Two Way Slab Best

7/28/2019 Two Way Slab Best

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Minimum Slab Thickness of Two-way Edge-supported Slabs

As specified by ACI Code 9.5.3.3 and 9.5.3.2, for slabs with beams spanning between the

supports on all sides, the minimum thickness shall be as follows:

a. For fm equal to or less than 0.20, minimum slab thickness is determined fromprovisions on slabs without interior beams spanning between the supports, shown in

Table (5), but not less than:

Slabs without drop panels 12.5 cm

Slabs with drop panels 10.0 cmTable (5): Minimum thickness of slabs without interior beams

Without drop panels With drop panels

Exterior panels Interiorpanels

Exterior panels Interiorpanels

Without

edge beams

With

edgebeams

Without

edgebeams

With

edgebeams

33/ln

36/ln

36/ln

36/ln

40/ln

40/ln

30/ln 33/ln

33/ln

33/ln

36/ln

36/ln

28/ln 31/ln

31/ln

31/ln

34/ln

34/ln

* 2y cm/kg2800f =

** 2y cm/kg4200f =

*** 2y

cm/kg5200f =

In Table (5), nl is the length of clear span in the long direction measured face-to-face of

supports in slabs without beams and face-to-face of beams or other supports in other cases.

For yf between the values given in the table, minimum thickness is determined by linear

interpolation. Moreover, for slabs with beams between columns along exterior edges, the

value of f for the edge beam is not to be less than 0.80.

b. For fm greater than 0.20 but not greater than 2.0, the thickness shall not be less than


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( )cm5.12

20.0536

14000/f80.0lh

fm

yn +

+=

(10)

c. For fm greater than 2.0, the thickness shall not be less thancm9

936

14000/f80.0lh

yn +

+=

(11)

d.

At discontinuous edges, an edge beam shall be provided with a stiffness ratio f not

less than 0.80 or the minimum thickness required by Eqn. (10) or Eqn. (11) shall be

increased by at least 10 % in the panel with a discontinuous edge.

where

h = overall thickness of slab

f = 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 beam determined

analytically as follows:

f =scs

bcb

IE

IE

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

Another way of determining f is graphically, using Figures 5 and 6.

fm = average value of f for all beams on edgs of a panel

= ratio of clear spans in long to short direction of slab.


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Figure 5: Beam Stiffness (Interior Beams)


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Figure 6: Beam Stiffness (Edge Beams)

ACI Code 318-89 (Revised 92) included three different equations for determination of

minimum slab thickness. Two of those equations will be given here for their simplicity, as

they require no prior evaluation of . The smallest thickness hmin is given as


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( )s

yn flh

++

+=

1536

14000/80.0min (12)

The largest thickness hmax is given as

36

14000/80.0max

yn flh

+= (13)

where s = ratio of length of the continuous edges to the total perimeter of the panel.

Summary of Design Procedure for Two-way Edge- supported Solid Slabs Using the

Simplified Methods

Once design compressive strength of concrete and yield stress of reinforcement are

specified, the next steps are followed:

1. Slab thickness is evaluated based on deflection control requirement.2. The total factored load supported by the slab is calculated. The dead load includes own

weight of the slab in addition to the covering materials it supports. The live load is

dependent on the intended use of the slab.

3. The loads distributed in the short and long directions of each panel are determined asdiscussed in earlier sections.

4. Representative 1 m wide design strips to span in the short and long directions areselected.

5. The shearing force and bending moment diagrams for each of the strips are drawn.6. Strength of each of the strips in shear is checked.7. Design the reinforcement at critical negative and positive bending moment sections for

each of the strips. Reinforcement in the short direction is placed closest to the concrete

surface in order to increase the effective depth din this direction.

8. Prepare design drawings showing the dimensions and the provided reinforcement.Example (1):

Using the Grashoff coefficients design the simply supported solid slab shown in Figure 7.a.

