Direct Design Method



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Direct Design Method

Flat slabs:

Concrete slabs may be carried directly by columns as shown in Fig.(1-1) without the use of beams or girders. Such slabs are described as flat plates and are commonly used where spans are not large and loads not particularly heavy.

µ §


Fig.(1-1) Slab supported directly on column.

At point (1) there is more (-M) and shearing stress, and col. try to punch the slab.

Flat slab construction shown in Fig.(1-2) is also beamless but incorporates a thickened slab region (drop panels) and column capitals.

µ §


Fig.(1-2) Flat slab.

Column capital:An element at the end of the column to give a wider support for the floor slab

Column capital and drop panel are used to reduce:

Stresses due to shear.

Negative bending around the columns.

Size of drop panel shall be in accordance with the following ACI-code (8.2.4 ¨CACI-2014). Drop panel shall be extended in each direction from center line of support a distance not less than one-sixth the span length measured from center to center of supports in that direction.

The side of the drop panel shall be at least µ §

Where


µ §

µ § Thickness of slab

µ §Thickness of slab with drop

µ §Thickness of drop

µ §

Bending moments in flat slab floors:



For purposes of design, a typical panel is divided into column strips and middle strips.

ACI-2014 (8.4.1.5)column strip is a design strip with a width oneach side of a column centerline equal to µ §orµ §, whichever is less. Column strip includesbeams, if any.

In all cases

µ §: is the span in the direction of the moment analysis (c. to c.).

µ §is the span in the lateral direction (transvers toµ §c.to.c).

µ §: clear span in µ §direction.

panel

panel


panel

µ §Panels (1,2,3,7,8,9) Exterior slabPanels (4,5,6)Interior slabµ §Panels (1,4,7,3,6,9) Exterior slabPanels (2,5,8)Interior slab

In case of flat slab the column strip is more critical than middle strip because it work as beam carrying the middle strip load to the column therefore column strip need more reinforcement. Plus that the beam takes 85% of the slab moment in the case of two-way slab with beams. µ §A similar requirement exists in the perpendicular direction.

Deflection control of two-way slab:

Design limits (ACI-code 2014):

Minimum slab thickness:

8.3.1.1For nonprestressedslabs without interior beams spanningbetween supports on all sides, having a maximum ratio of long-to-shortspan of 2, overall slab thickness h shall not be less than the limits in Table 8.3.1.1 andshall be at least the value in (a) or (b) unless the calculated deflection limits of 8.3.2 are satisfied:

(a) Slabs without drop panels asdefined in 8.2.4..............................125 mm.

(b) Slabs with drop panels as definedin 8.2.4 ..................................100 mm.

8.3.1.2 For nonprestressed slabs with beams spanning between supports on all sides, overall slab thickness h shall satisfy the limits in Table 8.3.1.2 unless the calculated deflection limits of 8.3.2 are satisfied.

Direct design method of two way slabs (8.10.2 ACI-2014):

Limitations:

Moments in two-way slab can be found using direct design method subject to the following restrictions:

8.10.2.1There shall be at least three continuous spans in each direction.

8.10.2.2Successive span lengths measured center-to-center of supports in each direction shall not differ by more than one-third the longer span.

8.10.2.3Panels shall be rectangular, with a ratioof longer to shorter panel dimensionsmeasured center-to-center of supports, not exceed 2.

8.10.2.4Columns offset shall not exceed 10percent of the span in direction of offset from eitheraxis between centerlines of successive columns.

8.10.2.5All loads shall be due to gravity only anduniformly distributed over an entire panel.

8.10.2.6Unfactored live load shall not exceed two times theunfactored dead load.

µ §


8.10.2.1For a panel with beams betweensupports on all sides, Eq. (8.10.2.7a) shall be satisfied forbeams in the two perpendicular directions

µ §


Where µ §andµ §are calculated by:

µ § (8.10.2.7b)

µ §=µ § in direction µ §

µ §=µ § in direction µ §

In the case of monolithic construction (two-way slab with beams)

Single side slab Symmetric slab

µ §

Panel


Panel

End span


Interior span

End span


For internal strip µ §

For external strip µ §

Total static moment for end span

µ §µ §µ §µ §µ §

For other direction:

End span


Int span

End span


Panel

Panel


For internal strip µ §

For external strip µ §

Total static moment for end span,µ §µ §

µ §: Clear span, circular or regular polygon shaped support shall be treated as square support with the same area.

