ACKNOWLEDGEMENT

Our first gratitude goes to the almighty God who gave us all the patience and
strength .Secondly we like to thank Civil Engineering department for giving us this project
specially our instructors Ato. Demelash ,Ato. Ephrem K.,Ato.Erkeno Y.,Ato. Sebsebe,
who helped us to enhance our knowledge of structural analysis and design and also for
providing us with all the necessary equipment’s that may be of use to us. We also like to
thank all the other instructors who stood beside us and helped us with moral support.
 OBJECTIVE

The project has a lot of aims (objectives) these are

 To help the students in order to revise what have learnt in time of studying.
 To work with the general concepts of designing methodology.
 It develops the habits of working together.
 To design a structure to fit to the environment it is constructed according to the given
manual.
 INTRODUCTION
Civil Engineering is a professional engineering discipline that deals with design, construction
and maintenance of the physical and naturally built environment.

This project is a structural design of a G+4 reinforced concrete building for learning purpose
.It is to be constructed in JIMMA UNIVERSITY around the hospital. The plan area covered
by the building is around 1030 m2 with total height of 26.72m above the reference level l (0
Level)

The serviceability limit state design method is used for the analysis of the entire building.
The design is made based on the ETHIOPIAN BUILDING CODE OF STANDARDS
(EBCS-1995) with additional reference books and lecture notes. ETABS 2009 computer
software is used for the frame analysis and SAP for truss analysis.

STRUCTURAL PLANING PREVIEW

Structural planning is the first step of structural design which includes positioning and orientation of
columns, positioning of beams, layout of staircases and selecting proper type of foundation.

POSITION OF BEAMS

As beam transfers load to columns, the position of beams should be aligned with the columns. So
beams are provided directly from column to column.

POSITIONING AND ORIENTATION OF COLUMN As much as possible the orientation of

columns should be provided in such a way that to avoid the projection of columns outside the wall
and orienting the columns depth on the major plane of bending

At ground floor, slabs are not provided and the floor will directly rest on ground. Therefore, only
ground beams passing through columns are provided as tie beams. The floor beams are thus absent in
the ground floor
SLAB

Selection of slab is on the basis of the load it carry, the intended use of the building and also the
intended sound insulation this project two types of slab, solid and ribbed slabs are used.

LAYOUT OF STAIRS

All of of the staircases in this project are designed as half turn stair.

TYPES OF FOUNDATION

The type of foundation depends up on the load carried by the columns and bearing capacity of the
supporting soil.

Material used

Concrete C-25

Steel S-300

Class I work

Cover 15mm and reinforcement of ϕ12

ϓ of concrete=24KN/m3

ϓ of walling material=14KN/m3

Thickness of plaster=2.5cm

CHARACTERSTIC STRENGTH

fcd =0.68fck/ϓc =0.68 * 25/1.5 =11.33Mpa

fyd =fyk/ϓs =300/1.15 =260.87Mpa

fctd=(0.21(fcy/1.25)2/3)ϓc =1.031Mpa
 CHAPTER 1
Design of truss
While designing a truss we should take in to consideration the effect of lateral
load such as wind load and earthquake. In the preceding section the design and
analysis of a single truss is dealt with only considering the effect of wind load.
The design process includes the design of EGA sheet, Purlin and Rafter.
According to the plan the dimension of the truss in centimeter is shown below.

step I : Design of EGA sheet

Height of truss = 1.57m.

Spacing of truss = 2.2 m.

Spacing of purlin = 1.25m.

Inclination of truss = 9.684o.

Type of EGA sheet selected = EGA 500, t=0.4.

Loading

 Dead load
 Weight of EGA =3.14 kg/m.
Coating = 0.7 kg/m.
Total load = 3.84 kg/m.
Effective width of EGA = 0.712m.

Load of EGA/m2 = =0.053KN/m2.

 Live load
As per EBCS – I, 1995, section 2.6.3.4.2
Distributed load (qk) = 0.25 KN/m2.
Concentrated load (Qk) = 1 KN.

Self weight of EGA = 0.053 KN/m2 *0.782m.

= 0.04 KN/m.
Live load 0.25 KN/m2 * 0.782m =0.2 KN/m.
1 KN at mid span =0.8 KN/m.
Wind load
Wind load is divided in to these are external load and internal loads. The
net effect of wind load can be computed by subtracting the internal wind
pressure from the external. This pressure depends on height of the
building, orientation of the building, height of other buildings around it,
etc…
EXTERNAL WIND PRESSURE
We=qref*ce(ze)*cpe-----------------------------From EBCS-1, 1995;page 53
Qref=1/2*ρ*vref2

Where qref – reference wind velocity

Ce (ze) - exposure coefficient
Cpe _external pressure coefficient
ρ_ air density
Vref=CDIR*CTEM*CACT*Vref-----------------------From EBCS-1, 1995; page
55
Where; CDIR- direction factor taken as 1
CTEM- temporary (seasonal)
factor=1
 CALT- altitude factor=1
 Therefore; Vref=1*1*1*22m/sec=22m/sec which is mostly taken.
 Altitude of Jimma (our site) =1950m above sea level.
 From EBCS-1,1995;page 55-table 3.1----------ρ=0.95Kg/m3

i.e. qref=1/2ρVref2=1/2*0.95*22=231.35N/m2
i. EXPOSURE COEFFICIENT ,Ce (Ze)

Ze=h=20.29m

Ce(Ze)=Cr2(Z)*Ct2(Z) [1+ ]

EBCS-1, 1995/PAGE 60
KT- terrain factor
Cr(z)= roughness coefficient
Ct(Z) = topographic coefficient
Cr(z) = KTln ( Z/ZO) , for Z min Z 200m.

Cr(z) =Cr(Zmin) }. For Z Zmin.

 From EBCS I 1995/table 3.2. urban area in which at least 15% of the
surface is covered in buildings and their average height exceeds 15m are
categorized as category IV. Therefore,
KT = 0.24 , Zo =1m, Zmin = 16m.
Since Zmin Z 200m.
Cr (z) = KT ln (Z/Zo)
= 0.24 ln (20.29/11) =0.722.
Ct (Z) = 1, site is not affected by topography (no escarpment and cliff)

Ce(20.29) = 0.7222 * 12 *

Ce (20.29) = 1.73

II External pressure coefficient (Cpe)

(a) General

Where; b-cross wind dimension

h=Ze and
e=b or 2h
 (b) Wind reaction

Fig key for Doupich Roofs

For 00

e Therefore; e=13.2m for 00

e=13.2

For 900

e e=18.4m
 i.e.

Internal wind pressure, Wi

Wint=qref*Ce (Ze)*Cpi ------------------ ----from EBCS1-1995/page 53

Where; Cpi-internal pressure coefficient

 For a building with internal partition the extreme values are

Cpi=0.8 and Cpi=-0.5.
 Cpe depends on size of loaded area (A), on the individual parts of
the building.

Cpe= Cpe,1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - for A m2

Cpe= Cpe,1 ( Cpe,10 Cpe,1) - - - - - - - - - - - -for 1m2 A 10 m2

Cpe= Cpe,10 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - for A 10 m2

For 00

Zone Area (m2) For 00 Cpe

Cpe 10 Cpe 1

F 4.356 - - 2.26 -1.67

1.33

G 8.712 -1.01 - 1.76 -1.05

H 208.03 -0.46 - 0.78 -0.46

J 17.42 -0.63 - 0.86 -0.63

I 208.03 - - 0.35 -0.35

0.35

For 90
o

Zone Area For 900 Cpe

Cpe 10 Cpe 1

F 7.54 - 1.46 - 2.1 - 1.45

G 9.38 - 1.3 -2 - 1.32

H 84.64 - 0.65 -1.2 - 0.65

I 36.8 - 0.5 - 0.5 - 0.5

Take Cpi = 0.8

For zero degree

Pitch F G H J I
angle

Cpe Cpi Cpe Cpi Cpe Cpi Cpe Cpi Cpe Cpi

9.684 - 1.67 0.8 - 1.05 0.8 - 0.46 0.8 - 0.63 0.8 - 0.35 0.8

Cpe - Cpi - 2.47 -1.85 - 1.26 - 1.43 - 1.15

Wnet = qref * Ce (Ze) (Cpe- Cpi)

q ref= 231.35 , Ce(Ze) = 1.73

 Zone W net

F -988.581

G -740.43

H -504.29

J -572.34

I -460.27

FOR 900

Pitch angle F Ci H I

9.684 -1.45 0.8 -1.32 0.8 -0.65 0.8 -0.5 0.8

Cpe-Cpi -2.25 -2.12 -2.45 -2.3

ZONE W net

F -900

G -848.5

H -580.34

I -520.3

Take W net ,max =-988.58 N/m2

Wind load =
LOAD COMBINATION FOR ULTIMATE LIMIT STATE

Combination from ESCP -1/1983

Cases: LL, DL & WL

Pd= 0.8(1.3DL+1.6LL+1.6WL)

Live load = 0.254KN/m+

= 1.24KN/m

Dead load=0.054KN/m

Therefore Pd =0.8{ 1.3(0.054)+1.6(1.24)+1.6(1.003)}

Pd=0.36 KN/M

Md= pdl2/8=0.07 KN.m

Md=

Load combination for ultimate limit state from ESCP.

Cases: LL, DL and WL.

Where LL= live load.

DL= dead load.

WL = wind load.

Pd = 0.8 {1.3DL + 1.6 LL + 1.6 WL}

 = 0.8 {1.3(0.04)+1.6(0.2+0.8)+1.6(0.773)}

= 2.3 KN/m

2.3 KN/m

1.25m
A B

2.30(1.25) = RA + RB.

RA = RB = 1.44KN.

Md = =

Check for flexure

Section property of EGA 500

Section modulus = 3756 mm2.

Moment of inertia =79972mm4.

Flexure (md) = 0.45KN.m.

Fbx =

fbx should be less than flexure allowable stress = 0.6 fyk.

Deflection

Allowable deflection all = = = 6.25 m, from EBCS – 1995/p105.

 ∆actual =

∆actual ˂ ∆allowable

Design of purlin.
Trial section 60*60*3 (a

1.44KN/2.2 m =0.65 KN/m.

DL = (0.65KN/m +0.04 KN/m) = 0.69KN/m

LL = 0.25 KN/m * 0.06 +1 KN/2.2 = 0.47 KN/m.

WL = 0.775 KN/m2 * 0.06 = 0.05KN/m.

Pd = 0.8 {1.3(0.69) + 1.6(0.47) – 1.6(0.05)} = 1.26 KN/m.

Θ1 = 42.482 Θ2 = 47.518

Θ3 = 47.14 Θ4 = 42.86

Θ5 = 52.62 Θ6 = 37.38

Θ7 =59.03 Θ8 = 30.06

Θ9 = 66.44 Θ10 =23.56

Θ11 =74.78 Θ12 = 15.22

Θ13 = 83.83 Θ14 =6.164

Load = 1.5cm *50cm*60cm

Load = 0.015*0.6*0.6*8 =0.0432KN

UDL=0.00235KN/m.
 CHAPTER TWO

2. 1 ANALYSIS AND DESIGN OF SOLID SLAB

Slabs are structural flat plates used to provide useful horizontal surface mainly for
roofs and floors of buildings and other structures. Based on load transferring mechanisms
there are two types of slabs. These are one way and two way slabs.

Slabs are composed of rectangular panels supported at all four edges by walls or beams
that are stiff enough to be treated as unyielding. In our case most of the slabs are two way
and need to be analysed based on the principle of two way actions. General procedure to
design two way slab system are described below.

DEPTH FOR DEFLECTION

The minimum depth of a slab for deflection requirement is computed by:

d ≥ (0.4+0.6fyk/400)le/βa

Where: le: effective length of the slab

fyk: characteristic tensile strength of reinforcement

LOAD DETERMINATITION

Dead and live loads are calculated depending on the service of the slabs and self weight.
Partition loads are distributed over the slab if they are not large enough to cause localized
effects. The design loads are factored according to the following formula that is given in
EBCS code.
Pd =1.3DL+1.6LL where: Pd=design load

DL=total dead load on slab

LL=Live load

ANALYSIS
The analysis of slab moments of two way slabs is accomplished by analysing individual
panel’s bending moments and shear forces. It accomplished by the formula:-

Mi =αipdLx2

Where: Mi=the design moment per unit width at the point of reference

αi= the coefficient given in table A-1 in EBCS-2,1995.

Pd = the design load

Lx= the shorter span of the panel

Ly= the longer span of the panel

The following symbols stand for:

S=support

f =span

x =direction of shorter span

y =direction of longer span

MOMENT ADJUSTMENTS

♦ Support adjustment
 For continuous support there will be two moments which are different in magnitude.
These moments are usually different in magnitudes and must be adjusted to make only one
moment.

If ∆M<20% of the larger moment, the design moment is the average of two.

If ∆M>20% then the unbalanced moment is distributed based on the stiffness.

Let MS - moment in the smaller

ML - moment in the larger

K - Stiffness of the slabs

Therefore the design moment Md can be calculated in the either of the following formula

Md = x ∆M, taking the larger; OR Where K=I/LX

Md=MS+KS/ (KL+KS) x ∆M, taking the smaller

♦ Span adjustment
If the moment in the adjusted support decreases, the span moment are increased to
compensate for the changes in the support moments. The design moments for the spans are
calculated as shown below.

MXd=Mxf +Cx ∆M Where ∆M=the change in moment in all supports.

Myd=Myf +CY ∆M

Cx and CY are coefficients for adjusting span moments in EBCS-2,table A

LOAD TRANSFER TO BEAM

Finally the design uniform loads are transferred to beam as shear. The shear is
calculated using the formulas from EBCS-2, 1995 as given below.

Vx=βVX*Pd*Lx

Vy=βVy*Pd*Ly

Where; βVX & βVy are load transfer coefficients given in EBCS-2, 1995 of table-3

Basic design constants

 ♦ Concrete

Unit weight = 24KN/m3

Strength C -25, class-1 work

Partial safety factor, δc = 1.5

♦ Concrete strength

fck = 0.8fcu = 0.8 25 = 20MPa

fctk = 0.21×fck 2/3

♦ Design strength

fcd = 0.85×(fck/ δc) = 0.85 x (20/1.5)= 11.33MPa

♦ Steel

Steel grade S-300

Partial safety factor, fc = 1.25, class 1-work

Characteristic strength

Yield strength, fyk =300MPa

Es= 200MPa

Design strength = fyk/δc = 300/1.15 = 260.87 MPa

The analysis and design of a 4 storey building located in side the campus of jimma university is
conducted in two separate parts. The building is composed of two separate buildings interconnected
by a stair. The analysis and design of these buildings is as shown below.

ANALYSIS AND DESIGN

PART -I

Design of 1st floor

Depth determination for deflection requirement

dmin= (0.4 + )( ) , where le ( shorter length)

d>= 0.85*le/Ba where le=Lx

Here the depth of panel 5 and pane 15 seem to be governing but designing with this depth will make
our design somewhat uneconomical, so we have to provide the maximum depth for these panels and
make the depth of other panels similar and the one which is maximum except the depth of the above
mentioned panels.

