型钢截面计算例题

1、【例5-1试确定截面尺寸及配钢如下图所示的型钢混凝土梁所能承受的最大极限弯矩。混凝土 C30,钢 筋为HPB235级钢筋，型钢Q235钢。习题4.1图解选择I22aAs/ =Z?xr = 110 x12.3 = 1353mm2= 7 5mmh5 = 220,= 4213?己216, A, = As = 402mm2fy = fy = 2 ION/，力=215N/ww2,/. = 14.3N/ mm2 as = (660 - 220 - 50)/?/? = 330mm. Ess = 2.06xl05 N / nwr计算受压区高度：当中和轴在型钢翼缘上通过时，有=fA+一 只A = fAf+fMs

2、X= =一而一77 = 18O.677215x1353 + 215x7.5x22014.3x250 x = 180.6 0.8q； = 0.8x330/?/? = 264属于第二种情况，即中和轴不通过型钢，此时人4 + 九&4 /A fb 215x421314.3x250nun = 2534 m0.86/0.003氏 s0.8x330 nrnun = 196 x0.8(/?-6/J _1 + L 1+。0.003E”0.003x2.06x105所以，不考虑型钢上翼缘的作用，重新计算x“/皿+叫 + & = 215x(1353 + 7.5x22。) =】806_fb14.3x250此截面所能承受

3、的极限弯矩虬=AA(/?-ar-as)+fshs(% + 号)+：A(4-) + fMs -1) = 210 x402x(600 35 330) + 215x220 x(1353+ 7；220)+ 210 x 402 x (330 -35) +14.3 x 250 x180.6 x(330-气当州. 7= 303kN,m【例5-2计算截面尺寸及配钢如下图所示的型钢混凝土梁所能承受的最大极限弯矩。混凝土 C30,钢筋 为HPB235级钢筋，型钢为Q235钢。习题4-2图解选择I40aAf =bxt = 142 xl6.5 = 2343m/?2, tw = L0.5mm9 hs = 400?，As

4、 ds = 50nun属于第一种情况，即中和轴通过型钢腹板，此时x_ (A$ 4)+ f 105.4S0.4/ bh0 =0.4xl4.3x 250 x 430N = 614900N = 614.9AN谿= 105.43 0.4力。【例54】有一型钢混凝土简支梁，计算跨度为9m,承受均布荷载，其中永久荷载设汁值为12.22kN/m(包 括梁自重)，可变荷裁设计值为16kN / nio由于空间高度限制，截面尺寸拟取为460 mm X 250 mm。经 正截面抗弯强度计算，拟配I36a普通热轧工字钢Q235,梁的上卜共配4 4)16架立钢筋。混凝土强度等级 为C30。试验算其斜截面抗剪承载力，并配

5、置钢箍。习题44图解 查C30混凝土强度得fc=143N/mm29fl=lA3N/mm2oI36a 工字钢 As = 7630/,/=360mm. % = 10，8t = 15.8/7/7?,/v = 215N / mm/? = (360-2 xl5.S)mm = 328.4hQ = (460-30) = 430mm梁上永久荷载设计值与可变荷载设计值之和为g + q = (12.22 +16)kN !m = 28.22妲V / in梁中最大剪力为1 , 1= -(g + Q)l = -x28.22x9AN=127.0ANOAfbh. = 0.4x14.3x250 x430 = 614900N

6、= 614.9RN= 127.0AN V= 127. OS剪切承载力满足要求，所以钢箍可按构造选配，拟配双肢箍48200。【例5-5】有一框架柱截面如习题4-5图所示，设计轴力N=1350kN,弯矩M=500 kN m,计算高度l=6m, 混凝土采用C30.钢筋为HPB235钢，型钢为Q235钢。验算其正截面强度。2电0习题4.5图4 = A = 628,九=14.3N / mm2 ,fy = 210N / mm2,Es = 210 x103A / mnr, 145a, As/ = Z;xr = 150 xl Smnr =2700?P,r”. = 11.5,Ass = 10245?/，九=21

7、5N / mnf, Exs = 2.06 x 10, N / mnr,as = ax = 50, q. =ar= 35mm,计算受压区高度：受压区界限高度为0.8(/? 外)0.8(550-50)2151 + r0.003x206x1()3nmi = 297先按大偏心计算v_ N + /X力 A + 2小1350 x103 + 215x550 x11.5 mm 14.3x400 + 2.5x215x11.5=227.1mm xh = 291mm1.2as x13503M = N/e0 = 1948x 1.126x0.37AN 7 = 82kN m 500kN m承载能力满足要求。【例5-6】框

8、架柱截面尺寸及配筋同习题4.5,设计轴力N=3200kN,弯矩M=300kN-m,计算高度lo=6m, 混凝土 C30,钢筋为HPB235钢，型钢为Q235钢，I45a ,验算正截面承载能力。解 A = A = 628 ,fc = 14.37V / mnr,=21 ON / nrnr,Es = 210 x103A / mm2.145a, Asf =Z?xr = 150 xl 3mm2 =2700,. = 11.5，Ass = 10245，九=215N / mnf, Exs = 2.06 x 10, N / nun1,as = as = 50mm, a r =dr= 35m?,、300 x 9血

