Embedded Files
Element H: Full physics and engineering analysis (M7)
Element H focuses on a more rigourous engineering analysis of the system, now that it has specific components with known values. The results of this analysis will help inform the next iteration, Revision B, in addition to more learning materials focusing on design for manufacturing. Revision B will be shown below the three engineering analyses.
Detailed design analysis 1: Torque analysis
The first design analysis that I will perform will be on the necessary torque to power my lifting mechanism. The lifting mechanism in my assembly is split into two halves, with the left half being anchored to the floor, housing the motor and gear assembly, and the right half being articulated to lift a ball 1 ft into the air. This articulation is circular in motion, so I believe that a torque analysis will be worthwhile to ensure my assembly can function.
The first step I will take is to determine the mass of the objects being moved. Using the measure function in Siemens Nx, my ABS right stair has a volume of 21.71 in^3, and a weight of 0.8235 lbs. The object being lifted is a 1 inch diameter steel ball. It has a volume of 0.5236 inches^3, and weight of 0.1481 lbs. Together, these components have a mass of 0.9716 lbs. I will round this up to 1 lb to add some redundancy to my system, and ease calculations. For the purposes of checking maximum torque, I will assume the ball is at its highest point. The center of mass of this subassembly is located at [ 0.0407, 1.37275, 0.5939 ], calculated from my local coordinate system in NX.
This load will be moved by connections to my driving arm, located at [ 0.0407, -4.6224, -4.3520] inches . The distance rarm between my center of mass and connection is therefore [0,5.995,4.9459] inches, rarm = 7.771 inches. The max distance from this connection to the gear’s center of rotation is [(1 + 4.269 + 5.995), 4.9459] inches, hypotenuse of 12.302 inches. Therefore, r = 12.302 inches. This will allow me to calculate torque
Torque = r * F
r = 12.302 in = 1.0252 ft
F = 1 lb
Therefore torque needed will be 1.0252 Feet pounds. This seems like a lot for a small motor like the one that I’m going to use, even with a favorable gear ratio working for me. I now will proceed to the next part of this analysis, the gear ratio:
Gear ratio = # driven /# driving = 12/8 = 1.5 to 1
So my actual amount is 0.6834 ft lbs, or 926 mNm. This still seems pretty high for my motor, but I will look at that in the next analysis, the motor analysis. A change that I could make to my design would be to increase the number of gears dramatically, change the motor, or alter the model to make the motor and center of gravity nearer.
Detailed design analysis 2: motor analysis
The motor that I provisionally selected is a pretty standard cheap hobby motor from a chain store, the xUmp 260. It's 1.5 - 6 V, max current 1.8 A, with 10000 RPM at 6V. My battery selection operates at 6V, and has 4.5 AH capacity, so I should have about 2.5 hours of operating time.
Calcs using lecture 5.4:
RPM/ V = 1666.67
Max voltage = 6 V
Max current = 1.8 A
Max watts = 10.8 W
(10000 Revs1 min ) (1 min 60s) (2π rad 1 rev) = 20000π60 =1047.1 rad/s
Torque = 10.8 W 1047.1 rad/s
Torque = 0.01031324 Nm
Output of my selected motor is 0.0076 ft lbs, way less than the 0.6834 ft lbs needed to rotate my assembly, even with gears.
This is too much for my current motor. After looking online, there are several options for other motors, but they will require at least a 12V battery, which means replacing my current 6 V selection. I have updated my bill of materials accordingly.
The new motor that I settled on operates at 12V, 160 RPMs, 1.6 A, and will fit in my previously designed motor brace, which is nice. The accompanying battery change has specs: 12 V, 2.1 A, and 7.2 AH.
RPM/ V = 13.333
Max voltage = 12 V
Max current = 1.6 A
Max watts = 19.2 W
(160 Revs1 min ) (1 min 60s) (2π rad 1 rev) = 320π60 =5.333 rad/s
Torque = 19.2 W 5.333 rad/s
Torque = 3.6 Nm
This converts to 2.655 Feet pounds, which is plenty of torque for my application. Now that I am confident that my assembly can rotate the stairs to get the ball 1 ft into the air, I will proceed to the next analysis.
