STRUCTURAL DESIGN OF B+G+7

BUILDING
SECTION ONE

Name ID.NO.

1. Kaleab shewadeg UGR/8328/12

2. Temesgen Aweke UGR/5812/12

3. Yeabsira Gettu UGR/5793/12

4. Yonas Seyoum UGR/7920/12

 Table of Content
Design and Analysis of Solid slab ------------------------------------------------------------- 3
Staircase design --------------------------------------------------------------------------------- 18
Wind load Analysis ----------------------------------------------------------------------------- 22
Center of mass ----------------------------------------------------------------------------------- 31
Earth-quake -------------------------------------------------------------------------------------- 36
Geometric Imperfections -----------------------------------------------------------------------41
Beam Analysis and Design -------------------------------------------------------------------- 47
Column Design and Analysis ----------------------------------------------------------------- 58
 Design and Analysis of Solid slab

Introduction

In reinforced concrete construction, slabs are utilized to create flat, functional surfaces. They
can be supported by reinforced concrete beams and are typically cast together with the beams.
Slabs can have support on two opposite sides only, resulting in one-way structural action,
where the load is mainly carried perpendicular to the supporting beams. Alternatively, beams
can be present on all four sides, allowing for two-way slab action. In cases where the length-
to-width ratio of a slab panel exceeds 2, the load is primarily borne in the shorter direction
towards the supporting beams, resulting in an effective one-way action, despite the presence
of supports on all sides.
In certain situations, concrete slabs may be directly supported by columns without the need
for beams or girders. These slabs, known as flat plates, are commonly used when spans are
not large and loads are not particularly heavy. Flat plate construction lacks beams but
incorporates a thickened slab area near the column, often with flared column tops. These
design elements, known as drop panels and column capitals, respectively, help reduce stresses
caused by shear and negative bending around the columns. A closely related slab type is the
two-way joist, also referred to as a grid or waffle slab. It involves creating voids in a
rectilinear pattern using metal or fiberglass form inserts, resulting in a ribbed construction that
reduces the dead load compared to solid-slab construction. Typically, the inserts are omitted
near the columns, allowing for the formation of solid slabs in those areas to better resist
moments and shear forces. In general, various types of slabs are used in construction projects.
For this particular project, the floor system predominantly consists of solid slabs. However,
there are also instances where one-way slabs and cantilever slabs are incorporated.

Analysis and Design Slab

Slabs with side ratio less than two are treated as two-way slabs and analysis can be made by
means of coefficients provide that the following conditions are satisfied.
The slab is composed of rectangular panels, supported at all four edges by walls or beams,
stiff enough to be treated as an unyielding.
This method is intended for slab subjected to uniformly distributed load. If the Slab is
subjected to uniform load or concentrated load, in addition to concentrated load, this can
generally be treated by considering them as equivalent uniform load using approximate rules,
provide that that the sum of non-uniform loads on a panel does not exceed 20 present of the
total load.
Type of slab
The first step to design the floor slab is distinguishing the type of slab based on Ly/Lx ratio.
This is defined as:
ly
If lx ≥ 2 the slab is called a one way slab
ly
If lx
< 2 the slab is called a two way slab

Where Ly- the longer length of the slab

Lx- the shorter length of the slab

Based on the following floor plan, we can classify the slabs as one way or two-way slab
Panel Lx Ly Ly/Lx Panel
Type
1 2 3.5 1.75 Two way
2 3.5 4.5 1.29 Two way
3 2.85 3.5 1.23 Two way
4 3.2 3.5 1.09 Two way
5 2.939 3.96 1.35 Two way
6 3.96 4.5 1.14 Two way
7 2.85 3.96 1.39 Two way
8 3.2 3.96 1.24 Two way
9 4 4.06 1.02 Two way
10 4.06 4.5 1.11 Two way
11 2.85 4.06 1.42 Two way
12 3.2 4.06 1.27 Two way
13 4 4 1 Two way
14 4 4.5 1.13 Two way
15 1 4 4 One way

Design for cover

Cmin=max Cmin, bond; Cmin, dur + ∆Cdur, γ − ∆dur, st − ∆Cdur, add; 10mm ………ES
EN-1992 section 4 Equation (4.2) where:
Cmin, bond Minimum cover due to bond requirement.
Cmin, dur Minimum cover due to environmental conditions.
∆Cdur, γ Additive safety element.
∆dur, st Reduction of minimum cover for use of stainless steel.
∆Cdur, add Reduction of minimum cover for use of additional protection.
Taking ∆c,dur, ∆c,duct,st and ∆cdur, add as zero
Cmin, bond
Cmin=max Cmin, dur
10mm
Assuming an exposure class of XC3 which is concrete inside the building with moderate air
humidity, from ES EN 1992-Table 4.1
Design life of the building =50 years
Normal quality control
Maximum aggregate size of 20mm
1HR fire protection
According to EN-1992, Table E.1N
For exposure class of XC3, a minimum concrete grade of C-20/25. Therefor we have used a
concrete grade of C-25/30 for the slabs.
Assuming a bar diameter of 10mm.
Cover for bond and durability
Type of steel ordinary
Cmin=Bar diameter =10mm……………….ES EN-1992-Table 4.2N
The recommended structural class for service life of 50 years is 4…from table 4.4N ES
EN1992-1-1
Cmin,dur=20mm……………………ES EN-1992, Table 4.4N
10mm
Cmin= max 20mm
10mm
Cmin=20mm
Cnom= Cmin+∆c,dev
∆ c,dev is recommended value by ES EN-1992-4.4.13(1) which accounts for the quality
control (5≤∆c,dev≤10mm)
Take ∆c,dev=10mm
Cnom=20mm+10mm
Cnom=30mm
Cover for 1 HR fire protection
For REI=60min, the minimum thickness of the slab should be 80mm according to ES EN-
1992-1-2, Table 5.8, but for our slab the calculated governing thickness for all slabs is
180mm which will satisfy this requirement. Then for panels (1,2,3,5,6,7,9,10,11,12,13 and 14)
Two-way slab 1.5<ly/lx≤2 amin is 15mm and for the one way slabs amn is 20mm. in both
cases the provided approve for durability and bond governs, there for the cover provided for
the slab should be 30mm.
Depth Determination
According to ES En-1992-1-1 the minimum depth of the slab should satisfy for the
serviceability requirement. The deformation of a member or structure shall not be such that it
adversely affects its proper functioning or appearance. Reinforced concrete beams or slabs in
buildings are dimension so that they comply with the limits of span to depth ratio. Their
deflections may be considered as not exceeding the limits set out in deflections that could
damage adjacent parts of the structure should be limited for the deflection after construction.
Other limits may be considered, depending on the sensitivity of adjacent parts.
The limit state of deformation may be checked by either:

by limiting the span/depth ratio, according to the equations 7.16.a or 7.16.b from ES EN
1992-1-1-2002 section 7.4.2 shown below
by comparing a calculated deflection, according to section 7.4.3 ES EN 1992-1-1-2002 of ,
with a limit value

The limiting span/depth ratio may be estimated using Expressions 1 and 2(ES EN 1992-1-1-
2002 article 7.4.2) below and multiplying this by correction factors to allow for the type of
reinforcement used and other variables

3
 
o  o 2 
l / d  k 11  1.5 fck  3.2 fck   1  if    o
     
 

 o 1 ' 
l / d  k 11  1.5 fck  fck  if    o
    ' 12 o 
Where
l/d is the limit span/depth
l/d is multiply by l/d*500/fyk because we used S-400.

K is the factor to take into account the different structural systems

ρ0 is the reference reinforcement ratio = fck ∗ 10−3
ρ is the required tension reinforcement ratio at mid-span to resist the moment due to the
design loads (at support for cantilevers)
ρ´ is the required compression reinforcement ratio at mid-span to resist the moment due to
design loads.
 Recommended values of K is given on table for each support over which the slab is
continuous, there will be two adjacent support moments. The difference may be distributed
between the panels at either side of support to equalize their moments, as in moment
distribution method of frames.
Effective depth calculation for slabs
Panel L Support K fck(Mpa) ρ ρ0 d
l/d l/d*500/fyk
condition
1 3500 End span 1.3 25 0.005 0.005 24.05 30.0625 116
2 3500 Interior 1.5 25 0.005 0.005 34.6875 101
27.75
span
3 3500 Interior 1.5 25 0.005 0.005 34.6875 101
27.75
span
5 4000 Interior 1.5 25 0.005 0.005 34.6875 115
27.75
span
6 4500 Interior 1.5 25 0.005 0.005 34.6875 130
27.75
span
7 3960 Interior 1.5 25 0.005 0.005 34.6875 114
27.75
span
9 4000 Interior 1.5 25 0.005 0.005 34.6875 115
27.75
span
10 4500 Interior 1.5 25 0.005 0.005 34.6875 130
27.75
span
11 4060 Interior 1.5 25 0.005 0.005 34.6875 117
27.75
span
12 4060 End span 1.3 25 0.005 0.005 24.05 30.0625 135
13 4000 End span 1.3 25 0.005 0.005 24.05 30.0625 133
14 4500 Interior 1.5 25 0.005 0.005 34.6875 130
27.75
span
15 4000 End span 1.3 25 0.005 0.005 24.05 30.0625 133

