Case Study in Reinforced Concrete adapted from Simplified Design of Concrete Structures, James Ambrose, 7 th ed.

Transcription

1 ARCH 631 Note Set 11 F015abn Case Study in Reinfored Conrete adapted from Simplified Design of Conrete Strutures, James Ambrose, 7 th ed. Building desription The building is a three-story offie building intended for speulative rental. Figure presents a full-building setion 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: Offie 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 Constrution Loads: Floor finish: 5 lb/ (0.4 kpa Ceilings, lights, duts: 15 lb/ (0.7 kpa Walls (average surfae weight: Interior, permanent: 10 lb/ (0.48 kpa Exterior urtain wall: 15 lb/ (0.7 kpa Materials Use f 3000 psi (0.7 MPa and grade 60 reinforement (f y 60 ksi or 414 MPa. Strutural Elements/Plan Case 1 is shown in Figure and onsists of a flat plate supported on interior beams, whih in turn, are supported on girders supported by olumns. We will examine the slab, and a four-span interior beam. Case will onsider the bays with flat slabs, no interior beams with drop panels at the olumns 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 ases, we will examine the exterior frames in the 3-bay diretion. 33

2 ARCH 631 Note Set 11 F015abn Case 1: Slab: The slabs are effetively 10 x 30, with an aspet ratio of 3, making them one-way slabs. Minimum depths (by ACI are a funtion of the span. Using Table for one way slabs the minimum is l n 8 with 5 inhes minimum for fire rating. We ll presume the interior beams are 1 wide, so l n minimum t (or h in / in 8 Use 5 in. (fire rating governs dead load from slab lb / in lb/ in / 1 total dead load ( lb/ + of lightweight onrete topping with weight of 18 lb/ (0.68 KPa (presuming interior wall weight is over beams & girders dead load lb/ live load (worst ase in orridor 100 lb/ total fatored 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, end span (integral with support for a 1 wide strip: M u (positive w u l 14 lb / n n wu 1 l 14 (80.6 (1 (9 14 1k 1000lb 1.6 k- 34

3 ARCH 631 Note Set 11 F015abn Maximum Negative Moments from Figure -5, end span (integral with support for a 1 wide strip: M u- (negative w u l 1 lb / n u 1 l n (80.6 w 1 (1 (9 1 1k 1000lb 1.89 k- The design aid (Figure an be used to find the reinforement ratio, ρ, knowing R n M n /bd with M n M u /φ f, where φ f 0.9 (beause ρ We an presume the effetive depth to the entroid of the reinforement, d, is 1.5 less than the slab thikness (with ¾ over and ½ of a bar diameter for a #8 (1 bar k in / k R n lb / psi in in ( 0. 9 ( 1 ( so ρ for f 3000 psi and f y 60,000 psi is the minimum. For slabs, A s minimum is bt for grade 60 steel. A s (1in(5in in / 35

4 ARCH 631 Note Set 11 F015abn Pik bars and spaing off Table 3-7. Use #3 1 in (A s 0.11 in. Chek the moment apaity. d is atually 5 in 0.75in (over ½ (3/8in bar diameter 4.06 in (OK a A s f y /0.85f b 0.11 in (60 ksi/[0.85(3 ksi1 in] 0. in 0. in 1 φm n φa s f y (d-a/ 0. 9( 0. 11in ( 60ksi ( 4. 06in ( 1.96 k- > 1.89 k- needed 1 in Maximum Shear: Figure -7 shows end shear that is w u l n / exept at the end span on the interior olumn whih sees a little more and you must design for 15% inrease: V u-max 1.15w u l n / 1. 15( lb/ ( 1 ( 9 V u at d away from the support V u-max w(d 145 lb 145 lb (for a 1 strip ( lb/ 1 ( 1 in / ( 4. 06in 1357 lb Chek the one way shear apaity: φ v V φ v λ f bd (φ v 0.75, λ1.0 for normal weight onrete: in in φ v V 0.75(( psi (1 ( lb Is V u (needed < φ v V (apaity? thiker. YES: 1357 lb 4003 lb, so we don't need to make the slab Interior Beam (effetively a T-beam: Tributary width 10 for an interior beam. dead load (100.5 lb/ ( lb/ (14.7 kn/m 36

