Reinforced Concrete

Report

2/2/2010
Presented to: Dr. Omar Asaad
Presented by: Mohammad Bader Al-Dein 2918
Table of Contents

Introduction 3
Given 4
I- Computation of Depth to Control Deflection 4
II- Estimation of Loads for Ribbed One-way Slab 6
III- Finding the Interval of Acceptance of r 8
IV- Design of Ribbed Slabs 9
Computation of ribbed slab’s loads 9
Design of slab R1 10
Design of slab R2 12
Temperature and shrinkage steel 14
V- Design of Beams 15
Computation of beam’s loads 15
Design of beam B1 16
Design of beam B2 19
General notes 22
VI- Design of Columns 23
Computation of column’s loads 23
Design of column C1 25
Design of column C2 27
Design of column C3 29
Design of column C4 31
VII- Design of Footings 33
Method of Calculation 33
Design of footing F1 34
Design of footing F2 38
Design of footing F3 42
Design of footing F4 46
VIII- Design of Tie Beams (Chainage) 50

Page 2 of 51
Introduction:

The aim of this project is to design a villa. This villa is located in a place where the soil has
a strong bearing capacity of 3.5 Kg/cm² which constitutes economical benefit.

The villa consists of two floors having each an area of 281 m2.
This villa was designed according to ACI code for reinforced concrete, which requires the
utilization of ultimate strength design with ultimate loads applied, taking into consideration
maximum safety and minimum cost.

To have the best design, we first start by estimating the thickness of slabs depending on
the largest span of between columns. Then, we proceed for ribs design and beams (according
to one-way slab design) that are lifted by the columns which transfer all the loads to footings,
then to soil.

The following pages reveal direct application of our theoretical study:

Page 3 of 51
 Given:

Compressive strength of concrete: f’c=280 Kg/cm² = 28 MPa

Tensile strength of reinforcement: fy= 4200Kg/cm² = 420 MPa.
γ concrete= 2500 Kg/m3
Soil bearing capacity: Bc= 3.5 Kg/cm²

I- Computation of depth to control Deflection:

Since deflection computations can take some time to carry out, the ACI code has
established a procedure to limit deflections by placing restrictions on the minimum depth of a
beam or ribbed one-way slab. This method eliminates the need to compute deflections and
controls deflections by requiring that depths not less than a specified fraction of span length.
Minimum depths for a variety of common support conditions are given in the following table:

Minimum thickness h

Simply One end Both end

cantilever
supported continuous continuous
Beams or
Ribbed one- L/16 L/18.5 L/21 L/8
way slab

Since beams carry 80% of the slab’s load, then deflection is studied according to the beams
spans.

Page 4 of 51
Span C1C3 for one end continuous:
l 550
hmin= = = 29.73 cm.
18.5 18.5

Span C3C3 for two ends continuous:

l 580
hmin= = = 27.62 cm.
21 21

So the critical hmin is 29.73 cm.

30 cm > h min ⇒ Ok
We design with h =
All deflections are controlled according to ACI code. ☺

Page 5 of 51
II-Estimation of Loads for Ribbed one-way slabs:

All structures must be proportioned so they will not fail or deform excessively under any
possible condition of service.
Therefore it is important that an engineer use great care in anticipating all the probable loads
to which a structure will be subjected during its lifetime, beneath it is very important to multiply
final loads by a factor according to its accuracy.

Loads = Dead Loads + Live Loads

Dead Load:

Slab cover: 0.06(Thickness of concrete) × 2500(Density of concrete) = 150kg/m2

(0.19+ 0.15)
Rib: × 0.24 × 2500 / 0.57 (Rib repetition) = 179 kg/m2
2
Hollow Block: 17(Weight of a block) / 0.57 × 0.2 (Depth of a block) = 150 kg/m 2

Plasters: = 45 kg/m2
Partitions: = 200 kg/m2
Tiles + sand + mortar: =200 kg/m2
∑D.L = 924 kg/m2 … Say = 925 kg/m2

Page 6 of 51
Live Load:

According to ACI code (Typical Live-Load values), minimum uniformly distributed Live loads
is 193.7 kg/m2 for an apartment building. Use 195 kg/m2 for such a villa.

Then ultimate Loads can be computed by multiplying loads by its safety factors according to
ACI code:

Wu = 1.4 D.L. + 1.7 L.L.

 Ultimate load: 1.4×925 + 1.7×195 = 1627 kg/m2

= 1.63 T/m2

Page 7 of 51
III- Finding the interval of acceptance of ρ:

Since in all the design we will use the same concrete compressive strength (f’c = 28 MPa)
and reinforcement tensile strength (fy = 420 MPa) for all elements, then it is important to
establish the interval of acceptances of ρ at the first of the design to simplify our work.

So we have to find rmin and ¾rb.

f’c = 28 MPa < 30.61MPa (4440lb/in2)

1.4 1.4
 rmin = = = 0.0033
fy 420
'
0.85 β1 f c 600
rb = = 0.0283
fy 600+ f y
 ¾rb = 0.0213

Where β 1 = 0.85 since fc’ < 30 MPa.

Then to avoid under reinforcement or over reinforcement the ratio ρ must belong to the
interval [0.0033, 0.0213].

Page 8 of 51
 IV- Design of Ribbed slabs:

 Rib width = 15 cm
 Width carried by rib = 15 + 21 + 21 = 57 cm
 Load/rib: PRib = 0.57  1.63 = 0.93 T/m.

Page 9 of 51
 Slab R1:

The final bending moment diagram and shearing forces diagram of slab R1 is illustrated in
the following figures, where these figures are determined using the WinBeam program.

Page 10 of 51
 {
M 1=18.82 KN . m
So M 2=25.55 KN . m
Vumax=28.73 KN

ρf y
Mu = frfybd2 (1 - ' )
1.7 f c
Take b = 15cm and d = 30 - 3(concrete cover) - 0.8(tie diameter) – 0.7(approximated bar
radius) = 25.5cm

For M1: (Bottom Reinforcement)

ρ=0.0054 => ρmin< ρ < ρmax
Then As = ρ b d = 2.07 cm2
Use 2T12 which provide an area of 2.27 cm2

For M2: (Top Reinforcement)

ρ=0.0075 => ρmin< ρ < ρmax
Then As = ρ b d = 2.87 cm2
Use 2T14 which provide an area of 3.08 cm2

For shear:
Use ties 5f8/m.

Slab R2:

Page 11 of 51
 The final bending moment diagram and shearing forces diagram of slabs R1 and R2 are
illustrated in the following figures, where these figures are determined using the WinBeam
program.

Page 12 of 51
 {
M 1=11.43 KN . m
So M 2=22.47 KN . m
Vumax=25.11 KN

ρf y
Mu = frfybd2 (1 - ' )
1.7 f c
Take b = 15cm and d = 30 - 3(concrete cover) - 0.8(tie diameter) – 0.7(approximated bar
radius) = 25.5cm

For M1: (Bottom Reinforcement)

ρ=0.0029 => ρmin> ρ so take ρ = ρmin = 0.0033
Then As = ρ b d = 1.34 cm2
Use 2T12 which provide an area of 2.27 cm2

For M2: (Top Reinforcement)

ρ=0.0065 => ρmin< ρ < ρmax
Then As = ρ b d = 2.49 cm2
Use 2T14 which provide an area of 3.08 cm2

For shear:
Use ties 5f8/m.

For temperature and shrinkage steel:

Page 13 of 51
The minimum required ratio of steel area to gross area is specified by ACI Code7.12 as

{
0.002if fy<300 Mpa
0.018 if fy=400 Mpa
As/Ag= 0.0018 (400)
if fy> 400 Mpa
fy
0.0018 ×400 0.0018 × 400 2
In this case As= bd= ×100 × 25.5=4.37 cm /m. = 2.50 cm2 /0.57m,
420 420

Use 1T14 /Block, providing an area of 1.54cm2 /0.57m.

In the opposite direction use 1T10 /30cm, providing an area of 1.50cm2 /0.57m.
So in both directions the area provided will be 3.04cm2 /0.57m.

Page 14 of 51
 V- Design of Beams:

 Load at beam B1: Pu = (2.55+ [6.6 × 1.1/16.8]) 1.63 = 4.87 T/m.

 Load at beam B2: Pu = (2.55+2.7) 1.63 = 8.56 T/m.