Table 1 depth determination

panel Ly(cm) Lx(cm) Ly/Lx Ba d min(m) d(mm)

S1 690 535 1.29 33.55 13.55 135.54

S2 690 420 1.64 31.8 11.23 112.26

S3 690 480 1.44 32.8 12.44 124.39

S4 690 420 1.64 31.8 11.23 112.26

S5 690 650 1.06 34.7 15.92 159.22

S6 460 535 0.86 35.7 10.95 109.52

S7 460 420 1.1 44 8.11 81.14

S8 460 480 0.96 45.4 8.61 86.12

S9 460 420 1.1 44 8.11 81.14

S10 460 650 0.71 36.45 10.73 107.27

S11 690 535 1.29 33.55 13.55 135.54

S12 690 420 1.64 38.6 9.25 92.49

S13 690 480 1.44 40.6 10.05 100.49

S14 690 420 1.64 38.6 9.25 92.49

S15 690 650 1.06 34.7 15.92 159.22

S16 630 115 5.48 18.96 5.16 51.56

S17 455 160 2.84 13.68 9.94 99.42

S18 1970 130 15.15 38.3 2.89 28.85

S19 1970 130 15.15 38.3 2.89 28.85

For panels 5 and 15

D=159.22 + 15 + 12/2 =181mm

Use D=200mm, d=200 – 15 – 12/2 =179mm

For other panels

D=135.54 + 15 + 12/2 =156.54mm

Use D=170mm, d=170 – 15 – 12/2 =149mm

2. LOADING

A. DEAD LOAD

1. Marble = 0.03 * 27 = 0.81KN/m

2. PVC tile =0.02 * 16 = 0.032KN/m

3. Cement screed = (0.03 * 20)*2 = 1.2KN/m

4. Partition wall = ((0.15*14) + (0.05*20))*3.6 /area =11.16/area

5. External wall = ((0.25*14) + (0.05*20))*3.6 /area =13.68/area

B. IMPOSED LOAD

From EBCS-2, 1995 page 46 the category for class room is C2 and for this category
the imposed load for class room and veranda is respectively 3& 4KN/m 2

Design load, p = 1.3 GK + 1.6qk

d
 Table 2-Summarized loadings on each panel is presented as follows

panel selfwt Cement partition PVC marble sum(DL) LL PD

screed load(KN/m) tile
(KN/m2) (KN/m2)
S1 5 1.2 0.81 7.01 4 15.513

S2 5 1.2 0.064 6.264 3 12.943

S3 5 1.2 2.19 0.064 8.454 3 15.79

S4 5 1.2 0.064 6.264 3 12.943

S5 5 1.2 0.064 6.264 3 12.943

S6 5 1.2 0.81 7.01 4 15.513

S7 5 1.2 0.578 0.81 7.588 4 16.264

S8 5 1.2 2.397 0.81 9.407 4 18.629

S9 5 1.2 0.578 0.81 7.588 4 16.264

S10 5 1.2 0.578 0.81 7.588 4 16.264

S11 - - - - 0 0

S12 5 1.2 0.064 6.264 3 12.943

S13 5 1.2 2.19 0.064 8.454 3 15.79

S14 5 1.2 0.064 6.264 3 12.943

S15 5 1.2 0.064 6.264 3 12.943

S16 5 1.2 0.81 7.01 4 15.513

S17 5 1.2 0.81 7.01 4 15.513

S18 5 1.2 0.064 6.264 3 12.94 point

load=13.68

3. DETERMINATION OF DESIGN INTERNAL FORCES

For a slab with Ly/Lx <2 and supported by beam, we use coefficient method which is specified on

EBCS 2, 1995 for flexure at span and support.

Types of support for coefficient of moment calculation

Fig. type of support condition

 panel support Ly/Lx PD(KN/m^2/m) Lx^2(m^2) αxs αys αxf αyf Mxs Mys Mxf Myf remark
Ta
bl condition e
3
S1 1 1.29 13.856 28.623 0.046 0.032 0.035 0.024 18.085 12.691 13.762 9.518

S2 1 1.64 11.286 17.64 0.056 0.032 0.042 0.024 11.149 6.371 8.362 4.778

S3 1 1.44 14.133 23.04 0.051 0.032 0.038 0.024 16.607 10.42 12.374 7.815

S4 1 1.64 11.286 17.64 0.056 0.032 0.042 0.024 11.149 6.371 8.362 4.778

S5 2 1.06 12.943 42.25 0.041 0.039 0.031 0.029 22.421 21.327 16.952 15.858

S6 1 0.86 13.856 28.623 0.025 0.039 0.018 0.029 9.915 15.467 7.139 11.501

S7 1 1.1 14.607 17.64 0.037 0.032 0.028 0.024 9.534 8.245 7.215 6.184

S8 1 0.96 16.972 23.04 0.03 0.032 0.022 0.024 11.731 12.513 8.603 9.385

S9 1 1.1 14.607 17.64 0.037 0.032 0.028 0.02 9.534 8.245 7.215 6.184

S10 2 0.71 14.607 42.25 0.025 0.039 0.017 0.029 15.429 24.069 10.491 17.897

S11 3 1.29 0 28.623 0.046 0.047 0.035 0.036 0 0 0 0

S12 1 1.64 11.286 17.64 0.056 0.039 0.042 0.029 11.149 7.764 8.362 5.773

S13 1 1.44 14.133 23.04 0.051 0.032 0.038 0.024 16.607 10.42 12.374 7.815

S14 1 1.64 11.286 17.64 0.056 0.032 0.042 0.024 11.149 6.371 8.362 4.778

S15 1 1.06 12.943 42.25 0.042 0.039 0.031 0.029 22.967 21.327 16.952 15.858

S16 4 13.856 10.258 M=Wl^2/2

S17 4 13.856 13.11 "

S18 4 pt=13.68,W=12.94 28.72 "

summarized value of support and span moments

Support moment adjustment

Length

115 535 420 480 420 650

K 0.828 0.172 0.238 0.208 0.238 0.154

DF 0.432 0.568 0.534 0.446 0.554 0.238 0.607 0.393

ΔM 6.936 5.458 5.458 11.272

Adjusted M 15.089 14.064 14.064 17.991

Length 535 420 480 420 650

K 0.172 0.238 0.208 0.238 0.154

DF 0.432 0.568 0.534 0.446 0.554 0.238 0.607 0.393

ΔM 5.993 2.979 2.979 14.535

Adjusted M 12.904 11.125 11.125 18.357

Length 690 460 690

K 0.145 0.217 0.145

DF 0.4 0.6 0.6 0,4

ΔM 5.898 5.898

Adjusted M 18.968 18.968

Coefficients for shear calculation

SPAN MOMENT ADJUSTMENT

Table_____ Adjusted span moment

Panel P.to be Moment Ly/ M Cx Cy Mxf Myf
adjusted condition Lx
(Mxf+M+Cx) (Myf+M+Cy)
1&2 1 1.29 2.996 0.370 0.351 14.871 10.570

273 3 1.44 2.543 0.407 0.323 13.409 8.636

3&4 3 1.44 2.543 0.407 0.323 14.444 9.457

4&5 5 1.06 4.430 0.300 0.376 18.281 17.524

6&7 6 0.86 2.563 0.414 0.364 8.200 12.434

7&8 8 0.96 1.338 0.389 0.304 9.123 9.792

8&9 8 0.96 1.338 0.389 0.304 9.643 10.199

9&10 10 0.71 5.712 0.449 0.454 13.056 20.490

5&10 5 1.06 2.359 0.366 0.244 19.144 18.099

10&15 15 1.06 2.359 0.366 0.244 17.815 16.444

 Fig. Adjusted moment

Check depth for flexure

Design strengths and constants

Design strengths and constants

max = , cu= 0.0035

yd = ,E=200GPa

= =1.304*10-3
 max =

=0.019

m= = =28.78

dmin =

d=76.029mm < d used OK!

Shear force calculation ……… …EBCS-2, 1995 page 114

Shear force transferred from the structure.

Vmax=66.743KN
Vc = 0.25*fctd*K1*1K2*bw*d

fctd = 1.031 , K1=1+50ρ

ρmin = (0.5/fyk) =0.5/300 = 1.67*

K2 = 1.6 – d = 1.6 -0.149 = 1.451m

Vc = 0.25*1.031*1.083 *1.451*1000*149

=60.35KN < Vmax = 66.743 KN…….NOT OK

Hence we have to provide shear reinforcement.

Vs=Vsd-Vc=66.743-60.35=6.393KN

S=Av*Fyd*d/Vs

= (2*π82 *260.87*149)/ (4*6.393*103) =4.1mm

Use minimum shear reinforcement, 6c/c 200mm

REINFORCEMENT DESIGN FOR FLEXURE

Depth for calculation of reinforcement for panel 5&15

-ds=200-15-6 = 179mm, de =200-15-12-6 =167mm

Depth for calculation of reinforcement for other panels

-ds=170-15-6 =149,dl=170-15-12-6=137mm

Provision of reinforcement in X direction

position Mmt in X- depth µ ẁ ₰ As S As

direction calculated required provided

S1 13.762 149 0.0547 0.0563 0.00245 364.31 310.45 ᴓ12C/C310

S2 8.362 149 0.0332 0.0338 0.00147 218.83 516.84 ᴓ12C/C350

S3 12.373 149 0.0492 0.0505 0.00219 326.56 346.34 ᴓ12C/C350

S4 8.362 179 0.023 0.0233 0.00101 181.19 624.22 ᴓ12C/C350

S5 16.952 149 0.0674 0.0698 0.00303 451.9 250.27 ᴓ12C/C250

S6 11.401 137 0.0536 0.0551 0.00239 328.05 344.77 ᴓ12C/C345

S7 7.125 149 0.0283 0.0287 0.00125 185.98 608.14 ᴓ12C/C350

S8 9.385 137 0.0441 0.0452 0.00196 268.66 420.97 ᴓ12C/C350

S9 7.125 149 0.0283 0.0287 0.00125 185.98 608.14 ᴓ12C/C350

S10 17.897 137 0.0842 0.088 0.00382 523.82 215.91 ᴓ12C/C215

S11 179 0 0 0 0

S12 8.362 149 0.0332 0.0338 0.00147 218.83 516.84 ᴓ12C/C350

S13 12.373 149 0.0492 0.0505 0.00219 326.56 346.34 ᴓ12C/C350

S14 8.362 149 0.0332 0.0338 0.00147 218.83 516.84 ᴓ12C/C350

S15 16.952 179 0.0467 0.0478 0.00208 371.93 304.09 ᴓ12C/C300

S16&S1 10.258 149 0.0408 0.0416 0.00181 269.52 419.63 ᴓ12C/C350

S1&S2 15.089 149 0.06 0.0619 0.00269 400.59 282.33 ᴓ12C/C280

S2&S3 14.064 149 0.0559 0.0576 0.0025 372.55 303.58 ᴓ12C/C300

S3&S4 14.064 149 0.0559 0.0576 0.0025 372.55 303.58 ᴓ12C/C300

S4&S5 17.991 149 0.0715 0.0743 0.00323 480.71 235.28 ᴓ12C/C235

S6&S7 12.904 149 0.0513 0.0527 0.00229 340.96 331.71 ᴓ12C/C330

S7&S8 11.125 149 0.0442 0.0453 0.00197 292.84 386.22 ᴓ12C/C350

S8&S9 11.125 149 0.0442 0.0453 0.00197 292.84 386.22 ᴓ12C/C350

S9&S10 18.375 149 0.0731 0.0759 0.0033 491.39 230.16 ᴓ12C/C230

S12&S13 14.064 149 0.0559 0.0576 0.0025 372.55 303.58 ᴓ12C/C300

S13&S14 14.064 149 0.0559 0.0576 0.0025 372.55 303.58 ᴓ12C/C300

S14&S15 17.991 149 0.0715 0.0743 0.00323 480.71 235.28 ᴓ12C/C235

Provision of reinforcement in Y direction

position Mmt in y- depth µ ẁ ₰ As S As

 direction calculated required provided

S1 9.518 137 0.0448 0.0458 0.002 272.56 414.95 ᴓ12C/C350

S2 4.778 137 0.0225 0.0227 0.001 135.23 836.37 ᴓ12C/C350

S3 7.815 137 0.0368 0.0375 0.0016 222.84 507.54 ᴓ12C/C350

S4 4.778 137 0.0225 0.0227 0.001 135.23 836.37 ᴓ12C/C350

S5 15.859 167 0.0502 0.0515 0.0022 373.65 302.69 ᴓ12C/C300

S6 7.139 179 0.0197 0.0199 0.0009 154.42 732.43 ᴓ12C/C350

S7 6.184 137 0.0291 0.0295 0.0013 175.62 643.99 ᴓ12C/C350

S8 8.603 179 0.0237 0.024 0.001 186.47 606.53 ᴓ12C/C350

S9 6.134 137 0.0288 0.0293 0.0013 174.18 649.32 ᴓ12C/C350

S10 10.491 179 0.0289 0.0293 0.0013 228.01 496.03 ᴓ12C/C350

S11

S12 5.773 137 0.0271 0.0275 0.0012 163.79 690.54 ᴓ12C/C350

S13 7.815 137 0.0368 0.0375 0.0016 222.84 507.54 ᴓ12C/C350

S14 4.778 137 0.0225 0.0227 0.001 135.23 836.37 ᴓ12C/C350

S15 15.859 167 0.0502 0.0515 0.0022 373.65 302.69 ᴓ12C/C300

S17&S1 13.11 149 0.0521 0.0536 0.0023 346.56 326.35 ᴓ12C/C325

S17&S2 28.72 149 0.1142 0.1216 0.0053 786.7 143.77 ᴓ12C/C140

S17&S3 28.72 149 0.1142 0.1216 0.0053 786.7 143.77 ᴓ12C/C140

S17&S4 28.72 149 0.1142 0.1216 0.0053 786.7 143.77 ᴓ12C/C140

S17&S5 28.72 149 0.1142 0.1216 0.0053 786.7 143.77 ᴓ12C/C140

S1&S6 12.699 149 0.0505 0.0518 0.0023 335.4 337.21 ᴓ12C/C335

S2&S7 7.308 149 0.0291 0.0295 0.0013 190.83 592.68 ᴓ12C/C350

S3&S8 11.076 149 0.044 0.045 0.002 291.52 387.97 ᴓ12C/C350

S4&S9 7.308 149 0.0291 0.0295 0.0013 190.83 592.68 ᴓ12C/C350

S5&S10 18.968 149 0.0754 0.0785 0.0034 507.92 222.67 ᴓ12C/C220

S7&S12 7.308 149 0.0291 0.0295 0.0013 190.83 592.68 ᴓ12C/C350

S8&S13 11.076 149 0.044 0.045 0.002 291.52 387.97 ᴓ12C/C350

S9&S14 7.078 149 0.0281 0.0285 0.0012 184.73 612.24 ᴓ12C/C350

S10&S15 18.968 149 0.0754 0.0785 0.0034 507.92 222.67 ᴓ12C/C220

S6&S11 12.699 149 0.0505 0.0518 0.0023 335.4 337.21 ᴓ12C/C335

S7&S12 7.308 149 0.0291 0.0295 0.0013 190.83 592.68 ᴓ12C/C350

S8&S13 11.076 149 0.044 0.045 0.002 291.52 387.97 ᴓ12C/C350

S9&S14 7.308 149 0.0291 0.0295 0.0013 190.83 592.68 ᴓ12C/C350

S10&S15 18.968 149 0.0754 0.0785 0.0034 507.92 222.67 ᴓ12C/C220

SOLID SLAB DETAILING
 DESIGN AND ANALYSIS OF PART TWO

Design of 1st floor.

Depth determination for deflection requirement

dmin= (0.4 + )( ) , where le ( shorter length)

d>= 0.85*le/Ba where le=Lx.

