1

NDIMUTO AUGUSTIN INGENIEUR EN CONSTRUCTION & CONSULTANT INDEPENDANT DIRECTEUR TECHNIQUE DU BUREAU D’ETUDE BETRAGEC

EXPERT IMMOBILIER AGREE PAR L’ORDRE DES EVALUATEURS DES BIENS IMMOBILIERS AU RWANDA (IRPV)

CERTIFICATE N° RC/IRPV/063/2011, RÉF. A/009/IRPV/2011 TEL.: 0788350775; E-MAIL: ndimutaug2009@yahoo.fr

B.P: 435 GISENYI / RUBAVU

PROJECT : REINFORCED CONCRETE STRUCTURAL DESIGN

OF TWO STORY RESIDENTIAL BUILDING LOCATED AT

RUBAVU DISTRICT; RWAZA CELL, RUGERERO SECTOR, PLOT

NUMBER : 2058

PROJECT OWNER : Mme UMUZIRANENGE Gisèle

DESIGN CODE : BS 8110 (Practice for structural use of

concrete 2013)

April 2018

2

TABLE OF CONTENTS

I. NOTATIONS AND ABBREVIATIONS ………………………………………….…... 2

II. MATERIAL STRENGTHS ………………………………………………….……...... 5

III. EXPOSURE CONDITIONS ………….…………………………………………….. 5

IV. REINFORCED CONCRETE DESIGN ………………………….……………….….…...6

4.1 DESIGN OF SLAB P12….………………………………….…….………………….…...6

4.2. DESIGN OF BEAM type Along E-E axis……………………….………….…….…...11

4.2.1. Dimensions of the beam …………………………………………….………..….…...12

4.2.2. Calculation of Areas of influence on the beam along E-E axis …………….…...14

4.2.3 Calculation of bending moments and shear forces

using CROSS’s Moment Distribution Method (MDM) ……………………….…. …...…...16

4.2.4. Required steel reinforcement in the beam ……. ……………..….………..….…... 22

4.2.5. Arrangement of steel reinforcement in the beam ………………………………….25

4.3. COLUMN DESIGN ANALYSIS …..………………………………..……………………26

4.3.1. Loads on the column C21…………………………………….………………………26

4.3.2. Ground floor part of the column ………………………………….………………….27

4.3.3. Required steel reinforcement ……………………………..………………………….27

4.4. DESIGN OF PAD FOUNDATION Footing under C21……………………………….29

4.4.1. Soil bearing capacity ………………………….………….………..………………….29

4.4.2. Characteristic load transmitted to the foundation…………………………………..29

4.4.3. Weight of the foundation ……………………………………………………………...29

4.4.4. Foundation base dimensions ………………………………………….……………...29

4.4.5. Checking of the punching shear ………………….…………………….……………..30

4.4.6. Required steel reinforcement for the foundation…………………………………….31

4.4.7. Steel Reinforcement Arrangement…………………………………………..……….32

4.5. DESIGN OF STAIR CASE……………………………………………………………….33

4.5.1 Calculation of load P …………………………………………...…………………….…34

4.5.2. Calculation of load P1……………………………………………………………..……35

4.5.3. Calculation of Maximum Bending moment for beam

P1 as simply supported ……………………………………………………………………..35

4.5.4. Calculation of Maximum Bending moment for beam ……………….……………..35

4.5.5. Calculation of steel reinforcement in the stairs……………………….…………… 36

4.5.6. Steel reinforcement arrangement in the stairs ……………………….…………….37

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I. NOTATION AND ABREVIATION

IS: Indian standard

BS: British Standard

As: Cross sectional area of tensile reinforcement

As’ : Cross sectional area of compressive reinforcement

Asv : Cross sectional area of shear reinforcement in the form of links

Acr: Distance from surface of crack to print if zero strain (crack width)

b : Width of any cross section

bw: Breadth of section width of web

d: Effective dept of section

fcu : Characteristic concrete cube strength

fs :Service strength of steel

fy :Characteristic strength of reinforcement

Gk: Characteristic dead load (Permanent load)

Qk :Characteristic live load (Live load) : Imposed load

Ht : Overall dept of section in the plane of bending :

ho :Effective depth of the beam

Le : Effective height of column :

M: Bending moment

Mu : Ultimate moment of resistance

N: Axial load

n: Total distributed load on the slab panel

n: Ultimate design load

n: number of legs ( branch ) of one stirrup

sv: Spacing links along member

V : Ultimate shear force

v :Design shear stress

vc: Designation concrete shear stress

Z:Lever arm

4

ɣm : Partial safety factor for strength

ɣf: Partial safety factor for load

- Dead load : 𝛾𝑓 = 1.40

- Imposed ( live load) : 𝛾𝑓 = 1.60

FEM : Fixed End Moment

DF : Distribution Factor

COM : Carryover moment

TOTM: Total moment (sum of moments)

COF : Carryover

Rxns : Reaction at supports

AuxV : Auxilliary shear

DiRV : Direct shear

TOT V : shear (algebric sum of shears ie DIRV and AUXV)

Ø: Bar diameter

S = xu = Location of neutral axis

So: Clear span

S: distance center to center between stirrups

Ps: Soil bearing capacity

e: eccentricity

Qf : Punching shear force in foundation

Nf : load transmitted by the column to footing of foundation

q : balanced soil pressure

Ab : Average lateral area of the punching pyramid

Um : Average perimeter of punching pyramid

NC : Characteristic load transmitted by the column to the foundation

ξR:0.559

ρrc: Specific weight of reinforced concrete

Factor

5

ρcp: Specific weight of cement plaster

ρmw : Specific weight of masonry wall

Vmax : Maximum shear force

GK: Dead load ( permanent load to be calculated

QK: (n) : Live load = 1.5 KN/m2 for residential house

Vmax : Maximum shear force

qsw : shear carried by stirrups

φbf: coefficient for the ordinary concrete

HA : HR : Hot rolled high yield bar

R: Mild steel

D: Overall

bf : Width of

hf: Depth of

As (prov) `: Area of steel

As ( reqd) : Area of steel

Asw: Area (cross section) of one leg of stirrup

C/c: Center-to-center

Rb: concrete compressive design strength ( = 1.40 KN/cm2)

Rbt: concrete tensile design strength ( = 0.09KN/cm2)

Rs: Steel design strength ( = 40 KN/cm2)

RSC: Design steel compressive force

Nb: Rb* Abc : Resultant compressive force carried by concrete

NS’: RSC*As’ : Resultant compressive force carried by reinforcement

NS: RSC * AS’ : Tensile force carried by reinforcement

Abt: concrete tensile area ( to be neglected)

