1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: (Beam theory) Deriving Virtual Work

  1. Apr 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Derive the equation used for calculating the deflection of a beam at some arbitrary point.

    Because I haven't seen anything like this on the web or in textbooks, I am asking for some feedback.

    2. Relevant equations
    [tex]v_{k} = \int\frac{\delta M(x)M(x)}{EI} dx[/tex]

    3. The attempt at a solution
    I decided I will be stubborn and derive it my way rather than what the textbook does, in a way that makes more sense.

    The work done on a system by external forces is equal to the change in energy of the system. What energy? In this case, elastic potential energy.

    [tex]W=\Delta E[/tex]

    We need to find the magnitude of this change in elastic potential energy, and I am doing that by energy density integrated over the entire beam:

    [tex]W=\int\int u _{ }dV[/tex]
    If we break the beam into cross sections over a length, the integral becomes.
    [tex]W=\int\int u _{ }dA _{ }dx[/tex]

    Inside the beam, only stress normal to the strain of the beam elements does work. Therefore, I only take into account the normal stresses.

    [tex]W=\int\int\int \sigma _{ }d\epsilon _{ }dV[/tex]
    We know from basic engineering that
    [tex]\sigma = E \epsilon[/tex]
    The integral becomes
    [tex]W=\int\int\int^{\epsilon_{f}}_{0} E \epsilon _{_{}}d\epsilon _{_{}}dA _{}dx[/tex]
    [tex]W=\int\int \frac {1}{2} E \epsilon _{f}^{2} dA_{ }dx[/tex]

    Or, since

    [tex]W=\int\int \frac {1}{2} \sigma \epsilon _{f}dA _{ }dx[/tex]

    I now found out the change in potential energy of an element dV inside the beam as a function of the stress and strain at that point. Big deal.

    But we know that in beam theory, there is a relationship between the bending moment of a cross section in the beam and the normal stress at any point within that cross section. Here it is, I am not deriving it.

    [tex]\sigma = \frac{My}{I}[/tex]
    where y is the distance from the neutral axis of the beam
    [tex]\epsilon = \frac{\sigma}{E} = \frac{My}{EI}[/tex]

    Putting the three together, we have
    [tex]W=\frac{1}{2}\int\int \frac{M^{2}y^{2}}{EI^{2}} dA _{ }dx[/tex]

    Throughout our cross section for dA, M and EI are constant, only y is changing, rearranging gives

    [tex]W=\frac{1}{2}\int \frac{M^{2}}{EI^{2}} \int y^{2}dA _{ }dx[/tex]
    [tex]\int y^{2}dA = I[/tex]
    [tex]W=\frac{1}{2}\int \frac{M^{2}I}{EI^{2}} dx[/tex]
    [tex]W=\frac{1}{2}\int \frac{M^{2}}{EI} dx[/tex]

    The total work done by real forces through a real displacement on a beam is a function of the moment at any point on the beam integrated throughout the beam.

    Now I apply a virtual load on the beam, on the point where I am interested in the deflection of the beam. The virtual load does an amount of work equal to the sum of the real and virtual deflections at that point.

    The bending moment throughout the beam will change by a virtual amount.

    [tex]W + \delta W =\frac{1}{2}\int \frac{(M + \delta M)^{2}}{EI} dx[/tex]
    [tex]W + \delta W =\int \frac{(M^{2} + 2M \delta M + \delta M^{2}}{2EI} dx[/tex]
    [tex]W + \delta W =\int \frac{M^{2}}{2EI} dx + \int \frac{2M \delta M}{2EI} dx + \int \frac{\delta M^{2}}{2EI} dx[/tex]

    I derived that
    [tex]W=\int \frac{M^{2}}{2EI} dx[/tex]
    so real work and the change in real energy cancel
    [tex]\delta W =\int \frac{M \delta M}{EI} dx + \int \frac{\delta M^{2}}{2EI} dx[/tex]

    Virtual work is work done by a virtual force through a real and virtual displacement.
    [tex]\delta F (v + \delta v) =\int \frac{M \delta M}{EI} dx + \int \frac{\delta M^{2}}{2EI} dx[/tex]
    [tex]\delta F v + \delta F \delta v =\int \frac{M \delta M}{EI} dx + \int \frac{\delta M^{2}}{2EI} dx[/tex]

    I am going to say that
    [tex]\delta F \delta v = \int \frac{\delta M^{2}}{2EI} dx[/tex]
    because a virtual force doing work through a virtual distance is going to change a purely virtual energy.

    [tex]\delta F v =\int \frac{M \delta M}{EI} dx[/tex]

  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted