the stress components at a point are represented as a matrix consisting of 3 normal stresses on the diagonal and three shear stresses off the diagonal. The condition of equilibrium requires that the matrix be symmetric. A typical stress matrix is written as follows: []= [ x xy xz xy y yz xz yz z ] The principal planes are defined as those planes on which only normal stresses act and are free of shear stresses. They also represent the maximum values of the normal stress at that point. The condition for a plane to be a principal plane and the stresses on it to be the principal stress is expressed in the form of a homogeneous equation: [ x− xy xz xy y− yz xz yz z−]{ l x l y lz} ={ 0 0 0} where is the principal stress and l x , l y and l z are the direction cosines of the outward normal to the principal plane. It is required to determine the principal stresses and the direction cosines of the principal planes. The characteristic equation of the above set of homogeneous equations is 3 −I1 2 I 2−I 3=0 . Being a cubic equation, it has three roots and thus there are three values of principal stresses. Corresponding to each principal stress is one principal plane represented by the direction cosines of its outward normal. In mathematics, this is an eigenvalue problem, the eigenvalues are the principal stresses and eigenvectors are the direction cosines of the outward normal to the principal planes. The principal stresses and principal planes can thus be obtained with the spec function in Scilab. As an example, let the stress matrix be as follows: []= [ 3 2.5 4 2.5 1 0 4 0 2] Then the solution can be obtained as follows: -->s = [3, 2.5, 4; 2.5, 1, 0; 4, 0, 2]; -->[x, p] = spec(s) p = -2.4316605 0. 0. 0. 1.2968769 0. 0. 0. 7.1347836 x = 0.6529665 -0.0979278 0.7510293 -0.4756928 -0.8246499 0.3060537 -0.5893651 0.5571020 0.5850523 Tutorial 18 – Principal Stresses | 51 The eigenvalues are on the diagonal of the matrix p. The columns of eigenvector x correspond to the eigenvalue on the diagonal. Thus the first column of x corresponds to the principal stress value -2.4316605. The eigenvalues are arranged in increasing order whereas the principal stresses are, by convention, arranged in decreasing order. Thus the first principal stress is 1=7.1347836 , second principal stress is 2=1.2968769 and the third principal stress is 1=−2.4316605 . The first principal plane is defined by the direction cosines l x=0.7510293 , l y=0.3060537 and l z=5850523 . The solution can also be obtained by an alternative way, using polynomials. In this approach the eigenvalues can be be obtained easily but obtaining the eigenvectors may involve a few additional manual calculations. But it has the advantage that we can obtain the characteristic equation, which is not available in the first solution. Here, let us first define a seed for a polynomial as p = poly(0, 'p') and use it to represent the homogeneous equations as follows: -->p = poly(0, 'p'); -->ss = s – eye(s) * p; -->ch = det(ss) ch = 2 3 -22.5 + 11.25p + p - p -->roots(ch) ans = 1.2968769 -2.4316605 7.1347836 To obtain the eigenvector corresponding to a chosen eigenvalue, we must substitute the value of the eigenvalue into the homogeneous equations, and choose any two of the three equations and solve them for two of the direction cosines in terms of the third. [ y−i yz yz z−i ]{ l y l z} =l x{ −xy −xz} The above equations can be solved to obtain l y and l z in terms of l x . Let l y=k y l x and l z=k z l x . Then using the fact that the sum of the square of the direction cosines is unity, all the three direction cosines can be obtained. lx 2 l y 2 l z 2 =1 or 1k y 2 kz 2 l z 2 =1 Rearranging and solving we get: l x= 1 1k y 2 kz 2 , l y=k y 1 1k y 2 k z 2 and l z=kz 1 1k y 2 kz 2 We could write our own functions to do these calculations, taking care to sort the eigenvalues before proceeding with the eigenvector calculations.
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หลักสำคัญในเทฤษฎีความยืดหยุ่น ติวเตอร์เรียล 18
Tutorial 18 – Principal Stresses
In the theory of elasticity,
the stress components at a point are represented as a matrix consisting of 3 normal stresses on the diagonal and three shear stresses off the diagonal. The condition of equilibrium requires that the matrix be symmetric. A typical stress matrix is written as follows: []= [ x xy xz xy y yz xz yz z ] The principal planes are defined as those planes on which only normal stresses act and are free of shear stresses. They also represent the maximum values of the normal stress at that point. The condition for a plane to be a principal plane and the stresses on it to be the principal stress is expressed in the form of a homogeneous equation: [ x− xy xz xy y− yz xz yz z−]{ l x l y lz} ={ 0 0 0} where is the principal stress and l x , l y and l z are the direction cosines of the outward normal to the principal plane. It is required to determine the principal stresses and the direction cosines of the principal planes. The characteristic equation of the above set of homogeneous equations is 3 −I1 2 I 2−I 3=0 . Being a cubic equation, it has three roots and thus there are three values of principal stresses. Corresponding to each principal stress is one principal plane represented by the direction cosines of its outward normal. In mathematics, this is an eigenvalue problem, the eigenvalues are the principal stresses and eigenvectors are the direction cosines of the outward normal to the principal planes. The principal stresses and principal planes can thus be obtained with the spec function in Scilab. As an example, let the stress matrix be as follows: []= [ 3 2.5 4 2.5 1 0 4 0 2] Then the solution can be obtained as follows: -->s = [3, 2.5, 4; 2.5, 1, 0; 4, 0, 2]; -->[x, p] = spec(s) p = -2.4316605 0. 0. 0. 1.2968769 0. 0. 0. 