Abc: Concrete compressive area

Ab: cross section area of the column

ho: Effective depth of the cross section :

x: 0.8 * s = compressive concrete depth :

S: Location of neutral axis

depth

flange

flange

provide

d require

d

6

Qsinc : Total vertical component of the shear force carried by all inclined bars at

the distance Co = Shear force carried by bent up bars

Co : Projection of stirrups

Qsw : shear force carried by stirrups = Σ Rsw * Asw

Rsw : 0.8 Rs: Design strength of the stirrups and the inclined bars

Qb : shear force carried by concrete in the compression area

Qmax : QD : Maximum shear force in the beam

ʎ : lo Slenderness ratio of column

φ : coefficient taking into account the slenderness ratio of column and the construction

inaccuracies

βSx : Short span coefficient in slab design

F/C : Footing under column

II. MATERIAL STRENGTHS

1. ( Cube strength of concrete (cu) = 25N/mm2

2. Density of concrete (concrete) = 24KN/m3

3. Characteristic strength of reinforcement (y) = 250N/mm2 ( Mild steel)

4. Characteristic strength of reinforcement (y) =460N/mm2 ( High yield steel)

III. EXPOSURE CONDITIONS (According to BS code of Practice for structural use

of concrete 2013)

1. Fire resistance of 1.5hrs for all members

2. Members in contact with soil : 50mm cover for very severe conditions

3. Members not in contact with soil : 30mm cover for very severe conditions

4. 25mm cover for staircase members.

5. 30mm cover for the beam

6. 50mm cover for the footing of foundation

7. 25mmcover for slab

a

7

IV. REINFORCED CONCRETE DESIGN

Ref Calculation Output

BS8110

2013

4.1 DESIGN OF SLAB critical Panel nº 14

Durability and fire resistance

Nominal cover for very moderate conditions of

Exposure = 25mm

Nominal cover for 1.5 hours fire resistance

=20 mm

Since 25>20, provide nominal cover 25mm

Provide nominal cover =

25mm

Preliminary sizing of slab

𝑙𝑥

40≤ ℎ𝑜 ≤

𝑙𝑥

25=

3.90

40≤ ℎ ≤

3.90

25

9.75 ≤ ℎ ≤ 15.60

Taken h = 15cm

Effective depth in all direction of the slab

ho = 15cm – 2.5cm = 12.5cm

Because in general the

range for the thickness of

slab ≤ ℎ ≤ 20cm;

12 cm ℎ =15cm

we take ℎ = 15 cm

ℎo=12.5 cm

REINFORCED CONCRETE DESIGN

Ref Calculation Output

Sketch

Lx: 3900

Ly = 5200

Ly = 5200mm Lx= 3900 mm

λ= Ly/Lx= 5200/3900= 1.33 < 2

Hence slab is designed as two ways span

with four edges continuous.

The critical panel (p14)

The chosen is the slab with the

largest side in order to obtain

the greatest thickness of the

slab ( h= 12 cm)

The panel is determined by

simultaneous parallel vertical

axis 2-2; 4—4 and parallel

horizontal axis B-B; C-C.

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Loading

-Self weight of slab

1.40*0.15*1*1*24= 5.04KN/m2

-Finishes =1.40*1.5=2.1KN/m2

Total dead load =5.04KN/m2 + 2.10 KN/m2

= 7.14 KN/m2

Total dead Load = 7.14KN/m2

REINFORCED CONCRETE DESIGN

Ref Calculation Output

BS Standard

8110

Design live load for residential house =

1.60*1.50KN/m2 = 2.40kN/m2

Design load (n) = 7.14KN/m2 + 2.40kN/m2=9.54KN/m2

For a 1m width, n=9.54 KN ( n=Total distributed load on

the slab panel)

Bending moment in simply slab supported slab

According to the moment coefficients related to the

design of slabs , cfr lecturer note table page 67

λ = Ly = 5.20 = 1.33 ~ 1.30 Lx 3.90 For panel P14 with four fixed sides( continuous edges);

Msx = ∝ sx * n Lx2

Mx- = 0.062 * 9.54 * 3.90 * 3.90 = 9.00 KNm

Mx+ = 0.027 * 9.54 * 3.90 * 3.90 = 3.92 KNm

Msy = ∝ sx * n Lx2

Mx- = 0.037 * 9.54 * 3.90 * 3.90 = 5.37 KNm

Mx+ = 0.016 * 9.54 * 3.90 * 3.90 = 2.32 KNm

Live Load:

Imposed load

Qk=

2.40KN/m2

n=

9.54KN/m2

9

Conclusion

Negative Mmax = 9.00KN/m ( For to used in design of

the required steel reinforcement at the top of slab

Positive Mmax : 3.92KNm ( For to use in design of

the required steel reinforcement the bottom of the

slab)

Reinforcement Analysis

Effective depth = ho = 15cm -2,5 cm = 12.50 cm

a. Required steel at the top

∝ 𝑚 = Mmax = 9.00KNm x 100 = 0.041

Rb* b * ho2 1.40*100*(12.50)2 ≅ 0.039 available in the table

∝ 𝑚 = 0.0.39 ξ = 0.04; = n = 0.980 ( see

table of coefficients relative to the design of members

subjected to bending moment page 65) of annex

RCDI

Ās= Mmax = 9.00*100 = 1.837cm2 n * Rs * ho 0.980*40*12.5

Ās = 1.66 cm2 =2 ø 12 = 2.26 cm2 ( not suffiscient)

Taken 5 ø12/m provide ø12/20cm. In general the

minimum bars required per meter the slab is taken as

5 bars ø 12

REINFORCED CONCRETE DESIGN

Ref Calculation Output

b) Required Steel at the bottom

∝m = Mmax = 3.92 x 100 = 0.018

Rb* b*ho2 1.40 *100*( 12.5)2

∝ 𝑚 = 0.031 ξ = 0.03; n= 0.985 Singly reinforced section

As+ = Mmax = 3.92 * 100 =0.792 cm2

n*RS*ho 0.990*40*12.5

As+= 1.010 cm2 ~ = 2 ø 8 ( not sufficient)

Taken 5ø12/m provide ø 12/20cm ( 5 bars min / m in slab)

