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第5章练习题参照答案第5章练习题参照答案【5-1】已知钢筋混凝土梁的截面尺寸为b250mm,600mm,混凝土保护层厚度c25mm,混凝土和钢筋资料的性能指标为fc23N/mm2,ft2.6N/mm2,Ec2.51104N/mm2;fy357N/mm2,Es1.97105N/mm2,受拉区配有325(As1472mm2)的纵向受拉钢筋。试计算(1)当截面所受的弯矩M50kNm时的s、c和;t(2)截面的开裂弯矩Mcr及相应的s、c和cr。t【解】(1)求当截面所受的弯矩M50kNm时的s、t和cEEs1.971057.849,EAs7.84914720.077,Ec2.51104bh250600A2EAs0.1540bh12EAsh10.1540600xcrbh321mm1EAs210.0772bhh0hcd60025562.5mm2252Mcrftb(hxcr)hxcr2xcr2EftAsh0xcr2332.6250(600321)600321232127.8492.6147250032123393506071Nmm93.506kNmM50kNmMcr93.506kNm,截面处于弹性阶段,此时采纳线性的物理关系tbtbt,scc,tsEcEcEs由几何变形方程:tbctsxnhxnh0xn将几何变形方程代入物理方程得:ttxnEc,ccEcsh0xnEs1)

bbhxnEcttEcsh0xnEs由受力均衡:X0,1ctxnbsAs1tb(hxn)b22代入(1)式,得:xn2Ec2sAs(hxn)2Ech0sbh0sbxnEsxnEsxn2Ecb2As(h0xn)Es(hxn)2Ecb2(AsEsbhEc)xn2Ash0Esbh2EcxnAsh0Es0.5bh2EcAsh0E0.5bh21472562.57.8490.52506002AsEsbhEcAsEbh14727.849319mm250600由力矩均衡:M0,MsAsh0xn1tb(hxn)b2h323xn(hxn)2EcbhsAsh0h0xnEss33解得:sxM(hx)250106(600319)218.27N/mm2bh1472562.5319250600Ash0nn33h0xn337.849562.5319EtxnEc31918.275828.133.05N/mm2ch0xnEss319)7.8491911.2315(562.5s(h0s(562.518.2718.271073.811071/mmh0xnxn)Es319)1.971054.79695(2)截面的开裂弯矩Mcr及相应的s、c和crt由前面计算可知,McrkNm93.506s2Eft27.8492.640.81N/mm2由截面静力均衡条件得tftb(hxcr)sAs2.6250(600321)40.811472241422.322c0.5bxcr0.5250321401256.02N/mmcrft2.6321)7.431071/mm0.5Ec(hxcr)0.52.51104(600【5-2】条件同练习题5-1,试计算1)截面的极限弯矩Mu及相应的u;2)按等效矩形应力争形的简化方法求截面的极限弯矩Mu3)与例5-2的计算结果进行比较。【解】(1)求截面的极限弯矩Mu及相应的uAs14720.981%min0.45ft0.452.60.328%bh250600fy357nb1.01.00.646fy3571150.0033Es0.00331.9710对强度等级不大于C50的混凝土,ccu0.0033,00.002,代入静力均衡方程,得tfc20.00414Es(1n)0,As1472nbh02501.05%562.523n20.004141.971051.05%(1n)020.37233n0.372330n0.372330.37233240.372330.9036n220.452nb1nt1.9710510.4520.0033788.17N/mm2fy357N/mm2sEsc0.452n混凝土压碎前,钢筋已折服(处于Ⅲa状态),以fy代人得fy1.2531.05%357n1.2530.204nbfc23MufyAsh0(10.412n)3571472562.5(10.4120.204)106270.75kNmt0.003328.761061/mmucnh00.204562.5(2)简化方法fy3570.16300.80.81.05%23bfy0.5161fc1.0113570.0033Es50.00331.9710MuAsfyh0(10.5)1472357562.5(10.50.163)106271.50kNm3)联合例5-2当As760mm2时,Mu当As1472mm2时,Mu