Table 1.2 depth determination

panel Ly(cm) Lx(cm) Ly/Lx Ba d min(m) d(mm)

P1 560 460 1.217 37.826 103.368

P2 560 460 1.217 37.826 103.368

P3 560 460 1.217 37.826 103.368

P4 690 420 1.6428 33.57 106.345

P5 460 420 1.09 44.1 80.95

P6 690 420 1.6428 33.57 106.345

P7 690 480 1.438 35.62 114.54

P8 480 460 1.043 44.57 87.73

P9 690 480 1.438 35.62 114.54

P10 690 420 1.6428 33.57 106.345

P11 460 420 1.095 44.05 81.04

P12 690 420 1.6428 33.57 106.345

P13 690 560 1.232 37.68 126.33

Canti.1 570 150 3.8 40 127.5

Canti.2 790 155 5.097 40 131.75

Canti.3 560 377 1.485 10.97 292.114

Governing Effective Depth for all panels and cantilever 1 & 2 = 131.75mm.

i.e. D = 131.75 + 15+ 6=152.75mm. take D= 180mm.

 for cantilever 3, d= 292.144mm.

D = 292.144 + 15 + 12 =319.144mm, Take D =350mm

Since D is sufficiently large, we have taken the cantilever x-section as a trapezoid as shown in the figure
below.

350m
180mm
m

4. LOADING

A. DEAD LOAD

1. Marble = 0.03 * 27 = 0.81KN/m

2. PVC tile =0.02 * 16 = 0.032KN/m

3. Cement screed = (0.03 * 20)*2 = 1.2KN/m

4. Partition wall = ((0.15*14) + (0.05*20))*3.6 /area =11.16/area

5. External wall = ((0.25*14) + (0.05*20))*3.6 /area =13.68/area

B. IMPOSED LOAD

From EBCS-2, 1995 page 46 the category for class room is C2 and for this category the
imposed load for class room and veranda is respectively 3& 4KN/m2

Design load, p = 1.3 GK + 1.6qk

d

Table 2.2 -Summarized loadings on each panel is presented as follows

panel selfwt Floor finish partition load(KN/m) sum(DL) LL PD
(Ceramic/PVC/Marble)
(KN/m2) + plastering (KN/m2)
KN/m2
P1 2.975 0.42 8.432 11.827 3.155 20.423
P2 3.23 0.302 2.07 5.602 3.22 12.432
P3 4.5 1.67 - 6.17 3.16 13.07
P4 4.5 1.18 - 5.68 3.14 12.408
P5 4.5 4.045 1.66 10.205 3.21 18.4
P6 4.5 1.18 - 5.68 3.14 12.408
P7 4.5 3.33 - 7.83 3.12 15.171
P8 4.5 4.045 4.418 12.963 3.18 21.94
P9 4.5 3.33 - 7.83 3.12 15.171
P10 4.5 1.18 - 5.68 3.14 12.408

panel selfwt Floor finish partition load(KN/m) sum(DL) LL PD

(Ceramic/PVC/Marble)
(KN/m2) + plastering (KN/m2)
KN/m2
P11 4.5 4.045 1.66 10.205 3.21 18.4
P12 4.5 1.18 - 5.68 3.14 12.408
P13

Cant.1 4.5 0.72 1.44 6.265 3.47 13.697

Cant.2 4.5 0.32 3.179 7.99 3.3 15.667

Cant.3 4.5

5. DETERMINATION OF DESIGN INTERNAL FORCES

For a two way slab with Ly/Lx <2 and supported by beam, we use coefficient method which is specified
on EBCS 2, 1995 for flexure at span and support.Types of support conditions encountered for the
determination of moment.

Fig. type of support condition

 Calculation of moment for cantilever 3.

Pd = 17.7 – 1.46x.

Pd @ x= 0 =17.7 KN

Pd @x=3.77=12.19 KN.

M =(3.77*12.19*3.77/2) + (0.5*5.51*3.77*3.72/3)

M = 99.7 KN.m.

 Calculation of moment for cantilever 2.

 Calculation of moment for cantilever 1.

REINFORCEMENT DESIGN FOR FLEXURE

Depth for calculation of reinforcement for all panels and cantilever 1 and 2.

Governing d=131.75.

D = 131.75 + 15+ 6=152.75mm. take D= 180mm.

Depth for calculation of reinforcement for other panels

-ds=180-15-6 =159mm.

-dl=180-15-12-6=147mm

Provision of reinforcement in X direction

Position Mmt in X- depth µ ẁ ₰ As S As
direction calculated required provided

s1 26.63 107 0.00447 0.0669 0.0101 1077 105 Ø12C/C 120

s2 9.29 107 0.00649 0.0805 0.0032 345.1 327.7 Ø12C/C 280

s3 12.95 107 0.00606 0.0779 0.0046 488.9 231.3 Ø12C/C 240

s4 10.58 119 0.00657 0.0811 0.003 352.3 321 Ø12C/C 280

s5 6.536 119 0.00695 0.0834 0.0018 214.7 526.8 Ø12C/C 280

s6 10.58 119 0.00657 0.0811 0.003 352.3 321 Ø12C/C 280

s7 12.16 119 0.00643 0.0802 0.0034 407.2 277.7 Ø12C/C 270

s8 6.768 107 0.00678 0.0824 0.0023 248.7 454.7 Ø12C/C 280

s9 12.16 119 0.00643 0.0802 0.0034 407.2 277.7 Ø12C/C 270

s10 10.58 107 0.00634 0.0796 0.0037 395.2 286.1 Ø12C/C 280

Position Mmt in X- depth µ ẁ ₰ As S As

direction calculated required provided

s11 6.536 107 0.00681 0.0825 0.0022 240 471.3 Ø12C/C 280

s12 10.58 119 0.00657 0.0811 0.003 352.3 321 Ø12C/C 280

S1 & Cant 36.6 119 0.00413 0.0642 0.0114 1354 83.5 Ø12C/C 80

S2 & Cant 36.6 119 0.00413 0.0642 0.0114 1354 83.5 Ø12C/C 80

Ø12C/C
S3 & S5 13.08 119 0.00634 0.0796 0.0037 439.4 257.4 250

Ø12C/C
S4 & S7 15.69 119 0.00609 0.0781 0.0045 532 212.6 210

Ø12C/C
S5 & S8 9.02 119 0.00672 0.082 0.0025 298.7 378.6 280

Ø12C/C
S6 & S9 15.69 119 0.00609 0.0781 0.0045 532 212.6 210

Ø12C/C
S7 & S10 15.69 119 0.00609 0.0781 0.0045 532 212.6 210

S8 & S 11 8.64 119 0.00676 0.0822 0.0024 285.8 395.7 Ø12C/C 280

Ø12C/C
S9 &S12 15.69 119 0.00609 0.0781 0.0045 532 212.6 210

Provision of reinforcement in Y direction.

Position Mmt in Y- depth µ ẁ ₰ As S As
direction calculated required provided
 Ø12C/C
s1 24.07 119 0.0053 0.0728 0.0071 842.9 134.2 130

Ø12C/C
s2 5.25 119 0.0071 0.0841 0.0014 171.7 658.7 280

Ø12C/C
s3 19.43 119 0.0057 0.0758 0.0056 668.2 169.3 160

Ø12C/C
s4 6.818 107 0.0068 0.0823 0.0023 250.6 451.2 280

Ø12C/C
s5 5.64 107 0.0069 0.0831 0.0019 206.3 548.2 280

s6 6.818 107 0.0068 0.0823 0.0023 250.6 451.2 Ø12C/C 280

Position Mmt in Y- depth µ ẁ ₰ As S As

direction calculated required provided
Ø12C/C
s7 7.369 107 0.0067 0.0819 0.0025 271.5 416.5 280

Ø12C/C
s8 7.248 119 0.0069 0.0830 0.0020 238.6 474.0 280

Ø12C/C
s9 7.369 107 0.0067 0.0819 0.0025 271.5 416.5 280

Ø12C/C
s10 6.818 119 0.0069 0.0832 0.0019 224.1 504.6 280

Ø12C/C
s11 5.64 119 0.0070 0.0839 0.0016 184.7 612.3 280

Ø12C/C
s12 6.818 107 0.0068 0.0823 0.0023 250.6 451.2 280

Ø12C/C
S1 & S2 20.26 119 0.0057 0.0753 0.0059 698.8 161.8 160

S4 & Cant 31.86 119 0.0046 0.0676 0.0097 1153.2 98.1 Ø12C/C 95

Ø12C/C
S4 & S5 7.52 119 0.0069 0.0828 0.0021 247.8 456.4 280

Ø12C/C
S5 & S6 9.17 119 0.0067 0.0819 0.0026 303.9 372.2 280

S6& Cant 31.86 119 0.0046 0.0676 0.0097 1153.2 98.1 Ø12C/C 95
S7 & Cant 31.86 119 0.0046 0.0676 0.0097 1153.2 98.1 Ø12C/C 95

Ø12C/C
S7 & S8 9.82 119 0.0066 0.0815 0.0027 326.1 346.8 280

Ø12C/C
S8 & S9 9.82 119 0.0066 0.0815 0.0027 326.1 346.8 280

S9 & Cant 31.86 119 0.0046 0.0676 0.0097 1153.2 98.1 Ø12C/C 95

CHECK FOR SHEAR

Vmax=89.59KNm
Vc=0.25fctdK1 K2 bw d

=0.25*1.032*1.64*1.44*1000*159

=96.87KNm

Vmax < Vc OK!

REINFORCEMENT DETAILING FOR THE SLAB

IN THE X- DIRECTION
IN THE Y- DIRECTION
 2.2 ANALYSIS AND DESIGN OF2ND, 3rd AND 4th FLOOR SLAB

RIBBED SLAB
INTRODUCTION
The functions of the floors are mainly for classrooms with live load of 3KN/m 2 and corridors
with live load of 4KN/m2. All slabs are designed as ribbed slab rather than solid slab as the live
load is moderately light and the span are somewhat long. This makes the design to be more
economical.
Most of the joists are arranged in the shorter direction of panels, but some joists are arranged in
longer direction to preserve the uniformity of joist arrangement. Since it is difficult to maintain
the outer cantilever of the slab by ribs, it can be maintained by solid slab.

Fig. Grouping of panel and direction of ribs

ANALYSIS AND DESIGN OF JOISTS

It is clearly shown in fig above that the length of the joists in the panels have variable length.
There are many joists which need separate analysis and design but this is very difficult task.
Therefore; it is decided to analyze one joist from each panel hav ing different dimensions. The
reinforcement is provided at support and middle (span of joists). This makes the design safer.
The analyses of joists are done manually. The bending moment diagram and shear force diagram
are shown in each joist design. The detail procedure for the design of joists is shown at the first
joist design and the remaining joists are shown in table
Table Determination of depth of joists for deflection requirement

Joist Span length(mm) βa le/βa Dmin=(0.4+0.6*fyk/400)*le/βa

1 5350 24 222.917 189.479
2 4200 28 150.000 127.500
3 4800 28 171.429 145.714
5 6500 24 270.830 230.208

Panel 5 is the governing panel for minimum depth from deflection requirement.
dmin = dj = 230.208
Assume 14 reinforcement for the joist and 15mm for cover.
Dj=230.208+15+14/2=252.208mm
Use Dj=260mm,
dj=260-15-14/2=238mm

Assume the hollow block dimension is 500*230mm as shown in the figure below.

Figure 2.15 hollow block dimension and topping

A=A1+3A2+2A3
A=0.02*0.5+3*(0.21*0.04) +2*(0.14*0.04) =0.0464m2
Requirements from EBCS-2 1995 pp 84

Thickness of slab (topping) Dt

Rib width bw 70mm

Over all depth Dj 4*bw+ thickness of topping (Dt)

Rib spacing is generally less than 1m.

Determination of thickness of topping

Dt

Dt=50mm

Dj = 210+50 = 260mm→dj = 260-15-14/2 = 238mm

Check for the requirement

Dj 4bw+ Dt = 4*100+50=450mm>260…..OK!
Total floor thickness=Dj+20mm=260+20=280mm
LOADING
SLAB DEAD LOAD CALCULATION
1. Class room

0.2cm pvc f.f (γ=16KN/m3)

4.8cm cement screed (γ =23KN/m3)

RC floor slab

3cm cement plastering (γ =23KN/m3)

2. Corridor

3cm marble tiles f.f (γ=27KN/m3)

2cm cement screed (γ =23KN/m3)

RC floor slab

3cm cement plastering (γ =23KN/m3)

Dead load from:

Pvc floor finish = 0.002*16*0.5 = 0.016KN/m
Marble tiles=0.03*0.5*27=0.405KN/m
Cement screed = 0.048*23*0.5 =0.552KN/m
Cement screed=0.02*0.5*23=0.23KN/m
Topping = 0.05*25*0.5 = 0.625KN/m
Load from HCB = γ*A = 14*0.0464 = 0.65KN/m,
Joist self-weight = bw *(Dj-Dt)* γ= 0.1*(0.26-0.05)*25 = 0.525KN/m
Plastering = 0.03*23 = 0.69 KN/m
Total slab dead load for class room = 3.058 KN/m
Total slab dead load for corridor=3.025KN/m
Therefore take total slab dead load=3.058KN/m
SLAB LIVE LOAD CALCULATION
For class room from EBCS 2, 1995 q=3KN/m2, Q=4KN
Therefore LL=3 + 4/A
For corridor from EBCS 2, 1995 q=4KN/m2, Q=4KN
Therefore LL=4 + 4/A
The detailed calculation is presented by the table below.

