Direct Design Method

Direct Design Method

Design of Two Way Floor System

for Slab with Beams

Figure-1 shows a two-way slab floor with a total area of 12,500 sq

ft. It is divided into 25 panels with a panel size of 25 ft x 20 ft.

Concrete strength is and steel yield strength is

fy=40,000 psi. Service live load is to be taken as 120 psf. Story

height is 12 ft. The preliminary sizes are follows: slab thickness is

6.5 in.; long beams are 14 28 in. Overall; short beams are 12

24 in. overall; columns are 15 15 in. The four kinds of panels

(corner, long-sided edge, short-sided edge, and interior) are

numbered 1, 2, 3, and 4 in Fig.-1. Determine the total factored

static moment in a loaded span in each of the four equivalent rigid

frames whose widths are designated A, B, C and D in Fig.2. 2

Given data [Problem-1]

psi3000fc

3

Given data

Fig.-1

4

For two way slab (with beams), the total factored static moment in

a loaded span in each of the four equivalent rigid frame whose

widths are designated A, B, C and D in Fig.2

Fig.-2

5

The factored load wu per unit floor area is

wu=1.2wD +1.6wL

=1.2(6.5)(150/12) +1.6(120)

=114 + 204 =318 psf

ACI states that Ln shall extend from face to face of columns,

capitals or walls.

For frame A, kipsft44825.12520318.08

1LLw

8

1M 22

n2uo

For frame B, Mo = 224 ft-kips

For frame C, kipsft34925.12025318.08

1LLw

8

1M 22

n2uo

For frame D, Mo = 175 ft-kips

6


For the two-way slab with beams design problem-1, Compute the

ratio of the flexural stiffness of the longitudinal beam to that of

the slab in the equivalent rigid frame for all the beams around

panels 1, 2, 3, and 4 in Fig.-3.

7

where

h=overall beam depth

t=overall slab thickness

bE=effective width of flange

bw=width of slab

For the two way slab (with beams), the ratio α of the flexural

stiffness of the longitudinal beam to that of the slab in the

equivalent rigid frame, for all the beams around panels 1, 2, 3,

and 4 in Figure.

(a) B1-B2, Referring to Fig.3, the effective width bE for B1-B2 is the smaller of 14 + 2 (21.5) = 57 and 14 + 8 (6.5) =66

thus bE = 57 in. Using Eq.11

Fig.-3

Fig.-4

In which

ht

bb

ht

bb

ht

ht

ht

bb

k

w

E

w

E

w

E

11

14641132

where

h = overall beam depth

t =overall slab thickness

bE =effective width of flange

bw = width of web

12

3hbkI w

b

11

43

b

w

E

.in400,4512

2814774.1I,774.1k

232.028

5.6

h

t,07.4

14

57

b

b

Using Eq.9, where Ecb=Ecs

27.85490

45400

IE

IE.in54905.6240

12

1I

scs

bcb43s

(a) B1-B2,

12

43

b

w

E

.in000,3812

2814484.1I,484.1k

232.028

5.6

h

t,54.2

14

5.35

b

b


83.132745

000,38

IE

IE.in27455.6120

12

1I

scs

bcb43s

(b) B3-B4, Referring to Fig.3, the effective width bE for B3-B4 is the smaller of 14 + 21.5 =35.5 in. and 14 + 4 (6.5) = 40; thus

bE = 35.5 in. Using Eq.11

(c) B5-B6, Referring to Fig.4, the effective width bE for B5-B6 is the smaller of 12 + 2 (17.5) = 47 in. and 12 + 8 (6.5) = 64

thus bE = 47 in. Using Eq.11

43

b

w

E

.in400,2412

2412762.1I,762.1k

271.024

5.6

h

t,92.3

12

47

b

b


55.36870

000,24

IE

IE.in68705.6300

12

1I

scs

bcb43s

14

43

b

w

E

.in500,2012

2412480.1I,480.1k

271.024

5.6

h

t,46.2

12

5.29

b

b


96.53435

500,20

IE

IE.in34355.6150

12

1I

scs

bcb43s

(d) B7-B8, Referring to Fig.4, the effective width bE for B7-B8 is the smaller of 12 +17.5 = 29.5 in. and 12 + 4 (6.5) = 38; thus

bE = 29.5 in. Using Eq.11

The resulting values for B1 to B8 around panels 1, 2, 3 and 4

are shown in Fig.4. For the design, the values vary between

3.55 and 13.83; thus the equivalent rigid frames have their

substantial portion along or close to the column lines, even

though their widths vary from 10 to 25 ft.

