Self-weight of slab = 0.120 m x 1.2875 m x 24 KNm−3 = 3.7080 KNm−1
Self-weight of beam = 0.225 m x 0.500 m x 24 KNm−3 = 2.70 KNm−1
Finishes of slab = 1.2875 m x 1 KNm-2 = 1.2875 KNm−1
Block wall load = 0.125 m x 1.00 m x 20 KNm−3 = 2.50 KNm−1
Total Dead load (Gk) = 3.708 + 2.70 + 1.2875 + 2.50 = 10.1955 KNm-1
 Imposed
Total imposed load (Qk) = 1.2875 m x 1.1 KNm−2 = 1.4163 KNm-1
 Ultimate design load
Total ultimate load, W is
W = (1.4 Gk + 1.6 Qk) × span
= (1.4 × 10.1955 + 1.6 × 1.4163) × 6.0 m
= 99.2387 KN
Beam - D
Beam- E
225 mm
6000 mm
h=500 mm
b=225 mm
SECTION OF BEAM
Load distribution
Direction
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11. b) MOMENT
 Design moment (M) =
𝑊 ×𝐿
8
=
99.2387 × 6.0
8
= 74.4290 KNm
c) EFFECTIVE DEPTH
 Assume diameter of main bars (Φ) = 20 mm
 Assume diameter of links (Φ′) = 8 mm
 Cover for beam Δc = 25 mm.
d = h − c − Φ′ − Φ/2
= 500 − 25 − 8 − 20/2 = 457 mm
d) ULTIMATE MOMENT OF RESISTENT
 Ultimate design moment (Mu) = 0.156𝑓𝑐𝑢 bd2
= 0.156 x 25 Nmm-2 x 0.225 m x 4572 mm2
= 183.265 KNm
Since M < Mu, no compression reinforcement is required.
e) MAIN STEEL
 K =
𝑀
𝑓 𝑐𝑢 b𝑑2 =
74.4290 × 106
25 × 225 × 4572 = 0.0634
 Z = 𝑑 [ 0.5 + √0.25 − 𝐾
0.9⁄ ]
= 457 [ 0.5 + √0.25 − 0.0634
0.9⁄ ]
= 457 [ 1.0661 ] mm = 487.1977 mm > 0.95𝑑 (= 434.150 mm)
Hence Z = 420.2828 mm
 As =
𝑀
0.95 × 𝑓𝑦 ×𝑍
=
74.4290 × 106
0.95 ×460 ×420.2828
= 405.2464 mm2
 Asmin = 0.13% Ac = 0.0013 x 225 x 500 = 146.250 mm2 < As
For detailing purposes this area of steel has to be transposed into bars of a given
diameter and spacing using steel area tables. Provide 20 mm diameter 2 bars .
 2H20 (As = 628 mm2)
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12. Effective span (clause 3.4.1.2, BS 8110)
The above calculations were based on an effective span of
6 m, but this needs to be confirmed. For a simply
supported beam this should be taken as the lesser of
(1) The distance between centers of bearings, A, or
(2) The clear distance between supports, D, plus the effective depth, d, of the beam
 Clear distance between supports = 6000 mm – 225 mm = 5775 mm
 Clear distance + effective depth = 5775 mm + 457 mm = 6232 mm
Assumed span length of 6 m is correct.
f) DEFLECTION
 Actual
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
=
6000
457
= 13.129
 Design services stress, 𝑓𝑠 =
2
3
× 𝑓𝑦 ×
𝐴𝑠 𝑟𝑒𝑞
𝐴𝑠 𝑃𝑟𝑜𝑣
=
2
3
× ×
405.2464
628.00
= 197.8910 Nmm-2
 Modification factor = 0.55 +
(477 − 𝑓𝑠)
120 × [0.9 +
𝑀
𝑏𝑑2]
= 0.55 +
(477−197.8910)
120 × [0.9+
74.4290 ×106
225 × 4572 ]
= 0.55 +
279.1090
120 × [0.9+ 1.5839]
= 0.55 + 0.9364
= 1.4864 < 2 , This factor satisfy for this case.
