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单向偏心受压柱基基底简便计算法
单向偏心受压独立柱基基底尺寸计算,目前有试算法和表格法两种。前者计算虽简单但误差较大,往往计算后需调整基底尺寸;后者计算虽准确但计算过程较繁,且需配合表格进行。现提出一种简捷计算方法。
一、基底尺寸B的计算
图1为一单向偏心受压独立柱基。由应力计算公式可知,其基底最大应力与最小应力分别为:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_1.jpg?sign=1739343404-SM0LevWHpWOTrsQWE7Qq5NN0PI7XkqlG-0-b10660b2b1ca5275cc305e438fed9270)
式(1)、(2)中
F——基底面积;
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_2.jpg?sign=1739343404-TKyIrqZngH5rbRBSuO5K6XChKrP1Fcn4-0-11d531c2f82add5ce097b034538c5b60)
W——基底抵抗矩。
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_3.jpg?sign=1739343404-d2ppXd6Yj6S1nSe2ZhcBquQcjYVBLhvs-0-791bbfaf7b0bb3324fadcac4d46a2887)
将式(3)、(4)代入式(1)、(2)得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_4.jpg?sign=1739343404-Tm1ucLUndHCfOrYC0zt75AMN3QBUwzAn-0-ca75b4ac904e79b948cc8d1c1e2dae8d)
将式(5)表示成B的二次方程得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_5.jpg?sign=1739343404-CooCACi7TBPqxWQgKfmF13u7VjOOX31n-0-ca7f4cb89d7ac83f5b0c23cb6e119b2a)
解式(7)二次方程得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_6.jpg?sign=1739343404-5gvLiQpaeUGfOK0PqIDzzh6gsmuBBujd-0-337512f9d20bccb0cba93f1c531fe4c1)
为(8)为基底尺寸B的直接计算公式。
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_7.jpg?sign=1739343404-bfuuL9goQs2STCcGXorGkk3QhAYUkWml-0-948520c66cda1e5aba9672df2ec3ed6d)
图1 单向偏心受压独立柱基示意图
二、基底尺寸A的计算
已知轴心受压矩形独立柱基的A轴计算公式为:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_8.jpg?sign=1739343404-KgQxJVfVuns4owzagniKnbPm7ev3WIUT-0-5e050e6e9c8d0f72f2500449d295b01f)
现令偏心受压与轴心受压基底短边之比为β,即:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_9.jpg?sign=1739343404-IINfwRHCkXMLhyTIMLLnlmhZ2tYWnhgd-0-159c19b90daae5a8baf728d22be21c55)
则偏心受压基底尺寸A偏的计算式为:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_10.jpg?sign=1739343404-FtA065FKMA0qgyVsueDhNlIVrbiAwSdh-0-cdfc74855fa35f1ecba5e53f93172e64)
式中 n——基底长宽之比;
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_11.jpg?sign=1739343404-2dEKKGBmGoFRIKD8lJo5c7OOPlraWnzK-0-71ae64049755209a9b5fbf54256bb9fa)
β——基础偏心修正系数。
现对n和β的取值介绍如下:
1.基底长宽之比n的取值。
(1)对不限制基底偏心程度的单向偏压独立柱基,在常用工程范围内,一般取n=1.25;
(2)对控制基底偏心程度的单向偏压独立柱基;
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_12.jpg?sign=1739343404-yZhnipyj4bpRNO8X2mjEGdgp8FZvkoTL-0-e3178f19432232289d006ed91f3bc82c)
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_13.jpg?sign=1739343404-4nnxBFNp7Bhy6pN9GVETLuxQCsFPSLVG-0-62ba285c01f7b56de8a73923412b34ba)
ξ——基础底面边缘的最小应力与最大应力之比,取值范围为:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_14.jpg?sign=1739343404-rx7j7cY0O1s2EvDXQ1I4pjhJ3TjRWU58-0-3dd7d3462e5ab64f4faf73e404c8c6ac)