Table live load calculation for each panel

panel S1 S2 S3 S4 S5 S6 S7 S8 S9 S10
2)
Area(m 34.51 26.80 30.82 26.80 42.21 22.66 17.60 20.24 17.60 27.72
LL(KN/m2) 3.12 3.15 3.13 3.15 3.09 4.18 4.23 4.20 4.23 4.14
LL(KN/m) 1.560 1.575 1.565 1.575 1.545 2.299 2.221 2.310 2.221 2.103

LOAD FROM PARTITION WALL

Partition load for each panel is shown in table___ and the sample calculation is show for the
third panel
Panel 3
Thickness of wall (t) =0.2m
Height of wall (h) =3.6m ,Length (l) =6.9m
Unit weight (γ) =17KN/m3
Area (A) =30.82m2
HCB wall
Areal load= (t*h*l* γ)/A= (0.2*3.6*6.9*17)/30.82=2.74KN/m2
Line load =2.74*0.5=1.37KN/m
Plaster
t=0.03*2=0.06m ,h=3.6m ,l=6.9m ,γ=23KN/m3
Areal load= (0.06*3.6*6.9*23)/30.82=1.112KN/m2
Line load=1.112*0.5=0.556KN/m
Total partition load =1.37 +0.556=1.926KN/m
Table___ calculated partition load

panel S1 S2 S3 S4 S5 S6 S7 S8 S9 S10
Par.load(KN/m) - - 1.926 - - - 4.107 3.664 4.107 2.617
Total design dead load =slab dead load + partition load
Table_____ total design dead load

panel S1,S2,S4,S5,S6 S3 S7,S9 S8 S10

DL(KN/m) - 1.926 4.107 3.664 2.617

LL(KN/m) 3.058 3.058 3.058 3.058 3.058

Pd(KN/m) 3.058 4.984 7.165 6.722 5.675

Design load(Pd)=1.3DL + 1.6LL

Table___calculation of design load

panel S1 S2 S3 S4 S5 S6 S7 S8 S9 S10
DL(KN/m) 3.058 3.058 4.984 3.058 3.058 3.058 7.165 6.722 7.165 5.675
LL(KN/m) 1.560 1.575 1.565 1.575 1.545 2.299 2.221 2.310 2.221 2.103
Pd(KN/m) 6.471 6.495 9.129 6.495 6.447 7.654 12.868 12.235 12.868 10.742

Analysis
The analysis for each joist is done manually and the sample calculation is shown for the first
panel as follows.
Panel 1
Pd=6.471KN/m,L=4.95m
Table______Calculated internal forces

panel Pd(KN/m) L(m) Msupport(KNm) Mspan(KNm) Vmax(KN) Transferred

load
1 6.471 4.95 14.414 9.9097 16.0157
2 6.495 3.8 8.5262 5.8617 12.3405
3 9.129 4.4 16.067 11.046 20.0838
4 6.495 3.8 8.5262 5.8617 12.3405
5 6.447 6.1 21.808 14.993 19.6633
6 7.654 4.95 17.049 11.721 18.9436
7 12.868 3.8 16.892 11.613 24.4492
panel Pd(KN/m) L(m) Msupport(KNm) Mspan(KNm) Vmax(KN) Transferred
load
8 12.235 4.4 21.534 14.804 26.9170
9 12.868 3.8 16.892 11.613 24.4492
10 10.742 6.1 36.337 24.982 32.7631
11 6.495 3.8 8.5262 5.8617 12.3405
12 9.129 4.4 16.067 11.046 20.0838
13 6.495 3.8 8.5262 5.8617 12.3405
14 6.447 6.1 21.808 14.993 19.66335

MA=MB=Pd L2/11=6.471*4.552/11=14.414KNm
Mspan= Pd L2/16=6.471*4.552/16=9.91KNm
VA=VB=P d L/2=6.471*4.55/2=16.016KN

Design for flexure (joist design)

Design constants

Fcd =11.33MPa fyd = 260.87MPa m = 28.78

For simplicity of work, both the support and the span moments which have approximate values are
grouped and designed on the maximum moment.
Mmax=36.337KN.m

dmin= = =141.24.5mm < 238…..ok!

i. Support reinforcement

For support Mmax = 36.337KN.m

ρ= *
 = 0.003524

A(area) = ρbd = 0.003524*100*238 = 419.386mm2

Use 3 14 bars → (A=462mm2)

ii. Span reinforcement Mmax = 24.982KN.m

→ be le-effective length of beam

be =1482mm

be = 500mm

ρ= *

= 0.003524

Neutral axis = x = ρmd =0.003524*28.78*238 = 24.14 < 50mm, so it is rectangular section

Note! 1. If 0.8x < hf, section is rectangular

2. If 0.8x > hf, sectionis analyzed as T-beam

A= ρbed = 0.003524*500*238= 419.386mm2. Therefore Use 3 14

Table________ Reinforcement provision for support

panel depth Msupport(KNm) µ ῳ ῥ As calculatedAs min As prov

1 238 14.414 0.0449 0.0460 0.0020 237.62029 198.33 2ᴓ14
2 238 8.5262 0.0266 0.0269 0.0012 139.20110 198.33 2ᴓ12
3 238 16.067 0.0501 0.0514 0.0022 265.60666 198.33 2ᴓ14
4 238 8.5262 0.0266 0.0269 0.0012 139.20110 198.33 2ᴓ12
5 238 21.808 0.068 0.0704 0.0031 364.07176 198.33 2ᴓ14
6 238 17.049 0.0531 0.0546 0.0024 282.30847 198.33 2ᴓ14
7 238 16.892 0.0526 0.0541 0.0023 279.63439 198.33 2ᴓ14
8 238 21.534 0.0671 0.0695 0.0030 359.32652 198.33 2ᴓ14
9 238 16.892 0.0526 0.0541 0.0023 279.63439 198.33 2ᴓ14
10 238 36.337 0.1132 0.1205 0.0052 622.78100 198.33 4ᴓ14
11 238 8.5262 0.0266 0.0269 0.0012 139.20110 198.33 2ᴓ12
12 238 16.067 0.0501 0.0514 0.0022 265.60666 198.33 2ᴓ14
13 238 8.5262 0.0266 0.0269 0.0012 139.20110 198.33 2ᴓ12
14 238 21.808 0.068 0.0704 0.0031 364.07176 198.33 3ᴓ14

Table________ Reinforcement provision for span

panel depth Mspan(KNm) µ ῳ ῥ As calculatedAs min As prov
1 238 9.9097 0.0309 0.0314 0.0014 162.15347 198.33 2ᴓ12
2 238 5.8617 0.0183 0.0184 0.0008 95.28941 198.33 2ᴓ12
3 238 11.046 0.0344 0.0350 0.0015 181.08378 198.33 2ᴓ12
4 238 5.8617 0.0183 0.0184 0.0008 95.28941 198.33 2ᴓ12
5 238 14.993 0.0467 0.0479 0.0021 247.40503 198.33 2ᴓ14
6 238 11.721 0.0365 0.0372 0.0016 192.36312 198.33 2ᴓ12
7 238 11.613 0.0362 0.0369 0.0016 190.55671 198.33 2ᴓ12
8 238 14.804 0.0461 0.0473 0.0021 244.20891 198.33 2ᴓ14
9 238 11.613 0.0362 0.0369 0.0016 190.55671 198.33 2ᴓ12
10 238 24.982 0.0779 0.0811 0.0035 419.38599 198.33 3ᴓ14
11 238 5.8617 0.0183 0.0184 0.0008 95.28941 198.33 2ᴓ12
12 238 11.046 0.0344 0.0350 0.0015 181.08378 198.33 2ᴓ12
13 238 5.8617 0.0183 0.0184 0.0008 95.28941 198.33 2ᴓ12
14 238 14.993 0.0467 0.0479 0.0021 247.40503 198.33 2ᴓ14

Design for shear

Vsd,= 30.36KN

At support :

Vc = 0.25*fctd*k1*k2*bw*d fctd = 1.031 ,

 k1 = 1+50ρmin = 1+50*0.0017 = 1.085,
k2 = 1.6-d = 1.6-0.238= 1.362
Vc = 9.057KN
1.1Vc=1.1*9.057 = 9.96KN <30.36KN…Not OK!

Hence shear reinforcement is required.

Vs=Vsd-Vc=30.36-9.96=20.40KN

S=Av*Fyd*d/Vs

=(2*π82*260.87*238)/(4*20.40*103)

=305.96mm

Vrd=0.25*Fcd*bw*d = 0.25*11.33*100*238 =67.41KN

2/3*Vrd=44.94KN>Vsd , Hence the maximum spacing to resist shear will be the minimum of:

S<=

Use 8c/c 115mm

SECOND PART
LOADING

PANEL A

0.2cm pvc f.f (γ=16KN/m3)

4.8cm cement screed (γ =23KN/m3)

RC floor slab

3cm cement plastering (γ =23KN/m3)

Corridor

2cm marble tiles f.f (γ=27KN/m3)

4.8cm cement screed (γ =23KN/m3)

RC floor slab
 3cm cement plastering (γ =23KN/m3)

Dead load from:

Pvc floor finish = 0.002*16*0.5 = 0.016KN/m
Marble tiles=0.03*0.5*27=0.405KN/m
Cement screed = 0.048*23*0.5 =0.552KN/m
Cement screed=0.02*0.5*23=0.23KN/m
Topping = 0.05*25*0.5 = 0.625KN/m
Load from HCB = γ*A = 14*0.0464 = 0.65KN/m,
Joist self-weight = bw *(Dj-Dt)* γ= 0.1*(0.26-0.05)*25 = 0.525KN/m
Plastering = 0.03*23 = 0.69 KN/m
Total slab dead load for class room = 3.058 KN/m
Total slab dead load for corridor=3.025KN/m
Therefore take total slab dead load=3.058KN/m
SLAB LIVE LOAD CALCULATION
For toilet from EBCS 2, 1995 q=2KN/m2, Q=2KN
Therefore LL=2 + 2/A
For corridor from EBCS 2, 1995 q=4KN/m2, Q=4KN
Therefore LL=4 + 4/A
The detailed calculation is presented by the table below.

Table live load calculation for each panel

panel q Q A Q/A q+Q/A(kn/m2) (q+Q/A)*0.5(kN/M)

A 4 4 25.76 0.155 4.115 2.077

C 2 2 25.76 0.078 2.078 1.039

D 2 2 25.76 0.078 2.078 1.039

Table_______ analysed design load

Panel Slab DL Parti.L LL Total DL 1.3Dl 1.6LL Pd

 A 2.666 - 2.077 2.666 3.460 3.323 6.783
C 2.670 1.443 1.039 4.113 5.346 1.662 7.000
D 2.670 2.288 1.039 4.958 6.445 1.662 8.110
Analysis

Sample calculation

MA=MB=Pd L2/11=6.783*5.22/11=16.614KNm

Mspan= Pd L2/16=6.783*5.22/16=11.463KNm

VA=VB=P d L/2=6.783*5.2/2=17.64KN

Table______Calculated internal forces

panel Pd(KN/m) L(m) Msupport(KNm) Mspan(KNm) Vmax(KN)

A 6.783 5.2 16.674 11.463 17.640
C 7.000 5.2 17.207 11.830 18.200
D 8.110 5.2 19.936 13.706 21.086
1 6.471 3.8 8.495 5.840 12.295
2 9.129 4.4 16.067 11.046 20.084
7 12.868 3.8 15.969 10.978 24.442
8 12.235 4.4 21.415 14.723 28.309

Table _____________support reinforcement provision

As
panel depth Msupport(KNm) µ ῳ ῥ calculated As min As prov
A 238 16.674 0.2598 0.3069 0.0133 317.23969 198.33 3ᴓ12
C 238 17.207 0.2681 0.3190 0.0139 329.73473 198.33 3ᴓ12
D 238 19.936 0.3106 0.3846 0.0167 397.54403 198.33 3ᴓ14
1 238 8.495 0.1324 0.1425 0.0062 147.32245 198.33 2ᴓ12
2 238 16.067 0.2504 0.2934 0.0127 303.27017 198.33 3ᴓ12
7 238 15.969 0.2488 0.2912 0.0126 301.03977 198.33 3ᴓ12
8 238 21.415 0.3337 0.4233 0.0184 437.50777 198.33 3ᴓ14

Table________ span reinforcement provision

panel depth Msupport(KNm) µ ῳ ῥ As As min As prov
 calculated
A 238 11.463 0.0357 0.0364 0.0016 188.04889 198.33 2ᴓ12
C 238 11.83 0.0369 0.0376 0.0016 194.18692 198.33 2ᴓ12
D 238 13.706 0.0427 0.0437 0.0019 225.68185 198.33 2ᴓ12
1 238 5.84 0.0182 0.0184 0.0008 94.93334 198.33 2ᴓ12
2 238 11.046 0.0344 0.0350 0.0015 181.08378 198.33 2ᴓ12
7 238 10.978 0.0342 0.0348 0.0015 179.94891 198.33 2ᴓ12
8 238 14.723 0.0459 0.0470 0.0020 242.83978 198.33 2ᴓ14

Design for shear

(28.309/2.4)=Vsd/(2.162)
Vsd,= 25.502KN

At support :

Vc = 0.25*fctd*k1*k2*bw*d ,
k1 = 1+50ρmin = 1+50*0.0017 = 1.085,
k2 = 1.6-d = 1.6-0.238= 1.362
fctd = 1.031
Vc = 9.065KN
1.1Vc=1.1*9.057 = 9.963KN <25.502KN…Not OK!

Hence shear reinforcement is required.

Extra shear force resisted by stirrups will be:

Vs=Vsd-Vc=25.502-9.963=15.54KN

S=Av*Fyd*d/Vs

=(2*π82*260.87*238)/(4*15.54*103)

=401.65mm

Vrd=0.25*Fcd*bw*d = 0.25*11.33*100*238 =67.41KN

2/3*Vrd=44.94KN>Vsd , Hence the maximum spacing to resist shear will be the minimum of:

S<=

S calculated is greater than maximum spacing,hence:

Use 8c/c 115mm

CHAPTER THREE
ANALYSIS AND DESIGN OF BEAM
The frame is analyzed by using software ETABS 2009.It is a finite element analysis software and
capable of analyzing structures under static and dynamic condition.

DESIGN OF BEAM
 Beams are flexural members which transfer the load from slab to columns. Basically beams
should be designed for flexure(moment).Further more it is essential to check & design the
section for shear..

In our case, we have designed a representative beam on axis k of first floor

ANALISIS & Design of the beam

PART ONE

Beam design along axis G

From L- beam section the effective width can be taken as:

Beff ≤ {bw+lo/10 =400+4200/10=820mm

{c/c distance =5350mm

Deff=500-(25+20/2)=465mm

Design for span AB for sample.

Assumme N.A lie within the flange

=0.074988

From chart for =0.074988 Kx= 0.12

X= Kx*d =0.12*465=55.8 ≤ 149 ok!

The area of reinforcement required is determined from the following equations

If µsd ≤ µus compression reinforcement is not required

As =
If μsd > µus compression reinforcement is required

As11 =

As12 =

As = = 982

# bars = ,

= = 3.127 ≈ 4 bars

Provide 4Ø20 bars on span AB.

Design for spans BC and CD are shown in the following table.

suppor As
t Msd fcd b(mm) d(mm) µsd μus* Kx X As min # bars As prov
A 120.71 11.33 500 565 0.067 0.295 0.152 85.88 872 372 2.776 3Ø20 bars
B 159.04 11.33 500 565 0.088 0.295 0.168 94.92 1157 372 3.682 4Ø20bars
C 158.02 11.33 500 565 0.087 0.295 0.168 94.92 1149 372 3.658 4Ø20bars
D 121.95 11.33 500 565 0.067 0.295 0.151 85.32 881 372 2.803 3Ø20 bars
Span
AB 137.84 11.33 820 565 0.046 0.295 0.12 67.8 982 790.5 3.127 4Ø20 bars
BC 7.29 11.33 850 565 0.002 0.295 0.038 21.47 50.2 790.5 2.516 3Ø20 bars
CD 137.64 11.33 850 565 0.045 0.295 0.12 67.8 981 790.5 3.122 4Ø20 bars

Design for shear

The critical shear will be occurred at d distance from face of the support

Design shear using ETABS is shown on the figure.

If Vsd > Vc provide shear reinforcement

If Vsd < Vc provide stirrups for ting purpose with maximum spacing

 Capacity of concrete for shear at section 1&6.

K1=(1+50ρ) ≤2 = =

K1=(1+50*2.6689*10-3=1.133

K2=1.6-d(m) ≥ 1 = 1.6-0.565=1.035 ok!

VC1=0.25*1.032*1.133*1.035*400*565=66.063 KN

 Capacity of concrete for shear at section (2&3 and 4&5 )

K1 = (1+50 =1.178 ≤ 2 OK!

VC =0.25*1.032*1.178*1.035*400*565 =71.09 KN
Vsd maX =162.57> VC , so shear reinforcement is required

 The capacity of concrete for diagonal compression resistance is

VRD= 0.25fcdbwd

VRD=0.25*11.33*400*565=640.15KN

Maximum spacing for Vsd < VRD

= for Vsd > VRD

VRD = *640.15= 426.77

Vsdmax < VRD use the first formula

Smax

 Smax=280 mm

Extra shear force resisted by stirrups;-

Vs=Vsd-Vc

For span AB , vsd =162.57-71.09KN=91.48KN

Vs =

S= = 647.89

S>Smax
Use S=Smax

Provide 8φc/c 280mm

Design of girder along axis C on story 3

Analysis result of ETABS

D=600mm ,d=600-25-20/2=565mm

b=400mm

By using the advantage of symmetry of the section we can design only for two supports.
I.e. Supports A&B are symmetric to support C&D .Span can also be designed in a similar
fashion.

