Recent content by steven10137

With the shaker tests we didn't actually measure the applied load, just the response signal from the strain gauges and accelerometers. I believe it was open-loop and it makes sense that this force would be amplified at resonance, and this was clearly visible in the pipe vibration during the...

OK Thanks, this is making me more confident with my results so far. The link element doesn't actually change things a great deal. I agree, the test setup is on the money and I believe that the strain/velocity data is quite accurate for the model. Thanks for clarifying in regards to the modal...

Cheers for the reply. Yeah the link is just sharing the 1 node. If you have a look at the attached deformed shape for the first mode (12Hz), there is a huge deflection which I'm not happy with- this node shouldn't deflect that much, if any (rigid support). I tried solving the problem by...

Hey guys, I am currently working on a project and need to model the tubing configuration as shown in the first attachment. Experimental testing has been completed and I have detailed data for the strain and vibration velocity as required. I need to model this in ANSYS to compare the...

Oh my god, I answered my own question. I changed all my angles to radians and put them into the the same equations, gave my answer in rad/s^2 and bingo! {\dot r = - 50\cos \left( {\frac{\pi }{{12}}} \right) = - 48.296\:ms^{ - 1} } {\dot \theta = \frac{{ - 50\sin \left( {\frac{\pi...

bump. Does anybody have any suggestions?

Yep that's correct, just an extension of the product rule.

Homework Statement http://img127.imageshack.us/img127/2695/coord2pq5gm6.jpg Homework Equations \underline v = \dot r\;\underline e _r + r\dot \theta \;\underline e _\theta \underline a = \left( {\ddot r - r\dot \theta ^2 } \right)\underline e _r + \left( {r\ddot \theta + 2\dot r\dot...

Well I was just curious as to whether or not there was another way to get the same result.

ok thanks :) Just to confirm my understanding of this, the tangential acceleration is defined as the rate of change of velocity, in the 't' direction, yeah? There is no other way to calculate this than to use kinematics?

OK thanks, I thought this might have been where I went wrong. How about? \begin{array}{l} a_t = \frac{{dv}}{{dt}} \Rightarrow dv = a_t .dt \\ \int {dv} = \int {a_t .dt} \Rightarrow v = \int\limits_0^2 {a_t .dt} = \int\limits_0^2 {6.7.dt} = \left[ {6.7t} \right]_0^2 = 13.4\,ms^{ - 1}...

Homework Statement A particle starts from rest at t=0s. It moves along a circular path of radius 18m and has an acceleration component along its path of 6.7m/s^2. What is the magnitude of the acceleration when t=2s Homework Equations \begin{array}{l} \left| a \right| = \sqrt {a_n ^2 +...

\begin{array}{l} \left| a \right| = \sqrt {a_n ^2 + a_t ^2 } \\ \Rightarrow \left| {a_n } \right| = \sqrt {\left| a \right|^2 - \left| {a_t } \right|^2 } = \sqrt {41.85^2 - 41.64^2 } = 4.23\;ms^{ - 2} \\ \end{array} Ahhhhhhh thanks so much mate, makes sense now :)

Of course, yes they were already .. argh I'm complicating it even more. I would prefer to use the latter method as it makes more sense to me: \begin{array}{l} \tan \alpha = \frac{{d^2 y/dt^2 }}{{d^2 x/dt^2 }} = \frac{{41.55}}{5} = 8.31 \Rightarrow \alpha = 83.14^ \circ \\ \left| a...

Indeed. Think about it, if for any two consecutive numbers, their corresponding function value is greater, how can the function be decreasing? You could also use the theorem that on any interval (a,b), if f '(x) is > 0, then the function is increasing on that range. Take the interval (1,2)...