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

Size: px

Start display at page:

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

Kristian Foster
5 years ago
Views:

1 ARCH 631 Note Set 11 F013abn Case Stdy Refored Conrete adapted from Simplified Design of Conrete Strtres, James Ambrose, 7 th ed. Bildg desription The bildg is a three-story offie bildg tended for spelative rental. Figre presents a fll-bildg setion and a plan of the pper 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 Wd: map speed of 80 mph (190 km/h; exposre B Assmed Constrtion Loads: Floor fish: 5 lb/ (0.4 kpa Ceilgs, lights, dts: 15 lb/ (0.7 kpa Walls (average srfae weight: Interior, permanent: 10 lb/ (0.48 kpa Exterior rta wall: 15 lb/ (0.7 kpa Materials Use f =3000 psi (0.7 MPa and grade 60 reforement (f y = 60 ksi or 414 MPa. Strtral Elements/Plan Case 1 is shown Figre and onsists of a flat plate spported on terior beams, whih trn, are spported on girders spported by olmns. We will exame the slab, and a for-span terior beam. Case will onsider the bays with flat slabs, no terior beams with drop panels at the olmns and an exterior rigid frame with spandrel (edge beams. An example of an edge bay is shown to the right. We will exame the slab and the drop panels. For both ases, we will exame the exterior frames the 3-bay diretion. 1

ARCH 631 Note Set 11 F013abn Case 1: Slab: The slabs are effetively 10 x 30, with an aspet ratio of 3, makg them one-way slabs. Mimm depths (by ACI are a fntion of the span.

2 ARCH 631 Note Set 11 F013abn Case 1: Slab: The slabs are effetively 10 x 30, with an aspet ratio of 3, makg them one-way slabs. Mimm depths (by ACI are a fntion of the span. Usg ln Table 3-1 for one way slabs the mimm is with 4 5 hes mimm for fire ratg. We ll presme the terior beams are 1 wide, so l n = 10-1 = 9 mimm t (or h Use / 4. 5 dead load from slab = lb / 5 = 6.5 lb/ / 1 total dead load = ( lb/ + of lightweight onrete toppg with weight of 18 lb/ (0.68 KPa (presmg terior wall weight is over beams & girders dead load = lb/ live load (worst ase orridor = 100 lb/ total fatored distribted load (ASCE-7 of 1.D+1.6L: w = 1.(100.5 lb/ + 1.6(100 lb/ = 80.6 lb/ Maximm Positive Moments from Figre -3, end span (tegral with spport for a 1 wide strip: M (positive = w 14 lb / n n w 1 14 (80.6 (1 (9 14 1k 1000lb = 1.6 k-

3 ARCH 631 Note Set 11 F013abn Maximm Negative Moments from Figre -5, end span (tegral with spport for a 1 wide strip: M - (negative = w 1 lb / n 1 n (80.6 w 1 (1 (9 1 1k 1000lb = 1.89 k- The design aid (Figre an be sed to fd the reforement ratio,, knowg R n = M n /bd with M n = M / f, where f = 0.9. We an presme the effetive depth to the entroid of the reforement, d, is 1.5 less than the slab thikness (with ¾ over and ½ of a bar diameter for a #8 (1 bar = R n = ( 0. 9 ( 1 k ( / lb / k psi so for f = 3000 psi and f y = 60,000 psi is the mimm. For slabs, A s mimm is bt for grade 60 steel. A s = (1(5 = / 3

ARCH 631 Note Set 11 F013abn Pik bars and spag off Table 3-7. Use #3 bars @ 1 (A s = 0.11. Chek the moment apaity. d is atally 5 0.75 (over ½ (3/8 bar diameter = 4.06 (OK a = A s f y /0.85f b = 0.

