Question

A bridge with an expected life of 50 years is designed for a flood of 10000 m\(^3\)/s corresponding to the return period of 100 years. The risk associated with this design is (rounded off to two decimal places).

Hint:

Risk equals one minus the probability that the design flood is not exceeded in any single year, raised to the power of the number of years of design life.

Answer 1

Correct Answer: 0.39

Step 1: Understanding the Question.
"Risk" here means the probability that a flood larger than the design flood occurs at least once during the useful life of the structure. We are given the design return period and the expected life, and asked to find this risk.

Step 2: Key formula for hydrologic risk.
If a flood has a return period \(T\) years, the probability that it is exceeded in any single year is \(p = 1/T\). Over \(n\) independent years, the probability that the flood is exceeded at least once, the risk, is:
\[ R = 1-(1-p)^n = 1-\left(1-\frac{1}{T}\right)^n \]

Step 3: Substitute the given values.
Here \(T = 100\) years and the design life \(n = 50\) years. The flood magnitude of \(10000\ \text{m}^3/\text{s}\) is not needed here, since risk depends only on \(T\) and \(n\).
\[ R = 1-\left(1-\frac{1}{100}\right)^{50} = 1-(0.99)^{50} \]

Step 4: Evaluate \((0.99)^{50}\).
Using logarithms: \(\ln(0.99) = -0.01005\), so
\[ \ln(0.99^{50}) = 50\times(-0.01005) = -0.5025 \]
\[ 0.99^{50} = e^{-0.5025} \approx 0.6050 \]

Step 5: Compute the risk.
\[ R = 1-0.6050 = 0.3950 \]
Rounded off to two decimal places, \(R \approx 0.39\).

Final Answer:
\[ \boxed{R \approx 0.39} \]