Matrices in Statics and Mechanics

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1 Matrices in Statics and Mechanics Casey Pearson 3/19/2012 Abstract The goal of this project is to show how linear algebra can be used to solve complex, multi-variable statics problems as well as illustrate another method to solve for principle stresses in mechanics. 1 Introduction to Linear Algebra Linear Algebra is a branch of mathematics that studies vectors.vectors are usually thought of as arrows that give magnitude and direction to something like force or torque, but they do not have to be that concrete. They can be more abstract, but in applications such as civil engineering problems, that is how they are used. These vectors are usually seen written in matrix form or on a linear equation of vectors. 1.1 System of Equations Vectors as matrices are used as the most efficient method to solve multi-variable systems of linear equations.the most general way to write a linear equation is in the form: a 1 x 1 + a 2 x a n x n = b Where b and a 1,...,a n are real or complex numbers that are given and x 1,...,x n are unknowns. A system of linear equations is merely a collection of these linear equations that involve the same variables (x 1,...,x n ). A solution to a system of linear equations is a list of numbers that when substituted for x 1,...,x n make each equation a true statement. 1.2 RREF In order to solve a system of linear equations the equations must be placed in a coefficient matrix where each column represents the coefficients attached to the corresponding x value, each row representing one equation and the final column is the b values. After the system of equations is in matrix form, the matrix can be put into reduced row echelon form or RREF. Using a combination of interchanging rows, multiplying rows by constants, and replacing rows with the 1

2 sum of itself plus a multiple of another row, you can get the matrix into a form where the leading number in each row is a one and above and below each one are zeros with the final column being constants. This form tells you the solution to your equation exactly where each row is the solution for the corresponding variable (row one is the solution for x 1, row two is solution for x 2 etc.). 1.3 Linear Independence The above solution will only work if a few very important qualifications are met. The first is that the number of linear equations in your system must match the number of unknowns or the equation may have infinitely many solutions. The second is that the vectors in your coefficient matrix must be linearly independent or, in other words, if the column of b values was replaced by zeros, the system must only have the trivial solution (all x values equal to zero). 2 Introduction to Statics Statics is a branch of mechanics that studies loads (forces and torques) on systems in static equilibrium. Since you know that the structure you are trying to analyze is not moving you can say that the sum of the forces equal zero and the sum of the torques equal zero. This assumption is the basis for all statics problems. 2.1 Forces Force is any influence that causes an object to undergo a change. It has magnitude and direction making it a vector quantity defined by the equation F = ma. As mentioned above in statics problems where the acceleration is zero because the object is not moving the force in every direction is equal to zero. At every point where forces are applied, the force in the x direction can be set up as an equation equaling zero as well as the forces in the y and z directions. 2.2 Torque Torque also known as a moment is the tendency of a force to rotate an object about an axis. While force is usually thought of the push or pull on an object torque can be thought of as the twist of an object. The magnitude of torque on an object is based on three things: the force applied, the length of the lever arm (distance from the force to the axis of rotation), and the angle between the force vector and the lever arm. Torque is represented by the equation τ = rf sinθ, where r is the length of the lever arm, F is the force acting on the object, and θ is the angle. 2

3 3 Example Problem The figure below shows a truss with known forces acting on it. This type of structure has forces working at every connection point to keep it at static equilibrium. Since we know the outside forces acting on it, we can solve for these internal forces at the joints. There will be two equations at every joint, one in each the x and y directions, so there will be a total of 10 equations. This means there must be ten unknowns as well that need to be solved for in order for the system of linear equations to work. We can see the three reaction forces, R 1 throughr 3, already on the figure and from statics we know there will be a torque on each of the seven braces giving you a total of ten unknowns. 3.1 Find Forces The two forces F 1 and F 2 pushing down on the truss at angles of θ 1 and θ 2 are given as F 1 = 10, 000lb at θ 1 = 75 and F 2 = 7, 000lb at θ 2 = 45. With this information we can write our equations for the forces working at each joint. 3.2 Write Equations Lower left joint x-direction Lower left joint y-direction Lower middle joint x-direction R 1 + T 1 cos 60 + T 2 = 0 R 2 + T 1 sin 60 = 0 T 2 T 3 cos 60 + T 4 cos 60 + T 5 = 0 3

