Landing Gear hydraulic system

Dipartimento di Ingegneria Aerospaziale Corso di impianti e Sistemi Aerospaziali a.a. 2016-2017 Landing Gear hydraulic system Matteo Tombolini Matricola 847166 Giugno 2017 1. Abstract Main landing gears and nose gear of a plane have a hydraulic retraction-extraction system working at 21 MPa. Gears have a 90° rotation during retraction around the hinge, which is the center of a Cartesian coordinates system. Draw the hydraulic system needed for actuators to work. Size the components, considering only weight and a time of retraction of 15 seconds. Figure 1. Main and nose landing gear wheels Main gear Main gear wheel Nose gear A B C D E Main gear Actuator attaching point Leg Wheel Nose gear Actuator attaching point Leg and wheel 2 Mass[kg] 30 20 20 X[mm] -110 0 320 Y[mm] 110 -300 -580 -100 0 100 -300 2. Index 1. Abstract/preface 2. Index 3. List of symbols 4. Problem description 5. Problem resolution 2 3 4 5 1. Ja k s force 2. Ja k s o po e t di e sio s 3. Pu p s po er 6,7 7 7 7 6. Problem initial data 1. Main Landing gear – extracted position 2. Nose landing gear –extracted position 3. Other data 7. Equations 8 1. Points coordinates in different positions 1. Main gear 2. Nose gear 2. Ja k s for e 1. Main gear 2. Nose gear 3. Ja k s di e sio s 1. Main gear 2. Nose gear 4. Pu p s po er 8.Results and deductions (same paragraphs as 7) 8,9 3 3. Symbols legend Symbol �, � � �′ F � W � �0 D’ D d' M Q S P α β γ Physical quantity Points coordinates Cylinder area Raised cylinder area Shaft area Minimized shaft area Difference between shaft and cylinder areas Ja k s for e Maximum extraction force Maximum retraction force Weight Lever arm of force applied in A Lever arm of force applied in D Pressure of the system Time of retraction Torque imposed by general force in P with respect to O (0,0) Total volume of oil needed for complete retraction for both gears Cylinder diameter Raised cylinder diameter Shaft diameter Minimized shaft diameter Mass Volumetric flow rate inside main gear jack Le gth of ja k s stroke Power Angle between positive vertical semi axis and appendix (considered >0 with clock-wise rotation) Angle between negative vertical semi axis a d leg s a is of s etr (considered <0 with clock-wise rotation) Angle between negative vertical semi axis and segment OC (considered <0 with clock-wise rotation) The su s ripts a d sta ds for ai a d ose 4 Unit of measurements mm N N N N mm mm MPa s N∙ mm mm mm mm Kg / mm W °, rad °, rad °, rad 4. Problem description A retractable landing gear consists of 2 main wheels and a nose one as shown in Figure 1. Oil is suctioned from a tank by a feed pump and arisen to a nominal pressure of 21 MPa. Downline of the po p there s a o -return valve which prevents oil from flowing back in event of malfunction causing it to lose pressure. A distributor valve controls then the linear actuators. These are jacks whose function is to make gears rotate around their hinge reaching retracted od extracted position. A line of return at low pressure brings oil back to the tank again passing through the same distribution valves. Between low and high pressure li es there s a e haust safety valve. In case of overpressure, it lets oil pass reducing instantly pressure to its nominal value. One of ja ks tips is constrained to a fixed point while the other is connected to an appendix of the leg 45° inclined. Jacks are necessary in case of retraction, because gravity would make the landing gear falling from wheel wells. Moreover, during landing the wheel must be perpendicular to the ground. This happens naturally when the gear is extracted if the center of the wheel is aligned with the body. There s the a ig difference between nose and main gears: this condition is always respected by the nose wheel whereas in main gear the centers of gravity of the wheel and the leg are not aligned when the gear is vertical. Therefore, jacks need to be double acting for main legs, right for adjusting them in both retracted and extracted position. O the other ha d, e d rather hoose a si gle a ti g ja k for the ose wheel. To be honest, real aircrafts use mechanical systems to freeze the position of gears, avoiding any problem connected to fluid compressibility especially when landing pressure is to its max. Main wheels are retracted into wheel wells sideways, while nose gear longitudinally towards the bow of the airplane. Although apparent wind helps nose gear to be extracted, it makes an opposition for