Assume that the beam webs are 25 cm wide. Weight of flooring materials is 200 kg/m2, and

the live load is 500 kg/m2. Use 2/250 cmkgfc = ,2/4200 cmkgfy = .


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(a) (b)

Figure 7: (a) Two way solid slab; (b) loads on beams

Solution:

1- Evaluate Slab thickness:nl = 5.0 0.25 = 4.75 m

Since r = 5/4 = 1.25 < 2.0, the slab is classified as being two way.

The smallest slab thickness is given by

( )

( )

( )cm

flh

s

yn

34.12)01(75.3/75.4536

14000/420080.0475

1536

14000/80.0min

=++

+=

+++=

The largest slab thickness is given by

( )cm

flh

yn

51.1436

14000/420080.0475

36

14000/80.0max

=+

=

+=

Slab thickness will be taken as 14.0 cm.

2- Determine the total factored load per unit area of the slab:Own weight of slab = 0.14 2.50 = 0.35 ton/m

2

( ) ( ) 2u m/ton46.150.060.120.035.020.1w =++=

3- Evaluate load distribution in both directions:Rectangularity ratio r= 5/4 = 1.25Using Grashoff coefficients, 71.0= , and .29.0=


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Load in the short direction ( ) 2s m/t03.146.171.0w ==

For a strip 1 m wide, mtonws /03.1=

Load in the long direction ( ) 2l m/t42.046.129.0w ==

For a strip 1 m wide, mtonwl /42.0=

4- Select representative 1 mwide design strips to span in the short and longdirections:

Two strips are used, one in the short direction and the other in the long direction.

5- Evaluate shear and bending moment at critical section locations:Maximum shear force in the short direction ( ) ton06.22/403.1V max,u ==

Maximum shear force in the long direction ( ) ton05.12/542.0V max,u ==

Maximum moment in the short direction ( ) m.t06.28/403.1M 2max,u ==

Maximum moment in the long direction ( ) m.t31.18/542.0M2

max,u ==

6- Check slab thickness for beam shear:a. Short Direction:Average effective depth d = 14 2 1.2 = 10.8 cm, assuming 12 mm bars in both

directions.

( ) ( )( ) ton79.61000/80.1010025053.075.0Vc ==

i.e. , slab thickness is adequate in terms of resisting beam shear in the shorter direction.

b. Long direction:cu VV


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Area of flexural reinforcementsA = 0.0049 (100) (10.8) = 5.29 cm

2/m

Use 12 mm @ 20 cm at the bottom side of the slab.

For main reinforcement, spacing between bars is not to exceed the smaller of 3(14) = 42 cm

and 45 cm, which is already satisfied.

b.Long direction:( ) ( )

( ) ( ) ( )003.0

2508.101009.0

31.1106353.211

4200

25085.0

2

5

=

=

Area of flexural reinforcements

A = 0.003 (100) (10.8) = 3.24 cm2/m

Use 12 mm@ 30 cm on top of the short direction reinforcement.

Design of beam B1:

cmlh 25.3116/50016/min ===

Try 25 cm

50 cm cross section.Factored own weight of beam = 1.2 (0.25) (0.5) (2.5) = 0.375 t/m

Maximum factored moment due to own weight of beam

= t.m172.18/)5(375.0 2 =

Maximum factored shear due to own weight of beam= ton937.02/)5(375.0 =

Loads from the slab, shown in Figure 7.c:

( ) ( ) m/t92.22/446.12/Sww u' ===

Equivalent load due to shear =( )

t/m752.125.12

1192.2 =

Equivalent load due to moment =( )

t/m297.225.13

1192.2

2=

Maximum factored shear= ton38.42/)5(752.1 =

Maximum factored moment= m.t178.78/)5(297.2 2 =

Total factored shear = 0.937 + 4.38 = 5.317 ton


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Total factored moment = 1.172 + 7.178= 8.35 ton.m

Assuming that 8 mm stirrups and 12 mm reinforcing bars are used,

d= 50 4.0 0.80 0.60 = 44.6 cm

( ) ( )

( ) ( ) ( )0046.0

2506.44259.0

35.810353.211

4200

25085.0

2

5

=

=

( ) ( ) 2

s

cm13.56.44250046.0A == , use 4 14 mm bars.

