ARCH 631 Note Set 11 S017abn Case Study in Reinforced Concrete adapted from Simplified Design of Concrete Structures, James Ambrose, 7 th ed. Building description The building is a three-story office building intended for speculative rental. Figure 17.37 presents a full-building section and a plan of the upper floor. The exterior walls are permanent. The design is a rigid perimeter frame to resist lateral loads. Loads (UBC 1994) Live Loads: Roof: 0 lb/ Floors: Office areas: 50 lb/ (.39 kpa) Corridor and lobby: 100 lb/ (4.79 kpa) Partitions: 0 lb/ (0.96 kpa) Wind: map speed of 80 mph (190 km/h); exposure B Assumed Construction Loads: Floor finish: 5 lb/ (0.4 kpa) Ceilings, lights, ducts: 15 lb/ (0.7 kpa) Walls (average surface weight): Interior, permanent: 10 lb/ (0.48 kpa) Exterior curtain wall: 15 lb/ (0.7 kpa) Materials Use f c =3000 psi (0.7 MPa) and grade 60 reinforcement (fy = 60 ksi or 414 MPa). 49
ARCH 631 Note Set 11 S017abn Structural Elements/Plan Case 1 is shown in Figure 17.44 (framing plan) and consists of a flat plate supported on interior beams, which in turn, are supported on girders supported by columns. We will examine the slab, and a four-span interior beam. Case will consider the bays with flat slabs, no interior beams with drop panels at the columns and an exterior rigid frame with spandrel (edge) beams. An example of an edge bay is shown to the right. We will examine the slab and the drop panels. For both cases, we will examine the exterior frames in the 3-bay direction. Case 1: Slab: The slabs are effectively 10 x 30, with an aspect ratio of 3, making them one-way slabs. Minimum depths (by ACI) are a function of the span. Using Table 7.3.1.1 for one way slabs the minimum is and there is a 5 inches minimum for fire rating. l n 8 (Found in Note Set 10.1) We ll presume the interior beams are 1 wide, so ln = 10 - ½ - ½ = 9 minimum t (or h) in/ 9 1 3.86in 8 Use 5 in. based on fire rating. dead load from slab (self weight) = t 3 lb/ 150 5 in/ 1 in = 6.5 lb/ 50
ARCH 631 Note Set 11 S017abn total dead load = (5 + 15 + 6.5) lb/ + of lightweight concrete topping with weight of 18 lb/ (0.68 KPa) (presuming interior wall weight is over beams & girders) dead load = 100.5 lb/ live load (critical case in corridor) = 100 lb/ total factored distributed load (ASCE-7) of 1.D+1.6L: wu = 1.(100.5 lb/ ) + 1.6(100 lb/ ) = 80.6 lb/ Maximum Positive Moments from Figure -3 (Note Set 8.1), end span (integral with support) for a 1 wide strip: Mu (positive) = w u 14 lb / n n wu 1 14 (80.6 )(1 )(9 14 ) 1k 1000lb = 1.6 k- Maximum Negative Moments from Figure -5 (Note Set 8.1), end span (integral with support) for a 1 wide strip: Mu-(negative) = w u 1 n w u 1 1 n (80.6 lb / )(1 )(9 1 ) 1k = 1.89 k- 1000lb 51
(tensile strain of 0.004) In Note Set10.1 ARCH 631 Note Set 11 S017abn The design aid (Figure 3.8.1 of Note Set 10.1) can be used to find the reinforcement ratio,, knowing Rn = Mn/bd with Mn = Mu/f, where f = 0.9 (because 0.005 = 0.0135). We can presume the effective depth to the centroid of the reinforcement, d, is 1.5 less than the slab thickness (with ¾ cover and ½ of a bar diameter for a #4 (1 ) bar) = 4. Rn = M bd 1.89 (0.9)(1 )(3.75 ) k u in/ lb/ k 1 1000 149.3 psi in in so for f c = 3000 psi and fy = 60,000 psi is the minimum. For slabs, As minimum is 0.0018bt for grade 60 steel. (See Note Set 10.1) As = 0.0018bt = 0.0018(1in)(5in) = 0.108 in / (Full table found in Note Set 10.1) 5
ARCH 631 Note Set 11 S017abn Pick bars and spacing off Table 3-7. Use #3 bars @ 1 in (As = 0.11 in ). Common chart of spaced bars based on bar areas. Check the moment capacity. d is actually 5 in 0.75in (cover) ½ (3/8in bar diameter) = 4.06 in a = Asfy/0.85f cb = 0.11 in (60 ksi)/[0.85(3 ksi)1 in] = 0. in Mn = Asfy(d-a/) = 0.in 1 0.9(0.11 in )(60 ksi)(4.06 in ) ( ) 1.96 k- > 1.89 k- needed 1 in (OK) Provide steel in the perpendicular direction as well, and make certain not to exceed the minimum spacings of 5t or 18" for shrinkage and crack control steel or 3t or 18" for flexure reinforcement (in found in Note Set 10.1). Maximum Shear: Figure -7 (Note Set 8.1) shows end shear that is wuln/ except at the end span on the interior column which sees a little more and you must design for 15% increase: lb/ 1.15(80.6 )( 1 )(9 ) Vu-max = 1.15wuln/ = = 145 lb (for a 1 strip) lb/ (80.6 )( 1 )(4.06 in) Vu at d away from the support = Vu-max w(d) = 145 lb in/ 1 = 1357 lb 53
ARCH 631 Note Set 11 S017abn Check the one way shear capacity: vvc = v f c bd (v = 0.75, =1.0 for normal weight concrete): in in vvc = 0.75()(1.0) 3000 psi (1 )(4.06 ) = 4003 lb Is Vu (needed) < vvc (capacity)? thicker. YES: 1357 lb 4003 lb, so we don't need to make the slab Interior Beam (effectively a T-beam): Tributary width = 10 for an interior beam. dead load = area load x tributary width = (100.5 lb/ )(10) = 1005 lb/ (14.7 kn/m) Reduction of live load is allowed (Note Set 3.4), with a live load element factor, KLL, of for an interior beam for its tributary width assuming the girder is 1 wide. The live load is 100 lb/ (corridors and lobbies): L = 15 lb 15 L (0.5 ) = 100 o (0.5 ) = 87.3 lb/ KLL A (30 1 )(10 ) T (Reduction Multiplier = 0.873 > 0.5) live load = 87.3 lb/ (10) = 873 lb/ (1.7 kn/m) Estimating a 1 wide x 4 deep beam means the additional dead load from self weight can be included. The top 5 inches of slab has already been included in the dead load (1005 lb/): dead load from self weight = lb 3 150 (1 in wide)(4 5 in deep) t tributary width = 1 1in = 37.5 lb/ (3.46 kn/m) wu = 1.(1005 lb/+37.5 lb/) + 1.6(873 lb/) = 888 lb/ (4.30 kn/m) 54
ARCH 631 Note Set 11 S017abn The effective width, be, of the T part is the smaller of Set 10.1): n 4, b w 16t, or center-center spacing (Note be = minimum{9 /4 = 7.5 = 87 in, 1 in + 16x5 in = 9 in, 10 = 10 in} = 87 in The clear span for the beam is ln = 30 - ½ - ½ = 9 Maximum Positive Moments from Figure -3 (Note Set 8.3), end span (integral with support): Mu (positive) = lb/ wu n 888 (9 ) 1k = 173.5 k- 14 14 1000lb Maximum Negative Moments from Figure -4 (Note Set 8.1), end span (integral with support): Mu-(negative) = w u 10 n 888 lb / (9 10 ) 1k = 4.9 k- 1000lb Figure 3.8.1 (5) can be used to find an approximate for top reinforcement if Rn = Mn/bd and we set Mn = Mu/f. We can presume the effective depth is.5 less than the 4 depth (for 1.5 cover and ½ bar diameter for a #10 (10/8) bar + #3 stirrups (3/8 more)), so d = 1.5. Rn = M bd f lb/ k k u 1000 4.9 in in 0.9 (1 )(1.5 ) 1 in/ = 584 psi so for f c = 3000 psi and fy = 60,000 psi is about 0.011 (and less than max-0.005 = 0.0135) Then we pick bars and spacing off Table 3-7 (pg. 53) to fit in the effective flange width in the slab. 55
