1. Design of One Way Slabs CE A433 – RC Design T. Bart Quimby, P.E., Ph.D. Spring 2007

2. Definition A One Way Slab is simply a very wide beam that spans between supports A One Way Slab is simply a very wide beam that spans between supports Span, L Width, b

3. Design for a 12” Width When you solve for A s, you are solving for A s /ft width. When you solve for A s, you are solving for A s /ft width.

4. Beam Profile Design variables: Thickness (h) and Reinforcing Design variables: Thickness (h) and Reinforcing

5. Solving for Thickness, h Thickness may controlled by either: Thickness may controlled by either: Shear Shear Flexure Flexure Deflection Deflection

6. Thickness Based on Shear Shear stirrups are not possible in a slab so all you have is  V c for strength. Shear stirrups are not possible in a slab so all you have is  V c for strength. ACI 318-05 11.5.6.1(a) exempts slabs from the requirement that shear reinforcement is required where ever V u exceeds  V c /2. ACI 318-05 11.5.6.1(a) exempts slabs from the requirement that shear reinforcement is required where ever V u exceeds  V c /2.

7. Thickness Based on Flexure Use the three equations that were presented earlier in the semester for computing bd 2 for singly reinforced concrete beams, using b = 12”. Use the three equations that were presented earlier in the semester for computing bd 2 for singly reinforced concrete beams, using b = 12”. Largest beam size (based on A smin as specified in the code) Largest beam size (based on A smin as specified in the code) Smallest beam size (based on the steel strain being.005) Smallest beam size (based on the steel strain being.005) Smallest beam size not likely to have deflection problems (c ~.375c b ) Smallest beam size not likely to have deflection problems (c ~.375c b )

8. Thickness Base on Deflection We haven’t covered deflection calculations yet. We haven’t covered deflection calculations yet. See ACI 318-05 9.5.2 See ACI 318-05 9.5.2 You must comply with the requirements of ACI 318-05 Table 9.5(a) if you want to totally ignore deflections You must comply with the requirements of ACI 318-05 Table 9.5(a) if you want to totally ignore deflections

9. Other Considerations For thinner multi-span slabs, it might be useful to put the steel at mid depth so that it can act as both positive and negative reinforcing. For thinner multi-span slabs, it might be useful to put the steel at mid depth so that it can act as both positive and negative reinforcing. Then h = d*2 Then h = d*2 Cover requirements are a bit different Cover requirements are a bit different See ACI 318-05 7.7.1(c) See ACI 318-05 7.7.1(c) You might need to make allowance for a wear surface You might need to make allowance for a wear surface

10. Flexural Steel Consider as a rectangular singly reinforced beam where b = 12” Consider as a rectangular singly reinforced beam where b = 12” M u <  A s f y (d-A s f y /(1.7f’ c b))  Solve for A s M u <  A s f y (d-A s f y /(1.7f’ c b))  Solve for A s The resulting A s is the req’d A s PER FOOT OF WIDTH. The resulting A s is the req’d A s PER FOOT OF WIDTH. Also consider min A s requirement ACI 318-05 10.5.1 Also consider min A s requirement ACI 318-05 10.5.1 All bars can provide this A s by selecting an appropriate spacing All bars can provide this A s by selecting an appropriate spacing Spacing = A b /(req’d A s /ft width) Spacing = A b /(req’d A s /ft width) Watch units!!!! Watch units!!!!

11. Spacing Limits ACI 318-05 7.6.5 has an upper limit on bar spacing ACI 318-05 7.6.5 has an upper limit on bar spacing S < min(3h, 18”) S < min(3h, 18”) The lower limit is as used in previous beam problems.. The lower limit is as used in previous beam problems.. The clear distance between bars > max(1”, max aggregate size/.75) The clear distance between bars > max(1”, max aggregate size/.75)

12. Typical Calculation Controlling Flexural Steel Requirement0.294in^2/ftw BarAbdbmax sUse sAct. AsAct dpMnMu/pMncStl Strain (in^2)(in) (in^2/ftw)(in)(ft-k/ftw) (in) #30.110.3754.494.500.2938.0610.360.9270.5070.04466 #40.200.5008.168.500.2828.009.900.9700.4880.04613 #50.310.62512.6513.000.2867.949.950.9650.4950.04510 #60.440.75017.9518.000.2937.8810.110.9500.5070.04355 #70.600.87524.4818.000.4007.8113.530.7100.6920.03087 #80.791.00032.2318.000.5277.7517.450.5500.9110.02252 #91.001.12840.8018.000.6677.6921.590.4451.1530.01699 #101.271.27051.8218.000.8477.6226.640.3611.4650.01260 #111.561.41063.6518.001.0407.5531.730.3031.7990.00958 #142.251.69391.8118.001.5007.4042.530.2262.5950.00556 #184.002.257163.2118.002.6677.1261.930.1554.6140.00163 Note: Check development lengths