Check section for shear:

( ) ( )( ) ton0.71000/6.442525053.075.0Vc ==

Use cmmm 20@8 as minimum shear reinforcement.

See Figure 7.c for reinforcement details.

(c) (d)

Figure 7: (continued); (c) Equivalent loads on beam B1; (d)

reinforcement details

Design of beam B2:

cmlh 0.2516/40016/min ===

Try 25 cm 50 cm cross section.

Factored own weight of beam = 1.2 (0.25) (0.5) (2.5) = 0.375 t/m

Maximum factored moment due to own weight of beam


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= m.t75.08/)4(375.0 2 =

Maximum factored shear due to own weight of beam

= ton875.02/)4(375.0 =

Loads from the slab, shown in Figure 7.e:

( ) ( ) m/t92.22/446.12/Sww u' ===

Equivalent load due to shear = 2.92 /2 = 1.46 t/m

Equivalent load due to moment = 2.92 (2/3) = 1.95 t/m

Maximum factored shear= ton92.22/)4(46.1 =

Maximum factored moment= m.t90.38/)4(95.1 2 =

Total factored shear = 0.75 + 2.92 = 3.67 ton

Total factored moment = 0.75 + 3.90 = 4.65 t.m

Assuming that 8 mm stirrups and 12 mm reinforcing bars are used,

d= 50 4.0 0.80 0.60 = 44.6 cm

( ) ( )

( ) ( ) ( )00254.0

2506.44259.0

65.410353.211

4200

25085.0

2

5

=

= < min

( ) ( ) 268.36.44250033.0 cmAs == , use 4 12 mm bars.

Check section for shear:

( ) ( )( ) ton0.71000/6.442525053.075.0Vc ==

Use cmmm 20@8 as minimum shear reinforcement.

See Figure 7.g for reinforcement details.

8- Prepare design drawings showing the dimensions and the provided reinforcement:Design drawings are shown in Figure 7.g.


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(e) (f)

(g)

Figure 7: (continued); (e) equivalent loads on beam B2; (f) beamreinforcement details; (g) slab reinforcement details

Example (2):

Design the two-way solid slab shown in Figure 8.a. The covering materials weigh 258

kg/m2, and the live load is 300 kg/m2. Also, calculate the factored load acting on beam C3-

C7-C11-C15 due to slab loads. Use 2/250 cmkgfc = and2

/4200 cmkgfy = . All beams

are 25 cm wide.


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Figure 8.a: Two-way solid slab

Solution:

1- Evaluate Slab thickness:The largest slab thickness is given by

( )

cm

flh

yn

51.1436

14000/420080.0475

36

14000/80.0max

=

+

=

+=

Use a slab thickness of 14.0 cm.

2- Determine the total factored load per unit area of the slab:( )[ ] 2u t/m21.1)30.0(6.1258.05.214.02.1w =++=

3- Evaluate load distribution in both directions:Load distributions, shown in Figure 8.b, are based on Egyptian code load coefficient given

in Table (3).


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Figure 8.b: Loads in both directions

For panels S1, S2 and S8 :

( )

( ).230.0475.0,25.1

87.00.4

87.00.5==== andr

For panels S3 and S6:

( )( )

.190.0517.0,33.187.00.3

87.00.4 ==== andr

For panel S4:

( )

( ).230.0475.0,25.1

76.00.4

76.00.5==== andr

For panel S5:

( )

( )

.174.0565.0,43.1

76.00.4

87.00.5==== andr

For panel S7:


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( )

( ).295.0396.0,092.1

87.00.4

76.00.5==== andr

4- Select representative 1 mwide design strips to span in the short and longdirections:

Six design strips are used, three in the short direction and three in the long direction.

