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预应力混凝土题解.doc

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1、P286.11-1解:截面参数: =471mm2, =1040 N/mm2, =1470 N/mm2 PApyfptkf=2.0 N/mm2(预应力筋) , =3.25 N/mm2sE510cE410=452 mm2, =300 N/mm2, =2.0 N/mm2(非预应力筋)yfs5= =6.15Esp4510.3=200150- 522+(6.15-1)452=30204mm 2 nA=0.65 =0.651470=956 N/mm2 conptkf1. 计算:JM12 锚具属有顶压的夹片式锚具 =5mm,1l a= = 2.0 =55.6 N/mm2 1lSEa8055102. 计算:

2、=0.0014,直线配筋 =02l=0.001418+0=0.02520.2x= ( )=9560.0252=24.1 N/mm2 2lcon= + =55.6+24.1=79.7 N/mm2 l1ll3. 计算:超张拉 =0.035 =0.035956=33.5 N/mm2 4l4lcon4. 计算:5l= = =13.66 N/mm2 pcnploA)(471302).956(/ =13.66/40=0.34150.5,公式可用uf= = =0.0153nsp23024571= = =106.2 N/mm2 5l18cupf153.8= + =33.5+106.2=139.7 N/mm2

3、l4ll= + =79.7+139.7=219.4 N/mm2 80 N/mm 2,可用。 lllP286.11-2解:截面参数: =699.8mm2, =1110 N/mm2, =1570 N/mm2 PApyfptkf=2.05 N/mm2(预应力筋) , =3.25 N/mm2sE510cE410=314 mm2, =300 N/mm2, =2.0 N/mm2(非预应力筋)sAyfsE510= =6.15, = =6.31Es4510.3Ep45.3=240200+314(6.15-1)+699.8(6.31-1)=53333mm 2 0=0.75 =0.751570=1178 N/mm

4、2 conptkf1. 计算:设 =1mm, = = 2.05 =4.0 N/mm2 1la1lSa505102. 计算: =2 =220=40.0 N/mm2 3L3Lt3. 计算:普通松弛超张拉4=0.44Lconptkconf)5.0(=0.40.9(0.75-0.5)1178=106.0 N/mm2 = + +50% =4.0+40.0+0.5106.0=97.0 N/mm2 l1l3L4L4. 计算:5l= = =0.009502AsP5318.69= =PC0plcon)( 538.69)7(=14.2 N/mm20.5 =0.5(0.7540)=15.0 N/mm2 公式可用cu

5、f= = =155.0N/mm2 5l 1/84p09.1/24=50% + =0.5106+155=208N/mm2 l4Ll= + =97+208=305N/mm2100 N/mm 2,可用。 llP305.12-1解:(1) = = =9.90N/mm2pcnslplcoA5)( 30245.167).96(=30204+6.15471=33101 mm2 0= = =9.9033101=327.7KNpoNpc0= =106.2 N/mm2 s5l= = - + =956-219.4+6.159.90=797.5 N/mm2pocnlEpc(2) =2.39N/mm2 tkf=( +

6、) =(9.90+2.39)33101=406.8KNcrNpctkf0A裂缝出现后瞬间有= - =106.2- =106.2-85.7=20.5N/mm2 s5lsptkf045271309.(说明非预应力筋受压)= + =797.5+85.7=883.2 N/mm2 (说明预应力筋受posptkAf0拉) (3) = + = 则 pospoNpyf= +( - ) ( + )yfoAs=327.7+(1040-797.5)(471+452)=551.5KN(4)图略P305.12-2解:(1)N=【(1.2300+1.4100) , (1.35300+1.40.7100) 】max=【50

7、0,503】 max=503KN+ =1040471+300452=625.4KN =503KNpyfAyfs N正截面强度满足要求(2)裂缝控制等级为二级, =30204+6.15471=33101 mm2 0A= = =9.90N/mm2 pcnslplco5)( 3045.167).296(= = =12.08N/mm2ck0AN310)(3= = =9.06N/mm2 cq0- =12.08-9.90=2.18N/mm2 =2.39N/mm2 ckp tkf- =9.06-9.90=-0.84N/mm20,裂缝控制满足要求 q(3) = = = =14.9 N/mm2cpnpcoA30

