Question

using figure 1 and information below find the forces and power needed to move the elevator(lift).

needed answers are:

1 finding the service loads

2 mechanical analysis to the shaft

3 power the motor required to move the system.

Project Introduction The system to be designed is a lifting system (elevator) for people located in a building using an electric drive system. There is one car that needs to be lifted as high as 15 metres. The arrangement of the system is shown in figure 1. Technical specifications In order to reduce the power requirements on the motor the elevator car has a counterweight which is intended to assist the system - The purpose of the counterweight is to reduce the load on the motor when lifting the weight of both the elevator car and the people being transported in the lift. The aim of this assignment is to design the lifting mechanism of the lift system incorporating the main electric motor drive system, cable drum and associated pulleys and shafting. Key dimensions are as follows: Elevator car platform size: 2m x 3 m and is 2.5 m in height The maximum working load of the elevator lift and people 9500kg of which: 3000kg can be considered to be the duty load i.e. the weight of the passengers Average Speed: 3 m/s Drive system supporting Structure working envelope: 4 m x 3m Lift counterweight: is required to have a mass of 80% of the empty elevator car.

Answer 1

Solution: Let mass of empty lift is M_e kg and mass of counter weight is m kg.

Maximum mass that lift can take M=9500 kg. Mass of the passangers in the lift is=3000kg

hence mass of the empty lift M_e=9500-3000=6500kg. Mass of counterweight=m = 80% of M_e

ie m=80x6500/100=5200kg.Let tension in the drive pully rope is T and tension in counterweight pulley is T' . Average speed of the lift is 3m/s. Let's assume that lift travels half of its distance with accleration a then rest half of the distance with retardation a. Let maximum speed atained by lift is V'. The total time taken by lift is t.

from Newton's second law 2T+T'-M.g = M.a ............1     [ take g=10m/s² ]

mg-T'= m.a .....................2 . Now adding equations 1 and 2

2T-(M-m).g=(M+m).a.........................3

Total travel time t= Total distance travelled/Average speed = 15/3=5 sec. The maximum speed will atained at the half of the total distance travelled. Hence applying equation of motion

S=u.t+1/2(a.t²) = 15/2 = 0x2.5 + 0.5 x a x 2.5²

7.5=0.5xax6.25 which gives a=7.5/(0.5x6.25)=2.4m/s². Putting the value of a in equation3.

2.T-(9500-5200)x10=(9500+5200)x2.4 which gives T=39140N and T'=m.(g-a)=5200x(10-2.4)=39520N

(1)Hence force required to lift the elevator=F=2T+T'=2x39140+39520=117800N=117.8KN. (ans)

(2) let shaft used is made of steel having maximum shear strength s and diameter d

then torque=T"=Fxd/2 and maximum shear strength s=16.T"/(pi)xd³. Let s is assumed to be 55MN/m²

Then 55X10⁶= 16xF.d/3.14xd³=16x117.8x10³/(3.14xd²) which gives diameter of the shaft=d=0.104m=10.4cm. (ans)

(3) Power required to drive the pulley P=F.V_av=117.8x3=353.4KW. (ans)