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38第四章受弯构件正截面承载力计算题参考答案1.已知梁的截面尺寸为b×h=200mm×500mm,混凝土强度等级为C25,fc=11.9N/mm2,2/27.1mmNft,钢筋采用HRB335,2/300mmNfy截面弯矩设计值M=165KN.m。环境类别为一类。求:受拉钢筋截面面积解:采用单排布筋mmh465355000将已知数值代入公式sycAfbxf1及)2/(01xhbxfMc得1.011.9200x=300sA165106=1.011.9200x(465-x/2)两式联立得:x=186mmAs=1475.6mm2验算x=186mm0hb0.55465=255.8mm2min200500200%2.06.1475mmbhAs所以选用325As=1473mm22.已知一单跨简支板,计算跨度l=2.34m,承受均布荷载qk=3KN/m2(不包括板的自重),如图所示;混凝土等级C30,2/3.14mmNfc;钢筋等级采用HPB235钢筋,即Ⅰ级钢筋,2/210mmNfy。可变荷载分项系数γQ=1.4,永久荷载分项系数γG=1.2,环境类别为一级,钢筋混凝土重度为25KN/m3。求:板厚及受拉钢筋截面面积As解:取板宽b=1000mm的板条作为计算单元;设板厚为80mm,则板自重gk=25×0.08=2.0KN/m2,跨中处最大弯矩设计值:39图1mKNlqgMkqkG.52.434.234.122.1818122由表知,环境类别为一级,混凝土强度C30时,板的混凝土保护层最小厚度为15mm,故设a=20mm,故h0=80-20=60mm,fc=14.3,ft=1.43,fy=210,b=0.618查表知,图20878.06010003.1411052.426201bhfMcs092.0211sa26037660954.02101052.4954.02115.0mmhfMAasysss选用φ8@140,As=359mm2(实际配筋与计算配筋相差小于5%),排列见图,垂直于受力钢筋放置φ6@250的分布钢筋。验算适用条件:⑴,满足。mmhmmhxb84.3660614.052.560092.000⑵2min120601000%2.0376mmbhAs3.某矩形截面简支梁,弯矩设计值M=270KN.m,混凝土强度等级为C70,22/8.31,/14.2mmNfmmNfct;钢筋为HRB400,即Ⅲ级钢筋,2/360mmNfy。环境类别为一级。求:梁截面尺寸b×h及所需的受拉钢筋截面面积As解:fc=31.8N/mm2,fy=360N/mm2,查表4-5,得α1=0.96,β1=0.76。假定ρ=0.01及b=250mm,则118.08.3196.036001.01cyff40令M=Mu可得:201015.012hbfxhbxfMccmmbfMhc564118.05.01118.02508.3196.0102705.01610由表知,环境类别为一类,混凝土强度等级为C70的梁的混凝土保护层最小厚度为25mm,取a=45mm,h=h0+a=564+45=609mm,实际取h=600mm,h0=600-45=555mm。115.05552508.3196.01027026201bhfMcs123.0115.0211211s939.0115.02115.02115.0ss2601439555939.036010270mmhfMAsys选配3φ25,As=1473mm2,见图3验算适用条件:⑴查表知ξb=0.481,故ξb=0.481ξ=0.123,满足。⑵2min390600250%26.01473mmbhAs,满足要求。图34.已知梁的截面尺寸为b×h=200mm×500mm,混凝土强度等级为C25,22/9.11,/27.1mmNfmmNfct,截面弯矩设计值M=125KN.m。环境类别为一类。求:(1)当采用钢筋HRB335级2/300mmNfy时,受拉钢筋截面面积;(2)当采用钢筋HPB235级2/210mmNfy时,受拉钢筋截面面积;(3)截面弯矩设计值M=225KN.m,当采用钢筋HRB335级mmNfy/3002时,受拉钢筋截面面积;解:(1)由公式得26204652009.110.110125bhfMcs=0.243283.00.243211211s41858.0)243.0211(5.0)2-1(15.0ss2601044465858.030010125/mmhfMAsys选用钢筋421017,18mmAs2min200500200%2.01044mmbhAs(2)采用双排配筋mmhh440600201/bhfMcs=271.04402009.110.11012526s211=323.0271.0211)2-1(15.0ss0.50.838271.0211()2601614440838.021010125/mmhfMAsys选用钢筋816As=1608mm22min270500200%27.01614mmbhAs(3)假定受拉钢筋放两排60ammh440605000201/bhfMcs=488.04402009.110.11022526s211=1-55.0845.0488.021故采用双筋矩形截面取bM)5.01(2011bbcbhf)55.05.01(55.04402009.110.12=183.7KNm266'0''9.339)35440(300107.18310225)(/mmahfMAys9.339300/4402009.110.155.0//''01yysycbsffAfbhfA42=2260mm2故受拉钢筋选用622As=2281mm2受压钢筋选用216A's=402mm2,满足最小配筋率要求。