15


For the two-way slab with beams problem-1, Determine the

minimum thickness requirement for deflection control; and

compare it with the preliminary thickness of 6.5 in.

16

The average ratios αm for panels 1, 2, 3, and 4 may be computed

from the α values shown in Fig.4; thus

90.7)83.1355.327.896.5(4

11 panel for mα

30.7)83.1355.327.855.3(4

12 panel for mα

51.6)27.855.327.896.5(4

13 panel for mα

91.5)27.855.327.855.3(4

14 panel for mα

The minimum slab thickness requirements are the same as for

slabs without interior beams.

For this case, )12()2.0(536

000,200

f8.0L

tMinm

yn

The minimum is not be less than 5 in.

Slabs supported on shallow beams where αm ≤ 0.2.

Slabs supported on medium stiff beams where 0.2 < αm < 2.0.

Slabs Supported on Beams. Minimum slab thickness for deflection control

Slabs supported on very stiff beams where αm > 2.0.

)13(936

000,200

f8.0L

tMin

yn

For this case,

The minimum is not to be less than 3.5 in.

18

Since the αm values for all four panels are well above 2, Eq.13

applies. The minimum thickness for all panels, using Ln=24 ft,

Sn=18.83 ft, and fy=40,000 psi, become

.in07.6

83.18249

36

0.11224

SL9

36

000,40

f2.08.0L

n

n

yn

min t

If a uniform slab thickness for the entire floor area is to be used,

the minimum for deflection control is 6.07 in., which compares

well with the 6.5 in. preliminary thickness.

19


For the two-way slab with beams problem-1, Show that the six

limitations of the direct design method are satisfied.

(1) There is a minimum of three continuous spans in each

direction.

(2) Panels must be rectangular with the ratio of longer to shorter

span center-to-center of supports within a panel not greater

than 2.0.

(7) Limitations of Direct Design Method

(3) The successive span lengths center-to-center of supports in

each direction do not differ by more than one-third of the longer

span.

(4) Columns are not offset more than 10% of the span in the

direction of the offset.

(5) The load is due to gravity only and is uniformly distributed over

an entire panel, and the service live load does not exceed two

times the service dead load.

(6) The relative stiffness ratio of L21/α1 to L2

2/α2 must lie between

0.2 and 5.0, where is the ratio of the flexural stiffness of the

included beam to that of the slab.

(7) Limitations of Direct Design Method

22

The first four limitations are satisfied by inspection. For the fifth

limitation,

psf 8112

1505.6 wload dead service D

psf 120 wload live service L

281

120=

w

w

D

L

For the sixth limitation, referring to Fig.3&4 and taking L1 and L2 in

the long and short directions, respectively, poll

Panel 1, 5.56)27.883.13(5.0

625L21

1α

0.84)55.396.5(5.0

400L22

2α

23

Panel 2,

Panel 3,

5.56)27.883.13(5.0

625L21

1α

7.11255.3

400L22 2α

6.7527.8

625L21 1α

0.84)55.396.5(5.0

400L22

2α

Panel 4 6.7527.8

625L21 1α

7.11235.3

400L22 2α

5. and 0.2 between lie L

to L

of ratios All2

22

1

21

24


For the two-way slab with beams problem-1, Determine the

longitudinal moments in frames A, B, C, and D. as shown in

Fig.2&5.

25

(a) Check the six limitations for the direct design method. These

limitations have been checked previously.

