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1、1,Chapter 3 Two-dimensional Problems in Rectangular Coordinates,2,第三章 平面问题的直角坐标解答,3,Chapter 3 Two-dimensional Problem in Rectangular Coordinates,Two-dimensional Problems in Rectangular Coordinates,3-1 Solution by Polynomials,3-2 Determination of Displacements,3-3 Bending of a Simply SupportedBeam by
2、 Uniform Load,3-4 Loading of a Wedge by Gravity and Hydraulic Pressure,3-6 Bending of a Simply Supported Beam by Arbitrary Lateral Load,3-5 Solution by Series,Exercises,4,第三章 平面问题的直角坐标解答,平面问题的直角坐标解答,5,1.The Stress Function in the form of a Polynomial of the First Degree,3-1 Solution by Polynomials,T
3、he Stress Components:,The Stress Boundary Condition：,Conclusion:（1）The linear stress function is corresponding to the state of no surface force and no stress.（2）Theres no effect to the stress to add a linear function to any stress function of two-dimensional problem.,2.The Stress Function in the for
4、m of a Polynomial of the Second Degree,Two-dimensional Problems in Rectangular Coordinates,。,6,一、应力函数取一次多项式,3-1 多项式解答,平面问题的直角坐标解答,应力分量：,应力边界条件：,结论：（1）线性应力函数对应于无面力、无应力的状态。,（2）把任何平面问题的应力函数加上一个线性函数，并不影响应力。,二、应力函数取二次多项式,1.对应于 ，应力分量 。,7,2.Corresponding to ,stress components,Conclusion: The stress functio
5、n can used to solute the problem of uniformly distributed shearing stresses rectangular plate.(Fig3-1b),Two-dimensional Problems in Rectangular Coordinates,8,平面问题的直角坐标解答,结论：应力函数 能解决矩形板在 方向受均布拉力（设 ）或均布压力（设 ）的问题。如图3-1（a）。,图3-1,（a）,（b）,（c）,9,Conclusion：The stress function（a) can be used to solute the p
6、roblem of pure bending of rectangular beam. The rectangular beam is shown in Fig3-2.,（a）,Two-dimensional Problems in Rectangular Coordinates,10,平面问题的直角坐标解答,3.应力函数 能解决矩形板在 方向受均布拉力（设 ）或均布压力（设 ）的问题。如图3-1（c）。,三、应力函数取三次式,对应的应力分量：,（a）,结论：应力函数 能解决矩形梁受纯弯曲的问题。如图3-2所示的矩形梁。,11,Specification as following:,In fi
7、g3-2,considering the unit width beam ,named the moment of double force on per unit width .The order of here isforcelength/length，the result is force .,On the left or right，the two-dimensional force should be combined to a double force，as the moment of double force is ，this request that：,Two-dimensio
8、nal Problems in Rectangular Coordinates,12,平面问题的直角坐标解答,具体解法如下:,如图3-2,取单位宽度的梁来考察,并命每单位宽度上力偶的矩为 。这里 的因次是力长度/长度，即力。,在左端或右端，水平面力应当合成为力偶，而力偶的矩为 ，这就要求：,前一式总能满足，而后一式要求：,代入式（a），得：,将式（a）中 的代入，上列二式成为：,13,The result is same with corresponding part in material mechanics,Two-dimensional Problems in Rectangular C
9、oordinates,14,平面问题的直角坐标解答,因为梁截面的惯矩是 ，所以上式可改写为：,结果与材料力学中完全相同。,注意：,对于长度 远大于深度 的梁，上面答案是有实用价值的；对于长度 与深度 同等大小的所谓深梁，这个解答是没有什么实用意义的。,15,3-2 Determination of Displacements,Take the pure bending of rectangular for example to explain how to determine the displacement by the stress components.,Two-dimensional
10、Problems in Rectangular Coordinates,16,3-2 位移分量的求出,平面问题的直角坐标解答,以矩形梁的纯弯曲问题为例，说明如何由应力分量求出位移分量。,一、平面应力的情况,将应力分量 代入物理方程,17,Then the distortion components are：,（a）,Then put (a) into the geometric equation：,and Is the arbitrary function。put（c）into the third equation of（b）,Two-dimensional Problems in Recta
11、ngular Coordinates,18,平面问题的直角坐标解答,得形变分量：,（a）,再将式（a）代入几何方程：,得:,前二式积分得：,（b）,（c）,其中的 和 是任意函数。将式（c）代入（b）中的第三式,19,so：,After the integral：,The arbitrary constants , , above must be obtained through the restrict conditions,Two-dimensional Problems in Rectangular Coordinates,On the left of the equal mark is
12、 the function of ,on the right of the equal is the function of .So both sides should equal to a constant ,and:,w,20,平面问题的直角坐标解答,得:,等式左边只是 的函数，而等式右边只是 的函数。因此，只可能两边都等于同一常数 。于是有：,积分以后得：,代入式（c），得位移分量：,其中的任意常数 、 、 须由约束条件求得。,（d）,21,1.The Simply Supported Beam,In Fig3-3（a），the restrict condition is：,Put in
13、to （d），then obtain the displacement of simply supported beam：,from（d）we have：,The flexibility equation of the beam axis is ：,Two-dimensional Problems in Rectangular Coordinates,22,平面问题的直角坐标解答,（一）简支梁,梁轴的挠度方程：,23,2.Cantilever,In Fig3-3（b），the restrict condition is：,Put into（d），then we get the displace
14、ment of cantilever：,The flexibility of beam axis is：,2、In the State of Two-dimensional Strain,Only need to instead and in distortion and displacement formula for the state of two-dimensional with and .,Two-dimensional Problems in Rectangular Coordinates,24,平面问题的直角坐标解答,（二）悬臂梁,二、平面应变的情况,只要将平面应力情况下的形变公
15、式和位移公式中的 换为 ， 换为 即可。,25,3-3 Bending of a SimplySupported Beam by Uniform Load,With the half-converse method. Suppose is only the function of ：,Two-dimensional Problems in Rectangular Coordinates,26,3-3 简支梁受均布载荷,平面问题的直角坐标解答,设有矩形截面的简支梁，深度为 ，长度为 ，受均布载荷 ，体力不计，由两端的反力 维持平衡。如图3-4所示。取单位宽度的梁来考虑，可视为平面应力问题。,用半逆解法。假设 只是 的函数：,则：,对 积分，得：,解之，得：,其中， 、 是任意函数，即待定函数。,（a),（b),27,Now lets check whether the stress function above meet the demand of consistent equation. So derivate the fourth degree of ：,