Negative and positive factored moment:

For interior span (8.10.4.1):

µ §µ §

For end span:



Use Table 8.10.4.2 (ACI-2014)

slab


beam

Interior


Exterior

8.10.4.5 Negative moment Mu shall bethe greater of the two negative Mu calculated for spans framing into a common support unless an analysis is made to distribute the unbalanced moment in accordance with stiffnesses of adjoining elements.

Slab with beams between supports

Factored moments in column strips

8.10.5.1The column strips shall resistthe portion of interior negative Mu in accordance with Table 8.10.5.1.

8.10.5.2The column strips shall resistthe portion of exterior negative Mu in accordance with Table 8.10.5.2.

The relative restrained provided by the torsional resistance of the effective transverse edge beam is reflected by the parameter µ §, defined as:

µ §


The constant C for T- or L- section is calculated by dividing the section into separate rectangular parts, each having smaller dimension (x) and larger dimension (y), and summing the values of C for each part.

µ §


The subdivision can be done in such away as to maximize C.

8.10.5.5The column strips shall resistthe portion of positive Mu in accordance with Table 8.10.5.5.

Factored moments in beams

8.10.5.7.1Beams between supports shall resist the portion of column stripMu in accordance with Table 8.10.5.7.1.

Direct loads on beams:- factored beam self-weight + factored wall weight

µ §µ §


Interior beam

µ §µ §


End beam (Use Table 8.10.4.2):

µ §µ §µ §

Design of moment reinforcement for slab

µ §µ §µ §

µ §

µ §µ §


Ext.

µ §


µ §

Ex\\ A two-way reinforced concrete building floor system is composed of slab panels measuring 6*7.5 m in plan supported by column-line beams as shown in figure below. Using concrete with fc’=27.6 MPa and steel having fy=414 MPa, design typical exterior panel to carry a service live load of 6.9 kN/m2 in addition to its own weight of the floor.

Sol:

Try h=17 cm



Either x=h-hf=500-170=330 mm=33 cm

Or x=4*hf=4*170=680 mm=68cm

Chose min. value (x=33 cm)

µ §


µ §µ §µ §µ §

µ §


µ §µ §µ §µ §

Beam µ §


µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §

Short span direction

µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §

µ §


µ §µ §µ §µ §µ §

µ §µ §


µ §µ §µ §

µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §

0.85*60= 51 kN.m beamµ §µ §

Beam


momentSlab C.S

momentM.S

momentInterior slab strip 6m span

Positive


Negative

97

179



17

32

53



99Exterior slab strip 6m span

Positive


Negative

51

94



9

17

28



52

Long span direction ( slab strip 6 m)

µ §µ §µ §µ §µ §

µ §µ §µ §µ §µ §µ §µ §µ §

Sec 8.10.5

µ §µ §µ §

Beam

momentSlab C.S



momentM.S

momentSlab - beam strip (7.5m span)

µ § (98)

µ § (349)

µ § (428)

77

240



295

14

42



52

7

66



81µ §µ §µ §µ §µ §µ §µ §

Required (d) for flexure:

µ §

In long direction



µ §

In short direction

µ §

µ §µ §µ §



µ §

µ §µ §µ §

locationµ §µ §b mmd mmµ §µ §µ §Number of bars7.5m spanTwo half

C.SExt. -M

+M

Int. -M14



42

522640


2640

2640132


132

1325.3


16

200.0025


0.0025

0.0031871

871

10808


8

10Two half

M.SExt. -M

+M

Int. -M7



66

813000


3000

3000132


132

1322.3


22

270.0025


0.0035

0.0043990

1386

17029


13

156 m spanExt. half

C.S-M

+M17


91320

1320144


14412.8

6.80.0023

0.0023437

4374


4M.S-M

+M99


534500

4500144


14422

11.70.0029

0.00231880

136117


14Int. half

C.S-M


+M16

8.51320


1320144

14412.1


6.40.0023

0.0023437

4374

4

Shear in slab system with beams



One-way shear (beam shear)