9、0 N 3200 x103计算受压区高度：界限受压区高度为-附一“，）-爵（55?5。）而=297顷215先按大偏心计算x值N + fth 3200 xl03 + 215x550 xll.5 feb + 2.5 fxtw 143x400 + 2.5x215x11.5故再按小偏心受压计算。型钢未屈服区高度J = 0.417x型钢受拉翼缘应力b矿= o.oo3q竺土LX=0.003x206x1038(2 ，0) _1 = 618(122一i)X受拉钢筋应力/=0.003氏雄，)一1X= 0.003 x 2 10 x1030秒。0 3)_1 = 630(ll 1)X3= 3x94 = 0 5121

10、h 550季15,故K = 1.0,则 h- = i+Ut()FY1400 Q 1h=1+(丝22)2x0.512x1 = 1.2594 5501400 x550承载能力1.25盼N = f(.bx+ fyAs + .& + (1.25x-ds-asAs -(rsf Asf0 417r=14.3 X 400 x + 210 X 628 + 2152700 + (1.25% - 50 - )x1 L5 4 TOC o 1-5 h z 630( -1)x628-618( -l)2700 + 上(550 50 1 .25Q x2经整理得 = 7.9xl03x+2.96xl06-8 92x10(a)X

11、Ng - ? +1.25x) = 0.75febx2 + fyAs (1.25x-ar) + fxAsf (1.25 4) +asAs(h-ar-1.25x) + asf Asf (h- as-1.25x) +(1.25i) tw-1 fsd + / j _ 1.254 zo5?/(1.25x94- + 1.25x) = 0.75 x 14.3 x 400 x2 + 210 x 628(1.25x -35) + 4215x 2700(1.25x 一 50) + 630( -l)x 628(550 一 35 -1 .25x) + x4001、618(1)x2700 x(550-50-1.25x)

12、4-x215(1.25x-50)-xl 1.5-x21x215(0417x)2x11.5+Ax618(-l)(550-50-1.25x)2xll.5 63 x经整理得N(1.25x -158) = 2447x2 + 6.37 x 10宇 3.89 x 10, + x联立式(a)及式(b)可解得x = 400代入式(a)可求得N = 3890S 32003归=N 乌=3890 x 0.094 = 367kNm 300RN m 故正截面承载能力满足要求。【例5-7】框架往净高3 m,截面尺寸为400mmX500mm在重力荷载与地震作用组合下的柱端设计剪力 V=380kN,轴向力为1 500 kN,

13、弯距为180 kNm。柱内配H型钢，翼缘为2-200 mmX6 mm.腹板为300mm X 8 mm,均为Q235钢。钢筋用HRB335钢，箍筋用HPB235钢.混凝土强度等级为C30.试计算其箍 筋用量。 解 查C30混凝土得/c=14.3N/ inin2 /=1.43 N/rnin2/?o=5OO-5O min=450inin0Afcbho=0A X14.3X 400 X 450 N = 1030 kN 380 kN 故截面符合要求。H 3000-/ = 3.33 2.5, 取人=2.52/?02 x 4500 070 7V=380 kN 一一 ftbhQ = 一-一 xl.43x400

14、x 450 N = 9kN A-0.5 1 0 2.5-0.5Q235 钢 w=215N/min2HPB235 钢 .=210N/nmr 取N=0.3/cZ?/?o=0.3X 14.3X400X450N= 1030 kN 380 kN TOC o 1-5 h z A r o 7i 2-f = V- f,bhQ - 0.056/V / fyvhQS L x -0.52 + 1.5=J|380000-90000-x8x215x300-0.056x7720001/ (210 x450)？U2.5+ 1.5JJ=0.837若采用6 8双肢箍，则Acv=2x50.3 nmf = 100.6 nun2c

15、100 lin S =nun = 119 imn0.837故配钢箍8100o【例5-8如习题4-8图所示，型钢混凝土柱截面尺寸bh = lOOOmmxlOOOfwn 9采用C40混凝土(fc = 19.5N/mm2 ), HRB400 ( fy = 360N/mm2 ),采用 Q235 型钢(fa = 315N/mm2 ),型钢采用H 型钢 HZ450a (440 x300 x11.5x21.0, Vx. = 2.896xl06/?/n A” = 1.78、10七?己)，对称配筋，承受内力设计值N = *10KN，试用平面假定的简化方法求其极限弯矩M值。解:360 + 315=0517由题意知

16、 ctl = 1.0,= 0.852 x 0.003E；2 x 0.003 x 2.06 x 10,1884x360+300 x21x315a = 1256x360 x50 + 628x360 x150 + 300 x21x315x(280 + 21/2)=空?hQ = h-a = 1000-237.7 = 762.3(mm)280 + 21ox = U.3”1762.3e 1000-280-21 -5. = 0.917-762.3%=岂庄)=涕戒-y-皿e。”假定/?0 = 1.25x, A - ANaw = 2.5 - (5； + )twhofa = 2.5 - (0.395 + 0.91

17、7) x 11.5 x 762.3 x 315= 6903579-3622998将己知数据带入平衡条件N = abx+f + faAat -弓A - o-aAaf + Ng ,解得* 0.579务x = #o = 0.579x762.3 = 441.4(0)，构件截面受力为小偏心受压。g =-1272* + 1017 = -1272x0.579 + 1017 = 280.5N/m/ fy= 1113 + 890 = 1113x 0.579 + 890 = 245.6 N/mm2 fa6山0 = 30bum 1.25x,符合假定。= |(5/ + 或)-(屯 + 如) + 2.5J(L25 打5注=-(