Letter of Transmittal

Transcription

1 Letter of Transmittal Date: October 17, 2014 To: Dr. Linda Hanagan From: Xiaodong Jiang Dear Dr. Hanagan, This is Xiaodong Jiang, This following report was submitted for Technical Report 3 for AE 481 W. This building is Xston Inn., a hotel building. This report was made for analyzing the lateral resistance system of Xyston Inn. The items included in this report are: Pre-modeling calculations and verifications Lateral System Introduction Center of Rigidity Calculation for typical floor plan Wind load calculation for chosen shear wall Seismic load calculation for chosen shear Load combination of wind and seismic cases Element size verification (shear wall verification) ETABS Computer modeling general information ETABS computer model result verifications Building torsion issue identification Building overturning discuss Thank you for reviewing this report. I m looking forward to discussing it with you in the near future. Sincerely, Xiaodong Jiang

2 Technical Report IV Xiaodong Jiang Structure Option Advisor: Dr. Linda Hanagan November 17, 2014

3 Advisor: Dr. Linda Hanagan Table of Contents Table of Contents 1 Excusive Summary 3 Building Abstract 4 Site and Location Plan 5 Reference Document 6 Pre-Modeling Analysis Lateral System Typical Floor Plan C.R. - Typical Floor Plan Wind Load Calculation Shear Wall B Seismic Load Calculation Shear Wall B Load Combination - Shear Wall B Element Size Verification 18 Computer Modeling Analysis Computer Modeling ETABS Modeling 21 Modeled Lateral System 21 Modeling Assumptions 21 Modeling Challenge and Possible Improvements Computer Model Result Verifications 24 Base Reactions 24 Center of Masss and Center of Rigidity 25 Force in Shear Walls 26 Story Drift and Overall Building Drift Building Torsion Issues Building Overturning Issue 31 1

4 Advisor: Dr. Linda Hanagan Conclusion 32 Appendix A - Additional Hand Calculation Information a-1 A4-3 Wind Shear in Shear Wall of the 3 rd floor a-1 A4-4 Seismic Shear in Shear Wall of the 3 rd floor a-5 A4-6 Shear Wall B Verification a-6 Appendix B - Additional Computer Modeling Information b-1 B4-7 Computer Modeling b-1 B4-8 Additional Computer Results b-2 Appendix C - Technical Report II c-1 Gravity Load Typical Roof Bay c-1 Typical Floor Bay c-2 Typical Exterior Wall c-3 Non-Typical Dead Load and Live Load c-5 Snow Load c-6 Snow Drift c-7 Lateral Load -Wind c-10 Lateral Load -Seismic c-26 2

5 Advisor: Dr. Linda Hanagan Executive Summary Xyston Inn is a 17 story hotel building that will be located in Brooklyn, New York. The design of Xyston Inn is inspired by the varied character of this still developing part of the city. This 65,000 sq. ft. hotel will provide a comfortable spot for the people visiting New York. To create a bright and commodious lounge space, the first floor is designed with a height of 20 feet. The building also includes 16 hotel function floors and a mechanical floor which all have heights of 10 feet. The overall building height is about 215 feet. Beside these the building also has a cellar level 12 feet below street level. The building structure system is two-way reinforced concrete building with a reinforced concrete shear wall lateral system. The building is built on a bedrock level which has great allowable bearing capacity values. Due to the height of the building, the mat foundation consists the varied thickness between three feet and four feet. To transfer the building load to the foundation, 20 x 24 reinforced columns are commonly used for the cellar level and first floor level. The lateral system employs a reinforced concrete shear wall system at several location, through the building. To balance the architectural needs and great torsional force and moment, the structure adopts a 1 x 75 shear wall on the west to east direction. The concrete compressive strength of the shear wall also varies along the height of levels. On the north to south direction, the lateral system consists of several separated shear walls which have varying lengths from 10 feet to 26 feet. To design the building structure system, the loads considered in this building design include live loads, building dead loads, snow loads, wind loads, seismic loads, and lateral soil loads. The live load, dead load and snow load are also known as the gravity load which is caused by gravity, and acts on vertical direction. The wind load and seismic load are known as the lateral load, which is the load caused by wind or ground motion; those loads are acting directly on the lateral direction, but they also have some effect on the vertical direction. The lateral soil load is also lateral load, but it only effects the structure underground. Structure design is based on the 2008 New York City Building Codes for general structure design. The details of concrete elements are based on ACI Building Code ; and the load calculations follows ASCE 7-10 requirements. 3

6 Advisor: Dr. Linda Hanagan 4

7 Advisor: Dr. Linda Hanagan Site Plan and Location Plan 5

8 Advisor: Dr. Linda Hanagan Reference Documents The codes and other documents used in preparation of this report are include: ASCE 7-10 Wind and Seismic Provisions ACI Edition. AISC Steel Manual 14 th Edition Material Unit Weight Table 6

9 Advisor: Dr. Linda Hanagan Pre-Modeling Analysis 7

10 Advisor: Dr. Linda Hanagan 4-1 Lateral System Typical Floor Plan Xyston Inn is an 18-story hotel building located in Brooklyn, NY. The building s gravity system is twoway reinforced concrete flab slab system. Building lateral system consists of several reinforced concrete shear walls erected on 3 -to-4 mat foundation. The figure shows the typical floor plan, 3 rd to 15 th floor, of this building. The columns shown on this floor plan are all the gravity columns. The shear walls are concentrated on the south side of this building to provide the maximum architectural, electrical, and mechanical flexibility of the building design. Fig rd to 15 th floor plan Due to the concentration of the west-to-east run shear walls, the building experiences extreme torsional irregularity. The shear wall B, highlighted on figure below, is designed to resist the direct lateral shear load as well as the great torsional shear load. Shear wall B s vertical reinforcement is (2) #7 grade O.C. The horizontal reinforcement is (2) #4grade O.C. The concrete strength of shear wall B from 2 nd floor to 8 th floor is 8,000 psi. Fig Shear Wall B Vertical Reinforcing Detail 8