7.1347836 x = 0.6529665 -0.0979278 0.7510293 -0.4756928 -0.8246499 0.3060537 -0.5893651 0.5571020 0.5850523 Tutorial 18 – Principal Stresses | 51 The eigenvalues are on the diagonal of the matrix p. The columns of eigenvector x correspond to the eigenvalue on the diagonal. Thus the first column of x corresponds to the principal stress value -2.4316605. The eigenvalues are arranged in increasing order whereas the principal stresses are, by convention, arranged in decreasing order. Thus the first principal stress is 1=7.1347836 , second principal stress is 2=1.2968769 and the third principal stress is 1=−2.4316605 . The first principal plane is defined by the direction cosines l x=0.7510293 , l y=0.3060537 and l z=5850523 . The solution can also be obtained by an alternative way, using polynomials. In this approach the eigenvalues can be be obtained easily but obtaining the eigenvectors may involve a few additional manual calculations. But it has the advantage that we can obtain the characteristic equation, which is not available in the first solution. Here, let us first define a seed for a polynomial as p = poly(0, 'p') and use it to represent the homogeneous equations as follows: -->p = poly(0, 'p'); -->ss = s – eye(s) * p; -->ch = det(ss) ch = 2 3 -22.5 + 11.25p + p - p -->roots(ch) ans = 1.2968769 -2.4316605 7.1347836 To obtain the eigenvector corresponding to a chosen eigenvalue, we must substitute the value of the eigenvalue into the homogeneous equations, and choose any two of the three equations and solve them for two of the direction cosines in terms of the third. [ y−i yz yz z−i ]{ l y l z} =l x{ −xy −xz} The above equations can be solved to obtain l y and l z in terms of l x . Let l y=k y l x and l z=k z l x . Then using the fact that the sum of the square of the direction cosines is unity, all the three direction cosines can be obtained. lx 2 l y 2 l z 2 =1 or 1k y 2 kz 2 l z 2 =1 Rearranging and solving we get: l x= 1 1k y 2 kz 2 , l y=k y 1 1k y 2 k z 2 and l z=kz 1 1k y 2 kz 2 We could write our own functions to do these calculations, taking care to sort the eigenvalues before proceeding with the eigenvector calculations.
the stress components at a point are represented as a matrix consisting of 3 normal stresses on the diagonal and three shear stresses off the diagonal. The condition of equilibrium requires that the matrix be symmetric. A typical stress matrix is written as follows: []= [ x xy xz xy y yz xz yz z ] The principal planes are defined as those planes on which only normal stresses act and are free of shear stresses. They also represent the maximum values of the normal stress at that point. The condition for a plane to be a principal plane and the stresses on it to be the principal stress is expressed in the form of a homogeneous equation: [ x− xy xz xy y− yz xz yz z−]{ l x l y lz} ={ 0 0 0} where is the principal stress and l x , l y and l z are the direction cosines of the outward normal to the principal plane. It is required to determine the principal stresses and the direction cosines of the principal planes. The characteristic equation of the above set of homogeneous equations is 3 −I1 2 I 2−I 3=0 . Being a cubic equation, it has three roots and thus there are three values of principal stresses. Corresponding to each principal stress is one principal plane represented by the direction cosines of its outward normal. In mathematics, this is an eigenvalue problem, the eigenvalues are the principal stresses and eigenvectors are the direction cosines of the outward normal to the principal planes. The principal stresses and principal planes can thus be obtained with the spec function in Scilab. As an example, let the stress matrix be as follows: []= [ 3 2.5 4 2.5 1 0 4 0 2] Then the solution can be obtained as follows: -->s = [3, 2.5, 4; 2.5, 1, 0; 4, 0, 2]; -->[x, p] = spec(s) p = -2.4316605 0. 0. 0. 1.2968769 0. 0. 0. 7.1347836 x = 0.6529665 -0.0979278 0.7510293 -0.4756928 -0.8246499 0.3060537 -0.5893651 0.5571020 0.5850523 Tutorial 18 – Principal Stresses | 51 The eigenvalues are on the diagonal of the matrix p. The columns of eigenvector x correspond to the eigenvalue on the diagonal. Thus the first column of x corresponds to the principal stress value -2.4316605. The eigenvalues are arranged in increasing order whereas the principal stresses are, by convention, arranged in decreasing order. Thus the first principal stress is 1=7.1347836 , second principal stress is 2=1.2968769 and the third principal stress is 1=−2.4316605 . The first principal plane is defined by the direction cosines l x=0.7510293 , l y=0.3060537 and l z=5850523 . The solution can also be obtained by an alternative way, using polynomials. In this approach the eigenvalues can be be obtained easily but obtaining the eigenvectors may involve a few additional manual calculations. But it has the advantage that we can obtain the characteristic equation, which is not available in the first solution. Here, let us first define a seed for a polynomial as p = poly(0, 'p') and use it to represent the homogeneous equations as follows: -->p = poly(0, 'p'); -->ss = s – eye(s) * p; -->ch = det(ss) ch = 2 3 -22.5 + 11.25p + p - p -->roots(ch) ans = 1.2968769 -2.4316605 7.1347836 To obtain the eigenvector corresponding to a chosen eigenvalue, we must substitute the value of the eigenvalue into the homogeneous equations, and choose any two of the three equations and solve them for two of the direction cosines in terms of the third. [ y−i yz yz z−i ]{ l y l z} =l x{ −xy −xz} The above equations can be solved to obtain l y and l z in terms of l x . Let l y=k y l x and l z=k z l x . Then using the fact that the sum of the square of the direction cosines is unity, all the three direction cosines can be obtained. lx 2 l y 2 l z 2 =1 or 1k y 2 kz 2 l z 2 =1 Rearranging and solving we get: l x= 1 1k y 2 kz 2 , l y=k y 1 1k y 2 k z 2 and l z=kz 1 1k y 2 kz 2 We could write our own functions to do these calculations, taking care to sort the eigenvalues before proceeding with the eigenvector calculations.
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