retraction. This side effect is not relevant, because maximum push is requested to the jack at the end of the process, when wet surface and then drag are at their least. For the sake of argument moments of inertia are not considered, so the only loads are components weight. An example of implant fitted for our conditions could be that shown in figure 2. Figure 2. Sample of landing gear implant 5 Figure 3.b. Retracted position Figure 3.a. Extracted position V Figure 5 Jack’s path Black: = ° Red: = ° Blue: = ° Figure 4 Linear actuator 5. Problem resolution 5. Jack’s force A Cartesian coordinate system with center in the hinge is defined for both types of gears. The sizing critical positions are that of extracted and retracted gear (see figure 3.a and 3.b). As we already stated the jacks are constraint; therefore, during their path they make a rotation and in retraction and e tra tio the re horizo tal as sho i figure 5. It s i porta t to oti e that ja k is ea t to e o the left of the wheel in our case. Anyhow except for aircraft stow problems it could also be installed on the right. The difference is that during retraction instead of pushing, it would be pulling, redu i g ja k s shaft instability due to buckling. A solution for our case could be oversizing shaft diameter. Co e tio all ja k s for e is headed to ards o strai t poi t V. It s the o sidered positive when pulling (made by oil pressure in right chamber against the differential area between the shaft and the cylinder) and negative while pushing (made by oil pressure in left chamber against the piston seal). Therefore, moments giving a clockwise rotation are assumed positive. We define 3 angles: • α: et ee the appe di a d positi e erti al axis • β: et ee the leg s a is of symmetry and vertical axis • γ: between the negative vertical semi axis and segment OC Fo usi g o ea h leg, e e aluate ja k s maximum force, equaling the overall moments with respect to the hinge equation to zero (the sum of moments caused by weights and opposed ja k s rea tio . The resolution is then quite the same for the two types of legs. Of course, we use the common formula = ⃗ ∧ ⃗⃗⃗� = � ‖ ⃗‖ ∙ ‖⃗⃗⃗� ‖ ∙ sin � where ⃗ is the position vector and � between ⃗ and ⃗⃗⃗� . We use then the angles introduced to evaluate all the moments. The main problem we encounter is when calculating the torque of ⃗ : we must set the coordinate x of point V o strai t poi t e ause it s ot spe ified. This values should be distant from the retracted position of 6 point A at least twice the stroke of the shaft (220 mm for main and 200 mm for nose gears), because we must consider thi k ess of parts, slo i g do zo e, o strai t joi t parts… For instance, we chose V as point of coordinates (-660,110) for main leg and (-600,100) for nose leg. We find then m and q , slope and y-intercept of the straight line passing through A and V (or D and V) when gear is retracted. Using the point-line distance explicit formula between this straight line and O (origin of the axis) − − we find the perpendicular component of the position vector of ⃗ , namely its lever arm: = . 2 √ + We are then able to evaluate ⃗ solving the equation. 5. Jack’s component dimensions Cylinder area � can be determined referring to retraction, as oil pushes against the piston seal: The minimum diameter of the cilinder is then =√ � � � . = � . On the other hand, the shaft area can be determined referring to extracted position. In fact, in this case oil pushes against the annulus differential area between the shaft and the cylinder: � = �− = E . Having already � we can calculate and then its diameter d. Again, the same procedure is used for both gears except for the fact we don t eed to al ulate dia eter of the shaft for ose o e e ause it s single acting. 5. Pump’s power To conclude we can evaluate the hydraulic power needed to complete retraction of both gears in = ∙ = � ∙ 6. Problem initial data 6.1 Main Landing gear – extracted position Component Actuator attaching point Leg Wheel Point A B C 6.2 Nose landing gear –extracted position Component Actuator attaching point Leg and wheel Point D E x(mm) -110 0 320 y(mm) 110 -300 -580 M(kg) / 30 20 x(mm) -100 0 y(mm) 100 -300 M(kg) / 30 6.3 Other data � = = � 7 �: 7. Equations 7.1 Points’ coordinates in different positions 7.1.1 Main gear = ̅̅̅̅ ∙ � = ̅̅̅̅ ∙ :{ :{ 7.1.2 Nose gear = ̅̅̅̅ ∙ � :{ = ̅̅̅̅ ∙ :{ 7.2 Jacks’ force 7.2.1 Main gear � = ∙ = ∙�∙ ��= ∙ = ∙�∙ − � = − � = − ∙ − = = ̅̅̅̅ ∙ � = −̅̅̅̅ ∙ = ̅̅̅̅ ∙ � = −̅̅̅̅ ∙ = ∙ 7.2.2 Nose gear � = = = − = = − − √ + Moments’ equilibrium equation: ∑ ��� = →� ∙ − − − � � =� = +� → = = = ∙ =� → =√ E = 7.3.2 Nose gear = � = � =√ → 7.4 Pump’s power = ∙ = � −� − = ∙ − → � −� +� , − ̅̅̅̅ = ̅̅̅̅ ∙ � = −̅̅̅̅ ∙ ∙�∙ 7.3 Jacks’ dimensions 7.3.1 Main gear � →� � √ + Moments’ equilibrium equation: ∑ ��� = →� :{ ��=0 =� � ∙ = � − � [ ( � ∙ = � → )+ � ∙ � →� = ] ∙ � −� − � 8 = → =√ � = − , 8 Results and deductions 8.1 Points coordinates in different positions 8.1.1 Main gear (mm) (mm) (°) (mm) (mm) -45 -110 -35 -89,2276 -25 -65,744 -15 -40,2628 -5 -13,5583 5 13,55825 15 40,26279 25 65,74397 35 89,22755 45 110 8.1.2 Nose gear (°) 110 127,4302 140,9884 150,2628 154,9715 154,9715 150,2628 140,9884 127,4302 110 0 10 20 30 40 50 60 70 80 90 0 -52,0945 -102,606 -150 -192,836 -229,813 -259,808 -281,908 -295,442 -300 -300 -295,44 -281,90 -259,80 -229,81 -192,83 -150 -102,60 -52,094 0 (°) (mm) (mm) (°) (mm) (mm) -45 -35 -25 -15 -5 5 15 25 35 45 -100 -81,116 -59,7672 -36,6025 -12,3257 12,32568 36,60254 59,76725 81,11596 100 100 115,8456 128,1713 136,6025 140,8832 140,8832 136,6025 128,1713 115,8456 100 0 10 20 30 40 50 60 70 80 90 0 52,09445 102,606 150 192,8363 229,8133 259,8076 281,9078 295,4423 300 -300 -295,442 -281,908 -259,808 -229,813 -192,836 -150 -102,606 -52,0945 0 8. Jacks’ force 8.2.1 Main gear (°) � � N∙mm -45 0 -35 -15331,4 -25 -30197 -15 -44145 -5 -56751,7 5 -67634,1 15 -76461,4 25 -82965,5 35 -86948,7 45 -88290 � N∙mm 62784 42069,7 20077,14 -2525,46 -25051,3 -46816 -67158,2 -85459,9 -101165 -113796 � 0 0,031 0,052 0,065 0,07 0,067 0,057 0,043 0,023 0 � (mm) 110 130,155 144,4167 152,8786 155,9147 154,0663 147,9478 138,1812 125,3543 110 Main gear Force 500 0 10 20 30 40 50 60 70 80 -500 -1000 -1500 -2000 28,88658 18,88658 8,886582 -1,11342 -11,1134 -21,1134 -31,1134 -41,1134 -51,1134 -61,1134 � (mm) 110 130,0943 144,2208 152,557 155,5388 153,724 147,7039 138,0554 125,3204 110 1000 0 γ(°) β 9 90 (N) 570,7636 205,5301 -70,169 -305,922 -525,933 -744,517 -972,348 -1219,98 -1501,06 -1837,15 (mm) (mm) 320 214,4 102,3 -12,87 -127,7 -238,6 -342,3 -435,6 -515,6 -580 -580 -626,756 -654,468 -662,295 -649,998 -617,951 -567,128 -499,073 -415,854 -320 = = � This graph shows the trend of the force ade ai gear s ja k to keep the gear steady. This force decrease monotonically from positive to negative values, confirming we need a double acting jack. It s then equal to 0 when reaches approximately 15-20 degrees, when the system is in balance, because its center of gravity and the hinge are aligned. It also underlines that the force has its maximum in modulus while the gear is retracted ( � =-1837,15 N). We can calculate analytically the position in which the main gear system is in balance and the force of the jack is not needed. This happens when the center of gravity of the system and the hinge are aligned. � +� ⃗⃗⃗⃗⃗ = → = → � + � = → � = −� → ∙�∙ = ∙�∙ − ��=0 → ∙ ̅̅̅̅ ∙ sin = ∙ ̅̅̅̅ ∙ sin (� ( )− ) ̅̅̅̅ → ∙ . ∙ sin = ∙ . ∙ sin . °− → . ∙ sin = sin . ° ∙ cos − cos . ° ∙ sin → . + cos . ° ∙ tg = sin . ° → ≈ . ° We can verify this result by evaluating the coordinate ̂ of the center of gravity of the system. =− � + �� ≈ then it s aligned with the hinge When = , ° we have: { → ̂= � +� = . 8.2.2 Nose gear (°) � � N∙mm � (mm) � (mm) -45 0 0 100 100 -35 10220,93 0,030538 118,3227 118,2676 -25 20131,31 0,052147 131,2879 131,1098 -15 29430 0,064968 138,9805 138,6881 -5 37834,48 0,069568 141,7407 141,3989 5 45089,38 0,066767 140,0603 139,7491 15 50974,26 0,057497 134,498 134,2762 25 55310,31 0,042699 125,6193 125,5049 35 57965,78 0,023264 113,9585 113,9277 45 58860 2,03E-17 100 100 (N) 0 -86,4221 -153,545 -212,203 -267,573 -322,645 -379,622 -440,702 -508,795 -588,6 = � 0 Nose gear Force -100 0 10 20 30 40 50 60 70 80 90 This graph is different to the previous one. The main difference is that force has only negative values, confirming jack is single acting. It shows again the maximum force in modulus is when the gear is completely retracted ( � =-588,6). -200 -300 -400 -500 -600 -700 β 8. Jacks’ dimensions 8.3.1 Main gear ≈ , → ≈ , mm It s re o e ded to increase this values to ′≈ mm to increase safety, being conservative oversizing the system to prevent risk of failure. The outer diameter of cylinder will be about twice, letting it tolerate oil stress on walls. ≈ , → ≈ , mm In this case, we reduce it in order to raise the annulus area, defining then ′ ≈ mm. � 8.3.2 Nose gear � ≈ , 8.4 Pump’s power ≈ , → ≈ , We raise it to 10 ′ =