受压构件承载力计算例题

受压构件承载力计算1(N1(1400 x 103 “c “JA'=-f A14.3 x 300 x 30010.9(p< 丿300(0.9 x 0.962丿1、某现浇框架柱，截面尺寸为300X 300，轴向压力设计值N = 1400 kN,计算 长度3.57 m，采用C30混凝土、II级(HRB335)钢筋。求所需纵筋面积。解: 10/b 二 3570300 二 11®，查得 2 = 0.9515,yA'1159.3二二=1159.3 =0.01288=1159.5mm2A 300 x 300> p '二 0.006min1(N1(1800 x 103“c “JA'=-f A9.6 x 400 x 4000.9c丿300(0.9 x 0.7875丿2、已知某正方形截面轴心受压柱，计算长度75 m,承受轴向压力设计值N = 1800 kN，混凝土强度等级为C20，采用II(HRB335)级钢筋。试确定构件截面 尺寸及纵向钢筋截面面积。二 3345.6mm2,解: 1 Jb 二 7500 400 二 18.75，查得 2 = 0.7875yA' s 二P - A 400 x 4003345.6=0.021> p '= 0.006min3、已知一偏心受压柱，bXh = 450 X 450，a = a z = 40C30，HRB335钢筋， gb = 0.55，承受纵向力N = 350 kN，计算弯距M = 220 kNm。柱计算长度 为10= 3.0 m，受压区钢筋A's = 402 (2#16)，求受拉区钢筋面积。解：(1) 设计参数a=，a = a z = 40 h0=410，£=14.3 N/mm2,e 0= 630，取 ° a=20，e i= e 0 + e a= e 0+20=648f' = 300N / mm 2y0.5 f Ac0.5 x 14.3 x 450 x 450350000二 4.1 取 Z 1=1=川-o.。1 ¥ =川-001 x 3000=1.08，取 Z 2=11耳=1 + 一 1400 h(2) 受压区高度02其=1 +12為x 曙2 x 1 x 1=1.02450ne i = 661> 0.3 h°按大偏压计算 e=661+(450/240)= 846，xNe =a f bx(h -) + f'A' (h 一a')1 c 02 y s 0350000 x 846 = 1 x 14.3 x 450x(410 - x2)+ 300 x 402(410 - 40)x = 110mm < x =g h - 0.55x410 = 225.5mmb b 0> 2a 80mm(3) 求钢筋面积N =a fbx + f'A' -f A1 c y s y sAs=1594mm2，取 4 # 22, As = 1520mm2(4) 验算配筋率A 1520p = - 0.75% >p 0.2%1 bh 450 x 450min垂直于弯矩作用方向的承载力验算10/b =6.6,可得申-L0N - 0.9叽fcbh + fy(A： + A,) - 3125kN>350kN 满足要求。4、已知矩形截面受压柱bXh = 300 X 500,柱上作用轴向力设计值N = 850 kN, 弯矩设计值M = 285 kN im,柱的计算长度L = 4.5 m,混凝土强度等级为C30, 采用HRB335钢筋，对称配筋，求柱所需的纵向钢筋截面面积。解：(1) 设计参数-1 - ,a = a ' = 40 h0=460 , fc=14.3 N / mm2, fy= 300N/mm2Me 0 = N = 335,取 7=20,=°0 + e a = 355, 1° / h = 9,fc=14-30.5 f AQ c1 N1.26 取 Z 1=10.5 x 14.3 x 300 x 500850000-川-°01 X-川-°01 x 葺0-1.06，取 Z 2=11400 eQQ13551400 x -460x (4500)2 x 1 x 1 -1.07500(2) 受压区高度 xne. = 379.8>0.3h0=138 大偏压。