From all panels’ given on the above table typical floor panels depth the governing effective
depth is d = 135mm, therefore, we consider this depth for all floor panels. Then the depth of
the panels will be:
bar diameter
D=d+ + cover
2
D=140+5+30 = 175mm
Take D = 180 mm
LOADING
The slab is loaded with both DL and LL. Dead load comes from self-weight of slab, floor
finish, cement screed, plastering and partition load. Live loads are either movable or moving
load without any acceleration or impact.
Dead and live loads are calculated depending on the service of the slabs and self- weight.
Ignoring any localized effects caused by concentrated load, the partition loads are distributed
over the area of the slab. The design loads are factored according to the following formula.
Pd=1.35Gk+1.5Qk
Where Pd - total factored design load
Gk - total dead load on slab
Qk- total live load on slabs

Unit weight plastic tile – 12KN/m³

Cement screed– 23KN/m³
Terrazzo tile –24 KN/m³
Concrete – 25KN/m³
Marple - 26 KN/m³

Dead load plastic tile= 12*0.02 = 0.24KN/m²

Terrazzo tile = 24*0.02 =0.480KN/m²
Cement screed = 23*0.03 =0.690KN/m²
Slab (concrete) = 25*0.180 =4.5 KN/m²
Plaster = 25*0.02 =0.5 KN/m²
Marple = 26x0.03 = 0.78KN/m²
Live load (from EBCS – 1 – 1995 tables 2.9 and 2.10)
Category B (Office, hospitals) =3KN/m²
Stairs = 4KN/m²
Partition wall of 3m height
PANEL 5
HCB wall weight =0.15*(3.439)3.06*23=36.3KN
5cm mortar =0.05*(3.439)3.06*23=12.126KN
Total weight =48.43KN
When distributed per unit area 48.43KN/(13.62m2)=3.556KN/m2

PANEL 6
HCB wall weight =0.15*(6.968)*3.06*23=73.56KN
5cm mortar =0.05*3.06*6.968*23=24.52KN
Total weight =98.1KN
When distributed per unit area 98.1KN/(17.82m2)=5.5KN/m2

PANEL 10
HCB wall weight =0.15(6)3.06*23=63.342KN
5cm mortar =0.05*3*6.86*23=21.114KN
Total weight =84.456KN
When distributed per unit area 84.456KN/(18.27)=4.62KN/m2

PANEL12
HCB wall weight =0.15(3.2)3.06*23=33.78KN
5cm mortar =0.05(3.2)3.06*23=11.261KN
Total weight =45.0KN
When distributed per unit area 45.0KN/(12.992)=3.467KN/m2

PANEL 14
 HCB wall weight =0.15(5.27)3.06*23=55.64KN
5cm mortar =0.05(5.27)3.06*23=18.55KN
Total weight =74.2KN
When distributed per unit area 74.2KN/(18)=4.12KN/m2

Floor Plastering Dead Design

Panel Cement screed finish Self wt. Partition load live load load
1 0.69 0.24 0.5 4.5 0 5.93 3 12.5055
2 0.69 0.24 0.5 4.5 0 5.93 3 12.5055
3 0.69 0.48 0.5 4.5 0 6.17 3 12.8295
4 0.69 0.24 0.5 4.5 0 5.93 3 12.5055
5 0.69 0.33 0.5 4.5 3.556 9.576 3 17.4276
6 0.69 0.33 0.5 4.5 5.5 11.52 3 20.052
7 0.69 0.48 0.5 4.5 0 6.17 3 12.8295
8 0.69 0.24 0.5 4.5 0 5.93 3 12.5055
9 0.69 0.24 0.5 4.5 0 5.93 3 12.5055
10 0.69 0.24 0.5 4.5 4.62 10.55 3 18.7425
11 0.69 0.48 0.5 4.5 0 6.17 3 12.8295
12 0.69 0.24 0.5 4.5 3.467 9.397 3 17.18595
13 0.69 0.78 0.5 4.5 0 6.47 3 13.2345
14 0.69 0.24 0.5 4.5 4.12 10.05 3 18.0675
15 0.69 0.78 0.5 4.5 0 6.47 3 13.2345
Slab support and span moment analysis
In this project two methods of slab analysis are used based on the type of the slabs( two way
or one way). These methods are:
 Using Coefficient method
 Using 1m Strip method
For the case of cantilever and one-way slabs, the slabs are analyzed as a beam by which a 1m
strip is taken and the slabs are analyzed as a beam, but for the two way slabs the coefficient
method of analysis is used.

Classification of slabs based on method of analysis

 Panel Type of slab Used method of analysis
1,2,3,4, 5, 6 , 7, 8,9, 10,11,12,13 and Two way Coefficient method
14
15 One way 1m Strip method

Analysis of two way slabs

Moment Calculation for two way slab is conducted using coefficient method
The first stage of design is to determine support and span moments for all panels.
The support and span moments are calculated as

Msi = PdBsilx2
Where
Msi.= Design moment per unit width of reference
Pd=Uniformly Distributed Design Load
Bsi=Coefficient given in table A-1of EBCS-2, 1995 as a function of ratio Ly/Lx and Support
condition.

Lx Ly/ Bsx Bsx Bsy Bsy Msx Msy Msy

Panel pd Lx support span support span support Mxs span support span
1 12.5055 2 1.75 0.087 0.065 0.045 0.034 4.35 3.25 2.25 1.7
2 12.5055 3.5 1.29 0.063 0.047 0.037 0.028 9.65 7.2 5.67 4.29
2.85 1.23 0.036
3 12.8295 0.0492 9 0.037 0.028 5.13 3.85 3.86 2.92
4 12.5055 3.2 1.09 0.056 0.042 0.045 0.034 7.17 5.38 5.76 4.35
2.93 1.35
5 17.4276 9 0.0535 0.04 0.037 0.028 8.05 6.02 5.57 4.21
6 20.052 3.96 1.14 0.0395 0.03 0.032 0.024 12.42 9.43 10.06 7.55
7 12.8295 2.85 1.39 0.05 0.037 0.032 0.024 5.21 3.86 3.33 2.5
8 12.5055 3.2 1.24 0.0588 0.044 0.037 0.028 7.53 5.63 4.74 3.59
9 12.5055 4 1.02 0.049 0.039 0.037 0.028 9.8 7.8 7.4 5.6
10 18.7425 4.06 1.11 0.056 0.042 0.032 0.024 17.3 12.98 9.89 7.41
2.85 1.42 0.037
11 12.8295 0.0506 6 0.032 0.024 5.27 3.92 3.33 2.5
12 17.18595 3.2 1.27 0.0672 0.049 0.045 0.034 11.83 8.76 7.92 5.98
 8
13 13.2345 4 1 0.04 0.036 0.045 0.034 8.47 7.62 9.53 7.2
4 1.13 0.037
14 18.0675 0.0511 8 0.037 0.028 14.77 10.93 10.7 8.09

For each support over which the slab is continuous there will thus generally be two different
support moments. The deference may be distributed between the panels on either side of the
support to equalize there moments, as in the moment distribution method for frames. Two
methods of differing accuracy are provided in EBCS 2, 1995-I,
Method1: When the differences between initial support moments are less than 20% of the
larger moment.

M1+M2
M adj= 2

Method 2: In this method consideration of the effects of changes of support moments is

limited to the adjacent span. The unbalanced moments are distributed using moment
redistribution technique.
 Lx : the shortest dimension of the panel
 Ly : the longest dimension of the panel
1 1
 Ki=LX Kj=LY
Ki Kj
 Di =Ki+Kj Dj= Ki+Kj

 Change of moment=M2-M1 where M2>M1

change of moment
 %change of moment= M large

Based on the %change of moment we can decide whether method 1 or method 2 to be used.
In this case we use method 2 to adjust the moment.

 Support moment adjustment of slabs.

Span Di Dj M1 M2 M2-M1 φ change% Msuport

adj
1,2 0.69 0.31 2.25 5.67 3.42 60 4.6098
2,3 0.61 0.39 3.86 5.67 1.81 32 4.9641
5,6 0.53 0.47 5.57 10.06 4.49 45 7.9497
6,7 0.61 0.39 3.33 10.06 6.73 67 7.4353
 9,10 0.53 0.47 7.4 9.89 2.49 25 8.7197
10,11 0.61 0.39 3.33 9.89 6.56 66 7.346
11,12 0.53 0.47 3.33 7.92 4.59 58 5.7627
13,14 0.53 0.47 9.53 10.7 1.17 11 10.1501
1,5 0.53 0.47 4.35 8.05 3.7 46 6.311
2,6 0.53 0.47 9.65 12.42 2.77 22 11.1181
3,7 0.53 0.47 5.13 5.21 0.08 2 5.1724
5.9 0.51 0.49 8.05 9.8 1.75 18 8.9425
6.10 0.51 0.49 12.42 17.3 4.88 28 14.9088
7,11 0.51 0.49 5.21 5.27 0.06 1 5.2406
9,13 0.5 0.5 8.47 9.8 1.33 14 9.135
10,14 0.5 0.5 14.77 17.3 2.53 15 16.035

 Span moment adjustment of slabs.

