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University of Salahaddin-Hewler College of Engineering Department of Civil Eng.

Prepared By:- -Dilshad Ismail Ahmad -Trifa Sabr Saeed - Kamaran Abdulla Suleyman

Supervised By:-

-Mr.Zrar Sddiq

(2006-2007)

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)32البقره(

They(Angels) said:"Glory be to You,we have no knowledge except what you have taught us.Verily it is you,the All-Knower,the All-Wise."

)32Baqarah -Al(

نا اال ما علمتنا انك قالوا سبحانك ال علم ل* *انت العليم الحكيم

صدق اهللا العظيم

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To:- Our parents whose learened us the way of life and dedicated their love to us, and our teachers for their useful advises and also for every word that they learned, and our project supervisor Mr.Zrar Sddiq who helped us to prepare this project of "Multistorey Design".

Dilshad,Trifa&Kamaran 27/5/2007

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Table Of Contents:

-CHAPTER ONE………………………....SLAB

-CHAPTER TWO………………...........BEAMS

-CHAPTER THREE………………COLUMNS

-REFERENCES

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CHAPTER ONE

Design of Slab using Direct Design Method(DDM)

Occupacy:-Supermarket Primary dimensions:

-Slab Thickness:160 mm -Beam Dimension:(300*560)mm -Column Dimension:(400*400)mm

:Limitations

1-Min. 3 panels in each direction. O.K. 2-l¹/l²=5.8/4.4=1.32<2 Two-way span O.K. 3-Difference in spans=4.8-4.4=0.4 <4.8/3 O.K. 4-Offsets of columns=0 <10% O.K. 5-Calculating α1 & α2 :-

be for T-section=bw+2(h-t)=<bw+8t =300+2(560-160)=<300+8*160 =1100=<1560 Be=1100 mm Be for L-section=bw+(h-t)=<bw+4t =300+(560-160)=<300+4*160 =700=<940 Be=700 mm

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7

-Beams 1&2:

, 12

3bhkb

I =

)ht1)(

wbeb

(1

]3)ht1)(

wbeb

(2)ht()

ht6()[4

ht1)(

wbeb

(1

−+

−++−−+=k

For T-section: K=1.72 For L-section: K=2.6434

α=Ib/Is=4.61

-Beams 3&4:

α=15.455

-Beams 5&6

α=3.8144

-Beams 6&7:

α=11.7244

for all panels:

0.2<((α1*l2^2)/(α2*l1^2))<5.0 O.K.

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6-Calculating L.L and D.L:

Min. uniformly distributed live load

for Supermarket,use 125 ksf

125 ksf=6.25 KN/m^2

Wd.l=wt. of slab+(tile+mortar)+gypsum plaster

:For roof

Wd.l=0.02*16+0.16*24+0.02*12+0.06*18+0.07*24

=7.16 KN/m^2

:For other floors

Wd.l=0.03*23+0.05*23+0.16*24+0.02*16=6.0 KN/M^2

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Wd.l/Wl.l =6.25/6.0=1.04<2 O.K.

.M can be applied.D.All limitations satisfied and D*

.3.3.59 ACI Code according toChecking thickness

4 3 2 1 Panel 4.21 6.19 6.92 8.9 αm

For all panels: αm>2 ,

Use Equ.(9-13)

t min.=(Ln(0.8+(Fy/1400))/(36+9*β))

Ln1=5.8-0.4=5.4 m Ln2=4.8-0.4=4.4 m

Ln=5.4

β=5.4/4.4=1.2273

t min.=0.114 m=114 mm t given=160 mm>114 mm

Use t=160 mm

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:Design Of Floor Slab Total factored static moment=1/8 * Wu*L2*Ln^2

-Long direction(L2=4.8 m,Ln=5.4 m) Wu=1.2*6.0+1.4*6.25 =17.342 KN/m^2 M=303.42 KN.m

-Short direction(L2=5.8 m,Ln=4.4 m)

M=243.41 KN.m

:ve moments in frames+& ve -

for exterior span according to ACI,13-6.3.3 table: Use case 2:

0.57M

0.16M

0.70M

0.35M

0.65M 0.65M

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-Long direction , M=303.42 KN.m

-Short direction , M=243.41 KN.m

-Column strip & Middle strip moments: C for the edge beam:

C=∑(1-0.63x/y)*((x^3 * y)/3)

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C,10^9 ∑(x^3 y)/3 10^9 mm^4

(1-0.63x/y)∑

Y,mm X,mm

0.956 0.856 700 160 1 2.7173 A1

3.6 0.5275 400 300 2 0.546 0.252 400 160 1 1.8384 A2

5.04 0.3375 560 300 2 C=2.7173 * 10^9 mm^4

:Distribution of moments into -Column strip -Middle strip -Beams

:ve moment at face of exterior support-

-long direction,column strip l2/l1=4.8/5.8=0.83 α1*l2/l1=4.61*0.83=3.83>1 Is=1.6384 * 10^9 mm^4 C=2.7173 * 10^9 mm^4

βt=C/(2*Is)=0.83 Using interpolation column strip shall be proportion to resist 93.7% of factored moment.

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Total moment of 48.55KN.m : -beam 85% -Col. st. 93.7% -slab 15% -M.st. 6.3%

Col.st. moment of beam=38.67 KN.m Col.st. moment of slab =6.82 KN.m M.st. moment = 3.06 KN.m

Short direction:

-beam 85% -Col.st. 94.75% -Total moment of 38.94KN.m -slab 15%

-M.st. 5.25%

Moment of beam=31.36 KN.m Moment of slab=5.53 KN.m Moment of M.st.=2.05 KN.m

:ve moment at exterior face of interior support-

-short direction Col st. 68.7%(beam 85% , slab 15%) M.st. 31.3%

of & ve moment at exterior face of first interior support -:typical interior supports

-long direction Col. st. 80.1%(beam 85% , slab 15%) M.st. 19.9%

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:interior spans& ve moment in exterior +

Same as –ve moment at interior support

Interior span Exterior span -ve +ve -ve -ve +ve -ve

197.22 106.2 197.22 212.4 173 48.55 134.28 72.31 134.28 144.61 117.8 38.67 23.7 12.76 23.7 25.52 20.8 6.82 39.25 21.13 39.25 42.27 34.43 3.06

T-M Slab Beam M.st

Width=4.8m M.st. =2.4m Col.st.=2.4 m

158.2 85.2 158.2 170.4 138.74 38.94 92.4 49.75 92.4 99.5 81 31.36 16.3 8.78 16.3 17.6 14.3 5.53 49.5 26.67 49.5 53.34 43.43 2.05

T-M Slab Beam M.st.

Width=5.8m Col.st.=2.9m M.st. =2.9m

:ρCalculation of

M=fy/(0.85f'c)=15.46 R=Mu/(Ф*b*d^2)

)211(1fymR

m −−=ρ

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Using Ф 12mm, 25mm concrete cover:

dL=d(long direc.)=160-25-(12/2)=129 mm

dS=d(short direc.)=129-12=117 mm

:Long direction -Col.st. & M.st:

ρ=0.0647(1-√(1-0.00316Mu))

:Short direction

-Col.st. ρ=0.0647(1-√(1-0.00318Mu))

-M.st.

ρ=0.0647(1-√(1-0.0027Mu))

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Area of steel bars

Ф12 mm ,As=113 mm^2 Smax.=2*t =2*160= 320 mm As=ρbd

-Long direction:

Interior span Exterior span -ve +ve -ve -ve +ve -ve

774 619 774 836 681 619 Col.st.As,mm^2 7 6 7 8 7 6 No. of bar Smax. Smax. Smax. 300 Smax. Smax. Spacing 1300.3 681 1300.3 1393.2 1114.56 619 M.st.As,mm^2 12 7 12 13 10 6 No. of bar 200 Smax. 200 184 240 Smax. Spacing

-Short direction:

Interior span Exterior span -ve +ve -ve -ve +ve -ve

678.6 678.6 678.6 678.6 678.6 678.6 Col.st.As,mm^2 6 6 6 6 6 6 No. of bar Smax. Smax. Smax. Smax. Smax. Smax. Spacing 1526.85 814.32 1526.85 1628.64 1323.27 678.6 M.st.As,mm^2 14 8 14 15 12 6 No. of bar 207 362.5 207 193 240 Smax. Spacing

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Long direction -Col.st.: -ve(top bar) Ф12@30 cm C/C……a +ve(bottom) Ф12@30 cm C/C…….b

-M.st.: -ve, Ф12@15 cm C/C…….c +ve, Ф12@15 cm C/C……d

Short direction -Col.st.: -ve , Ф12@30 cm C/C……e +ve , Ф12@30 cm C/C…...f

-M.st.: -ve , Ф12 @15 cm C/C……g +ve , Ф12@15 cm C/C……h

For corner use Ф10@20 cm C/C

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19

Factored shear strength

Floor area around B3 (most critical beam):

A=4*(0.5*204^2)+2*1 =16.32 m^2

Applied shear force=A*Wu =16.32*17.342=283.02 KN

Resisting shear force=(Ф/6)*√f'c *b*d =(0.75/6)*√20.7 *4.8*0.129 =352 KN

Resisting shear force>Applied shear force

-No shear reinforcement required

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Design Of Roof Slab

D.L=7.16 KN/m^2 L.L=2 KN/m^2

Wu=11.39 KN/m^2

Moment: -long direction=198.75≈200 KN.m -short direction=159.45≈160 KN.m

-Long direction:

Col.st. width=2.4 m M.st. width =2.4 m

d=129 mm , m=15.46

cf

fym

′= , 2

u

bdM

RΦ

=

Col.st: -ve As=619.2 mm^2 Use 8Ф12 @ 30cm C/C +ve As=619.2 mm^2 Use 8Ф12 @ 30cm C/C

M.st.: -ve As=897.84 mm^2 Use 8Ф12 @ 30cm C/C +ve As=743 mm^2 Use 8Ф12 @ 30cm C/C

Interior span Exterior span -ve +ve -ve -ve +ve Ex -ve 130 70 130 140 114 32 Total 25.87 13.93 25.87 27.86 22.69 2.016 M.st. 88.51 47.66 88.51 95.32 77.64 25.49 Beam 15.62 8.41 15.62 16.82 13.7 4.5 Slab

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-Short direction

-Col.st. width=2.9 m -M.st. Width =2.9 m

d=117 mm , m=15.46

Col.st.: -ve M , Use 8Ф12 @ 30 cm C/C +ve M , Use 8Ф12 @ 30 cm C/C

M.st.: -ve M , Use 12Ф12 @ 24 cm C/C +ve M , Use 9Ф12 @ 32 cm C/C

Interior span Exterior span -ve +ve -ve -ve +ve Ex -ve 104 56 104 112 91.2 25.6 Total

35 28.55 M.st. 65.4 53.26 Beam 11.54 9.4 Slab

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CHAPTER TWO

Design Of Beams

BL: Beam in Long direction BS: Beam in Short direction

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Floor Beams-

-Long direction:

Using Moment Distribution Method Of Analysis -:1Grid

=8.8 * 10^-3 212

bhI

3

b ×=

Ic=(0.4*0.4^3)/12= 2.13*10^-3 m^4

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LEI

4K =

Kb=129.78 Kc=60.73

DF(1-2)=129.78/(129.78+2*60.73)=0.52 DF(1-1`,1-1``)=0.24 DF(2-3)=0.34 DF(2-2` ,2-2``)=0.16

2ulW

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1FEM =

Wd.l=Wt. of slab+Wt. of beam+Wt. of wall =(7.92*0.16*24)+(0.3*0.4*5.8*24)+(7.7*5.8) =91.8 KN/5.8 m =15.8 KN/m≈16 KN/m