e =383.4+(500/240)=565- 238mmN 850000x a f b 1 x 14.3 x 3001c(3) 求钢筋面积由a 人 Ne -a fbx(h - 0.5x)850000 x 565 - 1 x 11.9 x 300 x 238(460 - 0.5 x 238)A = A' =1_c0=s sf' (h 一 a')300 x (460 一 40)y01382mm,取 4 #22, A = 1520mm2s(4)验算配筋率A 1520p= = 1% >p = 0.0021 bh300 x 500min垂直于弯矩作用方向的承载力验算10/b = 15,可得9 - 0.9N = 0.9叽f严 + fy(AS + As)二 2073kN>850kN满足要求5、已知矩形截面受压柱bXh = 300 X 500,柱上作用轴向力设计值N = 1650 kN, 弯矩设计值M = 285 kN im,柱的计算长度L = 4.5 m,混凝土强度等级为C30, 采用 HRB335 钢筋,对称配筋,求柱所需的纵向钢筋截面面积。f' = 300N / mm2y解：a 1 = 1-0 ,a = a 二 40 h0=460,£=14.3 N/mm2,e0= 173，ea=20，ei= e0 + ea =1930.5 f AQ =1 N6.5 取 Z 1=1=1.06，取 Z 2=10.5 x 14.3 x 300 x 500165000=1-15 一 °01 X=1J5 一 °01 x 鬻1耳=1 + 一 1400 红 h 02其=1 +12仁x (鄴x 1 x 1=1.13460(2)受压区高度ne. = 218 > 0.3 h0 = 139.5,e =n e. +- a' = 410N 1650000 qq八 八 “x = 384 > x = 0.55 x 460 = 253a f b 1 x 14.3 x 300b1c 按小偏压计算 戸N-ga fbhQ =b_1 c0-Ne 0.43a f bh2 1 c+ a (p -Q )(h 一a')1 b 0按大偏压计算16500001 fcbh0(3)求钢筋面积Ne -a f bx(h=1_c0f (h -a') y0A = A 'ss+ g =0.682, x = gxh0=317.13 b- 0.5 x)=1650000 x 410 -1.0 x 14.3 x 300 x 317.13 x (460 - 0.5 x 317)300 x (460 - 40)=2113，取 7 #20，As= 2200> p' bh = 0.002 x 300 x 460 = 276sm.n 0垂直于弯矩作用方向的承载力验算l /b = 15,可得= 0.90N = 0.9叽fbh + f (A + A )=2806kN>285kNc y s s满足要求6、已知矩形截面柱bXh = 300X 600,柱的计算长度为L = 6 m,柱上承受M =110 kNm, N = 1680 kN,混凝土强度等级为C30,采用HRB335钢筋，对称配筋,试确定该柱的配筋。解：(1)设计参数巴=1.0, a ' = a = 40mm,h0=560,巴=l.° , eo = 65.5,e = 20,e .=101 c 4 + a f b0ai85.5，l0 / h = 10，0.5 f A 0.5 x 14.3 x 300 x 6000 76i N1680000C = 1.15 - 0.01L = 1.15 - 0.01 x 6000 = 1.052CC = 1 +121耳=1 + 1400h02h600，取 Z2=11x (6000)2 x 1 x 1 = 1.461400 x 空 600560(2)受压区高度ne.= 125 < 0.3 h0=168按小偏压计算he =耳 e + 一 a = 385,i 2g =N 一乙WWo+ E = 0.773, x = g xh0=436.745Ne - 0.43a f b2 b0mm时，柱能承受的轴向压力设计值. 解：(1)判别大小偏心 a' =40,ho=36O,1 1耳=1 + 1400 红h0e o= 225 ,e-0Ih丿2其=1+1 21400 x=2 0,ei = e a+ e 0= 245，l0 / h = 7.5 >5,245x 7.52 = 1.05360a f bx(e 一 0.5h + 0.5x) = f A (qe + 0.5h 一 a) - f'A'ge - 0.5h + a')1 ciy s iy s ix = 242.03,g = 0.663 > 弭=0.55应按小偏压计算将已知数据代入下面三式N = a f bgh + f'A' -a A1 c 0 y ss sNe =a fbh亿(1 -0.5g) + f'A'(h 一a')'cy s 0g-卩fg -卩丿yb 1解得：(3)垂直于弯矩作用平面的承载力验算1 农=3000/= 10因0b300 ，查表5.1得申=°98x = 155.211, N = 676.473 kNN = 0.9p (f A + f'A') = 0.9x0.9814.3x300x400 + 300x2279 = 2116KNcy s比较计算结果得该柱所能承受的轴向力设计值为676.47KN8、某钢筋混凝土短柱，bXh = 300X 500, As = A's = 1570。