(b) Total Factored Static Moment Mo

The total factored static moments M0 for the equivalent rigid frames

A, B, C, and D have been computed previously; they are

M0 (frame A) = 448 ft-kips

M0 (frame B) = 224 ft-kips

M0 (frame C) = 349 ft-kips

M0 (frame D) = 175 ft-kips

26

27

Fig.-5

28

Fig.-6

Longitudinal Moments in the frames

The longitudinal moments in frames A, B, C, and D are computed

using Case-2 of Fig.21(DDM) for the exterior span and

Fig.19(DDM) for the interior span. The computations are as

shown below, and the results are summarized in Fig.-5&6.

30

For Frame A: M0=448 ft-kips

Mneg at exterior support = 0.16(448) =72 ft-kips

Mpos in exterior span = 0.57(448) =255 ft-kips

Mneg at first interior support = 0.70(448) =313 ft-kips

Mneg at typical interior support = 0.65(448) =291 ft-kips

Mpos in typical interior span = 0.35(448) =157 ft-kips

31

For Frame B: M0=224 ft-kips




Mneg at typical interior support = 0.65(224) =146 ft-kips


32

For Frame C: M0=349 ft-kips




Mneg at tupical interior support = 0.65(349) =227 ft-kips


33

For Frame D: M0=175 ft-kips




Mneg at tupical interior support = 0.65(175) =114 ft-kips


34


For the two-way slab with beams problem-1, Compute the tensional

constant C for the edge and interior beams in the short and long

directions

363.01

3 yx

y

xC

where

x = shorter dimension of a component rectangle

y = longer dimension of a component rectangle

and the component rectangles should be taken in such a way

that the largest value of C is obtained.

The torsional constant C equals,

36

Each cross-section shown in Fig.7 may be divided into component

rectangles in two different ways and the larger value of C is to

be used.

Fig.-7

or

in 400,16600,114800

3

)14(28

28

)14(63.01

3

)5.6(57

57

)5.6(63.01

beam

eriorintC

4

33

For long direction,

Usein 700,20500,17)1600(2beam

eriorintC 4

363.01

3 yx

y

xC

or

in 500,41600,112900

3

)14(5.21

5.21

)14(63.01

3

)5.6(5.35

5.35

)5.6(63.01

beam

edgeC

4

33

For long direction,

Usein 100,19500,171600

3

)14(28

28

)14(63.01

3

)5.6(5.21

5.21

)5.6(63.01

beam

edgeC

4

33

or

in 965057253925

3

)12(5.17

5.17

)12(63.01

3

)5.6(47

47

)5.6(63.01

beam

eriorintC

4

33

For short direction,

Usein 930,119470)1230(2beam

eriorintC 4

40

For short direction,

or

in 805057252325

3

)12(5.17

5.17

)12(63.01

3

)5.6(5.29

5.29

)5.6(63.01

beam

edgeC

4

33

Usein 700,1094701230

3

)12(24

24

)12(63.01

3

)5.6(5.17

5.17

)5.6(63.01

beam

edgeC

4

33

41


For the two-way slab with beams as described in problem-1,

Distribute the longitudinal moments computed for frames A, B,

C, and D [Fig.5&6] into three parts-- for the longitudinal beam,

for the column strip slab, and for the middle strip slab.

(a) Negative moment at the face of exterior support.

For Frame A,

98.0)5490(2

700,10

I2

C

in 549012

)5.6(240I

in 700,10C

61.6L

L

27.8

80.0L

L

st

43

s

4

1

21

1

1

2

43


Table-1 shows the linear interpolation for obtaining the column

strip percentages from the prescribed limits of Table-2(DDM).

The total moment of 72 ft-kips is divided into three parts, 92.6%

to column strip (of which 85% goes to the beam and 15% to the

slab since and 7.4% to the middle strip slab.

