长安大学结构设计原理课程设计.doc

长安大学结构设计原理课程设计1. Known:（1）Span of SSB: 28m；Computed span L=26.66m；（2）Design load：Vehicular Load -；Pedestrian Load=3.0kN/m2；Coefficient of Importance 0=1.0；（3）Environment：Bridge locates in a common field, environment ; 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 diameter15.2mm;Elastic modulus ；Strand tapered of group anchors.Nonprestressed reinforcement:HRB400,fsk=400MPa,fsd=330MPa.HRB335：(d=(Ms/W-0.7ftk)/(1/A+ep/W)=(3823106/239.927106-0.72.65)/(1/876000+1156/239927000) =2359158NThe control stress of pre-stressed rebar is .The pre-stress losses are estimated as 20% ,We can get the area of pre-stressed reinforcement is:Ap=Npe/(1-0.2)con=2359158/(0.81395)=2114(mm2)To use 3 bundle 6s15.24 strands, the area of pre-stressed reinforcement is :Ap=36140=2520(mm2). And use the strand tapered group anchors, metal corrugated pipe into the hole.2) Lay out pre-stressed reinforcement.(1)Arrangement of pre-stressed rebar in mid-span section:According to the requirement of The Road Bridge Gauge,arrangement of pre-stressed reinforcement in mid-span section shown Figure 3.(2)Lay out steel cable of anchor cross-section: Fig1. End of pre-stressed beam Fig.2 anchor position in the end Fig.3 cable position in mid-spanIn order to be convenient for construction,three steel cables anchor end of beam,which fits principle of homogeneous disperse and requirement of tension.It will offer much pre-shear force by bending more depth of N1 and N2 in end of beam. The all shown Fig1 and Fig2.(3):Location and angle of steel cable from other sections.Bending shape、angle and radius of bending.Obtain interpolate curve between straight lines.In order to make pre-applied force be perpendicular to anchor bearing plate,bending angle of N1, N2, N3 are . Radius of each steel cable is: .calculating location of steel stable control points. Fig.4Calculation step can be shown example of N3,arrangement figure of bending shown Fig.4Calculating horizontal distance between lead point and anchoring point,According to the formula Ld=ccot0,we can have Ld=400cot8=2846mmCalculating horizontal distance between lead point and bending point,According to the formula Lb2=Rtan(0/2),we can have Lb2=15000tan40=1049mmThe horizontal distance between bending point and anchoring point isLw=Ld+Lb2=2846+1049=3895mmThe horizontal distance between bending point and mid-span section isXk=26660/2+256-Lw=5796mmThe horizontal distance between bending end point lead point isLb1=Lb2cos0=2098cos8=2078mmThe horizontal distance between bending end point and mid-span section.X=Xk+Lb1+Lb2=5796+2078+2078=9972mmAccording to the same theory,we can get location of control point from N1 and N2,shown following table钢束号升高值c弯起角弯起半径R支点至锚固点的水平距离弯起点至中截面距离弯止点至跨中截面水平距离N11610.00 8.00 45000156.00 -408.845850.8N2900.00 8.00 30000256.00 5794.049967.13N3500.00 8.00 15000312.00 9746.0211832.6Calculating 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 i. c=100mm.when0，0， 100mm；0when0， 8， Position and angle of steel table from all cross section计算截面钢束编号xkLa+Lbxi-xkcai=a+c跨中截面xi=0N1-408.846259.64 为负值，未弯起0.00 0.00 100.00 N25794.044173.09 N39746.022086.55 L/4截面xi=6665N1-408.846259.64 7073.848.00 551.86 651.86 N25794.044173.09 870.9581.66 12.65 112.65 N39746.022086.55 为负值0.00 0.00 100.00 变化点截面xi=9630N1-408.846259.64 10038.88.00 968.35 1068.35 N25794.044173.09 3835.967.35 246.25 346.25 N39746.022086.55 为负值未弯起0.00 0.00 100.00 支点截面xi=13330N1-408.846259.64 13738.88.00 1488.09 1588.09 N25794.044173.09 7535.968.00 764.04 864.04 N39746.022086.55 3583.988.00 356.17 456.17 (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, N2、N3 must be bend from both sides to the middle line in the main beam lab. N2、N3 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 non-prestressed reinforcementTo be satisfied with the ultimate limit state, the number of non-prestressed reinforcement are:After deciding the reinforcement number, non- prestressed reinforcement is decided according to the normal cross-sections ultimate limit state.Assume the distance from prestressed and non-prestressed reinforcements resultant force point to the cross-section bottom is a=80mm,so Assume first kind of T shape beam, according to , we can get the depth of compressive zone x1.05934.5106=22.42200x(1720-x/2)x= 71.5mmlf截面不受反摩擦影响0N21358610361150.5637822N31364210377150.3373309L/4截面xi=8165N1832111854131.596903239.22431.64640158N2842110361150.563782228.192N3847710377150.337330927.522变化点截面xi=11630N1385611854131.596903288.79191.35958522N2395610361150.563782293.076N3401210377150.337330992.21