3D Stress Transformation Defining Stress states https://www.youtube.com/watch?v=w5cb0D1Ve-4 3D Stress Tensors https://www.youtube.com/watch?v=335geNKQJ8w

First a reminder of 2D Stress Transformation Given stress state in xy; find stress state in x’y’ ( orientation)

First a reminder of 2D Stress Transformation A closer look; observations!: Angles of x’ with x and y are  and  =90-  respectively. Therefore Cos  =Sin  Using the notation Cos  =l and Cos  =m: l2 + m2=1 There is only one independent angle ( ) as  =90- ! The sum of normal stresses is invariant! i.e. x +  y = x’ + y’ Transformation relations were obtained through equilibrium conditions!

First a reminder of 2D Stress Transformation For the specific case of interest, Principal stresses stress state: Confirmation that the sum of normal stresses is invariant! x +  y = x’ +  y’ =  1 +  2 Stress state in this specific case is diagonal with shear stress component zero! Can you present the relationship between the two stress states in matrix form?

Equivalent 3D stress states in different coordinate systems Transformation from a generic stress state (from x1, x2, x3 coordinates or x, y, z) To another generic coordinate system (x’, y’, z’ or 1, 2, 3) * or to a specific coordinate system ( x’1, x’2, x’3 or 1, 2, 3); i.e. principal stress state! *) Note: different notations here!

Principal stress state; a useful specific state for design practice! A special set of “transformed” axes exist on which there is no shear stress component, only normal ones! Such set represent Principal axes and corresponding normal stresses are known as Principal stresses. This is a very important state of stress for design purposes! For convenience we may re-order them from highest to lowest although this is not necessary! How can we identify this special set?!

General Transformation to a New Plane (ABC) New plane ABC is defined by the direction cosines of the normal to it, N(l,m,n) with direction cosines: l = cos a, m = cos b, and n = cos g. l2 + m2+ n2=1 This could be the new x’ plane, so the normal N(l,m,n) is the new x’ axis.

Normal stress components on the new plane This is the expression for normal stress for plane with normal I x’ direction! Similarly other terms of stress state in alternative coordinate system (x’, y’, z’) may be derived using the corresponding direction cosines of the other two planes (with normal axes y’ and z’)

Shear stress (t ) on the new plane Component forces per unit area (Sx, Sy, Sz) in direction of normal vector N(l,m,n) are:

Normal stress (s ) and shear stress (t ) on ABC

Direction cosines of shear stress (t ) Direction cosines for shear stress in the new plane are: Note: ls ≡ lt , ms ≡ mt , ns ≡ nt , (just in different notations!)

Worked example Normal stress (s ) and shear stress (t ) on a plane The state of stress at the weak point of a component is given as: Normal stresses in x, y, and z: 14, 10, and 35 MPa respectively; Shear stresses in xy, yz, and zx: 7, 0, and -7 MPa respectively; Find the normal and shear stresses for a plane defined by a normal vector having direction cosines proportional to 2, -1, and 3 with coordinate axes x, y, and z respectively. Find the direction of the shear stress acting on that plane

Maths reminder! eigenvalues, eigenvectors Stress: a symmetric matrix, eigen values and eigen vectors in stress analysis For a real symmetric matrix (like stress tensor!) the eigenvalues are real and eigenvectors are orthogonal 𝑀𝑥=𝜆𝑥 (𝑀−𝜆𝐼)𝑥=0 𝑀−𝜆𝐼 =0 Find real values of  Replace in matrix form to find eigenvectors, x Trying to find eigenvalues for a 33 matrix we end up solving a cube equation in terms of . There is no rules how to solve such equation! Scale of eigenvectors is not set! We can also write: 𝟓𝑴𝒙=𝟓𝝀𝒙. Note that the equation is obviously true for any other multiplier number! It means if  is an eigenvalue then is also an eigenvalue for any . We can use this to transform tensors and specifically to find stress states of interest!

A little trick?! 𝑀𝑥=𝜆𝑥 𝐷= 𝜆 1 0 0 0 𝜆 2 0 0 0 𝜆 3 Let’s build a matrix of eigenvalues, matrix D 𝑀𝑥=𝜆𝑥 𝐷= 𝜆 1 0 0 0 𝜆 2 0 0 0 𝜆 3 𝑋= 𝑥 11 𝑥 21 𝑥 31 𝑥 12 𝑥 22 𝑥 32 𝑥 13 𝑥 23 𝑥 33 𝑀𝑋=𝑋𝐷 It can be shown that matrix form D for eigenvalues is true and justified. Expand the matrix form to show XD represents all sets of three eigenvalues and three eigenvectors

Rotation tensor?! 𝑀𝑋=𝑋𝐷 𝐷= 𝑋 −1 𝑀𝑋 𝑀=𝑋𝐷 𝑋 −1 𝑀 𝑛 =(𝑋𝐷 𝑋 −1 )(𝑋𝐷 𝑋 −1 )(𝑋𝐷 𝑋 −1 )……., (n times) 𝑀 𝑛 =𝑋 𝐷 𝑛 𝑋 −1 The above matrix equations suggest that: D is actually the Rotation matrix (tensor when dealing with stress!) and is obtained by rotating M

Principal stresses in 2D using direct analysis In class activity Example of Principal stresses in 2D Using direct analytical method for 2D

Cramar’s rule may be used to find principal stresses!

Useful MATLAB Commands To create the matrix A = A = [8 -4 5; 3 2 -1; 4 6 7]; To multiply two matrices such that A = BC A = B * C; To transpose a matrix A so that B = AT, B = A’; To invert a matrix A to give B = A-1 B = inv(A); To produce a diagonal matrix D of eigen values and a full matrix V whose columns are the corresponding eigen vectors so that X*V = V*D. [V,D] = eig(X); The matrix V gives the eigen vectors as unit vectors. So when obtaining principal stresses the eigen vectors are the direction cosines of the three principal planes.

Principal stresses in 3D using transformation rule In class activity Example of Principal stresses in 3D

A numerical example eigen values, eigen vectors, and principal stresses Consider the stress tensor [s] = MPa Eigen values from MatLab are 70.7, 27.1 and -42.8 in descending order. Corresponding eigen vectors are Note that So the principal stresses are: s1 = 70.7 MPa, s2 = 27.1 MPa, s3 = -42.8 MPa T =

A numerical example (cont.) Direction cosines for Principal planes So the direction cosines for the principal axes x’, y’, z’ relative to the x, y, z axes are , and respectively. Note that these three vectors are orthogonal since their dot products are zero, i.e. the three new axes are at right angles as would be expected. This can be a useful check when analysing stress tensors.

Further useful information (normal stress invariant!) sxx + syy + szz = sx’x’ + sy’y’ + sz’z’ = s1 + s2 + s3 Check: In the case of our example: 50 + (-35) + 40 = 55 ✔ 70.7 + 27.1 + (-42.8) = 55 ✔ The “stress invariant”, verified earlier for 2D state, is also valid for the general case of 3D stress state Similarly, “strain invariant” is valid for strain states!). This can be used as a useful check that the principal stresses are correctly identified.

Summary of stress transformation (expected learning outcomes) Background knowledge of matrix algebra provides a useful tool for stress (or strain) analysis in its most generic form. The concept of eigenvalues and eigenvectors is the key to this procedure Generally we will need to solve a 3rd order equation to find the eigenvalues that is not straightforward. Using a matrix of eigenvalues provides a practical tool for stress analysis Of specific interest is to find the critical stress states like principal stress state Examples of 2D and 3D using matrix method led to obtaining the required solution Using a 2D example and solving in different methods the consistency of results was appreciated.