Detailed design analysis 3: lever analysis
The third and final analysis that I will perform will be the force analysis for the falling ball and lever. My system uses a falling ball to depress a lever, which in turn dislodges a tab. This means that my lever will have to be long enough to overcome the frictional force of my tab.
Here, F tension is equal to 4.45 N. This is because the rope is holding up a 0.454 kg hammer, and F = 0.454 * 9.81 = 4.45. Therefore, the lever must generate at least 3.56 N to overcome friction, assuming a frictional coefficient of 0.8 for steel that I found on an engineering website.
Ball weight: 1.481 lbs
1.481 lbs = 6.5878 N
Ball mass = 6.5878 / 9.81 = 0.672 kg
Fall distance: 6 inches
0.5 ft = 0.1524 Meters
Ball distance to fulcrum = 9 inches
9 in = 0.2286 m
Tab distance to fulcrum = 3 inches
3 in = 0.0762m
Although I have the ball fall from 1 ft, the lever is 6 inches in the air, so the ball only falls 6 inches.
I’ll start by finding the ball’s potential energy.
PE = m * g * h
PE = 0.672 kg * 9.81 m/s^2 * 0.1524 m = 1.002 J
W = F * d
F= W/d
F = 1.002 J / 0.0762 m
The force the ball exerts on the long side of the lever is F = 13.15 N
Next, since torque will be the same on both sides, I’ll calculate it to find force output.
T = F * d
Foutput = Finput * (d1/d2)
F out = 13.15 * (9/3) = 39.45 N
This is plenty of force to dislodge the tab, and I won’t even need to use the graphite powder lubricant to lower the friction for this part of the system. I can also shorten the lever if I want to , which would lower material costs at the price of redundancy.
Another set of considerations made during this phase of the design process were how components would be manufactured during high volume production. These considerations are as follows and can be expanded:
Thermoformed ABS components
Thermoformed ABS components
After reviewing my components, there are several that would make sense to thermoform. They each have a single face that needs complexity, with the rest being filler geometry leftover from designing. These components include my right stair, both gears, tab, and disc. My models have been updated accordingly.
Extrusion steel pins
Extrusion steel pins
Previously, my pins were made of PLA, but upon consideration of the forces that they will experience I have decided to instead make them from steel. This will let them undergo the shear stress with much less risk of failure. I thought that extrusion would be a good fit for manufacturing them, as they will have a simple cross section for their length. I have updated the part materials in Siemens NX, to account for the new center of mass and workload on the motor.
Reducing component count
Reducing component count
My original design included a frame which everything was mounted to. This was the left staircase. It was supported by two braces, which each had several fasteners. After watching the design for assembly video, I have come to the conclusion that I can integrate these into one part. Because of the added complexity this entails, I will injection mold this new component rather than thermoform like I preliminarily had intended. This should help reduce assembly time, as well as material cost.
Standardized fasteners
Standardized fasteners
In an effort to reduce the number of different components to manufacture, I have standardized the fastening pins that my assembly uses. They come in two forms now, a heavy duty lever pin, which is present in my lever’s fulcrum, and the standard fastener pin, which is present throughout the design
Injection molding
Injection molding
I have several complex, hard to manufacture parts in my design, which I believe would benefit from injection molding. These include the connecting arm, motor brace, lever, and left stair assembly. Each of these components have irregular geometry, and depth that would not allow for thermoforming like the rest of my plastic components will have. I have updated the components to have injection sites that will let the plastic reach all parts of the mold evenly before cooling.
PLA TO ABS
PLA TO ABS
After viewing the design for plastics manufacturing video, I reviewed my components and came to the conclusion that switching my plastic components from PLA to ABS would be beneficial to me because of increased strength and durability. This increased durability will be great for components like the connecting arm, which will experience some shear stress. I have updated my part materials in Siemens NX
Using all of the above considerations, I created a complete revision B, seen below:
Bill of Materials - Revision B
Page updated
Report abuse