36

Direct Design Method

The direct design method consists of certain steps for distributing moments to slab and beam sections to satisfy safety requirements and most serviceability requirements simultaneously. Three basic steps are involved:

a. Determination of the total factored static moment;

b. Distribution of the total factored static moment to negative and positive sections;

c. Distribution of negative and positive factored moments to the column and middle strips and to the beams if any.

Limitations

The direct design method was developed from theoretical procedures for determination of moments in slabs, requirements for simple design, construction procedures, and performance of existing slabs. Therefore, the slab system, to be designed using the direct design method, should conform to the following limitations as given by ACI Code 13.6.1:

1. There must be three or more spans in each direction.

2. Slab panels must be rectangular with a ratio of longer to shorter span, center-to-center of supports, not greater than 2.0.

3. Successive span lengths, center-to-center of supports, in each direction must not differ by more than one-third of the longer span.

4. Columns must not be offset more than 10 % of the span in the direction of offset from either axis between centerlines of successive columns.

5. Loads must be due to gravity only and uniformly distributed over the entire panel. The live load must not exceed 2 times the dead load.

6. For a panel with beams between supports on all sides, the relative

stiffness of beams in two perpendicular directions 212f

221f

l

l

α

α is not to be

less than 0.20 and not greater than 5.0.

37

where

fα = ratio of flexural stiffness of beam section to flexural stiffness of a

width of slab bounded laterally by centerlines of adjacent panels , if any, on each side of the beam

fα =scs

bcb

IEIE

cbE = modulus of elasticity of beam concrete

csE = modulus of elasticity of slab concrete

bI = moment of inertia about centroidal axis of gross section of beam

sI = moment of inertia about centroidal axis of gross section of slab

1fα = fα in direction of 2

l 3f'1f'1

αα += , as shown in Figure 10

2fα = fα in direction of 2

l 4f'2f'1

αα += , as shown in Figure 10

Figure 10: Panel bounded by four beams

Figure 11 shows dimensions of interior and edge beams that need to be considered in relative stiffness calculations.

38

(a) (b) Figure 11: Effective beam section; (a) interior beam; (b) exterior beam

Design Procedure

1. Determination of the total factored static moment:

Total factored static moment for a span is determined in a strip bounded laterally by centerline of panel on each side of centerline of supports, as shown in Figure 12.

(a)

39

(b)

Figure 12: (a) Column strip for l2 ≤ l1; (b) Column strip for l2 > l1

Absolute sum of positive and average negative factored moments in each direction is not to be less than

8/llqM 2n2u=o (14)

Where nl is clear span in the direction moments are being considered,

measured face-to-face of supports and not be less than 165.0 l , and uq is

factored load per unit area of slab, where the transverse span of panels on either side of the centerline of supports varies, 2l in Eq. (14) is taken as the

average of adjacent transverse spans. Besides, when the span adjacent and parallel to an edge is being considered, the distance from edge to panel centerline is to be substituted for 2l in Eq. (14).

2. Distribution of the total factored static moment to negative and positive moments:

According to ACI Code 13.6.3, in an interior span, total factored static moment is distributed in such a way that the negative factored moment at

40

face of support is taken as oM65.0 , and positive factored moment at

midspan is taken as oM35.0 .

In end spans, total factored static moment oM is distributed as shown in

Table 6 and Figure 13.

Table 6: Distribution of total static moment in end spans (1) (2) (3) (4)

No beams between interior supports

Exterior edge unrestrained

Beams between

all supports

Without edge beam

With edge beam

Exterior edge fully restrained

Interior negative factored moment

0.75

0.70

0.70

0.70

0.65

Positive factored moment

0.63

0.57

0.52

0.50

0.35

Exterior negative factored moment

0.00

0.16

0.26

0.30

0.65

(a) (b)

(c) (d)

(e)

Figure 13: (a) Exterior edge fully restrained; (b) exterior edge unrestrained; (c) beams between all supports; (d) No beams between interior supports and without edge beam; (e) No beams between interior supports and with edge beam

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3. Distribution of the positive and negative factored moments to the column and middle strips:

Definitions:

Column strip:

Column strip is a design strip with a width on each side of a column centerline equal to 125.0 l or 225.0 l , whichever is less, as shown in ل م ! خط أ

.یتم العثور على مصدر المرجع . Column strip includes beams, if any.

