斜截面受剪承载力计算例题.pdf

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1、斜截面受剪承载力计算例题114－1解：1）剪力图见书，支座剪力为V＝ql=××703.65=124.6kN0222）复合截面尺寸h454.5whw=h0=h-c-8-25/2=500-25-8-12.5=454.5==<2.34b2000.25βfbh=××××=0.251.09.6200454.5218.16kN>=V124.6kNcc0满足。3）验算是否按计算配置腹筋0.7fbh=×××0.71.1200454.5=69.993kN<=V124.6kNt0应按计算配置腹筋4）计算腹筋数量①只配箍筋Asv由Vf≤

2、+0.7bhfh得：ty00vs33nAsv12V−0.7fbht0124.610×−69.99310×2≥==0.572mm/mmsfh210454.5×yv0nA250.3×sv1选双肢φ8箍筋sm≤==175.87m0.5720.572取s=170mm验算最小配箍率nA250.3×f1.1sv1tρρ===0.00296>=0.24=×=0.240.0013svsv,minbs200170×f210yv满足仅配箍筋时的用量为双肢φ8@170②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

3、筋，将122的纵筋以45°角弯起，则弯筋承担的剪力：2Vf==0.8Asinα0.8380.1300×××=64.5kNsbysbs2nAVf−−0.7bhf0.8Asinα124.610×−3369.99310×−×64.5103sv1ty0sbs≥==负值sfh210454.5×yv0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，nA228.3×f1.1sv1tρρ===0.00142>=0.24=×=0.240.0013svsv,minbs200200×f210yvb.先选箍筋，再算弯筋先按

4、构造要求选双肢φ6@200的箍筋，nA228.3×f1.1sv1tρρ===0.00142>=0.24=×=0.240.0013svsv,minbs200200×f210yv满足要求。Asv33228.3×Vf−−0.7bhfh124.610×−×69.99310−210××454.5ty00vs2002A≥==162.6mm弯sb0.8fsinα2ys0.8300××22起122的纵筋，Asb=380.1mm5）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500-25-6-22/2-25-6-2

5、2/2+50=466mm，该处剪力V1=124.6-0.466×70=91.98kNAsv33228.3×Vf=+=×0.7bhfh69.99310+210××454.5=97.010×Ncst00yvs2003>91.98×10N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h557whw=h0=h-c-8-20/2=600-25-8-10=557==<2.794b2000.25βfbh=0.251.09.6200557××××=267.36kN>==VV180kNcc0maxA满

6、足。3）验算是否按计算配置腹筋V160V140集集A支座：==88%；B支座：==87.5%V180V160总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a1000AC段：λ===1.80h55701.751.75fbh=×1.1200557××=76.59kN3.593，取λ=3h55701.751.75fbh=×××=1.120055753.6kN>VC=50kNt0λ

7、++1.031按构造要求配置箍筋，选用φ6@350的箍筋。a2000DE段：λ===>3.593，取λ=3h55701.751.75fbh=×××=1.120055753.6kN

8、mmsfh210557×yv0nA250.3×sv1选双肢φ8箍筋sm≤==113.8m0.8840.884取s=110mm验算最小配箍率nA250.3×f1.1sv1tρρ===0.00457>=0.24=×=0.240.00126svsv,minbs200110×f210yv满足箍筋用量为双肢φ8@1101.75nAVfEt−bh03sv1λ+1.0(7