DESIGN OF TWO WAY SLAB (with beams) BY DDM

Fig 1: Two way slab with beams.

Problem:

A two way slab floor with a total area of 7500 sq ft. is divided into 25 panels with a panel size of 20 ft. x 15 ft.

fc′ = 3000 psify = 60000 psiService Live Load = 120 psfAll Column = 14” x 14”Slab Thickness = 6.5”Storey Height = 12’Long Beam = 14” x 28”Short Beam = 12” x 24”

Solution:

1. Calculation of Factored Load

Slab Thickness = 6.5”DL = (6.5/12) x 150 = 81.25 psfLL = 120 psf

Total factored load, W = 1.4 DL + 1.7 LL = (1.4 x 81.25) + (1.7 x 120) = 318 psf = 0.318 ksf

2. Total Factored Static Moment in Equivalent Rigid Frame

Fig 2: Equivalent rigid frame notations.

Total factored moment, M o=18

W l2 ln2

a. For Frame A and B:

c/c distance in short direction, l2 = 15’Clear span in long direction, ln = [20-(14/12)] = 18.83’

For frame A:

Mo.A = (1/8) x 0.318 x 15 x 18.832 = 212 kip-ft.

For frame B:

Mo.B = ½ Mo.A = ½ x 212 = 106 kip-ft.

b. For Frame C and D:

c/c distance in long direction, l2 = 20’Clear span in short direction, ln = [15-(14/12)] = 13.83’

For frame C:

Mo.C = (1/8) x 0.318 x 20 x 13.832 = 152 kip-ft.

For frame D:

Mo.D = ½ Mo.C = ½ x 152 = 76 kip-ft.

3. Relative Stiffness, α

Relative stiffness, α= Flexural rigidity of t h e beamflexural rigidity of t h e slab

=Ecb I b

Ecs I s

=I b

I s [Here, Ecb = Ecs]

The moment of inertia of a flanged beam section, I b=kbw h3

12

The moment of inertia of a slab, I s=b h3

12

Here,

k=

1+( bE

bw

−1)( th ){4−6( t

h )+4 ( th )

2

+( bE

bw

−1)( th )

3}1+( bE

bw

−1)( th )

Where,h = overall beam deptht = overall slab thicknessbE = effective width of flangebw = width of web

Fig 3: Computation of α value.

a. B1-B2: Long side T beam

The effective width,bE = bw +2(h – t) or bw + 2(4t) = 14 + 2(28-6.5) or 14 + 2(4 x 6.5) = 57” or 66”

Thus, bE = 57”

bE

bw

=5714

=4.07 ;th=6.528

=0.232

k=1+ (4.07−1 ) (0.232 )[4−6 (0.232 )+4 (0.232 )2+ (4.07−1 ) (0.232 )3]

1+ (4.07−1 )(0.232)=1.774

I b=1.77414×283

12=45400¿4

I s=(15×12)×6.53

12=4120¿4

α=I b

I s

= 454004120

=11.02

b. B3-B4: Long side edge beam

The effective width,bE = bw + (h – t) or bw + (4t) = 14 + (28-6.5) or 14 + (4 x 6.5) = 35.5” or 40”

Thus, bE = 35.5”

bE

bw

=35.514

=2.54 ; th=6.528

=0.232

k=1+ (2.54−1 ) (0.232 )[4−6 (0.232 )+4 (0.232 )2+(2.54−1 ) (0.232 )3]

1+ (2.54−1 )(0.232)=1.484

I b=1.48414×283

12=38000¿4

I s=(7.5×12)×6.53

12=2060¿4

α=I b

I s

=380002060

=18.45

c. B5-B6: Short side T beam

The effective width,bE = bw +2(h – t) or bw + 2(4t) = 12 + 2(24-6.5) or 12 + 2(4 x 6.5) = 47” or 64”

Thus, bE = 47”

bE

bw

=4712

=3.92 ;th=6.524

=0.271

k=1+ (3.92−1 ) (0.271 )[4−6 (0.271 )+4 (0.271 )2+(3.92−1 ) (0.271 )3]

1+(3.92−1 )(0.271)=1.762

I b=1.76212×243

12=24400¿4

I s=(20×12)×6.53

12=5493¿4

α=I b

I s

=244005493

=4.44

d. B7-B8: Short side edge beam

The effective width,bE = bw + (h – t) or bw + (4t) = 12 + (24-6.5) or 12 + (4 x 6.5) = 29.5” or 38”

Thus, bE = 29.5”

bE

bw

=29.512

=2.46 ; th=6.524

=0.271

k=1+ (2.46−1 ) (0.271 )[4−6 (0.271 )+4 (0.271 )2+(2.46−1 ) (0.271 )3]