ARCH 631 Note Set 11 S017abn For bottom reinforcement (positive moment) the effective flange is so wide at 87 in, that it resists a lot of compression, and needs very little steel to stay under-reinforced (a is between 0.6 and 0.5 ). We d put in bottom bars at the minimum reinforcement allowed and for tying the stirrups to. Maximum Shear: Vmax = wul/ normally, but the end span sees a little more and you must design for 15% increase. But for beams, we can use the lower value of V that is a distance of d from the face of the support (see Figure.7 on page 53): Vu-design = 1.15wuln/ wud= lb/ lb/ in 1.15(888 )(9 ) 888 (1.5 ) 1 in/ 4,983 lb = 43.0 k Check the one way shear capacity = vvc = v f c bd, where v 0.75 and = 1.0 in in vvc = 0.75()(1.0) 3000 psi (1 )(1.5 ) = 1,197 lb = 1. k Is Vu (needed) < vvc (capacity)? NO: 43.0 k is greater than 1. k, so stirrups are needed vvs = Vu -vvc = 43.0 k 1. k = 1.8 k (max needed) Using #3 bars (typical) with two legs means Av = (0.11in ) = 0. in. (Found in Note Set 10.1) To determine required spacing, use Table 3-8 (pg. 57). For d = 1.5 and vvs v4 v4 c f c bd (where f bd= = (1. k) = 4.4 k when λ=1.0), the maximum spacing is d/ = 10.75 in. or 4. V c srequired = Av f ytd Av f ytd V V V u c s = in ksi in 0.750. 60 1.5 k 1.8 = 9.75 in, so use 9 in. We would try to increase the spacing as the shear decreases, but it is a tedious job. We need stirrups anywhere that Vu > vvc/. One recommended intermediate spacing is d/3. 56
ARCH 631 Note Set 11 S017abn greater of and smaller of and (ACI 9.7.6.) (ACI 0...4) must also be considered (ACI 9.6.3.3) The required spacing where stirrups are needed for crack control (vvc Vu > ½vVc ) is srequired = smaller of Af v yt 0.in (60,000psi) 50b = in and Af v yt 0.in (60,000psi) 50(1 in) 0.75 fb 0.75 3000psi(1 in) w maximum spacing is d/ = 10.75 in. or 4. Use 10 in. c w = 6.8 in, and the A recommended minimum spacing for the first stirrup is in. from the face of the support. A distance of one half the spacing near the support is oen used. Vu 36.7 k 41.9 k to find distances when w is known: x = (peak crossing value)/w ex: 41.9 k 1. k)/.888 k/ = 7. 1. k 10.6 k 7. = 86 in 10.84 = 130 in C L 14.5 in 13.0 = 156 in 9.35 = 11 in 10.6 k 1. k x 14 @ 10 in 5 @ 10 in 13 @ 9 in 43.0 k 48. k END SPAN STIRRUP LAYOUT 57
spandrel girder ARCH 631 Note Set 11 S017abn Spandrel Girders: Because there is a concentrated load on the girder, the approximate analysis can t technically be used. If we converted the maximum moment (at midspan) to an equivalent distributed load by setting it equal to wul /8 we would then use: Maximum Positive Moments from Figure -3 (pg. 51), end span (integral with support): Mu+ = w u 14 n Maximum Negative Moments from Figure -4 (pg. 55), end span (column support): Mu- = w u 10 n (with w u 16 n at end) Column: An exterior or corner column will see axial load and bending moment. We d use interaction charts for Pu and Mu for standard sizes to determine the required area of steel. An interior column sees very little bending. The axial loads come from gravity. The factored load combination is 1.D+1.6L + 0.5Lr (Note Set 3.). The girder weight, presuming 1 x 4 girder at 150 lb/ 3 3 is 150 (1 )(4 ) = 600 lb/ lb Top story: presuming 0 lb/ roof live load, the total load for an interior column (tributary area of 30 x30 ) is: Lower stories: DLroof*: 1. x 100.5 lb/ x 30 x 30 * assuming the same dead load and materials as the floors = 108.5 k DLbeam 1. x 37.5 lb/ x 30 x 3 beams = 5.6 k DLgirder 1. x 600 lb/ x 30 = 1.6 k LLr: 0.5 x 0 lb/ x 30 x 30 = 9.0 k Total = 164.7 k DLfloor: 1. x 100.5 lb/ x 30 x 30 DLbeam 1. x 37.5 lb/ x 30 x 3 beams DLgirder 1. x 600 lb/ x 30 LLfloor: 1.6 x (0.873) x100 lb/ x 30 x 30 Total = 108.5 k = 5.6 k = 1.6 k = 15.7 k (includes reduction) = 81.4 k 58
