Chapter 13 Airworthiness 399 Chapter 14 Airframe loads 405
15.4 Prediction of aircraft fatigue life
S∞,m =Sa,m/(1+C/ Nm) (15.10) The number of cycles to failure at a mean stress of 90 N/mm2would have been, from the above
N = Sm
90Nm (15.11)
The corresponding fatigue limit stress would then have been, from a comparison with Eq. (15.10)
S∞ ,m =Sa,m/(1+C/√
N) (15.12)
The standard endurance curve for the component at a mean stress of 90 N/mm2is from Eq. (15.7)
Sa =S∞,m/(1+C/√
N) (15.13)
Substituting in Eq. (15.13) for S∞ ,m from Eq. (15.12) we have Sa = Sa,m
(1+C/√
N)(1+C/√
N) (15.14)
in which Nis given by Eq. (15.11).
Equation (15.14) will be based on a few test results so that a ‘safe’ fatigue strength is usually taken to be three standard deviations below the mean fatigue strength. Hence we introduce a scatter factor Kn(>1) to allow for this; Eq. (15.14) then becomes
Sa Sa,m Kn(1+C/√
N)(1+C/√
N) (15.15)
Kn varies with the number of test results available and for a coefficient of variation of 0.1, Kn=1.45 for 6 specimens, Kn=1.445 for 10 specimens, Kn=1.44 for 20 specimens and for 100 specimens or more Kn=1.43. For typical S–N curves a scatter factor of 1.43 is equivalent to a life factor of 3 to 4.
15.4 Prediction of aircraft fatigue life
We have seen that an aircraft suffers fatigue damage during all phases of the ground–air–
ground cycle. The various contributions to this damage may be calculated separately and hence the safe life of the aircraft in terms of the number of flights calculated.
In the ground–air–ground cycle the maximum vertical acceleration during take-off is 1.2 g for a take-off from a runway or 1.5 g for a take-off from grass. It is assumed that these accelerations occur at zero lift and therefore produce compressive (negative) stresses,−STO, in critical components such as the undersurface of wings. The maximum positive stress for the same component occurs in level flight (at 1 g) and is +S1g.
The ground–air–ground cycle produces, on the undersurface of the wing, a fluctuating stress SGAG=(S1g+STO)/2 about a mean stress SGAG(mean)=(S1g−STO)/2. Suppose that tests show that for this stress cycle and mean stress, failure occurs after NGcycles.
For a life factor of 3 the safe life is NG/3 so that the damage done during one cycle is 3/NG. This damage is multiplied by a factor of 1.5 to allow for the variability of loading between different aircraft of the same type so that the damage per flight DGAGfrom the ground–air–ground cycle is given by
DGAG =4.5/NG (15.16)
Fatigue damage is also caused by gusts encountered in flight, particularly during the climb and descent. Suppose that a gust of velocity uecauses a stress Suabout 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(Su); the damage caused by one cycle is then 1/N(Su). Therefore 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 sheets1present the data in two forms, as we have previously noted. First, l10 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≥3.05 m/s be encountered. It follows that 1/l10is the number of gusts encountered in unit distance (1 km) at a particular height. Secondly, gust frequency distribution curves, r(ue) against ue, give the number of gusts of velocity uefor every 1000 gusts of velocity 3.05 m/s.
From these two curves the gust exceedance E(ue) is obtained; E(ue) is the number of times a gust of a given magnitude (ue) will be equalled or exceeded in 1 km of flight.
Thus, from the above
number of gusts≥3.05 m/s per km=1/l10
number of gusts equal to ueper 1000 gusts equal to 3.05 m/s=r(ue) Hence
number of gusts equal to ueper single gust equal to 3.05 m/s=r(ue)/1000 It follows that the gust exceedance E(ue) is given by
E(ue)= r(ue)
1000l10 (15.17)
in which l10is dependent on height. A good approximation for the curve of r(ue) against uein the region ue=3.05 m/s is
r(ue)=3.23×105u−e5.26 (15.18) Consider now the typical gust exceedance curve shown in Fig. 15.3. In 1 km of flight there are likely to be E(ue) gusts exceeding uem/s and E(ue)−δE(ue) gusts exceeding
15.4 Prediction of aircraft fatigue life 437
Fig. 15.3Gust exceedance curve.
ue+δuem/s. Thus, there will beδE(ue) fewer gusts exceeding ue+δuem/s than uem/s and the increment in gust speedδue corresponds to a number−δE(ue) of gusts at a gust speed close to ue. 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−δE(ue)/2. Suppose that each cycle produces a stress S(ue) and that the number of these cycles required to produce failure is N(Su,e). The damage caused by one cycle is then 1/N(Su,e) and over the gust velocity intervalδuethe total damageδD is given by
δD= − δE(ue)
2N(Su,e) = −dE(ue) due
δue
2N(Su,e) (15.19)
Integrating Eq. (15.19) over the whole range of gusts likely to be encountered, we obtain the total damage Dgper km of flight. Thus
Dg= − ∞
0
1 2N(Su,e)
dE(ue)
due due (15.20)
Further, if the average block length journey of an aircraft is Rav, the average gust damage per flight is DgRav. Also, some aircraft in a fleet will experience more gusts than others since the distribution of gusts is random. Therefore if, for example, it is found that one particular aircraft encounters 50 per cent more gusts than the average its gust fatigue damage is 1.5 Dg/km.