5 ARCH 631 Note Set 11 F015abn Redution of live load is allowed, with a live load element fator, K LL, of for an interior beam for its tributary width assuming the girder is 1 wide. The live load is 100 lb/ : 15 lb 15 L L o ( ( lb/ ( 30 1 ( 10 K LL A T live load 87.3 lb/ ( lb/ (1.7 kn/m (Redution Multiplier > 0.5 Estimating a 1 wide x 4 deep beam means the additional dead load from self weight ( w γ A in units of load/length an be inluded. The top 5 inhes of slab has already been inluded in the dead load: dead load from self weight 150 lb 3 1 (1in wide(4 5in deep 1in w u 1.(1005 lb/+37.5 lb/ + 1.6(873 lb/ 888 lb/ (4.30 kn/m 37.5 lb/ (3.46 kn/m l n The effetive width, b E, of the T part is the smaller of 4, b w + 16t, or enter-enter spaing b E minimum{9 / in, 1 in+16x5 in 9 in, in} 87 in The lear span for the beam is l n Maximum Positive Moments from Figure -3, end span (integral with support: M u (positive w u l 14 n 888 lb / (9 14 1k 1000lb k- Maximum Negative Moments from Figure -4, end span (integral with support: M u- (negative w u l 10 n 888 lb / (9 10 1k 1000lb 4.9 k- 37

6 ARCH 631 Note Set 11 F015abn Figure an be used to find an approximate ρ for top reinforement if R n M n /bd and we set M n M u /φ f. We an presume the effetive depth is.5 less than the 4 depth (for 1.5 over and ½ bar diameter for a #10 (10/8 bar + #3 stirrups (3/8 more, so d 1.5. R n 1000 lb / k 0.9 (1 in 4.9 (1.5 k in 1 in / 584 psi so ρ for f 3000 psi and f y 60,000 psi is about (and less than ρ max Then we pik bars and spaing off Table 3-7 to fit in the effetive flange width in the slab. For bottom reinforement (positive moment the effetive flange is so wide at 87 in, that it resists a lot of ompression, and needs very little steel to stay under-reinfored (a is between 0.6 and 0.5. We d put in bottom bars at the minimum reinforement allowed and for tying the stirrups to. Maximum Shear: V max w u l/ normally, but the end span sees a little more and you must design for 15% inrease. But for beams, we an use the lower value of V that is a distane of d from the fae of the support V u-design 1.15w u l n / w u d lb/ lb/ in 1.15(888 (9 888 (1.5 1 in / 4,983 lb 43.0 k Chek the one way shear apaity φ v V φ v λ f bd, where φ and λ 1.0 v in in φ v V 0.75(( psi (1 (1.5 1,197 lb 1. k Is V u (needed < φ v V (apaity? NO: 43.0 k is greater than 1. k, so stirrups are needed φ v V s V u -φ v V 43.0 k 1. k 1.8 k (max needed Using #3 bars (typial with two legs means A v (0.11in 0. in. To determine required spaing, use Table 3-8. For d 1.5 and φ v V s φ v 4λ φ v 4λ f bd (where f bd (1. k 4.4 k, the maximum spaing is d/ in. or 4. φv s required φ A f d φ A f d v yt v yt V φv φv u s in k 1. 8 ksi in 9.75 in, so use 9 in. We would try to inrease the spaing as the shear dereases, but it is a tedious job. We need stirrups anywhere that V u > φ v V /. One reommended intermediate spaing is d/3. 38