8、24719560.8 =0.826.8=21.44 N/mm2满足要ckf求(4)JM12 锚具 D=100mm,垫板厚 20mm,按照 45 角扩散计算= 15394mm2,简化为矩形截面 20076计算, =15200 lA4)201( lAmm2=20076- 13076 mm2 ,预应力钢筋形心中心距离构件底边尺ln52寸为 125mm( +76)=114mm,所以 =200376=45600mm2 76bA= = =1.732, =1.0llbA3c=1.2 =1.2956471lFconp=540.3KN1.35 =1.351.01.73219.113076clflnA=584.0

9、KN满足要求间接钢筋采用 4片焊接钢筋网片,第一片保护层厚 25mm,钢筋网片间距为 60mm, HPB235规格 8 钢筋,两垂直方向各 4根,水平方向长=170mm,垂直方向长 =230mm,保护层=15mm。按照同心、对称原则取1l 2l=1701142=38760 mm2 =45600mm2 corAbA =15200 mm2 满足条件l= = =1.597corlcr15203876= =3.44%0.5%满足条件, =1.0,v14.4.4 有0.9( +2 )clfvcoryflnA=0.9(1.01.73219.1+21.03.44%1.597210)13076=660.9KN

10、 =540.3KN满足要求。 lF钢筋网片布置区域长 h=25+360=205mm =170mm,满足要求。1l图略。P305.12-4解:(1)N=【(1.2500+1.4160) , (1.35500+1.40.7160) 】max=【824,832】 max=832KN+ =1110699.8+300314=871KN =832KNpyfAyfs N正截面强度满足要求(2)裂缝控制等级为二级 = = =10.5N/mm2 pc05)Aslplcon( 533148.69)0178(= = =12.4N/mm2ck0N3)6(3= = =10.6N/mm2 cq0A510)4.(3- =1

11、2.4-10.5=1.9N/mm2 =2.39N/mm2 ckp tkf- =10.6-10.5=0.1N/mm20,裂缝控制不满足要求 q(3) = = =cp0)(Aplcon538.69)71(=14.2 N/mm20.8 =0.820.1=16.1 N/mm2ckf满足要求13.1解:材料 混凝土预应力钢筋(螺旋肋钢丝)1.材料及几何参数见下表其中: 、 取值为相应翼缘厚度再加上加腋高度的一半。fhf2.换算截面分 7个部分,按下表计算-腹板延伸上、下翼缘矩形; 1-压翼缘扣除腹板部分的两矩形; 2- 压翼缘加腋处两三角形; 3-拉翼缘扣除腹板部分的两矩形; 4-拉翼缘加腋处两三角形;

12、 5-受压预应力筋; -受拉预应力筋。 6 7编号 iA(mm 2)(mm1y)iiyAS(mm 3i0(mm)20)(iiyA(mm 4)(mm 4)iI对分部形心强度(N/mm 2)C40混凝土 =30cuf=19.1 =20.1cfk=1.71 =2.01t tf=26.8 tk=2.39ckf =1570ptkf=1110y=410pf弹性模量(N/mm 2)=3.25104cE=2.05105 =6.31sEE截面尺寸(mm)及面积(mm 2)=360 =80 =280fbfb=120 =125 =800 fhfh=750 =50 =200a36 5 =707 mm2HpA9 5 =

13、177 mm2距底边 ) 1 80800=64000400 25.610616 16.3841063413106 228080=22400760 17.024106-344 2650.73106121063 8021=11200693 7.762106-277 859.361064.0106 4 200100=2000050 1.0106366 2679.1210617.01065 5021=5000117 0.585106299 447.01060.7106 6 177(6.31-1) =940780 0.733106-364 124.55106 7 707(6.31-1)=375450 0

14、.188106366 502.87106 =1272940A52.8921067280.0141063446.71060y= 127948.56=416mm 0y=800-416=384mmp=416-50=366mm p=384-20=364mm=(7248+3447)106=1072710 6mm4 0I3.预应力损失计算= =0.75 =0.751570=1178N/mm2 conptkf=01l= =2 =220=40 N/mm2 3llt= =0.44ll conptkconf)5.0(=0.40.9(0.75-0.5)1178=106.0 N/mm2 = = +50% =40+0.