 Mxspan(adj) = Mxspan + Cx(Msupmax - Msup,adjx)
 Myspan(adj) = Myspan + Cy(Msupmax - Msup,adjx)
1 1
 Ki=LX Kj=LY
Ki Kj
 Cx =Ki+Kj Cy= Ki+Kj

Msup,adjx Msupmax Msupmax - Cx Cy

Panel Mxs span Msypan Msup,adjx Mx spanadj Myspanadj
2 7.2 4.29 4.6098 5.67 1.0602 0.69 0.31 7.931538 4.618662
5 6.02 4.21 6.311 8.05 1.739 0.47 0.53 6.83733 5.13167
6 9.43 7.55 7.44 10.06 2.62 0.47 0.53 10.6614 8.9386
7 3.86 2.5 5.17 5.21 0.04 0.47 0.53 3.8788 2.5212
9 7.8 5.6 8.94 9.8 0.86 0.49 0.51 8.2214 6.0386
10 12.98 7.41 7.346 9.89 2.544 0.49 0.51 14.22656 8.70744
11 3.92 2.5 5.24 5.27 0.03 0.49 0.51 3.9347 2.5153
12 8.76 9.53 5.76 7.92 2.16 0.49 0.51 9.8184 10.6316
14 10.93 8.09 10.15 10.7 0.55 0.5 0.5 11.205 8.365
Analysis of one way slabs

Figure : Rectangular slab with two edges fixed and two edges simply supported
 For a BMD for X-direction middle strips (section A-A) with constant moment
over the unloaded part, the following maximum moments are achieved

 Cantilever moment = the absolute of the negative moment at a support plus the
span moment

Now the ration of negative to positive moment X-directin middle strips is

Maximum positive moment

Negative support moment

 Usd=
As1=Msd/
design moment depth Msd/Fc Kz Z =d*Kz spacing sp pro
ZFyd
dbd^2
X-middle

Mxs 28.352 180 0.07 0.951 171.18 476.95 164.59 ɸ10c/c160

Mxf 14.144 180 0.04 0.971 174.78 233.04 336.85 ɸ10c/c335
X-edge
Mxs 3.544 180 0.01 0.985 177.3 57.56 1363.79 ɸ10c/c400
Mxf 1.768 180 0.01 0.99 178.2 28.57 2747.64 ɸ10c/c400
Y-middle
Mys 7.088 180 0.02 0.98 176.4 115.71 678.42 ɸ10c/c400
Myf 3.536 180 0.01 0.985 177.3 57.43 1366.88 ɸ10c/c400
Y-edge
Mys 3.544 180 0.01 0.985 177.3 57.56 1363.79 ɸ10c/c400
Myf 1.768 180 0.01 0.99 178.2 28.57 2747.64 ɸ10c/c400

Flexural reinforcement design of the slab

For the design of each panel we use design procedures and design table prepared by ato
Misgun. The design procedures are the following:
���
1. Evaluate ��� = �����2

2. Find Kz that corresponds to the value of Usd from design table prepared by ato
Misgun for concrete grade C12/15 –C50/60
���
3. Compute the area of reinforcement As1= ���

4. Check the calculated area of reinforcement As1for minimum and maximum area of
reinforcement required according to ES EN 1992-1-1 section 9.2.1.1
��∗1000
5. Compute the spacing of reinforcement bars required by using, spacing= ��1

6. And finaly check the spacing computed against for maximum spacing required
according to ES EN 1992-1-1 section 9.3.1.1
The slabs are designed using a concrete grade and steel grade of C25/30 and S400
respectively. For both secondary and primary reinforcements a ф10 diameter reinforcement is
used. According to ES EN 1992-1-1:

 The maximum and minimum reinforcement area required are computed using the
following formulas
 0.26∗����∗�∗�
Asmin= ���

Asmax=0.4 ∗ ��
 The maximum spacing is computed using the following formulas
 For primary reinforcement Smax=max(3�, 400��)
For secondary reinforcement Smax=max 3.5�, 450��

panel direction of design d Usd= Kz Z =d*Kz spacing

reinforcement span Msd/Fcdbd^2 .As1=Msd/Z provided
spacing
moment Fyd

1 short 3.25 150 0.01 0.99 148.5 63.02 1245.64 ɸ10c/c400

long 1.7 140 0.01 0.99 138.6 35.32 2222.54 ɸ10c/c400
2 short 4.62 150 0.01 0.99 148.5 89.59 876.21 ɸ10c/c400
long 7.93 140 0.03 0.98 137.2 166.44 471.64 ɸ10c/c330
3 short 3.85 150 0.01 0.99 148.5 74.66 1051.43 ɸ10c/c360
long 2.92 140 0.01 0.99 138.6 60.67 1293.88 ɸ10c/c400
5 short 5.132 150 0.02 0.985 147.75 100.02 784.84 ɸ10c/c300
long 6.84 140 0.02 0.985 137.9 142.84 549.57 ɸ10c/c320
6 short 8.94 150 0.03 0.98 147 175.13 448.24 ɸ10c/c380
long 10.66 140 0.04 0.976 136.64 224.66 349.42 ɸ10c/c320
7 short 2.52 150 0.01 0.99 148.5 48.87 1606.3 ɸ10c/c370
long 3.88 140 0.01 0.99 138.6 80.61 973.82 ɸ10c/c400
9 short 8.22 150 0.03 0.98 147 161.03 487.49 ɸ10c/c370
long 6.0386 140 0.02 0.985 137.9 126.1 622.52 ɸ10c/c400
10 short 8.71 150 0.03 0.98 147 170.63 460.06 ɸ10c/c390
long 14.23 140 0.05 0.971 135.94 301.44 260.42 ɸ10c/c260
11 short 3.935 150 0.01 0.99 148.5 76.31 1028.7 ɸ10c/c400
long 2.52 140 0.01 0.99 138.6 52.36 1499.24 ɸ10c/c400
12 short 9.82 150 0.03 0.98 147 192.37 408.07 ɸ10c/c280
long 10.63 140 0.04 0.976 136.64 224.03 350.4 ɸ10c/c350
13 short 8.09 150 0.03 0.98 147 158.48 495.33 ɸ10c/c400
long 10.93 140 0.04 0.976 136.64 230.35 340.79 ɸ10c/c340
14 short 8.365 150 0.03 0.98 147 163.87 479.04 ɸ10c/c400
long 11.2 140 0.04 0.976 136.64 236.04 332.57 ɸ10c/c330
 Staircase design

Stair case analysis and design is similar to one way slab analysis and design. It
involves the analysis steps followed for slabs. The inclined configuration is analyzed
by projecting the loads on a horizontal plane.
 Riser = 15cm Stair Width is 1.55m
 Thread = 25cm Waist = 20cm
 marble finish 3cm thick cement screed 3cm
 H = 0.25m*8 = 2m
 Angle=tan -1 (1.53/2.0) =37.42˚
 Inclined Length = Horizontal Length/cos =2/cos37.42 =2.52m

Load calculation per unit Length

Horizontal Projection of loads = Inclined load/cos

 Dead load of Waist

 Slab weight of waist 0.2*25*1m 5KN/m
 Plastering 0.03*23*1m 0.69KN/m
 Total dead load of waist 5.69KN/m
 Dead load of steps
 floor finish thread 0.03*27*0.25 0.2025KN/m
 floor finish riser 0.03*27*0.15 0.122KN/m
 cement screed thread 0.03*23*0.25 0.1725KN/m
 cement screed riser 0.03*23*0.15 0.1035KN/m
 own weight of step 0.5*0.25*0.15*25 0.4688KN/m
 Total dead load of steps 1.0693KN/m
 Total dead load 6.7593KN/m
 live load 4KN/m2
 Pd = 1.35 Gk + 1.5 Qk
 design dead load 9.13KN/m
 design live load 6KN/m2
 Pd 15.13KN/m
 Horizontal Projection of loads = Inclined load/cos
Horizontal Projection of load = 15.13KN/m/cos37.42 = 19.05KN/m
 Horizontal projection of Design Load = 19.05 KN/m
Take the slab of depth
D =140+30+5=175
= 180 mm

Dead load on landing

 Slab weight 0.18*25*1m 4.5KN/m
 floor finish 0.03*27*1m 0.81KN/m
 cement screed 0.03*23*1m 0.69KN/m
 Total dead load 6KN/m
 design dead load 8.1KN/m
 design Live load 6 KN/m

Pd = 1.35 Gk + 1.5 Qk
Pd =14.1KN/m
Analysis
 Solving the reactions and We get R1=29.366KN and R2=25.65 KN
 M1= (29.366x1) -(19.05x1x1/2) =19.84KNm
 Mmax=(29.366x1.54) -(19.05x1.54x1.54/2) =22.6 KNm
 M3=(29.366x2.6) -(19.05x2x1.6)- (14.1x0.6x0.3) =12.85 KNm
 Our design moment is Mmax=22.6 KNm