11 Advisor: Dr. Linda Hanagan 4-2 C.R. - Typical Floor Plan CR Fig rd to 15 th floor plan To calculate the center of rigidity, the stiffness of each shear wall should be determined first. To calculate the stiffness of each shear wall (fixed at top and bottom) the following equations are used: Since G 0.4 E, this for formula can be simplified as below: K = Example Calculation (shear wall B): Shear Wall B: E t ( h b ) + 3 (h b ) Height = 10 ft. b = 17 ft. t = 14 Concrete f c = 8000psi E = (8000) E = 5098 ksi K = ( 10 = 36,262 k/in 17 ) + 3 (10 17 ) 9

12 Advisor: Dr. Linda Hanagan By using the same equation and Excel spread sheet, the other shear walls stiffness can be determined. The Excel table of stiffness calculation is shown below: Table Shear Wall Stiffness Calculation for Typical Floor Plan Then, using the formulas below to calculate center of rigidity, CR. CRx = ( ) CRy = ( ) Table Center of Rigidity Calculating for Typical Floor Plan Therefore, the center of rigidity, CR, is (65, 4.48) ft. for 3 rd floor plan. 10

13 Advisor: Dr. Linda Hanagan 4-3 Wind Load Calculation Shear Wall B About wind cases: Because shear wall B runs on the west-to-east direction, X-direction, the effect of wind load from the south-to-north direction, Y-direction, only causes partially torsional shear to the shear wall B and would not be the critical wind load case for shear wall B. In following analysis, neglect the wind cases that the wind load is from Y-direction only (it means when the wind is loading on both direction, that case should be considered). Case 1-X: full wind load on x-direction, acts on the center of projected building surface. Case-2-X: 75% wind load acts to the floor diaphragm with 0.15 eccentricity (two sub-cases). Case-3: 75% of wind load from both direction, acts on the center of projected building surface. Case-4: 56.3% of wind load from both directions and acts to the floor diaphragm with 0.15 eccentricity of both direction (four sub-cases only two sub-cases are considered). Because when wind blows from west to east, left to right. The critical eccentricity of BY is positive, which would create additive torsional shear stress while negative eccentricity of BY would help shear wall to reduce the shear stress. Therefore, ex = +/-0.15 BX, ey = BY Fig Considered Wind Load Cases. 11

14 Advisor: Dr. Linda Hanagan The figure on the previous page shows the wind cases that are considered in this report. To calculate the wind loads act on the shear wall B, the following equations are used. Besides that, the Excel spread sheets are also adopted to process repeatable calculations. Direct Shear: Torsional Moment of Inertia: Torsion: Torsional Shear: Vi: Story shear Vx = 59.8 kips, and Vy = kips are from Technical Report 2. Therefore, the 100%, 75% and 56.3% of story is given by Table Table The following calculations use the example calculation of Case 1-X to explain the formulas used in Excel spread sheets. Besides the sample calculation, the summary of calculated wind loads on shear wall B and other details are also provided in appendix A Direct Shear: Rx = R A +R B + R C = 36, , ,570 = 194,094 V B = 36,262 / 194,094 * (59.8) = 11.2 kips Torsional Shear: Wind acts on the point: (49.85, 24.5). Center of rigidity: (65, 4.48) Eccentricity-y: Total torsion: ey = = ft T = 59.8 * (-20.02) = -1,197 k-ft ( - means it is clockwise torsion) 12

15 Advisor: Dr. Linda Hanagan Torsional Moment of Inertia J = 63,833,298 from calculation table below. Table Note: Xi Shear wall x-distance from point (0, 0) Yi - Shear wall y-distance from point (0, 0) CRx, CRy Center of rigidity coordinate Ri Center of rigidity of shear wall di distance between individual shear wall and the center of rigidity. Ri*di^2 moment of inertia due to each shear wall Therefore, the torsional shear on shear wall B is: V T,B = (1197) * (36,262 * 7.52) / (63,833,298) = 5.1 kips Total story shear of shear wall B: V = = 16.3 kips Similarly, for other cases, the total story shear of shear wall B can be calculated by Excel spread sheet provided on Appendix A The results of the table calculations can be summarized as below, table 4-3-3, and the critical shear is 16.3 kips. (See Appendix A4-3-1 for calculation details). Table Therefore, the critical story wind load on shear wall B is 16.3 kips. (Note: 16.3 kips is 27.3 % of 59.8 kips) 13

16 Advisor: Dr. Linda Hanagan Assume the wind load distributions of upper stories are the same as the 3 rd floors, 27.3% of story wind load is resisted by shear wall B. Total wind load above the 3 rd floor is = kips. Therefore, the simplified calculation of wind load on the 3 rd floor shear wall B V 3,B = 27.3% * = kips. (Note: story shear of 3 rd floor is resisted by 2 nd floor s shear walls) M 3,B = 39,260 k-ft (See Table A4-3-4 on Appendix) Therefore, the table beblow shows shear load and moment on each shear walls on W-E direnction based on the critical load case, case 1-X. Wind Case 1-X, 3 rd fl Shear, kips Moment, k-ft Total Shear Load ,808 Shear Wall A ,260 Shear Wall B ,260 Shear Wall C ,288 Table

