16 Reinforced Concrete Design Design of Slabs Types of Slabs Load Paths and Framing Concepts One-way Slabs Two-way Slabs Mongkol JIRAVACHARADET S U R A N A R E E UNIVERSITY OF TECHNOLOGY INSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERING
Types of Slab One-way slab One-way slab Two-way slab Flat plate slab Flat slab Grid slab
Load Path / Framing Possibilities L n = 3. m L n = 4.4 m L n = 8. m L n = 3.6 m Think we ll need some additional framing members???
Framing Concepts Let s use a simple example for our discussion Column spacing 8 m c-c Think about relating it to your design project. Plan
Framing Concepts We can first assume that we ll have major girders running in one direction in our one-way system
Framing Concepts We can first assume that we ll have major girders running in one direction in our one-way system If we span between girders with our slab, then we have a load path, but if the spans are too long
Framing Concepts We will need to shorten up the span with additional beams But we need to support the load from these new beams, so we will need additional supporting members
Framing Concepts Now let s go back through with a slightly different load path. We again assume that we ll have major girders running in one direction in our one-way system. This time, let s think about shortening up the slab span by running beams into our girders. Our one-way slab will transfer our load to the beams.
Two Load Path Options
Framing Concepts - Considerations For your structure: Look for a natural load path Identify which column lines are best suited to having major framing members (i.e. girders) Assume walls are not there for structural support, but consider that the may help you in construction (forming)
Example Condo Floor Plan
One-way Slab ก L S 1.0 m Main reinforcement Design of one-way slabs is like design of parallel 1m beams.
Design of One-way Slab (L > S) L 1 m ก 1 w S S Minimum Thickness (ACI) Simply supported One end continuous Both ends continuous Cantilever L/0 L/4 L/8 L/10 * multiplied by 0.4 + f y /7,000 for steel other than SD40
ACI Design Provision Shrinkage and temperature reinforcement For structural slabs only; not intended for soil-supported slabs on grade Ratio of reinforcement A s to gross concrete area A g : A s /A g RB4 (fy =,400 ksc)................. 0.005 DB30 (fy = 3,000 ksc)................. 0.000 DB40 (fy = 4,000 ksc)................. 0.0018 DB (fy > 4,000 ksc)................... 0.0018 4,000 0.0014 Spacing 5 t 45 cm f y Main Steel (short direction): A s 6 mm Max. Spacing 3 t 45 cm Min. Spacing f main steel 4/3 max agg..5 cm
Effect of column width b b A A B B wl 1 wl b/ L w b/ wl wl 1 Moment at A : = wl 1 wl = 1 + wl b wlb 4 + wb 8 ( / ) w b If A and B are fiexed against rotation, M = ( b) w L 1 = wl 1 wlb 6 + wb 1
Typical reinforcement in a one-way slab Top bars at exterior beams Top bars at exterior beams Bottom bars Exterior span Temperature bars Interior span (a) Straight top and bottom bars Bent bar Bent bars Bottom bars Exterior span Temperature bars Interior span (b) Alternate straight and bent bars
Example: Design one-way slab as shown below to carry the live load 500-kg/m f c = 10 kg/cm, f y =,400 kg/cm A G1 S1 S S3 A 3 @ 8 m = 4 m 0.4 + 400/7000 = 0.74 min h = 400(0.74)/4 = 1.3 cm USE h = 13 cm DL = 0.13 400 = 31kg/m 4 @ 1 m = 48 m w u = 1.4(31) + 1.7(500) = 1,86.8 kg/m clear span = 4-0.3 = 3.7 m M u = (1,86.8)(3.7) /10 = 1,76 kg-m ρ max = 0.75ρ b = 0.75(0.0454) = 0.0341
USE RB9 with cm covering: d = 13--0.45 = 10.55 cm R n M = φ bd 176 100 u = = 0.9 100 10.55 17.6 ksc ' 0.85 f c R n ρ = 1 1 = 0.0077 < ρ ' 0.85 max OK f y f c A s = ρbd = 0.0077(100)(10.55) = 8.16 cm /m Select RB9@0.07 (A s = 9.8 cm /m) Temp. steel = 0.005(100)(13) = 3.5 < 9.8 cm /m OK Select RB9@0.18 (A s = 3.53 cm /m)
L 1 4 L 1 3 Detailing of one-way slab Temp. steel L 1 8 L 1 RB9@0.18 1.0. RB9@0.10 1.3. 4.0. RB9@0.07.13. RB9@0.18 RB9@0.14.13. RB9@0.07 1.0. 1.3. 4.0.
