工程结构第五章习题解上课讲义.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,工程结构第五章习题解,1,习题（,P211212,）,5-1,已知矩形截面梁，,b=250mm,h=500mm,(,取,h,0,=465mm),，梁承受的弯矩值,M=180kN.m,。按下列条件计算梁的纵向受拉钢筋面积,A,S,。并分析混凝土强度及钢筋强度对受弯构件纵向受拉钢筋面积的影响。,（,1,）混凝土为,C20,级，纵筋为,HRB335;,（,2,）混凝土为,C20,级，纵筋为,HRB400;,（,3,）混凝土为,C40,级，纵筋为,HRB335,.,5-7,已知梁截面尺寸,bh=200mm500mm,，受拉纵筋为,3 16,钢筋，混凝土强度等级,C30,。问该截面能承受多大的弯矩,M,u,?,2,5-1【,解,】,(1),混凝土为,C20,级，纵筋为,HRB335,基本数据：由,P76,表,5.1.7,：,C20,砼,f,c,=9.6N/,2,f,t,=1.10N/,2,1,=1.0,；由,P71,表,5.1.1,：,HRB335,钢筋,f,y,=300N/,2,。,初步估计纵向受拉钢筋为单层布置，由,P97,知,a,s,=a+10=25,+10,=35,h,0,=h-a,s,=500-35=465,计算配筋,3,由表,5.2.1,查得,sb,=0.399,由公式,5.2.17,s,=M/(,1,f,c,bh,0,2,),=180,10,6,/(1.0,9.6,250,465,2,),=0.34725,且,d=28,满足要求。,4,（,2,）混凝土为,C20,级，纵筋为,HRB400,基本数据：由,P76,表,5.1.7:C20,砼,f,c,=9.6N/,2,f,t,=1.10N/,2,1,=1.0,；由,P71,表,5.1.1,：,HRB400,钢筋,f,y,=360N/,2,。,初步估计纵向受拉钢筋为单层布置，由,P97,知,a,s,=a+10=25,+10,=35,h,0,=h-a,s,=500-35=465,计算配筋,由表,5.2.1,查得,sb,=0.384,s,=M/(,1,f,c,bh,0,2,)=180,10,6,/(1.0,9.6,250,465,2,),=0.34725,且,d=25,满足要求。,6,（,3,）混凝土为,C40,级，纵筋为,HRB335,基本数据：由,P76,表,5.1.7:C40,砼,f,c,=19.1N/,2,f,t,=1.71N/,2,1,=1.0,；由,P71,表,5.1.1,：,HRB335,钢筋,f,y,=300N/,2,。,初步估计纵向受拉钢筋为单层布置，由,P97,知,a,s,=a+10=25,+10,=35,h,0,=h-a,s,=500-35=465,计算配筋,由表,5.2.1,查得,sb,=0.399,s,=M/(,1,f,c,bh,0,2,)=180,10,6,/(1.0,19.1,250,465,2,),=0.17425,且,d=25,满足要求。,分析：提高混凝土强度及钢筋强度可以减少受弯构件纵向受拉钢筋的用量，节约钢材。,8,5-7【,解,】,（,1,）验算截面最小配筋率,查得,f,c,=14.3N/,2,f,t,=1.43N/,2,1,=1.0,f,y,=360N/,2,A,s,=603,2,A,s,=603,2,min,bh=0.2%,0.45f,t,/f,y,max,bh,=0.002,0.0018,max,200,500,=0.002,200,500=200,2,满足第二个适用条件。,计算受压区高度,截面有效高度,h,0,=h-a,s,=500-(25+16/2)=467,由式,5.2.4 x=f,y,A,s,/(,1,f,c,b)=360,603/(1,14.3,200),=75.9,9,由表,5.2.1,b,=0.518,x=75.9,M=450kN.m,属于第一类,T,形截面,（,2,）求纵向受拉钢筋面积,12,s,=M/(,1,f,c,b,f,h,0,2,)=450,10,6,/(1.0,11.9,2500,640,2,),=0.037,由附录,D,附表,D-1,查得,s,=0.982,=0.037,则,A,s,=M/(f,y,s,h,0,),=450,10,6,/(300,0.982,640)=2389,2,选用,5 25,，,A,s,=2454,2,(3),验算适用条件,由表,5.2.1,b,=0.55,=0.038,min,=0.2%,0.45f,t,/f,y,max,=0.2%,0.19%,max,=0.2%,A,s,=2454,2,min,bh=0.002,300,700,=420,2,满足适用条件。,13,5-12【,解,】,预估配置双层纵向受拉钢筋，,a,s,=60,h,0,=550 -60 =490,f,c,=14.3N/,2,f,t,=1.43N/,2,f,y,=f,y,=360N/,2,sb,=0.384,b,=0.518,s,=M/(f,c,bh,0,2,)=300,10,6,/(14.3,250,490,2,)=0.350,s,min,bh=0.002,250550=275,2,满足第二条件。,14,此课件下载可自行编辑修改，仅供参考！感谢您的支持，我们努力做得更好！谢谢,15,