 Permissible
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
= Basic ratio x modification factor
= 20 x 1.4864
= 29.7279
Hence Actual
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
< Permissible
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
The beam therefore satisfies the deflection criteria in BS 8110.
460
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13. g) SHEAR REINFORCEMENT
Ultimate design load W = 104.093 KN
Since beam is symmetrically loaded
RA = RB =
𝑊
2
=
99.2387
2
= 49.6194 KN
 Ultimate shear force (V ) = 49.6194 KN
And shear stress, is
 =
𝑉
𝑏 𝑣 𝑑
=
49.6194 × 103
225 × 457
= 0.4826 Nmm-2 < Permissible = 0.8√𝑓𝑐𝑢
h) SPACING OF LINKS
All through the shear force will be maximum at the face of the support. The design shear
force for uniformly distributed load is at the section “d” from the face
 Calculate the design shear force
𝑉 𝑑
3000−𝑑
=
𝑉
3000
𝑉 𝑑
3000−457
=
49.6194
3000
Vdesign = 42.0607 KN
 Calculate the design shear stress, d is
d =
𝑉 𝑑
𝑏 𝑣 𝑑
=
42.0607 × 103
225 × 457
= 0.4091 Nmm-2
 Design concrete shear stress, c
c = {
100 𝐴𝑠
𝑏 𝑣 𝑑
}
1
3⁄
(
400
𝑑
)
1
4⁄
𝛾 𝑚
c = {
100 ×628
225× 457
}
1
3⁄
(
400
457
)
1
4⁄
1.25
c =
0.79 × 0.8484 ×0.9672
1.25
= 0.5186 Nmm-2
RBRA 6000 mm
W
V
V
[= 0.8√25 = 4 Nmm-2
]
VVdd
V Vdesing
3000 mm
0.79
0.79
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14.  0.5c = 0.5 x 0.5186 = 0.2593 Nmm-2
 c + 0.4 = 0.5186 + 0.4 = 0.9186 Nmm-2
Hence, 0.5 𝑐 <  𝑑 < (  𝑐 + 0.4 )
Design links required according to
𝐴 𝑠𝑣 ≥
0.4 𝑏 𝑣 𝑆 𝑣
0.95 𝑓𝑦𝑣
[Diameter of link 8 mm and 𝑓𝑦𝑣 = 250 Nmm-2]
 𝐴 𝑠𝑣 = 2 ×
𝜋
4
× 𝑑2
= 2 ×
𝜋
4
× 82
= 100.531 mm2
Hence 100.531 ≥ 0.4 × 225 × 𝑆 𝑣
0.95 ×250
𝑆 𝑣 ≤ 100.531 ×0.95 ×250
0.4 ×225
𝑆 𝑣 ≤ 265.29 mm
Maximum spacing of links is 0.75d = 0.75 × 457 = 342.75 mm. Hence provide H8
links at 200 mm centers.
i) REINFORMATION DETAILS
The sketch below shows the main reinforcement requirements for the beam “D”
(from Table 3.7 ,BS-88110)
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15. Design of a beam - E
E – Beam as shown below:
Self-weight of block work is 20 KN/m3.
Design the beam for 𝑓𝑐𝑢= 25 Nmm-2 and
cover for beam is 25 mm. [Assuming the
following material strength 𝑓𝑦 = 250
Nmm-2 ]
a) LOADING
 Dead
Self-weight of beam = 0.225 m x 0.300 m x 24 KNm−3 = 1.62 KNm−1
Block wall load = 0.125 m x 1.00 m x 20 KNm−3 = 2.50 KNm−1
Total Dead load (Gk) = 1.62 + 2.50 = 4.120 KNm-1
 Imposed
Total imposed load (Qk) = 0.00
 Ultimate design load
Total ultimate load, W is
W = (1.4 Gk + 1.6 Qk) × span
= (1.4 × 4.120 + 1.6 × 0.00) × 2.8 m
= 16.15 KN
b) MOMENT
 Design moment (M) =
𝑊 ×𝐿
8
=
16.15 ×2.8
8
= 5.653 KNm
c) EFFECTIVE DEPTH
 Assume diameter of main bars (Φ) = 10 mm
 Assume diameter of links (Φ′) = 6 mm
225 mm
2800 mm
h=300 mm
b=225 mm
SECTION OF BEAM
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16.  Cover for beam Δc = 25 mm.
