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Quiz-Fall2010-1_2

# Quiz-Fall2010-1_2 - = I1 7 I2 5 3 3 11 I3 5 Hence the cubic...

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EML 4507 Fall 2010 Quiz 1-2 If you want, write your name in the back. Solution

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Problem 3 : Matrix A given by [ ]= A 8442 and the eigen values of [ A ] are: 0 and 10. Determine if it is: (a) positive-definite; or (b) Positive semi definite; or (c) Negative definite. Solution Since the eigne values are greater than or equal to zero ( ≥ ) λ 0 , matrix A is positive semi definite. Illustration The quadratic form = f xy8442xy can be expanded as = + f 22x y2 . Then one can see f =0 for all 2 x + y =0 or y =-2 x . Thus, f =0 for combinations other than x = y =0. However, f >0 is always satisfied, because it is a perfect square. Problem 1 : Derive the (cubic) characteristic equation for determining the principal stresses of the stress matrix shown below. Do NOT solve the equation. Solution Calculate the invariants: = , = + + =
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Unformatted text preview: , = I1 7 I2 5 3 3 11 I3 5 Hence, the cubic equation for principal stresses is:-+-= σ3 I1σ2 I2σ I3 0 or-+- = σ3 7σ2 11σ 5 0--- = σ 5σ 1σ 1 0 Problem 2 : It is determined that σ n =5 is one of the principal stresses for the state of stress in Problem 1. (i) Write the three linear equations you will use to determine the corresponding principal direction. (ii) Solve the above equations to determine the principal direction. If there are multiple principal directions for σ n =5, then show them in a parametric form. Solution ( - )-- ( - ) ( - ) = 3 5 20 2 3 5 000 1 5 nxnynz 000 The first two equations are the same:--= → =-2nx 2ny 0 nx ny The last equation yields n z =0 Then the principal direction is given by ,- , 121 1 0...
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Quiz-Fall2010-1_2 - = I1 7 I2 5 3 3 11 I3 5 Hence the cubic...

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