CONCRETE STRUCTURES
CONCRETE STRUCTURES 
One-Way Slab 6'-12' 4" min. thickness thickness = span/30 for floor slabs thickness = span/36 for roof slabs
Problem: ======= A T-beam and one-way slab sytem carries a live load of 50 lb/sqft and a uniform dead load of 110 lb/sqft The dead load contains an allowance for the weight of the slab, beam stem, partitions etc. The T-beams are simply supported at each end and span 26 ft. 3,000 psi concrete fc' = 3 kips/sqin Grade 60 reabar exterior bay 11'-0" center to center interior bay 12'-0" center to center interior exposure Establish the dimensions of a typical interior beam and the slab and select the required reinforcement,
wl = 50 lb/sqft wd= 110 lb/sqft includes slab beam stem interior partitions interior exposure beam span 26 ft. interior bay 12'-0" center to center exterior bay 11'-0" center to center beam simple supported 3,000 psi concrete fc' = 3 kips/sqin Grade 60 reabar fy = 60 kips/sqin
Graphic Representation: ================== wl = 50 lb/sqft wd= 110 lb/sqft _________________ I I I I 11' 12' 12'
For interior bay only! Beam depth width Slab thickness reinforcement size reinforcement spacing
Step (#) by Step (#) =============== #1. Minimum Thickness of Slab Leet P 76 Table 9.5 A
#2. Minimum depth of beam: simply supported beam h min = l / 16 h min = 26 ft x 12 -------------- 16 h min = 19.5 inch h = 20 inches selected always even number of inches #3. Minimum depth slab: solid one-way slab both end continuous h min = l / 28 h min = 12ft x 12 --------------- 28 h min = 5.14 inches h min = 5 inches close enough #4. Try b w = 12 inches the width must be adequate to carry shear and allow proper spacing between reinforcing bars. #5. Compute the design load per square foot of slab Leet P. 9, Table 1.3 w u = 1.7 (wl) + 1.4 (wd) w u = 1.7 (50 lb/sqft) + 1.4 (110 lb/sqft) wu = 239 lb/sqft
#6. Typical 1 foot wide strip of slab graph
#7. Slab modeled as continuous beam graph
#8. Design Moments graph
#9. Design Moments over support A formula for continuous beam from P&A Ma = wu x l x l / 24 Ma = 0.239 x 10 x 10 /24 Ma = 1.00 ft kips #10. Design Moment over center of first bay Mb = wu x l x l / 11 Mb = 0.239 x 10 x 10 / 11 Mb = 2.17 ft kips #11. Moment over support C Mc = wu x l x l / 10 #12. l = average of the adjacent clear spans l = 10 + 11/2 l = 21/2 l = 10.5 #13. Mc = wu x l x l / 10 Mc = 0.239 x 10.5 x 10.5 / 10 Mc = 2.63 ft kips #14. Moment over midspan second bay Md = wu x ln x ln / 16 Md = 0.239 x 11 x 11 /16 Md = 1.81 ft kips #15. Moment over support E Me = wu x ln x ln / 11 Me = 0.239 x 11 x 11 / 11 Me = 2.63 ft kips # 16. As Compute As per foot of slab at the critical sections where moments have been evaluated For example: at the first interior support, top steel must carry Mc = 2.63 ft kips / ft #17. Minimum thickness of concrete cover to protect reinforcement Table 3.3 Condition Bar size Min. Cover not exposed to earth or weather slabs no. 14-18 1.5" no 11 and smaller 3/4" permanent contact with earth all 3" formed concrete exposed to earth or weather no. 6-8 2" no. 5 and under 1.5"
A minimum of 3/4" of cover for slab steel that is not exposed to weather or in contact with the ground is required per ACI Code 7.7.1. #18. If we assume 1/2" diameter bars d = h - (coverage + db/2) d = 5" - (3/4" + 1/2" / 2) d = 5" - (0.75" + 0.25") d = 5" - 1" d = 4" bottom slab to center bar h = 5" slab thickness # 19. Moment due to factored loads Mu = Phi x T ( d - a/2) #20. Phi Phi = safety factor for factored load Phi = 0.9 #21. a = area in tension a = 0.85 f c a = 0.4" guessed #22. T = As x fy #23. Moment due to factored loads Mu = Phi x T ( d - a/2) Mu = 0.9 x As x fy (d - a/2) Mu = previous calculated for various locations 2.63 (12) = 0.9 T ( 4 - 0.4/2) 31.56 = 0.9 T (4 - 0.2) 31.56 = 0.9 T (3.8) 31.56/3.8 = 0.9 T 8.3053 = 0.9 T 9.228 = T T = 9.23 kips #24. As = T / fy As = 9.23 kips / 60 kips/ sqinch As = 0.154 sqinch/ft #25. Table B 2 Bar Number 3 spacing 8 inches As = 0.17 sq inch / ft
#26.Maximum spacing for flexural steel 3 times slab thickness 18 inches #27. T = C #28. Tension force C by temperature creep/support C = a x b (0.85 fc) 9.23 kips = a x 12" (0.85 x 3) 9.23 kips = a x 12" x 2.55 9.23 kips = 30.6 a a = 9.23 / 30.6 a = 0.3016339 a = 0.3 inch #29. a selected > a calculated 0.4 > 0.3 a actual is smaller than a assumed O.K. #30. Temperature steel Temperature steel is designed on an emperical basis. The minimum required ratio of temperature steel area to gross area of the concrete slab for various of rebar is specified by ACI Code 7.12 as A s / A g = 0.02 for fy= 40 to 50 kips/sqin 0.0018 fy= 60 0.0018 (60,000)/ fy for fy > 60 kips/sqin #31. But in any case: As/Ag must be greater than 0.0014 Maximum spacing between bars 5 times slab thickness 18 inches #32. Temperature Steel: At = 0.0018 Ag At = 0.0018 ( 5" x 12") At = 0.0018 x 60 At = 0.108 sqinch/ft #33. Table B2 No 3 bars 12 inches on center As = 0.11 sq in /ft
#34. Graphic representation 5" slab thickness 4" edge to center rebar at mid-bay rebar over supports no 3 at 8" flexural steel in direction of slab span no 3 at 12" temperature steel perpendicular to slab span
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©Dr. Gruenwald 1996, 1997,1998
