混凝土结构计算例题

1、单筋矩形截面梁正截面受弯承载力计算例题1. 钢筋混凝土简支梁，讣算跨度/=5.4m,承受均布荷载，恒载标准值gk=10kN/m,活载标 准值=16kN/m,恒载和活载的分项系数分别为Y g=1.2, Yq=1.4o试确定该梁截面尺寸，并 求抗弯所需的纵向受拉钢筋£。解：选用材料混凝土C30, /；=14.3N/mm2,/； =1.43N/mm2 ； HRB400 钢筋，fy =360N/mnr,= 0.518确定截面尺寸1 1 1 1 h=/=x5400 =450 675mm , ® /? = 500mm112 8 丿 112 8 丿b = g" =(* *jx5

2、00 = 167 250mm,取b = 200mm力计算荷载设计值= +/Qf/k = 1.2x10 + 1.4x16 = 34.4kN/m跨中弯矩设计值 M =-ql2 =-x 34.4 x 5.42 = 125.4kN - m8 8(4)配筋计算布置一排受拉钢筋，取 =40nun ,则 /?0 = h 一=500 - 40 = 460 nun将已知值代入力0-勻，得 125.4x106 = 1.0xl4.3 x 200 q 460-y整理为 x2-920x + 87692 = 0解得x = 108nni < gJS =0.518 x460 = 238mn ,满足适筋梁要求4 = 85

3、8 mm2 > pminb/i = 0.002 x 200 x 500 = 200 mn2,满足最小配筋率要求选 3020. A=942mm，2. 已知钢筋混凝土梁截面尺寸为bXh=250mmX550mm,混凝土强度等级C30,已配纵 向受拉钢筋为6根直径20mm的HRB400级钢筋，承受的弯矩设计值M=250kN验算此 梁受弯承载力是否满足要求。解：由已知条件可得，As = 1884mm2 t =h-as = 550 -65 = 485nm根据基本公式，得a£b360x1884=189.7mm<b/70 =0.518x485 = 251mm ,满足适筋梁条件仏=ajcb

4、x= 1.0xl4.3x250xl89.7x 485189.7=264.6 x106N-mm = 264.6RN- m > M = 250kN m受弯承载力满足要求3条件同题1,试用系数法求该梁抗弯所需的纵向受拉钢筋。M% a£b 惊125,4xl062= 0.2072解:0.8826M125.4xl06严.= fy/A=858mrrf360x0.8826x460(1 + 71-2x0.2072)=双筋矩形截面梁正截面受弯承载力计算例题1.已知钢筋混凝土梁截面尺寸为bXh=200mmX500mm,混凝土 C30, HRB400级钢筋, 承受的弯矩设计值M=260kNm。试计算此

5、梁受弯所需的纵向受力钢筋。解：查表得 £=14.3N/mm2, fy=f；.= 36ON/mm2布置两排受拉钢筋，取込=65mm ,则代）=h-a3 = 500 - 65 = 435mmM260x10°Qf =S aj皿1.0xl4.3x 200x4352 曲。0384若按单筋截面设计，将成为超筋梁，所以应按双筋截面设计。取"；=40mm为使总用钢量最小，取天=则360x(435-40)=367mrrTdox(10xl4-3x20°x435x0.518+360x367 )=2157nun2260x106 - 0.384x 1 .Ox 14.3x 200x

6、 4352选受拉钢筋 6022 ( A =2281mm2),受压钢筋 2016 (/!；= 402mm2)2.已知数据同题1,但已配受压钢筋3018, A=763mmS试求受拉钢筋用量入。解：A/】仇/) = 260xl0&360x763x(43540)=151.5xl0°NmmM151.5X106a =! = 0 2799'ajhhl 1.0xl4.3x 200x4352/s =(l + Jl 2a、)=1(l + Jl 2x0.2799)=0.8317 2 2d M,I51.5X106*2A. = = 1163mmfy/A 360x0.8317x435As =As

7、l+A； = 1163+763 = 1926mnr A选用 2O22+4020 (As=760+1256 =2016nun2 )3. 某钢筋混凝上梁截面尺寸/?X/z=250X600mnn混凝± C30, HRB400钢筋，已配6根 直径22mm的受拉钢筋(As二228lmm，), 3根直径20mm的受丿E钢筋(二942mm?),(取 as=65mm, «=40mm)。该梁承受弯矩设汁值M=340kNmo验算该梁的正截面受弯承载力是否 满足要求。解：几=h 一 as = 600 -65 = 535 mmZA-/X 360x2281-360x942。< 狀=0.518x

