混凝土课程设计无错版(中南大学)

混凝土结构课程设计—1—一、钢筋砼连续梁设计一钢筋砼矩形截面两跨连续梁，一类工作环境，承受均布恒载标准值为gk(不含自重)，均布活载标准值qk，在每跨三分点处截面还承受集中恒载标准值GK，集中活载标准值QK，活载准永久值系数ψK=0.5，如图一所示。跨度、砼采强度等级，纵筋级别见表，箍筋采用HPB235级，按《混凝土结构设计规范（GB50010-2010）》设计该梁。要求：(1)进行正截面及斜截面承载力计算，并确定所需的纵筋、箍筋和弯起钢筋的数量。（要求支座处考虑纵向钢筋弯起抗剪）(2)绘制抵抗弯矩图和弯矩包络图，并给出各根弯起钢筋的弯起位置。(3)验算裂缝是否满足要求。(4)验算挠度是否满足要求。解：１、设计计算数据混凝土C302/1.20mmNfck；2/01.2mmNftk；2/3.14mmNfc；2/43.1mmNft；482.0b；0.11；8.01；mmc20；24/1000.3mmNEc纵筋HRB50022/410,/435mmNfmmNfyy；25/100.2mmNEs箍筋HPB235210yvf2/mmN2、截面尺寸、荷载及内力计算混凝土结构课程设计—2—跨度：mmbllno600024024060001mmlbllnn60241202406000025.1025.0202取两者的较小者mml60000取mmbmmlmmh250,4001560001570005.38.22507000.2bh，满足要求。;①恒载标准值计算mKNMk12.105660222.0610070.021mKNMMKK12.10512mKNMB88.164660333.0610125.0262.52KNVV117.48KNVV117.48KN60333.1610625.052.6260667.0610375.0ACBB左右左BAVKNV②活载满布时内力计算KN35.51VVKN65.96VV6.65950333.168625.035.5150667.068375.09.135650333.068125.076.8676.86650222.068070.0ACBB21221左右左KNVKNVmKNMmKNMMmKNMBABKKK③仅左跨作用活载时内力计算混凝土结构课程设计—3—KN374.1150167.068063.0VKN374.1150167.068063.0VKN374.8550167.168563.0626.6250833.068437.0244.68650167.068063.0016.37048.1113131048.111650278.068096.0CB21221右左BABKKKVKNVmKNMmKNMMmKNM④仅右跨作用活载时内力计算KN626.62VKN374.85VKN364.11374.11244.68CB右左BABVKNVmKNM3、内力组合求最不利荷载及控制截面⑴①+②KN914.146VVKN286.276VVKN286.2764.165.962.148.117914.1464.135.512.152.62116.3884.19.1352.144.168608.247608.2474.1*76.862.1*12.105ACBB121左右左BABVKNVmKNMmKNMMmKNM⑵①+③KN1004.594.1374.112.152.62VKN8996.1564.1374.112.148.117VKN4996.2604.1374.852.148.1177004.1624.1626.622.152.623976.2934.1244.682.188.164MKN3216.744.1016.372.112.105MmKN6112.2814.1048.1112.1105.12MCBB21右左BAVKNVmKNm⑶①+④mKN048.111MmKN016.37M2K1K混凝土结构课程设计—4—KNKNKNVKNVmKNMmKNMmMBAB7004.1624.1626.622.152.62V4996.2604.1374.852.148.117V8996.1564.1374.112.148.1171004.594.1374.112.152.623976.2934.1244.682.188.1646112.2814.1048.1112.112.105KN3216.744.1106.372.112.105CB21右左剪力和弯矩包络图如下：剪力图弯矩图由图可知剪力的控制截面在Ａ、Ｂ、Ｃ支座截面，弯矩的控制截面在１、２、Ｂ截面处。混凝土结构课程设计—5—4、验算控制截面尺寸按配置两排纵筋验算；取mmas60；mmhhw640607000456.2250640bhwmKNMmKNbhfKNVNbhfcbbcc116.3887.535106402503.140.1)482.05.01(482.0)5.01(286.2765720006402503.140.125.025.0max62201max0所以截面尺寸满足要求。5、根据正截面承载力计算纵向钢筋跨中最大正弯矩mKNM6112.281max，按单筋截面设计，采用一排布置。