Linear Elastic Systems We shall restrict attention to systems in which the displ

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Linear Elastic Systems We shall restrict attention to systems in which the displacements are small and linearly proportional to the applied loads, in which case the principle of linear superposition applies. The principle of linear superposition then permits us to write Where Cij is the displacement component due to n units force Fj acting alone. We shall call the coefficients Cij influence coefficients. They can also be regarded as the elements of a matrix C, which is called the compliance matrix. Strain Energy Suppose there are only row external forces F1, F2. We consider two scenarios. F1 applied first, then F1. During the first phase of the loading, F2 = 0 and F1 increases from zero to its maximum value. The work done is therefore We now hold F1 constant, while gradually applying F2. The additional work done during this phase of the process is Where Delta u1, Delta u2 are the changes in u1, u2, respectively during (and due to) the application of F2. Since the total work done during both phases of the loading process must be stored as strain energy, we have F1 applied first, then F1. Suppose we now consider a similar scenario, explicit that the load F2 is applied first, followed by F1. The argument is unchanged except that the suffixes 1, 2 are reversed, with the result for the final strain energy. Examination of the equations show that this will be the case if and only if C12 = C21. More generally, if there are N forces and displacement components, we must have Cij = Cij Prove the bounds on the influence coefficients below .

Explanation / Answer

The coefficient of determination can never be negative. it is bounded by 0 = r² = 1 and this is because the coefficient of determination is the square of the correlation coefficient r which is bounded by -1 = r = 1 The correlation coefficient, r, is a measure of the linear relationship between two variables. If the data is non-linear then the correlation coefficient is meaningless. r takes on values between -1 and 1. negative values indicate the relationship between the variables is indirect, i.e., on a scatter plot the data tends to have a negative slope. Positive values for r indicate the data tends to have a positive slope. if r = 0 we say the variables are uncorrelated. the closer the absolute value of r is to 1, the stronger the linear association between the two variables. there are many different formulas for calculating the value of r. if we let xbar and ybar be the means of two data sets. sx and sy are the standard deviations in the data sets and n = total sample size then: r = 1/(n - 1) * S( ((xi - xbar)/sx) * ((yi - ybar)/sy)) with the sum going from i = 1 to n r = Covariance(X,Y) / [(v(Var(X))v(Var(Y))] r = S(xi - xbar)(yi - ybar) / [ v(S(xi - xbar)²) v(S(yi - ybar)²)] the second equation shows that the correlation coefficient the ratio between the measure of spread between the variables and the product of the spread within each variable. r is unit less. r is not affected by multiplying each data set by a constant, and a constant to each data set or interchanging x and y. r is subject to outliers. r² is called the coefficient of determination. It is a measure of the proportion of variance in y explained by regression. Also note that correlation is not causation. Here is an example: the shoe size of grade school students and the student's vocabulary are highly correlated. In other words, the larger the shoe size, the larger the vocabulary the student has. Now it is easy to see that shoe size and vocabulary have nothing to do with each other, but they are highly correlated. The reason is that there is a confounding factor, age. the older the grade school student the larger the shoe size and the larger the vocabulary. you cannot compare models by comparing the r values. This is a long discussion, a full day lecture in the prob/stat courses I've instructed. Model comparison is a topic usually saved for high level under grad courses or graduate level courses.

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