OK, I've done some research and math.

Lake Mead has an area of 112 sq mi and a total volume of 28.5 million acre-feet.

If this were an ideal 112 sq mi square, the average depth would be 397.6 feet.

Water weighs about .433 psi per foot, therefore the average pressure would be about 172 psi.

Hoover Dam is 726.4 feet high, which equates to a pressure of 314.6 psi. This does not include the weight of the dam itself.

The Mississippi Great Flood of 1993 inundated an area of 30,000 square miles. If the volume of Lake Mead were spread over this area, it would be 1.48 feet deep. This is a pressure of .642 psi.

Even if the flood were 30 feet deep, that's only 13 psi - about 13.2 times less than the average of Lake Mead, and 24 times less than the maximum.

Now, I couldn't find any numbers on how much volume of water the flooded area was, but let's make a guess that on average it was 5 feet deep. That's a volume of 19.2 million acres, so 96 million acre-feet.

Lake Mead's 28.5 million acre-feet is about 1.24 trillion cubic feet and at 62.4 pounds per cubic foot, that's about 77.46 trillion pounds total weight.

Out hypothetical 5 foot flood of 96 million acre-feet is about 4.18 trillion cubic feet and nearly 261 trillion pounds. This is about 3.37 times the weight of Lake Mead, but remember, it's spread over nearly 268 times the area.

Even though a massive flood of the Mississippi may weigh more than Lake Mead, it's spread over so much more of an area that the unit area static pressure is far many times less.

All information derived from Wikipedia articles. My math has not been certified, verified, or rectified; although it may be putrefied, undignified, and ultimately nullified. :)

Brian