考虑到荷载偏心作用,可将轴心荷载下所得到的基底面积之增大30%得初选基底面积:
A?1.3Fk1.3?400??3.6m2fa??Gd164.6?20?1.0
取边长比n=1.5得基础宽度:
b?A3.6??1.55m,取b?1.6m。 n1.5l?nb?1.5?1.6?2.4m
Fk?Gk?400?20?1.6?2.4?1?476.8kN 验算偏心距:
e?Mk110l??0.231m??0.4m(可以)
Fk?Gk476.86?6e??1??l??476.8?6?0.231????1? ?1.6?2.4?2.4??195.9kPa?1.2fa?1.2?164.6?197.5kPapkmax?(可以)
(2)计算基底净反力设计值
Fk?GkAF1.35?400??140.6kPa A1.6?2.4F6M1.35?4006?1.35?110pjmax??2???237.3kPa 2blbl1.6?2.41.6?2.4F6M1.35?4006?1.35?110pjmin??2???43.9kPa 2blbl1.6?2.41.6?2.4pj?平行于基础短边的柱边Ⅰ-Ⅰ截面的净反力:
pj??pjmin?l?ac?pjmax?pjmin??43.9?2.4?0.3??237.3?43.9??152.7kPa 2l2?2.4(3)确定基础高度
22采用C20混凝土,ft?1.10Nmm,钢筋用HPB235级,fy?210Nmm。取基础高
度h?500mm,h0?500?45?455mm。
因bc?2h0?0.3?2?0.455?1.21m?b?1.6m,故按式(2-57)作冲切验算如下(以pjmax6
取代式中的pj):
2??lac??bbc??pjmax????h0?b????h0????22?????22?2??2.40.3??1.60.3???237.3?????0.455??1.6????0.455??
22??2????2???216.9kN0.7?hpft?bc?h0?h0?0.7?1.0?1100??0.3?0.455??0.455?264.5kN?216.9kN(可以) (4)确定底板配筋
对柱边Ⅰ-Ⅰ截面,按式(2-65)计算弯矩:
M???1?pjmax?pj???2b?bc???pjmax?pj??b?l?ac?248??12???237.3?152.7???2?1.6?0.3???237.3?152.7??1.6???2.4?0.3? 48?137.8kN?mMI137.8?106ASI???1602mm2
0.9fyh00.9?210?45522配钢筋15?12,As?1695mm?1602mm,平行于基底长边布置。
M??1122pj?b?bc??2l?ac???140.6??1.6?0.3??2?2.4?0.3??50.5kN?m 2424AS?M?50.5?106???587mm2 0.9fyh00.9?210?0.45522按构造要求配筋13?10,As?1021mm?587mm,平行于基底短边布置。
如图所示
3-4 以倒梁法计算例题3-1中的条形基础内力。 【解】 (1)用弯矩分配法计算肋梁弯矩
沿基础纵向的地基净反力为: bpj边跨固端弯矩为:
7
?F?6.4?10?l173?376.5KN/m
M21?中跨固端弯矩为: M23?112bpjl1??376.5?4.52?635.3KN?m 1212112bpjl2??376.5?62?1129.5KN?m 1212112bpjl0??376.5?12?188.2KN?m 221截面(左边)伸出端弯矩: Ml1? 节点 分配系数 固端弯矩 分配结果(kN·m) 1 0 1.0 0.5 2 0.5 0.5 3 0.5 1.0 4 0 188.2 -635.3 635.3 188.2 -238.2 1011
-1129.5 1129.5 -635.3 660.3 -188.2 -1011 1011 1011 238.2 -188.2 (2)肋梁剪力计算
1截面左边的剪力为: Vl1?bpjl0?376.5?1.0?376.5KN
计算1截面的支座反力
R1?1l11?1?1??2'2??bpl?l?M?M?376.5?5.5?1011?502???1051.9KNj011???2?4.5?2?1截面右边的剪力:
Vr1?bpjl0?R1?376.5?1051.9??675.4KNR2?bpj?l0?l1??R1?376.5?5.5?1051.9?1018.8kN'
取23段作为脱离体:
R''2?1?12''??bpjl2?M2?M3??l2?2?1?1?2??376.5?6?1011?1011??1162.5KN6?2?R2?R'2?R''2?1018.8?1162.5?2181.3KNVl2?R'2?1018.8KNVr2??R''2??1162.5KN按跨中剪力为;零的条件来求跨中最大负弯矩:
8
bpj?R1?376.5x?1043.8 x?1043.8/376.5?2.8m
所以M1max?11bpjx2?R1?1.8??376.5?2.82?1011?1.8??344.0KN?m22181823段对称,最大负弯矩在中间截面:
M2max??bpjl2?M2???376.5?62?1011??683.2KN?m 由以上的计算结果可作出条形基础的弯矩图和剪力图
683.2
344 344
188.2 188.2 238.2 238.2
弯矩图M(kN·m)
1011 1011 1018.8 1162.5 376.5 675.4
剪力图V(kN)
376.5 675.4 1162.5 1018.8
补充题:设一箱形基础置于粘性土(fk?300kPa)地基上,其横剖面上部结构及上部结构荷重如图,上部结构总重为48480KN,箱形基础自重为18000KN,箱形基础及设备层采用C20混凝土,上部结构梁、柱采用C30混凝土,框架柱0.5m×0.5m,框架梁0.25m×0.60m,求矩形基础纵向跨中的整体弯矩。
【解】 (1)箱型基础内力计算,按纵向整体弯曲计算,由静力平衡条件计算跨中最大弯矩为:
2Mmax?1281.46?6?21?972.61?6?15?900.59?6?9?885.35?6?3?3030?24?6060?18?6060?12?6060?6?22690kN/m(2)计算箱型基础刚度EFIF 箱型基础横截面惯性矩IF?
1334?? 12.5?3.55?(12.5?0.8)?2.77?26.3260m??129
箱基刚度EFIF?26.3260EF (3)上层结构折算刚度EBIB
纵向连续钢筋混凝土墙的截面惯性矩Iw?2?各层上下柱的截面惯性矩Iui?Ili?3?各层纵梁的截面惯性矩Ibi?3?1?0.3?2.23?0.5324m4 121?0.5?0.53?0.0156m4 121?0.3?0.53?0.0094m4 120.01560.0094各层上下柱、纵梁的线刚度Kui?Kli? 6?0.0056 Kbi??0.0012.86上部结构折算刚度
??Kui?KliEBIB???EbIbi(1?m2)??EwIw2Kbi?Kui?Kli1??0.0056?0.005648???7?Eb?0.0094??1??()2?6? ?2?0.0016?0.0056?0.0056n0.005648???Eb?0.0094??1??()2??Eb?0.53246??2?0.0016?0.0056?4.2658Eb(4)箱型基础所承担的整体弯矩MF(取EF?Eb)
MF?MEFIF26.3260EF?22690??19526kN?m
EFIF?EBIB26.3260EF?4.2658Eb3-1
(a)强柱弱梁 弱柱强梁
基础受约束较小,趋于自由变形,整体变形。 相当于叠合在一起,弯矩方向基础受到梁的约束 地基反力
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《基础工程》课后习题答案
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