Reinforcement provision

Minimum reinforcement=(0.5/fyk)*d*b=(0.5/300)*400*565=376.66mm2

At support A

Msd=6.42KNm

Km= =7.094

Km is less than the minimum value of Km for C-25,so we will take the minimum Km=19
so the value of Ks=3.95

As=Ks*Msd/d=3.95*6.42/.565=44.88mm2<Asmin=376.66mm2

Use As=Asmin=376.66 mm2

Use 2φ20

At support B

Msd=304.62KNm

Km= =48.84

Ks=4.38

As=Ks*Msd/d=4.38*304.62/.565=2362.02mm2

Provide 8φ20 in tow rows

Design for span

Msd=249.24KNm
 Check for T-beam

Beff< =

Beff=1780mm

Assume the neutral axis lies in the flange

µsd=Msd/fcd*b*d2=(249.24*106)/(11.33*1780*5652)=0.038

From the chart µsd =0.038 ,Kx=0.1

Then X=Kx*d=0.1*565mm=56.5mm which is greater than hf=50mm.Hence treat the

beam as T-beam.

As1*fyd=fcd*0.8*x*b

As1=11.33*0.8*56.5*400/260.87=785.24mm2

M2=fyd(d-0.4x)=260.87*(565-0.4*56.5)=111.108 KNm

Msd2=Msd1-M2=249.24-111.108=138.13KNm

As2=Msd2/260.87(565-d2)=999.05mm2,d2=35mm

Assuming the yielding of compression reinforcement As*fyd=As12*fyd

£yd=fyd/Es=1.04*10-3

€s2/x-d=0.0035/x

€s2=0.00133>£yd=1.04*10-3,so the compression is yielded.

As12=As2=99905mm2

Astotal=As1+As2=785.24+999.05=1784.29mm2

Provide 6φ20 tension reinforcement

Compression reinforcement=999.05/φ20

Provide 4φ20 on the compression side.

For span BC
 Check for T-beam

Beff< =

Beff=1320mm

Assume the neutral axis lies in the flange

µsd=Msd/fcd*b*d2=(57.72*106)/(11.33*1320*5652)=0.012

Then X=Kx*d=0.012*565mm=33.9mm which is less than hf=50mm.Hence treat the

beam as rectangular section.

As=Msd/fyd(d-0.4x)=57.72*1o6/260.87(565-0.4*33.9)=401.24mm2

Provide 2φ16

DESIGN FOR SHEAR

The critical shear occurs at d distance from the face of column.

If Vsd>Vc, provide shear reinforcement and if Vsd<Vc provide stirrups for tying purpose
with maximum spacing.The output result of analysis by ETABS is as follows.

Vc1=267.80 Vc4=203.93

Vc2=291.81 Vc1=291.62

Vc3=213.09 Vc1=268.03

 Capacity of concrete for shear at section 1&6.

K1= (1+50ρ) ≤2 = =

K1=(1+50*2.6689*10-3=1.133

K2=1.6-d(m) ≥ 1 = 1.6-0.565=1.035 ok!

VC1=0.25*1.032*1.133*1.035*400*565=65.72 KN

 Capacity of concrete for shear at section (2&3 and 4&5 )

K1 = (1+50 =1.355 ≤ 2 OK!

VC2= V3C= V4C= VC5(take Vsd max=291.62)

VC2 =0.25*1.032*1.355*1.035*400*565 =81.77 KN

In all cases the design shear force is greater the the capacity of the concrete for shear,therefore shear
reinforcement is required.The capacity of the concrete for diagonal compression resistance

VRD= 0.25fcdbwd

VRD=0.25*11.33*400*565=640.15KN

Maximum spacing for Vsd < VRD

= for Vsd > VRD

VRD = *640.15= 426.77

Vsdmax < VRD use the first formula

Smax

 Smax=280 mm
Extra shear force resisted by stirrups ;-

Vs=Vsd-Vc

For section 1&6 , vs =Vsd-Vc=268.03-65.72KN=202.31KN

Vs =
 S= = 73.24<Smax

Provide φ8 c/c 70mm.

For the span AB and CD

Take the maximum Vsd from the face

Vs=291.84-81.77=210.07KN

S= = 70.53<Smax

Provide φ8 c/c 70mm

For span BC

Vs=213.09-81.77=131.32KN

S= = 112.8<Smax

Provide φ8 c/c 110mm

PART TWO
(L beam on axis k 278.095)

Km= , As=

We used C25 and S300 for our design.

Fcd=11.33 N/mm2 fyd =261.87N/mm2

Support moment M1=276.333KN.m M2=271.008KN.m

M3=278.095KN.m M4=269.002KNm

For M1=276.233

 KM= =72.39

 Compression reinforcement is required because

Km*57.83
Km>Km*
72.39>57.8

= 0.1

Km/km* = 72.39/57.83 =1.252

Ks =
 Ks ’ =

Correction factors ρ and ρ’

For ρ=1.01 and ρ’=1.03

(i) Tension reinforcement

As= = =3456.3mm2

For Ø24 bars, as= 452.389mm2 and For Ø20 bars, a s=314.6mm2
for Ø24 As/ as =7.64≈ 8 bars and for Ø20 A s/ as =11.0 ≈12bars but using Ø24 is more
economical for our design.
Therefore provide 8Ø24

(ii) Compression reinforcement

As’ = = = 1166.3 mm2

for Ø24 As/ as =2.57 ≈ 3 and for Ø20 As/ as =3.7 ≈4

Therefore provide 4Ø20

For M2 =271.008KN.m
Km= =71.7 Km*= 57.33

Km > Km* →compression reinforcement is required.

Km /Km* =1.239 and d2/d= 0.1

Ks =

Ks ’ =

Correction factors ρ and ρ’

For ρ=1.01 and ρ’=1.03

(i) Tension reinforcement

As= = =3396.216mm2

For Ø24 bars, as= 452.389mm2 and For Ø20 bars, as=314.6mm2
for Ø24 As/ as =7.5≈ 8 and for Ø20 A s/ as =11.0 ≈11 bars, but using Ø24 is more
economical for our design.

Therefore provide 8Ø24

(ii) Compression reinforcement

As’ = = = 1104.25 mm2

for Ø24 As/ as =2.44 ≈ 3 and for Ø20 A s/ as =3.5 ≈4 bars, but using Ø20 is more
economical for our design.
 Therefore provide 4Ø20

For M3 =278.095KN.m

Km= = 72.64 Km*= 57.83

Km > Km* →compression reinforcement is required.

Km /Km* =1.256 and d2/d= 0.1

Ks =

Ks ’ =

Correction factors ρ and ρ’

For ρ=1.01 and ρ’=1.03

(i) Tension reinforcement

As= = =3474.2 mm2

For Ø24 bars, as= 452.389mm2 and For Ø20 bars, as=314.6mm2
for Ø24 As/ as =7.68 ≈ 8 and for Ø20 A s/ as =11.04 ≈11 bars, but using Ø24 is more
economical for our design.
Therefore provide 8Ø24

(ii) Compression reinforcement

As’ = = = 1186.78 mm2

for Ø24 As/ as =2.62 ≈ 3 and for Ø20 A s/ as =3.77 ≈4 bars, but using Ø20 is more
economical for our design.

Therefore provide 4Ø20

For M4 =269.062KN.m

Km= = 71.45 Km*= 57.83

Km > Km* →compression reinforcement is required.

Km /Km* =1.24 and d2/d= 0.1

 Ks =

Ks ’ =

Correction factors ρ and ρ’

For ρ=1.01 and ρ’=1.03

(i) Tension reinforcement

As= = =3391.3 mm2

For Ø24 bars, as= 452.389mm2 and For Ø20 bars, as=314.6mm2

for Ø24 As/ as =7.496 ≈ 8 bars

Therefore provide 8Ø24

(ii) Compression reinforcement

As’ = = = 1099.375 mm2

for Ø24 As/ as =2.43 ≈ 3 and for Ø20 A s/ as =3.49 ≈4 bars, but using Ø20 is more

economical for our design.

Therefore provide 4Ø20

DESIGN FOR SHEAR

Vrd = 0.25*fcd*b*d.

Vrd = 0.25 * 11.33 * 400 * 363 = 411.279 KN.

Vmax = 248.28 KN (from fig above).

Vc = 0.25*k1*k2*fctd*b*d.
 Where

Vc = 0.25*2*1.237*1.032*400*363.

Vc =92.67999KN.

 For span (8-6).

Vc ≤ Vd≤ Vrd.

92.7≤Vd≤274.189.

(i). Vd right of 8=208.84KN.

Vc<Vd.
Vs=203.84-92.67999.
Vs=111.2KN.
Taking φ 8  Av=50.26.

S= =85.601mm < Smax Provide φ 8 c/c 85mm.

(ii). Vd left of 6 = 202.33KN.

Vs =202.33 – 92.67999=109.65KN.

S= =86.81mm < Smax .

Provide φ 8 c/c 85mm

 For span (6-5).

Vc ≤ Vd≤ Vrd.

92.7≤Vd≤274.189.
(i) Vd right of 6=97.32KN.
Vc<Vd.
Vs=97.32-92.67999.
Vs=4.64KN.
Taking φ 8  Av=50.26.

S= =2051.47mm > Smax, take Smax = 180mm.

 Provide φ 8 c/c 180mm.
(ii). Vd left of 5 =96.23KN.
Vc<Vd.
Vs=96.23-92.67999=3.55KN.
Vs=3.55KN.

S= =2681.36mm > Smax, take Smax = 180mm.

Provide φ 8 c/c 180mm.

 For span (5-3).

Vc ≤ Vd≤ Vrd.

92.7≤Vd≤274.189.
(i) Vd right of 5 =204.4KN.
Vc<Vd.
Vs=204.4-92.679.
Vs=111.721KN.
Taking φ 8  Av=50.26.

S= =85.2mm < Smax

Provide φ 8 c/c 85mm.

(ii). Vd left of 3 =201.78KN.

Vc<Vd.
Vs=201.78-92.679=109.101KN.
Vs=109.101KN.

S= =87.25mm < Smax .

Provide φ 8 c/c 180mm

.BEAM DETAILING DEVELOPMENT LENGTH.

(i) ld = , fbd=2*1.032=2.064.

lbent a* ld*Ascal/Asprov.
 lbent≥ forbar on tension. lbent≥ forbar on compression.

(ii). ld= =ф(31.598).

24*31.598=756.51mm

provided for all sections of the beam.

Curtailment.

 SUPPORT A

X1 =1484mm.

Length = x1 + (lbreq +al).

= 1484 + (722.744 +0.743631).

= 2478.994 mm.

-lb = , as=0.75 (363).

= = 722.477mm.
Lbmin = , Lbmin =240mm˂ 722.477mm ok!!

Support B

X1=5528mm,X2=8069mm.
Length = (X2-X1)+2(lbreq+2l).
=(8069-5528)-2(709.92+0.75(363)).
=4505.34mm.
Lbreq=a*ld*Ascal/Aspro.

= =709.92mm.

Lb min =

Support C

X1=5480mm,X2=8015mm.

lbreq= = =726.23mm.

Lb min =

lbreq>lbmin=240mm ok!!
Length= (8015-5480)+2(726.23+0.75(363))
= 4531.96mm.

Support D

X1=1360mm.

Length = x1 + (lbreq +al).

 = 1360 + (708.89 +272.25).

= 2341.14 mm.

= = 708.89mm.

Lbmin =

lbmin =240mm.
lbreq>240mm …ok!!

Span

Ld = =φ*31.598 =756.61mm (for all).

For AB, lb= = =708.628mm.

Lbmin =

lbmin =240mm.
lbreq>240mm …ok!!

X1=1484mm, X2=5528mm.
Length = (5528-1484)+2(708.628+0.75(159)).
= 5699.756mm.
For BC

lb= = =335.137mm.
 Lbmin =

X1=1169mm,X2=3485mm.
Length= (3485-1169)+2(335.135+0.75(159)).
= 3224.77mm.

X1=1420mm,X2=5540mm.
Lbreq=a*ld*Ascal/Aspro.

= =709.48mm.

Lb min =

lbmin =240mm.
lbreq>240mm …ok!!
Length= (5540-1420)+2(709.48+0.75(159))
= 5777.46mm

CURTAILMENT DETAILING.
 CHAPTER FOUR

ANALYSIS &DESIGN OF COLUMN

DESIGN OF COLUMNS
A column is vertical member subjected to axial compressive force, with or without bending
moment and transmitting the load to the ground through the foundation According to EBCS
2.1995 the maximum and minimum longitudinal reinforcement that can be provided to a column
of cross-section Ac equals to 0.08Ac and 0.008Ac respectively

The effective length,Le, of the column is given by

a) Sway mode = ≥ 1.15

b) Non sway mode = ≥ 0.7

Where: 𝞪1= 𝞪2=

𝞪m=

K1 ,K2=are column stiffness coefficient ( )

Kc =the design column stiffness coefficient

Kij =is the effective beam stiffness coefficients
Design of isolated columns
1) Total eccentricity :
The total eccentricity to be used for the design of columns of constant cross-section
is given by:
etot =ee +ea + e2 ,
Where ee =is the equivalent first order eccentricity of the design axial load
ea =is additional eccentricity

ea = ≥ 20mm (Le – is the effective length mm)

e2 – is the second order eccentricity

For first order moment varying linearly along the length , the equivalent eccentricity is
the higher of the following two values:
etot =0.6e02 + 0.4e01 or
ee =0.4e02
Where e01 and e02 are first order eccentricities at the ends , e 02 being positive and greater
in magnitude than e01
For different eccentricities at the ends (negative and positive)

Tthe critical end section shall be checked for first order moments
 etot =e02 + ea

Second order eccentricity

Non-sway frames, the second order eccentricity e 2 of an isolated column may be

obtained as

e2 = ( )

K1 = -0.75 For 15 ʎ 35

K1 = 1.0 For ʎ

is the curvature at the critical section

= k2 ( )*10-3 ,where

d – is the column dimension in the buckling plane less the cover to the center of the

longitudinal reinforcement

K2 =

The second order eccentricity can be neglected when :