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ARRANGEMENT OF STEEL REINFORCEMENT IN THE SLAB

a) Transverse cross section

Ln/3 Ln/3 Ln/3 Ln/3

Ref Calculation Output

B. Plan view cross section

Ln Ln

5Ø12/m

5Ø12/m

025L Φ12/20cm

x

Φ12/20cm

Ly = 5.20m

Φ12/20cm

0,25L

Lx = 3.90m

y

y

Ø12/20 cm

Ø12/20 cm Ø12/20 cm

Ø12/20 cm

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REINFORCED CONCRETE DESIGN

Ref 4.2. DESIGN OF TYPICAL BEAM along C-C axis

Output

- The total height (ht) of the beam has to be in the

range below :

ℓy < ht < ℓy =520 < ht < 520 = 34.67 < ht < 65

15 8 15 8

Taken ht = 50 cm

- The Breadth of the section (bw) of the beam has to

be in the range below :

0.50 <bw < 1 = 0.50 = bw = b = 25 cm

ht 50

Taken : bw = 30 cm

- The flange (bf’) of the beam has to be the lesser of a) ℓy = 520 = 173.33cm ~ 175cm

3 3

b) ℓx = 390 = 195 cm 2 2

c) 12hf+b = ( 12 * 15) + 25 = 205 cm

d) Taken bf’ = 175 cm

ht : 50 cm

bw = 30 cm

bf’ = 180 cm

REINFORCED CONCRETE DESIGN

Ref Calculation Output

BSS8110

4.2.1. Dimensions of the beam

( T. section)

Sketch

bf

ht

bw

bf=175cm

bw = 30cm

hf = 15 cm

ht = 50cm

hf

12

4.2.2. STRUCTURAL PLAN SHOWING LOCATION OF COLUMNS AND BEAMS

4.2.3. Influence areas chart showing the beam type along c-c axis that is

considered as overloaded

11111

420 520 320 270

A B C D E

C

100

100

200 C

370

340 100 100 100

100 100 100

370

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4.2.4. Calculation of areas of influence on the typical longitudinal beam along C-C axis

A1a = 1.00+4.20 x 1.70 = 4.42m2

2

A1b = 1.00+4.20 x 1.85 = 4.81m2

2

A2a = 1.00+ 5.20 x 1.00 = 3.10m2

2

A2b = 1.00+ 5.20 x 1.85 = 3.74m2

2

A3a = 100+3.20 x1.00 = 2.10m2

2

A3b = 3.20+1.35 = 2.16m2

2

A4a = 1.00 + 2.70 x 1.00 = 1.85m2

2

A4b = 2.70+ 1.35 x 1.82 = 1.82m2

2

Summary

A1 = A1a + A1b = 4.42 m2 + 4.81 m2 = 9.23 m2

A2 = A2a + A2b = 3.10 m2 + 3.74 m2 = 6.84 m2

A3 = A3a + A3b = 2.10 m2 + 2.16 m2 = 4.26 m2

A4 = A4a + A4b = 1.85 m2 + 1.82 m2 = 3.67 m2

4.2.5. Calculation of dead load on the typical longitudinal beam C-C axis

Static calculation chart

A B C D E L 1 L 2 L 3 L 4

4.20m 5.20 m 3.20 m 2.70 m

14

a) Dead load on span AB

- Self-weight of slab on span AB = 9.23 m2 * 0.125m * 24KN/m3 = 27.69KN

- Finishes = 9.23m2 * 0.50KN/m2 = 4.62KN,where 0.50KN/m2 is permanent load for

finishes (Béton armé, guide de calcul page 25)

- Maconnery wall = 0.20m * 3.00m * 3.00m * 18KN/m3 = 32.40KN

- Plaster on the wall = 0.03m * 3.00m * 3.00m * 2 * 20KN/m3 = 10.80KN

- Self-weight of beam span AB = (0.50m – 0.125m) * 4.20 m * 0.30 m * KN/m3

= 0.375m * 4.20 * 0.30m * 24KN/m3

= 11.34KN

- Total dead load on span AB = 27.69KN+4.62KN+32.40KN+10.80KN+11.34KN

= 86.85KN

b) Dead load on span BC

- Self-weight of slab on span BC = 6.84 m2 * 0.125m * 24KN/m3 = 20.52KN

- Finishes = 6.84m2 * 0.50KN/m2 = 3.42KN

- Maconnery wall = 0.KN No wall above span BC ( beam crossing the living room)

- Plaster on the wall = 0 KN

- Self-weight of beam span BC = (0.50m – 0.125m) * 5.20 m * 0.30 m * 24KN/m3

= 0.375m * 5.20 * 0.30m * 24KN/m3

= 14.04KN

- Total dead load on span BC = 20.52KN+3.42KN+14.04KN= 37.98KN

C. Dead load on Span CD

- Self weight of slab on span CD = 4.26m2 x 0.125m x 24KN/m3 = 12.78 KN

- Finishes = 4.26m2 x 0.5 = 2.13 KN

- Maconnery wall = 0.20m * 3.00m * 3.20m * 18KN/m3=34.56KN

- Plaster on the wall = 0.03m * 3.00m * 3.20 m * 2 * 20 KN/m3= 11.52KN

- Self weight of beam span CD = ( 0.50 – 0.125) x 3.20m x 0.30m x 24KN/m3 = 14.58 KN

= 0.375m * 3.20m * 0.30m * 24KN/m3

= 8.64 KN

15

-Total dead load on span CD = 12.78 KN + 2.13 KN + 34.56 KN + 11.52 KN + 8.64 KN

= 69.63 KN

C. Dead load on Span DE

- Self weight of slab on span DE = 3.67m2 x 0.125 x 24KN/m3 = 11.01 KN

- Finishes = 3.67m2 x 0.50KN/m2 = 1.84 KN

- Maconnery wall = 0.20m * 2.70m * 3.00m * 18KN/m3 = 29.16KN

- Plaster on the wall = 0.03m * 2.70m * 3.00 m * 2 * 20 KN/m3= 9.72KN

- Self weight of beam span DE = ( 0.50 – 0.125) x 2.70 x 0.30m x 24KN/m3

= 0.375m * 2.70m * 0.30m * 24 KN/m3

= 7.29KN

- Total dead load on span DE = 11.01 KN + 1.84KN + 29.16KN + 9.72KN = 51.73 KN

4.2.6. Calculations of live load on the typical longitudinal beam along C –C axis

Because of the purpose of the building, ( Residential) we assume that the live load is taken

as 1.50 KN/m2 = QK

4.2.7. Calculation of dead load per meter on the beam type = GK

Span AB= 86.85KN / 4.20m = 20.68KN/m

Span BC= 37.98KN / 5.20m = 7.31KN/m

Span CD= 69.63KN / 3.20m = 21.76KN/m

Span DE= 51.73KN / 2.70m = 19.16KN/m

4.2.8. Calculations of combination of load ( Design load ) P= n = 1.40GK+1.60Qk

Span AB = (1.40 x 20.68) + ( 1.60 x 1.50) = 28.96 KN /m + 2.40KN/m = 31.36KN/m

Span BC = (1.40 x 7.31) + ( 1.60 x 1.50) = 10.24 KN /m + 2.40KN/m = 12.64KN/m

Span CD = (1.40 x 21.76) + ( 1.60 x 1.50) = 30.47 KN /m + 2.40KN/m = 32.87KN/m

Span DE = (1.40 x 19.16) + ( 1.60 x 1.50) = 26.83 KN /m + 2.40KN/m = 29.23KN/m

16

4.2.9. Calculation of bending moments and shear forces using CROSS’s moments

Distribution method (MDM)