Evaluate shear and bending moment at critical section locations:

Shear force and bending moment diagrams for each of the six strips are shown in Figures

8.c through 8.h.

cmdavg 8.102.1214 == , assuming 12 mm reinforcing bars.

5- Check slab thickness for shear:( )( ) ( ) ( ) ton79.61000/8.1010025053.075.0Vc == O.K

6- Design the reinforcement:Strip 1-1:

For mtMu .87.0=

( ) ( )

( )( )( ) ( )002.0

2508.101009.0

87.010353.211

4200

25085.0

2

5

=

=

( ) ( ) /mcm...As21628101000020 ==

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

Figure 8.c: S.F.D and B.M.D for strip 1-1


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For mtMu .49.0=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

Strip 2-2:

Figure 8.d: S.F.D and B.M.D for strip 2-2

For mtMu

.62.0=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

For mtMu .43.0=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

Strip 3-3:


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Figure 8.d: S.F.D and B.M.D for strip 3-3

For mtMu .82.0=

( ) ( )

( )( )( ) ( )00189.0

2508.101009.0

82.010353.211

4200

25085.0

2

5

=

=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

For mtMu

.51.0=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

Strip 4-4:

For mtMu .46.0=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm For mtMu .26.0=


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Figure 8.f: S.F.D and B.M.D for strip 4-4

( ) ( ) mcmAs /52.20.141000018.02

min, == , use .30@10 cmmm

Strip 5-5:

Figure 8.g: S.F.D and B.M.D for strip 5-5

For mtMu .95.0=

( ) ( )

( )( )( ) ( )0022.0

2508.101009.0

95.010353.211

4200

25085.0

2

5

=

=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm


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For mtMu .73.0=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

Strip 6-6:

Figure 8.h: S.F.D and B.M.D for strip 6-6

For mtMu .01.1=

( ) ( )

( )( )( ) ( )00234.0

2508.101009.0

01.1106353.211

4200

25085.0

2

5

=

=

( ) ( ) /mcm...As2

532810100002340 ==

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

For mtMu .70.0=

( ) ( ) /mcm...As,2

min 52201410000180 == , use .30@10 cmmm

7- Prepare design drawings showing the dimensions and the provided reinforcement:Design drawings are shown in Figure 8.i.


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Figure 8.i: Reinforcement layout

Load on beam C3-C7-C11-C15:

( ) ( ) mtwu /84.422/421.1 == , as shown in Figure 8.j.

Figure 8.j:Tributary loads on beam C3-C7-C11-C13

Summary of Design Procedure for Two-way Edge- supported Ribbed Slabs, Using the

Simplified Methods

Once design compressive strength of concrete and yield stress of reinforcement are

specified, the next steps are followed:


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1. Determine the overall slab thickness h based on deflection control requirement. Also,thickness of topping slab t, rib width b, and hollow block size are to be determined

based on code requirements.

2. Calculate the total factored load per unit area.3. Distribute the loads in the short and long directions of each panel, as discussed in earlier

sections.

4. Select representative design strips to span in the short and long directions. The width ofeach of these strips is equal to center-to-center spacing between ribs in each of the two

directions.

5. Draw shearing force and bending moment diagrams for each of the strips.6. Check rib width for beam shear in the two directions.7. Design the reinforcement at critical negative and positive bending moment sections of

the ribs in each of the two directions. Reinforcement in the short direction is placed

closest to the concrete surface in order to increase the effective depth din this direction.

8. Prepare design drawings showing arrangement of ribs and details of the reinforcement.Example (3):

Design the two-way ribbed slab shown in Figure 9.a. The covering materials weigh 150

kg/m2, equivalent partition load is equal to 73 kg/m

2, concrete hollow blocks are 40 cm 25

cm 17 cm in dimension, each 17 kg in weight and the live load is 400 kg/m2. Use

2/300 cmkgfc = ,2/4200 cmkgfy = . All beams are 30 cm wide.