15、5106.0=93.0 N/mm2 ll3l4l=( - ) +( - )poNconlpAconlpA=(1178-93)(707+177)=959140N=( - ) -( - ) /poeconlpyconlpypoN= =220mm95140)3647)(317( =pc00IyeNApoo= + =7.53+0.0197366=14.73 N/mm2(压)12794561723= =7.53-0.0197364=0.36 N/mm2 (压)pc00IyeApoo= = =0.0056 = = =0.00140s127940Asp1794= = =168 N/mm2 5l84cupf0

16、56.3718= = =47 N/mm2 5l120cupf14.24=50% + =0.5106.0+168=221 N/mm2 l4l5l=50% + =0.5106.0+47=100 N/mm2lll= + =93+221=314 N/mm2 100 N/mm 2,可用。 lll= + =93+100=193 N/mm2 100 N/mm 2,可用。 lll13.2解:内力计算 = (1.215+1.45.4)11.652=434KN.mM81= (1.215+1.45.4)11.35=145KNV2= (15+5.4)11.652=346KN.mk81= (15+0.55.4)11.6

17、52=300KN.mqM(1) = - =1178-193=985 N/mm2pocnl- =985-410=575 N/mm2(受拉)yf=xfcpypopbAA)(= =129mm =120mm 为第二类 T形3601.975fh=xbfbAAfcffcpypopy )()(1= 80.9120)836(.7501=160mm =40mm 满足公式条件a= - =1178-314=864 N/mm2, pocnl=1.0, =0.8, =0.003311cu= = =0.406bcuspoycuEf02. 03.15.286403.1.=0.406750=305 满足公式条件b0hx= +

18、 ( )-( - ) ( )uM1cf)2(0x1cf)2()0fff hb poyfpApah0=19.180160(750-0.5160)+19.1280120(750-0.5120)-575177(750-20)=164+443-74=533KN.m =434KN.m满足要求M配筋率要求 验算见下述(2)ucr(2) =( ) +( )poNlconpAlconpA=(1178-314)707+(1178-193)177=785193N=( ) -( ) /poelconpylconpypoN= =204mm78519336417)(60)31478( =pc00IyeNApoo= +

19、=12.38 N/mm2 1279485361724=42, =0.1560.2,查得 =1.35,计算bf/hf/ m=(0.7+120/800)1.35=1.15=( + tkf) =(12.38+1.152.39)crMpc0yI 416072=390KN.m =533KN.m纵筋配筋率满足要求u= = =13.42 N/mm2ck0Iy6107243= = =11.63 N/mm2 cq0IM6- =13.42-12.38=1.04N/mm2 =2.39N/mm2 ckp tkf- =11.63-12.38 =-0.75 N/mm2 0 裂缝控制满足要求 cqp(3) = =6.125

20、6.0bhw8049=145KN0.2 =0.21.019.18075010-3=229.2KNVcfb0h截面可用=785KN0.3 =0.319.112729410-3=729.4KN,取poNcf0A=729.4KN,另外有= =0.13 5=351mm = =325mmtyldftkpe01.29378al21350则 =0.05 ( / )=0.05729.4(325/351)=33.8KNpVpoNalty0.7 + =0.71.718075010-tfb0h3+33.8=105.6KN =145KNV需要计算配箍筋,箍筋采用 HPB235级钢筋 =210N/mm2 yvf/ =(

21、 -0.7 - )/(1.25 )svAtfb0hpyvf0h=(145-105.6)103/(1.25210750)=0.200mm2/mm选用双肢 8( 6亦可) , =100.6/0.200=503mm,满足构造取s8250= / =100.6/(80250)svAbs=5.0310-3 =0.241.71/210=1.9510-3 min,sv最小配箍率满足(4)略(5) =2 = =15.6mm2f028IEleNcpo642107105.35789=0.85 =0.853.2510410727106=2.9631014N.mmsB0IEc= =2.9631014 =1.587101

22、4N.mmskqBM30= = 30.8mm1fl24851426587.3= - =30.8-15.6=15.2mm /400=116400/400=29mm,满足f2f l(6)放张后上边缘(预拉区)= = -pc00IyeNApoo 127945061072384=7.534-0.01967384=-0.020 N/mm2(拉)放张后下边缘(预压区)= =7.534+0.01967416=15.717 N/mm2 (压)pc00IyeNApoo梁自重=(127294-3754-940)110 -625=3.07KN/m,设动力系数为 1.5,有负弯矩 = 1.53.0722=9.21KN.mkM21上边缘拉应力= - =-0.020- ctp0Iyk 61072384.9=-0.350N/mm2(“-”为受拉) =2.01N/mm2 ,可以tkf下边缘压应力= + =15.717+ cp0IyMk 610724.9=16.074 N/mm20.8 =16.08N/mm2 可以ckf= =0.144%0.2%,不满足。Aps9403751可在预拉区设置非预应力钢筋来满足构造要求,设置 2 10即可,=0.27%0.2%。ps

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