板式樓梯梯板厚度確定（The thickness of the plate staircase is determined）.doc

1、板式樓梯梯板厚度確定（The thickness of the plate staircase is determined）collection of documents : the thickness of the plate staircase is determinedThe thickness of the plate staircase is determinedTime:The plate thickness of the board is 1/25 1/30l, and the net step of the board stair is 4M, but many frame s

2、tructures are larger than 4M, and how should the thickness of the plate be determined? (only step on the side of the ladder beam, and on the other side of the rest platform)ASA:According to the actual plate span of 1/25, if the thickness of the plate is greater than 250mm, it is indicated that the s

3、tructure plan is changedHHH:It is said that the net step of the board stair is mainly from an economic point of view, and the actual project is 6 meters common. When the span is large, the thickness of the board is mainly controlled by deflection, and 1/25-1/30l is reasonableNewx:A 3,600-span stairc

4、ase, the thickness of the board should be made of 140, seems a little too thick,The following is attached with the slide book of the table: tb-2I. basic information:Project name: project 1The stairs type: plate type A () bearing condition: elastic on both endsThe horizontal projection length of the

5、ladder is Lo1 = 3600mm, which is equal to Lo1 = 3600mmNet width of ladder plate B = 1200mmSlab thickness h1 = 140mmHeight Hv = 1500mm staircase series n = 10Lower support width d1 = 200mm upper support width d2 = 200mmLinear constant load Pk = 0.00 kN/mUniform live load qk = 3.00 kN/m bits of c = 0.

6、7 bits of q = 0.6Thickness of surface layer c1 = 25mm ceiling thickness c2 = 20mmSurface, ceiling density gamma c = 20.00 kN/m stair weight density gamma b = 25.00 kN/mStrength grade of concrete: C25Fc = 11.94 ft = 1.27 FTK = 1.78 N/mm Ec = 27871 N/mmSteel bar intensity design value fy = 210 n/mm Es

7、 = 210000 n/mmThe longitudinal reinforcement points to the near edge distance as = 25mm2. Calculation results:1. Stair geometry parameters:Step height t = Hv/n = 1500/10 = 150.0 mmStep width w = Lo1 / (n - 1) = 3600 / (10-1) = 400.0 mmBench ramp Angle alpha = ArcTan (t/w) = ArcTan (150.0/400.0) = 20

8、.6 Ladder section of the length of the inclined plate Lx = Lo1 / cosine alpha = 3600 / Cos20.6 = 3845 mmLadder plate thickness to the vertical height of the h1 = h1 / cosine alpha = 140 / Cos20.6 = 150 mmT = (T + 2 * h1 ) / 2 = (150 + 2 * 150) / 2 = 225mmLadder plate effective height: h1o = h1 - as

9、= 140-25 = 115mm2. Load design value:(1) standard value of the uniformly distributed value:1.Gk1 = gamma b * T = 25.00 * 0.225 = 0.225 kN/m 2.Gk2 = gamma c * c1 * n * t + (n - 1) * w / Lo= 20.00 * 0.025 * 10 * 0.150 + (10-1) * 0.400 / 3.600 = 0.71 kN/m The top of the ceiling: gk3 = gamma c * c2 = 20

10、.00 * 0.020 = 0.40Lx/Lo = = gk3 gk3 * 0.40 * 3.845/3.600 = 0.43 kN/mGk = gk1 + gk2 + gk3 = 6.75 kN/m (2) design value of uniformly distributed load:Controlled by the live load load design value of p (L) = 1.2 + 1.4 * * gk qk = 12.30 kN/m By constant load control load design value of p (D) = 1.35 + 1

11、.4 * * gk bits of c * qk = 12.05 kN/m The most unfavorable load design value of p = Max p (L), p (D) = 12.30 kN/m 3. Calculation of shear bearing capacity of oblique section:Vmax is less than or equal to 0.7 * ft * b * hoVmax Lo * cosine alpha = = 0.5 * p * 0.5 * 12.30 * 3600 * Cos20.6 = 22137 * 0.9

12、363 = 20727 n7 * ft * b * ho = 0.7 * 1.27 * 1000 * 115 = 102311NV Max = 20727N, meet the requirements.4. Calculation of bending bearing capacity of normal section:Calculate span L = Min Lo + (d1 + d2) / 2, 1.05 * Lo= Min 3800, 3780 = 3780mmAcross the Mmax = p * L 2/10 = 12.30 * 12.30 2/10 = 17.572 k

13、N MAs = 541 mm rho = 0.47%() 12 175 (As = 646)Bearing Mmin = p * L 2/24 = 12.30 * 12.30 2/24 = 7.322 kN MAs = 311 mm rho = 0.27% rho min = As * = 381 mm 0.27%Phi 10 170 (As = 462)5. Cross-medium deflection calculation:Mk - - the bending moment value calculated according to the standard combination o

14、f the load effectThe moment value of the quasi-permanent composite calculated by the load effect(1) parameters of deflection calculation:Mk = 13.93 kN M Mq = 12.21 kN M(2) under the combination of the standard combination of the load effect, the short-term stiffness of the bending member:1. The diff

15、erential coefficient of tension of the longitudinal tensile steel reinforcement is calculated according to the following formula:Psi = 1.1-0.65 * FTK/rho te/sigma sk (concrete specification 8.1.2-2)Sigma sk = Mk / 0.87 / ho/As (concrete specification 8.1.3-3)Sigma sk = 13929097/0.87 / 115/646 = 215

16、n/mmRectangular cross section, Ate = 0.5 * b * h = 0.5 * 1000 * 140 = 70000 mm Rho te = As/Ate (concrete specification 8.1.2-4)Rho te = 646/70000 = 0.92%Psi = 1.1-0.65 * 1.78/0.00923/215 = 0.518The ratio of the elastic modulus of the steel to the concrete modulus isAlpha E = Es/Ec = 210000/27871 = 7

17、.53The ratio of the compression flange area to the effective area of the abdominal plate gamma f :The rectangular cross section, gamma f = 0The ratio of steel reinforcement to longitudinal tensile reinforcement is equal to As/b/ho = 646/1000/115 = 0.00562The Bs of the bending member of the reinforce

18、d concrete is calculated by the formula (8.2.3-1 of concrete specification) :Bs = Es * As * ho 2 / alpha bits of 1.15 + 0.2 + 6 * E * rho/(1 + 3.5 gamma f )Bs = 210000 * 646 * 115 m 2 / * 0.518 + 0.2 + 1.156 * 7.53 * 7.53 / (1 + 3.5 * 0.000) = 1708.86 kN M (3) considering the influence of long-term

19、effect combination of load on the deflection influence coefficient theta:According to section 8.2.5 of concrete specification, when rho = 0, theta = 2.0(4) the long-term stiffness B of the flexural member can be calculated according to the following formula:B = Mk / Mq * (theta - 1) + Mk * Bs (concr

20、ete specification 8.2.2)B = 13.93 / (2-1) + 12.21 * 12.21 * 1708.86 = 910.47 kN M (5), midspan deflection f = 3.0/384 * Qk * L 4 / BlF = 0.00781 * 9.75 * 9.75 * 1000 = 4/910.47 17.1 mmF/Lo = 1/211Waterdrop.I would like to ask you, are the stair stair suitable for the design of both ends? Should cons

21、ider the fixed end effect, from the textbook formulas: M = QL2/10 can be seen, so I calculate more commonly is calculate by elastic and increase bearing reinforcement in reducing cross to determine, I usually take 1/30 for ladder section thickness, that is enough, there is no need to big.Wj4112:How

22、much thickness is related to the load size. We made the stairs 4 kpa live load, span of 5.8 meters, the thickness of 250, span and large its very difficult. 1/25 1/30 is only reference, as a general rule, the larger the span, the greater the thickness growth. Usually beam according to the approximat

23、e hinge (1/10), according to the structure with plate of negative moment reinforcement and liang of the torsion bar, there is no need to calculated according to the fixed end. In fact, according to the fixed end calculating reinforcement instead be unsafe, so is insufficient bending of reinforcement

24、, negative bent bar usually wont pull through, internal force redistribution effect is limited.Wwily:The stair inclined board is calculated by two ends consolidation calculation is unfavorable, also did not see what book so calculateThe previous teaching material was to take 1/10 of the slanting boa

25、rd, which was to say that the constraint of the support was considered, but some of the new textbooks had been changed to 1/8, which is pure and simple. The reason is simple, the span of the stairs, the greater the restriction effect between the plate and beam, the more obvious, calculated on 1/10,

26、for small span board, is thought to be less than 3 meters is not considered before, but now common plate stairs span larger, was calculated at 1/10, so you should consider beam subjected to torsion. Because the field trip found a lot of cracks in the board, resulting in the decoration cracking.Board

27、 span 1/25 1/25, we generally take 1/30, because stair ramp is different from general tablet, southeast university has been published in journal of building structures ride man would be an experimental results, which consider the ladder board sawtooth effect on the deflection, prove that ladder plat

28、e deflection is more abundant.Goodwill:Its very common to have a level of over 4m,Are generally made of plate, the plate thickness is generally 1/30. Also didnt see whats the problem, the beam end consolidation effect is generally considered, because I was the textbooks are written so, calculate by

29、1/10 from bending, actual reinforcement is big on the board and board band, is supposed to be unreasonable, but with bigger can control cracks, so has been designed in this way.Yxm:When the rest platform and floor panel connected with the ladder beam are in the present casting, the ladle support sha

30、ll be elastic, and the ladle shall be thickT = ln / 25 ln / 30, ln is the horizontal projection length of the ladder.JGJG:The actual reinforcement is under the board and the board is the same size, should be unreasonable, but the bigger point can control the crack, so the design is always so.This is

31、 also a problem for me, thank you!Scoobydoo:More than 4m suitable for girder stairs, the structure is reasonable, and the brokerNhlyk:I also think the thickness of the plate is 1/30.flysky2004Please support Mmin = p * L 2/24 how come?stirrupI think it should be on both sides to consider as is consol

32、idation, so will get P * L 2/24QNXXPlate thickness is mainly deflection control (calculated according to Morgain)yaokdThe deflection of the board stair is not found in the general book data, the thickness of the plate will be very large according to the general simple support beam calculation. This is the case with Morgain, even if the elastic support has a large deflection. It is reported that yan laodi is studying this problem, the original solutio