Middle strip:

Middle strip is a design strip bounded by two column strips, shown in Figure 12.

a. Factored moments in column strips:

According to ACI Code 13.6.4, column strip moments, as percentages of total factored positive and negative moments, are given in Table 7.

In Table 7, tβ is defined as ratio of torsional stiffness of edge beam section

to flexural stiffness of a width of slab equal to span length of beam, center-to-center of supports = scscb IECE 2/ , where C is cross-sectional constant

to define torsional properties given by Eq. (15)

363.01

3 yxyxC ∑

−= (15)

where x is the shorter overall dimension of rectangular part of cross section and y is the longer overall dimension of rectangular part of cross section. The cross section is to be divided into separate rectangular parts and carrying out the summation given in Eq. (15) in such away to give the largest value of C, as shown in Figure 14.

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Table 7: Column strip factored moments

12 / ll 0.50 1.0 2.0

0=tβ 100 100 100 0l/lf 121 =α

5.2≥tβ 75 75 75

0=tβ 100 100 100

Exterior negative factored moment

1l/l 121f ≥α

5.2≥tβ 90 75 45

0l/l 121f =α 60 60 60 Positive factored moment 1l/l 121f ≥α 90 75 45

0l/l 121f =α 75 75 75 Interior negative factored moment

1l/l 121f ≥α 90 75 45

Figure 14: Torsional cross sectional dimensions for βt calculations

b. Factored moments in beams caused by slab loads:

As specified by ACI Code 13.6.5, beams between supports are to be designed to resist 85 % of column strip moments if 1l/l 121f ≥α . For

values of 121f l/lα between 1.0 and zero, proportion of column strip

moments resisted by beams is obtained by linear interpolation between 85 and zero. In addition to these moments, beams are to be proportioned to resist moments caused by factored loads applied directly on these beams.

c. Factored shears in beams caused by slab loads:

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As specified by ACI Code 13.6.8, beams with 1l/l 121f ≥α are to be

designed for shear caused by factored loads on tributary areas which are bounded by 45 degree lines drawn from the corners of the panels and the centerlines of the adjacent panels parallel to the long sides, as shown in Figure 4. For values of 121f l/lα less than 1.0, linear interpolation

assuming beams carry no load at 01f =α , is permitted to be used. In

addition to these shears, beams are to be proportioned to resist shears caused by factored loads applied directly on these beams.

Example (4): Design the two-way flat plate with no edge (spandrel) beams, shown in Figure 15.a, given the following: interior columns are 40 cm × 40 cm, exterior columns are 30 cm × 30 cm, covering materials weigh 229 kg/m2, equivalent partition load 50 kg/m2 and the live load is 300 kg/m2. Use

2/280 cmkgfc =′ and 2/4200 cmkgf y = .

44

Figure 15.a: Two-way flat plate

45

Solution: 1- Evaluate slab thickness:

cmln 66515207001 =−−=

cmln 66020207002 =−−=

cmln 67015157003 =−−=

For flat plates with no edge beams, minimum slab thickness cmh 33.2230/670min == , taken as 22 cm.

2- Check limitations for slab analysis by the direct design method:

The first five conditions are satisfied, while the sixth condition does not apply due to the nonexistence of beams.