4 ARCH 631 Note Set 11 F013abn Pik bars and spag off Table 3-7. Use #3 1 (A s = Chek the moment apaity. d is atally (over ½ (3/8 bar diameter = 4.06 (OK a = A s f y /0.85f b = 0.11 (60 ksi/[0.85(3 ksi1 ] = M n = A s f y (d-a/ = 0. 9( ( 60ksi ( ( 1.96 k- > 1.89 k- needed 1 Maximm Shear: Figre -7 shows end shear that is w l n / exept at the end span on the terior olmn whih sees a little more and yo mst design for 15% rease: 1. 15( V -max = 1.15w l n / = lb/ ( 1 ( 9 V at d away from the spport = V -max w(d = 145 lb = 145 lb (for a 1 strip ( lb/ ( 1 1 / ( = 1357 lb Chek the one way shear apaity: v V = v f bd ( v = 0.75: v V = 0.75( 3000 psi( 1 ( = 4003 lb Is V (needed < v V (apaity? thiker. YES: 1357 lb 4003 lb, so we don't need to make the slab Interior Beam (effetively a T-beam: Tribtary width = 10 for an terior beam. dead load = (100.5 lb/ (10 = 1005 lb/ (14.7 kn/m 4

5 ARCH 631 Note Set 11 F013abn Redtion of live load is allowed, with a live load element fator, K LL, of for an terior beam for its tribtary width assmg the girder is 1 wide. The live load is 100 lb/ : 15 lb 15 L = L o ( 0. 5 = 100 (. 0 5 = 87.3 lb/ ( 30 1 ( 10 K LL A T live load = 87.3 lb/ (10 = 873 lb/ (1.7 kn/m (Redtion Mltiplier = > 0.5 Estimatg a 1 wide x 4 deep beam means the additional dead load from self weight ( w A nits of load/length an be lded. The top 5 hes of slab has already been lded the dead load: dead load from self weight = 150 lb 1 (1 wide(4 5 deep = 37.5 lb/ (3.46 kn/m 1 3 w = 1.(1005 lb/+37.5 lb/ + 1.6(873 lb/ = 888 lb/ (4.30 kn/m n The effetive width, b E, of the T part is the smaller of 4, b w 16t, or enter-enter spag b E = mimm{9 /4 = 7.5 = 87, 1 +16x5 = 9, 10 = 10 } = 87 The lear span for the beam is l n = 30 1 = 9 Maximm Positive Moments from Figre -3, end span (tegral with spport: M (positive = w 14 n 888 lb / (9 14 1k = k- 1000lb Maximm Negative Moments from Figre -4, end span (tegral with spport: M - (negative = w 10 n 888 lb / (9 10 1k = 4.9 k- 1000lb 5

6 ARCH 631 Note Set 11 F013abn Figre an be sed to fd an approximate for top reforement if R n = M n /bd and we set M n = M / f. We an presme 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 stirrps (3/8 more, so d = 1.5. R n = 1000 lb / k 0.9(1 4.9 (1.5 k 1 / = 584 psi so for f = 3000 psi and f y = 60,000 psi is abot (and less than max = Then we pik bars and spag off Table 3-7 to fit the effetive flange width the slab. For bottom reforement (positive moment the effetive flange is so wide at 87, that it resists a lot of ompression, and needs very little steel to stay nder-refored (a is between 0.6 and 0.5. We d pt bottom bars at the mimm reforement allowed and for tyg the stirrps to. Maximm Shear: V max = w l/ normally, bt the end span sees a little more and yo mst design for 15% rease. Bt for beams, we an se the lower vale of V that is a distane of d from the fae of the spport V -design = 1.15w l n / w d= 1.15(888 lb/ lb/ (9 888 (1.5 / 1 4,983 lb = 43.0 k Chek the one way shear apaity = v V = v f bd, where v v V = 0.75( 3000 psi (1 (1.5 = 1,197 lb = 1. k Is V (needed < v V (apaity? NO: 43.0 k is greater than 1. k, so stirrps are needed v V s = V - v V = 43.0 k 1. k = 1.8 k (max needed Usg #3 bars (typial with two legs means A v = (0.11 = 0.. To determe reqired spag, se Table 3-8. For d = 1.5 and v V s v 4 v 4 f bd (where f bd= V = (1. k = 4.4 k, the maximm spag is d/ = or 4. s reqired = A V v f y d V Av f V s y d = k 1. 8 ksi = 9.75, so se 9. We wold try to rease the spag as the shear dereases, bt it is a tedios job. We need stirrps anywhere that V > v V /. One reommended termediate spag is d/3. 6