Page 15 of 51
 Beam B1:

Beam B1 have a span of 16.8 m and carry a uniform factored load wu = 4.87 T/m.
We want to consider only flexure and design a rectangular cross section that satisfies the
provisions of the ACI Code.

Page 16 of 51
 {
M 1=114.82 KN .m
So M 2=154.97 KN . m
Vumax =162.12 KN

ρf y
Mu = frfybd2 (1 - ' )
1.7 f c
Take b = 60cm and d = 30 - 3(concrete cover) - 0.8(tie diameter) – 1.2(approximated bar
radius) = 25cm

1- Bottom Reinforcement:

For M3: (At midpoint of span C3C3)

Using the method of three moment equation, we find the moment M3 at midpoint of span
C3C3, which was 29.18 KN.m.
ρ=0.0021 => ρ < ρmin
Then As = ρmin b d = 4.95 cm2
Use 4T14@14cm, along the entire beam, providing an area of 6.16 cm2, and with an
overlapping of Ls = 50 × db = 50 × 1.4 = 70cm at the position of M3.

For M1:
ρ=0.0088 => ρmin< ρ < ρmax
Then As = ρ b d = 13.20 cm2
But there is an overlapping of 4T14 having an area of 6.16cm2, then an area of 7.04cm2 is
sufficient.
Use 4T16@14cm (providing an area of 8.05 cm2), and over a distance of 5m.

2- Top Reinforcement:

Asmin = 4.95 cm2, use 6T14@8cm, along the entire beam, providing an area of 9.24cm2,
and with an overlapping of Ls = 50 × db = 50 × 1.4 = 70cm at the middle of the beam.

For M2:
ρ=0.0123 => ρmin< ρ < ρmax
Then As = ρ b d = 18.45 cm2
But there is an overlapping of 6T14 having an area of 9.24cm2, then an area of 9.21cm2 is
sufficient.
Use 6T14@8cm (providing an area of 9.24 cm2), and over a distance of 3m, at the
positions of M2.

Page 17 of 51
3- Shear:

Vu = 162.12 KN
fVc = 0.85 × √f’c b d / 6 = 116.95KN.
Vu−ɸ Vc
 ɸ V c <V u Vs = =53.15 KN
ɸ
Take 3f8 with Av = 6Ab = 3 cm2.

{
d
=13 cm Controls .
2
S= smaller of
dAv fy
=62.2 cm
Vs
Use 3f8@12 cm in L/3 from the front of supports and 3f8@24cm in the middle third.

Page 18 of 51
 Beam B2:

Beam B2 have a span of 16.8 m and carry a uniform factored load wu = 8.56 T/m.
We want to consider only flexure and design a rectangular cross section that satisfies the
provisions of the ACI Code.

Page 19 of 51
 {
M 1=201.81 KN . m
So M 2=272.39 KN . m
Vumax=284.93 KN

ρf y
Mu = frfybd2 (1 - ' )
1.7 f c
Take b = 80cm and d = 30 - 3(concrete cover) - 0.8(tie diameter) – 1.2(approximated bar
radius) = 25cm

1- Bottom Reinforcement:

For M3: (At midpoint of span C4C4)

Using the method of three moment equation, we find the moment M3 at midpoint of span
C4C4, which was 51.29 KN.m.
ρ=0.0028 => ρ < ρmin
Then As = ρmin b d = 5.60 cm2
Use 8T12@8cm, along the entire beam, providing an area of 9.05 cm2, and with an
overlapping of Ls = 50 × db = 50 × 1.2 = 60cm at the position of M3.

For M1:
ρ=0.0120 => ρmin< ρ < ρmax
Then As = ρ b d = 24.00 cm2
But there is an overlapping of 8T12 having an area of 9.05cm2, then an area of 14.95cm2 is
sufficient.
Use 8T16@8cm (providing an area of 16.09 cm2), and over a distance of 5m.

2- Top Reinforcement:

Asmin = 5.60 cm2, use 8T16@8cm, along the entire beam, providing an area of 16.09cm2,
and with an overlapping of Ls = 50 × db = 50 × 1.6 = 80cm at the middle of the beam.

For M2:
ρ=0.0170 => ρmin< ρ < ρmax
Then As = ρ b d = 32.00 cm2
But there is an overlapping of 8T16 having an area of 16.09cm2, then an area of 15.91cm2
is sufficient.
Use 8T16@8cm (providing an area of 16.09 cm2), and over a distance of 3m, at the
positions of M2.

Page 20 of 51
3- Shear:

Vu = 284.93 KN
fVc = 0.85 × √f’c b d / 6 = 155.92KN.
Vu−ɸ Vc
 ɸ V c <V u Vs = =151.78 KN
ɸ
Take 4f8 with Av = 8Ab = 4 cm2.

{
d
=13 cm Controls .
2
S= smaller of
dAv fy
=28.9 cm
Vs
Use 4f8@12 cm in L/3 from the front of supports and 4f8@24cm in the middle third.

Page 21 of 51
Since the two floors have the same frame then our design will be the same for all slabs,
except for the slab of the first floor, where there is a void at the center of the building due to
the existing of the stairs. This void is surrounded at two edges by beams B2, while at the other
edges a triple rib is to be constructed so that it holds the load of the stairs and transfer it to
beams. At the periphery of each slab, a double rib is constructed. Double and triple ribs are
nothing except two or three ribs constructed beside each other.

Page 22 of 51
 VI- Design of Columns:

Columns are structural elements that carry beam loads and transfer it to footings.
Using the Winbeam program we get the reactions at each support for all the beams,
this force represent the ultimate load at each column. Moment is zero at all columns
since the bending moment diagram is continuous. This load is multiplied by three, since
there are two floors in this villa. Then all the columns are designed as short axial tied
columns having an ultimate load Pu.

The following figures illustrate the reaction diagrams for all beams:

C1 C3 C3 C1

Beam B1

C2 C4 C4 C2

Beam B2

Page 23 of 51
Then the ultimate load Pu for each column is listed in the following table:

Reaction at Beams Ultimate Load Pu

Column
[KN] [KN]

C1 88.75 266.3

C2 185.88 557.7

C3 254.57 763.8

C4 533.17 1599.6

Page 24 of 51
Column C1:

Column C1 carry a uniform factored load Pu = 266.3KN

Pu
A g≥
0 . 45 [ f 'c + ρf y ]
266 . 3×10
¿
0 . 45 [ 28+0 . 01×420 ]
¿ 184 cm 2

Take Ag = 2060 = 1200 cm2

Ast = 0.011200 =12 cm2

Use 10T14 bars which provides an area of 15.40 cm2.

60−2 ( 3 )−2 ( 0.8 )−5 ( 1.4 )

Clear spacing between main bars = = 11cm = 4.3in < 6in, so there is
4
no need for extra tie.

For ties:

Use 2f8, with a spacing s between ties:

s smal est of¿{16×bar diameter ¿ {48×tiediameter ¿¿¿

s smal est of¿{16×1.4=2 .4 cm ¿ {48×0.8=38.4 cm ¿ ¿¿
¿
Use 2f8@10cm at column-beam junction and otherwise 2f8@20cm.

For splices: Ls = 50 × db = 50 × 1.4 = 70cm.

So column’s bars should be overlapped 70cm.

Page 25 of 51
Column C2:

Page 26 of 51
Column C2 carry a uniform factored load Pu = 557.7KN

Pu
A g≥
0 . 45 [ f 'c + ρf y ]
763 . 8×10
¿
0 . 45 [ 28+0 . 01×420 ]
¿ 385 cm2

Take Ag = 2050 = 1000 cm2

Ast = 0.011000 =10 cm2

Use 8T14 bars which provides an area of 12.32 cm2.

50−2 ( 3 )−2 ( 0.8 )−4 ( 1.4 )

Clear spacing between main bars = = 12cm = 4.7in < 6in, so there is
3
no need for extra tie.

For ties:

Use 2f8, with a spacing s between ties:

s smal est of¿{16×bar diameter ¿ {48×tiediameter ¿¿¿

s smal est of¿{16×1.4=2 .4 cm ¿ {48×0.8=38.4 cm ¿ ¿¿
¿
Use 2f8@10cm at column-beam junction and otherwise 2f8@20cm.

For splices: Ls = 50 × db = 50 × 1.4 = 70cm.

So column’s bars should be overlapped 70cm.

Page 27 of 51
Column C3:

Column C3 carry a uniform factored load Pu = 763.8KN

Page 28 of 51
 Pu
A g≥
0 . 45 [ f 'c + ρf y ]
763 . 8×10
¿
0 . 45 [ 28+0 . 01×420 ]
¿ 528 cm2

Take Ag = 2060 = 1200 cm2

Ast = 0.011200 =12 cm2

Use 10T14 bars which provides an area of 15.40 cm2.

60−2 ( 3 )−2 ( 0.8 )−5 ( 1.4 )

Clear spacing between main bars = = 11cm = 4.3in < 6in, so there is
4
no need for extra tie.

For ties:

Use 2f8, with a spacing s between ties:

s smal est of¿{16×bar diameter ¿ {48×tiediameter ¿¿¿

s smal est of¿{16×1.4=2 .4 cm ¿ {48×0.8=38.4 cm ¿ ¿¿
¿
Use 2f8@10cm at column-beam junction and otherwise 2f8@20cm.

For splices: Ls = 50 × db = 50 × 1.4 = 70cm.

So column’s bars should be overlapped 70cm.

Page 29 of 51
Column C4:

Column C4 carry a uniform factored load Pu = 1599.6KN

Page 30 of 51
 Pu
A g≥
0 . 45 [ f 'c + ρf y ]
763 . 8×10
¿
0 . 45 [ 28+0 . 01×420 ]
¿ 1104 cm2

Take Ag = 2070 = 1400 cm2

Ast = 0.011400 =14 cm2

Use 10T14 bars which provides an area of 15.40 cm2.

60−2 ( 3 )−2 ( 0.8 )−5 ( 1.4 )

Clear spacing between main bars = = 13cm = 5.1in < 6in, so there is
4
no need for extra tie.

For ties:

Use 2f8, with a spacing s between ties:

s smal est of¿{16×bar diameter ¿ {48×tiediameter ¿¿¿

s smal est of¿{16×1.4=2 .4 cm ¿ {48×0.8=38.4 cm ¿ ¿¿
¿
Use 2f8@10cm at column-beam junction and otherwise 2f8@20cm.

For splices: Ls = 50 × db = 50 × 1.4 = 70cm.

So column’s bars should be overlapped 70cm.

Page 31 of 51
 VII- Design of Footings:

Loads carried by columns should be transmitted to soil through foundations.

The geotechnical report produced gives the soil beneath foundations a bearing capacity B.C.
= 3.5Kg/cm2.

Page 32 of 51
Since there is no large loads, and the provided area is sufficient, all footings are designed as
an isolated square footing.

Footing F1:

Footing F1 is designed to carry the column C1, having an ultimate load Pu = 266.3KN.

Page 33 of 51
1- Footing Area:

Ultimate load
Total applied load = × 1.1 = 195.3KN
1.5

Total applied load

≤B .C .
Af
19530
≤3 .5
Af
A f ≥5580 cm2
2
take A f =100×100=10000 cm

2- Footing Depth:

To find the depth of the footing, two checks should be done, flexural shear check and
punching shear check.

Pu 26630
q u= = =2 . 67 kg /cm2
A f 10000

 Flexural Shear Check:

V u ≤ϕ .V C

V u =q u ×A eff
¿ 2 .67 (40−d )(100 )

ϕ .V C
√f '

=φ C bd=0 . 85× √
28×10
×100×d
6 6
V u ≤ϕ.V C

2 .67 (40−d )×100≤0 . 85×

√28×10 ×100×d
6
⇒ d≥10 .5 cm
Take d=30 cm
 Punching Shear Check:

V u ≤ϕ .V C

V u =q net × punshing area

¿ 2 .67×[ 100×100−( 20+ d ) (60+d ) ]
¿ 14685 kg

Page 34 of 51
 ϕ .V C =φ 4 √ f ' C . bd=127961kg

We have V u ≤ϕ .V C ⇒Ok

Then the depth of footing will be h = d + 9cm (Cover) + 1cm (Bar radius)
= 40cm.

3- Reinforcement:

2 b
M u =qu l
2
100
¿ 2 .67×( 402 )×
2
¿ 334 t . cm
M
ku= u2
bd
¿ 53 . 81
ρ=0 .0010≺ρ min fromtable
so take ρ=ρ min= 0 .0033
then A s =ρ bd=9 . 90 cm2

Use 7T14 bars, which provides an area of 10.78cm2.

Ab
Spacing s between centers of bars should be smaller than × b=15.5 cm
As
s should also be larger than { ¯
diameter=1.4 cm
Largest aggregate ¿ 1 inch=2.54 cm
= 100−2 ×8 ( cover ) =14 cm ⇒Ok
s 6

4- Development Length:

l required =
0. 04 A b f y
=
( 12.539
0 . 04×
.54 )
×60916
2
×2 . 54=10 cm
√f c'
√ 4061
l provided=b0 −clear cover=20−8=12 cm>l d ( required) ⇒OK

Page 35 of 51
5- Design of Dowels:

All forces are to be transferred from columns to footings, therefore reinforcement

are to transfer tensile forces, and bearing to transfer compression forces.

Nominal force Pn at column base = 0.85 × f’c × Ac

= 0.85 × 28,000 × 0.2 × 0.6
= 2856KN
Total Ultimate Applied Load Pu= 266.3KN

→ Pu < ΦPn = 0.7 × 2856 = 1999.2KN ⇒OK

Then use dowels with:

A s,min =0 . 005 Acolumn

¿ 0 . 005×20×60
¿ 6 . 0 cm2
Use 8T12 Bars providing an area of 9.05cm2

0 . 02 f y d b
l d (required )= =16 cm
√f 'c
l d req≺l d pro=d=30 cm ⇒ok

6- Temperature and Shrinkage Steel:

0.0018 ×400 × Ag
As = = 6.86cm²
420
Use 7T12@12cm which provides an area of 7.92cm2.

Page 36 of 51
Footing F2:

Footing F2 is designed to carry the column C2, having an ultimate load Pu = 557.7KN.

1- Footing Area:

Page 37 of 51
 Ultimate load
Total applied load = × 1.1 = 408.98KN
1.5

Total applied load

≤B .C .
Af
40898
≤3 . 5
Af
A f ≥11685 cm2
take A f =110×110=12100 cm2

2- Footing Depth:

To find the depth of the footing, two checks should be done, flexural shear check and
punching shear check.

Pu 55770
q u= = =4 . 61 kg/cm 2
A f 12100

 Flexural Shear Check:

V u ≤ϕ .V C

V u =q u ×A eff
¿ 4 .61( 45−d )(110)

ϕ .V C
√f '

=φ C bd=0 . 85× √
28×10
×110×d
6 6
V u ≤ϕ.V C

4 .61 (45−d )×110≤0 . 85×

√28×10 ×110×d
6
⇒ d≥17 . 1 cm
Take d=30 cm
 Punching Shear Check:

V u ≤ϕ .V C

V u =q net × punshing area

¿ 4 .61×[ 110×110−( 20+d ) (50+d ) ]
¿ 37341 kg

ϕ .V C =φ 4 √ f ' C . bd=140757 kg

Page 38 of 51
 We have V u ≤ϕ .V C ⇒Ok

Then the depth of footing will be h = d + 9cm (Cover) + 1cm (Bar radius)
= 40cm.

3- Reinforcement:

2 b
M u =qu l
2
110
¿ 4 .61×( 45 2 )×
2
¿ 803 t . cm
M
ku= u2
bd
¿ 117 . 61
ρ=0 .0022≺ρ min from table
so take ρ=ρ min= 0 .0033
then A s =ρ bd=10 . 89 cm2

Use 10T12 bars, which provides an area of 11.31cm2.

Ab
Spacing s between centers of bars should be smaller than × b=11.4 cm
As
s should also be larger than { ¯
diameter=1.2 cm
Largest aggregate ¿ 1 inch=2.54 cm
= 110−2× 8 ( cover ) =10 cm ⇒Ok
s 9

4- Development Length:

l required =
0. 04 A b f y
=
( 12.131
0 . 04×
.54 )
×60916
2
×2 .54=17 cm
√f c'
√ 4061
l provided =b0 −clear cover=30−8=22 cm>l d ( required) ⇒OK

5- Design of Dowels:
Page 39 of 51
 All forces are to be transferred from columns to footings, therefore reinforcement
are to transfer tensile forces, and bearing to transfer compression forces.