The results are shown in Table-2

0.161.6L

L

1

21

ASPECT RATIO L2/L1 0.5 1.0 2.0

Negative moment at α1L2/L1 = 0 βt=0 100 100 100

exterior support βt2.5 75 75 75

α1L2/L1 > 1.0 βt = 0 100 100 100

βt> 2.5 90 75 45

Positive moment α1L2/L1 = 0 60 60 60

α1L2/L1 > 1.0 90 75 45

Negative moment at α1L2/L1 = 0 75 75 75

interior support α1L2/L1 > 1.0 90 75 45

Table-:Percentage of longitudinal moment in column strip

0.5 0.8 1.0

βt=0 100% 100% 100%

βt=0.98 96.1% 92.6% 90.2%

βt≥2.50 90% 81% 75%

Table-1: Linear interpolation for column strip percentage of exterior negative moment-frame A

61.6L

L

1

21

1

2

L

L

Using the prescribed values in

Table-2(DDM), the proportion of

moment going to the column strip is determined to be 81% by linear interpolation.

0.161.6L

L

80.0L

L

1

21

1

2

0.5 0.8 1.0

90% 81% 75%

(b) Negative moments at exterior face of first interior support and at face of typical interior support.

Frame A

1

2

L

L

61.6L

L

1

21

47

Frame A

Total width = 20 ft, Column strip width=10 ft, Middle strip width = 10 ftEXTERIOR SPAN INTERIOR SPAN

EXTERIOR

NEGATIVE POSITIVE

INTERIOR

NEGATIVE NEGATIVE POSITIVE

Total Moment -72 +255 -313 -291 +157

Moment in beam -57 +176 -216 -200 +108

Moment in column strip slab

-10 +31 -38 -36 +19

Moment in middle strip slab

-5 +49 -60 -55 +30

Table-2: Transverse Distribution of Longitudinal Moments

-72+255

-313 -291+157

-291

Moments in A


Frame B,

Aframe for as same the

92.6%percentage moment strip column The

Aframe for as same the ,98.0

1.11L

L

83.13

80.0L

L

t

1

21

1

1

2

49

The proportion of

moment is 81% for column strip, the same as for frame A.

1.11L

L

80.0L

L

1

21

1

2


Frame B,

50

Frame B

TOTAL WIDTH = 10 ft, COLUMN STRIP WIDTH=5 ft, MIDDLE STRIP WIDTH = 5 ft

EXTERIOR SPAN INTERIOR SPAN

EXTERIOR

NEGATIVE POSITIVE

INTERIOR


Total Moment -36 +128 -157 -146 +78

Moment in beam -28 +88 -108 -101 +54


-5 +16 -19 -17 +9


-3 +25 -30 -28 +15

Table-3: Transverse distribution of longitudinal moments

-36

+128

-157 -146

+78

-146

Moments in B


Frame C,

39.1

)6870(2

100,19

)I2(

C

in 6870

12

)5.6(300I

in 100,19C

44.4L

L

55.3

25.1L

L

st

4

3

s

4

1

21

1

1

2

52


Table-4 shows the linear interpolation for obtaining the column

strip percentage from the prescribed limits of Table-2(DDM).

The total moment of 56 ft-kips is divided into three parts, 81.9%

to column strip (of which 85% goes to the beam and 15% to the

slab since and 18.1% to the middle strip slab.

The results are summarized in Table-5

L2/L1 0.5 0.8 1.0

βt=0 100% 100% 100%

βt=1.39 86.1% 81.9% 69.4%

βt≥2.50 75% 67.5% 45%

Table-4: Linear interpolation for column strip percentage of exterior negative moment-frame C

0.144.4L

L

1

21

44.4L

L

1

21

53

Using the prescribed values in

Table-2(DDM), the proportion of

moment going to the column strip is determined to be 67.5% by linear interpolation:

44.4L

L

25.1L

L

1

21

1

2


Frame C,

1.0 1.25 2.0

75% 67.5% 45%

1

2

L

L

44.4L

L

1

21

54

Frame C



EXTERIOR

NEGATIVE POSITIVE

INTERIOR


Total Moment -56 +199 -244 -227 +122

Moment in beam -39 +114 -140 -130 +70


-7 +20 -25 -23 +13


-10 +65 -79 -74 +40

Table-5: Transverse Distribution of Longitudinal Moments in Two-Way Slab with beams

-56

+199

-244 -227

+122

-227

Moments in C

55


Frame D,

C) frame for as same (the

%9.18percentage moment strip column The

C) frame for as same (the

39.1

45.7L

L

96.5

25.1L

L

t

1

21

1

1

2

56

The proportion of

moment is again 67.5% for column strip, the same as for frame C.

