第五版混凝土结构设计原理第四章

筋。级钢筋，求所需受剪钢采用筋，环境类别为一类，箍，承受剪力设计值混凝土为面尺寸为钢筋混凝土简支梁，截3001403040,5002001.4HPBKNVCmmammmmhbsmmhhw46040-5000验算截面尺寸）进行验算：（则属于厚腹梁，应按式17-443.2200460bhw则，故取不超过混凝土的强度等级为,150,30cCCKNVkNbhftc1409.3284602003.14125.025.00截面符合要求置箍筋验算截面是否按计算配KNVKNbhfc140029.9246020043.17.07.00故需按计算配置箍筋解：只配箍筋不配弯起钢筋有令,uVVmmmmhfbhfVsnAyvtSV/402.046027046020043.17.0101407.023001采用双肢箍筋8@200mmmmmmmmsnAsv/402.0/503.02003.502221箍筋的配筋率：%127.027043.124.024.0%25.02002003.502min,1yvtsvsvsvffbsAn可以，应该如何处理？及但，梁的截面尺寸等同上题KNVKNV280626.2.4可得：同上题，验证截面尺寸mmas4043.2200460,46040-5000bhmmhhww）演算属于厚腹梁，应按式（17-4150,30CCC，不超过混凝土的强度等级为KNVkNbhfcc629.3284602003.14125.025.00截面符合要求验算截面是否配置箍筋KNVKNbhfc62029.9246020043.17.07.00故需按构造配置箍筋假设配置8@300%127.027043.124.024.0%16.03002003.502min,1yvtsvsvsvffbsAnV=62kN时mmhhw46040-5000验算截面尺寸）进行验算：（则属于厚腹梁，应按式17-4则，故取不超过混凝土的强度等级为,150,30cCCKNVkNbhfcc2809.3284602003.14125.025.00截面符合要求置箍筋验算截面是否按计算配KNVKNbhft028029.9246020043.17.07.00故需按计算配置箍筋300)HRB配置箍筋（有令,uVV100320.7280100.71.432004601.513/270460SVtyvnAVfbhsfhmmmm270/yvfNmm剪力设计值V=60kN时采用双肢箍筋12@1201222113.11.885/1.513/120svnAmmmmmmmms箍筋的配筋率：%114.030043.124.024.0%706.01602001.1132min,1yvtsvsvsvffbsAn4.3,4-403040/(,CqKNmABB左右钢筋混凝土梁如图所示，采用级混凝土，均布荷载设计值包括自重），环境类别为一类，求截面受剪钢筋。4500A1800q=40kN/mB75.6kN72kN104.4kN189071.442kN·m64.8kN·m解：mmhhw60340-0400验算截面尺寸mmas4048.1200603bhw）进行验算：（则属于厚腹梁，应按式17-4150,30CCC，不超过混凝土的强度等级为KNVkNbhfcc104.44.2576032003.14125.025.00KNVKNV726.75截面符合要求)300HPB置箍筋（验算截面是否按计算配KNVKNbhft6.75072.7260320043.17.07.00mmmmhfbhfVsnAyvtSV/036.060302736020043.17.0106.757.023001有令,6.75kNVVu采用双肢箍筋6@200%127.027043.124.024.0141%.0002200328.2min,1yvtsvsvsvffbsAnmmmmmmmmsnAsv/036.0/283.02003.282221A支座)300HPB置箍筋（验算截面是否按计算配KNVKNbhft104.4072.7260320043.17.07.00有令,4.104kNVVummmmhfbhfVsnAyvtSV/0.33360302736020043.17.010104.47.023001采用双肢箍筋6@150mmmmmmmmsnAsv/333.0/377.01503.282221%127.027043.124.024.00.188%150200328.2min,1yvtsvsvsvffbsAn支座B左）当箍筋为（筋，求所需箍筋。利用现有纵筋为弯起钢（剪箍筋：）不设弯起钢筋时的受（求：，环境类别为一类，试包括自重），混凝土为的荷载弯矩设计值所示的简支梁，所承受如图32)130(/5041-4.4.4CmkNq8@200？