可知146.40kNm,u55.721061/mm270.75kNm,u28.761061/mm当As1520mm2时,Mu279.57kNm,u27.861061/mm跟着配筋量的增添,截面的抗弯承载力会提升,但截面的变形能力会降低。5-3】条件同练习题5-1,求当截面的纵向受拉钢筋分别为212和1028时截面的抗弯承载力。【解】(1)当截面的纵向受拉钢筋为212(As226mm2)时As2260.15%<min0.45ft0.452.60.328%0.2%bh250600fy357E7.849,A2EAs27.8492260.02365bh250600MuMcr0.292(12.5A)ftbh20.292(12.50.02365)2.6250600210672.37kNm(2)当截面的纵向受拉钢筋为1028(As6158mm2)时b0.80.80.516fy135710.0033Es0.00331.97105bb1fc0.5161.0233.324102fy357纵筋许多,需要分三层部署,从下往上钢筋的数目分别为4、4、2。4600252846002528282826002542828h0222516.2mm10As61584.772102b3.324102,超筋梁bh0250516.2sfy0.8,代入静力均衡方程,1fcbh0sAs,得b0.80.80.80.57831fcbh0(0.8b)11.023250516.21(0.80.516)fyAs3576158xh00.5783516.2299mmMu1fcbh02121.023250516.220.578310.57831062629.85kNm【5-4】已知某简支钢筋混凝土平板的计算跨度为l02.8m,板厚h90mm,配置10@200的钢筋,混凝土和钢筋资料的性能指标为fc11.9N/mm2,ft1.6N/mm2;fy280N/mm2,Es2.0105N/mm2。试求该板每m2承受的荷载是多少?【解】板的保护层厚度c15mm,取c15mm;h0hcd90151070mmd220.20.257,取min0.257%min45ft451.6fy280b0.80.80.5617fy280110.00332.01050.0033Esbb1fc0.56171.011.92.387102fy280取单宽即b1000mm,因配置10@200,因此在单宽范围内为410,As314mm2。As3140.4486%,minb,为适筋矩形截面。bh0100070由静力均衡得,xfyAs2803147.388mm1fcb1.011.91000由力矩平衡得,MufyAsh0x280314707.3881065.904kNm22M1qbl02Mu5.904kNm8q85.9046.024kN/m21.02.82【5-5】钢筋混凝土简支梁的截面尺寸为b220m,h500m,计算跨度为l06m,承受均布q24kN/m(包括梁的自重),配有受拉纵筋和钢筋资料的性能指标为fc

2222213N/mm,

20,混凝土2ft1.2N/mm;fy365N/mm2,Es1.97105N/mm2。问此梁的正截面能否安全?【解】(1)计算最小配筋率、界线配筋率min0.45ft0.451.21.479103fy3650.80.8bfy0.512113650.0033Es0.00331.97105bb1fc0.5121.0131.824102fy365(2)当纵筋为222220时As7606281388mm2As13881.262102min1.479103bh220500h05002522464mm2As13881.360102b1.824102bh0220464为适筋截面(3)截面承载力计算fyAs3651388x177mm1fcb1.013220截面的极限承载力:MufyAsh0x3651388464177190.24106Nmm190.24kNm22梁中遇到的最大弯矩为:Mmax1ql212462108kNmMu190.24kNm88此梁正截面安全。5-6】某简支梁的截面尺寸为b200mm,h500mm,混凝土和钢筋资料的性能指标为fc13N/mm2,ft1.2N/mm2;fy的弯矩分别为求相应的配筋论之。【解】

310N/mm2,Es1.97105N/mm2。当其所受M40,60,80,100,120,140,160,180,200,220kNm时,As为多少?画图表示MAs关系并讨1)计算最小配筋率、界线配筋率、适筋构件的最大抗弯承载力min0.45ft0.451.21.7421030.2%fy310min0.2%b0.80.80.542fy310110.0033Es0.00331.971051fc0.5421.0132.273102bbfy310s,maxb10.5b0.54210.50.5420.395因为不知道直径,取h050035465mmMu,maxs,max1fcbh020.3951.0132004652222.06106Nmm222.06kNm(2)当M40kNmMu,max222.06kNm时1fcbxfyAs由Mu1fcbxx得Mh02x34mm,As288mm2;As2.88103min2.0103bh故取As288mm2(3)当M60kNmMu,max222.06kNm时同理得x53mm,As441mm2(4)当M80kNmMu,max222.06kNm时同理得x72mm,As601mm2(5)当M100kNmMu,max222.06kNm时同理得x92mm,As770mm2(6)当M120kNmMu,max222.06kNm时同理得x113mm,As948mm2(7)当M140kNmMu,max222.06kNm时同理得x136mm,As1137mm2(8)当M160kNmMu,max222.06kNm时同理得x160mm,As1340mm2(8)当M180kNmMu,max222.06kNm时同理得x205mm,As1721mm2(注:配328,As1847mm2排能够放下)(9)当M200kNmMu,max222.06kNm时,此时与Mu,max