Nominal force Pn at column base = 0.85 × f’c × Ac

= 0.85 × 28,000 × 0.2 × 0.6
= 2380KN
Total Ultimate Applied Load Pu= 557.7KN

→ Pu < ΦPn = 0.7 × 2380 = 1666.0KN ⇒OK

Then use dowels with:

A s,min =0 . 005 Acolumn

¿ 0 . 005×20×50
¿ 5 . 0 cm2
Use 8T12 Bars providing an area of 9.05cm2

0 . 02 f y d b
l d (required )= =16 cm
√f 'c
l d req≺l d pro=d=30 cm ⇒ok

6- Temperature and Shrinkage Steel:

0.0018 ×400 × Ag
As = = 7.54cm²
420
Use 7T12@14cm which provides an area of 7.92cm2.

Page 40 of 51
Footing F3:

Footing F3 is designed to carry the column C3, having each an ultimate load Pu = 763.8KN.

1- Footing Area:

Ultimate load
Total applied load = × 1.1 = 560.12KN
1.5

Page 41 of 51
 Total applied load
≤B .C .
Af
56012
≤3. 5
Af
A f ≥16005 cm2
2
take A f =130×130=16900 cm

2- Footing Depth:

To find the depth of the footing, two checks should be done, flexural shear check and
punching shear check.

Pu 76380
q u= = =4 . 52 kg/cm 2
A f 16900

 Flexural Shear Check:

V u ≤ϕ .V C

V u =q u ×A eff
¿ 4 .52(55−d )(130 )

ϕ .V C
√f '

=φ C bd=0 . 85× √
28×10
×130×d
6 6
V u ≤ϕ.V C

4 .52(55−d )×130≤0 . 85×√

28×10
×130×d
6
⇒ d≥20 .7 cm
Take d=30 cm
 Punching Shear Check:

V u ≤ϕ .V C

V u =q net × punshing area

¿ 4 .52× [ 130×130−( 20+d ) (60+ d ) ]
¿ 56048 kg

ϕ .V C =φ 4 √ f ' C . bd=166349 kg

We have V u ≤ϕ .V C ⇒Ok

Page 42 of 51
 Then the depth of footing will be h = d + 9cm (Cover) + 1cm (Bar radius)
= 40cm.

3- Reinforcement:

2 b
M u =qu l
2
130
¿ 4 .52×( 552 )×
2
¿ 1390 t . cm
M
ku= u2
bd
¿ 170 .25
ρ=0 .0033=ρ min from table
so take ρ=ρ min= 0 .0033
then A s =ρ bd=12 .87 cm2

Use 9T14 bars, which provides an area of 13.86cm2.

Ab
Spacing s between centers of bars should be smaller than × b=15.5 cm
As
s should also be larger than { ¯
diameter=1.4 cm
Largest aggregate ¿ 1 inch=2.54 cm
= 130−2 ×8 ( cover ) =14 cm ⇒Ok
s 8

4- Development Length:

l required =
0. 04 A b f y
=
( 12.539
0 . 04×
.54 )
×60916
2
×2 . 54=24 cm
√f c'
√ 4061
l provided=b0 −clear cover=45−8=37 cm>l d (required ) ⇒OK

5- Design of Dowels:

All forces are to be transferred from columns to footings, therefore reinforcement

are to transfer tensile forces, and bearing to transfer compression forces.

Page 43 of 51
 Nominal force Pn at column base = 0.85 × f’c × Ac
= 0.85 × 28,000 × 0.2 × 0.6
= 2856KN
Total Ultimate Applied Load Pu= 763.8KN

→ Pu < ΦPn = 0.7 × 2856 = 1999.2KN ⇒OK

Then use dowels with:

A s,min =0 . 005 Acolumn

¿ 0 . 005×20×60
¿ 6 . 0 cm2
Use 8T12 Bars providing an area of 9.05cm2

0 . 02 f y d b
l d (required )= =16 cm
√f 'c
l d req≺l d pro=d=30 cm ⇒ok

6- Temperature and Shrinkage Steel:

0.0018 ×400 × Ag
As = = 8.92cm²
420
Use 8T12@14cm which provides an area of 9.05cm2.

