Analysis of flood data is done for Brahmaputra river at Assam covering the period of 1947 to 1997. Mean discharge and standard deviation obtained from above analysis yields the value of 12000 cumecs and 1000 cumecs respectively. It is decided to construct Bridge at this site so as to provide 90% assurance that the structure should not fail in the next 50 years. Use Gumbel’s method to estimate the peak value of discharge from the Given data. Take Y̅ n = 0.5362 S̅ n = 1.1750 If instead of Estimated discharge, design value of 25000 m 3 /sec is adopted, calculate the safety factor for the maximum flood discharge up to two decimel places.

Concept: Peak value of discharge is given by \({{\rm{X}}_{\rm{T}}}{\rm{\;}} = {\rm{\bar X}} + {{\rm{K}}_{\rm{T}}}{\rm{\sigma \;}}\) X T ⇒ Peak value of discharge = Mean of hydrological data σ = Standard deviation K T = Frequency factor \(\begin{array}{l} {{\rm{K}}_{\rm{T}}}{\rm{\;}} = \frac{{{{\rm{Y}}_{\rm{T}}} - {{{\rm{\bar Y}}}_{\rm{n}}}}}{{{{\rm{S}}_{\rm{n}}}}}\\ {{\rm{Y}}_{\rm{T}}} = {\rm{reduced\;variable}} = - {\rm{lnln}}\left( {\frac{{\rm{T}}}{{{\rm{T}} - 1}}} \right) \end{array}\) Safety Factor = \(\frac{{{\rm{Design\;value}}}}{{{\rm{Estimated\;value}}}}\) Calculation: X̅ = 12000 m 3 /sec σ = 1000 m 3 /sec \({\rm{T}} = {\rm{\;}}\frac{1}{{\rm{p}}}\) Reliability = (1 - p) n 0.90 = \({\left( {1 - \frac{1}{{\rm{T}}}} \right)^{50}}\) T = 476.19 years \({{\rm{Y}}_{\rm{T}}} = - {{\rm{l}}_{\rm{n}}}{{\rm{l}}_{\rm{n}}}\left( {\frac{{476.19}}{{476.19 - 1}}} \right)\) Y T = 6.16 \({{\rm{K}}_{\rm{T}}} = \frac{{6.16 - 0.5362}}{{1.1750}}\) K T = 4.786 X T = 12000 + 4.786 × 1000 X T = 16786.21 m 3 /sec \({\rm{Safety\;factor}} = \frac{{25000}}{{16786.21}} = 1.489\)