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Vector space and basis

  1. Jan 22, 2012 #1
    I'm currently doing a self study course on Linear Algebra.

    Can anyone give me an example of vector space and basis with reference to Structural Engineering?

    For example I have a displacement vector for a simply supported beam as:

    [thata_a theta_b]^T

    where; theta_a and theta_b represent rotations at two ends of the beam.

    What we call a vector space here and what is the basis here?
  2. jcsd
  3. Jan 23, 2012 #2


    Staff: Mentor

    Your vector is two-dimensional, with real coordinates, so the relevant vector space is R2. The standard basis for R2 is {<1, 0>T, <0, 1>T}.

    Every vector in R2 can be written as a linear combination of these two basis vectors.

    I don't imagine that structural engineers do much with linear algebra (I could be wrong, though). They are mostly interested in elements of the plane (i.e., R2) or solids in space (R3).

    Linear algebra deals with these spaces and many more of higher dimension.
  4. Jan 23, 2012 #3
    Pretty well all the methods and maths used in structural engineering uses linear methods.

    The general differential equations of a beam are linear so succumb to linear analysis for solution.

    Matrix methods, virtual work, tension coefficients, frame analysis, moments of inertia - the list goes on.

    This is because they are most methods are based on linear - elastic theory viz stress is proportional to strain and its components are individually proportional to the strain in the component direction. Unit vectors in these directions form the basis vectors you ask about.

    The main non linear analyses in structures are the use of plastic theory and fracture mechanics.

    go well
  5. Jan 23, 2012 #4


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    Actually they do a great deal of it, in vector spaces with dimensions much larger than 2 or 3.

    But they don't usually call it "doing linear algebra". The components of the basis vectors are more likely to be called "unit load cases", "vibration mode shapes", etc, than "basis vectors".
  6. Feb 1, 2012 #5
    AlephZero, another example where Civil Engineers use linear algebra- request your feedback

    Given a state of stress at a point, if we want to compute the principal stresses.Then, talking in the linear algebra terms; we need to transform the matrix to a cordinate system that correpsonds to the coordinate system of principal directions.

    We do this (or we can do this) by diagonalzing the stress matrix so that all off diagonal terms are zero and diagonal terms represent the eigen values which are the principal stresses

  7. Feb 1, 2012 #6
    AlephZero- what is the basis here?
  8. Feb 1, 2012 #7


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    Trying to figure how to invoke vector spaces in your example. Here's what I can come up with. The beam angle along the length of of the beam will be an angle valued function of the lateral coordinate.
    Imagine then [itex]\theta_x = \theta(x)[/itex] for [itex] a\le x \le b[/itex] being the angle at each point on the beam.

    The set of continuous functions on an interval is a vector space, you can add functions and multiply by scalars so you can call them vectors.

    Now the mechanics of the beam deflection will probably manifest as a complicated 2nd order differential equation on these functions and the solutions will be uniquely defined by two boundary conditions, for example the angle at the end points. The problem is that the manifold of solutions will probably not be a linear space of functions however for small deflections near the 0,0 case you can linearize, i.e. pick a flat plane tangent to this manifold of solutions. You'll then get a 2 dimensional vector space of linearized solutions.

    Think of it this way. For very small deformations of the beam, assume that increasing the deflection will occur proportionately i.e. the functions [itex]\theta(x)[/itex] will be linear:
    [tex]\theta(x) = \theta_a + (x-a)(\theta_b - \theta_a)/(b-a)=\frac{b-x}{b-a}\theta_a + \frac{x-a}{b-a}\theta_b [/tex]
    Note I've written the function as a linear combination of two functions with multipliers equal to your two end angles.

    So within our big space of functions we are considering functions of this linear form and the basis is the pair of functions:
    [tex][\mathbf{e}_1, \mathbf{e}_2] =\left[ \frac{b-x}{b-a} , \frac{x-b}{b-a}\right][/tex]
    and then the coordinates are:
    [tex]\left[\begin{array}{c}\theta_a \\ \theta_b \end{array}\right][/tex]
    The vector is then:
    [tex]\theta(x) = \left[\begin{array}{cc} \mathbf{e}_1 & \mathbf{e}_2 \end{array}\right]\left[\begin{array}{c} \theta_a \\ \theta_b\end{array}\right][/tex]
  9. Feb 2, 2012 #8


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    My experience is that Engineers rely a lot on differential equations- and the set of all solutions to an nth order linear differential equation is a vector space of dimension n.
  10. Feb 2, 2012 #9


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    nth order homogenous linear diff. eqn.
  11. Feb 2, 2012 #10


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    Yes, thank you. The set of all solutions to an nth order non-homogeneous linear differential equation is a linear manifold but not a vector space.
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