1+(2.46−1 )(0.271)=1.480

I b=1.48012×243

12=20500¿4

I s=(10×12)×6.53

12=2746¿4

α=I b

I s

=205002746

=7.47

Fig 4: Values of α

4. Check for Slab Thickness

α m=α 1+α2+α 3+α 4

4

β=ln

Sn

Here, ln = Clear span long side Sn = Clear span short side

When, 0.2 < αm ≤ 2.0

Minimum slab thickness, t min=

l n(0.8+f y

200000)

36+5β (αm−0.2)≥5∈¿

When, αm > 2.0

Minimum slab thickness, t min=

ln(0.8+f y

200000)

36+9 β≥3.5∈¿

For panel 1, αm = ¼(18.75+11.02+7.47+4.44) = 10.42For panel 2, αm = ¼(18.75+11.02+4.44+4.44) = 9.66For panel 3, αm = ¼(11.02+11.02+7.47+4.44) = 8.49For panel 4, αm = ¼(11.02+11.02+4.44+4.44) = 7.73

Here, for all panels, αm > 2.0

ln = 20-(14/12) = 18.83’Sn = 15-(14/12) = 13.83’

β=ln

Sn

=18.8313.83

=1.362

t min=ln(0.8+

f y

200000)

36+9 β=18.83(0.8+ 60000

200000)

36+9×1.362=0.429'=5.15<6.5

So, given thickness of the slab =6.5’’ (ok)

5. Check for Limitation of DDM

1. There shall be a minimum of three continuous spans in each direction.In this problem, there are five continuous spans in each direction.

2. Panels shall be rectangular, with a ratio of longer to shorter span center-to-center of supports within a panel not greater than 2.Here panels are rectangular and the ratio of longer (20’) to shorter (15’) span c/c of supports is (20/15)=1.33<2.0

3. Successive span lengths center-to-center of supports in each direction shall not differ by more than one-third the longer span.One-third of longer span is (1/3 x 20) =6.67’In both directions, span lengths are equal.

4. Offset of columns by a maximum of 10 percent of the span (in direction of offset) from either axis between centerlines of successive columns shall be permitted.In longitudinal direction, 10% of the longer span is (20x12) x 10% = 24” and column width in this direction is 14”, which is less than 24”.

In transverse direction, 10% of the shorter span is (15x12) x 10% = 18” and column width in this direction is 14”, which is less than 18”.

5. All loads shall be due to gravity only and uniformly distributed over an entire panel. Live load shall not exceed two times dead load.Service LL = 120 psf and service DL = 81.25 psfLL/DL = (120/81.25) = 1.477<2.0

6. The relative stiffness ratio of (l12/α1) to (l22/α2) must lie between 0.2 and 5.0 where α is the ratio of the flexural stiffness of the included beam to that of the slab.

Taking l1 and l2 in the long and short directions respectively,

Panel 1, l12

α1= 202

0.5 (18.75+11.02)=26.87

l22

α2= 152

0.5 (7.47+4.44 )=37.78

l12

α1l22

α2

=26.8737.78

=0.738

Panel 2, l12

α1= 202

0.5 (18.75+11.02)=26.87

l22

α2= 152

0.5 (4.44+4.44 )=50.68

l12

α1l22

α2

=26.8750.68

=0.530

Panel 3, l12

α1= 202

0.5 (11.02+11.02 )=36.30

l22

α2= 152

0.5 (7.47+4.44 )=37.78

l12

α1l22

α2

=36.3037.78

=0.961

Panel 4, l12

α1= 202

0.5 (11.02+11.02 )=36.30

l22

α2= 152

0.5 (4.44+4.44 )=50.68

l12

α1l22

α2

=36.3050.68

=0.716

All ratio of (l12/α1) to (l22/α2) lie between 0.2 and 5.0

This problem satisfies all the limitations imposed by ACI 13.6.1 for using DDM.

6. Longitudinal Distribution of Moment

Fig 5: Longitudinal Moment diagram for exterior span

Fig 6: Longitudinal moment diagram for interior span.

Fig 7: Equivalent rigid frames

Mo.A = 212 kip-ft.

Mo.B = 106 kip-ft.

Mo.C = 152 kip-ft.

Mo.D = 76 kip-ft.