47.7L

L

25.1L

L

1

21

1

2


Frame D,

57

Frame D



EXTERIOR

NEGATIVE POSITIVE

INTERIOR


Total Moment -28 +100 -123 -114 +61

Moment in beam -20 +57 -71 -65 +35


-3 +10 -12 -12 +6


-5 +33 -40 -37 +20

Table-6: Transverse distribution of longitudinal moments

-28+100

-123 -114+61

-114

Moments in D

58

(c) Positive moments in exterior and interior spans.

Since the prescribed limits for are the same for positive

moment and for negative moment at interior support. The

percentage of column strip moment for positive moments in

exterior and interior spans are identical to those for negative

moments as determined in part (b) above.

0.1L

L

1

21

59

EXTERIOR SPAN

INTERIOR SPAN

LINE

NUMBER

ITEM NEGATIVE MOMENT

POSITIVE MOMENT

NEGATIVE MOMENT

NEGATIVE MOMENT

POSITIVE MOMENT

NEGATIVE MOMENT

1 Moment, Table-2, line 3 (ft-kips)

2 Width b of drop or strip (in.)

3 Effective depth d (in.)

4 Mu/Ø (ft-kips)

5 Rn(psi)= Mu/(Øbd2)

6 ρ, Eq. or Table A.5a

7 As = ρbd

8 As =0.002bt*

9 N=larger of (7) or(8)/0.31

10 N=width of strip/(2t)

11 N required, larger of (9) or (10)

Table-4: Design of reinforcement in column strip

60


LINE

NUMBER

ITEM NEGATIVE MOMENT

POSITIVE MOMENT

NEGATIVE MOMENT

NEGATIVE MOMENT

POSITIVE MOMENT

NEGATIVE MOMENT

1 Moment, Table 3, line 4 (ft-kips)

2 Width b of strip (in.)

3 Effective depth d (in.)

4 Mu/Ø (ft-kips)

5 Rn(psi)= Mu/(Øbd2)

6 ρ

7 As = ρbd

8 As =0.002bt

9 N=larger of (7) or(8)/0.31*

10 N=width of strip/(2t)

11 N required, larger of (9) or (10)

Table-5: Design of reinforcement in Middle Strip

61


Investigate if the preliminary slab thickness of 6.5” in the two-way

slab with beams design example as described in problem-1, is

sufficient for resisting flexure and shear.

62

For each of the equivalent frames A, B, C and D, the largest

bending moment in the slab occurs at the exterior face of the

first interior support in the middle strip slab. From tables-

2,3,5,6, this moment is observed to be 60/10, 30/5, 75/15, or

40/7.5 ft-kips per ft. of width in frames A, B, C and D,

respectively. Taking the effective depth to the contact level

between the reinforcing bars in the two directions, and

assuming #5 bars.

averaged d =6.50-0.75-0.63=5.12 in.The largest Rn required is

psi254)12.5)(12(90.0

)12(6000

bd

MR

22u

n

Reinforcement ratio for this value of Rn is 0.0067, which is well

below Hence excessive deflection should not

be expected; this is further verification of the minimum

thickness formula given in ACI.

0067.0

000,40

254686.15211

686.15

1

686.15300085.0

000,40

f85.0

fm

f

mR211

m

1

c

y

y

n

.0139.0375.0 b

03712.0

000,4087000

87000

000,40

)3000)(85.0)(85.0(

87000

87000)85.0( 1

yy

c

b ff

f

The factored floor load wu is

wu=1.2wD+1.6wL=318 psf

Since all 1L2/L1 values are well over 1.0, take V from Eq.18(DDM)

as

65

kips 66.32

)20)(318.0(15.1

2

Sw15.1V u

u

kips 73.6)12.5)(12(30002dbf2V 10001

w'cc

ok kips 66.3Vkips 72.5)73.6(85.0V uc

Documents

Direct Design Method