单值较靠近,考虑部署二排钢筋,取h050060440mm同理得x241mm,As2018mm2(10)当M220kNmMu,max222.06kNm时,此时与Mu,max值较靠近,考虑部署二排钢筋,取h050060440mm同理得x284mm,As2381mm2(11)绘制MAs关系图(略)u,max议论:梁的正截面承载能力与钢筋的面积之间不是简单的线性关系,跟着配筋量的增添,承载能力增添的速度渐渐降落,并趋于M。【5-7】已知某简支混凝土平板的计算跨度为l01.92m,板厚h80mm,承受均布荷载q4kN/m2(包括梁的自重),混凝土和钢筋混资料的性能指标为fc13N/mm2,ft1.2N/mm2;fy270N/mm2,Es1.90105N/mm2。求板的配筋。【解】取单宽板带为剖析对象,则b1000mm,h0802060mm跨中最大弯矩为:Mmax1ql02141.9221.84kNm88b0.80.80.559fy127010.0033Es0.00331.90105s,maxb10.5b0.55910.50.5590.403Mu,maxs,max1fcbh020.4031.013100060218.86106Nmm18.86kNmMmax1fcbxfyAs由Mu1fcbxh0x得x2.4mm,As116mm2M2min0.45ft0.451.22.01030.2%fy270故min0.2%As1161.451030.2%bh100080按结构配筋,受力钢筋可选