From fig 5 (case 2) and fig 6,

Fig 8: Longitudinal moment distribution

7. Torsional Constant

Torsional constant, C=∑(1−0.63 xy )( x3 y

3 )Here,x = Smaller dimensiony = Larger dimension

Use larger value of C.

a. For long direction edge beam

C=(1−0.63 6.535.5 )( 6.53×35.53 )+(1−0.63 1421.5 )( 14

3×21.53 )=2900+11600=14500¿4

C=(1−0.63 6.521.5 )( 6.53×21.53 )+(1−0.63 1428 )( 14

3×283 )=1600+17500=19100¿4

Use C = 19100 in4

b. For long direction T beam

C=(1−0.63 6.557 )( 6.53×573 )+(1−0.63 1421.5 )(14

3×21.53 )=4800+11600=16400¿4

C=2(1−0.63 6.521.5 )( 6.53×21.53 )+(1−0.63 1428 )(14

3×283 )=2×1600+17500=20700¿4

Use C = 20700 in4

c. For short direction edge beam

C=(1−0.63 6.529.5 )( 6.53×29.53 )+(1−0.63 1217.5 )( 12

3×17.53 )=2325+5725=8050¿4

C=(1−0.63 6.517.5 )( 6.53×17.53 )+(1−0.63 1224 )( 12

3×243 )=1230+9470=10700¿4

Use C = 10700 in4

d. For short direction T beam

C=(1−0.63 6.547 )( 6.53×473 )+(1−0.63 1217.5 )( 12

3×17.53 )=3925+5725=9650¿4

C=2(1−0.63 6.517.5 )( 6.53×17.53 )+(1−0.63 1224 )( 12

3×243 )=2×1230+9470=11930 ¿4

Use C = 11930 in4

8. Transverse Distribution of Longitudinal Moment

a. Aspect Ratio (l2/l1):

For frame A and B: l2/l1 = 15/20 = 0.75For frame C and D: l2/l1 = 20/15 = 1.33

b. Calculation of βt:

β t=C2 I s

For frame A and B:

I s=(15×12)6.53

12=4120

Torsional Constant, C = 10700 in4

β t=C2 I s

= 107002×4120

=1.30

For frame C and D:

I s=(20×12)6.53

12=5493

Torsional Constant, C = 19100 in4

β t=C2 I s

= 191002×5493

=1.74

c. Calculation of Percentage of Moment in Column Strip:

Frame A B C DC 10700 10700 19100 19100Is 4120 4120 5493 5493

t=C/2Is 1.30 1.30 1.74 1.74 11.02 18.75 4.44 7.47

l2/l1 0.75 0.75 1.33 1.33 (l2/l1) 8.27 14.06 5.91 9.94

Column Strip Moment, Percent of Total Moment at Critical Section l2/l1

0.5 1.0 2.0

Interior Negative Moment (l2/l1) = 0 (l2/l1) ≥ 1.0

7590

7575

7545

Exterior Negative Moment (l2/l1) = 0 t = 0

t ≥ 2.510075

10075

10075

(l2/l1) ≥ 1.0t = 0t ≥ 2.5

10090

10075

10045

Positive Moment (l2/l1) = 0 (l2/l1) ≥ 1.0

6090

6075

6045

i. Percentage of Exterior Negative Moment

Frame A

(l2/l1) t (l2/l1)

8.27

0.5 0.75 1

0 100 100 100

1.30 90.9

2.5 90 82.5 75

Frame B

(l2/l1) t (l2/l1)

14.06

0.5 0.75 1

0 100 100 100

1.30 90.9

2.5 90 82.5 75

Frame C

(l2/l1) t (l2/l1)

5.91

1 1.33 2

0 100 100 100

1.7475.7

12.5 75 65.1 45

Frame D

(l2/l1) t (l2/l1)

9.941 1.33 2

0 100 100 100

1.7475.7

12.5 75 65.1 45

ii. Percentage of Positive Moment

Frame A

(l2/l1) 0.5 0.75 1 (l2/l1) =

8.2790 82.5 75

Frame B

(l2/l1) 0.5 0.75 1 (l2/l1) =

14.0690 82.5 75

Frame C

(l2/l1) 1 1.33 2 (l2/l1) =

5.9175 65.1 45

Frame D

(l2/l1) 1 1.33 2 (l2/l1) =

9.9475 65.1 45

iii. Percentage of Interior Negative Moment

Frame A

(l2/l1) 0.5 0.75 1 (l2/l1) =

8.2790 82.5 75

Frame B

(l2/l1) 0.5 0.75 1 (l2/l1) =

14.0690 82.5 75

Frame C

(l2/l1) 1 1.33 2 (l2/l1) =

5.9175 65.1 45

Frame D

(l2/l1) 1 1.33 2 (l2/l1) =

9.9475 65.1 45

d. Transverse Distribution of Longitudinal Moment

Fig 9: Middle strip and column strip diagram for frame A & B

For frame A &B:

0.25 l1 = 0.25(20 x 12) = 60”0.25 l2 = 0.25(15 x 12) = 45”

y = 45”

For frame A:Column strip = 2 x 45” = 90”Half middle strip = 2@ [(15 x 12)-90]/2 = 2@45”

For frame B:

Column strip = 45”Half middle strip = 45”

Fig 10: Middle strip and column strip diagram for frame C & DFor frame C & D:

0.25 l1 = 0.25(15 x 12) = 45”0.25 l2 = 0.25(20 x 12) = 60”

y = 45”

For frame C:Column strip = 2 x 45” = 90”Half middle strip = 2@ [(20 x 12)-90]/2 = 2@75”

For frame D:Column strip = 45”Half middle strip = 75”

e. Summary of Calculation

Equivalent Rigid Frame

A B C D

Total Transverse Width (in)

180 90 240 120

Column Strip Width (in) 90 45 90 45

Half Middle Strip (in) 2@45 45 2@75 75

Torsional Constant C (in) 10700 10700 19100 19100

Is (in4) in t 4120 4120 5493 5493

t = C/2Is 1.30 1.30 1.74 1.74

11.02 18.75 4.44 7.47

(l2/l1) 0.75 0.75 1.33 1.33

(l2/l1) 8.27 14.06 5.91 9.94

External (-ve) Moment, % to Column Strip

90.90 90.90 75.71 75.71

Positive Moment, % to Column Strip

82.5 82.5 65.1 65.1

Internal (-ve) Moment, % to Column Strip

82.5 82.5 65.1 65.1

9. Distribution of Factored Moment in Column Strip and Middle Strip

All the moments are divided into three parts, percentage to column strip (of which 85% goes to the beam and 15% to the slab) and rest to the middle strip slab.

Equivalent Rigid Frame A

Total Width = 180” Column Strip = 90” Middle Strip = 90” 

Moments at Vertical Section (kip.ft)

Exterior Span Interior Span-ve

Moment

+ve Momen

t

-ve Momen

t

-ve Momen

t

+ve Momen

t

-ve Momen

tTotal Moment in

Frame A-33.92 120.84 -148.4 -137.8 74.2 -137.8

% to Column Strip 90.90% 82.50% 82.50% 82.50% 82.50% 82.50%Moment in Beam -26.2 84.7 -104.1 -96.6 52.0 -96.6

Moment in Column Strip Slab

-4.7 15.0 -18.4 -17.1 9.2 -17.1

Moment in Mid Strip Slab

-3.1 21.1 -26.0 -24.1 13.0 -24.1

Equivalent Rigid Frame B

Total Width = 90” Column Strip = 45” Middle Strip = 45” 

Moments at Vertical Exterior Span Interior Span

Section (kip.ft) -ve Momen

t

+ve Momen

t

-ve Momen

t

-ve Momen

t

+ve Momen

t

-ve Momen

tTotal Moment in

Frame B-16.96 60.42 -74.2 -68.9 37.1 -68.9

% to Column Strip 90.90% 82.50% 82.50% 82.50% 82.50% 82.50%Moment in Beam -13.1 42.4 -52.0 -48.3 26.0 -48.3

Moment in Column Strip Slab

-2.3 7.5 -9.2 -8.5 4.6 -8.5

Moment in Mid Strip Slab

-1.5 10.6 -13.0 -12.1 6.5 -12.1

Equivalent Rigid Frame C

Total Width = 240” Column Strip = 90” Middle Strip = 150” 

Moments at Vertical Section (kip.ft)

Exterior Span Interior Span-ve

Moment

+ve Momen

t

-ve Momen

t

-ve Momen

t

+ve Momen

t

-ve Momen

tTotal Moment in

Frame C-24.32 86.64 -106.4 -98.8 53.2 -98.8

% to Column Strip 75.71% 65.10% 65.10% 65.10% 65.10% 65.10%Moment in Beam -15.7 47.9 -58.9 -54.7 29.4 -54.7

Moment in Column Strip Slab

-2.8 8.5 -10.34 -9.7 5.2 -9.6

Moment in Mid Strip Slab

-5.9 30.2 -37.1 -34.5 18.6 -34.5

Equivalent Rigid Frame D

Total Width = 120” Column Strip = 45” Middle Strip = 75” 

Moments at Vertical Section (kip.ft)

Exterior Span Interior Span-ve

Moment

+ve Momen

t

-ve Momen

t

-ve Momen

t

+ve Momen

t

-ve Momen

tTotal Moment in

Frame D-12.16 43.32 -53.2 -49.4 26.6 -49.4

% to Column Strip 75.71% 65.10% 65.10% 65.10% 65.10% 65.10%Moment in Beam -7.8 24.0 -29.4 -27.3 14.7 -27.3

Moment in Column Strip Slab

-1.4 4.2 -5.2 -4.8 2.6 -4.8

Moment in Mid Strip Slab

-3.0 15.1 -18.6 -17.2 9.3 -17.2

Fig: Moment Distribution