Landing Gear Stresses

3.73 X 750 = 2,800 pounds. This is divided between the two wheels so that each wheel takes. 2,800/2 = 1,400 pounds. This is applied at the center-line: of.
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Landing Gear Stresses Conrad J. Morgan

The first step in running the stress analysis of the landing gear is to find the load factors required and the height of the free drop in inches. By the Department of Commerce Bulletin 7-A, the factors are found as follows: level landing for airplanes of over one-thousand pounds gross weight, 9,000 n = 2.80 + ——————— where W is the gross weight of the W + 4,000 ship in pounds. For airplanes of less than one-thousand pounds gross weight, n = 3.00 -f 0.133 s, where s is the wing loading per square foot. The factor need not exceed 4.33 in either case. Three point landing shall be the same as level landing. Sideload landing introduces a new force. An applied load equal to the gross weight of the angles 9° and 40° on the drawairplane (less landing gear) ing. Then the load in the rear times one-tenth of the level is applied at the center-line: of strut C, is; landing load factor shall be as- the wheel. One side only need 1 sumed to act inward at the point be analyzed. Then to find the load in the of contact on one wheel with the sin [180° — (40° + 9°)] ground in a plane normal to the side struts, A on the drawing. 1 This is the load in both of these plane of symmetry. X sin 9° X 3,800 = ——————— The height of free drop in struts and is called the resultsin 131° inches shall be 0.36 times the ant of their loads, it will be brokX .1564 x 3,800 =1.325 X .1564 calculated stalling speed in miles en down into its individual X 3,800 = 788 pounds compresper hour. Need not be more loads later. Then: sion. The load in the front strut 1 than 18 inches for conventional ——— X sin (180°—62oxl400 = 1 airplanes. sin 19° A, is; ———————————————— Now let's analyze the landing sin [180° — (40° + 9°)] 1 gear in the level landing condiX sin 40° X 3,800 = 1,325 tion of a hypothetical airplane ——— X.8829 x 1400 = 3,800 Ibs. X .6428 x 3,800 = 3,240 Ibs. .3256 of 800 Ibs. gross weight and a The load in the shock strut B, It will be seen that the addition stalling speed of 40 mph. It of these loads is more than equal is: has 140 sq. ft. of wing area, a to the resultant load of 3,800 1 wing loading of 5.48 pounds per ———— X sin 43° X 1400 = pounds, this is because the struts square foot. The landing gear sin 19° are at an angle to the resultant. weighs 50 pounds. A drawing 1 The loads thus found are due of the landing gear is shown ——— X .6820 x 1400 = 2,920 Ibs. to the vertical load only, in adin the above figure; it is drawn .3256 dition to this we have a drag in the loaded condition. Level Then taking the loads in the component. This drag compoload (landing) factor is side legs, 3,800 pounds, and nent is due to the center of n = 3.00 + 0.133 X 5.48 = 3.73. breaking them up into their ingravity being to the rear of the Three point landing is also dividual loads we have; the rewheels as well as above them. 3.73. Side landing, the applied sultant, 3,800 pounds. This is In order that the ship may stay load is; in the same plane as the shock .10 X 3.73 X 750 = 280 pounds. strut, and the shock strut must in a level position, the resultant of the vertical and the drag comHeight of free drop; be at right angles to the fuselage, 0.36 x 40 = 14.4 inches. measured fore and aft, in order ponents must pass through the center of gravity. In measuring the angles be- that the axle will not twist when This resultant is the hypotenthe wheel moves up and down. tween the struts it is necessary that the landing gear is in the It follows then that the resultant use of a right triangle, the two loaded condition, with the shock meets the fuselage at an angle sides being the vertical distance absorbers in the full extended of 90°. So on the side drawing from the point of intersection condition, for it is this condition of the landing gear we draw a of the centerline of the struts we are interested in as the loads dotted line from the point of in- on the centerline of the tire to tersection of the two struts in the center of gravity, and the are much greater when the gear the center-line of the tire up to horizontal distance from this is compressed. Then in the level landing con- the fuselage, making a right point of intersection to the center of gravity. In our ship the dition the load on the gear is; angle to the fuselage, (D) on the vertical distance is 36 inches, as 3.73 X 750 = 2,800 pounds. drawing, this is the line of the shown on the drawing, and the resultant. This is divided between the two horizontal distance is 10 inches We measure our angles from wheels so that each wheel takes it and not between the struts, as shown. 2,800/2 = 1,400 pounds. This

Then, J 362 io-< - 37.36 in. which is the length of the resultant. Then the resultant force is

37.36/36 x

1,400 =

1,460 pounds. The drag component is; 10/37.36 x 1,460 = 375 pounds. This load acts back and at the point of intersection of the centerline of the struts and the centerline of the tire. Then the load in the front strut is; 1 —————— X sin 53° X 375 = sin 49° 1 ———— X .7986 X 375 = 397 .7547 pounds tension, it is tension because the action of the drag tends to swing the wheel down as well as back because of the position of the rear strut. Now the load in the rear strut is; 1 —————— X sin 78° X375 = sin 49° 1 ———— X .9781 X 375 = 485 .7547 pounds compression. We have the two separate loads in the struts caused by the vertical and drag components. Then the final load is the sum of these two loads. As the drag load in the front strut is tension, we subtract it from the load caused by the vertical component, we have; 3,240 — 397 = 2,843 pounds compression. As both loads in the rear strut are compression we add them; 788 +

485 =

1,273 pounds.

This completes the level landing condition. •