8、535 = 277mmaj；b1.0x14.3x250>2d： = 2x40 = 80mm(C 4 O = 1.0xl4.3x250xl34.8x 535- +360x942x(535-40)< 2丿= 393x106N mm = 393kN m >M = 340kN m满足承载力要求钢筋混凝土受弯构件斜截面受剪承载力计算例题1. 钢筋混凝土简支梁,截面尺寸Z?X/i=250X550mm, /io=485mm,净跨/n=5.76m,均布荷载作用，荷载设计值q =75kN （包括自重），混凝土 C30,箍筋HPB300,试讣算斜截面受剪 所需的箍筋用量。解：查表得 fc =14

9、.3N/mm2, / = 1.43N/mm2, fyv = 270N/mm2（1）求剪力设计值支座边缘处截面剪力值V=l/n=lx75x5.76 = 216kN2 2（2）验算截面尺寸0.25=0.25x14.3 x 250 x 485 = 433469 N = 433.5kN > V = 216kN截面尺寸满足要求（3）验算是否需要按计算配箍筋0.7= 0.7x 1.43 x 250 x 485 = 121371 N = 121.4kN <V = 216kN需按计算配置箍筋计算受剪所需的箍筋 也上0.7/网）_ 216000121371_07226 咄sfyJ270x485mm选

10、双肢 0 8 箍筋，s = ""J = 139mm,取 $ = 125mm0.7226实配双肢箍筋加125偏心受压构件正截面受压承载力计算例题1.钢筋混凝土柱截面/9X/?=300X400mm, s=«；=40mm.弯矩设计值M=220kNm,轴向压力设计值 N=400kN,混凝土 C30 (/c=14.3N/mnr), HRB400 钢筋(人= 360N/mm) 不 考虑P§效应的附加影响，求所需的纵向受力钢筋。解：/"=h -= 400 - 40 = 360 nunM因=30220=0.55m = 550mm400:兰巴=13.3mm &l

11、t; 20mm,所以取 q = 20mm30ei=eo+ea= 5 刃 + 20 = 570 iiini因= 570nm >0.3 =0.3x360 =108mn ,所以按大偏心受压情况计算h400e = e, dci° =570+40 = 730mm2 2取X =気他,则得_ 恥一冬/0处久(1一0.5気)_ 400X10'x730-1.0X14.3X300x360, x0.518x(1 -18) 厲 _仇-M) _360x(360-40)=682mm2 > p'mxJyh = 0.002x300x400 = 240mm24 =y(a1XZ?/b+/&g

12、t;X-A)=-x(1.0xl4.3x300x360x0.518 + 360x682-400xl03)=1793mm2选受拉钢筋 5022 ( As=1900nin2 ),受压钢筋 3018 (A； =763 mm2)2. 已知数据同题1,但已配受压钢筋3022, = 1140imi2,试求受拉钢筋用量人。解:M, =e-/；(/70-) = 400x1x730-360x1140x(360-40)= 160.7xl06N mms.max怎=豐需：6。2 = 02890< = 0384即满足XV轨的条件/s =1(1+ /l-2crs )=1(1 + 71-2 x 0.2890 )=0.8

13、248 <1- = 1-= 0.888922h360即满足x>2g：的条件人160.7xl06 s 2A. = = 1503mm/；/s/7()360x0.8248x360A=Ai+ A：-= 1503+11 4O-4O()X10- = 1532mm2A fy360选受拉钢筋422(4=1520nm2)3钢筋混凝上柱截ffi/?X/=35OX5OOninn依二/=40mim 弯矩设讣值M=420kNim轴向丿k 力设计值N=1100kN,混凝土 C35 (/c=16.7N/mnr), HRB400钢筋，不考虑效应的附加影 响，对称配筋，求所需的纵向受力钢筋。解:h) = /? as = 500 40 = 460 nun5 =0.382m = 382mmuh30 'N 1100500=16.7 mm<20mm, . eo = 20 mm30= eQ + ea = 382 + 20 = 402 mmh500/e = ©cr = 402