4.2.9.1. Calculation of Fixed End Moments ( FEM )

a) FEM for span AB and BA

- +

Mx- = WL12

12

Mx- = -31.36 * 4.20 * 4.20

12

Mx- = -46.10KN.m

b) FEM for span BC and CB

- +

Mx- = WL22

12

Mx- = -12.64 * 5.20 * 5.20

12

Mx- = -28.48KN.m

c) FEM for span CD and DC

Mx

31.36KN/m = w

L1= 4.20m

Mx- = +WL12

12

Mx+ = -31.36 * 4.20 * 4.20

12

Mx+= + 46.10KN.m

Mx

12.64KN/m = w

L2= 5.20m

Mx- = +WL22

12

Mx+ = +12.64 * 5.20 * 5.20

12

Mx+= + 28.48KN.m

Mx

Mx

32.87KN/m = w

17

- +

Mx- = WL32

12

Mx- = -32.87 * 3.20 * 3.20

12

Mx- = - 28.05KN.m

d) FEM for span DE and ED

- +

Mx- = WL42

12

Mx- = -29.23 * 2.70 * 2.70

12

Mx- = - 17.76KN.m

4.2.9.2. Calculation of Distribution Factors ( DF)

Considering that :

- Joints A and E are rigid, so that moments at these points are zero

- Joints B attaching span BA and span BC

- Joints C attaching span CB and CD

- Joints D attaching span DC and span DE

According to formulas for calculation of DF and K value below:

DF= 𝐾( 𝑣𝑎𝑙𝑢𝑒)𝑜𝑓 𝑟𝑒𝑙𝑒𝑣𝑎𝑛𝑡 𝑠𝑝𝑎𝑛

Ʃ𝐾(𝑣𝑎𝑙𝑢𝑒)𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑝𝑎𝑛 , where F is relative stiffness

K = 𝐼

𝐿, where I = moment of Inertia and = Length of relevant span

El taken as constant therefore I =1 and K 𝐼

𝐿 =

𝐼

𝐿 in the case of far end fixed

a) DF of span BA

Mx

L3= 3.20m

Mx- = +WL32

12

Mx+ = +32.87 * 3.20 * 3.20

12

Mx+= + 28.05KN.m

Mx

29.23KN/m = w

L4= 2.70m

Mx- = +WL42

12

Mx+ = +29.23 * 2.70 * 2.70

12

Mx+= + 17.76KN.m

Mx

18

4.20 = 4.20 = = = 0.558 1 1 1 1 0.24 + 0.19 0.43 + + 4.20 5.20 4.20 5.20

b) DF of span BC

5.20 = 0.19 = = 0.442 1 1 0.19 + 0.24 0.49 +

5.20 4.20

c) DF of span BC

5.20 = 0.19 = = 0.380 1 1 0.19 + 0.31 0.50 + 5.20 3.20

d) DF span CD

3.20 = 0.31 = = 0.620 1 1 0.31 + 0.19 0.50 + 3.20 5.20

e) DF of span DC

3.20 = 0.31 = = 0.456 1 1 0.31 + 0.37 0.68 + 3.20 2.70

f) DF of span DE

2.70 = 0.37 = = 0.544 1 1 0.37 + 0.31 0.68 + 2.70 3.20

4.2.9.3. Calculation of distribution moment or balanced moment ( BM)and

carryover of moment (COM)

(See on Moment Distribution Table below )

4.2.9.4. Calculation of Direct Shears (DIRV) and Auxiliary Shears ( AUXV)

l DF =

l 0.24

0.24

l DF =

0.19

l DF =

0.19

l DF =

0.31

l DF =

0.31

l DF =

0.37

19

(See on Moment Distribution Table below)

4.2.9.5. Calculation of TOT Moments ( TOTM), Total Shears ( TOTV) and Reactions

at Supports ( RXNS)

(See on Moment Distribution Table below)

