An elevator in a 20 story building is supported by a steel c

An elevator in a 20 story building is supported by a steel cable attached to a motor driven pulley on the roof. When the fully loaded elevator is located on the 1st floor, the cable is stretched 0.25 m from its unstructured length. a) If the upper end of the cable is suddenly stopped when the elevator is moving upward at its maximum velocity, there is the possibility that the elevator will continue upward sufficiently to cause the cable to become slack. If this happens, the elevator will be snapped when its subsequent downward motion loads the cable again, which would make the passengers uncomfortable. Determine the maximum upward velocity that the elevator may have without allowing the cable to become slack. Assume that the elevator is travelling up from the sub basement and is traveling at constant speed when the upper end of the cable is suddenly brought to rest as the elevator ranches the 1st floor. b) Consider the fully loaded elevator at rest the 1st floor when the upper end is given is given a constant speed upward to lift the elevator. Find the maximum deflection from unstressed in the cable if the speed of the upper end is that found is (s).

Solution

When the cable is stopped at the top the elevator is still continuing to move with velocity u. it will now be acted upon the \'g\' downwards, it will stop a distance (u²/(2x9.81)). Obviously this value should lower than the total elongation of 0.25m given as the input.

F, The force required to move the elevator of weight W, F= W(1+a/g) where a the allowable acceleration at the start.

The passengers will feel comfort if they do not feel any downward movement of the lift. This means that, from the instant of stopping of the cable at the top, the elevator shall come to stop in such distance that the cable will just hold the weight W (not F) as it reaches speed=0 in its upward journey. This will ensure that when the weight W tries to move downward, it will be resisted by equivalent cable force W. in addition the resultant extension of the cable due to only weight W shall also be equal to 0.25/(1+a/g).

0.25(1-1/(1+a/g)) = (u²/(2x9.81))

In this equation we have only two unknown variables. But we have practical guides to state that a max is 0.25 m/sec², u max 0.75m/s.

We shall try and match ‘a’ and ‘u’ by trial and error, such that the above equation gets satisfied. The following values satisfy

a=0.2, u=0.187m/sec and extension =0.207m

a= 0.25 u= 0.224 m/sec and extension =0.199m

a =0.3 u=0.26m/sec and extension =0.191m

Part (b)

The extension depends on the allowable acceleration. For various accelerations the deflection is calculated.

If a=0.2m/sec², deflection corresponding to forceF= W(1+a/g) =0.25m (of course the stiff ness of the cable is lower compared to the one at a=0.3m/sec²)

If a=0.3m/sec², deflection is 0.25m