Page 44 of 51
Footing F4:

Footing F4 is designed to carry the column C4, having an ultimate load Pu = 1599.6KN.

1- Footing Area:

Ultimate load
Total applied load = × 1.1 = 1173.1KN
1.5

Page 45 of 51
 Total applied load
≤B .C .
Af
117310
≤3 . 5
Af
A f ≥33517 cm 2
2
take A f =190×190=36100 cm

2- Footing Depth:

To find the depth of the footing, two checks should be done, flexural shear check and
punching shear check.

Pu 159960
q u= = =4 . 44 kg/cm 2
A f 36100

 Flexural Shear Check:

V u ≤ϕ .V C

V u =q u ×A eff
¿ 4 . 44(85−d )(190 )

ϕ .V C
√f '

=φ C bd=0 . 85× √
28×10
×190×d
6 6
V u ≤ϕ.V C

4 . 44 (85−d )×190≤0 .85×√

28×10
×190×d
6
⇒ d≥31 .7 cm
Take d=40 cm
 Punching Shear Check:

V u ≤ϕ .V C

V u =q net × punshing area

¿ 4 . 44×[ 190×190−( 20+d ) (70+ d ) ]
¿ 130980 kg

ϕ .V C =φ 4 √ f ' C . bd=324167 kg

We have V u ≤ϕ .V C ⇒Ok

Page 46 of 51
 Then the depth of footing will be h = d + 9cm (Cover) + 1cm (Bar radius)
= 50cm.

3- Reinforcement:

2 b
M u =qu l
2
190
¿ 4 . 44×( 852 )×
2
¿ 4766 t . cm
M
ku= u2
bd
¿ 227 . 32
ρ=0 .0044≻ρ min from table
so take ρ= 0 .0044
then A s =ρ bd=33 . 44 cm2

Use 17T16 bars, which provides an area of 34.18cm2.

Ab
Spacing s between centers of bars should be smaller than × b=11.4 cm
As
s should also be larger than { ¯
diameter=1.6 cm
Largest aggregate ¿ 1 inch=2.54 cm
= 190−2 ×8 ( cover ) =10 cm ⇒Ok
s 16

4- Development Length:

l required =
0. 04 A b f y
=
( )
0 . 04×
2 . 012
2 .54 2
×60916
×2 .54=20 cm
√f c'
√ 4061
l provided =b0 −clear cover=60−8=52 cm>l d ( required) ⇒OK

5- Design of Dowels:

All forces are to be transferred from columns to footings, therefore reinforcement

are to transfer tensile forces, and bearing to transfer compression forces.
Page 47 of 51
 Nominal force Pn at column base = 0.85 × f’c × Ac
= 0.85 × 28,000 × 0.2 × 0.7
= 3332KN
Total Ultimate Applied Load Pu= 1599.6KN

→ Pu < ΦPn = 0.7 × 3332 = 2332.4KN ⇒OK

Then use dowels with:

A s,min =0 . 005 Acolumn

¿ 0 . 005×20×70
¿ 7 . 0 cm2
Use 10T12 Bars providing an area of 11.31cm2

0 . 02 f y d b
l d (required )= =16 cm
√f 'c
l d req≺l d pro=d=40 cm⇒ ok

6- Temperature and Shrinkage Steel:

0.0018 ×400 × Ag
As = = 16.29cm²
420
Use 11T14@15cm which provides an area of 16.94cm2.

Page 48 of 51
VIII- Design of Tie Beams (Chainage):

Above foundations, the columns are joined by beams called “Tie Beams”.
To design this beams, we consider the maximum length of span between faces of footings,
which is to be 450cm.

450 d
Then the depth of tie beam d will be = 45cm, and b = = 22.5cm so take b = 25cm.
10 2

Since these beams are designed only to join the columns -so that they make equal
settlements- and not to hold any load, then taking r as rmin for computing As will be much
sufficient.

Page 49 of 51
As,min = rminbd = 0.0033 × 25 × 45 = 3.72cm2.

Use 2T16 for top and bottom reinforcement, which provide an area of 4.03cm2.

Page 50 of 51
Page 51 of 51