7 ARCH 631 Note Set 11 F015abn greater of and smaller of and NOTE: setion numbers are pre ACI The required spaing where stirrups are needed for rak ontrol (φ v V V u > ½φ v V is A f v yt 0.in (60,000psi s required in and the maximum spaing is d/ in. or 4. Use 10 in. 50b 50(1 in w A reommended minimum spaing for the first stirrup is in. from the fae of the support. A distane of one half the spaing near the support is oen used. Vu 36.7 k 41.9 k to find distanes when w is known: x (peak rossing value/w ex: 41.9 k 1. k/.888 k/ k 10.6 k in in C L 14.5 in in in 10.6 k 1. k x 10 in 10 in 9 in 43.0 k 48. k END SPAN STIRRUP LAYOUT Spandrel Girders: Beause there is a onentrated load on the girder, the approximate analysis an t tehnially be used. If we onverted the maximum moment (at midspan to an equivalent distributed load by setting it equal to w u l /8 we would then use: spandrel girder 39

8 ARCH 631 Note Set 11 F015abn Maximum Positive Moments from Figure -3, end span (integral with support: M u+ wl u 14 n Maximum Negative Moments from Figure -4, end span (olumn support: M u- wl u 10 n (with wl u 16 n at end Column: An exterior or orner olumn will see axial load and bending moment. We d use interation harts for P u and M u for standard sizes to determine the required area of steel. An interior olumn sees very little bending. The axial loads ome from gravity. The fatored load ombination is 1.D+1.6L + 0.5L r. The girder weight, presuming 1 x 4 girder at 150 lb/ lb/ Top story: presuming 0 lb/ roof live load, the total load for an interior olumn (tributary area of 30 x30 is: Lower stories: DL roof* : 1. x lb/ x 30 x 30 * assuming the same dead load and materials as the floors k DL beam 1. x 37.5 lb/ x 30 x 3 beams 5.6 k DL girder 1. x 600 lb/ x k LL r : 0.5 x 0 lb/ x 30 x k Total k DL floor : 1. x lb/ x 30 x 30 DL beam 1. x 37.5 lb/ x 30 x 3 beams DL girder 1. x 600 lb/ x 30 LL floor : 1.6 x (0.873x100 lb/ x 30 x 30 Total k 5.6 k 1.6 k 15.7 k (inludes redution 81.4 k nd floor olumn sees P u k 1 st floor olumn sees P u k Look at the example interation diagram for an 18 x 18 olumn (Figure 5-0 ACI using f 4000 psi and f y 60,000 psi for the first floor having P u 77.5 k, and M u to the olumn being approximately 10% of the beam negative moment 0.1*4.9 k- 4.3 k-: (See maximum negative moment alulation for an interior beam. The hart indiates the apaity for the reinforement amounts shown by the solid lines. For P u 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. 40

9 ARCH 631 Note Set 11 F015abn Lateral Fore Design: The wind loads from the wind speed, elevation, and exposure we ll aept as shown in Figure 17.4 given on the le in psf. The wind is ating on the long side of the building. The perimeter frame resists the lateral loads, so there are two with a tributary width of ½ [(30x(4 bays + for beam widths and ladding] 1/ 61 The fatored ombinations with dead and wind load are: 1.D +1.6L r + 0.5W 1.D + 1.0W + L + 0.5L r The tributary height for eah floor is half the distane to the next floor (top and bottom: 41

10 ARCH 631 Note Set 11 F015abn Exterior frame (bent loads: H lb / (61 11,895 lb 11.9 k/bent H 34 lb / 1000 (61 lb / k 14.3 k/bent H 3 7 lb / 1000 (61 lb / k 13.8 k/bent Using Multiframe, the axial fore, shear and bending moment diagrams an be determined using the load ombinations, and the largest moments, shear and axial fores for eah member determined. M 03.7 k- V 6.6 k P 54.7 k M k- V 8.0 k P 11.6 k M 18.8 k- V 45.4 k P 6.6 k M V 60.5 k P 1.9 k M 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 k- V 60.0 k P 6.4 k M 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 k- V 59.7 k P 10.4 k M k- V 6.0 k P 1.6 k M k- V 4.8 k P 53.5 k M 14.7 k- V 1. k P 10.3 k M k- V 1.6 k P 19.8 k M k- V 11.6 k P k M 95.5 k- V 10.3 k P k M 10.6 k- V 8.5 k P k (This is the summary diagram of fore, shear and moment magnitudes refer to the maximum values in the olumn or beams, with the maximum moment in the beams being negative over the supports, and the maximum moment in the olumns being at an end. Axial fore diagram: Shear diagram: y x Bending moment diagram: y x Displaement: y x y x 4