7 ARCH 631 Note Set 11 F013abn The reqired spag where stirrps are needed for rak ontrol ( v V V > ½ v V is s reqired = A v f 50b y w 0. (60,000psi 50(1 = and the maximm spag is d/ = or 4. Use 10. A reommended mimm spag for the first stirrp is. from the fae of the spport. A distane of one half the spag near the spport is oen sed. V 36.7 k 41.9 k to fd distanes when w is known: x = (peak rossg vale/w ex: 41.9 k 1. k/.888 k/ = k 10.6 k 7. = = 130 C L = = k 1. k x k 48. k END SPAN STIRRUP LAYOUT Spandrel Girders: Bease there is a onentrated load on the girder, the approximate analysis an t tehnially be sed. If we onverted the maximm moment (at midspan to an eqivalent distribted load by settg it eqal to w l /8 we wold then se: spandrel girder 7

8 ARCH 631 Note Set 11 F013abn Maximm Positive Moments from Figre -3, end span (tegral with spport: M + = w 14 n Maximm Negative Moments from Figre -4, end span (olmn spport: M - = w 10 n (with w 16 n at end Colmn: An exterior or orner olmn will see axial load and bendg moment. We d se teration harts for P and M for standard sizes to determe the reqired area of steel. An terior olmn sees very little bendg. The axial loads ome from gravity. The fatored load ombation is 1.D+1.6L + 0.5L r. The girder weight, presmg 1 x 4 girder at 150 lb/ 3 = 600 lb/ Top story: presmg 0 lb/ roof live load, the total load for an terior olmn (tribtary area of 30 x30 is: Lower stories: DL roof* : 1. x lb/ x 30 x 30 * assmg 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 30 = 1.6 k LL r : 0.5 x 0 lb/ x 30 x 30 = 9.0 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 (ldes redtion = 81.4 k nd floor olmn sees P = = k 1 st floor olmn sees P = = 77.5 k Look at the example teration diagram for an 18 x 18 olmn (Figre 5-0 ACI sg f = 4000 psi and f y = 60,000 psi for the first floor havg P = 77.5 k, and M to the olmn beg approximately 10% of the beam negative moment = 0.1*4.9 k- = 4.3 k-: (See maximm negative moment allation for an terior beam. The hart diates the apaity for the reforement amonts shown by the solid les. For P = 77.5 k and M = 4.3 k-, the pot plots below the le marked 4-#11 (1.93% area of steel to an 18 x 18 area. 8

9 ARCH 631 Note Set 11 F013abn Lateral Fore Design: The wd loads from the wd speed, elevation, and exposre we ll aept as shown Figre 17.4 given on the le psf. The wd is atg on the long side of the bildg. The perimeter frame resists the lateral loads, so there are two with a tribtary width of ½ [(30x(4 bays + for beam widths and laddg] = 1/ = 61 The fatored ombations with dead and wd load are: 1.D +1.6L r + 0.5W 1.D + 1.0W + L + 0.5L r The tribtary height for eah floor is half the distane to the next floor (top and bottom: 9

