非常感谢ggbbggb在这个问题上花费了很多笔墨。这里把他的回复及我的理解列在下面，望大家指正：
我的陈述：
Here by small, I mean that the virtual displacement should be small enough such that during the process of virtual displacement the external load would remain constant. But you don't buy it.
ggbbggb的答复：
You use "such that" indicating the external load remaining constant is a consequence of small virtual displacement .So I don't buy it. If I say" the virtual displacement should be large enough such that during the process of virtual displacement the external load would remain constant", how do you prove I am wrong? If I say " the sun rises in the west so that during the process of virtual displacement the external load would remain constant". Anyhow the external force is assumed to be constant during the variation of displacement and the actual external force can be non-constant. If assume in this case， external load F=F（u）， so that during the variation of u, F should not be taken constant (agree with this?). But in using WVP, we are calculating the virtual work as Fdeltu=F(u) delta u, although obviously F (u) is not constant during variation of u. This is an example showing the “constant” of F in Fdeltu=F(u) delta u is not due to the variation of u, since during variation of u, F is actually changing. We still treat it as a constant.
我的最新回复：
使用虚功原理我们可以直观的得到具有物理意义的弱形式，我们不必采用泛函然后应用变分，或者由控制方程采用分步积分得到弱形式。要使虚功原理具有物理意义，前提就是要求虚位移为无限小的可允许的位移，这样在虚位移发生的过程中，对于一个小的微元，应力才能假设为恒定，同样的体力和面力也假设为恒定，我们才能得到虚功或者虚应变能。假设在虚位移delta u发生的过程中，F的方向保持不变，如果我们有这样的情形： 在0~1/3*delta u，力的大小为F，在1/3*delta u ~ 2/3*delta u的过程中，力的大小为1/2*F，在2/3*delta u ~ delta u的过程中力的大小为 1/4*F，那么虚功为：F*1/3*delta u + 1/2*F*1/3*delta u + 1/4*1/3*delta u.
对于F = F(u)的情形，计算虚功时，之所以我们可以使用 F(u)*delta u，恰恰是因为虚位移delta u 无限小的原因，我们才可以假设在的delta u发生的过程中 F(u)保持恒定。如果这里的虚位移为任意大小，那么虚功的计算严格来讲应为： int [0, delta u] F(u) du. 可以看出对于这个积分，如果delta u 很微小，我们可以去掉积分符号得到F(u)*delta u。 在鹫津久一朗的书中，虚位移始终是无限小的。我这里附上他关于虚功原理的叙述。从虚功原理推导出最小势能原理要求虚位移必须是微小的可允许的位移，这样我们才可以假设外力的大小和方向保持不变。对于虚功原理不能把虚位移和有限元离散时的试函数混为一谈，如Belytschko所说，有限元离散中的力和试函数的乘积没有物理意义。
我的表述：
For this problem, the load is going to be deformation-dependent.Therefore to get the virtual work due to traction, we have to assume that thevirtual displacement has to be small enough such that during the virtualdeformation process we can stillcalculate the virtual work using the current configuration at t+dt. Otherwisewe cannot calculate the virtual work.
ggbbggb的回复：
There is no necessity to use the large deformation largerotation example at all. It is just VWP applied to the current configuration(unknown beforehand). If I do not assume a small virtual displacement, whycan't I use the WVP. Why can't I use the virtual work delta WF just = FdeltaU.