4.2.9.6. Moment Distribution Table using balanced Method

TOTM = End moments at supports

TOT V = shear forces at support

RXN’S = reactions at support s

SPAN AB BA BC CB CD DC DE ED

DF 1 0.558 0.442 0.380 0.620 0.456 0.544 1

FEM -46.10 +46.10 -28.48 +28.48 -28.05 +28.05 -17.76 +17.76

Bal # 1 +46.10 -9.83 -7.79 -0.16 -0.27 -4.69 -5.60 -17.76

Com # 1 0 +23.05 -0.08 -3.90 -2.35 -0.14 -8.88 0

Bal # 2 0 -12.82 -10.15 +2.38 +3.88 +4.11 +4.91

Com # 2 0 0 +1.19 -5.08 +2.06 +1.94 0 0

Bal # 3 0 -0.66 -0.53 +1.14 +1.87 -0.88 -1.06

Com # 3 0 0 +0.57 -0.27 -0.44 +0.94 0 0

Bal # 4 0 -0.32 -0.25 +0.27 +0.44 -0.43 -0.51

Com # 4 0 0 +0.14 -0.13 -0.22 +0.22 0 0

Bal # 5 0 -0.08 -0.06 +0.13 +0.22 -0.10 -0.12

Com # 5 0 0 +0.07 -0.03 -0.05 +0.11 0 0

Bal # 6 0 -0.04 -0.03 +0.03 +0.05 -0.05 -0.06 0

Com # 6 0 0 +0.02 -0.02 -0.03 +0.03 0 0

TOTM 0 +45.40 -45.38 +22.84 -22.89 +29.11 -29.08 0

DIRV 65.86 65.86 32.86 32.86 52.59 52.59 39.46 39.46

AUX V 10.81 10.81 4.33 4.33 1.94 1.94 10.77 10.77

TOT V 55.05 76.67 37.19 28.53 50.65 54.53 50.23 28.69

RXN’S 55.05 113.86 79.18 104.76 28.69

20

4.2.10. Calculation of bending moments at midspan

a) MAB = WL12 = 31.36 x 4.20 x 4.20 = 69.15 KN.m

8 8

b) MBC = WL22 = 12.64 x 5.20 x 5.20 = 42.72 KN.m

8 8

c) MBC = WL32 = 32.87 x 3.20 x 3.20 = 42.07KN.m

8 8

d) MBC = WL42 = 29.23 x 2.70 x 2.70 = 26.64 KN.m

8 8

4.2.11. Chart of shears force or supports reactions

A B C D E

L 1 =4.20m L 2=5.20m L 3=3.20m L 4=2.70m

55.05 KN 113.86 KN

79.68KN

28.69KN

104.76KN

21

Summary

Maximum hogging moment = Maximum at support = Mmax- = 45.40 KN.m

Maximum sagging moment = Maximum at midspan = Mmax+ = 69.15 KN .m

Maximum shear force = Vmax = 76.67KN

4.2.12. Required steel reinforcement in the beam type

Effective depth (ℎ𝑜) = 50 cm – 3.00 cm = 47.00 cm

a) Required steel at the top or support reinforcement

∝ 𝑚 = Mmax- 45.40 x 100

Rb * b * ho2 = 1.40 * 30 * 47 * 47

∝ 𝑚 = 0.049 ξ = 0.04 and ᵑ = 0.980

ξ = 0.03 < ξ R = 0.559 The T section is singly reinforced

Thus x = ξ * ℎ𝑜 = 0.03 x 47 < 29 cm = ( ℎ𝑡 − ℎ𝑓) = 0.47- 0.15)

Where ℎ𝑓 is the thickness of the flange of the 𝑇 section. For that the compression area is

reinforced, we have :

As- = Mmax- 45.40 x 100

𝑛 x 𝑅𝑆 x ℎ𝑜 = 0.980 x 40 x 47

Because of the minimum bar in the beam is Ø 12, we must use Ø16

Provide = 3 Ø 16 = 6.03 cm2

b) Required steel at the bottom or mid span reinforcement

∝ 𝑚= Mmax+ = 69.15 * 100 = 0.079

𝑅𝑏 ∗ 𝑏𝑓 ∗ ℎ𝑜 2 1.40 * 175 * 47 * 47

∝ 𝑚 = 0.77 ξ = 0.08 and 𝑛 = 0.960

= 0.049

= 2.464 cm2

= 0.079 cm2

22

ξ = 0.01 < ξ R = 0.559 The T section is singly reinforced

Thus x = ξ x ℎ𝑜 = 0.01 x 47.00 < (ℎ𝑓 − ℎ𝑓 = 15 cm ) ; The compression area is

reinforced

As = 69.15 x 100

0.960 x 40 x 47

Taken 3 Ø 14 = 4.62 cm 2

c) Design of stirrups or shear reinforcement

Vmax ( Maximum shear force) = 76.67 KN

qsw = shear force carried by stirrups

qsw = (Vmax)2

4φ𝑏𝑓 𝑥 𝑅𝑏𝑡 𝑥 𝑏𝑤 𝑥 ℎ𝑜2

Where φ𝑏𝑓 = 1.50

𝑅𝑏𝑡 = 0.09 KN/cm2

qsw = (76.67)2 5878.2889

4 * 1.50 * 0.09 * 30 * 47 * 47 35785.80

Set us use stirrups of Ø 8 Asw = 50.3 mm2 = 0.503 cm2

Rsw = 0.8 x Rs = 0.8 x 40 KN / cm2 = 3.20 KN / cm2

Distance between stirrups ( S)

S = Rsw x Asw x n , where n = number of legs for stirrup

qsw

S = 0.8 x 40 KN/cm2 x 0.503 cm2 x 2

0.164 KN/Cm2

Note: The distance between stirrups must be lesser than the three

following values

1) Smax = 0.75 φbf x Rbt x b x ℎo2

Vmax

= 0.75 x 1.50 x 0.09 x 30 x ( 47)2 = 6709.84 = 87.52 cm

Vmax 76.67

= 3.83 cm2

= = 0.164KN/cm2

= 196.29 cm

23

2) The width of the beam web = bw =30 cm

3) 30 cm

Thus S = Min 87.52cm; 30 cm; 30 cm = 30 cm Taken : Ø 8 @30 cm

4.2.13. Arrangement of steel reinforcement in the beam

a) Longitudinal section

Column Column A

≤ 5 cm 30 cm,30 cm, 30 cm ≤ 5 cm 30 cm 30 cm

Beam C- C

A

a) Cross section a- a

15 cm

≤ 5 cm

3 Ø 16HR

2 Ø 12 construction bars

35 cm

8 Ø 30 (Stirrup )

24

3 Ø 16 HR

Note: In order to respect the homogeneity and architectural appearance, all the beam have

to maintain the same cross section like above one.

4.2.14. DESIGN OF TIE BEAM or PLINTH BEAM ( See lecturer notes)

Without doing calculation, the theory assumes that the cross section of the tie beam

(plinth beam) have to be in the following range.

a) ℎ = 𝐿

15 ~

𝐿

10 where L = greatest span beam

b) b ≥ 20 cm where b= width of the tie beam

c) As

1% of the cross section of the tie beam therefore cross section

characteristics are:

a) ℎ = 4.40

15 ~

4.40

10≫ ℎ = 0.29m ~ 0.44m

Taken h = 0.30m

b) b ≥ 20 cm

Taken b = 0.25

c) As = 1% * 25 cm * 30 cm = 705cm2( minimum value

As = 7.50cm2 = 4 Ø14

Provide 6 Ø 14 = 9.24 cm2

SKECTH

Ø8@20cm 6Ø12 30cm

25cm

4.2.15. DESIGN OF THE LINTEL BEAM

25

Without calculation, the theory assumes that the cross section of the lintel beam have to be

in following range.

a) h>20cm

b) b=thickness of the wall maconery

c) As=1% of the cross section of the lintel beam.