mmas35；mmh665357000；mmhxbb53.320665482.00sytycscAmmbhAffmmfbxfAmmmmbhfMhx2minminmaxmaxmin21262010350700250002.0002.0]43543.145.0%,2.0[]45.0%,2.0[660.1039435504.1262503.140.153.320504.126)6652503.14106112.281211(665)211(满足要求，即在截面受拉区配置4根直径为22mm的二级钢筋21520mmAS中间支座为最大负弯矩处，mKNM116.388max，若按单筋截面设计：482.0314.0265.0211211265.06652503.140.110116.38826201bscsbhfM2min201350082.1716435665314.02503.140.1mmAmmfhbfAycs选用5根直径为22mm的二级钢筋，22082.17161900mmmmAs混凝土结构课程设计—6—6、根据斜截面承载力配置箍筋和弯起钢筋支座边缘处最大剪力设计值为KNV9164.159，而在集中荷载作用下支座边缘的剪力为集V115.0KN，%75688.0159.91645.011VV集。所以不需要考虑剪跨比。全长按一排配筋。mmh65910202227000KNNbhft502.27316491407565925043.17.07.00，需按计算配置腹筋选用双肢箍6@100,。（1）不配弯起钢筋。%163.021043.1*24.0%228.0100*250579164.1592437976591005721065925043.17.07.0min,00svsvsvsvyvtcssbANVNhSAfbhfV可得两端支座不需配置弯起钢筋。NVNhSAfbhfVsvyvtcs27050226351825.17.0100其中1V为中间支座处边缘剪力设计值，可见中间支座处需配置弯起钢筋。（2）中间支座处配置弯起钢筋。取45s；则211573.4045sin4358.0263518273502sin8.0mmfVVAosycssb在下边缘弯起一根22的纵筋，21.380mmAs钢筋弯起点距中间支座边缘的距离为mm800600200，弯起点处的剪力设计值为csVKNV942.2548.02.23502.2732。所以不需弯起第二排钢筋。7、裂缝验算该构件允许出现裂缝，按三级抗裂计算。（1）在中间支座边缘处mKNMk78.3009.13588.164钢筋应力为260/115.267190065987.01078.30087.0mmNAhMsKsk混凝土结构课程设计—7—矩形截面受弯件上半截面受拉01.00217.02507005.019005.0bhAAAsteste又因为882.0115.2760217.001.265.01.165.01.1sktetkf换算钢筋直径mmdvndndiiiiieq22220.1522522)08.09.1(maxteeqsskcrdcE)0217.02208.0209.1(100.2115.276882.09.15mmmm3.0275.0lim满足抗裂要求。（2）跨中抗裂验算KNMk168.226048.11112.105钢筋应力为：260/53.259152065987.010168.22687.0mmNAhMsKskk矩形截面受弯构件下半截面收拉01.00174.02507005.015205.0bhAAAsteste又因为811.053.2590174.001.265.01.165.01.1sktetkf换算钢筋直径mmdvndndiiiiieq22220.1522522)08.09.1(maxteeqsskcrdcE)0174.02208.0209.1(100.253.259811.09.15mmmm3.0278.0满足抗裂要求。混凝土结构课程设计—8—8、挠度验算（1）短期刚度sB667.610*0.3100.245csEEE0115.065925019000bhAs则fEssshAEB5.3162.015.12005.310115.0667.662.0882.015.16591900100.225213/10856.9mmN（2）挠度增大系数根据《规范》，取0.2（3）受弯构件的刚度B21313/10900.510856.9168.21639.150)10.2(168.216)1(mmNBMMMBskqk（4）跨中挠度mmBlMfk74.1310900.5600010168.21648513262允许挠度为mmfmmlf74.132425060002500lim故挠度满足要求。二、预应力混凝土简支梁设计一多层房屋的预应力混凝土屋面梁，构件及截面尺寸如图二所示。先张法施工时在工地临时台座上进行，在梁的受拉、受压区采用直径10mm的热处理45Si2Cr直线预应力钢筋。分别在梁的受拉、受压区采用锥形锚具一端同时超张拉钢筋。养护时预应力钢筋与张拉台座温差为250C，混凝土达到设计强度以后放松预应力钢筋，混凝土采用C40，非预应力钢筋采用HPB235钢筋。