10 ARCH 631 Note Set 11 F013abn Exterior frame (bent loads: H 1 = 195 / (61 =11,895 lb = 11.9 k/bent H = 34 lb / 1000 (61 lb / k 7 lb / =14.3 k/bent H 3 = lb / k 1000 (61 =13.8 k/bent Usg Mltiframe, the axial fore, shear and bendg moment diagrams an be determed sg the load ombations, and the largest moments, shear and axial fores for eah member determed. 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 = 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 smmary diagram of fore, shear and moment magnitdes refer to the maximm vales the olmn or beams, with the maximm moment the beams beg negative over the spports, and the maximm moment the olmns beg at an end kip kip Axial fore diagram: Px' Shear diagram: Vy ' kip kip y kip- y x Bendg moment diagram: Mz' x Displaement: kip- y y x x 10

11 ARCH 631 Note Set 11 F013abn Beam-Colmn loads for design: The bottom exterior olmns see the largest bendg moment on the lee-ward side (le: P = 19.8 k and M = k- (with large axial load The terior olmns see the largest axial fores: P = k and M = 95.5 k- and P = k and M = k- Refer to an teration diagram for olmn reforement and sizg. Case Slab: The slabs are effetively 30 x 30, makg them two-way slabs. Mimm thiknesses (by ACI are a fntion of the span. Usg Table 4-1 for two way slabs, the mimm is the larger of l n /36 or 4 hes. Presmg the olmns are 18 wide, l n = 30 (18 /(1 / = 8.5, h = l n /36 = (8.5x1/36 = 9.5 The table also says the drop panel needs to be l/3 long = 8.5 /3 = 9.5, and that the mimm depth mst be 1.5h = 1.5(9.5 = 1. For the strips, l = 30, so the terior olmn strip will be 30 /4+30/4 = 15, and the middle strip will be the remag 15. dead load from slab = lb / 9. 5 = lb/ / 1 11

12 ARCH 631 Note Set 11 F013abn total dead load = lb/ + of lightweight onrete 18 lb/ (0.68 KPa (presmg terior wall weight is over beams & girders total dead load = lb/ live load with redtion, where live load element fator, K LL, is 1 for a two way slab: 15 L = Lo( 0. 5 lb 15 = 100 (. K A 0 5 = 75 lb/ 1( 30 ( 30 LL total fatored distribted load: T w = 1.( lb/ + 1.6(75 lb/ = lb/ total panel moment to distribte: M o = ln w l lb / (30 8 (8.5 1k 1000lb = k- Colmn strip, end span: Maximm Positive Moments from Table 4-3, (flat slab with spandrel beams: M + = 0.30M o = 0.30(938.4 k- = 81.5 k- Maximm Negative Moments from Table 4-3, (flat slab with spandrel beams: M - = 0.53M o = 0.53(938.4 k- = k- Middle strip, end span: Maximm Positive Moments from Table 4-3, (flat slab with spandrel beams: M + = 0.0M o = 0.0(938.4 k- = k- Maximm Negative Moments from Table 4-3, (flat slab with spandrel beams: M - = 0.17M o = 0.17(938.4 k- = k- 1

13 ARCH 631 Note Set 11 F013abn Design as for the slab Case 1, bt provide steel both diretions distribtg the reforg needed by strips. Shear arond olmns: The shear is ritial at a distane d/ away from the olmn fae. If the drop panel depth is 1 hes, the mimm d with two layers of 1 diameter bars wold be 1 ¾ (over (1 ½(1 =abot 9.75 (to the top steel. olmn shear perimeter tribtary area for olmn drop panel The shear resistane is v V = v 4 f b o d, v where b o, is the perimeter length. The design shear vale is the distribted load over the tribtary area otside the shear perimeter, V = w (tribtary area - b 1 x b where b s are the olmn width pls d/ eah side. b 1 = b = / / = b 1 x b = ( ( = lb / 1k V = (84.7 ( = 54.7 k 1000lb Shear apaity: b o = (b 1 + (b = 4(7.75 = 111 v V = psi = 177,83 lb= ksi < V! The shear apaity is not large enogh. 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 olmn to shear. Defletions: Elasti allations for defletions reqire that the steel be trned to an eqivalent onrete E s material sg n =. E an be measred or allated 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 F013abn Defletion limits are given Table 9.5(b 14

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

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