I'm having some trouble understanding somethings with problem 8/72. First, it says the mass of the trailer is 500 kg, and that each 75 kg added to the load during the loading caused the trailer to sag 3 mm on its springs. Is the 500 kg the mass of the trailer itself (i.e. without any 75 kg additions to the load), and how many 75 kg additions were added to it? And how do each of those 3 mm amounts that the trailer sags from each addition affect the problem?
Also, how do we find omega for the trailer? Are we supposed to divide the speed of the trailer (25 km/hr) by the "wavelength" between two bumps on the road (1.2 m)? I would assume omega is necessary in solving the problem, I don't see how it wouldn't be if the contour of the road is a sinusoidal function.
15 comments:
I used the loading information to find the spring constant:
m*g = K * displacement
I also figured omega using the 1.2m information. I first found the frequency of the bumps then multiplied by 2pi to get omega.
The forced response I used I made from assuming the 50mm was a peak to peak measurement, its hard to tell if that is the case.
I'm looking for assistance on the FBD. Clearly the spring acts on the trailer. Is the force the weight of the trailer? Does this weight act like a sine wave?
Okay - so I just kept working until I got it, and I've figured it out.
FBD
This is an X, Xb type problem.
I don't want to give away too much information, please ask more questions if you're stuck.
I have everything figured out for X, but I'm not sure how to go about finding the critical speed. Any advice?
The critical speed will create omega_n.
w_n = sqrt(k/m)
Set w = w_n, then undo your original calulation to find omega the first time.
I found the omega by using the distance between the two bumps. I also have the spring constant figured out. But where does the velocity of the cart is used in equation?
You will need the velocity of the cart for finding the period, which will be used to find omega.
I am having trouble with the amplitude, it is the coefficient in front of the sine (or cosine) function right? I am not getting the right answer though for this. I made my EOM, and then just calculated the value of the coefficient, assuming sine to be one.
Kyle,
your right, the amplitude should just be the coefficient in front of sine I'm pretty sure. When I found the equation for amplitude, it worked out to be the standard one for an undamped system. You can find this on page 26 of our notes if you want to check that yours matches it. From there it should just be plugging in the values. If you found the critical velocity already, then you know your K and omega_n are correct.
--to J Biberstine--
Yes, the 50 mm is peak-to-peak (hard to see in the figure). Therefore, the amplitude for the "base motion" in this case is 25 mm.
--in general--
To get the base motion for this, write down the road height, y, as a function of the position z along the road: y(z) = 25*sin(2*pi*z/1.2).
You know that the trailer is going at a constant speed of v. Therefore, you can write z = v*t. From this the base motion (as a function of time) becomes: y(t) = 25*sin((2*pi*v/1.2)*t). This says that the excitation frequency is: omega = 2*pi*v/1.2
The critical speed, as J Biberstine says above, occurs when omega = omega_n.
I'm a little confused as to how to find the amplitude of vertical vibration. I've got my EOM and then i plug in the sin and cos part for the EOM then seperate them into sinwt and coswt accordingly. Then I find the constant. Does that sound right or am I forgetting something.
Yes, wccheng, that is correct, you'll solve for A and B
do we know how many loadings of 75kg were added? is this the 500kg?
ok never mind it doesn't matter, right?
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