Wl.l=6.25*7.92=49.5 KN/5.8m =8.53 KN/m≈9 KN/m

Wu=1.2*Wd.l+1.6Wl.l=1.2*16+1.6*9=33.6 KN/m

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FEM=94.19 KN.m

Joint 5 Joint 4 Joint 3 Joint 2 Joint 1

5-5 5-4 4-5 4-4 4-3 3-4 3-3 3-2 2-3 2-2 2-1 1-2 1-1

0.24 0.51 0.34 0.16 0.34 0.34 0.16 0.34 0.34 0.16 0.34 0.51 0.24 DF

0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 COF

94.19 -94.19 94.19 -94.19 94.19 -94.19 94.19 -94.19

FEM

-22.79

-48.04

0 0 0 0 0 0 48.04 22.79 DM

-24.02 24.02 0 COM

8.19 3.84 8.19 -8.19 -3.84 -8.19 DM

4.095 -4.095 -4.095 -4.095

COM

-0.995

-2.089

2.089 0.991 DM

-1.045 1.045 COM

0.36 0.168 0.36 -0.36 -0.167

-0.36 DM

0.18 0.18 -0.18 -0.18 COM

-23.78

48.34 -110.705

4.01 102.74 -89.915

89.915 -102.74

-4.01 110.7 -48.34

23.78 ∑

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According to ACI (8-3-3): Using Coefficient method for calculating moments:

-ve moment at exterior face of first interior support:

10l

WM2

nu=

-ve moments at other faces of interior supports:

11l

WM2

nu=

-ve moment at interior face of first exterior support:

16l

WM2

nu=

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-It is clear that the results of coefficient method are close to (greater or equal)moment dist. Method,so it is possible to use coeff. Method(ACI 8.3.3) to determine moments and shear in design of beams.

Long Direction- Beam BL1:

+ve moment =80.74 KN.m -ve moment =113 KN.m Shear =112 KN

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m=15.46

R=Mu/(0.9*b*d^2)

)272

M01402.046.15211(065.0 u×××

−−=ρ

0015.0f4f

f4.1

y

c

ymin =≥=ρ

-ve As=968.36 mm^2 (5Ф16)

+ve As=775.6 mm^2 (4Ф16) ,ρmin control

Spacing:

(300-2*20-2*10-5*160)/4=40 mm>25 mm O.K

Shear:

KN06.1122lw

15.1v nuu ==

1-If Vu>ФVc Shear reinforcement required

mm600or2d

S,dbc̀f31

Va wu ≤<−

mm300or4d

S,dbc̀f32

Vudbc̀f31

b ww ≤<<−

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2-If (ФVc/2)<Vu<ФVc Min. reinforcement required

75.0;dbc̀f6

Vc

2d

S,f

Sb31

Av

w

maxy

w

=ΦΦ=Φ

==

ФVc=87.7 KN

For beam BL1 : min. shear required S=257 mm

Use Ф10mm@25cm C/C

Beam BL2:

-Ve moment=102.76 KN.m

+Ve moment=70.64 KN.m

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:Floor Beams

Shear Reinf.

+Ve Reinf.

-Ve Reinf.

Shear Force,KN

+Ve M KN.m

-Ve M KN.m

Ф10@25cm 4Ф16 5Ф16 112.1 80.74 113 BL1 Ф10@25cm 4Ф16 5Ф16 97.44 70.64 102.76 BL2

Grid (1)

Ф10@25cm 4Ф20 6Ф20 177.42 127.83 178.97 BL`1 Ф10@25cm 4Ф20 5Ф20 154.3 127.8 162.7 BL`2

Grid (2)

Long Dir.

Ф10@25cm 4Ф16 4Ф16 85.87 47 65.71 BS1 Ф10@25cm 4Ф16 4Ф16 74.67 55.86 71.1 BS2

Grid (A)

Ф10@25cm 4Ф16 5Ф16 139.3 76.12 106.57 BS`1 Ф10@25cm 4Ф16 5Ф16 132.1 90.6 115.3 BS`2

Grid (B)

Short Dir.

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:Roof Beams

Shear Reinf.

+Ve Reinf.

-Ve Reinf.

