Two Way Slab Design (DRAFT)

7/28/2019 Two Way Slab Design (DRAFT)

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Two Way

Slab

Fc' 21 Mpa

Fy 420 Mpa

Stress Block depth () Coef. 0.850 0.85 Minimum 0 .65 Max 0.85

cbalance 0.593

a balance 0.504 *d

a maxallowed by ACI (0.75

a b) 0.378 *d =------------------------------------- 0.428

Z=d-a/2=d(1-a/2) 0.811 *d

Nc (Concrete Compressive Force) 0.321 f'c*d*b

M 0.261 f'c*b*d

Max RU by ACI =(0.9*M)*fc

*b*d 4.925 units N, mm

Max Moment Allowed in the

section (Ru*b*d) 4.925 b*d Mpa

max allowed as per ACI318 0.0161

Slab Thickness (h)

According to ACI 318-99 Chapter 9

For a panel with beams

between supports on all

sides, the realtive stifness of

beams in two perpendicular

directions shall between .2

and 5.0 m

Column Section mm C 0.35 m

Span in long Direction Ll 6.25 m

Span in short Direction Ls 5 m

Net Span in long Direction LLn' 5.9 m

Net Span in short Direction Lsn' 4.65 m

Ratio between net span in

long Direction to net span in

short direction ' 1.268817204Width of beam in long

direction b beam s 25 cm no needWidth of beam in short

direction b beam l 25 no needWidth of Slab in long

Direction blslab ???:Width of Slab in short

Direction bs slab ?????

Thickness of Slab Assume 10 cm no need

Thickness of beam assume 30 no needMoment of Inertia of beam in

long Direction Ibl 56250 cm4 no needomen o ner a o eam n

short Direction Ibs 56250 cm 4 no needomen o ner a o s a n

long direction Isl 520.8333333Moment of Inertia of slab in

short direction Iss 416.6666667

units are kg, cm

cb=d*612612+Fy becaus

While ACI equation iscb = d* 600

600+Fy because


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1

Ratio of flextural stifness of

beam section to flexural

stifness of slab 1 in direction Ls 1 in direction L L 2Ratio of length of continous

edges to total perimeter ofslab panel s 1 checkAverage for all sides m 1 check

Iss=

Isl=

LL/Ls 1.25

= 1

1/ 2 =

=ACI says no

need 1.5625 Ok.

ACI says no

need

= 0.154472727 m

20 mm

d=h-cover -5mm 129 mm

Loads

Dead Load DL = 3.75 KN /m2

Live Load LL = 5 KN /m2

W (Total Load )

1.4*DL+1.7*LL

= 13.75 kN/m2

MomentsLong Span ML (W*Ls*LLn' 2)/8 299.1484375 KN.m

1.25

from ACI 318 para 7.7.1 Concrete Cover is

LL*h12

Ls*h12

IssIsl

Ib s X IssIsl Ib L

1X L2 2XLI

Ibs

Ib L

Assuming the same

section of beam inboth diections

0.2 1X L2 5

2XLI

h= LLn'(0.8+fy/1500) ACI eq. no 9-1136+5 '(m - 0.2)

h= LLn'(0.8+fy/1500) ACI eq. no 9-1236+9'

if 1/2 < 0.2 the provsions of ACI 9.5.3.2 shall apply

if 0.2 < 1/2 < 2 ACI eq. no 9-11 shall apply . But h shall not taken less than 120 mmif 1/2 >2 ACI eq. no 9-12 shall apply . But h shall not taken less than 90 mm

take it as


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Total Negative Moment (M

neg.) 0.65*ML = -194.446484 KN.mTotal Positive Moment (M

pos.) 0.35*ML = 104.7019531 KN.m

Column Strip neg. M Fm*M neg.= -157.501652 KN.m

Middle Strip neg. M

M neg -Fm*M

neg. = -36.944832 KN.m

Column Strip pos. M Fm* M pos 84.80858203 KN.m

Middle Strip pos. M

M pos-Fm*M

pos 19.89337109 KN.m

Short Span Ms(W*LL*Lsn'

