混凝土结构 第四版 混凝土设计原理 上册 (程文瀼 颜德姮 著) 中国建筑

课后答案网您最真诚的朋友网团队竭诚为学生服务，免费提供各门课后答案，不用积分，甚至不用注册，旨在为广大学生提供自主学习的平台！课后答案网：视频教程网：课件网：解：214.3cfNmm=；21.43tfNmm=；2300yfNmm=；C30混凝土，一类环境，c=25mm，as=35mm，h0=500-35=465mm6221090100.1161.014.3250465scMfbhαα×===×××查表得：0.1230.55bξξ==2101.014.32500.123465681.6300csyfbhAmmfαξ××××===选216+214钢筋，2710sAmm=(满足构造要求)。min1.430.450.450.00210.002300tyffρ==×=22min0.0021250500262.5710sbhmmAmmρ=××==取2710sAmm=3.2解：219.1cfNmm=；21.71tfNmm=；2360yfNmm=C40混凝土，二类a环境，c=30mm，as=40mm，h0=450-40=410mm62210140100.218119.1200410scMfbhαα×===×××查表得：0.2490.518bξξ==2101.019.12000.2494101083.3360csyfbhAmmfαξ××××===选322钢筋，21140sAmm=(满足构造要求)。min1.710.450.450.00210.002360tyffρ==×=22min0.00212004501891140sbhmmAmmρ=××==取21140sAmm=3.4解：214.3cfNmm=；21.43tfNmm=；2360yfNmm=；C30混凝土，一类环境，c=25mm，as=25+16/2=33mm，h0=450-33=417mm01360804101.20.5184172161.014.3200ysbcfAxmmhmmfbξα×====×=××min1.430.450.450.00180.002360tyffρ==×=22min0.002200450180804sbhmmAmmρ=××==10101.2()1.014.3200101.2(417)1068422ucxMfbxhkNmMkNmα=−=××××−=⋅=⋅安全3.5解：211.9cfNmm=；2300yyffNmm′==；C25混凝土，一类环境，c=25mm，设受拉钢筋放两层，as=60mm，h0=h-as=500-60=440mm2622100(10.5)26010111.92004400.55(10.50.55)628()300(44035)cbbsysMfbhAmmfhaαξξ−−×−×××××−×′===′′−×−102111.92000.554403006282548300cbyssyfbhfAAmmfαξ′′+××××+×===sA选325+322，sA＝2613mm2；sA′选220，sA′＝628mm23.6解：1）219.1cfNmm=；21.71tfNmm=；2360yfNmm=，C40混凝土，二类a环境，c=30mm，设受拉钢筋放两层，as=65mm，h0=750-65=685mm1066100()1.019.1550100(685)22667.11050010fcffhfbhhNmmNmmα′′′−=××××−=×⋅×⋅属于第一类T形截面62210500100.101119.1550685scfMfbhαα×===′×××查表得0.1070.518bξξ==（可不验算）1021.019.15500.1076852139360cfsyfbhAmmfαξ′××××===选622钢筋，22281sAmm=(满足构造要求)。min1.710.450.450.00210.002360tyffρ==×=22min0.0021250750393.752281sbhmmAmmρ=××==取22281sAmm=2）227.5cfNmm=；22.04tfNmm=；2360yfNmm=，C60混凝土，二类a环境，c=30mm，设受拉钢筋放两层，as=65mm，h0=750-65=685mm，10.98α=1066100()0.9827.5550100(685)22941.21050010fcffhfbhhNmmNmmα′′′−=××××−=×⋅×⋅属于第一类T形截面62210500100.0720.9827.5550685scfMfbhαα×===′×××查表得0.0750.499bξξ==（可不验算）1020.9827.55500.0756852115360cfsyfbhAmmfαξ′××××===选818钢筋，22036sAmm=(满足构造要求)。min2.040.450.450.002550.002360tyffρ==×=22min0.00255250750478.1252036sbhmmAmmρ=××==取22036sAmm=3.7解：214.3cfNmm=；21.43tfNmm=；2360yfNmm=；C30混凝土，一类环境，c=25mm，设受拉钢筋放两层，as=60mm，h0=h-as=500-60=440mm106680()1.014.340080(440)22183.041030010fcffhfbhhNmmNmmα′′′−=××××−=×⋅×⋅属于第二类T形截面110680()()1.014.3(400200)80(440)2291.5210fucffhMfbbhhNmmα′′′=−−=××−××−=×⋅66213001091.5210uuMMM=−=×−×=208.48×106N﹒mm622210208.48100.377114.3200440uscMfbhαα×===×××查表得0.5040.518bξξ==21021.014.32000.5044401762360csyfbhAmmfαξ××××===121()114.3(400200)80636360cffsyfbbhAmmfα′′−××−×===21217626362398sssAAAmm=+=+=选525钢筋，22454sAmm=(满足构造要求)。