Therefore, cross section characteristics are:

a) h=20cm

) b=20cm

c) As=1%*20cm*20cm=4cm2

SKECTH

4Ø12

REINFORCED CONCRETE DESING

Ref Calculation Output

4.3. COLUMN DESIGN ANALYSIS C22 ( see on structural plan Clear height of ground floor column = 300 cm ( see

architectural plan)

End conditions Condition at top of the column

End of column is connected monolithically to beams on

either side and are at least as deep at the overall

20cm

20cm

26

BS 8110

Condition at bottom of the column

End of column is connected monolithically to beams or to

footing on either side and are at least as deep as the

overall

Dimension of the column ( minimum cross section of

column AB = 25 * 25 cm

a= 250 mm b = 250 mm

𝛽 = 0.7 ( braced column)

H = Total height of column

Effective height of column = ℓo = 𝛽 * H

ℓo = 0.70 * 3.0m = 210 cm

λ ( slenderness ratio) : ℓo = 210 cm = 8.40≈ 8 a 25 cm φ = 0.91 if λ = 8 ( page 25 on table of φ values in

RCDI)

Hence column is to be designed as short braced axially

loaded column ( short column )

Let us do design analysis of one internal column type.

4.3.1. Design analysis of column C22

4.3.1.1. Loads on the column nº C22

a) Column loading area = (2.10 x 4.10 ) + (2.10 x 5.10 ) + (5.10 x 3.80) + (3.80 x 4.10) = 13.58m2

2 2 2 2 2 2 2 2

b) Slab (permanent load) = 1.40 * 0.15m * 24KN/m3 * 13.58m2= 68.44KN

c) Live load from the slab = 1.50 KN / m2 x 1.60 m2 * 13.68m2= 32.59 KN

d) Load from beam 1.40*0.300* 0.375* 7.55 * 24 = 28.54KN

e) Load from the wall maconery = 1.40*0.20*3.00*5*18=75.60KN

S/total 28.54 KN + 75.60 KN = 104.14 KN

f) On floor of column = 1.40 x 0.25 x 0.25 x 3.00 x 24 = 6.30 KN

g) Load from the light roof ≅ Permanent load from slab = 68.44KN = 34.22KN 2 2

27

4.3.2. Ground floor part of the column

N1 = ( 68.44KN + 32.59KN + 104.14KN) x 1 +(6.30KN * 2) + 34.22KN

N1 = 205.17 KN + 12.60KN+34.22KN = 251.99KN ~ 252.00KN

4.3.3. Required steel reinforcement

ℓo = 0.7 * 3.00 = 2.10

λ = 2.10 = 8.40 ˂ 14.3 short column ( see RCDI, page 46)

0.25

If λ = 8.40 ≈ 8 φ = 0.91 (see RCDI, page 25)

AS = =

RS 40

AS = 276.92 – 875 = - 14.95 cm2

40

Negative sign indicate that compression steel reinforcement is not required because AS < 0

Therefore the theory assumes that the minimum percentage of steel reinforcement must be

evaluated as follows

Asmin = 0.004 Ab

Asmin = 0.004 x 25 x 25 = 2.5 cm2

Token = 4 Ø 12 = 3.14 cm2

But, because of the minimum diameter of bar in the column is assumed as Ø 12, we must

use 4 Ø12

Thus we arrange the same steel up to the top floor

Smin = distance between stirrups = 1 / 4 * 12 mm = 3mm

Smax = 12 * Ø max = 12 * 12 = 144 mm ~ 15 cm

S= 300mm

N1

φ

252.00KN – 1.40 * 625

0.91

-Rb * Ab

28

Taken 15 cm

Ref Calculation Output

Token S = 15 cm

Thus we use Ø 8 @ 15 cm

CROSS SECTION OF THE COLUMN C22

25 cm 8 @15cm

Ref Calculation Out

put

4.4. DESIGN OF PAD FOUNDATION footing nº 22

( see page 115 in B.S)

4.4.1. Soil bearing capacity

We assume that PS = 200 KN /m2

4.4.2. Characteristic load transmitted to the foundation

NC = 68.44 + 32.59 + 104.14 x 1 + 6.30 * 2 + 34.22 1.40 1.60 1.40 1.40 1.40

25 cm

6Ø12

29

NC = ( 48.89 + 20.37 +74.39 ) * 1 + 24.44 + 9.00

NC = 143.65 + 24.44 +9.00

NC = 177.09KN

4.4.3. Weight of the foundation

NC = 177.09 KN = 17.71 KN

10 10

4.4.4. Foundation base dimensions

Af = Area of footing = NC + NC / 10 = 177.09 + 17.71KN PS 200

Af = 0.974cm2

af x bf = √0.98𝑐𝑚2 = 0.99 cm ~ 1.00 m

af = bf =0.99m Af = 1.00m2

Because of seismic zone provide 1.20m * 1.20m

4.4.5. Checking of the punching shear

Condition of no punching shear:

Qf = Nf - ∆q ≤ Rbt X Ab

Where : Qf : Punching shear force

N1 = Nf = load transmitted by the column to the foundation

∆q = Balanced soil pressure

Ab = Average lateral area of the punching pyramid

Um : Average perimeter of the punching pyramid

Rbt = Concrete tensile design strength = (0,09 KN / cm2 )

P = pressure = Force = F = N1 = 252.00 KN= Area A Af 14400 cm2 P = 0.018 KN/ cm2 af = bf = sides of footing

30

ac = bc = dimensions of cross section of column

ℎ𝑜 = Effective depth of footing

Let us take ℎ𝑓 = 30 cm ℎ𝑜 = ℎ𝑓 - 5 cm( it is recommended

to take minimum value of ℎ𝑓 with respect the equilibrium of 𝑄𝐹

formula below

ℎ𝑜 = 30 cm – 5 cm = 35 cm

𝑈𝑚 = 2 (𝑎𝑐 + 𝑏𝑐 + 2 ℎ𝑜) = 2 (25 + 25 + 2 𝑥 25)

𝑈𝑚 = 200 cm

𝐴𝑏 = 𝑈𝑚 𝑥 ℎ𝑜 = 200 cm x 25 cm = 5000 cm2

∆q = 𝑃( 𝑎𝑐 + 2 ℎ𝑜 )( 𝑏𝑐 + 2 ℎ𝑜)

∆q = 0.013 𝐾𝑁/𝑐𝑚2 ( 25 + 2 𝑥20) ( 25 + 2 x 25)