混凝土强度等级为 C30, HRB335钢筋。试确定： 1、该柱所能承担的最大轴力是多少？此时的偏心距为多大？2、该柱所能承担的最大弯矩是多少？此时的偏心距和轴力是多大？解: =( )1、 轴心受压时承担最大轴力：Nu = 0.9p fcA + fy As + fyAs对于矩形截面稳定系数取决于dN =0.9x1x(14.3x300x460+300x1570+300x1570)=2623 KN此时：e = 09、某矩形截面钢筋混凝土单向偏心受压柱bXh = 400X 600，承受轴力设计值 N = 1200 kN,弯矩M = 600 kNm,柱的计算长度L = 4.2 m,采用C30混凝 土，HRB335钢筋。若采用对称配筋，求As = A's = ?解:(1)设计参数M=500 mm , a' - a - 40mm, h0=560mm,a 二 1.01f = 14.3 N / mm 2e° = Ne,0=520 fy =300,0.5 f A0.5 x 14.3 x 450 x 600ei = ea +Q =1 N=1.68 取 Z 1=11200000l4200:=1.15 - 0.01旳=1.15 - 0.01 x = 1.08，取 Z2=12h600211耳=1 + 一1400 h0-0Ih丿x 將x 1 x 1=1.03560(2) 受压区高度N 1200000x = 二一a f b 1 x 14.3 x 400 1 c按小偏压计算he =耳 e + 一 a =760 ,i 2(3) 求钢筋面积a a , Ne - a f bx(1 - 0.5x) As 二 As 二f ( - a J As= A's = 7125mm2二 209，x < 乜=0.55 x 560 二 3081200000 x 760 -1 x 14.3 x 400 x 209(1 - 0.5 x 209)300 x (560 - 40)=6639 ,选 7 #36,验算配筋率P < 5% P > Pmin二 0.2%垂直于弯矩作用方向的承载力验算101 b = 10.5 ,可得申二0.98N 二 0.9叽 fbh + f (A' + A )二 6797kN > 1200kNcy ss满足要求10、已知荷载作用下柱的轴向力设计值N = 300 kN,弯矩M = 180 kNm , 截面尺寸b = 300, h = 400, a = a' = 40；混凝土强度等级为C30,钢筋用HRB335 钢筋；10 / h = 7。求钢筋截面面积A's及As。解：(1)设计参数h0=360,e 0 =(l-0-l h丿N =600,e = 20,a、e i =e a +e 0= 620, lo / h= 7 >5,2匚匚=1.03121400 ,h(2)判断大小偏心ne i>0.3 h0按大偏心计算 ” =n 5 + 竺-a' = 785mm(3) 钢筋面积A = N-叮严咅.上空丄=541,选2 #20, A = 628 s广(h - a')sy0As = 1687,选 3 #18 +3 #20, As = 1704(p -g )( -a')1 c 01 b 0(3)求钢筋面积Ne-a f bh 2 3 4g(1-0.5g) 1680000 x 385 - 1.0 x14.3 x 300 x 5602 x 0.773 x (1 - 0.5 x 0.773)As = As =0=f ( - a')300 x (560 - 40)y0= 56取 As= P minb0 = 0.002x300x560=336取 3#12(4)垂直于弯矩作用方向的承载力验算10/ b =2U,可得申=°75N = 0.9叽fb + f (A' + A ) = 1873kN > 1680kNcy s s满足要求7、 已知矩形截面偏心受压柱截面尺寸 bXh = 300 X 400，柱的计算长度 L = 3m,柱采用C30混凝土，对称配筋，每侧配有3#22HRB335钢筋，求e0 = 2252、 x = x b=0.55xh0=255.75mm ,时弯矩最大M = 467.3 kNm, N = 736.56 kN,e 0 = 63mm