 For sway frames

λ Vd =
  For non-sway frames

λ ,where M1 and M2 are first order moments,

M2 being positive and greater than M1

M1 =

PART ONE

ETABS frame output result

Story Column Load Loc P V2 T M2 M3
STORY6 C1 COMB1 0 -54.1475 -8.48137 0.041861 -11.8607 -13.3462
STORY6 C1 COMB1 1.55 -49.8747 -8.48137 0.041861 -2.47257 -0.20043
STORY6 C1 COMB1 3.1 -45.6018 -8.48137 0.041861 6.928945 12.94639
STORY5 C1 COMB1 0 -99.8527 -6.13055 0.1056 -12.384 -11.0686
STORY5 C1 COMB1 1.55 -95.5798 -6.13055 0.1056 -0.82454 -1.57378
STORY5 C1 COMB1 3.1 -91.3069 -6.13055 0.1056 10.74353 7.937474
STORY4 C1 COMB1 0 -176.84 -13.8752 0.197486 -19.2324 -22.5587
STORY6 C2 COMB1 0 -67.7096 -5.08161 0.035246 -88.2295 -10.6241
STORY6 C2 COMB1 1.55 -63.4367 -5.08161 0.035246 -32.4186 -2.76204
STORY6 C2 COMB1 3.1 -59.1638 -5.08161 0.035246 23.61633 5.119084
STORY5 C2 COMB1 0 -462.126 -6.56894 0.104544 -80.7808 -10.8779
STORY5 C2 COMB1 1.55 -457.853 -6.56894 0.104544 0.648704 -0.70237
STORY5 C2 COMB1 3.1 -453.58 -6.56894 0.104544 82.04842 9.508535
STORY4 C2 COMB1 0 -893.616 -12.3112 0.112765 -143.768 -20.7726
STORY4 C2 COMB1 1.5 -886.264 -12.3112 0.112765 -20.0058 -2.36735
STORY4 C2 COMB1 3 -878.913 -12.3112 0.112765 104.3326 16.10619
STORY3 C2 COMB1 0 -1329.65 -11.1988 0.08221 -84.8421 -19.2832
STORY3 C2 COMB1 1.5 -1322.3 -11.1988 0.08221 -2.57543 -2.50765
STORY3 C2 COMB1 3 -1314.95 -11.1988 0.08221 79.80383 14.37567
STORY2 C2 COMB1 0 -1684.04 -11.6824 0.106301 -40.3386 -13.0063
STORY2 C2 COMB1 1.5 -1672.55 -11.6824 0.106301 14.42777 4.62317
STORY2 C2 COMB1 3 -1661.06 -11.6824 0.106301 68.87352 22.14991
STORY1 C2 COMB1 0 -1758.65 0.210581 -0.26698 -5.61225 -0.50048
STORY1 C2 COMB1 0.5 -1754.82 0.210581 -0.26698 -4.40952 -0.60513
STORY1 C2 COMB1 1 -1750.99 0.210581 -0.26698 -3.19529 -0.70821
STORY6 C7 COMB1 0 -122.062 -13.43 0.057473 -4.78107 -21.4613
STORY6 C7 COMB1 1.55 -117.789 -13.43 0.057473 0.104541 -0.64169
STORY6 C7 COMB1 3.1 -113.516 -13.43 0.057473 4.988814 20.1862
STORY5 C7 COMB1 0 -240.173 -9.2047 0.071952 -7.1032 -16.3031
STORY5 C7 COMB1 1.55 -235.9 -9.2047 0.071952 -1.29483 -2.05382
STORY5 C7 COMB1 3.1 -231.627 -9.2047 0.071952 4.546996 12.24852
STORY4 C7 COMB1 0 -409.491 -20.0079 0.170839 -13.8267 -32.2373
STORY4 C7 COMB1 1.5 -402.14 -20.0079 0.170839 -1.88854 -2.23742
STORY4 C7 COMB1 3 -394.789 -20.0079 0.170839 10.07432 27.79176
STORY3 C7 COMB1 0 -579.218 -17.3488 0.051151 -8.29606 -31.5466
STORY3 C7 COMB1 1.5 -571.867 -17.3488 0.051151 -0.25932 -5.54996
STORY3 C7 COMB1 3 -564.516 -17.3488 0.051151 7.782279 20.54989
STORY2 C7 COMB1 0 -782.656 -21.723 -0.08705 -2.63687 -25.7881
STORY2 C7 COMB1 1.5 -771.17 -21.723 -0.08705 0.671671 6.877746
STORY2 C7 COMB1 3 -759.684 -21.723 -0.08705 3.973318 39.47301
STORY1 C7 COMB1 0 -869.088 -8.75614 0.083398 1.8558 -6.62475
STORY1 C7 COMB1 0.5 -865.259 -8.75614 0.083398 0.265812 -2.24201
STORY1 C7 COMB1 1 -861.43 -8.75614 0.083398 -1.32452 2.143603
STORY6 C8 COMB1 0 -112.774 -3.54452 0.046136 33.50499 -6.97072
STORY6 C8 COMB1 1.55 -108.502 -3.54452 0.046136 14.41113 -1.49256
STORY6 C8 COMB1 3.1 -104.229 -3.54452 0.046136 -4.85372 4.003306
STORY5 C8 COMB1 0 -752.958 -4.51326 0.083802 29.58311 -7.80582
STORY5 C8 COMB1 1.55 -748.685 -4.51326 0.083802 -1.37238 -0.84843
STORY5 C8 COMB1 3.1 -744.412 -4.51326 0.083802 -32.2192 6.178365
STORY4 C8 COMB1 0 -1415.96 -9.75039 0.119272 52.61626 -16.27
STORY4 C8 COMB1 1.5 -1408.61 -9.75039 0.119272 7.731877 -1.73101
STORY4 C8 COMB1 3 -1401.26 -9.75039 0.119272 -37.5061 12.88732
STORY3 C8 COMB1 0 -2082.36 -6.3502 0.071204 32.31274 -8.96545
STORY3 C8 COMB1 1.5 -2075.01 -6.3502 0.071204 2.077879 0.647446
STORY3 C8 COMB1 3 -2067.66 -6.3502 0.071204 -28.2979 10.2178
STORY2 C8 COMB1 0 -2635.05 -4.00477 0.014282 13.98913 -5.87915
STORY2 C8 COMB1 1.5 -2623.56 -4.00477 0.014282 -8.01485 0.166866
STORY2 C8 COMB1 3 -2612.08 -4.00477 0.014282 -29.7401 6.207185
STORY1 C8 COMB1 0 -2722.3 -6.87018 -0.01577 -3.81364 -4.5895
STORY1 C8 COMB1 0.5 -2718.47 -6.87018 -0.01577 -0.251 -1.14411
STORY1 C8 COMB1 1 -2714.64 -6.87018 -0.01577 3.312642 2.305888
STORY6 C9 COMB1 0 -112.684 6.83E-05 0.035053 28.11107 0.286211
STORY6 C9 COMB1 1.55 -108.412 6.83E-05 0.035053 12.05794 0.280316
STORY6 C9 COMB1 3.1 -104.139 6.83E-05 0.035053 -4.13814 0.271098
STORY5 C9 COMB1 0 -800.514 0.130452 0.041165 25.47236 0.28654
STORY5 C9 COMB1 1.55 -796.241 0.130452 0.041165 -1.01632 0.046049
STORY5 C9 COMB1 3.1 -791.969 0.130452 0.041165 -27.42 -0.19843
STORY4 C9 COMB1 0 -1519.13 1.383115 0.066311 42.23584 2.820133
STORY4 C9 COMB1 1.5 -1511.78 1.383115 0.066311 5.676357 0.664774
STORY4 C9 COMB1 3 -1504.43 1.383115 0.066311 -31.1617 -1.52313
STORY3 C9 COMB1 0 -2232.32 -3.77898 0.072619 41.64035 -8.71421
STORY3 C9 COMB1 1.5 -2224.97 -3.77898 0.072619 7.955797 -3.17329
STORY3 C9 COMB1 3 -2217.62 -3.77898 0.072619 -26.3007 2.595084
STORY2 C9 COMB1 0 -2952.01 -8.74301 0.096691 27.11566 -10.0944
STORY2 C9 COMB1 1.5 -2940.52 -8.74301 0.096691 -12.8013 3.18023
STORY2 C9 COMB1 3 -2929.04 -8.74301 0.096691 -52.22 16.33117
STORY1 C9 COMB1 0 -3035.01 0.376296 0.026904 -6.59926 -0.54394
STORY1 C9 COMB1 0.5 -3031.18 0.376296 0.026904 -0.36345 -0.73092
STORY1 C9 COMB1 1 -3027.35 0.376296 0.026904 5.873985 -0.9146
STORY6 C10 COMB1 0 -116.237 -0.05197 0.035319 28.10473 -0.37613
STORY6 C10 COMB1 1.55 -111.964 -0.05197 0.035319 12.07622 -0.3053
STORY6 C10 COMB1 3.1 -107.692 -0.05197 0.035319 -4.10016 -0.23073
STORY5 C10 COMB1 0 -810.127 -0.59303 0.024881 25.70427 -0.95992
STORY5 C10 COMB1 1.55 -805.854 -0.59303 0.024881 -0.98514 -0.08418
STORY5 C10 COMB1 3.1 -801.582 -0.59303 0.024881 -27.5913 0.798975
STORY4 C10 COMB1 0 -1540.49 -1.28849 0.078038 41.9907 -2.19841
STORY4 C10 COMB1 1.5 -1533.14 -1.28849 0.078038 5.611426 -0.3588
STORY4 C10 COMB1 3 -1525.79 -1.28849 0.078038 -31.0472 1.498654
STORY3 C10 COMB1 0 -2267.14 2.634969 0.058622 42.1869 8.385346
STORY3 C10 COMB1 1.5 -2259.79 2.634969 0.058622 8.090903 4.62943
STORY3 C10 COMB1 3 -2252.44 2.634969 0.058622 -26.5957 0.536912
STORY2 C10 COMB1 0 -2974.24 7.378525 0.079289 28.15074 6.093689
STORY2 C10 COMB1 1.5 -2962.76 7.378525 0.079289 -12.6627 -5.10975
STORY2 C10 COMB1 3 -2951.27 7.378525 0.079289 -52.9795 -16.1127
STORY1 C10 COMB1 0 -3062.88 -2.95013 0.035399 -6.49519 -0.51093
STORY1 C10 COMB1 0.5 -3059.05 -2.95013 0.035399 -0.14558 0.965778
STORY1 C10 COMB1 1 -3055.22 -2.95013 0.035399 6.204685 2.438097
STORY6 C11 COMB1 0 -115.885 -0.75815 0.036338 33.89455 -0.44548
STORY6 C11 COMB1 1.55 -111.612 -0.75815 0.036338 14.25364 0.730056
STORY6 C11 COMB1 3.1 -107.339 -0.75815 0.036338 -5.56125 1.896682
STORY5 C11 COMB1 0 -883.795 0.724826 0.024092 31.0093 1.372997
STORY5 C11 COMB1 1.55 -879.523 0.724826 0.024092 -1.05182 0.201938
STORY5 C11 COMB1 3.1 -875.25 0.724826 0.024092 -33.0168 -0.98844
STORY4 C11 COMB1 0 -1674.22 3.493051 0.081812 52.07969 7.170659
STORY4 C11 COMB1 1.5 -1666.87 3.493051 0.081812 7.189652 1.884647
STORY4 C11 COMB1 3 -1659.52 3.493051 0.081812 -38.0892 -3.50299
STORY3 C11 COMB1 0 -2460.6 -9.36762 0.052265 47.23487 -24.2061
STORY3 C11 COMB1 1.5 -2453.25 -9.36762 0.052265 7.891688 -10.6535
STORY3 C11 COMB1 3 -2445.9 -9.36762 0.052265 -32.0769 3.7394
STORY2 C11 COMB1 0 -3259.85 -25.6949 0.062457 30.12942 -28.5268
STORY2 C11 COMB1 1.5 -3248.36 -25.6949 0.062457 -13.2128 10.54839
STORY2 C11 COMB1 3 -3236.88 -25.6949 0.062457 -55.9875 49.17062
STORY1 C11 COMB1 0 -3344.18 13.92237 0.030472 -6.3133 7.87003
STORY1 C11 COMB1 0.5 -3340.35 13.92237 0.030472 0.105013 0.88678
STORY1 C11 COMB1 1 -3336.52 13.92237 0.030472 6.522809 -6.10087
STORY6 C12 COMB1 0 -109.566 15.17978 0.04854 23.4997 24.16219
STORY6 C12 COMB1 1.55 -105.293 15.17978 0.04854 9.007183 0.614012
STORY6 C12 COMB1 3.1 -101.02 15.17978 0.04854 -5.58903 -22.9413
STORY5 C12 COMB1 0 -425.159 9.697305 0.022642 21.54002 17.19243
STORY5 C12 COMB1 1.55 -420.886 9.697305 0.022642 -0.29539 2.149414
STORY5 C12 COMB1 3.1 -416.613 9.697305 0.022642 -22.1178 -12.9927
STORY4 C12 COMB1 0 -776.502 20.21462 0.092021 37.11896 31.45562
STORY4 C12 COMB1 1.5 -769.151 20.21462 0.092021 4.821817 1.039593
STORY4 C12 COMB1 3 -761.799 20.21462 0.092021 -27.596 -29.4027
STORY3 C12 COMB1 0 -1126.3 29.3585 0.049694 35.59616 64.49963
STORY3 C12 COMB1 1.5 -1118.95 29.3585 0.049694 6.378196 20.82867
STORY3 C12 COMB1 3 -1111.6 29.3585 0.049694 -23.0715 -23.5979
STORY2 C12 COMB1 0 -1675.34 48.80981 0.060378 25.10062 48.75536
STORY2 C12 COMB1 1.5 -1663.85 48.80981 0.060378 -9.30959 -24.9281
STORY2 C12 COMB1 3 -1652.37 48.80981 0.060378 -43.514 -98.0602
STORY1 C12 COMB1 0 -1756.72 0.429663 0.033497 -3.19306 7.007693
STORY1 C12 COMB1 0.5 -1752.89 0.429663 0.033497 0.264647 6.783671
STORY1 C12 COMB1 1 -1749.06 0.429663 0.033497 3.721664 6.541996
STORY6 C20 COMB1 0 -67.4145 -0.52116 0.037932 90.63664 -0.65758
STORY6 C20 COMB1 1.55 -63.1416 -0.52116 0.037932 33.83037 0.152193
STORY6 C20 COMB1 3.1 -58.8687 -0.52116 0.037932 -23.2086 0.960924
STORY5 C20 COMB1 0 -521.155 -0.76735 0.125242 81.50184 -1.62751
STORY5 C20 COMB1 1.55 -516.882 -0.76735 0.125242 -1.05101 -0.42743
STORY5 C20 COMB1 3.1 -512.61 -0.76735 0.125242 -83.5481 0.79675
STORY4 C20 COMB1 0 -1010.31 -2.3022 0.087224 144.3222 -4.23605
STORY4 C20 COMB1 1.5 -1002.96 -2.3022 0.087224 18.9016 -0.78407
STORY4 C20 COMB1 3 -995.604 -2.3022 0.087224 -107.136 2.693448
STORY3 C20 COMB1 0 -1501.46 2.212459 0.091353 115.1461 7.160244
STORY3 C20 COMB1 1.5 -1494.1 2.212459 0.091353 15.68519 3.994043
STORY3 C20 COMB1 3 -1486.75 2.212459 0.091353 -84.5365 0.634927
STORY2 C20 COMB1 0 -2146.78 8.453311 0.027276 59.70449 8.3632
STORY2 C20 COMB1 1.5 -2135.29 8.453311 0.027276 -32.4721 -4.42315
STORY2 C20 COMB1 3 -2123.81 8.453311 0.027276 -123.728 -17.0841
STORY1 C20 COMB1 0 -2218.5 -7.47051 -0.00441 5.468602 -4.72595
STORY1 C20 COMB1 0.5 -2214.67 -7.47051 -0.00441 8.43905 -0.98202
STORY1 C20 COMB1 1 -2210.84 -7.47051 -0.00441 11.38175 2.765147
Method of calculation of area of reinforcement

(Mb)Msdy =P* etot

V =

µb =

Using biaxial chart no. (9&10)

By interpolation we can get the value of ῳ and after that the area of reinforcement can be
calculated by the formula :

Asmain = =

Check whether the the calculated area is within the limit

For type one column

Check Asmin = 0.008Ac = 0.008*500*500 =2000mm2

Asmax = 0.08Ac = 0.08*400*500 =20,000000mm2

Asmin < Asprovide < Asmax……………………ok

For type two column

Check Asmin = 0.008Ac = 0.008*400*400 =1280mm2

Asmax = 0.08Ac = 0.08*400*500 =12,800mm2

Asmin < Asprovide < Asmax……………………ok

For type three column

 Check Asmin = 0.008Ac = 0.008*300*300 =720mm2

Asmax = 0.08Ac = 0.08*400*500 =7,200mm2

Asmin > Asprovide …………..not ok

Provide minimum reinforcement.