 Design For Flexure

Design Values
 C 25/30, fcd = 14.167 MPa S-400, fyd =
347.826 MPa
 Fctm = 2.6 (EN 1992-1-1:2014 Table 1-1) D = 180mm
 d = D – Cover – Ø/2 = [180 – 30 – (10/2)] mm= 145 mm
 Msd = 22.6 KNm b=1m
For 10 mm diameter transverse and longitudinal reinforcement bar:
as =ᴨd 2 /4 =78.5mm2
Minimum Reinforcement
As, min= max 0.26 fctm∗ bd/ fyk
0.0013bd
As, min= max 0.26* 2.6 *1000*145/400 = 245 mm2
= 0.0013*1000*145 = 188.5 mm2
Take As, min= 245 mm2
Maximum Rienforcement
As, max = 0.04Ac = 0.04*180*1000 = 7,200mm2
Longitudinal bar
µsd= Msd/ [fcd * b * d 2] = 22.6* 106 / [14.17 * 1000 *(145)2] = 0.076
 from design chart we get
Kz=0.96=z/d
z=0.96*145=139.2mm
 Calculate the amount of reinforcement
Ast=Msd,s/zFyd =22.6* 106 / [139.2*347.826] =466.77 mm2
Ast> As,c min=245 (true)
Number of reinforcement= Ast/ as= 466.77/78.5=5.95≈6
S = (b *as) / Ast = (1000 *78.5) /466.77 = 168mm
Take 165 mm
Maximum spacing
Smax=min 3D = 3x145 = 435mm
400mm
Smax = 400 mm > Sprovided = 165 mm
Thus, use Ø 10 c/c 165mm
Transverse bar
As for transverse bar = 30% longitudinal bar 0.3 (466.77) = 140mm2 < Asmin =245
=245mm2 ...….. Provide minimum reinforcement. Thus, As, transverse = As, min =
245mm2
S = (b *as) / Ast = (1000 *50.26) / 245 = 205 mm
Thus, use Ø 8 c/c 205mm
 Wind load Analysis

3.1 Introduction
In the design of roof, the major loads which are applied on roof are wind load and live
load. Normal maintenance, repair, painting and minor repairs are the live loads acting
on the roof structure. The action of wind can be of the type of suction or pressure to
our structures both externally or internally. However, these effects are more
magnified for structures with more openings and large surface areas. Some of the
major factors that wind pressure depends on;
• Velocity of air
• Shape of the structure
• Density of air
• Stiffness of structure
• Angle of the induced wind
• Topography
To achieve the proper design (according to EBCS - 1 1995), the following conditions
have to be taken into account
• Turbulent wind acting over part or all of the structure
• Fluctuating pressures induced by the wake behind the structure
• Fluctuating forces induced by the motion of the structure
There are two methods for wind load analysis; simple procedures applied to the
structural properties that do not make them susceptible to dynamic excitation. The
other method is detailed dynamic analysis. This is applied to structures which are
likely to be susceptible to dynamic excitation. The choice of analysis from the above
two methods depend on the value of the structure‟s dynamic coefficient.The dynamic
coefficient depends on the type of structure, the height of the structure and its breadth.
The quasi-static method is used for structures whose Cd value is less or equal to 1.2.
In our case, the building variables are:
 Height of building =25.22 m
 Width of building = 14.55 m
 Length of the building=15.52 m
Wind load on the vertical wall
 Basic wind velocity (Vb)

where:
vb is the basic wind velocity, defined as a function of wind direction and time of year
at 10 m above ground of terrain category II vb,0 is the fundamental value of the basic
wind velocity = 22m/s. We assumed 22 m/s base velocity occurred in the region
where the building found.
 C-dir is the directional factor =1
 C-season is the season factor =1

Mean wind Variation with height

The mean wind velocity Vm (z) at a height z above the terrain depends on the terrain
roughness and orography and on the basic wind velocity, Vb, and should be
determined using the following expression.

Where
 Cr(z) is the roughness factor
 Co(z) is the orography factor

Terrain roughness
The roughness factor, cr(z), accounts for the variability of the mean wind velocity at
the site of the structure due to:
 The height above ground level.
 The ground roughness of the terrain upwind of the structure in the wind direction
considered.

Where
 Zo is the roughness length
 kr terrain factor depending on the roughness length Zo calculated using

Kr = 0.19*(zo/zo, II)0.07
Where
 ZO,II = 0.05m (terrain category II)
 Zmin is the minimum height =10m < 25.22m < Zmax is to be taken as 200 m

Table 1. 1 Terrain categories and terrain parameters

Note: for the table above the highlighted row shows this case is selected for this
project Therefore, Kr is calculated as
Kr = 0.19*(1/0.05) 0.07 = 0.2343
Cr(z)= kr*ln(z/zo) = 0.2343ln(25.22/1) = 0.756

Terrain orography
The effects of orography may be neglected when the average slope of the upwind
terrain is less than 3° in our case the roof type is flat.
Vm(z)= 0.756*1*22= 16.64m/s

Peak velocity pressure

The National Annex gives the rules below for the determination of qp(z)

The turbulence intensity Iv(z) at height z is defined as the standard deviation of the
turbulence divided by the mean wind velocity.
The recommended rules for the determination of Iv(z) are given as follow

Where:
 kI is the turbulence factor. The value of kI may be given in the National Annex.
The recommended value is kI = 1.0.
 co is the orography factor Co=1
 z0 is the roughness length (Z, given in table above) Accordingly
lv(z)= 1/(1*ln(25.22/1)) = 0.31
Qp(z)= (1+7*0.31)*1/2*1.25*16.642 = 549 pa or = 0.549Kpa

 For qp(b) : b=15.52m

Cr(b) = 0.2343ln(15.52/1) = 0.64
Vm(b)=0.64*1*22= 14.135m/s
lv(b)= 1/(1*ln(15.52/1)) = 0.365
Qp(b)= (1+7*0.365)*1/2*1.25*14.1352 = 0.444Kpa

 For qp(d): d=14.55m

Cr(d) = 0.2343ln(14.55/1) = 0.627
Vm(d)=0.613*1*22= 13.8m/s
lv(d)= 1/(1*ln(14.55/1)) = 0.37
Qp(d)= (1+7*0.37)*1/2*1.25*13.82 = 0.430Kpa

 Wind load along the longer direction (b=15.52m)

qp(Z)=qp(h=25.22)=0.549KPa
qp(Z)= qp(b=15.52)=0.444KPa
qp(Z)= qp(d=14.55)=0.430KPa

 External wind pressure coefficient b=15.22m

e=min (b,2h) h=25.22m
e =min (15.52, 2*25.22) e =15.52m
Since e=15.52 and d=14.55…...…for e≥d
For h/d=25.52/14.55=1.73
All the region area is greater than 10 m2 ,so we use Cpe,10.

Zone A B D E

h/d Cpe,10 Cpe,10 Cpe,10 Cpe,10

5 -1.2 -0.8 0.8 -0.7

1.73 -1.2 -0.8 0.8 -0.54

1 -1.2 -0.8 0.8 -0.5

We=qp(z)*Cpe
The lack of correlation of wind pressures between the windward(region D) and
leeward side(region E) may be considered as follows. For buildings with h/d ≥ 5 the
resulting force is multiplied by 1. For buildings with h/d ≤ 1, the resulting force is
multiplied by 0,85. For intermediate values of h/d, linear interpolation may be applied.
Region Z qp(z) A B D E

1 15.52 0.549 -0.659 -0.44 0.44 -0.296

2 25.22 0.444 -0.533 -0.355 0.355 -0.24

 Internal pressure coefficient (Cpi)

Cpi=0.2,-0.3…for closed building

Net pressure for region-1 will be

Zone We Cpi Wi Wnet= We-Wi
A -0.659 -0.3 0.2 -0.165 0.11 -0.494 -0.769
B -0.44 -0.3 0.2 -0.165 0.11 -0.275 -0.55
D 0.44 -0.3 0.2 -0.165 0.11 0.605 0.33
E -0.296 -0.3 0.2 -0.165 0.11 -0.131 -0.406

Net pressure for region-2 will be

Zone We Cpi Wi Wnet= We-Wi

A -0.533 -0.3 0.2 -0.133 0.089 -0.4 -0.622

B -0.355 -0.3 0.2 -0.133 0.089 -0.222 -0.444

D 0.355 -0.3 0.2 -0.133 0.089 0.488 0.266

E -0.24 -0.3 0.2 -0.133 0.089 -0.107 -0.329

Wind load on the roof

Flat roofs are defined as having a slope (α) of –5°< α< 5°

We have two possible critical wind directions, ϴ=00 and ϴ=900.

 For ϴ=00

External wind pressure coefficient

b=15.52m e=min (b,2h)

h=25.22m e =min (15.52, 2*25.22)
e =15.52m
Since e =15.52m and d=14.55m…..…for e≥d

For hp/h=1.5/15.52=0.099

All the region area is greater than 10 m2. except region F

Zone F G H I

hp/h Cpe 1 Cpe,10 Cpe,10 Cpe,10 Cpe,10

0.05 -2 -0.9 -0.7 0.2 -0.2

0.099 - -0.802 -0.7 0.2 -0.2

1.804
0.1 -1.8 -0.8 -0.7 0.2 -0.2

We=qp(z)*Cpe

qp(b)=0.549 F G H I

We -0.99 -0.44 -0.38 0.11 -

0.11

 Internal pressure coefficient (Cpi)

Cpi=0.2,-0.3…for closed building

 Net pressure will be

Zone We Cpi Wi Wnet= We-Wi

F -0.99 -0.3 0.2 -0.165 0.11 -0.825 -1.1

G -0.44 -0.3 0.2 -0.165 0.11 -0.275 -0.55

H -0.38 -0.3 0.2 -0.165 0.11 -0.215 -0.49

I(0.2) 0.11 -0.3 0.2 -0.165 0.11 0.275 0

I(-0.2) -0.11 -0.3 0.2 -0.165 0.11 0.055 -0.22

 6 5 4 3 2 1 Panel

A. Slab
3.96 2.939 3.2 2.85 3.5 2 Lx

4.5 3.96 3.5 3.5 4.5 3.5 Ly

C16
Design
20.052 17.4276 12.5055 12.8295 12.5055 12.5055
load

357 203 140 128 197 88 W(KN)