^2)/8 232.2729492 KN.m

Total Negative Moment (M

neg.) 0.65*Ms = -150.977417 KN.m

Total Positive Moment (M

pos.) 0.35*Ms = 81.29553223 KN.m

Column Strip neg. M Fm''*M neg.= -101.909756 KN.m

Middle Strip neg. M

M neg-Fm*M

neg = -49.0676605 KN.m

Column Strip pos. M Fm'' * M pos 54.87448425 KN.m

Middle Strip pos. M

M pos-Fm*M

pos= 26.42104797 KN.m

Column Strip Coef. Long

Direction

Find Fm Ll/Ls 0.8

From Table 1 Fm 0.81

Column Strip Coef. Short

Direction

Ls/LL 1.25

Find Fm'' 0.675

Sample Caculation ---------------

- Moments

KN.m

+Moments

KN.m

- Moments

KN.m

+ Moments

KN.m

beam -133.876 72.087 -86.623 46.643

Slab (half) -11.813 6.361 -7.643 4.116

-36.945 19.893 -49.068 26.421

2.50 1.25

Column

Strip

Middle Strip

Long Direction Short direction

Kind of Strip

m m


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1.25

3.75 6.25

5

LsFig 2 Design Strip

-14.78

-13.08

Km (Ru)= 0.62 units N , mm

Check Ru max from Table 4.93 units N, mm

2.50 m 3.75 m

Mu (KN.m) -36.94 19.89 -49.07 26.42

M= Mu/0.9 (KN.m) -41.05 22.10 -54.52 29.36d = h- cover -0.5rebar dia

(mm) 129 129 129 129

As(min)=0.0018*b h mm2 695 695 1043 1043

Min No. of bars where S=2hUse

931 501 1236

Short Span

Strip Width ( b) m

Long Span

666

m

m

m

m

KN.m

KN.m

Mb d

The Positive Moment ??????

As= M*1000000 (mm2)Fy*Z

a=0.1d is sutiable in such

case and Z=0.95d

On the long Span M= M/Strip width =

On the short Span M= M/Strip width =

Since Ru Actual is Ru mthickness 150 mm is ok

The middle strip Negative Moment on the short span is normally considered the most crtitcal witregard to slab depth of flexture

Calcualtion and selection of reiforcement for middle strip


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Short Span

1.10 m 1.10 m

Mu (KN.m) -11.81 6.36 -7.64 4.12

M= Mu/0.9 (KN.m) -13.13 7.07 -8.49 4.57

d = h- cover -0.5rebar dia

(mm) 129 129 129 129

As(min)=0.0018*b h mm2 306 306 306 306

Min No. of bars where S=2hse

assuming Width of all beams 300 mm

298 160 193 104

Net half Strip Width ( b) mLong Span

As= M*1000000 (mm2)Fy*Z

a=0.1d is sutiable in suchcase and Z=0.95d

Calcualtion and selection of reiforcement for part of slab column strip


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Design the beams (B1)


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section 9.5

e Es=2040000

Fy =200000 Mpa


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154 mm

IssIsl


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ax , the slab

h


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Important Notes

E Modulaus of Elasticty for all grades of Steel is the same = 2040 ton/cm2 = 200'000 Mpa

Important Units

From to X from to X from to X

Ib (force ) Kg kg/cm2 Mpa 1/10 Kg.m KN.m 1/100Ton KN 10 kg/m2 Mpa Ib.ft N.m 1.4

Kibs KN 4.45 psi (lbf/in2) Mpa 0.007 N.mm KN.m

Kg KN 1/100 KN/m2 Mpa 1/1000 Kg.cm KN.m

Ib (force ) N 4.5 Psf (lbf/ft2) pa 48

Fc' MPa 21 22 25 28 30 35 40 45 50

depth ()Coef. 0.85 0.85 0.85 0.85 0.84 0.80 0.76 0.73 0.69

Force Stress Moment

Constant

0.65

0.66

0.67

0.68

0.69

0.7

0.71

0.72

0.73

0.74

0.75

0.76

0.77

0.78

0.79

0.8

0.81

0.82

0.83

0.84

0.85

21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93 97 101 105 109 113 117 121 125 129 133 137 141 145 1

WhitneyCoef.

