Recent content by Axecutioner

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    Deflection vs. Bending in a Beam

    Thank you cmmcnamara, that was extremely helpful. I got it figured out now. All starting to make sense. http://cache.gtpla.net/forum/images/smilies/thumbsup.gif
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    Deflection vs. Bending in a Beam

    And how/where could I do that? I'd like to teach myself before taking the class because I'll need to use this stuff before then.
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    Deflection vs. Bending in a Beam

    Sorry, here: http://i.imgur.com/jfqU6.png P(B-A), the area would be WH but there is more moment at the edges of the beam than towards the center, so it would be: P(B-A)(H/2) but then it would be on a line of material and not an area. So P(B-A)(H/2)/(W) ? Moment of force P on the outer fibers...
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    Deflection vs. Bending in a Beam

    I have the beam tables, and I've gone through Calc 4 so throw anything relevant my way and I'll see if I can get it. Here's what I have: http://uploading.com/files/ab5m1512/Beam_Deflection_Formulae.pdf/
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    Deflection vs. Bending in a Beam

    Problem: What is the equation that relates properties of a metal (modulus of elasticity, yield strength, elongation, etc) to how far it can deflect before permanently being bent. I can calculate how far a beam will deflect under different loading and support but I just don't know what the...
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    Solving an ODE-45 from Euler-Lagrange Diff. Eqn.

    Here's what I've done so far, before I posted the OP: http://i.imgur.com/c1Sdb.jpg Some of the K values are the same but just keep each one different, since this entire double pendulum case is a simplification of part of a project I'm working on. And Rotational Kinetic energy and Kinetic...
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    Solving an ODE-45 from Euler-Lagrange Diff. Eqn.

    I need to find the equation of motion of a double pendulum, as shown here: I've gotten as far as the two euler-lagrange differential equations, simplified to this: K1\ddot{θ}1 + K2\ddot{θ}2cos(θ1 - θ2) + K3\dot{θ}22sin(θ1 - θ2) + K4sin(θ1) = 0 K5\ddot{θ}2 + K6\ddot{θ}1cos(θ1 - θ2) +...
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    What Is the Kinetic Energy Equation for a Mass on a Double Pendulum?

    Sorry, misunderstanding. Please delete this post.
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    What Is the Kinetic Energy Equation for a Mass on a Double Pendulum?

    That doesn't help at all. I know K = 1/2*I*w^2 but what would the omega be? Just both angular velocities added together?
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    What Is the Kinetic Energy Equation for a Mass on a Double Pendulum?

    You can treat the cable as another rigid bar in this case, both massless. I just need the equation for the Kinetic Energy of the mass on the end. Is there an equation for if the main bar is torqued at a given τ(t), and any other necessary information is given, to find the position of the mass...
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    Strange parabolic calculations - need simplification help.

    I went ahead and used your original suggestion to solve for A using: C=(fac(a-c)+bce(c-e)+dea(e-a))/((a-c)(c-e)(a-e)) And let me tell ya, it's amazing how complicated it got, then how simple the answer is. I'll scan my work tomorrow and post it - way too long to type out. Again, thank you for...
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    Strange parabolic calculations - need simplification help.

    Interesting. If you ever find the title or author please let me know. Here's a different version of the C equation: \frac{ade^2 - bce^2 + bc^2e - a^2de - ac^2f + a^2cf}{ae^2 - ce^2 + c^2e - a^2e - ac^2 + a^2c} I wonder if the top could factor out similarly to the way you got it to in the...
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    Strange parabolic calculations - need simplification help.

    Bill Simpson, thanks! I'm surprised the denominator could factor into (a-c)(c-e)(a-e) but it does work. How'd you figure that out? I like the numerator you set up too, will play around with it and see what works out. Mentallic, is there a way to separate that fraction out, in any way? Also...
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    Strange parabolic calculations - need simplification help.

    Dodo, I know how to do that given points. I'm trying to find an equation for the parabola given any three points as I stated in the OP. I know how to do it. All I'm asking for is simplifying the C= equation I put in the OP if it can be done and nobody has posted anything related to what I...
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    Strange parabolic calculations - need simplification help.

    I have no idea what linear regression is. :P and that's interesting about the Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 formula. But given 3 points and 6 unknowns, wouldn't it be impossible to solve? Meaning there are many different parabolas that fit those 3 points given variable axis slope?
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