Asminbh0.2%100080160mm26@150散布钢筋,单位宽度上的配筋不宜小于单位宽度上受力钢筋的15%,且配筋率不宜小于0.15%;散布钢筋的直径不宜小于6mm,间距不宜大于250mm。故散布钢筋可选6@200。【5-8】已知某双筋矩形截面梁b400mm,h1200mm,配置的受压钢筋428,1228的受拉钢筋,混凝土和钢筋资料的性能指标为fc14.3N/mm2,ft1.43N/mm2,Ec3.0104N/mm2;fy310N/mm2,Es1.97105N/mm2,试计算(1)截面的开裂弯矩Mcr及相应的cr(2)截面的极限弯矩Mu及相应的u(一般方法)(3)截面的极限弯矩Mu及相应的u(简化方法)【解】压区钢筋面积As2463mm2,受拉钢筋的面积As7390mm2EEs1.971056.567Ec3.01042EAs26.5677390Abh0.20240012002EAs26.5672463Abh0.0674001200(1)截面的开裂弯矩计算Mcr0.29212.5A0.25Aftbh20.29212.50.2020.250.0671.4340012002366.00106Nmmcr2tu4ft41.431.59107hEch3.01041200(2)受拉钢筋双排部署,h0hcd28141130mmd1200282ascd282842mm22nb110.677fy310110.0033Es0.00331.97105As73901.635102bh04001130As24635.449103bh04001130n1.253fyfy1.2531.6351023105.4491033100.292fcfc14.314.3nb,受拉钢筋能折服xnnh00.2921130330mm2.5as105mm,受压钢筋能折服asMufyAsh010.412nfyAsh00.412nh03107390113010.4120.292310246311300.412420.29211302349.01106Nmm2349.01kNmucu0.00331.001051/mmnh00.2921130(3)As2fyAs2463mm2,As1AsAs2739024634927mm2fyMufyAsh0as3102463113042830.72106Nmm830.72kNmfyAmx1.014.31fcb400b0.80.80.542fy310111050.00331.970.0033Es2as24284mmx267mmbh00.5421130612mm,为适筋截面Mu1fyAs1h0x310492711302671522.02106Nmm1522.02kNm22MuMuMu1830.721522.022352.74kNmu1cu0.80.00339.891061/mmx267【5-9】简支梁如图5-48所示,混凝土和钢筋材料的性能指标为fc16N/mm2,ft1.5N/mm2;fy365N/mm2,Es1.97105N/mm2。求所能承受的均布荷载(包含梁的自重)q。【解】受拉钢筋配筋As942mm2,受压钢筋配As628mm2fyAs365628628mm2As2fy365As1AsAs2942628314mm2c25mm,h0hcd3502520315mm,ascd35mm222MufyAsh0as3656283153564.18106Nmm64.18kNmb0.80.80.512fy365110.0033Es0.00331.97105bh00.512315161mm,2as23570mmfyAs136531436mmx1.0162001fcb2as,受压钢筋不可以折服MufyAsh01as36594231513596.27106Nmm96.27kNmh0315MuM1ql02q8Mu896.2743.66kN/m8l024.22【5-10】图5-49所示的两跨连续梁,混凝土和钢筋的资料性能指标为fc13N/mm2,ft1.2N/mm2;fy370N/mm2,Es1.97105N/mm2。求所能承受的荷载P值。【解】对称配筋,AsAs1520mm2c25mm,h0hcd5002522464mm,ascd252236mm2222此时,As1AsAs20mm2,xfyAs10mm,x2as,受压钢1fcb筋不可以折服MufyAsh01as3701520464136240.71106Nmm240.71kNmh0464作出弯矩图(略),跨中的最大弯矩为M0.156Pl,中支座的最大负弯矩为M0.188PlMuMu0.188PlP214kN0.188l【5-11】已知钢筋混凝土简支梁的截面尺寸为200mm,h500mm,混凝土和钢筋资料的性能指标为fc16N/mm2,ft1.5N/mm2;fy320N/mm2,Es1.97105N/mm2。承受的弯矩M250kNm,求所需受拉受压钢筋As及As。【解】(1)计算参数确立h050035465mm0.80.8bfy0.5363200.0033Es10.00331.971052)求As1和M1xbh0249mmAs11fcbx1.0162002492490mm2fy320M1Mu1As1fyh00.5x24903204650.5249271.31106Nmm271.31kNmMu1M250kNm,仅配受拉钢筋即能知足受弯要求。(3)结构配筋(略)【5-12】某钢筋混凝土梁的截面尺寸为250mm,h600mm,受压区已配有322的纵筋,混凝土和钢筋资料的性能指标为f1.2N/mm;fc13N/mm22tfy310N/mm2,Es1.97105N/mm2。承受的弯矩M330kNm,求所需受拉钢筋As。【解】(1)计算参数确实定受压区配置h0h6060060540mmacd252236mms22(2)计算As2和MAs2Asfy/fyAs1140mm2MAs2fyh0as114031054036178.11106Nmm178.11kNmMu1MMu330178.11151.89kNm1fcbxfyAs1即1.013250x310As1Mu11fcbxh0xMu11.013x解得x95mm22250x540As1995mm2b0.80.80.542fy131010.0033Es0.00331.97105bh00.542540293mm2asxbh0AsAs1As299511402135mm2【5-13】某钢筋混凝土梁的截面尺寸为250mm,h600mm,受压区已配有214的纵筋,混凝土和钢筋资料的性能指标为f1.2N/mm;fc13N/mm22tfy370N/mm2,Es1.97105N/mm2。承受的弯矩M390kNm,求所需受拉钢筋As。【解】As308mm2,h0565mm,as25732mm,取As2As308mm2MufyAsh0as3703085653260740680Nmm60.74kNmMu1MMu39060.74329.26kNm1fcbxfyAs1Mu1xfyAs1x1fcbxh0h022解得,224mmAs11963mm20.80.8bfy0.510370110.0033Es0.00331.971052as23264mmx224mmbh00.510565288mmAsAs1As230819632271mm2【5-14】已知T形截面梁bf280mm,hf120mm,b180mm,450mm,计算跨度l05m,配置325的纵筋,混凝土和钢筋资料的性能指标为fc13N/mm2,ft1.2N/mm2;fy370N/mm2,Es1.97105N/mm2。试求截面的抗弯承载力。【解】bf280mm,bf280mml0/35000/31667mmbf280mmb12hf180121201620mm切合要求。受拉区,As1473mm2,minft1.230.450.451.45910fy370As,minminbh1.459103180450118mm2,As,minAs1fcbfhf1.013280120436800N436.8kNfyAs3701473545010N545.01kN属于第二类T形截面。h04502525412.5mm2Muf1fcbfbhfh0hf1.0132801801201206mm2412.554.9910N21fcbx1fcbfbhffyAs解得:x166mm0.80.80.510bfy370111050.0033Es0.00331.97x1660.402h0b412.5Mu11fcbxh0x1.013180166412.5166127990980Nmm127.99kNm22MuMu1Muf127.9954.99183kNm【5-15】已知T形截面梁bf2500mm,hf180mm,b400mm,h1200mm,计算跨度l012mm,配置1225的纵筋,混凝土和钢筋资料的性能指标为fc14.3N/mm2,ft1.4N/mm2;fy370N/mm2,Es1.97105N/mm2。试求截面的抗弯承载力。【解】bf2500mm,bf2500mml0/312000/34000mmbf2500mmb12hf400121802560mm切合要求。min0.45ft451.40.170%0.2%fy370Asminminbh0.2%4001200960mm2As5890mm2Asmin,h012002525251137.5mm21fcbfhf1.014.325001806435000N6345kNfyAs2179300N2179.3kN第一类xfyAs370589061.0mm1fcbf1.014.32500MufyAsh0x37058901137.561.01062412.

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