µ §µ §


µ §µ §µ §µ §µ §

Shearing stress in flat slabs

One-way shear (beam shear)

Vu=Wu(ln/2-d), (ln=clear span of long direction)

µ §µ §µ §

µ §


Unless a more detailed calculation is made in accordance with Tab 22.5.5.1

More design, it is convenient to assume that the second term in expression (a) and (b) of Table 22.5.5.1 equalsµ § use

µ §

Punching shear (two-way shear)



Failure may occur by punching shear, with potential diagonal crack following the surface of a truncated cone or pyramid around the column, capitals or drop panel.

The failure surface extends from the bottom of the slab at the support diagonally upward to the surface

The critical section for shear is taken perpendicular to the plane of the slab at a distance µ § from the support.

µ §


c= column diameterµ §Area defined by the critical shear perimeter.

µ §= perimeter along the critical section.

µ §

µ §For Int. column



µ §

For drop panel check punching shear:

Inside the drop

Outside the drop

µ §

µ §


µ §

µ §


µ §

c=column capital diameter.

µ §Shall be the smallest of

µ §µ §µ §

µ §is the ratio of long side to short side of the column concentrated load reaction area.

µ §Are 40 for interior column, 30 for edge column, 20 for corner column.

µ §

For column with non-rectangular cross section



Design of shear reinforcement (Bent bar reinforcement)

Special shear reinforcement is used at the support for flat plates and sometimes for flat slab as well. It may take several forms. The bent-bar arrangement is suited for use with concrete column.

If shear reinforcement in the form of bars is used Vc is reduced to

µ §µ §µ §

µ §µ §

µ §


µ §=45°

Where inclined shear reinforcement is all bent at the same distance from the support

µ §µ §

Successive sections at increasing distances from the support must be investigated and reinforcement provided where µ §. Only the center three-quarters of the inclined portion of the bent bars can be considered effective in resisting shear.



Ex: A flat plate floor has thickness (h=191 mm) and support a 458mm square column spaced 6.1m on centers each way. The floor will carry a total factored load of 14.4 µ § check punching shear at a typical interior column and provides shear reinforcement if needed using bent bars. Use d=152mm fy=414 MPa, fµ §=28 MPa.

Solution

The first critical section at µ § from face of column

µ §µ §µ §µ §µ §µ §µ §µ §

µ §µ §

Bar area Av



µ §

Four bars will be used (two in each direction)

µ §

With bars bent at µ § and effective through the center three-fourths of the inclined length the next critical section is at µ § from the first critical section,



µ §µ §µ §

µ §no additional bent bars are needed the bar will be extended along the bottom of slab to the full development length of 381mm

Ex\The R.C flat slab shown in fig. carries a total service dead load of (6.8 µ §) (including its self-weight) and service live load of (4.14 µ §). The following design data is given, slab thickness(µ §), all edge beams are (300*600 mm), interior circular columns are (300 mm) in diameter, columns along slab edges are (300*300 mm) square µ § and µ §, use µ § bars for slab reinforcement.

Check the slab for shear requirements

Using ACI direct design method, design the exterior top reinforcement of end span of edge beam.

Sol\


µ §

Check the slab for shear requirements

One-way shear

µ §µ §µ §µ §

Two-way shear around circular column: (inside the drop)

µ §µ §µ §µ §µ §µ §µ §µ §µ §Check shear around drop panel (outside the drop)

µ §µ §µ §µ §

The design the exterior top reinforcement of end span of edge beam.

µ §µ §

µ §µ §



µ §

µ §µ §µ §µ §

µ §µ §µ §

µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §µ §

Opening in slab system without beams

OpeningLimiting dimensionCode reference1µ §µ § 2µ §µ § 3µ §µ § 1




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