ACI318 轨道基础计算书.docx

1、ACI318 轨道基础计算书封面 COVERCONTENTCONTENT I1 GIST OF CALCULATION 12. MATERIAL AND DESIGN LOAD 12.1 MATERIAL 12.2 DESIGN LOAD 23 Analysis Modal 34 Analysis Result 44.1 PROJECT : 44.2 ANALYZE CAPACITY OF MAJOR AXIS BENDING OF RC-WALL. 54.3 ANALYZE CAPACITY OF MINOR AXIS BENDING OF RC-WALL 64.4 ANALYZE SHEA

2、R CAPACITY OF RC-WALL 71 GIST OF CALCULATION(1)Building Code Requirements for Structural Concrete(ACI318-11) and Commentary(2)International Building Code(2006)(3)Minimum Design Loads for Buildings and Other Structures（ASCE/SEI7-10）(4) Principles of Foundation Engineering(Braja M.Das)(5) DESIGN ENTRU

3、STING DOCUMENT AND THE THE OWNERS DESIGN REQUIREMENTS(6) DESIGN INFORMATION PROVIDED BY RELATED DESIGN DISCIPLINES2. MATERIAL AND DESIGN LOAD 2.1 MATERIALCONCRETE IS SELECTED FROM IRANIAN LOCAL STANDARDS，CONCRETE MATERIAL IS SELECTED FROM THE LOCAL IRANIAN .THE MATERIAL SHOULD BE ANTICORROSIVE IN SO

4、IL COMPLY WITH LOCAL CRITERIONS.BASE CUSHION ADOPT C15(2100PSI) LEVER CONCRETE(fc=15MPa)THE MAIN STRUCTURE C35(5000PSI) LEVER CONCRETE(fc=35MPa)AMONG THEM, THE CONCRETE STRENGTH GUARANTEE RATE IS 91%.REINFORCED STEEL IS ADOPTED OF ERANIAN STANDARDS,REAINFORCED MATERIAL IS SELECTED FROM THE LOCAL IRA

5、NIAN .THE REINFORCED STEEL IS ADOPTED AIII LEVER CONCRETE (fy=400MPa (50000PSI),fu=600MPa(85000psi) SYMBOLS USE THE NON-REINFORCED STEEL IS ADOPTED AII LEVER CONCRETE (fy=300MPa (42000PSI),fu=500MPa(7000psi) SYMBOLS USE SECONDARY GROUT:HIGH INTENSITY GROUTING MATERIAL.ANCHOR BOLTS MATERIAL SHALL BE

6、USE AII REINFORCED STEEL FOR BAR DIAMETER LESS THAN 32mm. FOR DIAMETERS LARGER THAN 32mm OR EQUAL TO 32mm ,AIII OR ST-52 REINFORCED STEEL COULD BE APPLIED.EMBEDDED PARTS :ST37-2 STEEL.2.2 DESIGN LOAD2.2.1 STANDARD VALUE OF UNIFORMLY DISTRIBUTED LIVE LOAD ON FLOOR AND ROOF PLATFORM:3.5KN/m22.2.2 BASI

7、C WIND VELOCITY :30M/S2.2.3 BASIC SNOW PRESSURE:0.25kN/m22.2.4 SEISMIC LOADS:1) THE SOIL CLASS IS B,ACCORDING TO IBC 2006.2) S1 IS CONSIDERED AS 0.8,AND Ss IS EQUAL TO 1.3.3) USE IBC 2006 AS THE BASE FOR SEISMIC LOADING.(SOIL DEPTH OF 30 M SHEAR WAVE VELOCITY RANGED BETWEEN 375m/s AND 750m/s)PICTURE

8、2.1 THE SKETCH MAP OF RESPONSE SPECTRUM2.2.4 DISTRIBUTION OF WHEEL PRESSURE：PICTURE2.2 WHEEL PRESSURE DISTRIBUTION OF RECLAIMER (ACT ON QU100 RAIL) 3 Analysis Modal The programme for analysis is MIDAS-GEN.PICTURE3.1 Full-page proof of analysis modalPICTURE3.2 Active earth pressure for earthquake con

9、ditions (kN/m2)PICTURE3.3 WHEEL PRESSURE DISTRIBUTION OF RECLAIMER (kN)4 Analysis Result4.1 PROJECT : *.DESIGN CODE : ACI318-05, *.UNIT SYSTEM : kN, m *.MEMBER : WALL ID = 1, WALL MARK = wM0001 STORY = B2, LCB = 2, POS = BOT *.DESCRIPTION OF WALL DATA Wall Height (HTw) = 0.500 m. Wall Length (Lw) =

10、21.600 m. Wall Thickness (hw) = 0.600 m. Concrete Cover to C.O.R. (dw) = 0.051 m. Concrete Cover to C.O.R. (de) = 0.051 m. Concrete Strength (fc) = 34473.797 KPa. Modulus of Elasticity (Ec) = 27789387.580 KPa. V-Rebar Strength (fy) = 413685.566 KPa. H-Rebar Strength (fys) = 275790.378 KPa. Modulus o

11、f Elasticity (Es) = 199948023.746 KPa.4.2 ANALYZE CAPACITY OF MAJOR AXIS BENDING OF RC-WALL. ( 1). Compute design parameters. -. Ast = SUM Asi = 0.0579 m2. -. Ag = Lw*hw = 12.9600 m2. -. Rhot = Ast/Ag = 0.004469 -. esu = fy/Es = 0.002069 -. beta1 = MAX 1.05-0.05*(fc/1000), 0.65 = 0.8000 (2 ). Comput