ARCH 631 Note Set 11 S017abn Summary: nd floor column sees Pu = 164.7+81.4 = 446.1 k 1 st floor column sees Pu = 446.1+81.4 = 77.5 k Look at the example interaction diagram for an 18 x 18 column (Figure 5-0 ACI 318-0) using f c = 4000 psi and fy = 60,000 psi for the first floor having Pu = 77.5 k, and Mu to the column being approximately 10% of the beam negative moment = 0.1*4.9 k- = 4.3 k-: (See maximum negative moment calculation for an interior beam.) The chart indicates the capacity for the reinforcement amounts shown by the solid lines. For Pu = 77.5 k and Mu = 4.3 k-, the point plots below the line marked 4-#11 (1.93% area of steel to an 18 in x 18 in area). 59
ARCH 631 Note Set 11 S017abn Lateral Force Design: The wind loads from the wind speed, elevation, and exposure we ll accept as shown in Figure 17.4 given on the le in psf. The wind is acting on the long side of the building (see pg. 49). The perimeter frame resists the lateral loads, so there are two with a tributary width of ½ [(30)x(4 bays) + for beam widths and cladding] = 1/ = 61 The factored combinations with dead and wind load are: 1.D +1.6Lr + 0.5W 1.D + 1.0W + L + 0.5Lr The tributary height for each floor is half the distance to the next floor (top and bottom): Exterior frame (bent) loads: lb/ H1 = 195 (61 ) =11,895 lb = 11.9 k/bent H = lb/ 34 (61 ) =14.3 k/bent lb/ k H3 = 1000 lb/ 7 (61 ) =13.8 k/bent lb/ k 1000 Using Multiframe, the axial force, shear and bending moment diagrams can be determined using the load combinations, and the largest moments, shear and axial forces for each member determined. 60
ARCH 631 Note Set 11 S017abn M = 03.7 k- V = 6.6 k P = 54.7 k M =190.8 k- V = 8.0 k P = 11.6 k M = 18.8 k- V = 45.4 k P = 6.6 k M = 33.4 - V = 60.5 k P = 1.9 k M = 338.5 k- V = 60.9 k P = 6.5 k M = 39.3 k- V = 4.7 k P = 91.0 k M = 70. k- V = 9. k P = 15.5 k M = 37.8 k- V = 45.5 k P = 8.9 k M = 330.9 k- V = 60.0 k P = 6.4 k M = 349.5 k- V = 61.0 k P = 4.1 k M = 3.4 k- V = 3.7 k P =91.0 k M = 51.1 k- V = 7.7 k P = 15.7 k M = 36.3 k- V = 46.6 k P = 30.8 k M = 317.9 k- V = 59.7 k P = 10.4 k M = 347.9 k- V = 6.0 k P = 1.6 k M = 183.9 k- V = 4.8 k P = 53.5 k M = 14.7 k- V = 1. k P = 10.3 k M = 165.1 k- V = 1.6 k P = 19.8 k M = 104.4 k- V = 11.6 k P = 340.6 k M = 95.5 k- V = 10.3 k P = 341.7 k M =10.6 k- V = 8.5 k P = 185.7 k (This is the summary diagram of force, shear and moment magnitudes refer to the maximum values in the column or beams, with the maximum moment in the beams being negative over the supports, and the maximum moment in the columns being at an end.) Axial force diagram: Shear diagram: Bending moment diagram: Displacement: 61
ARCH 631 Note Set 11 S017abn Beam-Column loads for design: The bottom exterior columns see the largest bending moment on the lee-ward side (le): Pu = 19.8 k and Mu = 165.1 k- (with large axial load) The interior columns see the largest axial forces: Pu = 341.7 k and Mu = 95.5 k- and Pu = 340.6 k and Mu = 104.4 k- Refer to an interaction diagram for column reinforcement and sizing. Case Slab: The slabs are effectively 30 x 30, making them two-way slabs. Minimum thicknesses (by ACI) are a function of the span. Using Table 4-1 (Note Set 8.1) for two way slabs, the minimum is the larger of ln/36 or 4 inches. Presuming the columns are 18 wide, ln = 30 ½(18 in)/(1 /in) ½(18 in)/(1 /in) = 8.5, h = ln/36 = (8.5x1)/36 = 9.5 in The table also says the drop panel needs to be l/3 long = 8.5 /3 = 9.5, and that the minimum depth must be 1.5h = 1.5(9.5 in) = 1 in. 6
