Question

A bridge has an expected design life of 50 years. It is designed for a flood discharge of 1000 m\(^3\)/s, which corresponds to a return period of 100 years. Determine the risk (probability) that the design flood will be equalled or exceeded at least once during the design life of the bridge. (Enter the numerical value of risk in decimal form, correct up to three decimal places.)

Hint:

For hydrological risk problems, remember the formula \(R = 1 - (1 - 1/T)^n\). A useful approximation for small \(P\) (or large \(T\)) is \(R \approx n/T\). In this case, \(50/100 = 0.5\), which is a rough estimate.

Answer 1

Step 1: Understanding the Question:
The question asks for the 'risk' of a hydrological event. Risk is the probability that an event of a given magnitude (or greater) will occur at least once in a specified period (the design life).
Step 2: Key Formula or Approach:
The probability (\(P\)) of an event with a return period \(T\) occurring in any given year is \(P = 1/T\).
The probability of the event not occurring in any given year is \(q = 1 - P\).
The probability of the event not occurring for \(n\) consecutive years is \(q^n\).
The risk (\(R\)) of the event occurring at least once in \(n\) years is 1 minus the probability of it never occurring. \[ R = 1 - q^n = 1 - (1 - P)^n = 1 - \left(1 - \frac{1}{T}\right)^n \] Step 3: Detailed Explanation:
Given values are:

• Design life, \(n = 50\) years
• Return period, \(T = 100\) years

First, calculate the annual probability of the design flood: \[ P = \frac{1}{T} = \frac{1}{100} = 0.01 \] Now, substitute the values into the risk formula: \[ R = 1 - \left(1 - \frac{1}{100}\right)^{50} \] \[ R = 1 - (0.99)^{50} \] Calculating the value of \((0.99)^{50}\): \[ (0.99)^{50} \approx 0.605006 \] Finally, calculate the risk: \[ R = 1 - 0.605006 = 0.394994 \] Rounding to three decimal places, we get: \[ R \approx 0.395 \] Step 4: Final Answer:
The risk that the design flood will be equalled or exceeded at least once during the bridge's 50-year design life is 0.395.