Design of Two-way Slab (L < S) S Min. Thickness: t 9 cm Perimeter/180 = (L+S)/180 L Reinforcement Steel: A s φ 6 mm Temp. steel Max. Spacing 3 t 45 cm Min. Spacing φ main steel 4/3 max agg..5 cm
Load transfer from two-way Slab S D A 45 o 45 o 45 o 45 o C B Short span (BC): Floor load = w kg/sq.m Tributary area = S /4 sq.m Load on beam = ws/4 ws/3 kg/m L Long span (AB): Span ratio m = S/L Tributary area = SL/ - S S /4 = m sq.m 4 m ws 3 Load on beam kg/m 3 m
Moment Coefficient Method ก ก -M s +M -M s L -M +M L L -M s S/4 S/ S/4 L/4 L/ L/4 Middle strip moment: M M = CwS Column strip moment: M C = M M /3
( C ) ก m 1.0 0.9 0.8 0.7 0.6 0.5 - - ก ก 0.033-0.05 0.040-0.030 0.048-0.036 0.055-0.041 0.063-0.047 0.083-0.06 0.033-0.05 - - ก ก 0.041 0.01 0.031 0.048 0.04 0.036 0.055 0.07 0.041 0.06 0.031 0.047 0.069 0.035 0.05 0.085 0.04 0.064 0.041 0.01 0.031 - - ก ก 0.049 0.05 0.037 0.057 0.08 0.043 0.064 0.03 0.048 0.071 0.036 0.054 0.078 0.039 0.059 0.090 0.045 0.068 0.049 0.05 0.037
( C ) ก m 1.0 0.9 0.8 0.7 0.6 0.5 - - ก ก 0.058 0.09 0.044 0.066 0.033 0.050 0.074 0.037 0.056 0.08 0.041 0.06 0.090 0.045 0.068 0.098 0.049 0.074 0.058 0.09 0.044 - - ก ก - 0.033 0.050-0.038 0.057-0.043 0.064-0.047 0.07-0.053 0.080-0.055 0.083-0.033 0.050
Bar detailing in slab L 1 /3 L /3 L 1 /7 L 1 /4 L /4 L 1 L Bar detailing in beam L 1 /3 L /3 L 1 /8 L 1 /8 L 1 L
ก ก L/5 ก L/5 L = ก
Example: Design two-way slab as shown below to carry the live load 300-kg/m f c = 40 kg/cm, f y =,400 kg/cm 4.00 3.80 0.10 0.50 0.0 0.0 5.00 4.80 Cross section Min h = (400+500)/180 = 10 cm DL = 0.10(,400) = 40 kg/m w u = 1.4(40)+1.7(300) = 846 kg/m Floor plan m = 4.00/5.00 = 0.8 A s,min = 0.0018(100)(10) = 1.8 cm /m
Short span Moment coeff. C -M( ) +M -M( ) 0.03 0.048 0.064 Max. M = C w S = 0.064 846 4.0 = 866 kg-m/1 m width d = 10 - (covering) - 0.5(half of DB10) = 7.5 cm R n M u 86, 600 = = = 17.11 kg/cm φbd 0.9 100 7.5 0.85 f c R n ρ = 1 1 0.0045 f = y 0.85 f c A s = 0.0045(100)(7.5) = 3.36. > A s,min ก ก DB10@0.0 (A s =3.90. )
Long span Moment coeff. C -M( ) +M -M( ) 0.05 0.037 0.049 Max. M = C w S = 0.049 846 4.0 = 663 kg-m/1 m width d = 10 - (covering) - 1.5(half of DB10) = 6.5 cm R n M u 66,300 = = = 17.44 kg/cm φbd 0.9 100 7.5 0.85 f c R n ρ = 1 1 0.0046 f = y 0.85 f c A s = 0.0046(100)(6.5) =.97. > A s,min ก ก DB10@0.0 (A s =3.90. )
ก ก V u = w u S/4 = (846)(4.0)/4 = 846 กก./. ก ก φv c = 0.85(0.53) (100)(7.5) 40 = 534 กก./. OK
DB10@0.40 DB10@0.0 DB10@0.40 0.50 0.55 0.95 0.0 3.80 0.0 0.95 1.30 0.10 DB10@0.40 DB10@0.0 DB10@0.40 0.50 0.0 0.70 1.0 4.80 1.0 1.60 0.0 0.10