3- Calculate the factored load on the slab:

( )[ ] 2u t/m475.1)30.0(6.105.0229.05.222.02.1q =+++=

4- Check slab thickness for shear:

cmd avg 182222 =−−= , assuming φ 20 mm reinforcing bars.

a. Interior columns, shown in Figure 15.b:

Figure 15.b:Beam shear and punching shear

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Punching shear:

( ) cmb 23218404 =+=o

475.1=uV [ ( ) ( )258.067 − ] ton45.61=

( ) ( ) ( ) uc Vton55.551000/1823228075.006.1V >==Φ

Beam Shear:

Section 1-1:

( ) tonVu 61.2718.02

40.00.76475.1 =

−

−

=

( ) ( ) ( ) uc Vton84.711000/1860028075.053.0V >==Φ

Section 2-2:

( ) tonVu 05.2718.02

40.00.67475.1 =

−

−

=

( ) ( ) ( ) .K.Oton81.831000/1870028075.053.0Vc ==Φ

b. Corner columns, shown in Figure 15.c:

Figure 15.c: Beam shear and punching shear

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Punching shear:

( ) cmb 782/18302 =+=o

475.1=uV [ 2)39.0()15.3(65.3 − ] ton73.16=

( ) ( ) ( ) uc Vton68.181000/187828075.006.1V >==Φ

Beam Shear:

Section 1-1:

( )[ ] tonVu 73.1430.018.065.315.3475.1 =−−=

( ) ( ) ( ) uc Vton71.371000/1831528075.053.0V >==Φ

Section 2-2:

( )[ ] tonVu 37.1430.018.015.365.3475.1 =−−=

( ) ( ) ( ) uc Vton70.431000/1836528075.053.0V >==Φ

c. Edge columns, shown in Figure 15.d and Figure 15.e:

Figure 15.d: Beam shear and punching shear

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Figure 15.e: Beam shear and punching shear

In Figure 15.d,

Punching shear:

( ) cmb 12618302/18302 =+++=o

( ) ( ) ( )[ ] tonVu 25.3248.039.00.715.3475.1 =−=

( ) ( ) ( ) uc Vton17.301000/18126280785.006.1V >==Φ

Beam Shear:

Section 1-1:

( ) tonVu 73.1418.02

30.00.715.3475.1 =

−

−

=

( ) ( ) ( ) uc Vton71.371000/1831528075.053.0V >==Φ

Section 2-2:

( )[ ] tonVu 57.2730.018.015.30.7475.1 =−−=

( ) ( ) ( ) uc Vton81.831000/1870028075.053.0V >==Φ

In Figure 15.e,

Punching shear

cmb 1261830)2/1830(2 =+++=o

[ ] tonVu 03.32)48.0)(39.0()6(65.3475.1 =−=

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uc Vton17.301000/)18)(126(280)75.0(06.1V >==Φ

Beam shear

Section 1-1:

[ ] tonVu 05.2818.030.065.3)6(475.1 =−−=

uc Vton84.711000/)18)(600(280)75.0(53.0V >==Φ

Section 2-2:

tonVu 37.1418.02

3.00.6)65.3(475.1 =

−

−

=

uc Vton70.431000/)18)(365(280)75.0(53.0V >==Φ

5- Calculate the factored static moment:

Figure 15.f: Column and middle strips in the short direction

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Strip in the shorter direction:

Width of intermediate strip = 700 cm and width of column strip is the smaller of )2/( 1l and )2/( 2l , taken as (600/2) = 300 cm, as shown in

Figure 15.f.

Total factored static moment:

Clear span for exterior panels = 600 – 15 – 20 = 565 cm

Clear span for interior panels = 600 – 20 – 20 = 560 cm

The larger of the two values will be used in moment calculations.

( ) ( ) mtM .20.418/65.57475.1 2 ==o

6- Distribute the total factored static moment into positive and negative moments:

The results are shown in the following table.

7- Distribute the positive and negative moments to the column and middle strips:

The results are shown in the same table.

Negative and positive factored moments:

End spans Interior spans Slab moment

(t.m) Exterior negative

Positive Interior negative

Positive Negative

Total moments ( TM )

0.26 oM = 10.71

0.52 oM = 21.42

0.70 oM = 28.84

0.35 oM = 14.42

0.65 oM= 26.78

Column strip

moments

1.0 TM = 10.71

0.60 TM = 12.85

0.75 TM= 21.63

0.60 TM= 8.65

0.75 TM= 20.08

Middle strip

moments

0.00 8.57 7.21 5.77 6.70

Bending moment diagrams are shown in Figure 15.g and Figure 15.h.

