由《混凝土规范》表 查得,截面一侧纵向钢筋r min =,截面全部纵向钢筋r min =. 钢筋混凝土偏心受压柱,截面尺寸 b=400mm,h=500mm,计算长度l0=,截面承受轴压力设计值 N=324kN,弯矩设计值 M1=M2=95kN﹒m,选用 C30 混凝土,HRB400 级纵向钢筋,钢筋混凝土保护层 c=20mm,取'as = 40mm; as = 40mm 截面的受压区已配置受压钢筋 3 18,。求受拉钢筋截面积As。【解】1. 确定基本数据'由《混凝土规范》表 -1 查得,纵筋 f y = f y =360 N / mm2 ;由《混凝土规范》表 -1 查得, fc = N / mm2 ;由《混凝土规范》第 条查得,a1 = , b1 = ;按《混凝土规范》第 条计算, xb = ;'''由《混凝土规范》附录 A 查得, As =763 mm2'取 as = 40mm; as = 40mm , h0 = h - as = 500 - 40 = M(根据规范 ;)95 I 400 ´ 5003 /12M1M 2= = , i = = = A 500 ´ 400= = > 34 -12lciM1M 2= 22z c = = = > 取z c = = Max{20mm, h / 30}= Max{20mm,500 / 30 = }= fc A ´´ 500 ´ 400N 324 ´103Cm = + M1M 2= + ´1 = 1( c )2z chns = 1+1 l1300(M 2 / N + ea ) / h0 h( ) ´ = (95´10 / 324 ´10 + 20) / 460 500= 1+6 31 8400 21 1 0cf bha 3324 10 617 763 360 (460 40)´ ´ - ´ ´ - 0( ) 360 (460 40)y sf h a- ´ - 763 603S SA A+ + 0 387e mm= = = 407 40 617i se e a mm= + - = + - = ( ) ' ' 407 40 197i se e a mm= - + = - + = 2422sA mm= = =M = Cmhns M 2 = 1´ ´ 95 = × ei 判别偏压类型M ´106N 324 ´103ei = e0 + ea = 387 + 20 = 407mmei = 407mm > = ´ 460 = 138mm故按大偏心受压构件计算。
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