d = h − c − Φ′ − Φ/2
= 300 − 25 − 6 − 10/2 = 264 mm
d) ULTIMATE MOMENT OF RESISTENT
 Ultimate design moment (Mu) = 0.156𝑓𝑐𝑢 bd2
= 0.156 x 25 Nmm-2 x 0.225 m x 2642 mm2
= 61.158 KNm
Since M < Mu, no compression reinforcement is required.
e) MAIN STEEL
 K =
𝑀
𝑓 𝑐𝑢 b𝑑2 =
5.653 × 106
25 ×225 ×2642 = 0.014419
 Z = 𝑑 [ 0.5 + √0.25 − 𝐾
0.9⁄ ]
= 264 [ 0.5 + √0.25 − 0.014419
0.9⁄ ]
= 264 [ 0.9837 ] mm = 259.6968 mm > 0.95𝑑 (= 250.80 mm)
Hence Z = 0.95𝑑 (= 250.80 mm)
 As =
𝑀
0.95 × 𝑓𝑦 ×𝑍
=
5.633 × 106
0.95 ×250 ×250.80
= 138.216 mm2
 Asmin = 0.24% Ac = 0.0024 x 225 x 300 = 162.00 mm2 > As
For detailing purposes this area of steel has to be transposed into bars of a given
diameter and spacing using steel area tables. Provide 10 mm diameter 3 bars .
 3H10 (As = 236 mm2)
Effective span (clause 3.4.1.2, BS 8110)
The above calculations were based on an effective span of 2.8 m, but this needs to be
confirmed. For a simply supported beam this should be taken as the lesser of
(Asmin from table 3.25,BS-8110)
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17.  Clear distance between supports = 2800 mm – 225 mm = 2575 mm
 Clear distance + effective depth = 2575 mm + 264 mm = 2839 mm
Assumed span length of 2.8 m is correct.
f) DEFLECTION
 Actual
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
=
2800
264
= 10.6061
 Design services stress, 𝑓𝑠 =
2
3
× 𝑓𝑦 ×
𝐴𝑠 𝑟𝑒𝑞
𝐴𝑠 𝑃𝑟𝑜𝑣
=
2
3
× ×
138.216
236.00
= 97.6102 Nmm-2
 Modification factor = 0.55 +
(477 − 𝑓𝑠)
120 × [0.9 +
𝑀
𝑏𝑑2]
= 0.55 +
(477−97.6102)
120 × [0.9+
5.653 ×106
225 × 2642]
= 0.55 +
379.3898
120 × [0.9+ 0.3605]
= 0.55 + 2.5082
= 3.0582 > 2 , This factor not satisfy for this case.
Hence,
This beam cross-section (225 mm x 300 mm) is not satisfy for this load distribution
So we can change the dimension, assume this cross-section and again design the
beam-E
 Cross-section = 225 mm x 180 mm
g) LOADING
 Dead
Self-weight of beam = 0.225 m x 0.180 m x 24 KNm−3 = 0.972 KNm−1
Block wall load = 0.125 m x 1.00 m x 20 KNm−3 = 2.50 KNm−1
Total Dead load (Gk) = 0.972 + 2.50 = 3.472 KNm-1
 Imposed
Total imposed load (Qk) = 0.00
250
h=180 mm
b=225 mm
SECTION OF BEAM
(Table 3.10 for BS-8110)
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18.  Ultimate design load
Total ultimate load, W is
W = (1.4 Gk + 1.6 Qk) × span
= (1.4 × 3.472 + 1.6 × 0.00) × 2.8 m
= 13.610 KN
h) MOMENT
 Design moment (M) =
𝑊 ×𝐿
8
=
13.610 × 2.8
8
= 4.7646 KNm
i) EFFECTIVE DEPTH
 Assume diameter of main bars (Φ) = 10 mm
 Assume diameter of links (Φ′) = 6 mm
 Cover for beam Δc = 25 mm.
