2022年英语公式用语 .pdf

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1、名师推荐精心整理学习必备4.轴对称圆板的大挠度方程通过物理方程与几何方程，可以将平衡方程组中的内力素用位移来表示。薄膜力由虎克定律可得：By Hook s law, the membrane forces can be calculated as1()1()rrrNNEsNNEs(6) 将几何方程（ 3）代入公式（ 6）并联立求解rN与N可得：Substituting Eq. (3) into (6) acquires the simultaneous solutions rNandNas follows: 2222221()()211(7)()() 211rrrEsEsduduNdrdrrE

2、sEsududNrdrdr用位移来表示的弯曲力表达式为：The bending force can be expressed by displacement, as in Eq. (8). 2222()()1()()rrrddMDDrdrdrddMDDr drdr(8) 将上式代入平衡方程（5）中可得剪力公式如下：3232211()dddQDrdrdrdrr2()dDdr(9) 式中2为在轴对称情况下的拉普拉斯算子22211()ddddrr drr drdrdrSubstituting Eq. (8) into (5) yields 3232211()dddQDrdrdrdrr2()dDdr,

3、 (9) where 2is the Laplace operator under axial symmetry, i.e.,名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 1 页，共 4 页 - - - - - - - - - 名师推荐精心整理学习必备22211()ddddrr drr drdrdr引入应力函数：Introducing the stress function, we have 221rdNr drdNdr(10) 则大挠度方程组为：Then the large de

4、flection equation is 2221()()()2dddDdrr drdrdEs ddrrdr(11) 设轴对称圆板具有微小初挠度0， 则微弯柔性圆板的方程为：Suppose that the axisymmetric circular plate is located at a small initial deflection0. Thus the equation for the micro-bending flexible circular plate can be written as 202201()()1()()2sddddDdrrdrdrdrddEsdddrrdrd

5、rdr(12) 其中为在荷载q作用下所产生的挠度。where is the deflection of the disc spring under load q. 202201()()1()()2sddddDdrr drdrdrddEsdddrrdrdrdr, (12) 5.基于能量法的载荷变形特性解析解名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 2 页，共 4 页 - - - - - - - - - 名师推荐精心整理学习必备总能量为：The total energy is gi

6、ven by bmpVUUU(13) 其中：bU为弯曲变形能 ; mU为伸缩变形能 ; pU为势能 . where bUrepresents the bending deformation energy, mUthe stretching deformation energy, and pUthe potential energy. 设0表示圆板的初挠度，表示在荷载作用下的附加挠度，0f与f分别为在两种情形下的最大挠度。设初挠度和附加挠度均按同样的规律变化如下：Let 0be the initial deflection of the circular plate, represent the

7、 additional deflection under load, and 0fand fdenote the maximum deflections of 0and respectively. Suppose the initial and additional deflections vary by the same rule as follows: 00(2 )fDrDd；(2 )fDrDd将初挠度和附加挠度代入上述方程组积分两次得：Substituting the initial and additional deflections into Eq. (20) and doing i

8、ntegration by parts twice give 0122221()112(ln)242(1)fffCCFCs(21) 式中的积分常数1C、2C可按以下的边界条件求得：where the integral constants 1Cand2Ccan be obtained in 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 3 页，共 4 页 - - - - - - - - - 名师推荐精心整理学习必备light of the limit conditions, i.e., 2()0drddr；2()0Drddr22012221()(12ln)22(1)(1)fffCCCCCCs2022221()ln22(1) (1)fffCCCCCs0122221()112(ln)242(1)fffCCFCs, (21) 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 4 页，共 4 页 - - - - - - - - -