F'c

Whitney Rectangualr Depth


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7844.523 7844.52

1 2 3 4 5 6 7 8

6 0.22 0.283 0.566 0.85 1.13 1.41 1.70 1.98 2.26

8 0.39 0.503 1.006 1.51 2.01 2.51 3.02 3.52 4.0210 0.62 0.786 1.57 2.36 3.14 3.93 4.71 5.50 6.29

13 1.04 1.328 2.66 3.98 5.31 6.64 7.97 9.30 10.62

16 1.58 2.01 4.02 6.03 8.05 10.06 12.07 14.08 16.09

19 2.23 2.84 5.67 8.51 11.35 14.18 17.02 19.86 22.69

22 2.98 3.80 7.61 11.41 15.21 19.01 22.82 26.62 30.42

25 3.85 4.91 9.82 14.73 19.64 24.55 29.46 34.38 39.29

28 4.83 6.16 12.32 18.48 24.64 30.80 36.96 43.12 49.28

32 6.31 8.05 16.09 24.14 32.18 40.23 48.27 56.32 64.37

38 8.90 11.3 22.7 34.04 45.38 56.73 68.07 79.42 90.77

1 2 3 4 5 6 7 8

5 0.15 0.196 0.393 0.59 0.79 0.98 1.18 1.38 1.57

7 0.30 0.385 0.770 1.16 1.54 1.93 2.31 2.70 3.08

12 0.89 1.131 2.26 3.39 4.53 5.66 6.79 7.92 9.05

14 1.21 1.540 3.08 4.62 6.16 7.70 9.24 10.78 12.32

18 2.00 2.55 5.09 7.64 10.18 12.73 15.27 17.82 20.37

20 2.47 3.14 6.29 9.43 12.57 15.71 18.86 22.00 25.14

24 3.55 4.53 9.05 13.58 18.10 22.63 27.15 31.68 36.21

26 4.17 5.31 10.62 15.93 21.25 26.56 31.87 37.18 42.49

30 5.55 7.07 14.14 21.21 28.29 35.36 42.43 49.50 56.57

34 7.13 9.08 18.17 27.25 36.33 45.41 54.50 63.58 72.66

36 7.99 10.2 20.4 30.55 40.73 50.91 61.10 71.28 81.46

Common shape Areas

One Way Slab Bending Moments Coef (K)

Multi Spans of 1 way slab or beams

mmWeight

Kg/m

Area of Cross Section in cm2

mmWeight

Kg/m

Area of Cross Section in cm2

Column

neg M=-1/16

+ M=1/14The Most critical(+)

neg M =-1/10

The most Critical ( -)

Pos M =+1/16

neg M =-1/11

Pos M =+1/16


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Two Spans of 1way Slab or beams

Shearing force coefOne Way Slab =K1

Shearing force* =K1*w*Ln

*note it is not Stress , it is a force

Ll/Ls Fm

0.5 0.90.555 0.88

0.625 0.86

0.67 0.85

0.71 0.84

0.83 0.8

1 0.75

1 0.75

1.2 0.69

1.4 0.63

1.5 0.6

1.6 0.57

1.8 0.51

2 0.45

Table 1

Shear Reinforcement

Av=2*leg area

the steel strength increase , the moment capacity of the section decreases?????? Why

Design double reinforceed beam in separate sheet

Coef. Of moments for short Span in column strip

Coef. Of moments for long Span in column strip

-M =-1/16

+ M=1/14 + M=1/14

-M =-1/16-M =-1/9

1/2 1/21.15/2 1/2

2* area for 1 bar from the table


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Fc' 21 Mpa

Fy 420 Mpa

Stress

Block

depth ()Coef. 0.850 0.85Minimum0 .65 Max 0.85

cbalance 0.593

a balance 0.504 *d

a maxallowed

by ACI

(0.75 a b) 0.378 *d---------------

---------------

------- 0.428

Z=d-

a/2=d(1-a/2) 0.811 *d

Nc(Concrete

Compressive

Force) 0.321 f'c*d*b

M 0.261 f'c*b*d

Max RU

by ACI

=(0.9*M)*f

c *b*d 4.925

units N,

mm

Max

Moment

Allowed in

the

section(Ru*b*d) 4.925 b*d Mpa

maxallowed

as per

ACI 318 0.0161

As min 0

max 0As max 0 cm2

for T-Beams, Ru table and the follwing formulas not applied beacsue the

section is T not recatngualr

cb=d*612612+Fy because Es=2040000

While ACI equation iscb = d* 600

600+Fy because Fy =200000 Mpa

As min(Rectangualr Section)=1.4*b*d

As max (Rectangualr Section)

