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: Same risk formula, evaluated a different way.
\[ R = 1-\left(1-\frac{1}{T}\right)^n = 1-(0.99)^{50} \]
Instead of using logarithms, evaluate $(0.99)^{50}$ directly by repeated squaring, which avoids log tables and is easy to check step by step.

Step 2: Build up the power by successive squaring.
\[ 0.99^2 = 0.9801 \]
\[ 0.99^4 = (0.9801)^2 = 0.960596 \]
\[ 0.99^5 = 0.960596\times0.99 = 0.950990 \]
\[ 0.99^{10} = (0.950990)^2 = 0.904382 \]
\[ 0.99^{20} = (0.904382)^2 = 0.817907 \]
\[ 0.99^{40} = (0.817907)^2 = 0.668972 \]

Step 3: Combine to get the 50th power.
\[ 0.99^{50} = 0.99^{40}\times0.99^{10} = 0.668972\times0.904382 \approx 0.605 \]

Step 4: Compute the risk.
\[ R = 1-0.605 = 0.395 \]
Rounded to two decimal places, this is $R \approx 0.39$, confirming the earlier logarithmic calculation.

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