Fig___ stiffness factor of beam and column along axis c

Fig____ stiffness factor of beam and column along axis 3
DESIGN OF COLUMN ON AXIS C
Table_____Calculation of effective length

column Kb1 Kb2 Kc K1 K2 ἁ1 ἁ2 ἁm L,m Le,m

C19 3.038 1 1.635
0.785 1 3.038 1.013 2.269 1 0.836
C29 0.785 1 1.013 3.038 2.269 1.359
0.785 1 1.013 0.415 0.8 3 2.444
C39 0.785 1 0.415 1.013 0.8 0.633
0.785 1 0.415 0.415 0.465 3 2.163
C49 0.785 1 0.415 0.415 0.465 0.385
0.785 1 0.415 0.127 0.304 3 2.1
C59 0.785 1 0.127 0.415 0.304 0.316
0.785 1 0.127 0.127 0.328 3.1 2.17
C69 0.785 1 0.127 0.127 0.328 0.246
0.785 1 0.127 0.164 3.1 2.17
Table____Check for mode of the column
story N H L δx1 δy1 δx2 δy2 Δx Δy ΔN/HL remark
6 689.18 58.45 3.1 0.000561 0.00038 0.000561 0.00038 0.000246 0.000034 0.000936 non
sway
5 3912.42 92.13 3.1 0.000315 0.000346 0.000315 0.000346 0.000297 0.000168 0.004069 non
sway
4 7335.79 138.05 3 0.00018 0.000178 0.00018 0.000178 0.00001 0.000002 0.000177 non
sway
3 10747.94 116.76 3 0.000028 0.000176 0.000028 0.000176 0.000004 0.000099 0.003038 non
sway
2 14279.15 121.22 3 0.000032 0.000077 0.000032 0.000077 0.000018 0.000072 0.002827 non
sway
1 14790.17 54.99 1 0.000014 0.000005 0.000014 0.000005 0.000014 0.000005 0.003765 non
sway
Table____check for consideration of second order effect

50-
25(M1/ 2nd order
column Le(m) b(h) A(m^2) I i λ M1(KNm) M2(KNm) M2) effect
not
C19 0.836 0.5 0.25 0.005208 0.144338 5.791978 2.31 4.59 35.6528 considered
not
C29 2.444 0.5 0.25 0.005208 0.144338 16.93253 5.88 6.21 30.21767 considered
not
C39 2.163 0.4 0.16 0.002133 0.11547 18.73213 8.97 10.22 30.19361 considered
not
C49 2.1 0.4 0.16 0.002133 0.11547 18.18653 12.89 16.27 28.05773 considered
not
C59 2.17 0.3 0.09 0.000675 0.086603 25.057 6.18 7.81 26.3285 considered
not
C69 2.17 0.3 0.09 0.000675 0.086603 25.057 4 6.97 37.4183 considered

Table___Design moment determination for non-sway mode

column le Nsd M1 M2 ea(mm) eo1 eo2 eo etot(mm) Msd(KN-

M)
C19 0.836 2722.3 2.31 4.59 20 0.000849 0.001686 0.001351 20.00135 54.44968
C29 2.444 2635.05 5.88 6.21 20 0.002231 0.002357 0.002307 20.00231 52.70708
C39 2.163 2082.36 8.97 10.22 20 0.004308 0.004908 0.004668 20.00467 41.65692
C49 2.1 1415.96 12.89 16.27 20 0.009103 0.01149 0.010536 20.01054 28.33412
C59 2.17 752.96 6.18 7.81 20 0.008208 0.010372 0.009506 20.00951 15.06636
C69 2.17 112.77 4 6.97 20 0.03547 0.061807 0.051273 20.05127 2.261182
DESIGN OF COLUMN ON AXIS 3
Table_____Calculation of effective length

column Kb1 Kb2 Kc K1 K2 ἁ1 ἁ2 ἁm L,m Le,m

C19 3.329 1 2.573 1 0.881
0.0666 1 3.329 1.109 4.145
C29 0.0666 1 1.109 1.109 4.415 2.941 3 2.679
0.0666 1 1.109 0.454 1.466
C39 0.0666 1 0.454 0.454 1.466 1.159 3 2.387
0.0666 1 0.454 0.454 0.852
C49 0.0666 1 0.454 0.454 0.852 0.704 3 2.202
0.0666 1 0.454 0.139 0.556
C59 0.0666 1 0.139 0.139 0.556 0.47 3.1 2.17
0.0666 1 0.139 0.139 0.385
C69 0.0666 1 0.139 0.139 0.385 0.289 3.1 2.17
0.0666 1 0.139 0.192
Table____Check for mode of the column
story N H L δx1 δy1 δx2 δy2 Δx Δy ΔN/HL remark
6 689.18 58.45 3.1 0.000561 0.00038 0.000561 0.00038 0.000246 0.000034 0.000936 non
sway
5 3912.42 92.13 3.1 0.000315 0.000346 0.000315 0.000346 0.000297 0.000168 0.004069 non
sway
4 7335.79 138.05 3 0.00018 0.000178 0.00018 0.000178 0.00001 0.000002 0.000177 non
sway
3 10747.94 116.76 3 0.000028 0.000176 0.000028 0.000176 0.000004 0.000099 0.003038 non
sway
2 14279.15 121.22 3 0.000032 0.000077 0.000032 0.000077 0.000018 0.000072 0.002827 non
sway
1 14790.17 54.99 1 0.000014 0.000005 0.000014 0.000005 0.000014 0.000005 0.003765 n. sway
 50- 2nd order
column Le(m) b(h) A(m^2) I i λ M1(KNm) M2(KNm) 25(M1/M2) effect
C19 0.881 0.5 0.25 0.005208 0.144338 6.103747 3.31 3.81 46.3806 n.considered
C29 2.679 0.5 0.25 0.005208 0.144338 18.56066 13.99 29.74 27.04842 not consid.
C39 2.387 0.4 0.16 0.002133 0.11547 20.67203 28.3 32.31 32.17883 not consid.
C49 2.202 0.4 0.16 0.002133 0.11547 19.06988 37.51 52.62 28.10275 not consid.
C59 2.17 0.3 0.09 0.000675 0.086603 25.057 29.58 32.22 38.23974 not consid.
C69 2.17 0.3 0.09 0.000675 0.086603 25.057 4.85 33.5 28.28084 not consid.

Table___Design moment determination for non-sway mode

column le Nsd M1 M2 ea(mm) eo1 eo2 eo etot(mm) Msd(KN-

M)
C19 0.881 2722.3 3.31 3.81 20 0.001216 0.0014 0.001326 20.00133 54.44961
C29 2.679 2635.05 13.99 29.74 20 0.005309 0.011286 0.008895 20.0089 52.72444
C39 2.387 2082.36 28.3 32.31 20 0.01359 0.015516 0.014746 20.01475 41.67791
C49 2.202 1415.96 37.51 52.62 20 0.026491 0.037162 0.032894 20.03289 28.36578
C59 2.17 752.96 29.58 32.22 20 0.039285 0.042791 0.041389 20.04139 15.09036
C69 2.17 112.77 4.85 33.5 20 0.043008 0.297065 0.195442 20.19544 2.27744

Calculation of reinforcement requirement with a combined effect

colum Nsd Msd(KN- Msd(KN b(h),m diamet µh=Mh/ µb=Mb/ V=N/ ω As As(mi As Reinforceme
M)y-y -M)x-x m er of fcd*Ac*h fcd*Ac*b fcd*Ac n) nt provided
n
bar
C19 2722.3 54.4496 54.4496 500 24 0.038446 0.07509 0.961094 0.4 4343.15 2000 4343. 9.600
8 1 9 2
C29 2635.0 52.7070 52.7244 500 24 0.037216 0.037228 0.930291 0.38 4126.00 2000 4126. 9.120
5 8 4 1 0
C39 2082.3 41.6569 41.6779 400 20 0.057448 0.057477 1.148698 0.36 2501.66 1280 2501. 7.963
6 2 1 7
C49 1415.9 28.3341 28.3657 400 16 0.039075 0.039119 0.78109 0.1 694.905 1280 1280. 6.366
6 2 8 5 0
C59 752.96 15.0663 15.0903 300 12 0.049251 0.049329 0.738413 0.001 4.69061 720 720.0 6.366
6 6 2 2
C69 112.77 2.26118 2.27744 300 12 0.007392 0.007445 0.110591 0.002 7.81768 720 720.0 6.366
2 7
provided reinforcement from bottom to top respectively are:10,10,8,8,8,8

column As calc As eff φ Ib Lb Ib net Lo min Io Io

min provided
C19 4343.16 4523.89 24 782.6 470 751.3445 360 1051.882 1055
C29 4126 4523.89 24 782.6 470 713.7773 360 999.2882 1000
C39 2501.66 2513.27 20 652.2 391 649.1622 300 908.8271 910
C49 1280 1608.5 16 521.7 313 415.1863 240 581.2609 585
C59 720 904.78 12 391.3 300 311.3902 200 435.9462 440
C69 720 904.78 12 391.3 300 311.3902 200 435.9462 440
Lateral reinforcement

For basement and ground floor,

Use

Spacing

Use c/c290mm

For 1st and 2nd floor

Use

Spacing

Use c/c
Between 1st and second floor

Use

Spacing
 Use c/c 9 mm
Between 2nd and 4th floor

Use

Spacing

Use c/c mm

Sketch of cross-sectional column details

PART TWO
TABLES FOR DESIGN OF COLUMNS

 Y-Y DIRECTION

 To check whether the columns in each story are sway or nonsway we used the

formula shown bellow in the table; i.e. Sway if 0.1.

Column δ N H L Remark

CF 0.003 1646.64 50.25 3.6 0.098 Non sway

CG 0.0165 1597.05 54.68 3.6 0.134 Non sway

CFR 0.013 1077.68 86.11 3.6 0.0452 Sway

CS 0.041 734.13 53.08 3.6 0.1575 Non sway

CT 0.036 401.15 71.92 3.6 0.05577 Sway

CFO 0.006 68.94 34.19 3.6 0.00336 Non sway

  Table for Calculation of effective length.

column kb1 kb2 kc k1 k2 α1 α2 αm L(m) Le(m)

CF - - 3.0348 - - 1 - 5.0295 3.6 3.3533

- 0.67 3.0348 - 3.0348 - 9.059

CG - 0.67 3.0348 3.0348 - 9.059 - 9.059 3.6 10.338

- 0.67 3.0348 - 3.0348 - 9.059

CFR - 0.67 3.0348 3.0348 - 9.059 - 7.695 3.6 3.43

- 0.67 3.0348 - 1.2075 - 6.332

CS - 0.67 1.2075 3.0348 - 6.332 - 6.332 3.6 8.866

- 0.67 1.2075 - 1.2075 - 6.332

CT - 0.67 1.2075 1.2075 - 6.332 - 6.332 3.6 3.4

- 0.67 1.2075 - 1.2075 - 6.332

CFO - 0.67 1.2075 1.2075 - 6.332 - 6.332 3.6 3.4

1.2075 - - - 6.332

 Table to check for consideration of second order effect

25 or
50- Second
COLUM A(m 25(M1/ order
2) 15/
N Le b(h) I i Kl λ M1 M2 Vd M2) effect

CF 3.35 0.5 0.25 0.005 0.1414 0.026 5.618 0.76 13.9 0.484 48.64092 NOT

CONSID
CG 10.3 0.5 0.25 0.005 0.1414 0.026 5.618 15.6 14.6 0.469 25 ERED

CFR 3.43 0.5 0.25 0.005 0.1414 0.026 5.618 22.6 24.2 0.32 26.60455 NOT

CONSID
CS 8.87 0.4 0.16 0.002 0.1118 0.01 7.303 8.95 7.45 0.337 26.516 ERED

CT 3.4 0.4 0.16 0.002 0.1118 0.01 7.303 5.88 6.36 0.22 26.88679 NOT

CFO 3.4 0.4 0.16 0.002 0.1118 0.01 7.303 -0.85 1.21 0.032 67.56198 NOT

 FOR SWAY CASE

 COLUM Le, Nsd,K M1,KN M2,KN ea,m eo1,m eo2,m eo,m δs*eo,m eto Msd,kN
N m N m m m m m m δs m t m

CG 10.3 1597. 15.62 14.55 34.46 9.78 9.11 9.378 1.16 10.8315 z 72.554

10.53 1.18
CS 8.8 734.1 8.95 7.45 29.55 11.71 9.75 4 7 12.5038 42. 32.158

 FOR NON SWAY CASE

COLUM Le, Nsd,K M1,KN M2,KN ea,m eo1,m eo2,m Msd,kN

N m N m m m m m eo,mm etot m

CF 3.35 1646.6 0.76 13.98 20 0.4615 8.49 5.2786 25.278 41.625

CFR 3.43 1077.6 22.6 24.15 20 20.97 22.41 21.834 41.834 45.083

CT 3.4 401.15 5.88 6.36 20 14.657 15.85 15.373 35.342 14.177

CFO 3.4 68.94 -0.85 1.21 20 -12.369 17.55 5.5824 25.582 1.763

 X-X DIRECTION

Column Δ N H L Remark

CF 0.02 1646.64 14.84 3.6 0.4 Non sway

 CG 0.033 1597.05 20.75 3.6 0.11 Non sway

CFR 0.017 1077.58 16.72 3.6 0.343 Sway

CS 0.035 734.13 26.75 3.6 0.19 Non sway

CT 0.02 401.15 46.88 3.6 0.31 Sway

CFO 0.056 68.94 27.9 3.6 0.3 Non sway

col k col
um b um
n 1 kb2 kc k1 k2 α1 α2 αm L(m) Le(m) n kb1 kb2

CF - - 2.756 - - 1 - 1.878 3.6 5.57 CF - -

1 2.756 - 2.756 - 2.756 1 -

CG 1 - 2.756 2.756 - 5.5 - 5.5 3.6 8.37 CG 1 -

1 - 2.756 - 2.756 - 5.5 1 -

CF CF
R 1 - 2.756 2.756 - 5.5 - 4.699 3.6 7.8 R 1 -

1 - 2.756 - 1.142 - 3.898 1 -

CS 1 - 1.142 2.756 - 3.89 - 3.091 3.6 6.63 CS 1 -

1 - 1.142 - 1.142 - 2.284 1 -

CT 1 - 1.142 1.142 - 2.28 - 2.284 3.6 6.053 CT 1 -

1 - 1.142 - 1.142 - 2.284 1 -

CF CF
O 1 - 1.142 1.142 - 2.28 1.142 3.6 4.694 O 1 -

1 - 1.142 - - - - 1 -

COLU Le b( A(m I i Kl ʎ M1 M2 Vd 50- Second

MN order
 h) ^2) 25 or 15/ 25(M1/M2) effect

CF 5.6 0.5 0.25 0.005 0.1414 0.026 5.6183 2.14 -35.4 0.484 25 - Consider

CG 8.3 0.5 0.25 0.002 0.1414 0.026 5.6183 0.55 14.7 0.469 25 - Consider

CFR 7.8 0.5 0.25 0.005 0.1412 0.026 5.6183 28.3 24.56 0.32 26.5165 - Consider

CS 6.63 0.4 0.16 0.002 0.1118 0.01 7.30296 19.4 13.6 0.337 25.8390 - Consider

CT 6.05 0.4 0.16 0.002 0.1118 0.01 7.30296 30.4 25.19 0.22 31.98010 - Consider

CFO 4.69 0.4 0.16 0.002 0.1118 0.01 7.30296 17.0 29.75 0.032 83.85254 - Consider

 SWAY MODE

COLUMN Le,m Nsd,KN M1,KNm M2,KNm ea,mm eo1,mm eo2,mm eo,mm δs δs*eo,mm etot Msd,kNm

CF 5.57 1646.6 2.14 -35.44 18.56 1.299 21.522 13.43 1.43 19.18203 37.7420 62.147550

CG 8.37 1597.0 0.55 14.7 27.9 0.344 9.2 5.657 1.43 8.079052 35.9790 57.460346

CFR 7.8 1077.6 28.3 24.56 26 26.26 22.79 24.17 1.23 29.73894 55.73894 60.068740

CS 6.63 734.1 19.4 13.6 22.1 26.46 18.47 21.66 1.52 32.97565 55.0756 40.432688

CT 6.053 401.1 10.28 7.85 20.18 25.6 19.57 21.98 1.124 24.70776 44.8877 18.006728

CFO 4.694 68.94 0.75 1.25 15.66 10.87 18.13 15.22 1.67 25.38174 41.0417 2.8294176

 We can now provide steel bars for the above moments obtained; using

Biaxial chart. The reinforcements obtained are summarized by the following table. The development
lengths are also shown in the following table.