斜截面受剪承载力计算例题4－1解：1）剪力图见书，支座剪力为V＝011703.6522ql=××=124.6kN2）复合截面尺寸hw=h0=h-c-25/2=500-30-12.5=457.5457.52.34200whb==00.250.251.09.6200457.5219.6124.6ccfbhkNVkNβ=××××==满足。3）验算是否按计算配置腹筋00.70.71.1200457.570.5124.6tfbhkNVkN=×××==应按计算配置腹筋4）计算腹筋数量①只配箍筋由000.71.25svtyvAVfbhfhs≤+得：3312000.7124.61070.5100.4511.251.25210457.5svtyvnAVfbhsfh−×−×≥==××mm2/mm选双肢φ8箍筋1250.3223.10.4510.451svnAsmm×≤==取s=200mm验算最小配箍率1,min250.31.10.00250.240.240.0013200200210svtsvsvyvnAfbsfρρ×=====×=×满足仅配箍筋时的用量为双肢φ8@200②即配箍筋又配弯筋a.先选弯筋，再算箍筋根据已配的225+122纵向钢筋，将122的纵筋以45°角弯起，则弯筋承担的剪力：20.8sin0.8380.130064.52sbysbsVfAkNα==×××=3330100.70.8sin124.61070.51064.5101.251.25210457.5tysbssvyvVfbhfAnAsfhα−−×−×−×≥==××负值按构造要求配置箍筋并满足最小配箍率要求选双肢φ6@200的箍筋，1,min228.31.10.001420.240.240.0013200200210svtsvsvyvnAfbsfρρ×=====×=×b.先选箍筋，再算弯筋先按构造要求选双肢φ6@200的箍筋，1,min228.31.10.001420.240.240.0013200200210svtsvsvyvnAfbsfρρ×=====×=×满足要求。33002228.30.71.25124.61070.5101.25210457.5200118.50.8sin20.83002svtyvsbysAVfbhfhsAmmfα×−−×−×−×××≥==××弯起122的纵筋，Asb=380.1mm25）验算弯起钢筋弯起点处的斜截面抗剪承载力弯起钢筋弯起点距支座边缘500-30-22/2-30-20/2+50=469mm，该处剪力V1=124.6-0.469×70=91.77kN3300228.30.71.2570.5101.25210457.5104.510200svcstyvAVfbhfhNs×=+=×+×××=×91.77×103N故不需要弯起第二排钢筋或加大箍筋用量。4－2解：1）剪力图见书2）复合截面尺寸hw=h0=h-c-20/2=600-30-10=5605602.84200whb==0max0.250.251.09.6200560268.8180ccAfbhkNVVkNβ=××××===满足。3）验算是否按计算配置腹筋A支座：V16088%V180==集总；B支座：V14087.5%V160==集总梁的左右区段均应按集中荷载作用下的独立梁计算将梁分为AC、CD、DE、EB段来计算斜截面受剪承载力AC段：010001.79560ahλ===01.751.751.120056077.31.01.791tfbhkNλ=×××=++VA=180kN应按计算配置腹筋CD段：020003.573560ahλ===，取3λ=01.751.751.120056053.91.031tfbhkNλ=×××=++VC=50kN按构造要求配置箍筋，选用φ6@350的箍筋。DE段：020003.573560ahλ===，取3λ=01.751.751.120056053.91.031tfbhkNλ=×××=++VE=70Kn应按计算配置腹筋EB段：010001.79560ahλ===01.751.751.120056077.31.01.791tfbhkNλ=×××=++VB=160Kn应按计算配置腹筋4）计算腹筋数量AC段：30101.75(18077.3)101.00.873210560AtsvyvVfbhnAsfhλ−−×+≥==×mm2/mm选双肢φ8箍筋1250.3115.20.8730.873svnAsmm×≤==取s=110mm验算最小配箍率1,min250.31.10.004570.240.240.00126200110210svtsvsvyvnAfbsfρρ×=====×=×满足箍筋用量为双肢φ8@110DE段：30101.75(7053.9)101.00.137210560EtsvyvVfbhnAsfhλ−−×+≥==×选双肢φ8箍筋1250.3734.30.1370.137svnAsmm×≤==取s=250mm验算最小配箍率1,min250.31.10.002010.240.240.00126200250210svtsvsvyvnAfbsfρρ×=====×=×满足箍筋用量为双肢φ8@250EB段：30101.75(16077.3)101.00.703210560BtsvyvVfbhnAsfhλ−−×+≥==×选双肢φ8箍筋1250.3143.10.7030.703svnAsmm×≤==取s=140mm验算最小配箍率1,min250.31.10.0035