Plan View of Floor System. beam spacing = 18 ft b w of beam = 14 in. f'c = 5000 psi fy = 60,000 psi. w SD = 10 psf w LL = 100 psf

Transcription

1 CE 4, mmer 01 lab Design Example 1 / 5 Design a one ay slab for an interior bay of a mlti story offie bilding sing the information speified belo. Neglet ompression reinforement. Assme partitions annot be damaged by defletions. Rond slab thikness p to the nearest ¼. Use a spreadsheet to design eah setion and sho yor reslts ith a sketh (elevation vie). bmit hand allations mathing the spreadsheet for one setion. Inlde nits on all nmbers. Exterior pan Interior pan Exterior pan paing paing paing slab span paing L beam L beam L beam Plan Vie of Floor ystem f' = 5000 psi fy = 60,000 psi D = 10 psf LL = 100 psf beam spaing = 18 ft b of beam = 14 in

2 CE 4, mmer 01 lab Design Example / 5 lab Thikness, t. et h = h min for defletion. Use ACI Table 9.5(a): h min = L / 8, L = distane from enter of spport to enter of spport = beam spaing = 16 ft h min = (16 x 1 / ) / 8 = 6.857, se h = 7.0 Chek shear strength: V = 1.15 L n / at exterior fae of 1 st interior spport (ACI 8., pg 109) self D L 1.( ) 1.6 self (0.150 kf 1" 7.75" )( )( ) klf 1"/' 1"/' 1.[ klf (0.010 ksf )(1')] 1.6(0.100 ksf )(1'), 0.88klf l n learspan beamspaing b of beam 16' 16" 1"/1, l n ft V 1.15(0.88klf )(14.67 ft)/.79k V (0.75)( f ' b d) Use #4 bars for flexral reinforement (revise if needed) over = 0.75 in (ACI 7.7.1, pg 9) 1 4 d h over bar 7.75" 0.75" ( )", 8 Use #4 bars for flexral reinf. d 6.75" V ( 0.75)( 4000 psi(1in)(6.75in) 8. 59k V k V, OK. Design Flexre Reinforement at eah etion Min. Reinforement: A s _ min bh (ACI10.5.4) A s_min (0.0018)(1")(7.75"), A s _ min in ft idth Max. Reinforement paing (ACI ): slabs min[18in,h] min[18in,(7.75in)] 18in max_

3 CE 4, mmer 01 lab Design Example / 5 Distribtion of reinforement (ACI ): max_ distrib h d max_ distrib max min[ min[15.5 f y bar 7.75in 6.75in min[15.5(0.75in), 1 ] (60,000 psi) (60,000 psi) max_ slabs, max_ distrib, 1 ] f y ] min[18in, 0.75in 1in], max 1in Design of Bar paings: l n Elevation Vie of Floor lab We an design one half of the slab length de to symmetry. Design etions M Coeff. (ACI 8.) 1/4 1/14 1/10 1/11 1/16 1/11 M, k ft A s_req d, in s, in A s_prov'd, in a, in s M n, k ft Callations for Colmn provided belo. Callations for Colmns throgh 7 performed by spreadsheet (attahed)

4 CE 4, mmer 01 lab Design Example 4 / 5 Typial als. for first olmn of above table: M A s_req d : et M n M ln (16.8 ft) (0.88klf ) k - ft A s f y (d a ) M (Assme steel yields at ltimate strength) Assme: d- a 0.95d A s _ req'd (.40k-ft) (1 in M f y 0.95d ft ) (0.90)(60ksi)(0.95)(6.75in) in ft idth A s _min A s _ req'd in ft idth Rebar spaing, s: s A bar 0.0in A s_req'd in " idth s 1in 14.4in max 1in train in Reinforement, s 1 A s _ prov'd A bar s From C = T: 0.85 f ' ab A s f y in bar (0.0 )(1 bar 1in )( 0.85(5ksi) a (1in) (0.0in )(60ksi), a 0.5in y t a 1 0.5in in s, y t d s in 0.00 s 6.75in s 0.004=ACI min. for beams and slabs, OK s so =0.90as assmed, OK 1in ft idth ) 0.0 in ft idth M n : M n A s f y (d a ) (0.90)(0.0in )(60ksi)(6.75in 0.5in ) 1ft 1in M n 5.97k ft

5 CE 4, mmer 01 lab Design Example 5 / 5 Design bar toffs: Extension of top bars past fae of spport (from Text Figre A 5) = ln 176" 44" '8" at end beams l n 0.(176") 5" 4' 5" at interior beams Temperatre & hrinkage teel: Use #4 bars (not bent by orkmen alking on) Also, A s _ req'd for T& reinforement = A s _min for slabs = s s max Abar As _ req' 1in d 0.0in 0.167in 1in 14.4in, min[18",5 h] min[18in, 5(7.75in)] 18in in ft idth T & Reinf : 1in min (Typ.) lear over = 0.75 top, bottom and sides lab Details

6 CE 4, Reinfored Conrete Design mmer 01 Example lab Design 1 / Material Properties Geometry ompressive strength of onrete f ' 5,000 psi span length of slab (beam spaing L 18 ft enter to enter oeffiient for depth of stress blok idth of spporting beam b spport 14 in yield strength of reinforement f y 60,000 psi lear span L n 16.8 ft = L - b spport /1 nit eight of reinf. on. UW 150 pf modls of elastiity of onrete E 4,00,000 psi = * QRT( f' ) minimm thikness h min in = L / 8 * 1 modls of rptre of onrete f r 50 psi = 7.5 * QRT( f' ) depth of slab h 7.75 in modls of elastiity of steel E s 9,000,000 psi Loads Reinforement self eight of beam s klf = UW/1000 * h/1 * 1 size of flexral reinforement Bar_size 4 sperimposed dead load DL 10 psf diameter of steel reinf. Bars bar 0.5 in live load LL 100 psf A bar 0.0 in^ niform distribted fatored load 0.88 klf lear over over 0.75 in = 1. * (s + DL/1000 * 1) * LL/1000 * 1 maximm effetive depth d max 6.75 in =h - over - bar / effetive depth d 6.75 in hear OK = IF( V n >= V, "OK", "NG" ) max shear fore V.79 k = 1.15 * * L n / shear strength of onrete V n 8.59 k = 0.75 * * QRT( f ' ) * 1 * d /1000 Flexre min. reinforement A s_min in = * 1 * h max spaing for slabs s max_slabs in = MIN( 18, * h ) lear distane from reinf. to ten. fae in = h - d - bar / max spaing for distribtion s max_distrib 1.00 s max 1.00 in = MIN( s max_slabs, s max_distrib )

7 CE 4, Reinfored Conrete Design mmer 01 Example lab Design / oeffiient from ACI 8. Moef moment de to fatored loads M k-ft As_req'd for Mn to eqal M A s_req'd in s req'd for #4 bar to provide As_req'd s req'd in design bar spaing s As for #4 bar at spaing "s" A s_prov'd in depth of stress blok a in strain in steel reinforement at lt. s strength redtion fator available flexre strength M n k-ft here M = * L n^ / Moef A s_req'd = MAX( A s_min, M * 1000 / (0.9 * f y * 0.95 * d ) ) s req'd = MIN( s max, A bar / ( A s_req'd / 1 ) ) a = A s_prov'd * f y / (0.85 * f ' * 1) s = 0.00 / (a / 1 ) * d =IF( s < 0.00, 0.65, IF( s > 0.005, 0.9, * s ) ) M n = * A s_prov'd * f y * ( d - a/ ) /1000