As used is within the


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Concrete densityv

kg/m3 Kn/m3 From to X

ton/m3 KN/m3 in2 mm2 645

ft2 m2 0.093

55 60 65 70 100 120 130 150 200

0.66 0.65 0.65 0.65 0.65 0.65 0.65 0.65 0.65

Area

Constant

49 153 157 161 165 169 173 177 181 185 189 193 197


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9 10

2.55 2.83

4.53 5.037.07 7.86

11.95 13.28

18.10 20.11

25.53 28.36

34.23 38.03

44.20 49.11

55.44 61.60

72.41 80.46

102.11 113.46

9 10

1.77 1.96

3.47 3.85

10.18 11.31

13.86 15.40

22.91 25.46

28.29 31.43

40.73 45.26

47.80 53.11

63.64 70.71

81.75 90.83

91.65 101.83

neg M =-1/11 neg M =-1/11

Pos M =+1/16

Continues

- = * - = * Total


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1/2

+M=W*Ln8

-8

-8

Extend Total Asto 0.25 Spanthen bend 1/2

s


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Load Distribution to Beams

6.25

Section C-

C

50 cm

25 cm

m

m

B

B2 B

c c

b =

h


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50

25

Fc' 21 Mpa

Fy 420 Mpa

concrete Density 25

Stress Block depth

() Coef. 0.850 0.85Minimum

0 .65 Max 0.85



Span in short

Direction Ls 5 m

Net Span in long

Direction LLn' 5.9 m

Net Span in short

Direction Lsn' 4.65 m

h' 50 cm

b' 25 cm

Concrete cover 5 cm

d'= 45 cm

Load Calculations

Wufrom slab to B1

Slab Load 10.4 KN/m

Wufrom slab to B1 16.10388 KN/mif concrete Density 25

B1 Self weight

3.13 KN/m

Ultimate self

weight(B1) 4.38 KN/m

Ultimate Total Load

(B1) 20.48 KN/m

The edge beams will be designed as a rectangualr section beam , while all interior ones desinged as T -Beams

B1

B2

c

c

Short Beam design (B1) as Rectangualr Section

KN/m3

for the purpose of load calcualtions, dimensions of beamsection will be assumed as follwing:

d'= h'-concrete cover

=0.333*Slab Load*short Beam Span

try to link this value to the slab Load calcs based on selecting1way or 2 way slab , if not possible you can enter manually

= h*b*concrete density

KN/m3

Section C-Ccm

cm

(S 1)

Edge Slab


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M(-ve)

Km (neg) 1/10

M(-ve) -44.28 KN.m

Mu(-ve) KN.m

Mu(-ve) -49.20051 KN.m

M(+ve)

Km (neg) 1/14

M(+ve) 31.63 KN.m

Mu(+ve) KN.m

Mu(+ve) 35.143221 KN.m

Max RU by ACI

=(0.9*M)*fc *b*d 4.925 b*d

units N,

mm

therfore

44280458.28 4.925 b*dd

d 0 *1/ b for check with Page 24 of the red book

Min total Thickness b (cm) h (cm)

20 0

25 0

30 0

35 0

40 0

0.3780523 *d

a max 17.012355 cmZ=d(1-a/2) 0.8109738 *d

Z= 36.493823 cm

Moment Reinforcement

3.2099698 cm2

2.2928356 cm2

As min 3.75 cm2

max 0.0160672As max 18.075627 cm2

=- Km(neg)*W*Lsn

From the ACI 318 ---------- the Moment Coef; Km as the follwing

=M(-ve)/0.90

=M(+ve)/0.90

=Km(pos)*W*Lsn

since neg Moment is higher than positive one, -M will beused for reinforcement clacualtions

Mu(-ve)=Ru*b*d

(M/Ru)b

Lsn/??

(N.mm)

a max allowed byACI (0.75 a b)

=As*fy0.85*f'c*b

As(-)= (-M)Fy*Z

As(+)= +MFy*Z

use 2 dia 16 at the top the botom

As min(Rectangualr Section) =1.4*b*dfy

As max (Rectangualr Section) = max*b*d

As used is within the limit

*


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3.0211481 cm 17.01 cm OK

Shear Reinforcement

K1 1/2

Vu(Force)= 47.61 KN

47.61 KN

0.45 m2.325 m

47.61 KN

Shear force at the

critical Section (Vuc)

Vuc 38.40 KN

Concrete shear strength 87.64 KN

The shear Force must

Carried by Stirrups

Vst 64645.56 N

64.64556 KN

The Clear Spacing of bars in layer must not be less than nominal bar Dia or 4/3 of aggreragtesize or 2.5 cm

ac ua = s y

0.85*f'c*bless than

according to ACI code the critical section for shear is at a distance equal to distance (d) from support .Shear force at the supporting point =K1*w*Ln

