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Two L $76 \times 76 \times 6.4-\mathrm{mm}$ angles are welded to a $\mathrm{C} 250 \times 22.8$ channel. Determine the moments of inertia of the combined section-with respect to centroidal axes respectively parallel and perpendicular to the web of the channel.

Physics 101 Mechanics

Chapter 9

Distributed Forces: Moments of Inertia

Section 2

Parallel-Axis Theorem and Composite Areas

Moment, Impulse, and Collisions

Cornell University

Simon Fraser University

Hope College

University of Sheffield

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

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In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

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So we're told that two l 76 by 76 by 6.4 millimeter angles are well, the two a c 2 50 by 22.8 channel. And we want to determine the moments the area moments of the combined section with respect to the central axis expectedly parallel per predict with the web of the channel. So these are axes, and our century is going to be somewhere on here. I would have to figure that out, but we can look up our the properties for our to our are different sections that we've well, well together. Um And so for the area for the sea channel, um is this and for the l brackets is this And so the distance here is 254 millimeters. So that's the distance between the outer flanges of the two See channel and then also the distance here because thes air lined up. Now we know that X bar is gonna be zero, because the axes I've to find here Okay, so I've defined these axes that at the top of the see Channel Web, um, and we'll figure out where the area moment is with respect to that. But we know and I've defined the Y axes as just at the center of this. And so we know, because of the symmetry, that X bar is gonna be zero. And why bar we can figure out OK, so we have just got to remember that we have two of these guys. We need twice this, and we can look up this and we can look up this. And so when we do that, we find that why one bar is 16.1 millimeters. But it's in the negatives down here, so it's a negative. And then this is, um, 21.2 millimeters up from, um, this part of the L bracket. So up here and then this one is also up here. We plug in all of those numbers and we get that the central it'll why coordinate is minus 1.5 millimeters. So down here somewhere. Um, now we can look up our central our section central. It'll properties what goes up in the book. So for the See channel, we have these area moments about its century and for the l brackets, we have these area moments about its century. So then we need to use the parallel access there to shift things. So we find the area moment about he sent the sports section Central relaxes is, um this area moment for the Sea Channel is the area moment about the sea channel centrally times area, the sea channel times the distance from this centre Troy of the whole body to the central to see channel squared. And we get 1.56 times center the 16 millimeters to the fourth, do the same thing for the l the l shapes and we get our the l angles and we get that thous give us 0.991 time center, the six millimeters to the fourth. So we have all these values and we know we can get the sum, the total area moment about this central axis. By adding these things up and remembering that we have two of these. So we get 3.54 times 10 of the six millimeters the fourth again, we can do everything the same for why, Okay, but we have to be a little careful because we need to use the parallel access the arm to get the that the area moment about the central. Why access? Because it's this sense that sent Detroit Y or X coordinate of these. Els is not on this axis, although this one is. So, um, I Y two is the i area moment of the wide about the y axis about. It's the l bracket, the L Angles Central, the Y axis. So then we have, um, the area of the l angle and D over two. Okay, so the distance here, that's the distance shifted to this point. But then we know that the, um the central right in the X direction is also this value. So we kept to come back, and so that's the distance from here to here Is this value here and we know all this so we can plug in and we get 11.4 times 10 to the 16 millimeters of the fourth. Um, and now we know that we know that the area moment of this of the Sea Channel about our about the central wax y axis is the same as the sense area moment about the sea Channel Central. It'll y axis. So we have that, um and so then we can plug in. And actually, this should not be a bar on top of here. And so we plug in and we get that the area moment about the central it'll y Axis is the is 49.8 times 76 millimeters, uh, to the fourth.

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