4 Lower middle joint y-direction Lower right joint x-direction Lower right joint y-direction Upper left joint x-direction Upper left joint y-direction Upper right joint x-direction Upper right joint y-direction T 3 sin 60 + T 4 sin 60 = 0 T 5 T 6 cos 60 = 0 T 6 sin 60 + R 3 = 0 T 1 cos 60 + T 3 cos 60 + T 7 F 1 cos θ 1 = 0 T 1 sin 60 T 3 sin 60 F 1 sin θ 1 = 0 T 4 cos 60 T 7 + T 6 cos 60 F 2 cos θ 2 = 0 T 4 sin 60 T 6 sin 60 F 2 sin θ 2 = Put into Matrix 1 1/ / /2 1/ /2 3/ / / / / / / /2 0 1/ /2 0 3/2 0 0 R 1 T 1 T 2 R 2 T 3 T 4 T 5 T 6 R 3 T 7 = F 1x F 1y F 2x F 2y F 3x F 3y F 4x F 4y F 5x F 5y 4

5 3.4 Solve Using RREF we can solve for our unknown variables R T T R T T 4 = T T R The above matrices show the solution to every unknown with no extra work or simplification needed. The force in the direction of R 1 is N (the negative simply tells you the force is in the opposite direction of what was written in your equations), the force in the direction of T 1 is N and so on down the column matrix. The negative tells you that the torque is rotating in the opposite direction of how it was written in the original equation. This is a handy thing to know because it means you can write your forces in any direction in your equations and in your answer it will tell you which direction they are actually pointing, saving time and headache. 4 Mechanics Application using Eigenvalues and Eigenvectors An eigenvector of an n x n matrix A is a nonzero vector x such that Ax=λx for some scalar λ. A scalar λ is called an eigenvalue of A if there is a nontrivial solution x of Ax=λx; such an x is called an eigenvector corresponding to λ. Eigenvalues and eigenvectors can be used in mechanics to find the principle stress at a given point A on a body at equilibrium. 4.1 Normal and Shear Stresses Stress is defined as force over area so normal stress can be defined as force on a plane perpendicular to an axis where as shear stress is simply stress parallel with that plane or perpendicular to an axis. Normal stress σ is written with a subscript that identifies that axis with which the stress is aligned. For example a stress on the plane perpendicular to x is denoted σ x. Shear stress τ has two subscripts. The first subscript names the face on which the stress acts and the second gives the direction on the stress on that face. For example a stress that acts on the face of x in the direction of the y-axis is denoted τ xy. T 7 5

6 4.2 Principle Stress Eigenvectors and values can be used in mechanics to find the principal stresses at any point on a body. The principal stresses are the maximum and minimum normal stresses on a body.the minimum stress is the smallest stress that will be found anywhere on the body and the maximum stress is the largest. To do this the normal and shear stresses must be put into a matrix. We will use the assumption that at equilibrium the stresses in the negative x direction must equal the stresses in the positive x direction and the same in the y and z. In order for the body to keep from spinning we must also assume that τ xy must equal τ yx, τ xz must equal τ zx, and τ yz must equal τ zy. The resulting matrix can be seen below. σ x τ xy τ xz τ yx σ y τ yz τ zx τ zy σ z For the purpose of illustration let us assume that be are given the values of the above matrix for a given point on a body We will then decompose this matrix into two matrices, one will be a matrix of eigenvectors and the other will be a diagonal matrix that contains the principal values. The values in the diagonal matrix will give you the maximum and minimum normal stresses at this point. The first step is to find the eigenvalues of this matrix. These values are found by taking the roots of the polynomial found by calculating det(a λ I) where λ is the eigenvalue variable, A is your original matrix and I is your identity matrix of size A λ 0 0 det( λ 0 ) λ To get the eigenvalues from this you must factor the resulting polynomial. This can be done by hand or by MATLAB. Since this is an messy polynomial to factor MATLAB is probably the most efficient route. The factored polynomial gives you the eigenvalues, 17.49, 75.39, and the corresponding eigenvectors , , From these values we can gather that the max stress on the body at the given point is while the minimum stress is The direction of these stresses is given by the corresponding eigenvectors. 6

7 5 Conclusion Linear Algebra has many useful applications in solving real life problems. Statics and mechanics problems are some of the simplest examples showing how Matrices can turn an hour long substitution problem into a very simple plug and chug solution. 7

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