Linear Actuator Motion Profile

[alzabroo]

Homework Statement

The problem deals with a linear rail system not unlike the one displayed in the image below

The relevant variables include:
Load = 70 Lbs
Load travel = 2.2 m
Load speed required = 2.3 m/s
Cycle time = 9 sec
Capacity of actuator = 33,000 N

With the given information:
A. determine the values for the graph below

B. ...and the life of the actuator in meters
C. ...and the life of the actuator in hours

Homework Equations

Eq. 1

Where:
L₁₀= actuator lifetime (in millions of revolutions)
Cr = Dynamic Load Rating
Pr = Equivalent Load Factor
Eq. 2

Where:
n = rotation speed (min^-1)
this second equation results in unit of hours
Eq. 3
V_m = (V₁q₁+V₂q₂+...+V_nq_n)/100%
Where:
V₁,V₂,...V_n = speed at phases of motion profile (in m/min)
q₁,q₂,...q_n = % of time for each phase

Eq. 4
L_{10 (hrs) var}=(L₁₀*10⁵)/(60*V_m)
Where:
L_{10 (hrs) var} = actuator life in hours modified for variable speed
V_m = equivalent speed

The Attempt at a Solution

There are a few assumptions I've made in my solution attempt to this problem, but I don't know where else to start.

First I assume that the dynamic load rating (Cr) is synonymous with the load capacity of the actuator
Second I assume that the Equivalent Load factor (Pr) is synonymous with the load provided (because no details of load distribution were provided)

To determine the life of the actuator in (millions of revolutions) I input the given values into Eq. 1:

L₁₀=(33,000N/311.5N)³=1.2*10⁶

Eq. 2:

L_{10 (hrs)}=(10⁶/(60*n))*(L₁₀)=(2.0*10¹⁰)/n

I'm very unsure of how to determine n, but since the cycle time is 10 sec:
n = 6(min^-1) ?

Assuming this, then:
L_{10 (hrs)}=3.3*(10⁹)

In order to use Eq. 3 account for the variable speed of the actuator, I believe I need to determine the values (time increments) of the graph given, but I'm unclear on how to do this with the given info.

I realize this a very long post, but I wanted to be as clear as possible.
Any pointers on how to where I have gone wrong in my process and/or how to proceed would greatly appreciated.
Thank you

 

[KataKoniK]

for your time and help.

Thank you for bringing this problem to our attention. It seems that you have made some good progress in your attempt to solve it. However, there are a few things that need to be clarified in order to provide a complete and accurate solution.

Firstly, it is important to note that the dynamic load rating (Cr) and the load capacity of the actuator are not necessarily the same thing. The dynamic load rating is a measure of the maximum load that the actuator can handle without failure, while the load capacity is the actual load being applied in the system. These two values may be different depending on the specifications of the actuator.

Secondly, the Equivalent Load Factor (Pr) is not the same as the load provided. It is a value that takes into account the type of load being applied (e.g. static, dynamic, shock) and the type of motion (e.g. continuous, intermittent) in order to calculate the equivalent load on the actuator. This value is typically provided by the manufacturer and is specific to the actuator being used.

In order to calculate the life of the actuator in (millions of revolutions), you will need to know the actual values of Cr and Pr for the specific actuator being used. These values can then be plugged into Eq. 1 to calculate L10.

To determine the rotation speed (n), you will need to use Eq. 2 and solve for n. This will give you the rotation speed in revolutions per minute (min-1). You can then use this value to calculate the life of the actuator in hours using Eq. 3 and 4.

To use Eq. 3, you will need to determine the values (time increments) of the graph given. This can be done by breaking down the motion profile into different phases and determining the speed and time for each phase. For example, if the actuator starts at 0 m/min, then accelerates to 2.3 m/s over a period of 2 seconds, then maintains a constant speed of 2.3 m/s for 5 seconds, and finally decelerates to 0 m/min over a period of 2 seconds, then the values for Eq. 3 would be:

V1 = 0 m/min, q1 = 2 sec
V2 = 2.3 m/s, q2 = 5 sec
V3 =