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The following example illustrates the design methods presented in the PCA book Simplified Design -Reinforced Concrete Buildings of Moderate Size and Height third edition. Unless otherwise noted, all

referenced table, figure, and equation numbers are from that book.

Example Building

Below is a partial plan of a typical floor in a cast-in-place reinforced concrete building. In this example,an interior strip of a flat plate floor system is designed and detailed for the effects of gravity loadsaccording to ACI 318-05.

Design Data

Materials Concrete: normal weight (150 pcf), 3/4-in. maximum aggregate, fc = 4,000 psi

Mild reinforcing steel: Grade 60 (fy = 60,000 psi)

Loads

Superimposed dead loads = 30 psf

Live load = 50 psf

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Two-Way Slabs

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The following example illustrates thedesign methods presented in the articleTimesaving Design Aids for ReinforcedConcrete, Part 2: Two-way Slabs, byDavid A. Fanella, which appeared in theOctober 2001 edition of StructuralEngineer magazine. Unless otherwisenoted, all referenced table, figure, andequation numbers are from that article.

Example BuildingBelow is a partial plan of a typical floor in acast-in-place reinforced concrete building. Ithis example, an interior strip of a flaplate floor system is designed and detailefor the effects of gravity loads accordinto ACI 318-99.

20?-0I 20?-0I 20?-0I

24?-0I

24?-0I

20Ix 20I (typ.) 24Ix 24I (typ.)

Designstrip

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Minimum Slab Thickness

Longest clear spanl

n = 24 (20/12) = 22.33 ftFrom Fig. 4-3, minimum thickness h per ACI Table 9.5(c) = ln/30 = 8.9 in.

Use Fig. 1-8 to determine h based on shear requirements at interior column assuming a 9 in. slab:

qu = 1.2(112.5 + 30) + 1.6(50) = 251.0 psf

A = 24 x 20 = 480 ft2

A/c12 = 480/22 = 120

From Fig. 1-8, d/c1 0.22

d = 0.22 x 24 = 5.3 in.

h = 5.3 + 1.25 = 6.55 in.

Try preliminary h = 9 in.

Design for Flexure

Use Fig. 4-4 to determine if the Direct Design Method of ACI Sect. 13.6 can be utilized to compute thebending moments due to the gravity loads:

3 continuous spans in one direction, more than 3 in the other O.K.

Rectangular panels with long-to-short span ratio = 24/20 = 1.2 < 2 O.K.

Successive span lengths in each direction are equal O.K.

No offset columns O.K.

L/D = 50/(112.5 + 30) = 0.35 < 2 O.K.

Slab system has no beams N.A.

Since all requirements are satisfied, the Direct Design Method can be used.

Total panel moment Mo in end span:

Mo =

= 248.5 ft - kips

Total panel moment Mo in interior span:

Mo =

= 244 ft - kips

For simplicity, use Mo = 248.5 ft-kips for all spans.

Division of the total panel moment Mo into negative and positive moments, and then column and middlestrip moments, involves the direct application of the moment coefficients in Table 4-2.

q

80 251 24 18 0

8u 2 n

2 2l l

=

. .

q

80 251 24 18 167

8u 2 n

2 2l l

=

. .

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Two-Way Slabs

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fMu = 0.62 x 64.6 = 40 ft-kips

Required As = 40/(4 x 7.75) = 1.29 in.2

Number of No. 4 bars = 1.29/0.2 = 6.5, say 7 bars

Must provide 7-No. 4 bars within an effective slab width = 3h + c2 = (3 x 9) + 20 = 47 in.

Provide the required 7-No. 4 bars by concentrating 7 of the column strip bars (11-No. 4) within the 47 in.slab width over the column.

Check bar spacing:

For 7-No. 4 within 47 in. width: 47/7 = 6.7 in. < 18 in. O.K.

For 4-No. 4 within 120 47 = 73 in. width: 73/4 = 18.25 in. > 18 in.Add 1 additional bar on each side of the 47 in. strip; the spacing becomes 73/6 = 12.2 in. < 18 in. O.K.

Reinforcement details at this location are shown in the figure on the next page.

Check the combined shear stress at the inside face of the critical transfer section.