5.已知梁的截面尺寸为b×h=250mm×450mm;受拉钢筋为4根直径为16mm的HRB335钢筋,即Ⅱ级钢筋,2/300mmNfy,As=804mm2;混凝土强度等级为C40,22/1.19,/71.1mmNfmmNfct;承受的弯矩M=89KN.m。环境类别为一类。验算此梁截面是否安全。解:fc=19.1N/mm2,ft=1.7N/mm2,fy=300N/mm2。由表知,环境类别为一类的混凝土保护层最小厚度为25mm,故设a=35mm,h0=450-35=415mm2min293450250%26.0804mmbhAs则,满足适用条件。55.0121.01.190.13000077.01bcyff,安全。mKNMmKNbhfMcu.89.49.93121.05.01121.04152501.190.15.0122016.已知梁的截面尺寸为b×h=200mm×500mm,混凝土强度等级为C40,22/1.19,/71.1mmNfmmNfct,钢筋采用HRB335,即Ⅱ级钢筋,2/300mmNfy,截面弯矩设计值M=330KN.m。环境类别为一类。求:所需受压和受拉钢筋截面面积解:fc=19.1N/mm2,fy’=fy=300N/mm2,α1=1.0,β1=0.8。假定受拉钢筋放两排,设a=60mm,则h0=h-a=500-60=440mm446.04402001.1911033026201bhfMcs55.0671.0211bs这就说明,如果设计成单筋矩形截面,将会出现超筋情况。若不能加大截面尺寸,又不能提高混凝土等级,则应设计成双筋矩形截面。取,由式得bmKNbhfMbbc9.29455.05.0155.04402001.190.15.0122011266'0''9.28835440300109.29410330mmahfMAsys432''014.33703003009.2883004402001.190.155.0mmffAfbhfAyysycbs受拉钢筋选用7φ25mm的钢筋,As=3436mm2。受压钢筋选用2φ14mm的钢筋,As’=308mm2。7.已知条件同上题,但在受压区已配置3φ20mm钢筋,As’=941mm2求:受拉钢筋As解:6'0'''103.11435440941300ahAfMsyKNm则6661'107.215103.11410330MMMKNm已知后,就按单筋矩形截面求As1。设a=60mm、h0=500-60=440mm。292.04402001.190.1107.21526201'bhfMcs,满足适用条件。55.0355.0292.0211211bs823.0292.02115.02115.0ss260'11986440823.0300107.215mmhfMAsys最后得2210.29279411986mmAAAsss选用6φ25mm的钢筋,As=2945.9mm28.已知梁截面尺寸为200mm×400mm,混凝土等级C30,2/3.14mmNfc,钢筋采用HRB335,2/300mmNfy,环境类别为二类,受拉钢筋为3φ25的钢筋,As=1473mm2,受压钢筋为2φ6的钢筋,A’s=402mm2;要求承受的弯矩设计值M=90KN.m。求:验算此截面是否安全解:fc=14.3N/mm2,fy=fy’=300N/mm2。由表知,混凝土保护层最小厚度为35mm,故5.4722535amm,h0=400-47.5=352.5mm由式,得sysycAfAfbxf''1mmammhmmbfAfAfxbcsysy8040221945.35255.03.1122003.140.14023001473300'''01代入式44,安全。mmNmmNahAfxhbxfMsycu.1090.1087.132405.35240230023.1125.3523.1122003.140.1266'0''01注意,在混凝土结构设计中,凡是正截面承载力复核题,都必须求出混凝土受压区高度x值。9.已知梁的截面尺寸b=250mm,h=500mm,混凝土为C30级,22/3.14,/43.1mmNfmmNfct,采用HRB400级钢筋,2/360mmNfy,承受弯距设计值M=300kN·m,试计算需配置的纵向受力钢筋。解:(1)设计参数由表查得材料强度设计值,C30级混凝土MPafc3.14,HRB400级钢筋MPaffyy360,518.0,384.0max,bsa,等级矩形图形系数0.1。初步假设受拉钢筋为双排配置,取mmh440605000。(2)计算配筋384.0433.04402503.1410300max,2620scsabhfM故需配受压筋,取mma40。226020m
本文标题:混凝土结构-第4章计算题及答案
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