6.21 1.98 10.925 7.925 4.25 1 Xi(m)

Center of mass

Center Of Mass ( First floor To tenth floor)

6.04 6.04 9.77 9.77 9.77 9.77 Yi(m)

2217 402 1530 1014 837 88 WXi

2156 1226 1368 1251 1925 860 WYi

 15 14 13 12 11 10 9 8 7

1 4 4 3.2 2.85 4.06 4 3.2 2.85

4 4.5 4 4.06 4.06 4.5 4.06 3.96 3.96

13.2345 18.0675 13.2345 17.18595 12.8295 18.7425 12.5055 12.5055 12.8295

2922 53 325 212 223 148 342 203 158 145

9 6.25 2 12.95 9.925 6.25 2 12.91 9.885

-2 -2 -2 2.03 2.03 2.03 2.03 6.04 6.04

7 19394 477 2031 424 2888 1469 2138 406 2040 1433

4 11295 -106 -650 -424 453 300 694 412 954 876
 B.COLUMN (50X50)
 Concrete self-weight = 0.5*0.5*25= 6.25 KN/m
 Plastering = 0.03*23*1.6= 1.104KN/m

 cement screed = 0..03*23*1.6= 1.104KN/m

 Total dead load of column = 8.45 KN/m

Design dead load of column =11.41KN/m

column W(KN) Xi Yi WXi Wyi

C1 34.91 2 11.52 69.82 402.16

C2 34.91 4 11.52 139.64 402.16
C3 34.91 8.5 11.52 296.74 402.16
C4 34.91 11.35 11.52
396.23 402.16
C5 34.91 14.55 11.52 507.94 402.16
C6 34.91 1.061 8.02
37.04 279.98
C7 34.91 4 8.02
139.64 279.98
C8 34.91 8.5 8.02 296.74 279.98
C9 34.91 11.35 8.02 396.23 279.98
C 10 34.91 14.55 8.02 507.94 279.98
C 11 34.91 0 4.06 0 141.73
C 12 34.91 4 4.06 139.64 141.73
C 13 34.91 8.5 4.06 296.74 141.73
C 14 34.91 11.35 4.06 396.23 141.73
C 15 34.91 14.55 4.06 507.94 141.73
C 16 34.91 0 0 0 0
C 17 34.91 4 0 139.64 0
C 18 34.91 8.5 0 296.74 0
C 19 34.91 11.35 0 396.23 0
C 20 34.91 14.55 0 507.94 0
C 21 34.91 4 -4 139.64 -139.64
C 22 34.91 8.5 -4 296.74 -139.64
SUM 768.02 5905.44 3840.07
Center of mass(x, y) 7.69 5
C.Beam(35x25)
 Concrete self-weight = 0.35*0.25*25= 2.188 KN/m
 Plastering = 0.03*23*1.2 = 0.828KN/m

 cement screed = 0.03*23*1.2 = 0.828KN/m

 Total dead load of beam = 3.844 KN/m

 Design dead load of beam = 5.1894 KN/m

wall w(KN)
beam length load Pd Xi Yi Wxi Wyi
5-AB 2 14.076 5.1894 38.53 3 11.52 115.59 443.87
5-BC 4.5 14.076 5.1894 86.69 6.25 11.52 541.81 998.67
5-CD 2.85 14.076 5.1894 54.91 9.75 11.52 535.37 632.56
5-DE 3.2 14.076 5.1894 61.65 12.95 11.52 798.37 710.21
4-AB 2.939 0 5.1894 15.25 2.53 8.02 38.58 122.31
4-BC 4.5 10.557 5.1894 70.86 6.25 8.02 442.88 568.3
4-CD 2.85 0 5.1894 14.79 9.75 8.02 144.2 118.62
4-DE 3.2 10.557 5.1894 50.39 12.95 8.02 652.55 404.13
3-AB 4 10.557 5.1894 62.99 2 2.03 125.98 127.87
3-BC 4.5 10.557 5.1894 70.86 6.25 2.03 442.88 143.85
3-CD 2.85 0 5.1894 14.79 9.75 2.03 144.2 30.02
3-DE 3.2 10.557 5.1894 50.39 12.95 2.03 652.55 102.29
2-AB 4 14.076 5.1894 77.06 2 0 154.12 0
2-BC 4.5 0 5.1894 23.35 6.25 0 145.94 0
2-CD 2.85 14.076 5.1894 54.91 9.75 0 535.37 0
2-DE 3.2 14.076 5.1894 61.65 12.95 0 798.37 0
1-BC 4.5 14.076 5.1894 86.69 6.25 -2 541.81 -173.38
A-54 3.5 14.076 5.1894 67.43 1.06 9.77 71.48 658.79
A-43 3.96 14.076 5.1894 76.29 0 6.04 0 460.79
A-32 4.06 14.076 5.1894 78.22 0 2.03 0 158.79
B-54 3.5 10.557 5.1894 55.11 4 9.77 220.44 538.42
B-43 3.96 10.557 5.1894 62.36 4 6.04 249.44 376.65
B-32 4.06 10.557 5.1894 63.93 4 2.03 255.72 129.78
B-21 4 10.557 5.1894 62.99 4 -2 251.96 -125.98
C-54 3.5 10.557 5.1894 55.11 8.5 9.77 468.44 538.42
C-43 3.96 10.557 5.1894 62.36 8.5 6.04 530.06 376.65
C-32 4.06 10.557 5.1894 63.93 8.5 2.03 543.41 129.78
C-21 4 14.076 5.1894 77.06 8.5 -2 655.01 -154.12
D-54 3.5 0 5.1894 18.16 11.35 9.77 206.12 177.42
D-43 3.96 10.557 5.1894 62.36 11.35 6.04 707.79 376.65
D-32 4.06 10.557 5.1894 63.93 11.35 2.03 725.61 129.78
E-54 3.5 14.076 5.1894 67.43 14.55 9.77 981.11 658.79
E-43 3.96 14.076 5.1894 76.29 14.55 6.04 1110.02 460.79
E-32 4.06 14.076 5.1894 78.22 14.55 2.03 1138.1 158.79
SUM 1986.94 14925.28 9279.51
Center of mass(x, y) 7.51 4.67

Total Center of mass of G+7B

Type W WXi WYi
structure
Beam 1986.94 14925.28 9279.51
Column 768.02 4338.27 2821.04
slab 2922 19394 11295

SUM 5676.96 38657.55 23395.55

center of mass(Xc, Yc) 6.81 4.12

 Earth-quake

An earthquake is the shaking of the surface of the Earth resulting from a sudden
release of energy in the
Earth's lithosphere that creates seismic waves. So, we have to know that how much is
the intensity of this seismic wave. In order to insure that Human lives are protected;
Damage is limited; and Structures important for civil protection remain operational.
Property And Location of the building
Type 3 Zone 4
Soil type C Ductility Class L(low)

Peak Ground Acceleration

Buildings are classified in 4 importance classes, depending on the consequences of
collapse for human life, on their importance for public safety and civil protection in
the immediate post-earthquake period, and on the social and economic consequences
of collapse. Ordinary buildings, not belonging in the other categories. Are category II.
The value of I for importance class II shall be, by definition, equal to 1.0. The
Seismic hazard map in Ethiopia is divided into 5 zones, where the ratio of the design
bedrock acceleration to the acceleration of gravity for the respective zones is
indicated in Table
For Seismic Zone of 4, ag = 0.15g* I = 0.15*9.81*1 = 1.47
Fundamental period of vibration
For buildings with heights of up to 40 m the value of T1 (in s) may be approximated
by the following expression:
where Ct is 0.075 for moment resistant space concrete frames and H is the height of
the building 31+1m Foundation = 32 m, from the foundation.
T = Ct⋅ H3/4 = 0.075*26.223/4 = 0.87sec
Lateral force of method analysis
This type of analysis may be applied to buildings whose response is not significantly
affected by contributions from modes of vibration higher than the fundamental mode
in each principal direction. They have fundamental periods of vibration T in the two
main directions which are smaller than the following values

4x0.6= 2.4
Therefore, We can use Lateral force method of analysis, also the building has regular
elevation.
For Low Ductility Class the behavior factor q is 1.5 ( EN_1998 -1: 2004(E))
Design spectrum for elastic analysis
Where;
TB- is the lower limit of the period of the constant spectral acceleration branch;
TC- is the upper limit of the period of the constant spectral acceleration branch;
TD- is the value defining the beginning of the constant displacement response range
of the spectrum;
S- is the soil factor;
- is the damping correction factor with a reference value of η= 1 for 5% viscous
damping,
ag- is the design ground acceleration (ag = γI agR)
Sd(T)- is the design spectrum;
q is the behavior factor; The adoption of values for q greater than 1.5 in the vertical
direction should be justified through an appropriate analysis.
β is the lower bound factor for the horizontal design spectrum, The recommended
value for β is 0,2
In our case TC ≤ T ≤ TD which is 0.4 ≤ 0.936 ≤ 2.0 (in Second). Therefore, We use
the following formula to determine the design spectrum.