ARCH 631 Note Set 11 S017abn For the strips, l = 30, so the interior column strip will be 30 /4+30/4 = 15, and the middle strip will be the remaining 15. lb/ in 150 9.5 dead load from slab = t = 118.75 lb/ in/ 1 3 total dead load = 5 + 15 + 118.75 lb/ + of lightweight concrete topping @ 18 lb/ (0.68 KPa) (presuming interior wall weight is over beams & girders) total dead load = = 5 + 15 + 187.5 + 18 = 156.75 lb/ live load with reduction (Note Set 3.4), where live load element factor, KLL, is 1 for a two way slab: 15 L = (0.5 ) lb 15 Lo = 100 (0.5 ) = 75 lb/ K A 1(30 )(30 ) LL T total factored distributed load: wu = 1.(156.75 lb/ ) + 1.6(75 lb/ ) = 308.1 lb/ total panel moment to distribute: Mo = lb/ wul ln 308.1 (30 )(8.5 ) 1k = 938.4 k- 8 8 1000lb Column strip, end span: Maximum Positive Moments from Table 4-3 (Note Set 8.1), (flat slab with spandrel beams): Mu+ = 0.30Mo = 0.30(938.4 k-) = 81.5 k- Maximum Negative Moments from Table 4-3 (Note Set 8.1), (flat slab with spandrel beams): Mu- = 0.53Mo = 0.53(938.4 k-) = 497.4 k- 63
ARCH 631 Note Set 11 S017abn Middle strip, end span: Maximum Positive Moments from Table 4-3 (Note Set 8.1), (flat slab with spandrel beams): Mu+ = 0.0Mo = 0.0(938.4 k-) = 187.9 k- Maximum Negative Moments from Table 4-3 (Note Set 8.1), (flat slab with spandrel beams): Mu- = 0.17Mo = 0.17(938.4 k-) = 159.5 k- Design as for the slab in Case 1, but provide steel in both directions distributing the reinforcing needed by strips. 64
ARCH 631 Note Set 11 S017abn Shear around columns: The shear is critical at a distance d/ away from the column face (Note Set 10.1). If the drop panel depth is 1 inches, the minimum d with two layers of 1 diameter bars would be 1 ¾ (cover) (1 ) ½(1 ) =about 9.75 in (to the top steel). (Note Set column shear perimeter tributary area for column drop panel The shear resistance is vvc = v4 f c bod, 0.75 v and =1.0 where bo, is the perimeter length. The design shear value is the distributed load over the tributary area outside the shear perimeter, Vu = wu (tributary area - b1 x b) where b s are the column width plus d/ each side. b1 = b = 18 + 9.75 / + 9.75 / = 7.75 in b1 x b = 1 (7.75 in) ( ) = 5.35 1in Vu = lb/ 1k (84.7 )(30 30 4.97 ) = 54.7 k 1000 lb Shear capacity: bo = (b1) + (b) = 4(7.75 in) = 111 in in in vvc = 0.75 4(1.0) 3000 psi 111 9.75 = 177,83 lb= 177.8 ksi < Vu! The shear capacity is not large enough. The options are to provide shear heads or a deeper drop panel, or change concrete strength, or even a different system selection... There also is some transfer by the moment across the column into shear. 65
ARCH 631 Note Set 11 S017abn Deflections: Elastic calculations for deflections require that the steel be turned into an equivalent concrete material using n = E E For normal weight concrete (150 lb/ 3 ): s c. Ec can be measured or calculated with respect to concrete strength. Ec = 57,000 3000 psi = 3,1,019 psi = 31 ksi E c 57, 000 f (Note Set 10.1) c n = 9,000 psi/31 ksi = 9.3 Deflection limits are given in Table 4.. 66