ANALYSIS OF COMPOSITE CONCRETE SLAB ON METAL DECK

Transcription

1 ANALYSIS OF COMPOSITE CONCRETE SLAB ON METAL DECK Assignment #5 Select a Vulcraft composite deck to support 150 psf service live load on a 10 ft clear span. The steel deck is to be used on a three-span condition and shoring is not permitted. Use normal weight concrete with f'c = 3000 psi Selected Deck: VLI17 - Note: If you used a different deck, use mathcad sheet to enter your information and check answer. Allowable Live Load given in table on pg 46 = 157 psf. This is greater than the specified 150 psf The construction clear span for a 3-span condition is 13'-0" which is greater than the specified 10 ft. Deck is OK Properties : D:= 4 in t := in W1 := 39 psf L := 10 ft S p := in^3 / ft S n := in^3/ft I p := in^4/ft I n := in^4/ft Fy := 40 ksi Fb allow := 0.6 Fy Fb allow = 4 ksi or 36 ksi (whichever is less) #1) Check construction SDI criteria for the steel deck selected Define loads W1 = slab weight + deck weight W1 = 39 psf Chart on pg 46 W = 0 psf construction load W := 0 psf Sec 3.a P1 = 150 lb concentrated load P1 := 150lb Sec 3.a For the 3-span Stress check- 3 cases on page 56 Case 1 positive moment M1 := ( 0.0 P1 L) W1 L M1 = lb*ft Positive Moment Stress M1 1 fb 1 := fb S p = ksi fb 1 = ksi < Fb allow = 4 ksi OK Case positive moment M := ( W1 + W) L M = lb*ft Positive Moment Stress M 1 fb := fb S p 1000 = 11.3 ksi 1

2 fb = ksi < Fb allow = 4 ksi OK Case 3 negative moment M3 := ( W1 + W) L M3 = lb*ft negative Moment Stress M3 1 fb 3 := fb S n = ksi fb 3 = ksi < Fb allow = 4 ksi OK Check Deflection for 3-span (pg 56) E := psi W1 L 4 1 := = 0.53 in EI p L1 Allowable Deflection allow := 180 allow = in which is less than 3/4" 1 = 0.53 in < allow = in Therefore deflection is OK #) Determine the maximum stress (forwork stress plus composite section stress) in the steel deck under service load Find area of steel in 1 ft section t s := in From given geometry on pg 46 l s :=.5 in +.4 in + 5in l s = in Area of steel A s := t s l s A s = in Mod. Ratio n := 9 Given on pg 46 I p y bar of steel y c.g.s := 1 in y S c.g.s = in p b Transform cracked concrete section to equivalent steel b := 1 in b eff := b n eff = in Sum areas about the cracked neutral axis to determine location of y cracked set initial value for yc to use in Given/Find function y c := 1in Given y c b eff y c A s ( D y c y c.g.s) = 0 ( ) y c := Find y c

3 y c = in Determine I cracked of section ( Parallel Axis Therom) 3 b eff y c I cr := 3 + I p + A s D y c.g.s y c ( ) I cr = 3.66 in 4 Calculate maximum tensile stress in deck due to live load only From table on pg 46 w ll := 157 lbf per foot ft w ll L M ll := M 8 ll = lbf ft M ll = 3550 lbf in M ll D y c f deckll := I cr f deckll ( ) = ksi < Fb allow = 4 ksi OK Check the maximum tensile locked formwork stress in the deck due to dead load W1 = 39 lbf per foot ft M dl := 0.08 W1 L Maximum Moment equation for 3-span beam with only concrete weight uniformly distributed ( see pg 56 ) M dl = 31 lbf ft M dl f deckdl := f S deckdl = ksi p Total stress on deck f total := f deckll + f deckdl f total = 3.4 ksi ~= Fb allow = 4 ksi OK #3) Determine the necessary temperature and shrinkage reinforcement. See section 5.5 for SDI criteria ( pg 54 ) A req := t 1 in A req = in per foot Can not be less than area of 6x6 W1.4 x W1.4 A 6x6 := in A 6x6 = 0.08 in per foot Must use minimum of 6x6 W1.4 x W1.4 to meet criteria This section is also recommended on table pg 47 3

4 #4) Design negative moment reinforcement to provide continuity and control cracking Live Load factor ll := 1.7 ll w ll L Negative Moment over support for 3 span M uneg := M 1 uneg = 4. lbf ft Find moment capacity of slab (per foot width) Flexure Negative Moment - Compression on Bottom, Tension on Top Yield of Steel fy := 60 ksi Conc comp. strength f c := 3 ksi width b comp := 5in area of concrete on bottom in compression ( see geometry of deck) Assume #3 bars π d b d b :=.375 in A bar := 4 A bar = 0.11 in per foot width Cover c :=.75 in Distance to compression face d1 := D c d b Sum of forces in horizontal direction = 0 C=T C = 0.85*f.c*a*b Find value of a Sum of moments = 0 T = As*fy A bar fy a1 := a1 = 0.5 in compression zone in concrete 0.85 f c b comp Sum moments about C a1 Force T1 := A bar fy Distance z1 := d1 z1 =.803 in Determine minimum required area of steel for negative moment M uneg = 4. lbf ft Mn1 := T1 z1 Mn1 = lbf ft Factored Nominal Capacity φ := 0.9 ACI 318 factor for flexure A sreq := 1in Given a1 φ A sreq fy d1 A sreq := Find( A sreq ) = M uneg 4

5 A sreq = in per foot width # of bars per foot width A sreq = A bar Need # 3 bars at 6" spacing 5

6 psf := lbf ft 6

7 I p := I p 1 in 4 ksi := 1000 psi L := L1 ft W1 := W1 lbf ft S p := S p in 3 t := tin 7

8 Pg 8 Area of steel (per foot) As := in^ 8