Column As,Calc,mm2 As,eff.,mm2 φ,mm lb,mm lb,net,mm lo,mm lo,prov.,mm

CF 1172.998 1258.64 20 631.953 588.95 824.53 830

CG 1172.998 1258.64 20 631.953 588.95 824.53 830

CFR 1172.998 1258.64 20 631.953 588.95 824.53 830

CS 750.72 834.132 16 505.562 455 637 640

CT 750.72 834.132 16 505.562 455 637 640

CFO 709.012 834.132 16 505.562 429.73 601.622 610

LATERAL REINFORCEMENT DETERMINATION

 Lateral reinforcement or stirrup is provided, if shear force carrying capacity concrete(V c) is less
than shear force exerted by loads(V cu). There is also
additional shear force carried by the concrete(Vcn).

VC=0.25fctdK1K2bwd where:K1=1.5+50ρ 2

K1=1.5+(50*0.0056)=1.78 2.0

K2=1.6*0.45 1.15

fctd=0.21fcu2/3/1.5=1.0315 & fcu=0.8fck=20

Vc for CF, CG &CF that is column one with 500*500mm x-section will be
0.25*1.0315*1.78*1.15*500*450*1000=118.8K
Vc for column 2 with 400*400mm x-section will be

Ρ=834.132/(400*360)=0.00579; K1=1.79 & K2=1.24 that is; Vc=82.46KN

Vcn=0.1* bwd.*Nsd/Ac ; & so the total shear force resisted by the concrete(Vtot) is

Vtot= Vc+ Vcn

Total shear for columns of each stories is shown bellow.

COLUMN Vtot

CF 148.19

CG 143.73

CFR 96.99

CS 66.07

CT 36.1

CFO 6.204

Maximum shear force =86.11KN which is on first floor @ which Nsd=1077.58.

CHECK FOR SHEAR

 Limiting value of ultimate shear force ;i.e. VRD =0.25*fcd* bwd=265KN

 Vtot first floor =215.79 Scince the shear resisted is greater than the maximum

S Therefore ; provide 8 c/c 192 mm nominal reinf.

Length of lap lo a1lb,net lo,min &

lo,min= 0.3aa1lb

Or 200mm

lb,net = a lb* b.min

For bar in tension; b.min=0.3 b

Or 300mm

For bar in compression ; b.min =0.6 b

Or 200mm where; a=1 for straight bar&

 b is anchorage length

Basic average b= /4*fyd/fbd= /4*(260.87/2.064)=31.6

fbd=2*fctd=2.064

- From EBCS2,1995/Art 2.165 it can be seen that a1=1.

 DEVELOPMENT LENGTH

Column As,Calc,mm2 As,eff.,mm2 φ,mm lb,mm lb,net,mm lo,mm lo,prov.,mm

CF 1172.998 1258.64 20 631.953 588.95 824.53 830

CG 1172.998 1258.64 20 631.953 588.95 824.53 830

CFR 1172.998 1258.64 20 631.953 588.95 824.53 830

CS 750.72 834.132 16 505.562 455 637 640

CT 750.72 834.132 16 505.562 455 637 640

CFO 709.012 834.132 16 505.562 429.73 601.622 610

Column Detailing
 CHAPTER FIVE

DESIGN OF STAIR CASE

Stairs are used to connect to floors that found on different elevation. Stair case analysis and
design is similar to slabs. The inclined configuration is analysed by projecting the loads on a
horizontal plane.

DESIGN OF STAIR CASE

Introduction
Staircase is structural element which provides movement from one floor to another.

Mainly there are two types of Staircases based on carrying and transferring of load to support.
 1. Transverse staircase:-supported on the shorter sides. They may be supported on wall,
beams or cantilever.
2. Longitudinal staircase:-supported on longer sides .They transfer load to the support in the
longer direction
Generally design of stair case is mainly as one way solid slab.
In our project, of the above two types, the second type (longitudinal stair case) is used and
the detail design as follow.

Figure - parts of stair

MODELING

Number of rising=number of going +1

Taking rising=15cm, and thread=30cm

Number of rising= (H/2)/18cm= (3.6/2)/15cm=10

Number of thread=10-1=9

Hence depending on the dimension of the building and the calculated number of
rising and thread the model of the stair case and its dimension will be as follows.
 Angle of inclination of the stair (θ)

Θ=tan-1(1.8/2.7) =33.69o

Depth for deflection requirement

d = le = 5350mm ßa = 24

d= (0.4+0.6fyk/400)*Le/Ba=0.85* Le/Ba

d = 189.48mm

AssuminØ-14 main reinforcement and cover 25mm.

D = d + cover + Ø/2 = 189.48+25+14/2 = 221.48mm

Use D = 225mm d = 225-25-14/2 = 193mm

LOADING

DEAD LOAD

1. INCLINED PART
1. Step

 concrete

= =10.141KN/m

 plastering

= =0.829KN/m

 steps

=0.5*rise*width*unit weight
 =0.5*0.18*1.5*25=3.375KN/m

 marble finish

=1.944KN/m

Total dead load for inclined part=10.14+0.829+3.375+1.944=16.289

2. LANDING
Concrete =depth*width*unit weight
=0.225*1.5*25=8.438KN/m
Plaster = depth*width*unit weight
=0.02*1.5*23=0.69KN/m
Marble = depth*width*unit weight
=0.03*1.5*23=1.215KN/m
Total dead load for landing = 10.343KN/m
LIVE LOAD
Live load=qk + Q*width/A =3+2*1.5/ (1.5*5.35) = 3.374KN/m

DESIGN LOAD

Pd=1.3DL+1.6LL
Pd1=1.3*16.289 + 1.6*3.374 =26.574
Pd2=1.3*10.343 + 1.6*3.343 =18.844
 Case 1

Case 2

Case і M support = (wl2)/8 =18.844*(5.35)2/8=67.42KNm

M span= 9/128*wl2=9/128*18.844* (5.35)2=37.924KNm
2 2 2
Case іі, і M support=wc/(16l)*(3l -c )=(7.73*1.325)/(16*5.35)*[3*(5.35) -

(1.325)2]=10.064KNm
To calculate M span:
MA=0
→RB*5.35-MB-7.73*2.7[(2.7/2+1.325] =0
RB=[10.064+20.871*2.675]/5.35=12.317KN
Mx= RB*x-MB-[w(x-1.325)2]/2
=-3.865x2+22.559x-16.849
d Mx/dx = 0 →x=2.92m
M(2.92)=16.1KNm
→ M span=16.1KNm
Msd span=37.924+16.1=54.024KNm
Msd support=67.42+10.064=77.484KNm
Check depth

d= = =124.32mm <dused …..ok!

Reinforcement provision

Span Msd=54.024KNm

μ= =0.085

Z= = 184.38
 As = = = 1123.18mm2

Spacing, S= = =205.583mm , as= *𝛑 =153.39mm2 , b=1500mm

Smax ≤ , As min =

Smax = 350mm>S= 205.583mm……..ok!

Provide Ø14 c/c 200mm

Support Msd= 77.484KNm

μ= = 0.122

Z= = 180.36

As = = 1646.818mm2

Spacing, S= = =140.179mm , as= *𝛑 =153.39mm2 , b=1500mm

→ S < Smax -------- ok!

Provide Ø14 c/c 140mm

 CHAPTER SIX

FOUNDATION

A foundation is the under lying structure which safely transmits the structural load to beneath ground
surface.

To choose this foundation there are factors which have to be considered like
 Bearing capacity of the soil
 Function of the structure
 Type of load it must carry
 Substructure condition
 Cost
In order to be economical it must be checked whether the total area of footing is greater than
half of the total area of the building or not

Assume soil is medium stiff clay with bearing capacity 200kpa. qall = 200kpa
 qult = F.S* qall = 1.5*200 = 300kpa

To determine the type of foundation, proportioning of footing pad is made for several
columns to estimate the total area of footing using the following formula

qult ≥ {1 ± ± }

Assume overburden pressure of soil & weight of footing comprises 10% of column load

ANALISIS & DESIGN OF FOUNDATION

PART ONE
DESIGN OF FOUNDATION

Foundations are building components which transfer building loads to the underlying soil. They
consist concrete slab located under each structural column and a continuous slab under load bearing
walls.

The following points has to be considered while selecting the appropriate type of foundation

 Engineering parameters of the given soil

 Location of ground water

 Limit of settlement

 Type of structure and the load from superstructure.

There are two basic types of foundation.

 1. Shallow foundation:-are types of foundation which transfer building loads to the soil at the
base of the structure.
Shallow foundations have the following systems

a. Spread footing

b. Mat/Raft foundation
2. Deep foundation:-they transfer structural loads far below the sub-structure. These type of
foundations extend several dozen of feet below the building. They have the following types.
a. Piles
b. Piers
c. Caissons

Considering the following advantages we selected square footing of shallow foundation .The
advantages are:-
 Affordable cost
 Simple construction procedure
 Mostly availability of construction material
 Less need of experts
We used Software which depends on EBCS 2/1995 for the design of the foundation.

The output of the analysis is shown as below

Reinforcement provision and detailed sketch is presented as follows
PART TWO
C25 class 1

Axial force=1646.64

Mx=13.98KN.m My=35.44KN.m

Fcd=11.33KN.m fyd=260.87N/mm2

ξcu =0.0035, ξyd=0.0013

Step1 proportioning

 Assuming it is a square footing (L=B)

δmax =

ey = ex= =0.022m

δmax=

=1646.64

If L=2.0 δmax=485.14KPa
L=2.6 δmax=277.03KPa<300...........ok!
Therefore take L=2.6m
Over burden pressure=unit wt of soil*volume
=16*1.5*2.62
=162.34KN
Reinforced overburden=1.3*(162.24)=210.9
P=1646.64+210.9=1857.54KN

δmin=

=210.14....................ok!
Stress distribution
 δ=

δ =277.03

δ (+,-) =219.695
δ (-,-) =210.143
δ (-,+) =267.148

δavg= /4

=243.586KN/m2

Step 2 depth determination

VP UP

Where vp is the shear due to presenting stress

Vup section resistance

we assume the critical section for wide beam shear to occur

1.5d distance from the face of the column.

Vp =pi+ *Aup,

Aup =CXCY+2(CX*1.5d)+2(CY*1.5d)+∏d2

=0.52+2(0.5*1.5d)*2+7.07d2

=0.25+-3d+7.07d2.........1

Vp =1646.64+ *

=1646.64-212.386(0.25+3d+7.07d2)

=963.543-637.158d-1504.57d2

Vup=0.25fctd*k1k2ud*103

Where, Ud =(2CX+2CY+3∏d)d

=2d+9.42d2
FCtd =

= =1.032

K2=1.6d take K2=1

VUP0.25*1.032*1.08*1*(2d+9.42d2)

=0.5572d+2.624d2.........2

Equate equation 1 and 2

0.5572d+2.624d2=1604.57d+963.543-637.158d

Using quadratic equation

d=601.34mm

D=d+ Ø/2+C

=601.34+14/2+50

=658.34

Take D=670mm, d=670-50-7=613mm

Check whether the punching area is within the pad

1.5*0.613=0.919

1.1695<1.3............ok!
Wide beam shear

In this case the critical section is assumed to occur d distance from face of the column

For wide beam shear VW<VUW,

Where Vw=presenting shear at the section due to wide beam

Along x-x

,
Using similarity of triangles

δx =214.919+47.7

=264.62KPa

Vwx= Ly=296.55

Vwx=0.25fctd*k1*k2bwd

Where, Fctd =1.032, k1=1.08, k2=0.987 take k2=1

Vwx =0.25*1.032*1.08*1*2.6*0.613*106

=444.096KN

Vwx ................ok!

Along Y-Y

By using similarity of triangle we get =238.812+7.986

= 246.79KN

Vwy
 = 245.85 KN

Vuwy = 0.25* 1.032*1.08*1*2.6*0.613*106

= 444.096KN

Vwy > Vuwy OK!

STEP 3. FLEXURAL DESIGN

While designing for flexure we assume the critical section to occur at the face of the column. We
provide reinforcement using the maximum moment resisted.

Along x-x (along y-axis)

By using similarity of triangle we get =249.1KN

=217.9

Mmax =

=128.626KNm

Along y-y (along X-axis)

Mmax =

=117.604|+1.406

=119.01KN

Check depth for flexure

d=

0.019

Msd(KNm) b (m) dcal dprov. remark

m=

X’ 128.628 1m 28.78 0.019 0.0013 697.867 613mm OK!

Y’ 119.01 1m 28.78 0.019 0.00123 597.77 613mm OK!

Spacing by using design table

as= 153.93

Msd(KNm) b (m) d (m) Ks Sprov

As s
Km

X’ 128.628 1m 0.613 18.5 3.95 829.8 185.5 Φ14 c/c 180

Y’ 119.01 1m 0.613 17.79 3.95 766.86 200.72 Φ14 c/c 200

STEP 4 DEVELOPMENT LENGTHS

Lb =0.7φ/4 *

For the column bars l bc =0.7

Bent length =885mm-613mm

=272mm

For the footing l b =0.7

In x-direction 0.61933< [1.3-0.25]

In y- direction 0.619< [1.3-0.25] therefore there is no need to bend the bars.

STEP 5 REINFORCEMENT DETAILING

BAR SCHEDULE

item Schedule spacing

A Φ14 C/C 180

B Φ 14C/C 200

Conclusion

This project was a great exposure for the students to apply the theoretical background which
is gained in the courses in the real world.

The project work enables us to identify

 What basic steps & ideas are gained to be followed in designing the
structure
 What software should be used in doing the project.
 This project work introduces us with the works that are done in
design office.

Generally, this project enabled us to build our confidence as a future civil engineer.