Example Building. Two-Way Slabs

Transcription

1 Page 1 of 7 The following example illustrates the design methods presented in the article Timesaving Design Aids for Reinforced Concrete, Part : Two-way Slabs, by David A. Fanella, which appeared in the October 001 edition of Structural Engineer magazine. Unless otherwise noted, all referenced table, figure, and equation numbers are from that article. 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 loads according to ACI Design strip x 0 (typ.) 4 x 4 (typ.)

2 Page of 7 Design Data Materials Concrete: normal weight (150 pcf), ¾- in. maximum aggregate, f c = 4,000 psi Mild reinforcing steel: Grade 60 (f y = 60,000 psi) Loads Superimposed dead loads = 30 psf Live load = 50 psf Minimum Slab Thickness Longest clear span l n = 4 (0/1) =.33 ft From Fig. 1, minimum thickness h per ACI Table 9.5(c) = l n /30 =.9 in. Use Fig. to determine h based on shear requirements at edge column assuming a 9 in. slab: w u = 1.4( ) + 1.7(50) = 4.5 psf A = 4 x [( )/] = 60 ft From Fig., d/c d = 0.39 x 0 = 7.0 in. h = = 9.05 in. Try preliminary h = 9.0 in. Design for Flexure Use Fig. 3 to determine if the Direct Design Method of ACI Sect can be utilized to compute the bending 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 = 4/0 = 1. < O.K. Successive span lengths in each direction are equal O.K. No offset columns O.K. L/D = 50/( ) = 0.35 < O.K. Slab system has no beams N.A. Since all requirements are satisfied, the Direct Design Method can be used. A/c 1 = 60/1.67 = 93.6

3 Page 3 of 7 Total panel moment M o in end span: Mo w u l l = n = = ft - kips Total panel moment M o in interior span: For simplicity, use M o =. ft-kips for all spans. Division of the total panel moment M o into negative and positive moments, and then column and middle strip moments, involves the direct application of the moment coefficients in Table 1. Mo w u l l = n = ft = - kips Slab End Spans Int. Span Moments (ft-kips) Ext. neg. Positive Int. neg. Positive Total Moment Column Middle Note: All negative moments are at face of support.

4 Page 4 of 7 Required slab reinforcement. Span Location M u (ft-kips) b* (in.) d** (in.) A s (in. ) Min. A s (in. ) Reinforcement + End Span Ext. neg No. 4 Column Positive No. 4 Middle Int. Neg No. 4 Ext. neg No. 4 Positive No. 4 Int. Neg No. 4 Interior Span Column Positive No. 4 Middle *Column strip width b = (0 x 1)/ = 10 in. *Middle strip width b = (4 x 1) 10 = 16 in. **Use average d = = 7.75 in. A s = M u /4d where M u is in ft-kips and d is in inches Min. A s = 0.001bh = 0.016b; Max. s = h = 1 in. or 1 in. (Sect ) + For maximum spacing: 10/1 = 6.7 spaces, say bars 16/1 = 9.3 spaces, say 11 bars Positive No. 4 Design for Shear Check slab shear and flexural strength at edge column due to direct shear and unbalanced moment transfer. Check slab reinforcement at exterior column for moment transfer between slab and column. Portion of total unbalanced moment transferred by flexure = γ f M u

5 Page 5 of 7 b 1 = 0 + (7.75/) = 3.75 in. b = = 7.75 in. b 1 /b = 0.6 From Fig. 5, γ f = 0.6* γ f M u = 0.6 x 73.4 = 45.5 ft-kips Required A s = 45.5/(4 x 7.75) = 1.47 in. Number of No. 4 bars = 1.47/0. = 7.4, say bars Must provide -No. 4 bars within an effective slab width = 3h + c = (3 x 9) + 0 = 47 in. Provide the required -No. 4 bars by concentrating of the column strip bars (1-No. 4) within the 47 in. slab width over the column. Check bar spacing: For -No. 4 within 47 in. width: 47/ = 5.9 in. < 1 in. O.K. For 4-No. 4 within = 73 in. width: 73/4 = 1.5 in. > 1 in. Add 1 additional bar on each side of the 47 in. strip; the spacing becomes 73/6 = 1. in. < 1 in. O.K. Reinforcement details at this location are shown in the figure on the next page (see Fig. 6). The provisions of Sect may be utilized; however, they are not in this example.

6 Page 6 of Column strip No. 4 -No. 4 3-No. 4 Check the combined shear stress at the inside face of the critical transfer section. vu V = u Ac γ M + v u J / c Factored shear force at edge column: V u = 0.5[(4 x 10.3) (1.99 x.31)] V u = 7. kips When the end span moments are determined from the Direct Design Method, the fraction of unbalanced moment transferred by eccentricity of shear must be 0.3M o = 0.3 x. = 4.7 ft-kips (Sect ). γ v = 1 γ f = = 0.3 c /c 1 = 1.0 c 1 /d = 0/7.75 =.5 Interpolating from Table 7, f 1 = 9.74 and f = 5.53 A c = f 1 d = 9.74 x 7.75 = 55.0 in.

7 Page 7 of 7 J/c = f d 3 = x 5.53 x = 5,14 in. 3 vu vu 7, 00 = , , 14 = = psi Determine allowable shear stress φv c from Fig. 4b: b o /d = (b 1 + b )/d b o /d = [( x 3.75) ]/7.75 = 9.74 β c = 1 φv c = 15 psi > v u = psi OK Reinforcement Details The figures below show the reinforcement details for the column and middle strips. The bar lengths are determined from Fig of ACI No No No. 4 Standard hook (typ.) 6 -No No. 4 Class A tension splice Column strip Standard hook (typ.) 14-No No No. 4 7-No No. 4 7-No. 4 Middle strip 0-0