17 Advisor: Dr. Linda Hanagan 4-4 Seismic Load Calculation Shear Wall B The seismic load is caused by building s rigidity. It acts on the center of mass. From Technical Report 2, the story shear is defined as Fx = 0.01 Wx = 6.84 kips. (Technical Report 2 is attached in Appendix C) Therefore, the story seismic load is 6.84 kips. Figure CM Dividing this floor plan into three rectangular regions to calculate the center of mass. 1: 53.5 x 49, A1 = 53.5 x 49 = 2622 ft 2, mass center: (26.8, 24.5) ft. 2: 23.1 x 36, A2 = 23.1 x 36 = 832 ft 2, mass center: (65.1, 18) ft. 3: 23.1 x 25, A3 = 23.1 x 25 = 578 ft 2, mass center: (88.2, 12.5) ft. The formulas used to calculate center of mass are: X = ; Y = Table Center of Mass Calculation for Typical Floor Plan Therefore, the center of mass is (43.5, 21.4) ft. 15

18 Advisor: Dr. Linda Hanagan Seismic Eccentricity The floor system is two-way reinforced concrete flat slab; therefore a good assumption of floor diaphragm is rigid diaphragm. For rigid diaphragm additional 5% eccentricity is required by ASCE % * BY = 0.05 * 49 = 2.45 ft Story Seismic Shear Recall: CM (43.5, 21.4) ft CR (65, 4.48) ft Total eccentricity = ( ) +/ = ft or ft +5% Eccentricity -5% Eccentricity T = 6.84 * = k-ft T = 6.84 * = 99.0 k-ft R B * d B / J = 272,688 / 63,833,298 = R B * d B / J = 272,688 / 63,833,298 = V T,B = * = kips V T,B =99.0 * = kips V D,B = 6.84 *(36,262 / 194,094) = 1.28 kips V D,B = 6.84 *(36,262 / 194,094) = 1.28 kips V B = = kips V B = = kips By following the same calculation steps, the seismic load acts on S-N (Y-direction) can also be found and listed below: Table Therefore, story seismic load of shear wall B on 3 rd floor is kips / 6.84 = % of story shear is distributed to shear wall B. Assume upper stories have the same distribution of seismic load. Accumulative story shear = 105 kips Accumulative story moment = 8,783 k-ft Then, V B = 0.27 * 105 = kips M B = 0.27* 8,783 = 2,371.4 k-ft 16

19 Advisor: Dr. Linda Hanagan Therefore, the table beblow shows shear load and moment on each shear walls on W-E direnction based on the critical seismic case, seismic-x. Shear, kips Moment, k-ft Total Seismic Load on 3 rd fl 105 8,783 Shear Wall A ,371 Shear Wall B ,371 Shear Wall C ,041 Table Note: 1. More details of calculation Excel spread sheet is shown on Appendix A The seismic design category, SDC, of this building is SDC-A, no additional calculations are required for seismic load identification. 3. Since this building is SDC-A building, the horizontal and vertical irregularity identifications are not required. 4-5 Load Combination of Shear Wall B (3 rd Fl.) Combinations from ASCE 7-10: Basic Combinations: D D+ 1.6L + 0.5(Lr or S or R) D+ 1.6(Lr or S or R) + (L or 0.5W) D+ 1.0W+ L+ 0.5(Lr or S or R) D+ 1.0E+ L+ 0.2S D+ 1.0W D+ 1.0E Lateral Force on Shear Wall B: Wind: Vu = (1.0)* (360.3) = kips Controls! Seismic: Vu = (1.0) * (28.36) = kips Therefore, the lateral load is controlled by wind load case 1-X, which is full wind load acts on the center of projected building surface. 17

20 Advisor: Dr. Linda Hanagan 4-6 Element Size Verification Hand Calculation From previous calculations, Vu = kips, Mu = 39,260 k-ft From Technical report 2, DL = 130 psf, LL = 80 psf. Assume DL = 130 psf and LL = 80 psf for all stories above. Then calculate gravity load, Pu, of shear wall B. Supporting area Effective Area Fig Gravity Supporting Zone and Effective Zone of Shear Wall B 34 x 7.75 Supporting Zone: 34 x 7.75 A = x 15 = ft 2 Effective Zone: 49 x 15.5 A eff = 760 x 15 = ft 2 Therefore, LL Min = 0.4 x 80 = 32 psf LL = 80 x ( / (11400) ^ 0.5) = 31.2 psf LL = 31.2 psf P LL = 32 x = lbs = kips PDL = PDL, fl + PDL, rf = 130 x 15 x x = 548,080 lbs = kips P SL = 20 x = 5,270 lbs = 5.3 kips Load Combination: 1.2 D W + L +0.5 S Pu = 1.2 x x x 5.3 = 787 kips Shear Wall Properties lw = 17 hw = 10 h = t = 14 f c = 8000 psi Vertical Reinforcing: (2) #7 grade O.C. Horizontal Reinforcing: (2) #4 grade O.C. 18

