Equating the total energy at the start and end (potential and kinetic) yields: () which is a quadratic in * and can be rearranged as follows: () Given the values of W = 5 lbs, h = feet and k = 20 lbs/ft, *S = 5/20 = ft. Substituting this in the above equation gives for the maximum deformation * = feet or 6 *S. Drop Test Using a Spring Having Finite Weight Let us repeat the drop test, but now assume that the bar has appreciable mass, Wb, as shown in Fig. . The uniform bar-mass model is. | Equating the total energy at the start and end potential and kinetic yields T U T2 U2 - r- 0 h - ô2 which is a quadratic in and can be rearranged as follows 62 _ 2 g Wh 0 0 k k W Define the static deflection of the spring . 6 k Substituting gives . 62 - 26s 6 - 265A yielding . Ô Ô 1 1 2 ÔJ Given the values of W 5 lbs h feet and k 20 lbs ft S 5 20 ft. Substituting this in the above equation gives for the maximum deformation feet or 6 S. Drop Test Using a Spring Having Finite Weight Let us repeat the drop test but now assume that the bar has appreciable mass Wb as shown in Fig. . The uniform bar-mass model is represented by a spring-mass model shown in Fig. . The spring mass is uniformly distributed along the length of the spring and has a finite weight Ws Wb . Fig. Vertical Bar in Drop Test Fig. Equivalent Spring in Drop Test First it should be noted that the spring mass will contribute to the kinetic energy during deflection of the spring since each differential element of mass along the length will move downward with some kinetic energy during the deflection. This will decrease the potential energy of the system. The uppermost element of mass will move the same amount as the block since it is immediately below it. The next element will move slightly less etc. The lowest element will not move at all since it is contiguous with the base of the spring. Second the block upon falling through a height and gaining speed now impacts another mass the spring and the change in speed upon impacting the spring needs to be ascertained using the principle of conservation of momentum 1 . 2002 by CRC Press LLC The derivation of the equations involved in this process is covered in an article by Spotts 2 . Spotts develops an equivalent mass of the spring accounting for the different amounts of movement of the mass elements in the spring during deflection. This equivalent mass is 1 3 of the spring mass and it is then .