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(g)

(h)

Figure 15: (continued); (g) Bending moment diagram for column strip; (h) bending moment diagram for middle strip

8- Design the reinforcement:

Column strip reinforcement:

Design sections at maximum positive and negative moments as rectangular

sections where, d = 18 cm, and b = 300 cm, 2/280 cmkgfc =′ , and 2/4200 cmkgf y = .

Reinforcement M, t.m ρ sA

Straight Bent-up 10.71 0.0030 16.20 6 φ 12 mm 9 φ 12 mm 12.85 0.0036 19.44 9 φ 12 mm 9 φ 12 mm 21.63 0.0062 33.48 15 φ 12 mm 9 φ 12 + 6φ 12 8.65 0.0024 12.96 6 φ 12 mm 6 φ 12 mm

Half middle strip reinforcement:

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Design sections at maximum positive and negative moments as rectangular

sections where, d = 18 cm, and b = 400 cm, 2/280 cmkgfc =′ , and 2/4200 cmkgf y = .

Reinforcement M, t.m ρ sA

Straight Bent-up 8.57 0.0018 15.84 7 φ 12 mm 8 φ 12 mm 7.21 0.0018 15.84 --- 8 φ 12 + 8 φ 12 5.77 0.0018 15.84 7 φ 12 mm 8 φ 12 mm

9- Prepare design drawings:

Figure 15.i shows reinforcement detail.

Continue from step 4 to calculate for the Strip in the longer direction

5- Calculate the factored static moment:

Strip in the longer direction:

Width of intermediate strip = 600 cm and width of column strip is the smaller of )2/( 1l and )2/( 2l , taken as (600/2) = 300 cm.

Total factored static moment:

Clear span for exterior panels = 700 – 15 – 20 = 665 cm

Clear span for interior panels = 700 – 20 – 20 = 660 cm

The larger of the two values will be used in moment calculations.

( ) ( ) mtM .92.488/65.66475.1 2 ==o

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Figure 15.i: Reinforcement layout

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6- Distribute the total factored static moment into positive and negative moments:

The results are shown in the following table.

7- Distribute the positive and negative moments to the column and middle strips:

The results are shown in the following table, also given in Figure 15.j.

Figure 15.j: Column and middle strips in the long direction

Negative and positive factored moments:

End spans Interior spans Slab moment (t.m) Exterior

negative Positive Interior

negative Positive Negative

Total moments

( TM )

0.26 oM= 12.72

0.52 oM = 25.44

0.70 oM = 34.24

0.35 oM = 17.12

0.65 oM = 31.80

Column strip moments

1.0 TM = 12.72

0.60 TM= 15.26

0.75 TM= 25.68

0.60 TM= 10.27

0.75 TM= 23.85

Middle strip moments

0.00 10.18 8.56 6.85 7.95

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Bending moment diagrams are shown in Figure 15.k and Figure 15l.

(k)

(l)

Figure 15: (continued); (k) Bending moment diagram for column strip; (l) bending moment diagrams for middle strip

8- Design the reinforcement:

Column strip reinforcement:

Design sections at maximum positive and negative moments as rectangular

sections where, d = 18 cm, and b = 300 cm, 2/280 cmkgfc =′ , and 2/4200 cmkgf y = .

Reinforcement M, t.m ρ sA

Straight Bent-up 12.72 0.00357 19.28 8 φ 12 mm 10 φ 12 mm 15.26 0.00431 23.27 11 φ 12 mm 10 φ 12 mm 25.68 0.00747 40.34 19 φ 12 mm 7 φ 12 + 10 φ 12 23.85 0.00690 37.26 19 φ 12 mm 7 φ 12 + 7 φ 12 10.27 0.00286 15.46 7 φ 12 mm 7 φ 12 mm

Half middle strip reinforcement:

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Design sections at maximum positive and negative moments as rectangular

sections where, d = 18 cm, and b = 300 cm, 2/280 cmkgfc =′ , and 2/4200 cmkgf y = .