时，弯起钢筋应为多少）计算截面最大剪力：（1解：kNql14421mmhhw56040-6000验算截面尺寸mmas4042.24250605bhw）进行验算：（则属于厚腹梁，应按式17-4150,30CCC，不超过混凝土的强度等级为KNVkNbhfcc144500.56052503.14125.025.00截面符合要求)300HPB置箍筋（验算截面是否按计算配KNVKNbhft144140.1460525043.17.07.00有令,144kNVVu故需按计算配置箍筋mmmmhfbhfVsnAyvtSV/025.060502756025043.17.0101447.023001mmmmmmmmsnAsv/036.0/402.02503.5022218@250采用双肢箍筋%127.027043.124.024.00.16%250250350.2min,1yvtsvsvsvffbsAn上弯起。）设其中的一根纵筋向（2KNfAVysbsb97.99223609.4908.0sin8.0则kNVVVsbAcs03.447.99144)300HPB置箍筋（验算截面是否按计算配00.70.71.43250560140.1444.03tfbhKNVKN则需要按构造配筋8@300采用双肢箍筋%127.027043.124.024.00.129%300250350.2min,1yvtsvsvsvffbsAn228.3141.14+270197.19140.14150VKNkN在纵筋的弯起点处符合要求，不必在弯起点再弯起钢筋当配置3)(8@200时：KNKNhSAfbhfVSVyvtcvcs14419.2165602003.50227056025043.17.000则则不需要配弯起钢筋。7.如图所示钢筋混凝土简支梁，采用C30混凝土，纵筋为热轧钢筋HRB400级钢筋，箍筋为HPB300级钢筋，如果忽略梁自重及架立钢筋的作用，环境类别为一类，试求此梁所能承受的最大荷载设计值F，此时该梁为正截面破坏还是斜截面破坏？2222=14.3/,=1.43/,=360/,f=270/ctyyvNmmfNmmfNmmNmm查表f02min001=40,=-=550-65=485,=22812281===2.14%=2204851.43550=0.45=0.45=0.2%360485ssssotyammhhammAmmAhbhhfhfh当梁达到最大弯矩值时0550.2%=0.2%=0.22%485hh同时012102360==2.14%=0.539=0.5181.014.3==0.518=1-0.5=1.014.32205100.5181-0.50.518=314.8kNm0.8=314.8,F=392.6kNybcbucbbffMfbhF取所以则021200===2.47485ah剪力达到最大值时1001.75==0.504+1=0.504+250.3=0.5041.43220485+270485150=164.7kN2V,247.083cvsvutyvunAVfbhfhsFFkN则=.综上，求得此梁所能承受的最大荷载设计值F24708kN此时该梁的皮怀属于斜截面破坏，可以%141.030043.124.024.0%335.01500023.502yvtmin,sv1svsvffbsnAbsAsv，属于厚腹梁455.2002/051//0wbhbh364650N＝510×200×14.3×1.0×0.25＝25.00ccubhfV49.205115000ha167.367kN0511503.50230005100243.1149.275.1175.1sv1yv0tuhsnAfbhfV不满足要求%,141.030043.124.03388.505130000，取haNbhfV6381305100243.11375.1175.10tu,min228.3====0.1415%=0.24200200svsvtsvsvyvAnAfbsbsf6.如图所示简支梁，环境类别为一类，混凝土为C30，求受剪箍筋。(1)如图7所示为该梁的计算简图和剪力图（2）验算截面条件1属厚腹梁424.225040600c0w，，bhbhNVNbhfk2075k.5000562503.14125.025.0max0cc截面尺寸符合要求。（3）确定箍筋数量（箍筋采用HPB335级钢筋）该梁既受集中荷载，又受均布荷载，但集中荷载在两支座截面上引起的剪力值均小于总剪力值的75％故梁的左右两半区段均应按均布荷载下的斜截面受剪承载力计算公式计算。由于梁所受的荷载是对称分布的，配筋亦是对称布置的，因此，可将梁分为AC、CD两个区段来计算斜截面受剪承载力。AC段：必须按计算配置箍筋7.075%,7%.50207105.cvVVBA总集中支座：NVNbhfk207k14.14005625043.117.07.0max0thsnAfbhfVsv1yv0t25.17.0所以选配箍筋8@200，实有CD段0.7ftbh0＝140140N＞VC＝86000N仅需按构造配置箍筋，选用由于此梁对称配筋，所以DE段选配箍筋，EF段选配箍筋。398.00563000140142070007.000sv1hfbhfVsnAyvt398.0503.02003.502sv1snA可以114%,.030043.124.024.0201%.020025023.50yvtminsv,svff8@3008@200