基础工程课 程 设 计 报 告柱下钢筋混凝土独立基础设计计算书姓 名： 张圣 学 号： 201208130357 指导教师：李晓明 曲文婷学 院： 城市与建筑工程学院 专 业： 土木工程 完成日期：2015年1月11日一、设计资料1.工程地质条件拟建建筑场地地形起伏不大，较为平整。
场地土层分布如下：①杂填土：厚约0.5m，含部分建筑垃圾；②粉质粘土：厚1.2m，软塑，潮湿，承载力特征值f ak=135KN/m2；③粘土：厚1.8m，可塑，稍湿，承载力特征值f ak=180KN/m2；④全风化砂质泥岩：厚2.7m,承载力特征值f ak=230KN/m2；⑤强风化砂质泥岩：厚3.0m,承载力特征值f ak=310KN/m2；⑥中风化砂质泥岩：厚4.0m,承载力特征值f ak=600KN/m2。
地下水位位于地表以下2.0m，且地下水对混凝土结构无腐蚀性。
3.上部结构资料拟建建筑物为多层全现浇框架结构房屋，框架柱截面尺寸为500×500mm，室外地坪标高同自然地面，室内外高差450mm。
拟采用柱下钢筋混凝土独立基础，上部结构作用在各柱柱底的荷载效应标准组合值见表2；上部结构作用在各柱柱底的荷载效应基本组合值见表3。
4.材料：混凝土的强度等级C20~C30，钢筋采用HPB235、HRB335级。
二． 柱下独立基础设计1. 按持力层的承载力特征值计算所需的基础底面尺寸 （1） 轴心荷载作用下的基础假设取基础埋深d=2m3/41.1923.04.192.1205.018m KN m =⨯+⨯+⨯=γ (1)()()KPa d k f f m d a a 58.2265.0241.196.11805.0=-⨯⨯+=-+=γη · (2)由： (3)287.722058.2261433m =⨯- (4)取：基础底面尺寸面积为9，且l=b=3mKN Ah G G 5.400225.2920=⨯⨯==γ (5) (6)m G F M e 612.05.4001433297=+=+= (8)KPa f KPa p a k 49.2802.17.269max == ································9 KPa f KPa p a k 58.2267.203== ·····································10 m lm e 333.06612.0==··············································11 故满足条件，假设成立。
2. 基础高度的确定（1） 计算基底净反力值KPa bl F p j 207331863=⨯==··············································12 净偏心距：m F M e 207.01863/386/0===··········································13 KPa l e lb F p j 7.2923207.061331863610max =⎪⎭⎫⎝⎛⨯+⨯⨯=⎪⎭⎫ ⎝⎛+=····················14 KPa l e lb F p j 3.1213207.061331863610min =⎪⎭⎫⎝⎛⨯-⨯⨯=⎪⎭⎫ ⎝⎛-=·····················15 （2） 基础高度的确定①柱变截面假设取基础高度mm h 800=，取保护层厚度mm c 45=，mm c h h 7550=-=m b h bc 301.2755.025.020=<=⨯+=+·······································16 满足条件,选用C20混凝土，HRB335型钢筋，查得300,1100==y t f f 因偏心受压，计算时j p 取max j p该式左边：⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛--200max 2222h b b b h a l p cc j (17)=KN 95.362755.025.0233755.025.0237.2922=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛---⨯⎪⎭⎫ ⎝⎛--⨯. 该式右边：()007.0h h b f c t hp +β (18)()()可以KN KN 95.3626.729755.0755.05.011000.17.0>=⨯+⨯⨯⨯=基础分两级，下阶m5.1b m 5.1l mm 355h mm 400h 11011====，，取，②变阶处截面m m h b 321.2355.025.12011<=⨯+=+ (19)冲切力：⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛---⎪⎭⎫ ⎝⎛--2011011max 2222h b b b h l l p j=KN 19.301355.025.1233355.025.1237.2922=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛---⨯⎪⎭⎫ ⎝⎛--⨯抗冲切力：()00117.0h h b f t hp +β (20)()()可以KN KN 19.301507355.0355.05.111000.17.0>=⨯+⨯⨯⨯= 3.配筋计算()min max min 2j j cj J p p l a l p I P -++=·································20 =()3.2213.1217.292325.033.121=-⨯++()()()[]()2m a x m i n 2481c jI j c jI j I a l b p p b b p p M --+++=···················21 =()()()[]()25.0333.2217.2925.0323.2217.292481-⨯-++⨯⨯+ =462.9KN ·m2608.22707553009.0109.4629.0mm h f M I A y I S =⨯⨯⨯==··························22 ()min max 1min 2j j j J p p l l l p III P -++=······································23 =()KPa 85.2493.1217.292325.133.121=-⨯++()()()[]()21max 1min 2481l l b III p p b b III p p M j j j j III --+++=···············24 =()()()[]()25.13385.2497.2925.13285.2497.292481-⨯-++⨯⨯+ =196.8KN ·m2608.22707553009.0108.1969.0mm h f M III A y III S =⨯⨯⨯==························25 比较I A S 和III A S ，应按2270.8mm ，采用1612钢筋，@180。