2.基础偏心修正系数β的取值。
已知偏心受压基底面积为:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_15.jpg?sign=1739343404-tmv3Hr0TqBuUgDkgpd6IS5iqcmbv74SR-0-1389e3370176a3b86c5299db32b22c8f)
轴心受压基底面积为:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_16.jpg?sign=1739343404-gCJRyyjhMAgPiezfFd2IVP4RD9MksfJe-0-d627550bc89d220943d2f2a7615c0fac)
又由式(12)得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_17.jpg?sign=1739343404-y09EhD05aFUOELnM4U5Qriu5OaYHneyX-0-f8b3b78a5190656475f7d071b6918118)
将式(14)代入式(20)得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_18.jpg?sign=1739343404-7BUHoeW5ET8clTE2wnWKM5abzF2EBTQx-0-5eaeada9c5528bbffcb7ce6997db854f)
又将式(18)、(19)代入式(21)得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_19.jpg?sign=1739343404-tR8vprIf0WmHMTLX4cBvwaUG32VmpUhZ-0-cfbb41d7dc4cd5cd8f95463b7b125ca1)
当N、M、R、H、n确定后,在1<n<2范围内;
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_20.jpg?sign=1739343404-oP546TibnReo82JOEfg2871Cx59zsfUV-0-c52f63a3cfa889e452b532b672bbfb9b)
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_21.jpg?sign=1739343404-S0scOIABKBedoOjgemucPFriwhQ9xuLn-0-bfa9ea644e023bfd2d71bb8111db49f6)
所以β的取值为:
(1)当不控制基底偏心程度时,取
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_22.jpg?sign=1739343404-1RzmCBlPiSYgV4THtnoKjxPOGPrMTQXx-0-be6f086c61212f828cee1798a2b09315)
(2)当控制基底偏心程度时,取
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_23.jpg?sign=1739343404-0XGXEEt9cM5xQssbEBqIAnZ8WrxX64Lk-0-37f1b7ececffd00eae163524a5b7c8bf)
应用本文公式计算单向偏压独立柱基基底尺寸,其最大误差一般在±5%以内。
三、计算实例
已知N=250kN,M=100kN·m,R=250kN/m2,H=2.0m,γ=20kN/m3,要求〔ξ〕=0,求基底尺寸。
解:(1)求n
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_24.jpg?sign=1739343404-kE18WjYlvDpq2H2sp9lNjxjtf9sH50Sd-0-ae6baa0a9c0606df92fdf6633eba0b9d)
将以上数字代入式(13)或式(26)得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_25.jpg?sign=1739343404-dncgVE5B7hLI5eqPgS4xvu7KrDZFwm9d-0-d660280a41866c3bd6a82564c4d33096)
(2)求基底尺寸A
将以上数字代入式(11)直接求解得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_26.jpg?sign=1739343404-tTmK9MlhPXtFllnFgDDdpd1rhqQJH7wu-0-4206a0c5e6241d8adfccf7b3d1048f9d)
(3)求基底尺寸B
将以上数字代入式(8)直接求解得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_27.jpg?sign=1739343404-aN74gC3xc3EBfkIQye6Tr6XFb1WozNv2-0-e4eb3e3e3e34610afd8d3bea5252eff3)
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_28.jpg?sign=1739343404-fHqJo01zx0YCrzQ7IxWlZGr5hAMJCbtC-0-67b9aa2d87c2e24a4c01cf1faab91d5b)
(4)验算
将以上数字代入式(5)、(6)得:
![](https://epubservercos.yuewen.com/B4465B/28343064702441906/epubprivate/OEBPS/Images/txt005_29.jpg?sign=1739343404-aeZo9Ca2hH9C35BaaBAkvH6SuCHMbQh1-0-0fe870c8a4505b2f93e01cdc34e01c1e)
(胡信芳)