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Figure 9.a: Two-way ribbed slab

Solution:

1- Evaluate overall slab thickness and key ribbed slab components:The smallest slab thickness is given by

( )

( )

( ) ( )cm

flh

s

yn47.19

5.10.1536

14000/420080.0770

1536

14000/80.0min =+

+=

++

+=

The largest slab thickness is given by

( )cm

flh

yn53.23

36

14000/420080.0770

36

14000/80.0max =

+=

+=

Use an overall slab thickness of 23.0 cm.

Topping slab thickness = 23 17 = 6 cm

Let width of web be equal to 12 cm

Area of shrinkage reinforcement ( ) ( ) /mcm.As2

08.1610000180 ==

Use mmm /65 in both directions.

2- Determine the total factored load on the slab:Load per rib, shown in Figure 9.b:


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Total volume (hatched) = 0.52 0.62 0.23 = 0.074152 m3

Volume of hollow blocks = 0.4 0.17 0.5 = 0.034 m3

Net concrete volume = 0.074152 - 0.034 = 0.040152 m3

Weight of concrete /m2 = 0.040152 (2.5) /(0.52) (0.62) = 0.311 ton/ m2

Weight of hollow blocks /m2 = 17 (2)/(0.52) (0.62) (1000) = 0.105 ton/ m2

( ) ( ) 2u m/ton407.140.060.115.0073.0105.0311.020.1w =++++=

Figure 9.b:Loads per rib

3- Determine load distributions in the two principal directions:Load distribution according toEgyptian Code coefficients:

r= (8 0.87) / (8 0.87) = 1.0

35.0==

( ) 221 /492.0407.135.0 mtww ===

Load per rib in the first direction:

=ribwu / 0.492 (62/100) = 0.305 t/m

Load per rib in the second direction:

=ribwu / 0.492 (52/100) = 0.256 t/m

4- Select representative design strips in the two principal directions:One design strips is selected (with the larger loads).

5- Draw shear force and bending moment diagrams:


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Analyzing the strip shown in Figure 9.c using theBENARIstructural analysis software, one

gets the shown shear force and bending moment diagrams.

(c)

Figure 9.c: Internal forces

6- Check web width for beam shear:For reinforcement in the first direction (the closest to the outside surface of concrete):

Effective depth d= 23 2 0.60 0.8 = 19.6 cm, assuming 16 mm reinforcing bars and

6 mm stirrups.

( )( ) ( ) ( ) uc Vton78.11000/6.191230053.075.01.1V1.1 >==

Though shear reinforcement is not required, mm64 U-stirrups per meter run are to be

used to carry the bottom flexural reinforcement.

7- Design rib reinforcement:Positive moment reinforcement:

Assuming that a < 6 cm, compressive force in concrete is:

( ) ( ) ( )abfC ec= 85.0

C= 0.85 (300) (62) (a) /1000 = 15.81 a ton

( ) ( )2/adCMM nu ==

( ) ( ) 100/2/6.1981.159.037.1 aa=


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025.192.392 =+ aa

Figure 9.d: Equivalent section

Solving this equation gives a = 0.5 cm, which means that the assumption made before is

valid, and the other root is too large to be considered.

From equilibrium, C= T

Area of flexural reinforcement

( ) 288.1

4200

)1000(50.081.15cmAs ==

Use mm101 and mm121 reinforcing bars in each rib.

Negative moment reinforcement:

Since the factored negative moment creates compression in the web, the section at

corresponding location is to be designed as a rectangular section.

For mtMu .44.2=

( ) ( )

( )( )( ) ( )0161.0

3006.19129.0

44.2106353.211

4200

30085.0

2

5

=

=

( ) ( ) 279.36.19120161.0 cmAs == , use mm162 per rib.

8- Prepare design drawings:Figure 9.e shows required rib reinforcement and concrete dimensions.


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Figure 9.e: Reinforcement layout

Two Way Slab Best

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