∆q = 0.013 𝑥 75 𝑥 75

∆q = 73.13 𝐾𝑁

Thus : 𝑄𝑓 = 𝑁𝑓 − ∆q ≤ 𝑅𝑏𝑡 𝑥 𝐴𝑏

𝑄𝑓 = 252.00 𝐾𝑁 − 73.13 𝐾𝑁 ≤ 0.9 𝑥 5000

𝑄𝑓 = 178.87 𝐾𝑁 < 4500 𝐾𝑁

The condition is satisfaction ; thus No punching shear

4.4.6. Required steel reinforcement for the foundation

𝑀𝑎𝑓 = 𝑀𝑏𝑓 = (𝑃 𝑥 𝑎𝑓

2) (

𝑏𝑓− 𝑏𝑐

2)2

Where: 𝑀𝑎𝑓 : Bending moment about side 𝑎𝑓 of the

Footing

𝑀𝑏𝑓 : Bending moment about side bf of the footing

Thus : 𝑀𝑎𝑓 = 𝑀𝑏𝑓 = 0,013 𝑥 120

2

120− 25

2

= 0.78 x 2256.25 = 1759.88 KN.cm

As = 𝑀𝑚𝑎𝑥

0.9 𝑥 𝑅𝑠 𝑥 ℎ𝑜=

1759.875 𝐾𝑁.𝑐𝑚

0.90 𝑥 40 𝑥 25 = 1.96 cm2

As = 1.96 cm2 because of seismic zone

provide 7Ø14/m provide Ø14@ 20cm

2

31

Cross section

af = 120 cm

4.4.7. Steel Reinforcement Arrangement

Lap length

120cm

15 Dowels

30

bf = 120cm

Ø 14 @ 20 cm

25

25

Stirrup

15

15

15

15

15

32

4.3.2. Design analysis of column C2

4.3.2.1. Load Design analysis of column C2 at intersection of I-I nad 2-2 axis on structural plan

a) Column loading are :

S1 = 2.10 * 1.60 = 3.36m2

S2 = 1.20 * 1.60 = 2.00m2

S/Total = 5.36m2

b) Slab( permanent load) = 1.40*15*24KN/m3 * 5.36m2 = 27.01KN

c) Live load from the slab = 1.50KN/m2 * 1.60*5.36m2= 12.86KN

d) Load from beams = 0.30*( 0.50-0.125) *4.95m*24KN/m3*1.40

= 0.30*0.375m*4.95m*24KN/m3*1.40= 18.71KN

e) Load from maconery wall = 1.40*0.20m*4.95m*3.00m*18KN/m3= 74.85KN

f) One floor of column = 1.40*0.25m*0.25m*3.00m*24KN/m3 = 6.30KN

g) Load from the light roof = permanent load from slab = 27.01KN= 13.51KN

2 2

4.3.2.2. Ground floor part of the column

N1=(27.01KN+12.86KN+74.85KN+18.71KN)+(6.30KN*2)

KN N1=133.43KN+12.60KN+13.51KN=159.54KN

4.3.2.3. Required steel reinforcement

Lo = 07 * 3.00 = 2.10m

2.10m = 8.40 ˂ 14.3 short column

7Ø14

+ 13.51

λ =

33

0.25m

If λ = 8.40 = 8 𝑄 = 0.91 (see RCDI, page 25)

N1 – Rb * Ab

AS = = = = 175.32- 875 = -17.49

RS 40 40

Negative sign indicates that compression steel reinforcement is not required because AS ˂ 0.

Therefore that theory assumes that the minimum percentage of steel reinforcement must be

evaluated as follows

Asmin = 0.004 * Ab

Asmin = 0.004 * 25 * 25 * = 2.5m2

Taken 6 Ø12 = 6.79cm

Thus, we arrange the same steel up to the top floor

Smin = distance between stirrup = ¼ * 12mm = 3mm

Simax = 12 * Ømax = 12 * 12mm = 144mm2≈15cm

S= 300mm

We chouse S= 15cm and we use Ø 8@15cm

Cross section of the column C2 is

4.5. DESIGNA OF PAD FAUNDATION FOOTING Nº2

4.5.1. Soil bearing capacity

We assume that Ps = 200 KN/m2

4.5.2. Characteristic load transmitted

NC = (27.01

1.40+

12.86

1.60+

93.56

1.40)*(

6.30∗2

1.40)+

13.51

1.40

NC = 19.29 + 8.06 + 66.83 + 9 + 9.65

NC = 112.81KN

φ 159.54 – 1.40 * 625

* 625

0.91

6Ø12 25m

25m

34

4.5.3. Weight of the foundation

𝐴𝑓 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑜𝑡𝑖𝑛𝑔 = 𝑁𝑐+

𝑁𝑐

10

𝑃𝑠 =

112.81+11.28

200

𝐴𝑓 = 124.09

200 = 0.62cm2

𝑎𝑓 ∗ 𝑏𝑓 = √0.62= 0.78m, taken 1.00m*1.00m

𝑎𝑓 = 𝑏𝑓 = 1.00m

4.5.4. Checking of the punching shear

Condition of no punching shear

Qƒ = ∆q ≤ Rbt * Ab

Where Qƒ = Punchimg shear force

∆ = Balanced soil pressure

Ab = Average lateral area of the punching pyramid

Um=Average perimeter of the punching pyramid

Rbt = Concrete tensile design strength = 0.9KN/cm2

P= Pressure on the soil =𝐹𝑜𝑟𝑐𝑒

𝐴𝑟𝑒𝑎 =

𝐹

𝐴 =

𝑁1

𝐴𝑓=

159.54𝐾𝑁

10000𝑐𝑚2

P= 0.016 KN/cm2

𝑎𝑓 = 𝑏𝑓= sides of footing

𝑎𝑐 = 𝑏𝑐 = 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑜𝑓 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛

ℎ𝑜 = 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑜𝑡𝑖𝑛𝑔

Let us take ℎ𝑓 = 25 𝑐𝑚 = ℎ𝑜 = ℎ𝑓 − 5 𝑐𝑚

ℎ𝑜 = 25 𝑐𝑚 − 5𝑐𝑚 = 20𝑐𝑚

um = 2(act+bc+2ho) = 2 (25+25+2*20)

Um = 120cm

Ab = Um *ho = 180cm* 20cm = 3600cm2

∆𝑞 = 𝑃(𝑎𝑐 + 2ℎ𝑜)(𝑏𝑐 + 2ℎ𝑜)

∆𝑞 = 0.14𝐾𝑁

𝑐𝑚2 * (25 + 2 ∗ 20) (25 + 2 ∗ 20)