Sd(T) = { = 1.47*1*(2.5/1.5)*(0.6/ 0.87) = 1.69

≥0.2*1.47 = 0.294
Therefore, the design spectrum Sd(T) is 1.69 .
Base shear force
The seismic base shear force Fb, for each horizontal direction in which the building is
analysed, shall be
determined using the following expression:
Fb = Sd (T)*m* λ

Where Sd (T) is the ordinate of the design spectrum at period T;

T- is the fundamental period of vibration of the building for lateral motion in the
direction considered;
m- is the total mass of the building, above the foundation.
λ- is the correction factor, the value of which is equal to: λ= 0.85 if T1< 2 TC and the
building has more than two storeys, or λ= 1.0 otherwise. And λ= 0.85 for our case
b/se T < 2TC.
Combinations of seismic action with other action

where E,i is the combination coefficient for variable action i

The combination coefficients E,i take into account the likelihood of the loads Qk,i
not being present over
the entire structure during the earthquake. These coefficients may also account for a
reduced participation
of masses in the motion of the structure due to the non-rigid connection between them.
Ei = ɸ* 2i

For stories with correlated occupancies is 0.8.

Structural design of building

For Domestic and Residential areas 2i is 0.3.(EN 1990 : 2002(E) table A1.1)
Therefore, E = 0.8*0.3 = 0.24
For shopping Areas 2 = 0.6, E will be 0.48(Ground )
For roof 2 = 0, we don’t consider live load on the roof.

level W(KN)
Basement 5261.15
Ground 5261.15
1st 5473.15
2nd 5473.15
3rd 5473.15
4th 5473.15
5th 5473.15
6th 5473.15
7th 5473.15
Roof 5261.15
sum 54095.5

Fb = Sd (T)*m* λ = 0.169x54095.5x0.85 = 7,770.82 KN

 14.4% of the total weight of the building.
(mi*Zi)/∑
level Mi Zi Mi*Zi Fi
zi*mi

Basement 5261.15 2.8 14,731.22 0.02 119.59

Ground 5261.15 6.4 33,671.36 0.04 273.34

1st 5473.15 10.2 55,826.13 0.06 453.19

2nd 5473.15 13.26 72,573.97 0.08 589.15

3rd 5473.15 16.32 89,321.81 0.09 725.10

4th 5473.15 19.38 106,069.65 0.11 861.06

5th 5473.15 22.44 122,817.49 0.13 997.02

6th 5473.15 25.5 139,565.33 0.15 1,132.97
7th 5473.15 28.56 156,313.16 0.16 1,268.93
Roof 5261.15 31.62 166,357.56 0.17 1,350.47
Sum 957,247.67

1,350.47

1,268.93

1,132.97

997.02

861.06

725.1

589.15

453.19

273.34
 Geometric Imperfections

The unfavorable effects of possible deviations in the geometry of the structure and the
position of loads shall be taken into account in the analysis of members and structures.
i.e. Geometrical Imperfection

Imperfections shall be taken into account in ultimate limit states in persistent and
accidental design situations.

The following provisions apply for members with axial compression and structures
with vertical load, mainly in buildings. Numerical values are related to normal
execution deviations. prEN 1992-1-1

ɑh= 2 /√25.22=0.398 but ɑh should be equal or between 2/3 and 1, therefor ɑh=2/3
ɑm= √0.5(1 + ( 1/22))) =0.72 Ө0 = 1/200 = 0.005
Therefore, Өi=0.005 ∗ 2/3 ∗ 0.72 = 0.0024
The definition of l and m depends on the effect considered, on bracing system: l =
height of building, m = number of vertical members contributing to the horizontal
force on the bracing system.
Then Lateral Load H is calculated using the formula, H= Θi*N where N is the Axial
force. And Өi is an inclination made by imperfection.
Level Nb-Na Өi H= Θi*N
Axial load
roof 1772.267 1772.267 0.0024 4.3
7th 4980.697 3208.43 0.0024 7.7
6th 8189.127 3208.43 0.0024 7.7
5th 11397.557 3208.43 0.0024 7.7
4th 14605.987 3208.43 0.0024 7.7
3rd 17814.417 3208.43 0.0024 7.7
2nd 21022.847 3208.43 0.0024 7.7
1st 24231.277 3208.43 0.0024 7.7
Ground 27439.707 3208.43 0.0024 7.7
Basement 30648.137 3208.43 0.0024 7.7
Etabs model

Fig. 3D model B+G+7B

Fig. Roof floor plan
Fig. Typical floor plan
Fig.Basement and Ground floor plan
 Beam Analysis and Design

A beam is a structural element that is capable of withstanding load primarily by

resisting bending.

Reinforced concrete beam analysis and design consists primarily of producing

member details which will adequately resist the ultimate bending moments, shear
forces and also torsional moments if necessary. At the same time serviceability
requirements must be considered to ensure that the member will behave satisfactorily
under working loads. It is difficult to separate these two criteria; hence the design
procedure consists of a series of interrelated steps and checks. These steps may be
condensed into three basic design stages which are preliminary analysis and member
sizing, detailed analysis and design of reinforcement and serviceability calculations.
The materials in the design of beams depends on the theory and design specification
of concrete and reinforcement. And the design of beams is in accordance with ES EN
1992-1.1:2015.
Basic principles and assumptions
Although the method used in the analysis and design of reinforced concrete beams
are different from those used in the design of homogeneous beams such as structural
steel, the fundamental principles are essentially the same. Accordingly, the basic
equations for flexural design of beams are derived based on the following basic
principles and assumptions at ultimate limit state described on ES EN 1992-
1.1:2015 section 6.1(2) P.
Internal stress resultants such as bending moments, shear forces etc. at any section
of the member are in equilibrium with the external action effects.
Plane sections before bending remains plane after bending.
The strain in bonded reinforcement or bonded prestressing tendons, whether in
tension or in compression, is the same as that in the surrounding concrete.
The tensile strength of concrete is ignored.
The stresses in the concrete in compression are derived from the design stress/strain
relationship given in ES EN 1992-1.1:2015 section 3.1.7 stress-strain relations for
the design of cross-sections.
Figure possible Strain diagram at Ultimate Limit State ES EN-1992-1.1:2015
Preliminary analysis and beam sizing

The layout and size of members are very often controlled by architectural details
and clearances for machinery and equipment. Therefore, it is necessary to either
check that beam sizes are adequate to carry the loading, or alternatively, decide on
the sizes that are adequate. The preliminary analysis need only provide the
maximum moments and shears in order to ascertain reasonable dimensions. Beam
dimensions required are: -
Cover to the reinforcement
Breadth (b)
Effective depth (d)
Overall depth (h)
Design for cover
The recommended procedure for the determination of the nominal concrete cover is
the same as the procedure used for slab. The only difference is that, nominal concrete
cover for beam sections is designed for both the longitudinal reinforcement and the
shear reinforcement (stirrup). Then the governing nominal concrete cover is the
minimum nominal concrete cover found by comparing the longitudinal and shear
reinforcement (stirrup) nominal concrete cover.
the nominal concrete cover is designed to meet requirements of durability, bond and
fire resistance according to ES EN 1992- 1.1:2015 section 4.4.1 and ES EN 1992-
1.2:2015 section 5.6, therefore the nominal concrete cover for the beam is designed for
a design service life of 50 years, normal quality control, maximum aggregate size of
20mm, 1HR fire resistance and exposure class of XC3.

Longitudinal reinforcement =max { 20mm

20mm
10mm
Cmin=16mm,allowing for in design deviation , ∆c,dev=10mm
The nominal concrete cover, ���� = ���� + ∆���� = 20�� + 10�� = 30��
Shear reinforcement, ���� = ��� {15mm

10mm
���� = 15��, allowing for in design deviation, ∆c,dev=10mm
The nominal concrete cover, ���� = ���� + ∆���� = 15�� + 10�� = 25��
It can be seen from the above calculation that the nominal concrete cover
for the shear reinforcement governs. Therefor the provided nominal cover
for our beam is, ���= 25��.

Check for fire resistance

According to ES EN 1992-1.2:2015 table 5.6 for standard fire resistance of
R60, the recommended bmin =200mm and a (nominal cover) =12mm.
therefore the nominal concrete cover, Cnom=25mm provided is also
satisfactory for R60 fire resistance.

Depth and width

As it has been mentioned, the minimum depth of the slab should satisfy for
the serviceability requirement in accordance with ES EN 1992-1-1:2015
section 7.4. the deformation of a beam member should be in way that does
not affect its appearance and functionality. This can be checked by
limiting the span/depth ratio, according to the ES EN 1992- 1-1:2015.