21 Advisor: Dr. Linda Hanagan Mu, top = 39,260 k-ft From Appendix A4-6, pg. a-6, the element size checks are given as below: Vertical Reinforcing: a = 5.0 ft Mu = M u, top + Vu * a = 41,060 k-ft A v, required = 0.39 in 2 per 12 A v, design = 1.2 in 2 > 0.39in 2 (Pass, and it is overdesigned for serviceability considerations) ρ l = > limit 1 and limit 2 (Pass) ρ l, limit1 = ρ l, limit2 = S 1 = 12 < limit 1, limit 2, and limit 3 (PASS) S 1, limit 1 =lw /3 = 68 in S 1, limit 2 = 3*h = 42 in S 1, limit 3 =18 in ρ t = A hor. / (S 2 * t) = 0.4 / (12 x 14) = < (Not Pass!) Note: (2) #4 horizontal reinforcing 12 O.C. and shear wall thickness is 14 in. S 2 = 12 < limit 1, limit 2, and limit 3 (PASS) S 2, limit 1 = lw / 5 = in S 2, limit 2 = 3*h = 42 in S 2, limit 3 =18 in Therefore, the design is adequate to resist the load it was designed for. However, the horizontal reinforcing ratio requirement of shear wall B did not pass. 19

22 Advisor: Dr. Linda Hanagan Computer Modeling Analysis 20

23 Advisor: Dr. Linda Hanagan 4-7 Computer Modeling ETABS Modeling To efficiently analyze this 18-story hotel building, ETABS 2013 was adopted. This structural modeling program was introduced in AE 530, Computer Modeling of Buildings. To be able to analyze the building model with the calculated loads from Technical Report 2, a simplified building model was made and analyzed. The detail information of elements can be found in Appendix-b, B4-7-1 pg. b-1. Fig shows the 3D view from north face of building. Fig shows the 3D view from south face of building. Note: the gravity columns were modeled with zero lateral effect approaches. Fig Fig Modeled Lateral System 1. Columns in this project are all gravity columns and excluded from lateral system. Because the gravity columns were originally not designed for lateral capacity. 2. Additional columns were created to be attached to shear walls. Because the membrane shear wall model might not be able to carry and transfer the shear and moment from all directions. 3. Coupling beams were modeled and included in lateral system. Because they were designed to transfer lateral shear and moment between the separated shear walls A and B. (Note: on hand calculation, the effect of coupling beams were excluded). 4. Shear walls were modeled with strength reduction factor of Modeling Assumptions 1. The gravity columns were modeled with zero lateral strength effect to get the ideal and conservative results of lateral system. 2. Floors were modeled as membrane and rigid diaphragm, to be able to transfer lateral loads to the reinforced concrete shear wall lateral resistance system. The building floor system is two-way reinforced concrete flab slab, which performs almost the same as a rigid diaphragm. 21

24 Advisor: Dr. Linda Hanagan 3. The only beams in this project are the ones designed to connect the separated shear walls on the west-to-east direction. Those beams are assumed to be cracked and modeled with strength reduction factor of The cracked beam moment of inertia is recommended to be 0.35 according to ACI Bottom boundary condition of columns were modeled to be pinned to mat foundation conservatively. The reason of pin connection is that those columns were gravity columns and not designed laterally. 5. Bottom boundary condition of shear walls were modeled to be fixed to mat foundation. Fig Bottom Boundary Condition (pined columns and fixed shear wall) 6. The floor and shear wall mesh were set to be auto mesh with a maximum size limit of 2 ft x 2 ft. 7. Elements weights were set to be zero. Because the weight of building could affect the lateral system behaviors and might make lateral analysis too complex to get the expected results. For More detail information of elements, please check Appendix B4-7 on page b-1. 22

25 Advisor: Dr. Linda Hanagan Modeling Challenge and Possible Improvements Challenge: Grid: The original drawing of two-way slab floor plan doesn t have a grid system. Some of the dimensions modeled in computer model were the results of approximation. It was challenging to start an ETABS model without a defined grid system. In the ETABS model, two separated grid systems were made to try to get the correct location and dimensions of elements. Material: This building includes five different strength concretes. And the material varies when story changes. To try to model a correct building model with varying material strength was a big challenge. Connection and constraint: the connection and constraint between elements were challenging. Especially, when a connection and constraint mistake occurred between floor diaphragms and shear walls, the results of element stress and reactions were totally wrong. The best way to correct it was to deleted and re-model the wrong elements. Eventually, 8 floors of elements were deleted and remodeled with additional constraint setting. In addition, the continuous shear wall sections with different thickness were not able to have correct results of shear reactions. Eventually, the thicker shear wall section was modeled with the same thickness as the thinner sections. The results of force reaction showed that the shear reactions of the continuous shear walls were correct, but the horizontal drift and building deflection became larger due to the reduced thickness of shear wall. Improvements: The results of the simplified ETABS model were acceptable. To improve the computer model, the wind load need to be re-calculated more accurately by using the detailed roof section. The mesh size of floor and shear wall might also be sized down and the 2 nd order effect also need to be included. 23