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vu

=

Factored shear force at edge column:

Vu = 0.251[(24 x 10.83) (1.99 x 2.31)]

= 64.0 kips

When the end span moments are determined from the Direct Design Method, the fraction of unbalancedmoment transferred by eccentricity of shear must be 0.3Mo = 0.3 x 248.5 = 74.6 ft-kips (Sect. 13.6.3.6).

v = 1 f = 1 0.62 = 0.38

c2 /c1 = 1.0

c1 /d = 20/7.75 = 2.58

From Table 4.9

Ac = (2b1 + b2 d) = 585.7 in.2

J/c = [2b1d(b1 + 2b2) + d3(2b1 + b2)/b1]/6 = 5,139 in.

3

vu =

vu = 109.4 + 66.2 = 175.6 psi

Determine allowable shear stress vc from Fig. 4-13:

bo /d = (2b1 + b2)/d

bo /d = [(2 x 23.875) + 27.75]/7.75 = 9.74

c = 1

vc = 190 psi > vu = 175.6 psi OK

9 slab is OK

Reinforcement Details

The figures below show the reinforcement details for the column and middle strips. The bar lengths aredetermined from Fig. 13.3.8 of ACI 318-05.

64 000585

0 38 74 6 12 0005139

,

,+

. . ,

V

A

M

J cu

c

v u+

/

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The PCA computer program pcaSlab can be used to expedite the design of different slab systems.The program covers wide range of two-way slab systems and can be used for more complex slab layouts.The output of the program for the slab in the example is shown in the following pages.

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X

Y

Z

pcaSlab V1.10 - PCA User, Portland Cement Association

File: C:\Program Files\PCA\pcaSlab\TSDA-Example.slb

Project: TSDA Example

Frame: Engineer:

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pcaSlab V1.10 - PCA User, Portland Cement Association

File: C:\Program Files\PCA\pcaSlab\TSDA-Example.slb

Project: TSDA Example

Frame: Engineer:

Column Strip Flexural Rebars

Middle Strip Flexural Rebars

5-#4

5-#4

13-#4

12-#4

16-#4c

13-#4

12-#4

13-#4

12-#4

10-#4c

13-#4

12-#4

5-#4

5-#4

16-#4c

13-#4

13-#4

7-#4c 6-#4

13-#4

13-#4

7-#4c 6-#4

13-#4

13-#4

7-#4c 6-#4

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02-04-2005 - pcaSlab V1.10 - Portland Cement Association - Page 108:08:10 AM - Licensed to: PCA User, Portland Cement Association - TSDA-Example

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=============================================================================pcaSlab V1.10 (TM)

A Computer Program for Analysis and Design of Slab Systems=============================================================================

Copyright 2000-2003, Portland Cement AssociationAll rights reserved

Licensee stated above acknowledges that Portland Cement Association(PCA) is not and cannot be responsible for either the accuracy oradequacy of the material supplied as input for processing by thePCA-Slab computer program. Furthermore, PCA neither makes any warrantyexpressed nor implied with respect to the correctness of the output prepared by the PCA-Slab program. Although PCA has endeavored toproduce PCA-Slab error free the program is not and cannot be certifiedinfallible. The final and only responsibility for analysis, design andengineering documents is the licensees. Accordingly, PCA disclaims allresponsibility in contract, negligence or other tort for any analysis,design or engineering documents prepared in connection with the use of

the PCA-Slab program.

=============================================================================================[2] DESIGN RESULTS=============================================================================================

Top Reinforcement:==================

Units: Width (ft), Mmax (k-ft), Xmax (ft), As (in^2), Sp (in)Span Strip Zone Width Mmax Xmax AsMin AsMax SpReq AsReq Bars---- ------ ------ -------- ------------ -------- -------- -------- -------- -------- -------

1 Column Left 10.00 8.40 0.833 1.836 14.631 12.000 0.277 10-#4 Middle 10.00 0.00 9.917 0.000 14.631 0.000 0.000 ---Right 10.00 138.22 19.000 1.836 14.631 4.800 4.801 25-#4

Middle Left 14.00 -0.00 0.833 2.570 20.483 12.923 0.000 13-#4 Middle 14.00 0.00 9.917 0.000 20.483 0.000 0.000 ---Right 14.00 46.08 19.000 2.570 20.483 12.923 1.535 13-#4

2 Column Left 10.00 118.88 1.000 1.836 14.631 4.800 4.097 25-#4 Middle 10.00 0.00 10.000 0.000 14.631 0.000 0.000 ---

Right 10.00 118.88 19.000 1.836 14.631 4.800 4.097 25-#4

Middle Left 14.00 39.63 1.000 2.570 20.483 12.923 1.318 13-#4