13. Temperature & Shrinkage Steel ACI 318-05 7.12 ACI 318-05 7.12 Req’d A s /ft width = (12”)h  Req’d A s /ft width = (12”)h   = 0.0020 for f y < 60 ksi  = 0.0020 for f y < 60 ksi  = 0.0018 for f y = 60 ksi  = 0.0018 for f y = 60 ksi This steel is placed TRANSVERSE to the flexural steel. This steel is placed TRANSVERSE to the flexural steel. ACI 318-05 7.12.2.2 ACI 318-05 7.12.2.2 Spacing < min(5h,18”) Spacing < min(5h,18”)

14. T&S Calculation

15. Layout Flexural Steel Temperature & Shrinkage Steel

16. Example Problem Materials: f’ c = 3 ksi, f y = 60 ksi Materials: f’ c = 3 ksi, f y = 60 ksi Imposed Loads: Live = 100 psf, Dead = 25 psf Imposed Loads: Live = 100 psf, Dead = 25 psf

17. Finding “h” At this point, we have enough information to determine h using ACI 318-05 Table 9.5a: At this point, we have enough information to determine h using ACI 318-05 Table 9.5a: Cantilevers: h > L/10 = 24”/10 = 2.4” Cantilevers: h > L/10 = 24”/10 = 2.4” Main Spans: h > L/24 = 120”/28 = 4.29” Main Spans: h > L/24 = 120”/28 = 4.29” We still need to check shear and flexure requirements… but need more info! We still need to check shear and flexure requirements… but need more info!

18. Determine Loads Consider only a 1 ft width of beam (b = 12”) Consider only a 1 ft width of beam (b = 12”) w LL = 100 psf = 100 plf/ft width w LL = 100 psf = 100 plf/ft width w DL = 25 psf + weight of slab w DL = 25 psf + weight of slab Make a guess at a slab thickness or write the equations of shear and moment in terms of slab thickness… Let’s try h = 6”… we will need to fix this later if it turns out to be greater. Make a guess at a slab thickness or write the equations of shear and moment in terms of slab thickness… Let’s try h = 6”… we will need to fix this later if it turns out to be greater. w DL = 25 psf + (150 pcf)*.5 ft = 100 psf = 100 plf/ftw w DL = 25 psf + (150 pcf)*.5 ft = 100 psf = 100 plf/ftw w u = 1.2(100 plf/ftw) + 1.6(100 plf/ftw) = 280 plf/ftw w u = 1.2(100 plf/ftw) + 1.6(100 plf/ftw) = 280 plf/ftw

19. An Almost Arbitrary Decision We will place the steel at mid-depth of the slab so that it handles both positive and negative moments We will place the steel at mid-depth of the slab so that it handles both positive and negative moments This means that we only need to design for the worst case moment (positive or negative) along the span. This means that we only need to design for the worst case moment (positive or negative) along the span. As a result, d = h/2 As a result, d = h/2 This is a good choice for a short relatively thin (less than 8”) slab. This is a good choice for a short relatively thin (less than 8”) slab. This makes things pretty simple. Only have to design one set of flexural steel! This makes things pretty simple. Only have to design one set of flexural steel!

20. Determine Maximum Shears Use ACI 318-05 8.3 (the slab meets the criteria!) to compute internal forces (or you can do a full elastic analysis) Use ACI 318-05 8.3 (the slab meets the criteria!) to compute internal forces (or you can do a full elastic analysis) The cantilevers are exempt from 8.3 since they are statically determinant (i.e. don’t meet the criteria to use 8.3) The cantilevers are exempt from 8.3 since they are statically determinant (i.e. don’t meet the criteria to use 8.3) V u = w u *L n = (280 plf/ftw)*(1.5 ft) = 420 lb/ftw V u = w u *L n = (280 plf/ftw)*(1.5 ft) = 420 lb/ftw The two center spans are the same The two center spans are the same V u = w u *L n /2= (280 plf/ftw)*(9 ft)/2 = 1260 lb/ftw V u = w u *L n /2= (280 plf/ftw)*(9 ft)/2 = 1260 lb/ftw

21. Determine Req’d h Based on Shear For our choice: For our choice: d = h/2 > V u /[  2sqrt(f’c)bw)] d = h/2 > V u /[  2sqrt(f’c)bw)] d > (1260 lb/ftw)/[.75(2)sqrt(3000)(12”)] d > (1260 lb/ftw)/[.75(2)sqrt(3000)(12”)] d > 1.28 in d > 1.28 in h > 2.56 in h > 2.56 in Deflection criteria (Table 9.5a) still controls!!! Deflection criteria (Table 9.5a) still controls!!!

