正截面第8至第14题答案

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题4.8矩形截面梁,梁宽b=200mm,梁高h=500mm,混凝土等级为C35,钢筋采用HRB335级,HRB400级和HRB500级,环境类别为一类,截面所承受的弯矩值180MkNm,试分别计算所需的纵向受拉钢筋截面面积sA,并分析sA与钢筋抗拉强度设计值的关系。解:环境类别为一类,取c=20mm,11.0混凝土强度等级为C35,216.7/cfNmm,21.57/tfNmmHRB335:21300/yfNmm,10.236%,10.550bHRB400:22360/yfNmm,20.2%,20.518bHRB500:23435/yfNmm,30.2%,30.482b按只有一排受拉钢筋考虑,40samm,则050040460hmmmmmm622210180100.2551.016.7/200(460)scMNmmfbhNmmmmmm1231121120.2550.300,,sbbb,都满足条件22101211.016.7/2000.3004601536300/csyfbhNmmmmmmAmmfNmm22102221.016.7/2000.3004601280360/csyfbhNmmmmmmAmmfNmm22103231.016.7/2000.3004601060435/csyfbhNmmmmmmAmmfNmm可知,随着钢强度等级加强,sA减小,变化较显著。4.9题矩形截面梁,梁宽b=200mm,梁高h=400mm,'40ssaamm,混凝土等级为C30,钢筋采用HRB400级,环境类别为一类。求下列情况下截面所能抵抗的极限弯矩M。(1)、单筋截面,2942sAmm(2)、双筋截面,2'2942,226ssAmmAmm(3)、双筋截面,2'2942,628ssAmmAmm(4)、双筋截面,'2942ssAAmm解:查表可得HRB400:'2360/yyffNmm,20.2%,20.518b混凝土强度等级为C30,214.3/cfNmm,21.43/tfNmm环境类别为一类,取c=20mm,11.0,0.518bmin0.451.43max{,0.2%}0.2%360(1)、单筋截面,2942sAmm2minmin0.2%200400160ssAbhmmmmmmA,满足最小配筋率的要求040,40040360sammhmmmmmm22210360/9420.3291.014.3/200360ysbcfANmmmmfbhNmmmmmm(10.5)0.329(10.50.329)0.275s222100.2751.014.3/200(360)101.9scMfbhNmmmmmmkNm(2)、双筋截面,2'2942,226ssAmmAmm2minmin0.2%200400160ssAbhmmmmmmA,满足最小配筋率的要求040,40040360sammhmmmmmm''2222210360/942360/2260.2501.014.3/200360sysybcfAfANmmmmNmmmmfbhNmmmmmm则02/0.2220.250sbh,满足公式适用条件(10.5)0.25(10.50.25)0.219s2'''1002222()0.2191.014.3/200(360)360/226(36040)107.2ssscyMfbhfAhNmmmmmmNmmmmmmmmkNm(3)、双筋截面,2'2942,628ssAmmAmm2minmin0.2%200400160ssAbhmmmmmmA,满足最小配筋率的要求040,40040360sammhmmmmmm''2222210360/942360/6280.1101.014.3/200360sysybcfAfANmmmmNmmmmfbhNmmmmmm则02/0.2220.110sh,采用近似公式计算'022()360/942(36040)108.52sysMfAhNmmmmmmmmkNm(4)、双筋截面,'2942ssAAmm2minmin0.2%200400160ssAbhmmmmmmA,满足最小配筋率的要求040,40040360sammhmmmmmm''2222210360/942360/94201.014.3/200360sysybcfAfANmmmmNmmmmfbhNmmmmmm则02/0.2220sh,采用近似公式计算'022()360/942(36040)108.52sysMfAhNmmmmmmmmkNm题4.10双筋矩形截面梁,梁宽b=250mm,梁高h=700mm,,,混凝土等级为C30,钢筋采用HRB400级,'40ssaamm,受压区已有配筋216,并且在计算机中考虑其受压作用。截面所承受的弯矩设计值195MkNm,试计算所需的纵向受拉钢筋面积。解:HRB400:2360/yfNmm,0.2%,0.518b混凝土强度等级为C30,214.3/cfNmm,21.43/tfNmm2402sAmm0'''62202221()19510360/402(66040)0.06761.014.3/250(660)ssyscMfAhNmmNmmmmmmmmfbhNmmmmmm1121120.06760.0701sb02/0.1210.0701sh采用近似公式'0622()19510874360/(36040)ssyMAfhNmmmNmmmmmm题4.12双筋矩形截面梁,梁宽b=250mm,梁高h=700mm,混凝土等级为C30,钢筋采用HRB400级,'40ssaamm,受压区已有配筋220,并且在计算机中考虑其受压作用。