長(zhǎng)安大學(xué)結(jié)構(gòu)設(shè)計(jì)原理課程設(shè)計(jì)-wenkub
2023-07-10 04:21:57 本頁(yè)面
【正文】 水平距離N1 45000 N2 30000 N3 15000 ③Calculating location and angle of steel cable from all kinds of sections.According to the Figure4,the distance point i to bottom of beam ai=a+ci, angle θi. c=100mm.when≤0，＝0， ＝100mm；＝0when0≤， when， ＝＝8176。cotθ0,we can have Ld=400cot8176。 bf180。(2) : Average distance of the two adjacent beams, to the middle beam is 2200mm。 TD Jack。 Relative humidity (average of per year)= 75％；（4）Materials：Tendon: low relaxation strands(17)；Standard value of tension strength ；Design value of tension strength；Nominal area= 140mm2 ,nominal 。Elastic modulus ；Strand tapered of group anchors.Nonprestressed reinforcement:HRB400,fsk=400MPa,fsd=330MPa.HRB335：(d12mm)normal value of tension strength ；design value of tension strength；Elastic modulus=105MPa。 Castinsite wet connection joint with width of 400mm。(3) : (b+6bh+12hf180。=2200mm。=2846mmCalculating horizontal distance between lead point and bending point,According to the formula Lb2=R Position and angle of steel table from all cross section計(jì)算截面鋼束編號(hào)xkLa+Lbxixkθcai=a+c跨中截面xi=0N1 為負(fù)值，未彎起 N2 N3 L/4截面xi=6665N1 N2 N3 為負(fù)值 變化點(diǎn)截面xi=9630N1 N2 N3 為負(fù)值未彎起 支點(diǎn)截面xi=13330N1 N2 N3 (4)The position and angle in flat bend zone N1, N2, N3 are in the same plane at the middle span, but at anchor terminal they are all in the middle lane, to get this result, NN3 must be bend from both sides to the middle line in the main beam lab. NN3 are take the same to bend up, and the position in flat bend as Fig. Shows. There are two curve arc, each angle is . 3)Requirement for nonprestressed reinforcementTo be satisfied with the ultimate limit state, the number of nonprestressed reinforcement are:After deciding the reinforcement number, non prestressed reinforcement is decided according to the normal crosssection’s ultimate limit state.Assume the distance from prestressed and nonprestressed reinforcement’s resultant force point to the crosssection bottom is a=80mm,so Assume first kind of T shape beam, according to , we can get the depth of pressive zone x106=2200x(1720x/2)x= As=(fcdbf? x fpdAp)/fsd=Select 5Φ18mm HRB400 As= ,lay out one line,space is 75mm. as = 45mm,shown following Figure.6. Calculation of geometric features of main beam crosssection . 跨中截面 分塊面積Aai至梁頂?shù)木嚯xyi對(duì)梁頂端的面積距自身慣性矩yuyiix=ai*(yuyi)2截面慣性矩804000469700000 01800000 170019617150014788136703 450082850 14575888316 804000469700000 018000 2590963026 170002056102257 00000 489629913 4647215515 876000544476180000 01800000 1700015266416602 0 000 496109913 15459759855 L/4截面 分塊面積Aai至梁頂?shù)木嚯xyi對(duì)梁頂端的面積距自身慣性矩yuyiix=ai*(yuyi)2截面慣性矩804000469700000 01800000 010220921733 10074224810 804000469700000 018000 0 9800063214 000000 9942962254 876000544476544000 018000 0 10693202823 000 0 10836309593 變化點(diǎn)截面 分塊面積Aai至梁頂?shù)木嚯xyi對(duì)梁頂端的面積距自身慣性矩yuyiix=ai*(yuyi)2截面慣性矩804000469700000 018000 0 6003355917` 5917192079 804000469700000 018000 0 5756160650 000 00 5840093765 876000544476544000 018000 0 6440815423 000 0 6527012637 支點(diǎn)截面 分塊面積Aai至梁頂?shù)木嚯xyi對(duì)梁頂端的面積距自身慣性矩yuyiix=ai*(yuyi)2截面慣性矩1128000 018000 0 1116461 1128000 018000 0 000 00 1139723 1128000695640000 018000 0 000 00 1139723 各控制界面不同階段的截面幾何特性匯總 AyuybepIWu=I/yuWb=I/ybWp=I/ep 1116461 1900695504 1139723 194419
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