00:04
For problem 5 .80, we are asked to find the resultant ground reaction forces along the surface a to b for a one foot deep concrete dam shaped as follows in this diagram.
00:30
The dimensions of the dam for this problem are shaped with a triangle starting from point a and that spans nine feet to the right, its base.
00:55
The next part could be seen as a block, a simple rectangular shape that has a width of six feet, and followed by another triangle, 18 feet tall and six feet at the base, that is holding the volume of water also span.
01:29
18 feet deep.
01:33
I propose to divide up the concrete structures just described and have these dotted lines here showing the boundaries that the shapes will consider.
01:44
The specific capacity or volume density as i call it of water in imperial units is 62 .4 pounds per cubic foot.
01:56
For concrete the specific capacity is 150 pounds per cubic foot.
02:02
And again, we're only considering the cross -section here.
02:05
So for ease of calculations, we'll say that the thickness of this dam, or the dimension of the dam into and out of the page, is one foot thick.
02:20
For part a, again, we're asked to determine the resultant of the reaction forces on the ground.
02:29
The first thing i propose to do is figure out the pressure of the water, against the inner wall of the dam.
02:42
Water or any basic liquid has the property that the pressure increases linearly with depth.
02:54
So in order to find a load from pressure, we need to integrate across the depth of the water.
03:07
And we do that inserting the pressure of the water, which is the volume, the depth dimension we're calling b, times the specific capacity of the water, times the height of the water.
03:23
And then we're going to integrate that over increasing depth, so multiply it by the infinitesimal dh, integrate from 0 to 18.
03:35
Symbolically, the integral is b times gamma times 8 squared over 2, plugging in numbers inside this expression, we have the total load pressure is one -half times one foot, times 62 .4 pounds per cubic foot times 18 -foot depth squared, and that comes out to 10 ,109 pounds.
04:03
We also can find the locus of that load in terms of its vertical position of the centroid, and since we're dealing with a triangle force here, we know that, and that would be drawn as so.
04:26
We know that the centroid of a triangle is one -third up from its base, in this case.
04:37
So we're going to express that as, and this again is the y position of the centroid, is one -third h.
04:44
H is 18 feet, so that comes to six feet.
04:48
Six feet above the above ground, and we're calling the ground surface here, is the vertical location of the pressure load centroid.
04:59
Next, we're going to calculate the load of shape 1, which is the triangle to the far left.
05:08
It has a base of 9 feet and the height of 15, i'm sorry, 15 feet, so 9 times 15.
05:15
And since it's a triangle, we divide by 2 times the specific capacity of concrete.
05:19
We're left with a load, force load for that shape, w sub 1 of 10 ,125 pounds.
05:29
We can also find the horizontal location of the centroid.
05:35
It's a triangle, and from point a, that's going to be two -thirds of the base span of that shape.
05:43
So two -thirds times the base, the base is nine feet.
05:47
Two -thirds of nine feet is six feet.
05:49
Now we'll calculate shape 2, which is just a simple rectangle.
05:56
Base times height.
05:58
The base is 6 feet.
06:00
The height is 18 feet, and again, the thickness is 1 foot, times the specific capacity of concrete of 150 pounds per cubic foot, leaves us with a load for that shape of 16 ,200 pounds.
06:15
Finding the centroid reference from point a will do one, and that's the centroid in the horizontal region of that shape, the horizontal position of the centroid that is.
06:31
It will be half of the width of that shape plus the offset of that shape.
06:36
The width is six feet, half of that is three feet plus an offset of nine feet comes out to 12 feet.
06:45
Location of the centroid for the second shape is, and that's the horizontal location, is at 12 feet.
06:53
Next, we'll calculate the load for the third shape, which is also a triangle, this one taller.
07:02
The triangles are 18 feet high by 16 feet wide.
07:08
Divide that by 2, and multiply by the specific capacity of concrete leaves us with 8 ,100 pounds.
07:18
Again, we'll calculate the centroid in the same fashion, since the triangle is oriented the other way.
07:25
This time it's going to be one -third times the base from the left, plus the offset of that shape.
07:34
One -third times six is two, plus 15 to 17 feet.
07:40
Next, we're going to find the load and horizontal centroid location of the volume of water vertically above shape 3.
07:51
We're calling that shape 4.
07:55
This time, same dimensions for the triangles, 6 times 18, divided by 2, 1 foot depth.
08:02
But this time, the specific capacity of the material we're talking about is less.
08:09
It's water instead of concrete, so we're going to plug in 62 .4 pounds per cubic foot, and that gives us a total load of 3 ,370 pounds, calculating the horizontal location of its centroid.
08:23
This time it's the flipped orientation compared to the third shape that we previous considered.
08:31
So it is two -thirds the base plus the offset location from point a.
08:36
Two -thirds time six plus 15 is 15 plus 4, which is 19 feet.
08:44
Now that we have all the forces and moment times of those forces with respect to a, the centroid locations, both horizontal and vertical, we can sum together components of the forces and the torques in order to solve for the resultant forces on the ground, reaction forces from the ground along surface ab.
09:13
So the only horizontal force you have to consider is that from the pressure of the water that we calculated at 10 ,109 pounds.
09:23
The reaction force is going, and that pressure is going leftward, so the reaction force has to be equal and opposite...