Reinforcement M, t.m ρ sA

Straight Bent-up 10.18 0.0028 15.12 7 φ 12 mm 7 φ 12 mm 8.56 0.0024 12.96 ----- 6 φ 12 + 7 φ 12 7.95 0.0022 11.88 ----- 6 φ 12 + 6 φ 12 6.85 0.0019 11.88 5 φ 12 mm 6 φ 12 mm

9- Prepare design drawings:

Figure 15.m shows reinforcement detail.

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Figure 15.m: Reinforcement details

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Example (5): For the two-way solid slab with beams on all column lines, shown in Figure 16.a, evaluate the moments acting on any of the internal beams, using the direct design method. All columns are 30 cm × 30 cm in cross section, all beams are 30 cm × 60 cm in cross section, slab thickness is equal to 14 cm, covering materials weigh 183 kg/m2 and the live load is 400 kg/m2. Use

2/280 cmkgfc =′ and 2/4200 cmkgf y = .

Figure 16.a: Two-way solid slab with beams

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Solution: 1- Evaluate the constants α and tβ :

Beam sectional properties are shown in Figure 16.b.

Calculation of relative beam stiffness α :

a. Internal beams:

Using Figure 5, relative stiffness of beam is calculated as follows:

Figure 16.b: Beam sectional properties for relative stiffness calculations

for a/h = 60/14 = 4.285 and b/h = 30/14 = 2.14, f = 1.775.

( ) ( ) 982.6775.1285.460030 3

3

2

=

=

= f

ha

lb

α

Calculation of torsional constant tβ (Figure 16.c):

−= ∑ 3

63.013 yx

yxC

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Figure 16.c: Beam sectional properties for relative stiffness calculations

( )( ) ( ) ( ) ( )( ) ( ) ( )4

33

3.305347

3/463046/3063.013/761476/1463.01

cm

CA

=

−+−=

( )( ) ( ) ( ) ( )( ) ( ) ( )4

33

3.403907

3/461446/1463.013/603060/3063.01

cm

CB

=

−+−=

4max 3.403907 cmCC B ==

( ) ( )472.1

12/1460023.4039073 ==tβ

( )( ) ( ) 2u m/t28.140.06.1183.05.214.02.1q =++=

2- Determine the total factored static moment (internal strip):

Total factored static moment:

( ) ( ) m.t19.318/7.50.628.18/llqM 22n12u ===o

Negative and positive factored moments:

0.1982.6l/l 121f >=α

For ,0.0=tβ exterior negative moment is 100 % of total strip moment

For ,5.2≥tβ exterior negative moment is 75 % of total strip moment

By linear interpolation, for ,472.1=tβ exterior negative moment is 85.28

% of total strip moment.

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3- Evaluate the moments in one of the internal beams:

The results are shown in the following table:

End spans Interior spans Bending moment

(t.m) Exterior negative Positive Interior

negative Positive Negative

Total moments ( TM )

0.16 oM = 4.990

0.57 oM = 17.778

0.70 oM = 21.833

0.35 oM = 10.916

0.65 oM = 20.273

Column strip mom. ( CM )

0.85 TM = 4.241

0.75 TM= 13.333

0.75 TM= 16.375

0.75 TM= 8.187

0.75 TM= 15.205

Beam moments 0.85 CM = 3.60

0.85 CM =11.33

0.85 CM =13.92

0.85 CM =6.96

0.85 CM = 12.92

Problems

P.1 Design the slab shown in Figure P.1, using the approximate coefficient method (solid). All columns are 40 cm × 40 cm in cross section, all beams are 40 cm × 60 cm in cross section, covering materials weigh 200 kg/m2, equivalent partition load is equivalent to 75 kg/m2 and the live load is 300

kg/m2. Use 2/350 cmkgfc =′ and 2/4200 cmkgf y = .

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Figure P.1

P.2 Design a ribbed slab for the plan shown in Figure P.1 using the relevant approximate method.

P.3 Design the slab shown in P.1 using the direct design method.

P.4 Design the slab shown in P.1 using the direct design method with edge beams only. Design one of these beams.

P.5 Design the slab shown in P.1 using the direct design method, with interior beams only (no edge beams). Also, design one of these beams.