22. Determine Maximum Moments Main spans: L n = 9 ft Main spans: L n = 9 ft Can use ACI 318-05 8.3: Can use ACI 318-05 8.3: Max positive M u = w u *L n 2 /16 = 1,418 ft-lb/ftw Max positive M u = w u *L n 2 /16 = 1,418 ft-lb/ftw Max negative M u = w u *L n 2 /11 = 2,062 ft-lb/ftw Max negative M u = w u *L n 2 /11 = 2,062 ft-lb/ftw Cantilevers are statically determinate: L n = 1.5 ft. Cantilevers are statically determinate: L n = 1.5 ft. M u = w u *L n 2 /2 = 315 ft-lb/ftw M u = w u *L n 2 /2 = 315 ft-lb/ftw Design for M u = 2,062 ft-lb/ftw Design for M u = 2,062 ft-lb/ftw

23. Select “h” Based on Flexure Can use the equations derived for choosing the size of rectangular singly reinforced beams earlier in the semester. Can use the equations derived for choosing the size of rectangular singly reinforced beams earlier in the semester. Use b = 12” and solve for d. Use b = 12” and solve for d. Try solving the equations for both max and min size to bracket the possibilities. Try solving the equations for both max and min size to bracket the possibilities. Max size (based on min reinforcing): h = 6.41 in Max size (based on min reinforcing): h = 6.41 in Min size (based on stl strain = 0.005): h = 3.40 in Min size (based on stl strain = 0.005): h = 3.40 in

24. Now… Make a Choice! I choose to use h = 5”… it is in the range for flexure and meets Table 9.5a deflection criteria and Shear Strength criteria I choose to use h = 5”… it is in the range for flexure and meets Table 9.5a deflection criteria and Shear Strength criteria Other choices that meet the limits computed are also valid Other choices that meet the limits computed are also valid No real need to go back and fix the “h” that our load estimate since they are close and the assumption was conservative, but can do it to refine the design if we want to. No real need to go back and fix the “h” that our load estimate since they are close and the assumption was conservative, but can do it to refine the design if we want to.

25. Determine the Flexural Steel Solve the flexural design inequality for A s : Solve the flexural design inequality for A s : M u <  A s f y (d-A s f y /(1.7*f’ c b)) M u <  A s f y (d-A s f y /(1.7*f’ c b))  A s > 0.199 in 2 /ftw  A s > 0.199 in 2 /ftw Watch those units!!! Watch those units!!! Also check to make sure that the minimum A s is met Also check to make sure that the minimum A s is met A s > min(200,3sqrt(f’ c ))*b w d/f y = 0.100 in 2 /ftw A s > min(200,3sqrt(f’ c ))*b w d/f y = 0.100 in 2 /ftw The larger value controls The larger value controls Use A s > 0.199 in 2 /ftw Use A s > 0.199 in 2 /ftw

26. Select the Flexural Steel Use #4 @ 12” O.C. Use #4 @ 12” O.C.

27. Consider T&S Steel For our case,  = 0.0018 For our case,  = 0.0018 Req’d A s > 0.0018(12”)(5”) = 0.108 in 2 /ftw Req’d A s > 0.0018(12”)(5”) = 0.108 in 2 /ftw Max allowed spacing = min(18”,5h) = 18” Max allowed spacing = min(18”,5h) = 18” Compute some spacing and choose a bar: Compute some spacing and choose a bar: For #3 bar: For #3 bar: s < 0.11 in 2 / (0.108 in 2 /ftw) = 1.02 ft = 12.2 in s < 0.11 in 2 / (0.108 in 2 /ftw) = 1.02 ft = 12.2 in For #4 bar: s < 22.2 in … use 18” For #4 bar: s < 22.2 in … use 18” #3 is the better choice! #3 is the better choice! Use #3 @ 12” O.C. for T&S steel Use #3 @ 12” O.C. for T&S steel

28. Final Design Slab Thickness = 5” Slab Thickness = 5” Longitudinal Steel = #4 @ 12” O.C. @ mid-depth Longitudinal Steel = #4 @ 12” O.C. @ mid-depth Transverse Steel = #3 @ 12” O.C. Transverse Steel = #3 @ 12” O.C.