11 ARCH 631 Note Set 11 F015abn Beam-Column loads for design: The bottom exterior olumns see the largest bending moment on the lee-ward side (le: P u 19.8 k and M u k- (with large axial load The interior olumns see the largest axial fores: P u k and M u 95.5 k- and P u k and M u k- Refer to an interation diagram for olumn reinforement and sizing. Case Slab: The slabs are effetively 30 x 30, making them two-way slabs. Minimum thiknesses (by ACI are a funtion of the span. Using Table 4-1 for two way slabs, the minimum is the larger of l n /36 or 4 inhes. Presuming the olumns are 18 wide, l n 30 (18 in/(1 /in 8.5, h l n /36 (8.5x1/ 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. For the strips, l 30, so the interior olumn strip will be 30 /4+30/4 15, and the middle strip will be the remaining 15. dead load from slab lb / in lb/ in / 1 total dead load lb/ + of lightweight onrete 18 lb/ (0.68 KPa (presuming interior wall weight is over beams & girders 43

12 ARCH 631 Note Set 11 F015abn total dead load lb/ live load with redution, where live load element fator, K LL, is 1 for a two way slab: 15 L Lo( lb (. K A lb/ 1( 30 ( 30 LL T total fatored distributed load: w u 1.( lb/ + 1.6(75 lb/ lb/ total panel moment to distribute: M o w ulln lb / (30 8 (8.5 1k 1000lb k- Column strip, end span: Maximum Positive Moments from Table 4-3, (flat slab with spandrel beams: M u+ 0.30M o 0.30 (938.4 k k- Maximum Negative Moments from Table 4-3, (flat slab with spandrel beams: M u- 0.53M o 0.53 (938.4 k k- Middle strip, end span: Maximum Positive Moments from Table 4-3, (flat slab with spandrel beams: M u+ 0.0M o 0.0 (938.4 k k- Maximum Negative Moments from Table 4-3, (flat slab with spandrel beams: M u- 0.17M o 0.17 (938.4 k k- Design as for the slab in Case 1, but provide steel in both diretions distributing the reinforing needed by strips. 44

13 ARCH 631 Note Set 11 F015abn Shear around olumns: The shear is ritial at a distane d/ away from the olumn fae. If the drop panel depth is 1 inhes, the minimum d with two layers of 1 diameter bars would be 1 ¾ (over (1 ½(1 about 9.75 in (to the top steel. olumn shear perimeter tributary area for olumn drop panel The shear resistane is φ v V φ v 4λ f b o d, φ and λ 1.0 where b o, is the perimeter length. v The design shear value is the distributed load over the tributary area outside the shear perimeter, V u w u (tributary area - b 1 x b where b s are the olumn width plus d/ eah side. b 1 b / / 7.75 in 1 1in b 1 x b ( 7. 75in ( 5.35 V u (84.7 Shear apaity: lb / ( k 1000lb b o (b 1 + (b 4(7.75 in 111 in 54.7 k in in φ v V ( psi ,83 lb ksi < V u! The shear apaity is not large enough. The options are to provide shear heads or a deeper drop panel, or hange onrete strength, or even a different system seletion... There also is some transfer by the moment aross the olumn into shear. Defletions: Elasti alulations for defletions require that the steel be turned into an equivalent onrete E s material using n. E an be measured or alulated with respet to onrete strength. E For normal weight onrete (150 lb/ 3 : E 57, 000 f E 57, psi 3,1,019 psi 31 ksi n 9,000 psi/31 ksi

14 ARCH 631 Note Set 11 F015abn Defletion limits are given in Table