Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

2.72. A harbor breakwater is made of massive tanks which are floated into place

ID: 2925157 • Letter: 2

Question

2.72. A harbor breakwater is made of massive tanks which are floated into place over a shallow trench

scraped out of the harbor floor and then filled with sand. There is concern over the possibility of breakwater

sliding under the lateral pressure of a large wave in a major storm. It is difficult to predict the lateral sliding

capacity of such a system. What is the reliability (the probability of satisfactory performance) of this system

with respect to sliding if the engineer judges the following?

(a) That the sliding resistance has a value 100, 120, or 140 units, with the middle value twice as likely as

the low value and twice as likely as the high value.

(b) That the lateral force under the largest wave in the economic lifetime of the breakwater X, has an

exponential distribution with parameter = 0.02; that is,

Fx(x) = 0.02e^(-0.02x) ,   x> or equal to 0

The units of sliding resistance and lateral force are the same. Resistance and force are independent.

Explanation / Answer

Sol:

The reliability of the system is the probability of satisfactory performance and satisfactory performance is when the sliding resistance is greater than the lateral force.

From (a), three possible sliding resistances are 100, 120 and 140 units, and 120 being twice likelly than the other two. Therefore, we can connclude that if P(SR=100) = P(SR=140) =p, then P(SR=120) = 2p. This implies p=0.25 and P(SR=100) = P(SR=140) =0.25, then P(SR=120) = 0.5

From (b), fx(x)= 0.02e-0.02x (assuming given function in the question is of porbability density function, pdf), therefore CDF is given by F(x)= 1-e-0.02x, for all x>0. Therefore P(LF<100) = 1-e-0.02*100 and vice versa.

Therefore from (a) and (b), Reliability = P(SR=100)*P(LF<100) + P(SR=120)*P(LF<120) + P(SR=140)*P(LF<140)

Reliability = 0.25*(1-e-0.02*100) + 0.5*(1-e-0.02*120) + 0.25*(1-e-0.02*140) = 0.9056.

Therefore, the reliability of the system is 90.56%