12、e concentric axial load capacity. -. Po = (0.85*fc)*(Ag-Ast) + fy*Ast = 402025.57 kN. -. Maximum Axial Load : Pomax = 0.80*Po = 321620.46 kN. -. Maximum Axial Tension : Pt = -fy*Ast = -23959.34 kN.(3 ). Compute capacity of concrete stress block.-. a = beta1*c = 21.382 m.-. Cc = 0.85*fc*a*hw = 375932

13、.38 kN.-. MnCc = Cc*(Lw/2-a/2) = 40957.06 kN-m.-. Ps = SUM Fsi(k) = 18620.88 kN. -. MnPs = SUM Fsi(k)*(Lw/2-di(k) = 34111.39 kN-m.(4 ). Compute nominal capacity(Pn,Mn) of given neutral axis. -. Pn = Cc + Ps = 394553.26 kN. -. Mn = MnPs + MnCc = 75068.45 kN-m. -. Pn Pomax - Pn = Pomax = 321620.46 kN.

14、 (5). Compute strength reduction factor. -. et = 0.00000 -. et_min = 0.00207 -. et_max = 0.00500 -. et phi =0.650 (6 ). Compute axial load and moment capacity(phiPn,phiMn). -. phiPn = phi*Pn = 209053.30 kN. -. phiMn = phi*Mn = 48794.49 kN-m. (7 ). Check ratios of axial load and moment capacity. -. R

15、at_P = Pu/phiPn = 0.023 O.K. -. Rat_M = Mu/phiMn = 0.019 O.K.4.3 ANALYZE CAPACITY OF MINOR AXIS BENDING OF RC-WALL(1 ). Compute design parameters. -. Ast = SUM Asi = 0.0579 m2. -. Ag = Lw*hw = 12.9600 m2. -. Rhot = Ast/Ag = 0.004469 -. esu = fy/Es = 0.002069 -. beta1 = MAX 1.05-0.05*(fc/1000), 0.65

16、= 0.8000( 2). Compute concentric axial load capacity. -. Po = (0.85*fc)*(Ag-Ast) + fy*Ast = 402025.57 kN. -. Maximum Axial Load : Pomax = 0.80*Po = 321620.46 kN. -. Maximum Axial Tension : Pt = -fy*Ast = -23959.34 kN.(3). Compute capacity of concrete stress block. -. a = beta1*c = 0.043 m. -. Cc = 0

17、.85*fc*a*Lw = 27353.96 kN. -. MnCc = Cc*(hw/2-a/2) = 7615.10 kN-m. -. Ps = SUM Fsi(k) = -10943.74 kN. -. MnPs = SUM Fsi(k)*(hw/2-di(k) = 3243.49 kN-m.(4 ). Compute nominal capacity(Pn,Mn) of given neutral axis. -. Pn = Cc + Ps = 16410.22 kN. -. Mn = MnPs + MnCc = 10858.59 kN-m.( 5). Compute strength

18、 reduction factor. -. et = 0.02750 -. et_min = 0.00207 -. et_max = 0.00500 -. et et_max - phi =0.900 (6 ). Compute axial load and moment capacity(phiPn,phiMn). -. phiPn = phi*Pn = 14769.20 kN. -. phiMn = phi*Mn = 9772.73 kN-m.( 7). Check ratios of axial load and moment capacity. -. Rat_P = Pu/phiPn

19、= 0.331 O.K. -. Rat_M = Mu/phiMn = 0.336 O.K.4.4 ANALYZE SHEAR CAPACITY OF RC-WALL(1 ). Check permissible maximum shear strength. ( LCB = 5, POS = J ) -. Applied axial force : Pu = 3236.74 kN. -. Applied bend. moment : Mu = 2240.24 kN-m. -. Applied shear force : Vu = 1747.09 kN. -. d = 0.8*Lw = 17.2

20、80 m. -. Vn = 10*SQRT(fc)*hw*d = 50547.43 kN. -. phi = 0.75 -. phiVn = phi*Vn = 37910.58 kN. -. Vu O.K. (2 ). Compute shear strength of concrete. -. Term1 = Lw*(1.25*SQRT(fc) + 0.2*Pu/(Lw*hw) = 81325.64 (by lb, in). -. Term2 = Mu/Vu-Lw/2 = -374.71 (by lb, in). (0.6*SQRT(fc) + Term1/Term2)*hw*d - Not

21、 applied because Term2 is smaller than 0. -. Vc = MAX 3.3*SQRT(fc)*hw*d + Pu*d/(4*Lw), 0 = 17328.00 kN.( 3). Compare factored shear force with phiVc/2. -. phi = 0.75 -. phiVc/2 = phi*Vc/2 = 6498.00 kN. -. Vu Minimum shear reinforcement. -. Allowable maximum spacing of reinforcement. Vertical : s1 =

22、MIN 3*hw, 18 in = 0.457 m. Horizontal : s2 = MIN 3*hw, 18 in = 0.457 m.( 4). Check required ratios of vertical and horizontal shear reinforcement. -. RhoN = 0.0025 -. RhoH = 0.0025 -. Rhot = Ast/Ag = 0.0045 RhoN - O.K.PICTURE4.1 THE RELATIONSHIP BETWEEN AXIAL FORCE(P) WITH MOMENT(Mx or My)PICTURE 4.2 THE RELATIONSHIP BETWEEN AXIAL FORCE WITH MOMENT OF MAJOR AXISPICTURE4.3 THE RELATIONSHIP BETWEEN AXIAL FORCE WITH MOMENT OF MINOR AXIS