26 Advisor: Dr. Linda Hanagan 4-8 Computer Model Result Verification Base Reaction Recall the building base shear and overturning moment information from Technical Report 2. W-E direction: V = kips OTM = 173,696 k-ft S-N direction: V = 2802 kips OTM = 332,318 k-ft Seismic: V = 113 kips OTM = 11,051 k-ft The reactions from ETABS model are: Load FX (W-E), kips % V (W-E) kips FY (S-N), kips % V (S-N) kips Close to hand Calculation? Wind Case 1-X % V Yes Wind Case 2-X % V Yes BY Wind Case 2-X % V Yes BY Wind Case % V % V 2102 Yes Wind Case 4-X %V %V 1578 Yes BY Wind Case 4-X % V % V 1578 Yes BY Seismic -X % Vseis Yes Seismic -Y % Vseis 113 Yes Table Base Shear Reaction of Overall Building. Load MX k-ft % M k-ft MY k-ft % M k-ft Close to hand Calculation? Wind Case 1-X , % M 173,696 Yes Wind Case 2-X ,299 75% M 130,272 Yes BY Wind Case 2-X ,299 75% M 130,272 Yes BY Wind Case 3 249,207 75% M 249, ,299 75% M 130,272 Yes Wind Case 4-X 187, %M 187,095 97, %M 97,791 Yes BY Wind Case 4-X 187, % M 187,095 97, % M 97,791 Yes BY Seismic -X % Mseis 11,051 Yes Seismic -Y % Mseis 11, Yes Table Base Overturning Moment of Overall Building. 24

27 Advisor: Dr. Linda Hanagan Center of Mass & Center of Rigidity Comparing the center of mass and center of rigidity between hand calculation and ETBAS model about the 3 rd floor. The results is given below: Center of Mass: Hand Calculation: CM-x = 43.5 ft; CM-y = 21.4 ft (Refers to Section 4-4) Computer Evaluation: CM-x = 43.5 ft; CM-y = 20.8 ft Between the hand calculation and computer evaluation, the % difference of center of mass location on x- direction is 0%. And the % difference of center of mass location on y-direction is 2.9%. The % difference is very small; therefore, the center of mass calculation is verified. Center of Rigidity: Hand Calculation: CR-x = 65.0 ft; CR-y = 4.48 ft (Refers to Section 4-2) Computer Evaluation: CR-y = 64.3 ft; CR-y = 4.75 ft Between the hand calculation and computer evaluation, the % difference of center of rigidity location on x- direction is 1.1%. And the % difference of center of rigidity location on y-direction is 5.7 %. The % difference is small; therefore, the center of rigidity calculation is verified. 25

28 Advisor: Dr. Linda Hanagan Force in Shear Walls (W-E Direction) Wind Recall the table 4-3-4, then the ETABS result of shear wall A, B, and C under the wind load case 1-X is also given in table Wind Case 1-X, 3 rd fl Shear, kips Moment, k-ft Total Shear Load ,808 Shear Wall A ,260 Shear Wall B ,260 Shear Wall C ,288 Table Hand Calculation Result of Wind Case 1-X, 3 rd Floor Wind Case 1-X, 3 rd fl Shear, kips Moment, k-ft ETABS Story Shear ,613 Shear Wall A ,440 Shear Wall B ,446 Shear Wall C ,727 Table ETABS Story Shear of Wind Case 1-X, 3 rd Floor In section 4-3-4, the wind load included in hand calculation was the wind load acts on the stories above the 3 rd floor. But in ETABS analysis, the story shear of 3 rd floor was also included. If the story shear, 59.8 kips, of 3 rd floor was excluded from ETABS model, the story shear would be exactly kips. ETABS also gave the larger values of the shear in shear wall A and B than hand calculation. The reason of that is the hand calculation excluded the rigidity increase by coupling beam. Therefore, the hand calculation had less values of rigidities of shear wall A and B. The result of less values of rigidity was that the shears in shear wall A and B were also less than exact values. Seismic Shear, kips Moment, k-ft Total Seismic Load on 3 rd fl 105 8,783 Shear Wall A ,371 Shear Wall B ,371 Shear Wall C ,041 Table Hand Calculation Result of Seismic X, 3 rd Floor Shear, kips Moment, k-ft Total Seismic Load on 3 rd fl 105 8,783 Shear Wall A ,518 Shear Wall B ,518 Shear Wall C ,747 Table ETABS Story Shear of Seismic X, 3 rd Floor Again, because the coupling beam effect was excluded on hand calculation. The less evaluated rigidities of shear wall A and B caused the smaller values of the shear in shear wall A and B. 26

29 Advisor: Dr. Linda Hanagan Story Drift and Overall Building Drift Wind Drift According to ASCE 7-10 the maximum drift of entire building is H/400. The building height of Xyston Inn is 215 ft. It implies that the maximum allowable roof displacement is ft, or 6.45 in. The table below shows the maximum roof drift of the left end joint of shear wall B at roof level. Drift X (in) Less than 6.45 in? Drift Y (in) Less than 6.45 in? W- Case 1-X 4,23 Pass 1.71 Pass W- Case 2-X 3.89 Pass 1.67 Pass BY W- Case 2-X 2.55 Pass 0.55 Pass BY W- Case Not Pass 13.1 Not Pass W- Case 4-X 4.13 Pass 8.94 Not Pass BY W- Case 4-X BY 3.12 Pass 8.40 Not Pass Table Roof Drift of the Left End Joint of Shear Wall B. A graph of maximum story displacement under wind load case 1-X was also provided by ETABS Fig Maximum Story Displacement From the table and figure 4-7-1, building maximum displacements did not pass the H/400 limit. The reason of that was the conservative judgments of wind load determination, exclusion of the lateral effects of gravity columns, and the reduced thickness of the continuous shear walls. The result also turned out that, there was a large story displacement shift between the 15 th story and 16 th story, because of the changed floor layout. Originally, the 16 th floor diaphragm was set as rigid diaphragm, but since the floor layout changes from 15 th story to 16 th story, the 16 th floor diaphragm should have been set as semi-rigid diaphragm to get more reasonable displacement results. 27