d = h − c − Φ′ − Φ/2
= 180 − 25 − 6 − 10/2 = 144 mm
j) ULTIMATE MOMENT OF RESISTENT
 Ultimate design moment (Mu) = 0.156𝑓𝑐𝑢 bd2
= 0.156 x 25 Nmm-2 x 0.225 m x 1442 mm2
= 18.1958 KNm
Since M < Mu, no compression reinforcement is required.
k) MAIN STEEL
 K =
𝑀
𝑓 𝑐𝑢 b𝑑2 =
4.7646 × 106
25 ×225 ×1442 = 0.040849
 Z = 𝑑 [ 0.5 + √0.25 − 𝐾
0.9⁄ ]
= 144 [ 0.5 + √0.25 − 0.040849
0.9⁄ ]
= 144 [ 0.9523 ] mm = 137.1371 mm > 0.95𝑑 (= 136.80 mm)
Hence Z = 0.95𝑑 (= 136.80 mm)
 As =
𝑀
0.95 × 𝑓𝑦 ×𝑍
=
4.7646 × 106
0.95 ×250 ×136.80
= 146.6482 mm2
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19.  Asmin = 0.24% Ac = 0.0024 x 225 x 180 = 97.20 mm2 < As
For detailing purposes this area of steel has to be transposed into bars of a given
diameter and spacing using steel area tables. Provide 10 mm diameter 2 bars .
 2H10 (As = 157.00 mm2)
l) DEFLECTION
 Actual
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
=
2800
144
= 19.4444
 Design services stress, 𝑓𝑠 =
2
3
× 𝑓𝑦 ×
𝐴𝑠 𝑟𝑒𝑞
𝐴𝑠 𝑃𝑟𝑜𝑣
=
2
3
× ×
146.6482
157
= 155.6775 Nmm-2
 Modification factor = 0.55 +
(477 − 𝑓𝑠)
120 × [0.9 +
𝑀
𝑏𝑑2]
= 0.55 +
(477−155.6775)
120 × [0.9+
4.7646 ×106
225 × 1442 ]
= 0.55 +
321.3225
120 × [0.9+ 1.0212]
= 0.55 + 1.3937
= 1.9437 < 2 , This factor satisfy for this case.
 Permissible
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
= Basic ratio x modification factor
= 20 x 1.9437
= 38.8749
Hence Actual
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
< Permissible
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
The beam therefore satisfies the deflection criteria in BS 8110.
j) SHEAR REINFORCEMENT
Ultimate design load W = 12.9752 KN
Since beam is symmetrically loaded
RA = RB =
𝑊
2
=
13.610
2
= 6.8050 KN
250
RBRA 2800 mm
W
V
V
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20.  Ultimate shear force (V ) = 6.8050 KN
And shear stress, is
 =
𝑉
𝑏 𝑣 𝑑
=
6.8050 × 103
225 × 144
= 0.2100 Nmm-2 < Permissible = 0.8√𝑓𝑐𝑢
k) SPACING OF LINKS
All through the shear force will be maximum at the face of the support. The design shear
force for uniformly distributed load is at the section “d” from the face
 Calculate the design shear force
𝑉 𝑑
1400−𝑑
=
𝑉
1400
𝑉 𝑑
1400−144
=
6.8050
1400
Vdesign = 6.1051 KN
 Calculate the design shear stress, d is
d =
𝑉 𝑑
𝑏 𝑣 𝑑
=
6.1051 × 103
225 × 144
= 0.1884 Nmm-2
 Design concrete shear stress, c
c = {
100 𝐴𝑠
𝑏 𝑣 𝑑
}
1
3⁄
(
400
𝑑
)
1
4⁄
𝛾 𝑚
c = {
100 ×157
225× 144
}
1
3⁄
(
400
144
)
1
4⁄
1.25
c =
0.79 × 0.7854 ×1.2910
1.25
= 0.6408 Nmm-2
 0.5c = 0.5 x 0.6408 = 0.3204Nmm-2
design < 0.5c , shear reinforcement may not be necessary.