Shear force =Critical shear forceX X-x''

The concrete shear strength according to the ACI-99=0.17*(F'c)*b*d

4 16

8@ 25 cm

Vst=Av*Fy*d*0.85s

for shear , assume dia 8mm U stirrups @ 25 cm will be enough ,

= Vuc-Concrete shear Strength


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D (Long span/short span ) 1.61

Wfrom slab to B2 19.154295 KN/MBeam Self

weight(Ultimate ) 4.38 KN/M

Total Ultimate load 23.53 KN/M

M(-)

M ultimate (-)

M(+)

M(-ve)

Km (neg) 1/10

M(-ve) -81.91 KN.m

Mu(-ve) KN.m

Mu(-ve) -91.00609 KN.m

M(+ve)

Km (neg) 1/14

M(+ve) 58.50 KN.m

Mu(+ve) KN.m

Mu(+ve) 65.004347 KN.m


5.9374749 cm2

4.2410535 cm2

As min 3.75 cm2

max 0.0160672As max 18.075627 cm2

5.5882117 cm 17.01 cm OK

3 16Reinforcment Details of B1

Long Beam design (B2)

WB2=(0.333*Slab Load*Slab Short Span*(1.5- 0.5 )D

=- Km(neg)*W*LLn


=M(-ve)/0.90

=M(+ve)/0.90

=Km os *W*Lsn


As(-)= (-M)Fy*Z

As(+)= +MFy*Z

use 4 dia 16 at the top 3 @ the botom

As min(Rectangualr Section) =1.4*b*dfy

As max (Rectangualr Section) = max*b*d

As used is within the limit


a actual =As*fy0.85*f'c*b

less than


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Shear Reinforcement

K1 1/2

Vu(Force)= 69.41 KN

69.41 KN

0.45 m

2.95 m

69.41 KN

Shear force at the


Vuc 58.82 KN



Carried by Stirrups

Vst 64.65 KN

according to ACI code the critical section for shear is at a distance equal to distance (

d) from support .Shear force at the supporting point =K1*w*Ln



4 16

3 16Reinforcment Details of B1

8@ 25 cm

Vst=Av*Fy*d*0.85s

for shear reinforcement , assume dia 8mm U stirrups @ 25 cm will beenough , Av =1.006 cm2



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4

5

6 6

0

4 m 3 m

13

50

20 6

13

50

4

20



Span in short

Direction Ls 5 m

Net Span in long

Direction LLn' 5.9 m

Net Span in short

Direction Lsn' 5.8 m

Slab thickness 13 cm

calcualtionFlange width for short beam

228 cm

595 cm

145 cm

145 cm

assume total depth (t)= 50 cm

m

m

B

B2 B4

c c

(S 2)intenal Slab

(S 3)Edge Slab

(S1)Edge Slab

Distribution of loads from slab to beams

B3

m m

m

m

m

m

a

a

section a-a

es gn o n erna eams , as eams

for Symitrical monolithicaly cast T-Beams , the ACI code 318, section.. Limits theflange width not to exceed 1/4 the span of the beam and the effective overhangingslab width on each side of the web not to exceed.1-8* slab thickness

2- 1/2 clear distance to the next webTherefore width of flange must be the smallest value of

2*8*slab thickness+b web

cm

cm

half clear distance to the next web

Span/4

The flange width will be

cm

(S 4)Edge Slab

(S 5)edge Slab

bb

section b-b

cm

cm


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43.2 cm

13 cm

38.88 cm

36.7 cm

Take the larger 38.88 cm

Load Calculations

Slab Load 10.4 KN/mWufrom (Slab to B3)

Wufrom S2 to B3 20.08656 KN/m

Wufrom S4 to B3 20.08656 KN/m

if concrete Density 25B1 Self weight

2.50 KN/m

Ultimate self

weight(B1) 3.50 KN/m

Ultimate Total Load

(B1) 43.67 KN/m

M(-ve)

Km (neg) 1/8

M(-ve) -183.65 KN.m

Mu(-ve) KN.m

Mu(-ve) -204.0505 KN.m

M(+ve)

Km (neg) 1/14

M(+ve) 104.94 KN.m

Mu(+ve) KN.m

Mu(+ve) 116.6003 KN.m

12.49575 cm2

As (+) 7.140426 cm2

3364.725 KN

524.8213 KN

Nc> Nt OK

d= h-cover- Stirrups- .5main steel

The initial value of arm should be obtained from the follwing two equations which ever giveslarger.