30 Advisor: Dr. Linda Hanagan Seismic Story Drift ASCE 7-10 specifies the seismic story drift on section The allowable story drift limits are provided in table , which is also shown below: The risk category of Xyston Inn is II, which refers to the allowable story drift of h sx. Therefore the allowable story drift of 1 st floor is 0.02 (20) = 0.4 ft = 4.8 in, and the allowable story drifts of other stories are 0.02 (10) = 0.2ft = 2.4 in. The ETABS result is shown below: Table Story Displacement v.s. Allowable Story Drift Seismic W-E 28

31 Advisor: Dr. Linda Hanagan Fig Maximum Story Displacement Seismic-W-E Table Story Displacement v.s. Allowable Story Drift Seismic S-N 29

32 Advisor: Dr. Linda Hanagan Fig Maximum Story Displacement Seismic-S-N Xyston Inn passes all the seismic story drift requirements based on ASCE The fact behind the table values and figures is that Xystion Inn was a wind-load-controlled building, the lateral wind has much more effects on building displacements than seismic load. 30

33 Advisor: Dr. Linda Hanagan 4-9 Building Torsion Issue Seismic The building structural design category, SDC, is A. Therefore, the seismic horizontal torsional irregularity is not required to be identified. Because the seismic load is too small compares to lateral wind load. Note: SDC of B, C, D, E, and F require seismic torsional irregularity identifications. Wind Xyston Inn has different center of mass locations and center of rigidity locations among floors. Because the W-E direction shear walls are concentrated on the south side of building. The center of rigidity s location is a lot different from the center of mass s location as well as the center of wind load. Therefore, the building experiences large torsional shears when lateral wind load is acting on the building. The large torsional shear also affects building serviceability. It causes large torsional drift on the building floor, especially on the 16 th floor where the floor plan s size and concrete strength are less than the floor below Building Overturning Issue The building lateral load is controlled by wind load. Therefore, the controlling load combination is 0.9 D +1.0 W for overturning issue. Building Overturning Moment = 332,318 k-ft (on short direction, S-N) Building Total Weight = 11,338 kips kips Moment Arm = 21.4 ft M resist = x 21.4 = 218,366 k-ft F.S. = 218,366 / 332,318 =0.66 < 1 (Not Pass) Due to the large value of building overturning moment, building self-weight is not able to resist the overturning moment. The building needs to be tightened down to mat foundation. On the structural drawing, the columns reinforcing is continuous to the mat foundation, which would help building to resist the overturning moment. Note that, there were several conservative assumptions to amplify the lateral wind load. One of the result of the conservative assumptions is that the building overturning safety factor could be reduced. 31

34 Advisor: Dr. Linda Hanagan Conclusion This technical report went through the lateral system spot checks and computer modeling verifications of lateral system strength and performance. The overall design of lateral system is good for strength requirements. However, the drift and deflection behaviors were not acceptable. The reasons of the unacceptable building drift and deflection behaviors can be divided into three parts. First of all, the assumptions made in this technical report might not be the same as the one made by the project structure engineer. Secondly, the analysis was done with conservative considerations, which were intended to amplify the lateral load and deflection. Finally, the building s structural details were simplified, the real building is much more complex and there were information loss when simplification approaches were made. The computer model, ETABS model, is not perfectly accurate. The ideal boundary conditions, element constraints, and material properties can never be exactly the same as the ones of real building. In addition, as a result of simplification of continuous shear walls with varying thickness, the internal stress and base shear might be correct, but the deflections of shear walls are amplified since the thicknesses of thicker shear wall sections are reduced. The overall outputs of ETABS model analysis were acceptable except the lateral deflections of shear walls exceeded the limit of H/400. In addition to building seismic analysis, the building seismic horizontal irregularity and seismic torsion were not the controlling issues. However, the torsion due to the lateral wind load was still significant and it caused the deflection performance of shear walls to fail by exceeding the allowable story drift limit of H/400. The weight of two-way reinforced concrete slab system and shear wall system provides large values of building weight, which could help building to resist overturning moment. It implies that the best choice of gravity system might be two-way reinforced concrete slab, and the best lateral resisting system might also be reinforced concrete shear wall. However, cast-in-place concrete takes too much time to construct and might cause delay of schedule due to weather issues. For a hotel building, the shorter time schedule of construction, the sooner owner can start making economic profits. Therefore, to design the building structure system with shorter time schedule of construction, one possible option of that is to design a steel frame or steel-concrete frame building with steel plate shear wall system. 32

35 A4-3 Shear in Shear Wall of the 3 rd floor Appendix a Case 1-X: full wind load on x-direction, acts on C.M. Case-2-X: 75% wind load acts to the floor diaphragm with 0.15 eccentricity (two sub-cases). Case-3: 75% of wind load from both direction, acts on C.M. Case-4: 56.3% of wind load from both directions and acts to the floor diaphragm with 0.15 eccentricity of both direction (four sub-cases only two sub-cases are considered). Because when wind blows from west to east, left to right. The critical eccentricity of BY is positive, which would create additive torsional shear stress while negative eccentricity of BY would help shear wall to reduce the shear stress. Therefore, ex = +/-0.15 BX, ey = BY Fig. A4-3-1 a-1