我的最新回复：
我的理解是要使虚功原理具有物理意义，虚位移必须为无限小的量。如果不是无限小的量，在大虚位移发生的过程中，力大小或者方向都可能发生改变，在这种情况下还使用 F*delta U 计算虚功，有物理意义吗？使用虚功原理后，我们得到弱形式，在等号左右我们总是可以乘以一个大常数，把实际的虚位移变成大的变量，但是积分号里的乘积不具有物理意义。比如悬臂梁受均匀横向载荷，如果我们取单位长度，假设相当大的虚位移，然后我们仍然使用 p*delta u，在我看来这没有实际的物理意义，因为至少在delat u 的发生过程中，均布载荷的方向是在改变的。
我的表述：
"Here by small, I mean thatthe virtual displacement should be small enough such that during the process ofvirtual displacement the external load would remain constant. But you don't buyit."
ggbbggb的回复：
You use “should be small enough”,indicating that if the virtual displacement is large, then VWP does not hold?Is this what you mean? If the above answer is yes, then canI demonstrate to you for large virtual displacement VWP still holds, asmentioned in my previous reply ? By saying this, we have to agree first on “what is WVP”. We do not need complex physics or maths here at all. We only needone particle moving along x direction only in a horizontal plane. Apply two x-direction forces to the particle. From physics, we haveF1+F2=0. The WVP says, if we apply an arbitrary virtual displacement to thisparticle (which is in equilibrium), the total virtual work F1deltu+F2deltu=0; and if we apply an arbitraryvirtual displacement to this particle, and if we always get F1deltu+F2deltu=0 ,then the particle is in equilibrium. This isall my understanding of VWP for this particle (nothing else). But you can seethere is no indication that deltu should be small. Let’s consider two cases,where deltu is small in case 1 and deltu is large in case 2. You can repeat theabove statements in black (which includes all about the VWP). So this showsthat if we apply WVP by using small deltu or large deltu, they make nodifference at all. Then how can one claim that in a VWP, deltu must be small.You are claiming the “constant” of F1 and F2 is due to the “smallness” ofdeltu. But in this example, the “constant” of F1 and F2 is a prescribed given condition if I want. For example, F1=10,F2=-10, I give this condition always. Then how can you say that “ a given prescribed”condition is a consequence of “smallness” of deltu. Whatever value of deltu isirrelevant to the given prescribed condition. “thevirtual displacement should be small enough such that a prescribed givenexternal force remain constant” make no sense. Agree? If you are not happy about this “givenprescribed constant of F1 and F2”, let’s move it to a condition whereF1=3u (prescribed) and F2 unknown to bedetermind. Here F2 to be determined, and also if you apply a variation of u,then you also have a variation of F1. But in using the VWP, we still use F1deltu+F2deltu=0=>F2=-3u .Is there any problem with this example at all? If you want you can use VWP indynamics by taking the D’Alembert force as one special external force. Andobviously, this D’Alembert force is also not constant in general during theprocess of virtual displacement, since it is a function of u. But we can stillapply VWP, by taking the D’Alembert force unchanged with respect to deltu incalculating its contribution to the total virtual work. I think the paragraph related to “givenprescribed constant of F1 and F2” is not due to smallness of deltu concludes all.If you apply large deltu, a “given prescribed constant of F1 and F2” shouldchange?
我的最新回复：
我应该这么说：如果虚位移不是小变量的话，虚功原理不具有物理意义。取微小的虚位移，物理上非常直观，由于虚位移比较微小，体力面力在虚位移发生的过程中大小方向保持不变，应力在虚应变发生过程中也保持不变，我们可以直接计算虚功，直接推导出有限元离散化需要的等式。如上面我提到过的，因为由于我们得到的是个等式，虚位移取的大小不会妨碍我们，但是等式是否具有物理意义就不好说了。
具体到你提到的例子，你这里有两个假设一个是你提到的F1,F2大小保持不变，另一个隐含的假设是它们的方向保持不变，如果F1， F2的方向发生变化的话，你的写法从物理意义还是有待商量。
如果我说在粒子平衡状态，施加一个微小的位移，因为虚位移微小，所以F1和F2的大小方向保持不变，我们得到虚功（F1+ F2)*delta u = 0，我这个说法应该说的过去。如果虚位移是很大量的话，那我们计算虚功是否应该为：int[0, delta u] (F1 + F2) du = 0.
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发表于 2014-11-28 22:28:15
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