截面所承受的弯矩设计值195MkNm,试计算所需的纵向受拉钢筋面积。解:解:HRB400:2360/yfNmm,0.2%,0.518b混凝土强度等级为C30,214.3/cfNmm,21.43/tfNmm,11.02628sAmm0'''62202221()19510360/628(66040)0.0351.014.3/250(660)ssyscMfAhNmmNmmmmmmmmfbhNmmmmmm1121120.06760.036sb02/0.1210.036sh采用近似公式'0622()19510874360/(36040)ssyMAfhNmmmNmmmmmm题4.12T型截面梁,''500,100,200,500ffbmmhmmbmmhmm,环境类别为一类,混凝土等级为C30,钢筋采用HRB400级。求下列情况下截面所能抵抗的极限弯矩M。(1)、纵向受拉钢筋942,40ssAmmamm(2)、纵向受拉钢筋1884,65ssAmmamm解:解:HRB400:2360/yfNmm,0.2%,0.518b混凝土强度等级为C30,214.3/cfNmm,21.43/tfNmm,11.0min0.451.43max{,0.2%}0.2%360(1)、纵向受拉钢筋942,40ssAmmamm2minmin0.2%200500200ssAbhmmmmmmA,满足最小配筋条件22''21360/9423391201.014.3/500100715000yscfffANmmmmNfbhNmmmmmmN属于第一类T型截面0460shhmm22210360/9420.1031.014.3/500460ysbcfANmmmmfbhNmmmmmm(10.5)0.103(10.50.103)0.098s222100.0981.014.3/500(460)148.3scMfbhNmmmmmmkNm(2)、纵向受拉钢筋1884,65ssAmmamm2minmin0.2%200500200ssAbhmmmmmmA,满足最小配筋条件22''21360/18846782401.014.3/500100715000yscfffANmmmmNfbhNmmmmmmN属于第一类T型截面0435shhmm22210360/9420.2181.014.3/500435ysbcfANmmmmfbhNmmmmmm(10.5)0.218(10.50.218)0.194s222100.1941.014.3/500(135)262.5scMfbhNmmmmmmkNm题4.13T型截面梁简支梁,''500,100,200,500ffbmmhmmbmmhmm,环境类别为一类,混凝土等级为C25,钢筋采用HRB400级。试确定下列情况下所需的纵向受拉钢筋截面面积sA。(1)、弯矩设计值120MkNm,预计一排钢筋;(2)、弯矩设计值290MkNm,预计二排钢筋;解:HRB400:2360/yfNmm,0.2%,0.518b混凝土强度等级为C25,211.9/cfNmm,21.27/tfNmm,11.0min0.451.43max{,0.2%}0.2%360取040,460ssmmhhmm'''210100()1.011.9/500100(460)243.9522fcffhmmfbhhNmmmmmmmmkNmM属于第一类T型截面。06'2221120100.0951.011.9/500(460)scfMNmmfbhNmmmmmm1121120.0950.1sb则'2102min21.011.9/5000.1460760360/cfssyfbhNmmmmmmAmmAfNmm,满足最小配筋率要求。题4.14T型截面梁,''500,100,200,500ffbmmhmmbmmhmm,环境类别为一类,混凝土等级为C30,钢筋采用HRB400级。受压区已有配筋220('2'628,40ssAmmamm),并且在计算中考虑其受压作用。截面所承受的弯矩设计值290MkNm,试计算所需的受拉钢筋面积sA。解:HRB400:2360/yfNmm,0.2%,0.518b混凝土强度等级为C30,214.3/cfNmm,21.43/tfNmm,11.0min0.451.43max{,0.2%}0.2%360因题目给出'sA,故可以判断受拉亚为2排,65smm0435shhmm(1)、判别T型截面类型'''210100()1.014.3/500100(435)275.322fcffhmmfbhhNmmmmmmmmkNmM故为第二类T型截面(2)、双筋T型计算思路类似于双筋,采用分解法分为三个部分:钢构件+翼缘(挑出)+单筋○1、由图○1知,'21628ssAAmm'''2210()628360/(43540)89.3yssMfAhmmNmmmmmmkNm○2、由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