36 The torsion due to the each case can be determined by the table below: Table A4-3-1 Wind Torsion Calculation Note : Wind load = windward wind load + leeward wind load 0.15 BY = 0.15 * 49 = BX = 0.15*99.7 = 15.0 eccentricity = C.R. coordinate - wind acting point coordinate, keep "+" or "-" to calculate the Total Torsion Torsion: "-" means the torsion is clockwise, "+" mean torsion is counter-clockwise. Torsional Moment of Inertia: Table A4-3-2 Torsional Moment of Inertia Calculation Therefore, the RB * db / total torsional Moment of inertia = 272,688 / 63,833,298 = a-2

37 Direct and Torsional Shear Direct shear: V B = V * R B / (R A + R B + R C) = V * (36,262 * ) = * V Torsional shear: V T,B = T * (Ri * di) / J = *T Total Shear = Direct shear + Torsional shear Therefore, the total shears of shear wall B of the different cases can be determined as below. Table A4-3-3 Total Shear Calculation of Different Cases Therefore, the critical story wind load of shear wall B is 16.3 kips. Which is 27.3% of the story wind load on x-direction. Assume the wind load distributions of upper stories are the same as the 3 rd floors. Total wind load above the 3 rd floor is kips. Therefore, the simplified calculation of wind load on the 3 rd floor shear wall B Accumulative V 3,B = 27.3% * = kips. Accumulative M 3,B = 39,260 k-ft (from table A4-3-4) Note: The loads above 3 rd floor is resisted by 3 rd floor lateral system. a-3

38 Table A4-3-4 Therefore the wind shear load on shear wall B of the 3 rd floor is 360 kips, and moment is 39,260 k-ft. a-4

39 A4-4 Assume the 27.0% of seismic load is distributed to shear wall B. Therefore, table A4-4-1 shows the Calculation of seismic shear load on shear wall B through the whole building. Table A4-4-1 a-5

40 A 4-6 a-6

41 a-7

42 Note: all the requirements pass except the horizontal reinforcing ratio, ρt, does not pass. a-8

43 Appendix b B4-7 Computer Modeling General Information Materials: Concrete: Concrete Strength: 4000psi, 5000psi, 6000psi, 7000psi, and 8000 psi Crack state: concrete was modeled with crack factors Mass: Structural element s masses were excluded to create a building model that had only lateral effect. Beam: The beam strength was modeled as same as the slab strength. Beam size is 12 x 30 with strength factors I22 & I33 = 0.35 based on ACI Column: all columns are gravity columns. To exclude the gravity columns lateral resistance, the strength factors of I22 and I33 were set to be zero. The reason of setting strength factors is that if complex model would be required in the further analysis, it is easier to change the strength factors than remodel the gravity columns. Shear Walls: Three sizes of column were modeled in this building model. The sizes are 21x21, 20 x 24, and 24x20. Each column s concrete strength was modeled to be the same as the shear wall strength at each level. Shear walls are the lateral system of this project, it is extremely important to model the shear walls correctly. This building modeling includes three sizes of shear wall thickness, which are 10, 12 and 14. The strength of shear wall concrete varies along the vertical direction. The table below shows the strengths of concrete corresponding to the story number. SW/ Col Concrete Story Number Strength 8,000 psi Ground to 8 th floor 7,000 psi 9 th floor to 15 th floor 5,000 psi 16 th floor and above Table Shear Wall and Column Concrete Strength The shear walls were modeled as membrane element with strength factors f11 and f22 equaled to 0.70 (un-cracked strength factor). b-1

44 Floor: Building floors were set to be rigid diaphragm. The strengths of floor concrete are shown on the table below: Floor / Beam Concrete Story Number Strength 6,000psi 1 st to 8 th floor 5,000 psi 9 th floor to 15 th floor 4,000 psi 16 th floor and above B4-8 Computer Results Fig. B4-8-1 Story Shear Wind Load Case 1-X b-2

45 Fig. B4-8-2 Story Moment Wind Load 1-X b-3

46 Fig & Elevation View of Grid Line 1 (left) and Grid Line 2 (right) b-4

47 Fig B4-8-5 Shell Forces/ Stress 3D View of Wind Load Case 1-X F11 Fig B4-8-6 Shell Forces/ Stress 3D View of Wind Load Case 1-X F22 b-5

48 Technical Report 2 Gravity Load

49 Technical Report II Xyston Inn. NY C-1

50 Technical Report II Xyston Inn. NY C-2

51 Technical Report II Xyston Inn. NY C-3

52 Technical Report II Xyston Inn. NY C-4

53 Technical Report II Xyston Inn. NY C-5

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57 Technical Report II Xyston Inn. NY C-9

58 Lateral Load -Wind Technical Report II Xyston Inn. NY C-10

59 Lateral Load Wind There are several wind load cases need to be considered. Case 1: Overall Lateral windward force. Case 2: Uplift force on green roof. (20 ft. above ground level). Case 3: Uplift force on the roof of 14 th story (153 ft. above ground level). Case 4: Uplift force on the roof above 18 th story (215 ft. above ground level). Case 5: Overall Lateral leeward force. Note: wind forces acting on side ward walls cancel each other out. The internal pressures acting on the internal side of exterior wall also cancel each other out. Only the windward force and leeward force can cause building to overturn. Technical Report II Xyston Inn. NY C-11

60 Technical Report II Xyston Inn. NY C-12

61 Technical Report II Xyston Inn. NY C-13

62 Wind Load (Excel Calculation & Details-01, N-S direction) Technical Report II Xyston Inn. NY C-14

63 ********This Excel sheet might only work for risk category II ********* Technical Report II Xyston Inn. NY C-15

