The base of a dam for a lake is designed to resist up to 124

The base of a dam for a lake is designed to resist up to 124 percent of the horizontal force of the water. After construction, it is found that silt (that is equivalent to a liquid of density p_s = 1.76 times 10^3 kg/m^3) is settling on the lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of the dam, determine the number of years until the dam becomes unsafe. (Round the final answer to the nearest whole number.) The number of years for which the dam will remain safe is

Solution

We know that the dam can take 124 percent of the horizontal force due to water. We would first determine this value for a unit area and equate it with the pressure that a given thickness of silt would result into.

Now, for the dam of heigh 6.6 m, we have:

Pressure at the bottom = pgH = 1000 x 9.81 x 6.6 = 64746 N/m^2

Hence the force on a unit area would 64746 N. Also the dam can take 124 percent of this value.

That is tha maximum force the dam can take is: 64746 x 1.24 = 80285.04 N

Now, let us assume that after x years, the dam becomes unsafe.

That is, the thickness of the silt = 12x mm

The net force on unit area due to water and silt would be:

F = pg[6.6 - 0.012x] + 1760 * 9.81 *0.012x

F = 64746 - 117.72 x + 207.187 x = 64746 + 89.467 x

Now for the dam to become unsafe this force must be equal to 80285.04

Hence 80285.04 = 64746 + 89.467 x

or x = 173.6.

Therefore the dam will remain safe for 173 years