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264  Airworthiness and airframe loads

                 Fatigue damage is also caused by gusts encountered in flight, particularly during the
                 climb and descent. Suppose that a gust of velocity u, causes a stress S, about a mean
                 stress corresponding to level flight, and suppose also that the number of stress cycles
                 of this magnitude required to cause failure is N(S,); the damage caused by one cycle is
                 then l/N(S,). Thus, from the Palmgren-Miner  hypothesis, when sufficient gusts of
                 this and all other magnitudes together with the effects of all other load cycles produce
                 a cumulative damage of  1.0, fatigue failure will occur. It is therefore necessary to
                 know the number and magnitude of gusts likely to be encountered in flight.
                   Gust data have been  accumulated over a  number  of  years from accelerometer
                 records from aircraft flying over different routes and terrains, at different heights
                 and at different seasons. The ESDU data sheets7 present the data in two forms, as
                 we have previously noted. First Zlo  against altitude curves show the distance which
                 must be flown at a given altitude in order that a gust (positive or negative) having
                 a velocity  2 3.05m/s be encountered. It follows that  l/Zlo is the number of  gusts
                 encountered in unit distance (1 km) at a particular height. Secondly, gust frequency
                 distribution  curves, r(ue) against  u,,  give  the  number  of  gusts of  velocity  u,  for
                 every 1000 gusts of velocity 3.05 m/s.
                   From these two curves the gust exceedance  E(ue) is obtained; E(u,) is the number
                 of times a gust of a given magnitude (u,) will be equalled or exceeded in 1 km of flight.
                 Thus, from the above
                       number of gusts 2 3.05m/s per km = l/Zlo
                       number of gusts equal to u, per 1000 gusts equal to 3.05m/s  = r(ue)
                 Hence
                    number of gusts equal to ue per single gust equal to 3.05m/s  = r(u,)/1000

                 It follows that the gust exceedance E(u,) is given by

                                                                                    (8.50)
                 in which llo is dependent on height. A good approximation for the curve of  r(ue)
                 against u,  in the region u,  = 3.05 m/s is
                                               = 3.23  105~i5.26                    (8.51)
                 Consider now the typical gust exceedance curve shown in Fig. 8.18. In 1 km of flight
                 there are likely to be E(u,) gusts exceeding u, m/s and E(ue) - SE(u,) gusts exceeding
                 ue + Sue m/s. Thus, there will be SE(u,)fewer gusts exceeding u, + Sue m/s than ue m/s
                 and the increment in gust speed Sue corresponds to a number -SE(u,)  of gusts at a
                 gust speed close to u,. Half of these gusts will be positive (upgusts) and half negative
                 (downgusts) so that if it is assumed that each upgust is followed by a downgust of
                 equal magnitude the number of complete gust cycles will be  -SE(u,)/2.  Suppose
                 that each cycle produces a stress S(ue) and that the number of these cycles required
                 to produce failure is N(S,:,). The damage caused by  one cycle is then  l/N(Su,e)
                 and over the gust velocity interval Sue the total damage SD is given by

                                                                                    (8.52)
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