Example 1. Examples for walls are available on our Web page: Columns

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1 Portlan Cement Association Page 1 o 9 Te ollowing examples illustrate te esign metos presente in te article Timesaving Design Ais or Reinorce Concrete, Part 3: an Walls, by Davi A. Fanella, wic appeare in te November 001 eition o Structural Engineer magazine. Unless oterwise note, all reerence table, igure, an equation numbers are rom tat article. Te examples presente ere are or columns. Examples or walls are available on our Web page: Example 1 In tis example, an interior column at te 1 st loor level o a 7-story builing is esigne or te eects o gravity loas. Structural walls resist lateral loas, an te rame is nonsway. Materials Concrete: normal weigt (150 pc), ¾-in. maximum aggregate, c 5,000 psi Mil reinorcing steel: Grae 60 ( y 60,000 psi) Loas Floor raming ea loa 80 ps Superimpose ea loas 30 ps Live loa 100 ps (loor), 0 ps (roo) Builing Data Typical interior bay 30 t x 30 t Story eigt 1 t-0 in. Te table below contains a summary o te axial loas ue to gravity. Te total actore loa P u is compute in accorance wit Sect. 9..1, an inclues an estimate or te weigt o te column. Live loa reuction is etermine rom ASCE Moments ue to gravity loas are negligible. Floor DL (ps) LL (ps) Re. LL (ps) P u (kips) Cum. P u (kips) , , ,499

2 Portlan Cement Association Page o 9 Use Fig. 1 to etermine a preliminary size or te tie column at te 1 st loor level. Assuming a reinorcement ratio ρ g 0.00, obtain P u /A g 3.0 ksi ( c 5 ksi). Since P u 1,499 kips, te require A g 1,499/ in. Try a x in. column (A g 484 in. ) wit a reinorcement ratio ρ g greater tan Ceck i slenerness eects nee to be consiere. Since te column is part o a nonsway rame, slenerness eects can be neglecte wen te unsupporte column lengt is less tan or equal to 1, were is te column imension (Sect ). 1 1 x 64 in. t > 1 t story eigt, wic is greater tan te unsupporte lengt o te column. Tereore, slenerness eects can be neglecte. Use Fig. 1 to etermine te require area o longituinal reinorcement. For a x in. column at te 1 st loor level: P u /A g 1,499/ ksi From Fig. 1, require ρ g 0.06, or A s 0.06 x x 1.58 in. Try 8-No. 11 bars (A s 1.48 in. ) Ceck Eq. (10-) o ACI : φp n(max) 0.80φ[0.85 c (A g A st ) + y A st ] φp n(max) 1,54 kips > 1,499 kips O.K. From Table 1, 5-No. 11 bars can be accommoate on te ace o a -in. wie column wit normal lap splices an No. 4 ties. In tis case, only 3-No. 11 bars are provie per ace. Use 8-No. 11 bars (ρ.58%). Determine require ties an spacing. Accoring to Sect , No. 4 ties are require wen No. 11 longituinal bars are use.

3 Portlan Cement Association Page 3 o 9 Accoring to Sect , spacing o ties sall not excee te least o: 16 long. bar iameters 16 x long. bar iameters.6 in. 48 tie bar iameters 48 x tie bar iameters 4 in. Least column imension in. (governs) Ceck clear spacing o longituinal bars: Clear space in Since te clear space between longituinal bars > 6 in., cross-ties are require per Sect Reinorcement etails are sown below. See Sect. 7.8 or aitional special reinorcement etails or columns. 8-No. 11 No. 4

4 Portlan Cement Association Page 4 o 9 Example In tis example, a simpliie interaction iagram is constructe or an 18 x 18 tie column reinorce wit 8-No. 9 Grae 60 bars (ρ g 8/ ). Concrete compressive strengt 4 ksi. Use Fig. 3 to etermine te 5 points on te interaction iagram. Point 1: Pure compression φpn(max) 0.80φAg[0.85c + ρg(y 0.85c )] [(0.85 4) (60 (0.85 4))] 871 kips Point ( 0) Layer 1: 1 C (1) 1 Layer : C Layer 3:.44 1 C Since 1 C ( 3 / 1 ) > 0.69, te steel in layer 3 as yiele. Tereore, set 1 C ( 3 / 1 ) 0.69 to ensure tat te stress in te bars in layer 3 is equal to 60 ksi. No. 3 tie 3-No. 9 -No (typ.) No. 9

5 Portlan Cement Association Page 5 o 9 n P C b 87 A 1 C i φ n φ si i {( ) + 87[(3 0) + ( 0.4) + (3 0.69)]} 0.70 ( ) 744 kips Point 3 ( -0.5 y ) Layer 1: 1 C (1) Layer : C φmn β 0.5C b 1 1 φ 1 1 C n 87 A 1 C i + si i 1 1 i / 1 Layer 3:.44 1 C Use { 0.70 [( ) [(3 0)( ) + ( 0.4)(9 9) + (3 0.69)( (1, t - kips 1,181.4) / 1.44)]} / 1 n P C b 87 A 1 C i φ n φ si i {( ) + 87[(3-0.34) + ( 0.3) + (3 0.69)]} 0.70( ) 514 kips

6 Portlan Cement Association Page 6 o 9 φmn β 0.5C b 1 1 φ 1 1 C n 87 A 1 C i + si i 1 1 { 0.70 [( ) [(3-0.34)( ) + ( 0.3)(9 9) + (3 0.69)(9 i.44)]} / 1 / 1 Layer 3:.44 1 C Use 0.69 n P C b 87 A 1 C i φ n φ si i {( ) + 87[(3-0.69) + ( 0.0) + (3 0.69)]} 0.70 ( ) 338 kips 0.70 (, t - kips Point 4 ( - y ) 1,763.5) / 1 Layer 1: 1 C (1) Layer : C φmn β 0.5C b 1 1 φ 1 1 C n 87 A 1 C i + si i 1 1 { 0.70 [( ) [(3-0.69)( ) + ( 0.0)(9 9) + (3 0.69)(9.44)]} / (, ,36.8) / 1 80 t - kips i / 1

7 Portlan Cement Association Page 7 o 9 Point 5: Pure bening Use iterative proceure to etermine φm n. > E s 9,000 ( 8) ksi -60 ksi, use 60 ksi Try c 4.0 in. T A ( 60) 10 kips c 1 c c 1 c T E 9,000 ( ) > - 60 ksi, use A s 3 ( 60) c 1 c ksi 51.4 ksi 180 kips C Cc E 9, ksi A s kips 0.85c ab (0.85 4) kips Total T (-180) + (-10) -300 kips Total C kips Since T C, use c 4.0 in.

8 Portlan Cement Association Page 8 o 9 Mn T 1 18 ( 180) / t - kips Mn T ( 10) / 1 Compare simpliie interaction iagram to interaction iagram generate rom te PCA computer program PCACOL. Te comparison is sown on te next page. As can be seen rom te igure, te comparison between te exact (black line) an simpliie (re line) interaction iagrams is very goo. Mn C / t - kips Mn 3 0.5Cc( a) + Mnsi i 1 [ (18 3.4)] / t - kips φ M n t - kips

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