Beam span l(mm) ρ(slightly stressed) ρ0= (√��� ∗ 10−3) K Check l/d
AB 2000 0.5% 0.5% 1.3 Ok 30.0625

BC 4500 0.5% 0.5% 1.3 Ok 34.6875

CD 2850 0.5% 0.5% 1.3 Ok 34.6875
DE 3200 0.5% 0.5% 1.3 Ok 34.6875

Required depth (H) = 130mm+20mm/2+25mm+8mm = 173mm

Provided depth (H) = 350mm > 173mm…………………………………..ok
Required minimum width for fire resistance (W) = 200mm

Provided width(W) = 250mm > 200mm……………………………. ok

Design of beam section for ultimate limit state
During construction operation of beams, concrete is placed in the beams and slabs in a
monolithic pour. As a result, the slab serves as the top flange of the beams, as
indicated by the shading such a beam is can be either a T beam or an inverted L beam
depending whether the beam is an interior beam which has flange on both sides or an
exterior beam which has flange on one side only.
Figure T beams and inverted L beams

Effective width of flange

The typical beam section on axis B of floor beams is an interior T beam which
means that it will have flanges on both sides. In T beams the effective flange width,
over which uniform conditions of stress can be assumed, depends on the web and
flange dimensions, the type of loading, the span, the support conditions and traverse
reinforcement (ES EN 1992-1.2:2015 section 5.3.2.1(1) P.
According to ES EN 1992-1.1:2015 section 5.3.2.1(1) P, the effective width of a
flange should be based on the distance lo between points of zero moments, which
are obtained from ES EN 1992- 1.1:2015
Figure Definition of l, for calculation of flange width
The effective flange width with beff for a T beam is driven by ES EN 1992-1.1:2015
expression 5.7: -

���� = ∑����, � + �� ≤ �

Where

����, � = 0.2�� + 0.1�� ≤ 0.2�� and

����, � ≤ ��

Figure. Effective flange width parameters

for span AB and BC

Lo=0.85 �1=0.85 ∗2=1.7m

Lo=0.7 �2=0.7 ∗4.5=3.15m

for support B

Lo= 0.15(�1 + �2) = 0.15 ∗ (2+4.5) = 0.98m

For Span AB and BC

����, 1=0.2((2-0.5)/2)+0.1*1.7 =0.32 ≤0.2*1.7=0.34………………………………ok
����, 1=320mm

����, 2=0.2((4.5-0.5)/2)+0.1*1.7 =0.57 ≤0.2*3.15=0.63………………………………ok

����, 2=570mm
���� = ����, 1 + ����, 2 +�� ≤ �

����=320+570+500≤2000

����=1390mm≤2000mm…………………………………………………..OK

Design for flexure

At the ultimate limit state, it is important that member sections in flexure should be
ductile and that failure should occur with the gradual yielding of the tension steel
and not by a sudden catastrophic compression failure of the concrete. Also, yielding
of the reinforcement enables the formation of plastic hinges so that redistribution of
maximum moments can occur, resulting in a safer and more economical structure.
To ensure that a beam member is ductile enough, the ratio of the neutral axis to
effective depth (x/d) should not be greater than 0.45m for concrete grades C50 or
below. If the ratio of neutral axis to effective depth (x/d) greater than 0.45, then the
member should be resized by increasing depth or width of the member and also be
providing a compression reinforcement in the case of beams in addition to the
tensile reinforcement.
For T beams and inverted L beams, when the beams are resisting sagging moments,
the slab acts as a compression flange and the members may be designed as T or L
beams or as rectangular beams depending on the position of the neutral axis. If the
neutral axis is within the flange depth, then the T or L beam is designed as
rectangular beam. But if the neutral axis is beyond the flange width then they are
designed as T or L beams using rectangular stress-strain distribution curve.
Tensile reinforcement bar, ∅ = 20mm
Effective depth, d = 350mm – 25mm – 8mm – 20/2mm = 307mm
Effective width, beff = 1390mm
Effective web, bw = 500mm
Rectangular beam section is considered because it is at the support. But according to
ES EN 1992- 1.1:2015 section 9.2.1.2(2) an intermediate supports of continuous
beams, the total area of tension reinforcement As1 can be spread over the effective
width of the flange and part of it may be placed on the web. But this may not be
necessary if the width of the web is enough to place the tensile reinforcement with
adequate spacing.
Design load (0 - 2) =38.53KN/m and
Design load (2 – 6.5)= 86.69KN/m
Design load ( 6.5-9.35)= 54.91KN/m
Design load (9.35-12.55)= 61.65KN/m

Using moment distribution method to balance support moment for BA and BC.
 Location: 0 (m), Force Reaction = 45.8994 (kN)
Location: 0 (m), Moment Reaction = 43.4429 (kN-m)
Location: 2 (m), Force Reaction = -320.7737 (kN)
Location: 6.5 (m), Force Reaction = -301.8856 (kN)
Location: 9.35 (m), Force Reaction = -131.9627 (kN)
Location: 12.55 (m), Force Reaction = -112.216 (kN)
Location: 12.55 (m), Moment Reaction = 67.089 (kN-m)

Span moment AB on axis 5

Rectangular beam section is considered because it is at the support. But according to ES EN
1992-
1.1:2015 section 9.2.1.2(2) an intermediate supports of continuous beams, the total area of
tension reinforcement As1 can be spread over the effective width of the flange and
part of it may be placed on the web. But this may not be necessary if the width of the
web is enough to place the tensile reinforcement with adequate spacing.
 2
��� = ���/(��� ∗ � ∗ � )

��� =(121∗ 10^6��� )/(14.16*250*3072)

��� =0.362>=0.295
This indicates that the section is designed as doubly reinforced beam.or Provide
compression reinforcement.
Calculate M * and Ast
M *=
���* ��d*b*� 2
M *=0.295*14.167*250*(307)^2=98.47KNm

Ast= (M */Zfyd)+(M- M *)/(d-d2)fyd

for ��� = 0.295,Kz=0.88

Z=d*Kz=307*0.88

Ast=(98.47*10^6/307*0.88*347.83)+((121-98.47)*10^6/{(307-48)347.83)}

Ast=1,047.89+250=1298mm2
For Double reinforced part(compression reinforcement)

A2=(M- M *)/(d-d2)fyd

A2=((121-98.47)*10^6/(307-48)*347.83))
A2=250mm2
A1=1298mm2
Check for maximum and minimum reinforcement limits

AS,min=max{ ( 0.26*2.6N/ mm2*250*307mm)/400=130 mm2

0.0013*250*307mm=100mm2

AS,min=130 mm2<A1&A2…………………Ok
AS,max=0.04*250*350=3500mm2

AS,max>A1&A2……………………………….Ok
There Fore As1=1298mm2
&As2=250 mm2
N1=As1/as=A1=1298mm2/(3.14*10^2) =4.13=5 Number of bars
N2=As2/as=250/(3.14*10^2)=0.8=2 Number of bars

Provide 5∅ 20tensile reinforcement

Provide2∅ 20compretion reinforcement
Check for longitudinal reinforcement spacing
According to ES EN 1992-1.1:2015 section 8.2(1) P, the spacing of bars shall be in
such that concrete can be placed and compacted satisfactory for the development of
adequate bond. The clear distance between individual parallel or horizontal layers of
parallel bars should not be less than the maximum of k1 *bar diameter, (dg +k2) or
20mm where dg is the maximum size of aggregate (in our case 20mm). the
recommended value of k1 and k2 are 1 and 5mm respectively.
250=2*25+2*8+5*20+4*S1
S1=21
250=2*25+2*8+2*20+1*S2
S2=144
Smin=max{ K1*bar diametrmm
dg+K2mm
20mm
The recommended values of k1 and k2 are 1 and 5 mm respectively
Smin=max{ 1*20=20
20+5=25mm
20mm
Smin=25>S1 and 25< s2
so we Provide s1=25mm and s2 = 144mm

Support moment A on axis 5

Design moment, Msd = 41.8 KNm
Tensile reinforcement bar, ∅ = 20m
Effective depth, d = 350mm – 25mm – 8mm – 20/2mm = 307mm Effective width,
beff = 1390mm
Effective web, bw = 500mm
Since the beam at this span is resisting a sagging bending moment, the first thing to
do is to check whether the beam should be designed as rectangular or T beam. To do
this we have to determine the neutral axis using a simplified stress-strain
distribution called rectangular stress-strain distribution curve as shown in Figure 8-6

Figure 8- Rectangular stress-strain block

Cc=0.8xfcd*beff
Ts=Asfyd
Z=d-0.4x
��� = 0.8� ∗ ��� ∗ ���� ∗ (� − 0.4�)
41.8*10^6=0.8x*14.167*1390*(307-0.4x)

=9mm<tf=200mm, the beam is designed as a rectangular beam

��� = ���/(��� ∗ � ∗ �2).

��� = (41.8 ∗ 10^6���)/(14.167�/��2 ∗ 250�� ∗ (3

07��)2)

��� =0.125 < 0.295, This indicates that the section is designed as single
reinforced beam.