Shear Force,KN

+Ve M KN.m

-Ve M KN.m

Ф10@25cm 4Ф16 4Ф16 78.7 57 79.39 BL1 Ф10@25cm 4Ф16 4Ф16 68.44 56.7 73 BL2

Grid (1)

Ф10@25cm 4Ф16 5Ф16 114.6 83 115.6 BL`1 Ф10@25cm 4Ф16 4Ф16 99.6 82.56 105 BL`2

Grid (2)

Long Dir.

Ф10@25cm 4Ф16 4Ф16 60.72 33.2 46.46 BS1 Ф10@25cm 4Ф16 4Ф16 57.6 33.2 51 BS2

Grid (A)

Ф10@25cm 4Ф16 4Ф16 78.43 43 60 BS`1 Ф10@25cm 4Ф16 4Ф16 74.41 43 65 BS`2

Grid (B)

Short Dir.

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CHAPTER THREE

Design of Columns

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:Roof

m027.0h03.015e .min =+= Axial load=∑Reactions + Wt. of column

:Ground floor

Case ey,m ex,m P,KN COL. Biaxial 0.1411 0.1313 500.31 C1 Uniaxial

.mine 0.1237 861.72 C2

Biaxial 0.1332 0.0357 839.42 C3 Uniaxial

.mine 0.0319 1525 C4

Uniaxial 0.125 0 894.42 C3` Axial

.mine 0 1566.24 C4`

Case ey,m My,KN.m ex,m Mx,KN.m P,KN COL. Biaxial 0.38 60 0.46 50 131.57 C1

Uniaxial .mine 60 0.274 6.39 218.94 C2

Biaxial 0.32 15.5 0.07 72.2 224.8 C3

Uniaxial .mine 22.5 0.061 10.6 368.5 C4

Uniaxial 0.283 --- --- 72.2 255 C3` Axial 0.029 --- --- 10.6 369.5 C4`

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:Design of Columns for Ground Floor

(C1): P=500.31 KN , ex=0.1313m , ey=0.1411m (Biaxial condition)

ex/h=0.33 ey/h=0.353 Assume ρ=0.01 (ρmin.) Ag=0.16 m^2 = 248 in^2 Ast=1600 mm^2

Using Interaction Diagrams:

8.0400

)20*2(400 =−=γ

Ksi60Mpa420f

Ksi4Mpa28cf

y ====′

From reciprocal method:

0nynxn P1

P1

P1

P1

Φ−

Φ+

Φ=

Φ

.K.OKN31.500KN7.803P

KN5.2295Kip8.515PP

KN1205Kip8.270PP

KN44.1176Kip37.264PP

n

u0

uyny

uxnx

>=Φ===Φ

===Φ===Φ

Use 8Ф16 Ast=1600 mm^2

Ties: -16*Фbar=256 mm -48*Фtie=480 mm -least column dimension=400 mm

Use Ф10@25 cm C/C

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(C`4): P=1566.24 KN (Axial condition)

]fAst)AstAg(cf85.0[8.0P yn +−′Φ=Φ

Use ρmin.=0.01 Ast=0.0016m^2=1600 mm^2 ФPn=2309.8 KN> 1566.24 O.K.

Use 8Ф16 mm with ties of Ф10@25 cm C/C

(C2): P=861.72 KN ex=0.1237 m (Uniaxial condition)

e/h=0.31

Pu=193.64 kip

Controls. minρ.……………… 3.0248*4*65.0

64.193cAgf

Pu ==′Φ

*:-In all columns of Roof and Floors ρmin. Controls

Use 8Ф16 mm and Ties of Ф10 mm@25cm C/C

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References

-Arthur H.Nilson,David Darwin,Charls W.Dolan, (Design of concrete structures,13th eddition)

-Jack C.Mc Cormac,James K.Nelson (Design of reinforced concrete,6th eddition)

-Building Code requirements for structural concrete and commentry(ACI 318M-05)

-Arthur H.Nilson,George Winter (Design of concrete structures,10th eddition)

-Lectures of Reinforced Concrete Design by Dr.Omer Qarani

THE END.