64 Technical Report II Xyston Inn. NY C-16

65 Table: Wind Pressure Calculation Note: Kzt = 1 Calculated in step 3-c Gf = 3.31 Calculated in step 3-d Cp From Step 6 Technical Report II Xyston Inn. NY C-17

66 Table: Story Shear and Overturning Moment - Windward Note: Story Shear = Involved Height * B * Pressure Accumulative Shear = sum of the shear from the stories above Moment = Story Shear * Height Technical Report II Xyston Inn. NY C-18

67 Table: Story Shear and Overturning Moment - Leeward Note: Story Shear = Involved Height * B * Pressure Accumulative Shear = sum of the shear from the stories above Moment = Story Shear * Height Technical Report II Xyston Inn. NY C-19

68 Wind Load (Excel Calculation & Details-02, W-E direction) Technical Report II Xyston Inn. NY C-20

69 ********This Excel sheet might only work for risk category II ********* Technical Report II Xyston Inn. NY C-21

70 Technical Report II Xyston Inn. NY C-22

71 Table: Wind Pressure Calculation Note: Kzt = 1 Calculated in step 3-c Gf = 3.31 Calculated in step 3-d Cp From Step 6 Technical Report II Xyston Inn. NY C-23

72 Table: Story Shear and Overturning Moment - Windward Note: Story Shear = Involved Height * B * Pressure Accumulative Shear = sum of the shear from the stories above Moment = Story Shear * Height Technical Report II Xyston Inn. NY C-24

73 Table: Story Shear and Overturning Moment - Leeward Note: Story Shear = Involved Height * B * Pressure Accumulative Shear = sum of the shear from the stories above Moment = Story Shear * Height Technical Report II Xyston Inn. NY C-25

74 Therefore, the base shear and overturning moment due to two orthogonal direction wind load is listed on the table below. The wind load of this project causes large numbers of story shear, base shears and overturning moments. The evaluation of lateral system shall use the shears values to determine whether the lateral system of this project, shear wall system, are adequate to carry the lateral load. And the evaluation of building structure system might also need to consider whether the structure of this project needs any design details to tight the structure down due to the large values of overturning moment. To determine the controlling lateral loading condition, the seismic force calculation is also required. The flowing pages show the calculation of seismic base on ASCE 7-10 requirements. Technical Report II Xyston Inn. NY C-26

75 Lateral Load- Seismic Technical Report II Xyston Inn. NY C-27

76 Lateral Load - Seismic Technical Report II Xyston Inn. NY C-28

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81 Seismic Load Determination and Tables Base on ASCE 7-10 Technical Report II Xyston Inn. NY C-33

82 Note: Wt is from Hand Calculation with conservative assumptions story shear = Fx = 0.01 Wx Accumulated Shear = Sum of story shear above Moment = Story shear * Height Therefore, the base shear is 111 kips due to seismic load, and overturning moment is 10,834 k-ft. due to seismic load. It comes out that wind load controls. Technical Report II Xyston Inn. NY C-34

83 Note: Forces shown on the right are not equal, and only depended on story weight (base on F = 0.01 Wx) The table below shows the comparisons between wind load and seismic load base on the base shear and overturning moment. The result shows that wind load is the controlling lateral load condition. According to base shear results, the base shear due to wind is about 1300% to 2600% of the base shear due to seismic. Technical Report II Xyston Inn. NY C-35

84 Conclusion Gravity Load: Typically, roof bay s dead load is 130 psf. And floor bay s dead load is 130 psf. The maximum possible live load in this project is 100psf. Design flat roof snow load is 20 psf, and design drift snow load can reach 55 psf; therefore, the gravity design of roof slab shall be proved to be able to carry 55 psf snow load. Lateral Load: Wind load controls the lateral loading condition. The base shear shall be designed for 2,970 kips (un-factored), and the design overturning moment is about 333,000 k-ft (un-factored) on the north-to-south direction. The base shear shall be designed for about 1,519 kips (un-factored), and the design overturning moment is about 174,000 k-ft (un-factored) on the west-to-east direction. Due to the high uplift pressures, the roof bay might need to be designed for uplift pressure. Even though the self-weight of concrete slab seems to be able to help, the uplift pressure could cause the slab to fail by surface tensile stress. Therefore, the shear reinforcement of roof slab should consider the wind uplift pressure effects. Due to the high value of overturning moment, the building might need to be tighten down. The tension in lower floors LFRS elements and non- LFRS elements shall also be considered. Technical Report II Xyston Inn. NY C-36

85 Appendix Assumption Wind Load: 1. Simplified rectangular boxes are considered as the design building shape. 2. Assume Cp = -0.9, or for uplift wind loads on both N-S and W-E directions on level line H = 20 feet. 3. Assume Cp = -0.8, or for uplift wind loads on both N-S and W-E directions on level line H = 153 feet. 4. Assume Cp = -1.3, or for uplift wind loads on both N-S and W-E directions on level line H = 215 feet. 5. Winds acts on the orthogonal directions of building exterior wall. Seismic Load: 6. Storage weight portion is negligible. 7. Partition load is 12 psf on floor from the structure drawing. 8. Use 75 psf mechanical room s live load for the equipment load in mechanical rooms. 9. Use 57 x rectangular shape for roof plan area calculation. 10. Use 140 psf for the roof over 14 th floor. 11. Building damping ratio is 5%. Technical Report II Xyston Inn. NY C-37