The gust damage predicted by Eq. (15.20) is obtained by integrating over a complete gust velocity range from zero to infinity. Clearly there will be a gust velocity below which no fatigue damage will occur since the cyclic stress produced will be below the fatigue limit stress of the particular component. Equation (15.20) is therefore rewritten
Dg= − ∞
uf
1 2N(Su,e)
dE(ue)
due due (15.21)
in which uf is the gust velocity required to produce the fatigue limit stress.
We have noted previously that more gusts are encountered during climb and descent than during cruise. Altitude therefore affects the amount of fatigue damage caused by gusts and its effects may be determined as follows. Substituting for the gust exceedance E(ue) in Eq. (15.21) from Eq. (15.17) we obtain
Dg= − 1 1000l10
∞
uf
1 2N(Su,e)
dr(ue) due due or
Dg= 1
l10dgper km (15.22)
in which l10is a function of height h and dg= − 1
1000 ∞
uf
1 2N(Su,e)
dr(ue) due due
Suppose that the aircraft is climbing at a speed V with a rate of climb (ROC). The time taken for the aircraft to climb from a height h to a height h+δh isδh/ROC during which time it travels a distance Vδh/ROC. Hence, from Eq. (15.22) the fatigue damage experienced by the aircraft in climbing through a heightδh is
1 l10dg V
ROCδh
The total damage produced during a climb from sea level to an altitude H at a constant speed V and ROC is
Dg,climb=dg V ROC
H
0
dh
l10 (15.23)
Plotting 1/l10against h from ESDU data sheets for aircraft having cloud warning radar and integrating gives
3000
0
dh l10 =303
6000
3000
dh l10 =14
9000
6000
dh l10 =3.4 From the above9000
0 dh/l10=320.4, from which it can be seen that approximately 95 per cent of the total damage in the climb occurs in the first 3000 m.
An additional factor influencing the amount of gust damage is forward speed. For example, the change in wing stress produced by a gust may be represented by
Su,e =k1ueVe (see Eq. (14.24)) (15.24) in which the forward speed of the aircraft is in equivalent airspeed (EAS). From Eq.
(15.24) we see that the gust velocity ufrequired to produce the fatigue limit stress S∞is
uf =S∞/k1Ve (15.25)
The gust damage per km at different forward speeds Veis then found using Eq. (15.21) with the appropriate value of uf as the lower limit of integration. The integral may be
15.4 Prediction of aircraft fatigue life 439 evaluated by using the known approximate forms of N(Su,e) and E(ue) from Eqs (15.15) and (15.17). From Eq. (15.15)
Sa=Su,e= S∞,m
Kn (1+C/ N(Su,e)) from which
N(Su,e)= C
Kn 2
S∞,m Su,e−S∞ ,m
2
where Su,e=k1Veueand S∞,m=k1Veuf. Also Eq. (15.17) is E(ue)= r(ue)
1000l10 or, substituting for r(ue) from Eq. (15.18)
E(ue)= 3.23×105u−e5.26 1000l10 Equation (15.21) then becomes
Dg= − ∞
uf
1 2
Kn C
2
Su,e−S∞ ,m S∞,m
2
−3.23×5.26×105u−e5.26 1000l10
due Substituting for Su,eand S∞,mwe have
Dg= 16.99×102 2l10
Kn C
2 ∞
uf
ue−uf uf
2
u−e6.26due or
Dg= 16.99×102 2l10
Kn C
2 ∞
uf
u−e4.26
u2f − 2u−e5.26
uf +u−e6.26
due from which
Dg= 46.55 2l10
Kn C
2
u−f5.26 or, in terms of the aircraft speed Ve
Dg= 46.55 2l10
Kn C
2 k1Ve S∞ ,m
5.26
per km (15.26)
It can be seen from Eq. (15.26) that gust damage increases in proportion to Ve5.26so that increasing forward speed has a dramatic effect on gust damage.
The total fatigue damage suffered by an aircraft per flight is the sum of the damage caused by the ground–air–ground cycle, the damage produced by gusts and the damage due to other causes such as pilot induced manoeuvres, ground turning and braking, and landing and take-off load fluctuations. The damage produced by these other causes can be determined from load exceedance data. Thus, if this extra damage per flight is Dextra the total fractional fatigue damage per flight is
Dtotal =DGAG+DgRav+Dextra or
Dtotal=4.5/NG+DgRav+Dextra (15.27) and the life of the aircraft in terms of flights is
Nflight =1/Dtotal (15.28)