�x = x/d= 9/307= 0.029<0.448 ……………………………………………0k

= (�-0.4x) = (307-0.4*23) = 303��

For singly reinforcement

AS1= M */fyd(d-0.4x)
AS1=41.8*10^6/347.83(303)=396.6 mm2

Check for maximum and minimum reinforcement limits

AS,min=130 mm2<As1…………………Ok
AS,max=0.04*250*350=3500mm2

AS,max>A1…………………………………………………………………...………
….Ok
There Fore As1=397mm2
N1=As1/as=397/(3.14*10^2) =1.26=2 Number of bar
Provide 2∅ 20tensile reinforcement
Check for longitudinal reinforcement spacing
250=2*25+2*8++2*20+1*S1
S1=144
Smin=25<S1……………………………..…………..…………………Ok
R
e
Tensile reinforcement

i
spacing required

spacing provided

reinforcement
spacing avail
As,min, mm2

Compression

n
N1 =As1/as
N2 =As2/as
Msd,KNm

A2 mm^2
As1,mm2
Msd,KNm*
position

d2(mm)

f
Z,mm
d,mm

μsdlim
μsd

Kz

o
r
c
e
m
 support -Vesuppt support support span span span span
moment
112.12 C moment
125.42 B moment
125.42 B momen
41.8 momen
35 momen
99.4 momen
100.28 momen
121
307 307 307 307 307 307 307 307
0.336 0.376 0.376 0.125 0.105 0.298 0.3 0.362
t
e
n

Double Double Double singly singly Double Double Double

0.295 0.295 0.295 0 0 0.295 0.295 0.295
48 48 48 0 0 48 48 48
98.49 98.49 98.49 0 0 98.49 98.49 98.49
0.88 0.88 0.88 0.93 0.943 0.88 0.88 0.88
289.50
270.16 270.16 270.16 285.51 270.16 270.16 270.16
1
1058.2 1067.9 1297.9
1199.399775 1347.033249 1347.033249 623 526
04452 72682 70095
151.3 298.93 298.93 0 0 10.1 19.87 249.87
144.67 144.67 144.67 144.67 144.67 144.67 144.67 144.67
3.82 4.29 4.29 1.98 1.68 3.37 3.4 4.13
0.48 0.95 0.95 0 0 0.03 0.06 0.8
25 25 25 25 25 25 25 25
35/144 21/144 21/144 144 144 35/144 35/144 21/144
35/144 25/144 25/144 144 144 35/144 35/144 25/144
2∅ 20 2∅ 20 2∅ 20 0 0 2∅ 20 2∅ 20 2∅ 20

4∅ 20 5∅ 20 5∅ 20 2∅ 20 2∅ 20 4∅ 20 4∅ 20 5∅ 20
 support -Vesupp support -Vesuppt
moment
67 E moment
23.5 D moment
23.5 D moment
112.12 C
307 307 307 307
0.201 0.07 0.07 0.336

singly singly singly Double

0 0 0 0.295
0 0 0 48
0 0 0 98.49
0.884 0.962 0.962 0.88
271.388 295.334 270.16
971 358 220 1199.399775
0 0 0 151.3
144.67 144.67 144.67 144.67
3.09 1.14 0.7 3.82
0 0 0 0.48
25 25 25 25
35 144 144 35/144
35 144 144 35/144
0 0 0 2∅ 20

4∅ 20 2∅ 20 2∅ 20 4∅ 20
 Column Design and Analysis

COLUMN DESIGN
Columns are upright members which carry the loads from beams and slabs dawn to the
foundation and therefore they are primary compression members, although they may also
have to resist bending forces due to some eccentricity or due to the continuity of the structure.
For the purpose of design calculations, structural members may be classified as Sway or
Nonsway depending on their sensitivity to second order effect due to lateral displacements.
According to EBCS 2 1995 section 4.4.4.2
1. A frame may be classified as non-sway if its response to in-plane horizontal forces is
sufficiently stiff for it to be acceptably accurate to neglect any additional internal forces or
moments arising from horizontal displacements of its nodes.
2. Any other frame shall be classified as a sway frame and the effects of horizontal
displacements of it nodes shall be taken in to account in its design
3. A frame may be classified as non-way for a given load case if the critical load ratio
Nsd/Ncr for that load case satisfies the criterion:
Nsd/Ncr ≤0.1
Where Nsd is design value of the total vertical load Ncr is its critical value for failure in a
sway mode
4. Beam and column type plane frames in building structures with beams connecting each
column at each story level may be classified as non-sway for a given load case, when first-
order theory is used, the horizontal displacements in each story due to the design loads (both
horizontal and vertical), plus the initial sway imperfection satisfy the criterion of the
following equation.

Where: is the horizontal displacement at the top of story, relative to the bottom of the story L
is the story height H is the total horizontal reaction at the bottom of the story N is the total
vertical reaction at the bottom of the story

DESINGPROCEDURE
1. To design a column in a particular frame first the frame is classified weather it is sway or
non-sway.
2. To determine the nature of the frame we substitute the beams and columns by one
substitute frame.
3. The value of the axial force on each substitute frame column is obtained by adding the
axial load each column for the story including self-weight.
4. The value of the stiffness coefficients of the substitute frame is given by
For beam = 2* Kbi
For column = Kci
Where: Kbi = stiffness coefficient of beam and Kci = stiffness coefficient of column
5. The effective length of the substitute frame is computed for each story assuming as sway
frame as shown below. The effective length buckling Le of a column in a given plane is
obtained from the following approximate equation provided that certain restriction is
compiled with.
a. Non-sway mode
Le/L = (αm + 0.4) / (αm + 0.8) ≥ 0.7
b. Sway mode

Slender
= Le/i
Where Le = effective buckling length
i = radius of gyration of the gross concrete section in the plane of buckling.
i = I/A

LIMITS OF SLENDERNESS
According to EBCS 2 1995 section 4.4.6
1. the slenderness ratio of concrete column shall not exceed 140
2. the second order effects in compressive members need not be taken in to account in the
following cases:
a. for sway frames, the greater of
λ ≤ 25

Where d = Nsd / fcdAc

b. for non-sway frames
 G 1 2 3 4 5 6 7 8 9 c

1970.59 1804.61 1566.7 1403.14 1229.4 986.37 793.45 658.02 446.05 275.4 Ned

2250 2250 2250 2250 2250 2250 2250 2250 2250 2250 lo

275000 275000 275000 275000 275000 275000 275000 275000 275000 275000 A
69322916 69322916 69322916 69322916 69322916 69322916 69322916 69322916 69322916 69322916
I
67 67 67 67 67 67 67 67 67 67
158.7713 158.7713 158.7713 158.7713 158.7713 158.7713 158.7713 158.7713 158.7713 158.7713 i

14 14 14 14 14 14 14 14 14 14 λ
n=Ned/A
0.4 0.4 0.3 0.3 0.3 0.2 0.2 0.1 0.1 0.1
cfcd
17 17 20 20 20 24 24 34 34 34 λ lim
curvature and negative if bent is double curvature .

M=eo*Ne
39.4118 36.0922 31.334 28.0628 24.588 19.7274 15.869 13.1604 8.921 5.508
d
0.24 0.22 0.19 0.17 0.15 0.12 0.1 0.08 0.05 0.03 N/bhfck

0.009 0.008 0.007 0.006 0.005 0.004 0.003 0.003 0.002 0.001 M/bh2fck

0.55 0.5 0.35 0.25 0.2 0.1 0.1 0.1 0.1 0.1 w f.graph

9075 8250 5775 4125 3300 1650 1650 1650 1650 1650 AS
453.0091 414.8528 360.1609 322.5609 282.6206 226.7517 182.4022 151.2689 102.5402 63.31034
ASmin
954 736 195 195 897 241 989 655 299 483
21 19 13 10 8 4 4 4 4 4 N of bar

24 27 39 51 63 126 126 126 126 126 spacing

positive and greater in magnitude than M1, and M1 being positive if member is bent in single
Where: M1 and M2 are the first order (calculated) moments at the ends, M2 being always
 Conclusion
In our analysis and design of the B+G+9 mixed-use building, we utilized our
extensive knowledge and experience in various key areas. Let's explore the details:
 Analysis and design of the SOLID slab using the coefficient method: We applied
the coefficient method to analyze and design the SOLID slab, taking into account
the applied loads and material properties. By understanding the loads and
structural behavior, we ensured that the slab design could effectively support the
anticipated loads.
 Meticulous analysis and design of the stair: We carefully analyzed and designed
the stair for the building, considering safety and structural requirements. Factors
such as dimensions, material strength, and imposed loads were taken into account.
Our design adhered to building codes and regulations to ensure a safe structure.
 Accurate computation of wind load according to Eurocode standards: We
accurately computed the wind load for the roof and building face, following
Eurocode standards. Factors such as location, height, shape, and wind exposure
were considered. Our calculations ensured that the building could withstand wind
loads as per industry standards.
 Precise computation of the center of mass: We precisely computed the center of
mass for the building, using our knowledge of structural dynamics. This
assessment of stability, load distribution, and seismic response was crucial and
relied on calculating the weighted average position of all mass elements within
the structure.
 Accurate computation of earthquake load: We accurately computed the
earthquake load for the assigned parameters, drawing on our understanding of
seismic design principles. Factors such as seismic zone and design spectrum were
considered, ensuring that the building could withstand anticipated seismic forces
based on Eurocode standards and seismic design codes.
 Calculation of geometric imperfections as lateral loads: We diligently computed
geometric imperfections as lateral loads, following Eurocode standards. These
imperfections accounted for deviations from the ideal shape of the building,
inducing additional lateral loads during wind or seismic events. Our analysis
aimed to ensure the structural integrity of the building.
 Building structure modeling using ETABS software: We utilized the ETABS
software to model the building structure. This powerful software allowed us to
create a detailed computer model that incorporated various structural elements
and their properties. The ETABS model facilitated simulations of the building's
behavior under different load conditions, aiding in informed design decisions.
 Thorough checking of lateral stability for ULS and Damage Limitation: We
rigorously checked the lateral stability of the building at the ultimate limit state
(ULS) and the damage limitation state. Advanced analysis techniques were
employed to ensure the building could effectively resist horizontal loads, such as
wind and earthquakes. Measures like bracing systems or shear walls were
incorporated to enhance lateral stability.
 Hand calculation-based design of a typical beam and column: We designed a
typical beam and column from foundation to top using hand calculations. This
involved analyzing internal forces, selecting appropriate dimensions and
 reinforcement based on material properties and design requirements. Our
meticulous approach ensured a robust and efficient design.
 Comparison of hand calculation design with ETABS output: When comparing the
design results obtained from hand calculations with the output from the ETABS
software, significant discrepancies were observed due to the aforementioned
reasons.