l) REINFORMATION DETAILS
The sketch below shows the main reinforcement requirements for the beam “E”
[= 0.8√25 = 4 Nmm-2
]
V
Vdd
V Vd
1400 mm
0.79
0.79
C–25mm
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21. REINFORCEMENT
BASES
BS 8110 : 1997
PART – 1 , SECTION 3 , CLAUSE 3.11 [ Bases ]
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22.  Design of a reinforced foundation ( BS 8110 )
A 225 mm square column carries a load transmit to foundation, safe bearing capacity
of the soil is 150 KNm-2. Design a square pad footing to resist the load.𝑓𝑐𝑢= 25 Nmm-2
[Assuming the following material strength 𝑓𝑦 = 250 Nmm-2]
a) LOADING
 Dead
Self-weight of slab = 0.120 m x 1.2875 m x 2.8875 m x 24 KNm−3 = 10.7069 KN
Self-weight of beam-E = 0.225 m x 0.180 x 1.2875 m x 24 KNm−3 = 1.2515 KN
Self-weight of beam-D = 0.225 m x 0.500 m x 2.8875 m x 24 KNm−3 = 7.796 KN
Self-weight of column = 0.225 m x 0.225 m x 3.5 m x 24 KNm−3 = 4.2525 KN
Finishes of slab = 1.2875 m x 2.8875 m x 1 KNm-2 = 3.7177 KN
Block wall load = 0.125 m x 1.00 m x 4.50 m x 20 KNm−3 = 11.250 KN
Total Dead load (Gk) = 10.7069 + 1.2515 + 7.796 + 4.2525 + 3.7177 + 11.250
= 38.9746 KN
 Imposed
Total imposed load (Qk) = 1.2875 m x 2.8875 m x 1.1 kNm−2 = 4.0894 KN
 Ultimate design load
Total ultimate load, W is
W = (1.4 Gk + 1.6 Qk)
= (1.4 × 38.9746 + 1.6 × 4.0894)
= 61.1075 KN
 Serviceability load
Design axial load =
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑙𝑜𝑎𝑑
1.45
= 61.1075
1.45
= 42.1431 KN.
Beam - D
Beam- E
Design
Area
3.0 m
1.4 m
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23.  Total weight
Total weight including footing weight = Serviceability load + 8% of Axial load
= 42.1431 +
8 ×42.1431
100
= 45.5145 KN.
b) PLAN AREA
Plan area of base =
𝑁
𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑖𝑙
=
45.5145
150
= 0.3034 m2 = 303430.00 mm2
Hence provide a 600 mm square base (plan area 0.36 m2)
c) SELF-WEIGHT OF FOOTING
[Assume the overall depth of footing (h) = 150 mm]
 Self-weight of footing = Area x h x density of concrete
= 0.36 x 0.15 x 24 KNm-3
= 1.296 KN
 Actual total load = 42.1431 + 1.296 = 43.4391 KN
 Actual earth pressure (Ps) =
𝑊
𝑃𝑙𝑎𝑛 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑠𝑒
=
43.4391
0.36
= 120.6642 KNm-2
Ps < Bearing capacity of soil
d) DESIGN MOMENT
 Total ultimate load (W) = 1.4 Gk + 1.6 Qk
= 61.1075 KN
 Earth pressure (Pe) =
Ultimate load
Area
=
61.1075
0.36
= 169.7431 KNm-2
0.1875 m 0.1875 m
0.225 m
168.9317 KNm-2
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24. davg
cover
Φ
 Maximum design moment (M) = Resultant force x length
(Occurs at face of column) = ( 169.7431 x 0.6 x 0.1875 ) x
0.1875
2
= 19.0961 x 0.0938
= 1.7912 KNm
e) ULTIMATE MOMENT
Base to be cast against blinding, hence cover (c) to reinforcement = 50 mm (Assume
10 mm diameter (Φ) bars will be needed as bending reinforcement in both directions.