∆𝑞 = 0.14 ∗ 65 ∗ 65 = 59.15𝐾𝑁

35

Thus 𝑄𝑓 = 𝑁𝑓 − ∆𝑞 ≤ 𝑅𝑏𝑡 ∗ 𝐴𝑏 = 159.54𝐾𝑁 − 59.15𝐾𝑁 ≤ 0.9 ∗ 3600𝑐𝑚

𝑄𝑓 = 100.39𝐾𝑁 < 3240𝐾𝑁

The condition is satisfaction; therefore no punching shear

4.5.4. Required steel reinforcement for the foundation

𝑀𝑐𝑓 = 𝑀𝑏𝑓 = (𝑃∗𝑎𝑓

2) (

𝑏𝑓−𝑏𝑐

2)2

Where 𝑀𝑎𝑓 = Bending moment about side 𝑎𝑓 of the footing

𝑀𝑎𝑓 = Bending moment about side 𝑏𝑓 the footing

Thus : 𝑀𝑎𝑓 = 𝑀𝑏𝑓 = (0.14∗100

2) (

100−25

2) = 0.70*1406.25 = 984.376KN.cm

𝐴𝑠= 𝑀𝑚𝑎𝑥

0.9∗𝑅𝑠∗ℎ𝑜 =

984.375𝐾𝑁

0.90 .40∗20 = 1.37cm2

Because of sismic zone, we provide 6Ø14/m

Cross section

4.5.5. Steel reinforcement arrangement

25cm

af = 100 cm

Ø14@20cm

6Ø14

bf = 100 cm

Stirrup = Ø

8@15cm

25 cm

25 cm

36

100cm

Ref Calculation Output

BS8110

4.6.DESIGN OF STAIR CASE ( Reinforced

concrete)

Durability and fire resistance

Nominal cover for very moderate condition of exposure =

25mm

Nominal cover for 1.5 hours fire resistance =20mm

Since 25>20,provide nominal cover =25mm

Therefore durability and fire resistance are satisfactory

Preliminary sizing of staircase members

Height from ground floor slab to first floor

slab=3000mm

Height from ground floor landing=3000/2=1500mm

Provide

nominal

Cover=25mm

R=183mm

G=278mm

a) Plan view

120cm

250cm

490cm

120cm

37

c) Horizontal equivalent slab

4.6.1 Calculation of load P

- Tg∝ = H/2 = 150 = 0.600 ∝ = 30º 96

L 250

- Thickness of horizontal equivalent slab

ℎ= dℓ + 2 𝐻 1 = 18 + 2 x 18.33 = 33.21 cm

Cos∝ 3 0.85749 3

Where dl = thickness of slab of stair case and h = waist of slab of stair case and H1 = rise of stair

- Self load = 1.40 * 0.332m *1m * 24KN/m3 = 11.16 KN/m

- Finishes = 1.40 * 1.50 = 2.10 KN /m

- Live load = 1.60 * 3KN/m2*1m = 4.80KN/m

b) Vertical cross

Vertical cross section

1.50 m = H/2 =30º96º

L = 2.50m

1.20

L = 2.50

1.20

L = 4.90m

P

P1 P1

120 120

L= 4.90

38

Calculation of load P = 11.16 + 2.10 + 4.8 = 18.06 KN /m

4.6.2. Calculation of load P1

5. Self weight = 1.40 * 0.18 * 1 * 1 * 24 KN/m3 = 6.05 KN/m

6. Finishes = 1.40 x 1.50 = 2.1 KN/m

7. Live load = 1.60 x 3 = 4.80 KN/m

Total load P1 = 6.05KN/m + 2.10KN/m + 4.80 KN/m = 12.95 KN/m

4.5.3 Calculation of Maximum Bending moment for beam P1 as simply supported

P1 = 12.95 KN/m

L = 4.90m

Mmax 1 = P1L2 = 12.95 * 4.90 * 4.90 = 38.87 KN.m 8 8 4.5.4. Calculation of Maximum Bending moment for beam

P2 = 18.06KN/m – 12.95KN/m= 5.11KN/m

A 2.50 B

L/2 L/2

L= 4.90 m

P = 18.06KN/m

P 1= 12.95KN/m

P1 = 12.95KN/m

1.20 1.20

39

* RA = RB = PL = 5.11*2.50 = 6.39KN

2 2

Mx x RA

Mx – 6.39 * 0 = Mx 6.39 * in the range ( 0 ˂ X ˂1.00)

Mx

RA = 6.39KN

Mx – RA *X + P2 ( X – 1.00)2

2

Mx = RA * X - P2 ( X – 1.00)2 in the range ( 0 ˂ X ˂ 3.10)

2

Mx = 6.39 X – 5.11 ( X – 1.00)2

2

We know that d Mx = TX( Shear force) = 6.39 – 5.11 ( X – 1.00)

dx = 6.39 – 5.11x + 5.11

Tx = 11.50 – 5.11 -5.11x +11.50

X = 11.50 = 2.25m - 5.11 b) Mmax2 = RA *X – P2 ( X-1.00)2 = 6.39 * 2.25 – 5.11 * (2.25 – 1.00)2 2 2 Mmax2 = 14.38 – 5.11 * 0.36

Mmax2 = 14.38 – 3.22 = 11.16KN.m

4.6.4. Calculation of Mmax

Mmax = Mmax1 + Mmax2

Mmax = 38.87KN.m + 11.16 KN.m

Mmax =50.03 KN.m

40

4.6.5 . calculation of steel reinforcement in the stairs

Ho = h -2.5cm = 33.21cm – 2.50cm = 30.71cm

∝m= Total Mmax = 50.03 x100 = 0.038

Rb x b xh2o 1.40 x100 x 30.71x 30.71 From the table of coefficients related to the design of members subjected to bending moment

∝m = 0.038 n = 0.980

Main steel reinforcement . AsM = Total Mmax = 50.03 x 100 = 4.16cm/m = 3Ø14 n x ho x Rs 0.980 x 30.71 x 40

Because of we have to use at least 5 Φ 14/ m in the slab, therefore we take the

minimum Provide 1 Φ14 @ 20cm as main steel reinforcement, it means 1 Ø14@20cm

Distribution steel reinforcement

Provide 1 Φ14@20cm as main steel reinforcement Distribution steel reinforcement

AsD = AsM * ⅕ = 7.70cm2÷⅕ = 1.54cm2

For the same reason, we choose the minimum such as 5 Φ 12 = 5. 65 cm2,

thus, Provide 1Φ12 @ 20cm as distribution steel reinforcement

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4.6.6. Steel reinforcement arrangement in the stairs

18.33cm

27.78cm 5ø12/m

5ø14/m

5ø12/m 1.20 2.50 1.20

R= Rise = 18.33cm

T= Tread = 27.78 cm

W = Waist slab = 33.21cm

Dl = Thickness of slab = 18 cm

Design by : Augustin NDIMUTO

Certified civil Engineer