assume a = thickness of flange(slab thickness)

Z=0.9*d or

Z=d- a/2

=0.333*Slab Load*short Beam S an

= h*b*concrete densityKN/m3

=- Km(neg)*W*Lsn


=M(-ve)/0.90

=M(+ve)/0.90

=Km(pos)*W*Lsn

since neg Moment is higher than positive one, -M will beconsidered for reinforcement clacualtions

Short Beam design (B3) as T-beam

First Trial

As(-)= (-M)Fy*Z

Nc = 0.85*F'c*a*bflange

Nt = As*Fy

If Nc > Nt , the beam can be desinged as rectangualr


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2.027707 cm

Assume a = 1.927707

Z=d-a/2 42.23615 cm

As 11.50281 cm2

Check a for this As 1.866583 cm

a is close to assumed one

Shear Reinforcem (B3)

K1 1/2

Vu(Force)= 126.65 KN

126.65 KN

0.432 m

2.9 m

126.65 KN

Shear force at the


Vuc 107.79 KN



Carried by Stirrups 40.48 KN

Vst 62.06 KN

Vst > Required OK

Far from assumption , 2nd trial is required

a =As*fy0.85*f'c*b

Second Trial

use rebards --------------

according to ACI code the critical section for shear is at a distance equal to distance (d) from support .Shear force at the supporting point =K1*w*Ln



Vst=Av*Fy*d*0.85s

for shear , assume dia 8mm U stirrups @ 25 cm will be enough ,

Vuc-Concrete shear Strength =


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calcualtionFlange width for long T beam228 cm

440 cm

147.5 cm

147.5 cm

assume total depth (t)= 50 cm

43.2 cm

13 cm

38.88 cm

36.7 cm

Take the larger 38.88 cm

D (Long span/short span ) S2 1.03

Wfrom S2 to B4 20.424125 KN/M

D (Long span/short span ) S3 4

Wfrom S3 to B4 14.2857 KN/M

Beam Self weight(Ultimate )

3.50

Total Ultimate load 38.21 KN/M

M(-ve)

Km (neg) 1/10M(-ve) -133.01 KN.m

Mu(-ve) KN.m

Mu(-ve) -147.7871 KN.m

M(+ve)

Km (neg) 1/14

M(+ve) 95.01 KN.m

Mu(+ve) KN.m

Mu(+ve) 105.56222 KN.m

Long Beam design (B4) as T-beam

Long Beam design (B4)

WB4=(0.333*Slab Load*Slab Short Span*(1.5- 0.5 )D

=- Km(neg)*W*LLn


=M -ve /0.90

=M(+ve)/0.90

=Km os *W*Lsn


= h*b*concrete density

KN/mB4 self wt = =

2*8*slab thickness+b web

half clear distance to the next web

Span/4

The flange width will be

d= h-cover- Stirrups- .5main steel

The initial value of arm should be obtained from the follwing two equations which ever giveslarger.

assume a = thickness of flange(slab thickness)

Z=0.9*d or

Z=d- a/2

Load Calculations


63/72


9.050259 cm2

As (+) 6.464471 cm2

3422.738 KN

380.1109 KN

Nc> Nt OK

1.44371 cm

Assume a = 1.34371

Z=d-a/2 42.52814 cm

As 7.44652 cm2

Check a for this As #DIV/0! cm

a is close to assumed one

Shear Reinforcement

K1 1/2

Vu(Force)= 112.72 KN

112.72 KN

0.432 m

2.95 m

112.72 KN

Far from assumption , 2nd trial is required

use ------------ at the top and ---------@ the botom


according to ACI code the critical section for shear is at a distance equal to distance (d) from support .Shear force at the supporting point =K1*w*LLn

First Trial

As(-)= (-M)Fy*Z

Nc = 0.85*F'c*a*bflange

Nt = As*Fy

If Nc > Nt , the beam can be desinged as rectangualrsection beam with Flange width

a =As*fy0.85*f'c*b

Second Trial


64/72

Shear force at the


Vuc 96.21 KN



Carried by Stirrups 28.90

Vst 71.83 KN

Vst > Required OK



4 16

3 16Reinforcment Details of B 1

8@ 25 cm

Vst=Av*Fy*d*0.85s

for shear , assume dia 8mm U stirrups @ 25 cm will be enough , Av=1.006 cm2



65/72

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Two Way Slab Design (DRAFT)

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