Thanks Matt, OK well maybe I'm not picturing this correctly in my head. I know that a moment = force x distance (from point of interest), but in my example there is an applied moment to the end of the beam (irrespective of the length of the beam). As I stated above, the bending stress is M*y /...
if you have a beam or something, anchored at one end and apply a moment at the other end, then the bending stress is given by M*y / I, where I is the moment of inertia of the beam. what affect does the length of the beam (i.e. the distance between the anchor and the end of the beam, where the...
Thanks, no that wasnt clear to me, that's what I'm having trouble processing and picturing in my head. So if it bends in XY axis then that is a moment about Z? And so I take it applying a moment about one axis then results in bending in the other two axis'? So if I applied a moment about Y, then...
Thanks for replying. I know this is pretty basic stuff! Yeah in particular its the resulting direction of the bending moments within a beam. For example, which direction would you need to apply a force to get a moment about Z etc?
I'm having trouble fully understanding bending moments. I get the calculations, the force times distance (lever arm), and how to calculate the bending stress (M*y/I). Its the orientation/direction that I am having trouble picturing in my head. I know this is pretty basic, but I just haven't...
yeah I've been doing some reading over night, i think it might be due to the effective force being overly conservative. there will be 'feed in' to it, which will reduce the compressive force. thanks
I'm calculating the natural frequency of a piping system that is spanning between two points, the formula is the square route of various terms, including the effective axial force in the piping. This is taken from a design code, see below:
f = C * SQRT[(E.I/M.L^4)(1+Seff/Pcr+C.(d/D)^2]
C =...
Ok so to clarify...
I only changed Mtotal (the total mass), the submerged weight stayed the same as I didnt actually change the mass, I just increased Mtotal (from submerged weight+added mass, to mass in air+added mass). And to calculate the terminal velocity you set dv/dt=0 so Mtotal is...
The drag force equation is: 0.5 p * (Cd* A)
where p = density of liquid
Cd = drag coefficient for object
A = drag area
So this doesn't change. Only thing I change is Mtotal.
Yeah the equation of motion is:
Mtotal dv/dt + Fdrag v(t)^2 - Wtotal = 0
Where M is the total mass, Fdrag is the drag force (proportional to the square of the velocity) and Wtotal is the submerged weight.
The surface area of the object did not increase at all, I only increased Mtotal...
I have a question regarding the terminal velocity of an object falling through a liquid. Now the speed will increase up to the terminal veloity in a given time. What I do not understand is in this calculation I'm performing (solving the equation of motion for the speed, when acceleration equals...
Thanks, so what would one of these be?
http://www.copleycontrols.com/Motion/Products/Motors/stb.html
Also, for matching up requirements, say you had an electric motor with a torque of 10Nm. Could you divide this by the length of the lead screw (or whatever type of shaft your using to...