Hence, average effective depth of reinforcement, davg is
 davg = h – cover – Φ
= 150 – 50 – 10
= 90 mm
 Ultimate design moment (Mu) = 0.156𝑓𝑐𝑢 bd2
= 0.156 x 25 Nmm-2 x 0.6 m x 902 mm2
= 18.9540 KNm
Since M < Mu, no compression reinforcement is required.
f) MAIN STEEL
 K =
𝑀
𝑓 𝑐𝑢 b𝑑2 =
1.7912 × 106
25 ×600 × 902 = 0.0147
 Z = 𝑑 [ 0.5 + √0.25 − 𝐾
0.9⁄ ]
= 90 [ 0.5 + √0.25 − 0.0147
0.9⁄ ]
= 90 [ 0.9834 ] mm = 88.5052 mm > 0.95𝑑 (= 85.50 mm)
Hence Z = 0.95𝑑 (= 85.50 mm)
 As =
𝑀
0.95 × 𝑓𝑦 ×𝑍
=
1.7912 × 106
0.95 ×250 × 85.5
= 88.2093 mm2
 Asmin = 0.24% Ac = 0.0024 x 600 x 150 = 216.00 mm2 > As
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25. For detailing purposes this area of steel has to be transposed into bars of a given
diameter and spacing using steel area tables. Provide 10 mm diameter 3 bars.
 3H10 (As = 236 mm2)
g) DISTRIBUTION OF REINFORCEMENT (See clause 3.11.3.2, BS 8110)
 𝑙 𝑐 = 600/2 = 300 mm

3𝑐
4
+
9𝑑
4
=
3 × 225
4
+
9 ×90
4
= 371.25 mm
 𝑙 𝑐 <
3𝑐
4
+
9𝑑
4
Hence, provide 3H10 (As = 236 mm2) distributed uniformly across the full width of the
footing parallel to the x–x and y–y axis.
h) PUNCH SHEAR
 Critical perimeter, pcrit, is
[ pcrit = column perimeter + 8 × 1.5d ] OR [ pcrit = 4 ( 225 + 3d ) ]
pcrit = ( 4 x 225 ) + ( 8 x 1.5 x 90 )
= 1980 mm
 Area within perimeter is = ( 225 + 3d )2
= ( 225 + 3 x 90 )2
= 0.2450 x 106 mm2
 Ultimate punching force, V, is
V = load on shaded area
= 169.7431 × ( 0.36 – 0.245) = 19.5205 KN
 Design punching shear stress,  , is
 =
𝑉
𝑃 𝑐𝑟𝑖 ×𝑑
=
19.5205 ×103
1980 ×90
= 0.1095 Nmm-2 < Permissible = 0.8√𝑓𝑐𝑢
 Design concrete shear stress, c
c = {
100 𝐴𝑠
𝑏 𝑣 𝑑
}
1
3⁄
(
400
𝑑
)
1
4⁄
𝛾 𝑚
1.5d
225 + 3d
0.79
[= 0.8√25 = 4 Nmm-2
]
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26. c = {
100 ×88.2093
600 × 90
}
1
3⁄
(
400
90
)
1
4⁄
1.25
c =
0.79 × 0.5466 ×1.4520
1.25
= 0.5016 Nmm-2
Since c >  , punching failure is unlikely and a 150 mm depth of slab is acceptable.
i) REINFORMATION DETAILS
The sketch below shows the main reinforcement requirements for the foundation
0.79
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27. REINFORCEMENT
COLUMN
BS 8110 : 1997
PART – 1 , SECTION 3 , CLAUSE 3.8 [ Columns ]
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28.  Design of a reinforced column ( BS 8110 )
A column may be considered to be unbraced if the lateral loads are resisted by the
sway action of the column [Assume this is unbraced column]
a) CLASSIFICATION OF A CONCRETE COLUMN
 For bending in the “Y” direction:
End condition at top of column = 2
End condition at bottom of column = 2
Hence from Table 3.20,BS-8110, βx = 1.5
𝑙 𝑒𝑦
ℎ
=
𝛽 𝑦 𝑙 𝑜
ℎ
=
1.5 × 3320
225
= 22.1333 > 10
 For bending in the “X” direction:
End condition at top of column = 1
End condition at bottom of column = 2
Hence from Table 3.20,BS-8110, βy = 1.3
𝑙 𝑒𝑥
𝑏
=
𝛽𝑥 𝑙 𝑜
𝑏
=
1.3 × 3000
225
= 17.3333 > 10
Since both
𝑙 𝑒𝑥
ℎ
and
𝑙 𝑒𝑦
𝑏
are both greater than 10, the column is SLENDER.
180
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29. Design the longitudinal steel and links for a 225 mm square, slender-Unbraced column
which supports the following axial loads: 𝑓𝑐𝑢= 25 Nmm-2 [Assuming the following
material strength 𝑓𝑦 = 460 Nmm-2 and 𝑓𝑦𝑣= 250 Nmm-2 and Diameter of main steel (Φ)
= 12 mm, diameter of links (Φ′) = 6 mm]
b) LOADING
 Dead
Self-weight of slab = 0.120 m x 1.2875 m x 2.8875 m x 24 KNm−3 = 10.7069 KN
Self-weight of beam-E = 0.225 m x 0.180 x 1.2875 m x 24 KNm−3 = 1.2515 KN
Self-weight of beam-D = 0.225 m x 0.500 m x 2.8875 m x 24 KNm−3 = 7.796 KN
Self-weight of column = 0.225 m x 0.225 m x 3.5 m x 24 KNm−3 = 4.2525 KN
Finishes of slab = 1.2875 m x 2.8875 m x 1 KNm-2 = 3.7177 KN
Block wall load = 0.125 m x 1.00 m x 4.50 m x 20 KNm−3 = 11.250 KN
Total Dead load (Gk) = 10.7069 + 1.2515 + 7.796 + 4.2525 + 3.7177 + 11.250
= 38.9746 KN
 Imposed
Total imposed load (Qk) = 1.2875 m x 2.8875 m x 1.1 kNm−2 = 4.0894 KN
 Ultimate design load
Total ultimate load, W is
W = (1.4 Gk + 1.6 Qk)
= (1.4 × 38.9746 + 1.6 × 4.0894)
= 61.1075 KN
c) LONGITUDINAL STEEL
Since column is axially loaded, use equation 28, BS-8110.
N = 0.4 𝑓𝑐𝑢 𝐴 𝑐𝑜𝑛 + 0.8 𝑓𝑦 𝐴 𝑠𝑐
61.1075 x 103 = 0.4 x 25 x (2252 - Asc) + 0.8 x 460 x Asc
61107.5 = 506250.0 – 10 Asc + 368 Asc
358 Asc = - 445142.50
Hence, no require for reinforcement steels for this column.
Asc Minimum = 0.13%Ac = 0.0013 x 225 x 255 = 65.8125 mm2
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30. For detailing purposes this area of steel has to be transposed into bars of a given
diameter and spacing using steel area tables. Provide 12 mm diameter 1 bar.
e) SIZE AND MINIMUM NUMBER OF BARS
From clause 3.12.5.3, BS 8110 . Columns with rectangular cross-sections should be
reinforced with a minimum of four longitudinal bars and each of the bars should not be
less than 12 mm in diameter.
 Hence, provide 12 mm diameter 4 bars.
 4H12 (As = 452 mm2)
f) LINKS - (clause 3.12.7, BS 8110)
 Size of links. Links should be at one-quarter of the size of the largest
longitudinal bar or 6 mm, whichever is the greater. That is
=
1
4
× 12 = 3 mm , but not less than 6mm diameter.
 The spacing of the links, is the lesser of
(a) 12 times the diameter of the smallest longitudinal bar, that is,
12 × 12 = 144 mm, or
(b) The smallest cross-sectional dimension of the column (= 225 mm).
 Provide H6 links at 125 mm centers.
g) REINFORMATION DETAILS
The sketch below shows the main reinforcement requirements for the column
PLAN
SECTION
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31. CARPARK
ISOMATRIC
VIEW
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[Assumefoundation
Depthfromground
Level=600mm]
255
0

32. THANK YOU
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