楊成永 馬文輝 韓薛果 程霖
摘? ?要:以矩形板的Navier解為基礎,采用帶補充項的傅里葉級數作為撓度函數,研究了局部均布荷載作用下四邊支承矩形薄板的彎曲問題. 推導了確定待定系數的線性代數方程組,給出了簡支邊和固支邊不同組合條件下的統(tǒng)一計算公式. 討論了帶補充項法級數解的收斂速度,并與疊加法級數解及有限元數值解分別進行了精度和計算量的對比. 結果表明,帶補充項法的級數解達到收斂的級數項數約為40項. 帶補充項法的級數解與疊加法級數解具有同樣的求解精度. 有限元解隨網格的細分,計算結果逐漸接近級數法解. 級數解法的計算量與有限元解法相比是微不足道的. 研究成果適于進行構筑物頂板受局部均布荷載作用的結構計算.
關鍵詞:矩形板;四邊支承;局部均布荷載;級數解;求解精度
中圖分類號:U411? ? ? ? ? ? ? ? ? ? ? ? ? ?文獻標志碼:A
Internal Force Calculation of Four Edges Supported Rectangular
Plates under Local Uniformly Distributed Load
YANG Chengyong,MA Wenhui?,HAN Xueguo,CHENG Lin
(School of Civil Engineering,Beijing Jiaotong University,Beijing 100044,China)
Abstract:On the basis of Naviers solution to rectangular plates, the bending problem was studied for the four edges supported thin plates under local uniformly distributed load, where the double Fourier series with additional terms was adopted as the deflection function of the plates. Linear algebraic equations for solving the undetermined coefficients were derived. A unified solution was obtained to the rectangular plates with clamped and simply supported edges. The rate of convergence was discussed on the solution of the series method with additional terms. The proposed method was compared both with superposition series method on accuracy, and with finite element numerical method on computational cost. The results show that 40 terms should be employed for a convergence of the series. The method with additional terms shows the same accuracy of solution as superposition series method does. The solution by finite element method gradually approaches that by the series method as the mesh gets finer and finer. In comparison with finite element method, the computational time by the series method is negligible. This work is applicable for structural analysis of the top plates of underground buildings under truck wheel pressure.
Key words:rectangular plate;four supported edges;locally uniformly distributed load;series solution;solution accuracy
地鐵、熱力和燃氣等地下工程中,地下構筑物的頂板多為四邊支承的薄板,板上常承受局部均布荷載如汽車輪壓作用. 為了確定像汽車輪壓這類荷載在板內產生的最大撓度和內力,需要進行任意位置局部均布荷載作用下?lián)隙群蛢攘Φ挠嬎?
對四邊支承的矩形薄板問題,可以從四邊簡支板的Navier解出發(fā),采用疊加方法[1-2]或加補充項的方法[3-4]解決. 如:蔡長安等[5-6]以帶附加補充項的Fourier級數作為撓度函數,求解了Winkler地基及Pasternak地基上自由邊矩形板的彎曲問題. 許琪樓等[7-8]采用一種能滿足自由角點條件的撓度表達式,解決了二鄰邊支承二鄰邊自由矩形板和二鄰邊及對角點支承矩形板的彎曲問題. 他們還采用疊加方法[9-10],提出了四邊支承矩形板及一邊固定一角點或二角點支承的矩形板的統(tǒng)一求解方法. 岳建勇等[11-12]采用一種雙三角級數形式的撓度函數,得到了三邊固定一邊自由及兩對邊固定兩對邊自由矩形板的精確解. 鐘陽等[13]在辛幾何空間中利用分離變量法推導出了四邊固支彈性矩形薄板的精確解析表達式. 于天崇等[14]假定矩形板的抗彎剛度沿板的寬度方向按照一般冪函數形式變化,研究了四邊簡支一對邊受彎作用下面內變剛度矩形板的彎曲問題. 肖閃閃等[15]采用載荷疊加法研究了集中載荷下四邊固支正交各向異性矩形板的線性彎曲,并討論了經典Kirchhoff薄板假設對于集中載荷的適用性.
4.4.2? ?與文獻[16]對比
文獻[16]列出了四邊簡支板中央受局部均布荷載作用時彎矩的計算系數. 為與其對比并避免查表計算中的插值,取泊松比μ = 0,輪壓x方向分布長度c = 2 m,y方向分布長度d = 1 m,其余參數采用4.2節(jié)的數據,板四邊均簡支.
根據 =? = 1.4, =? = 0.4, =? = 0.2,按文獻[16] 中表4-29查得計算彎矩Mx的系數為0.148 0,My的系數為0.130 8. 然后有
計算結果列于表3.
由表3看出,兩種方法的結果,前3位有效數字相同. 由于文獻[16]表格的有效數字是4位,可以認為表3中兩種結果是一致的.
4.5? ?計算結果與有限元對比
采用4.2節(jié)的計算參數,板左邊(x = 0邊)及前邊(y = 0邊)固支,其余兩邊簡支. 按相同的參數和邊界條件采用ANSYS軟件SHELL63號單元,劃分3種粗細不同網格進行計算. 計算結果列于表4.
由表4可看出:
1)隨有限元網格的加密,計算結果逐漸趨于本文的級數解. 由此可說明,本文級數解是四邊支承板變形問題的理論解或精確解. 當有限元網格細到5 mm × 5 mm時,撓度及彎矩有5位有效數字與級數解相同. 可以認為這時數值解與級數解基本一致.
2)表4中的軟件運行所用時間是從數據輸入到輸出全部的計算機運行時間. 要達到較高的精度,有限元需要花費的計算機時間大大高于級數解. 就本算例來說,相差達10萬倍以上. 需要注意的是,本算例中,有限元在5 mm × 5 mm網格時,需要求解的方程組的階數,不少于(x方向節(jié)點數5/0.005) × (y方向節(jié)點數7/0.005) × 6個自由度 = 840萬;而級數解取40項時需要求解的方程組的階數僅為40 × 2個固支邊 = 80.
5? ?結束語
采用帶補充項的撓度函數,研究了四邊支承矩形薄板的彎曲問題. 給出了局部均布荷載作用下簡支邊和固支邊不同組合條件下的統(tǒng)一計算公式.
對比計算表明,以Navier解為基礎帶補充項的傅里葉級數解,達到收斂的級數項數約為40項. 該級數解與其他采用疊加法得到的傅里葉級數解,具有同樣的求解精度. 與有限元數值法相比,級數解的計算量十分微小.
值得一提的是,式(3)若換成滿布荷載、線荷載和集中力相應的傅里葉系數,本文方法也適用.
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