You are involved in the design of an elevator for a two stor

You are involved in the design of an elevator for a two storey building with an distance between the storey\'s h = 4 m (see figure 1.1). The fully loaded elevator has a mass m = 1.5 metric tonnes. The acceleration profile chosen to perform the upward journey is qualitatively sketched in figure 1.2. It is composed of a first part where the acceleration increases linearly from 0 m/s^2 (at time 0 s) to a_1 = 6 m/s^2 (at time t_1 = 1 s). Then, acceleration is dropped instantaneously to 0 m/s^2 and a constant velocity motion is kept between times t_1 and t_2. The final part of the motion (from t_2 to t_3) is allowed for the mass to slow down with the motor not pulling the elevator to reach the top. This is executed since the cable cannot transmit any compression (i.e. downward) force to slow down the mass in the final stage of the upward journey. A. Calculate the values t_3 (total time for lifting) and the time t_2 (when the motor should stop pulling). B. In the choice of the cable, which is the nominal maximum cable load (in kN) that you will require from your cable provider? C. Which minimum motor power and torque is required to perform the journey calculated in point A, considering the double pulley system presented in figure 1.1 (with r_1 = 1 m, r_2 = 0.5 m)? D. Regulations require the design of safety brakes at the bottom. If the lift cable breaks in the top position (h) and falls towards the ground, the braking system is designed to act in the last I m (at the bottom) before the elevator hits the ground floor. This braking system is based on friction and applies a total normal force of 300 kN. Calculate the minimum friction coefficient required to stop the elevator just before hitting the ground.

Solution

A)

At t= 0, a = 0, v = 0, h = 0.

From t=0 to t1,

a(t) = (a1/t1)*t = 6t

v(t) = 3t^2 (integrating a(t) and putting v(0) = 0)

h(t) = t^3 (integrating v(t) and putting h(0) = 0)

At t1,

v = 3 m/s , h = 1 m

From t1 to t2,

a(t) = 0

v(t) = v(t1) = 3 m/s

h(t) = h(t1) + v(t1)*(t - t1)

At t = t2,

v(t2) = 3 m/s

h(t2) = h(t1) + v(t1)*(t2-t1) = 1 + 3*(t2-1) = 3t2 - 2

From t2 to t3,

Motor is stopped.

a = -g = -10 m/s^2

v(t) = v(t2) + a*(t-t2)/2 = 3 - 5(t-t2) ------------------ (1)

h(t) = h(t2) + v(t2)*(t-t2) + a*(t-t2)^2 = 3t2 -2 + 3*(t-t2) - 5*(t-t2)^2 -------------------------(2)

At t3,

v(t3) = 0 = > t3-t2 = 10/3 (From (1))

h(t3) = 4 = 3t2 - 2 + 3*(t3 -t2) - 5*(t3 - t2)^2 (From (2))

=> 4 = 3t2 - 2 + 3*(10/3) - 5*(10/3)^2

=> t2 = 17.18 sec

Hence, t3 = 20.518 sec

_______________________________

B) Maximum acceleration is 6 m/s^2

max mass = 1.5 tonnes = 1500 kg

Max nominal load = m*(g + a) = 1500 * 16 N = 24 kN

_______________________________

C) At t2, force and speed are maximum.

Hence, Maximum power is at t2.

Max power = Force*speed = 24 kN * 3 m/s = 72 kW

Assuming pulley to be massless, torque is totally transmitted to pull the lift.

Hence max torque is T = F*r2 = 24*0.5 = 12 kN-m

_________________________________

D)

If the cable breaks, the lift will experience free fall from h=4 m to h = 1m.

The velocity from h = 4 to h = 1 is given by

v = (2*g*dh)^0.5 (downward)

At h = 1m, the velocity is:

v1 = (2*10*3)^.5 = 7.75 m/s^2 --------- (3)

In last 1 m fall (dh1), let the effective deceleration be a3 and velocity at bottom is 0.

hence,

dh1 = v1^2/(2*a3)

=> a3 = v1^2/(2*dh1) = 30 m/s^2

Hence, the frictional force is

f = m*(g + a3) = 1.5 * 40 kN

The coefficient of friction required is

mu = f/R = 1.5*40/300 = 0.2

(In the solution, g is assumed as 10 m/s^2 which is a conservative rounding of the more accurate value of 9.8 m/s^2)