## 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.
Landing Gear Stresses Conrad J. Morgan

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. •