Henry Kurniadi

What your purpose? Additional length means additional applied force and bending moment. You need to check the working stress on the screwed connection.

You can always change/transfer to other division. But never mention this to your recruiter/hiring manager. The ideal answer would be: the position that you offering is the best position for me and I would love to learn further and do the job at that position.

For a very simple analysis, You need to maintain working stress if you using the same shaft material. Stress = applied moment x shaft radius / shaft inertia Applied load (moment, M) are in scale with R^3. (cube of blade radius, could be approximated by the ice cream space) shaft radius, r define the dimension. shaft inertia, I defined as pi x r^3 x t for thin hollow cylinder shaft. t = thickness. Let's say initially for a 1 m x 1 m x 1 m, you use 100 mm diameter shaft with thickness 5 mm. If you want to make the ice cream space larger to 2 m x 2 m x 2 m, i.e. 8 times the volume, you need 283 mm diameter shaft, with thickness 5 mm. You need to consider further whether 5 mm thickness suitable for larger shaft (consider buckling analysis)

Identify the load. In this case, wind load. Wind load could be calculated from ASCE 7 or EN 1991-1-4 or other relevant codes. The wind load (essentially pressure) multiply by effective area (5m x 2m of signboard in your drawing) to get a working wind force. Multiply this with 3 m height, to get working moment at the base. Let say using structural steel with yield strength of 240 MPa, safety factors 3, you get allowable working stress of 80 MPa. Check the cross-sectional properties (second moment area, radius, thickness) to get this allowable working stress. Stress = M x r / I For a thin cylinder, I = pi x r^3 x thickness. Then you get the dimension needed for the pole/support.

Chill down, the deformation often could be ignored in static simulation. Just make sure it not over than the safety stress limit. (material strength divided by a safety factor designated by standard/code/common practices).

You are limited by the measurement equipment sensitivity/accuracy/limit. Ruler limited to 1 mm, for example. I f someone measure with ruler and report it as 7.1 mm... that's bull.

Thin materials (sheet) have higher strength (albeit minor) than same thicker ones (rod or bulk). Strength is stress/pressure, i.e. force per area. You thin material is stronger stress-wise but actually weaker force-wise. Depends on what's your material (sheet, rod or bulk), you make design based on the listed value.

Electrical motor not really my thing. But check this for torque vs rpm relations for an electrical motor. http://lancet.mit.edu/motors/motors3.html You could conclude that that maximum output power occurs at [T] = ½, and [W] = ½ repectively. First you need to check the catalog/information of each motor, for stall torque and no-load speed () Say you use the BEMONOC = 12000 rpm motor ( = ???), it will perform the best at 6000 rpm. The maximum power is the product of ½ x ½ (roughly) Greartisan motor have = 200 rpm? ( = 0.2157463 N.m?), it will perform the best at 100 rpm. The maximum power is the product of ½ x ½(roughly) Generally, Find the one with the highest maximum power. In this case, it seems that the BEMONOC is better for your purpose, although a bit worry about the high rpm. (Make sure what's the load needed to reel, you don't want to flip the mechanism) ===== Unless you found that the manufacturers defined it differently in the catalog. For example I use this DC motor DMN-29 BA ( = 5000 rpm, = 30 mN.m) The catalog mention a specific rated torque and rpm, instead of 1/2 of the defined stall torque and no-load speed. The catalog said that the rated power at 0.42 A, should be the product of 3700 rpm and 7.8 mN.m = 3.02 W. Regarding speed reduction, the gear ratio will change the torque-rpm curve. See example from TG-201A-EU DC motor below. Gear ratio 1: 12.1 -> rpm 358 rpm, T 19.6 mN.m Gear ratio 1: 129 -> rpm 34.8 rpm, T 147 mN.m Again, compare which motor has higher power (½ x ½)

Learn to write properly? American companies won’t give a job to someone not American, with likely half-decent English, weely neely. Did you see how many unemployed people in the street? Sorry to say this bluntly.

https://www.thefabricator.com/article/stamping/ask-the-expert-how-do-i-calculate-press-tonnage-for-swaging- Force = Final flat area x Ultimate tensile strength x 1.2. This example part uses 1-inch-diameter ASTM A 500 tube (assumption thickness 0.063 inch). The nominal tensile strength of this material is 45,000 pounds per square inch. Following is the process for completing the calculations: Flattened area of rod = 2.5 sq. in. (again, assumption!) Tensile stress = 45,000 PSI Compressive strength = 1.2 x Tensile strength = 1.2 x 45,000 = 54,000 PSI Flattening force = Area x Compressive strength = 2.5 x 54,000 = 135,000 lbf. Flattening force in tons = 135,000/2,000 = 67.5 US tons Never use a press that is rated for the exact amount of required tonnage you have calculated. Always add a minimum of 20 percent for safety. In this case, the minimum press tonnage should be 81 to 91.

The enthalpy reading must use the same source with the questions. i.e. question from book X must use enthalpy reading from book X. Book Y will give different reading.

This means "pressure vessel", right? You need to design the pressure vessel based on your usage (dimension, materials, fluid density, corrosive-ness, etc.) The design shall follow ASME Section VIII. The output is what is the wall thickness, welding location/throat, flange dimension/thickness, openings and other accessories (faucets, valve, etc.). I had done one design for a client company that plan to use it for sewage water treatment a few years ago. Service comes at a cost, though.

99.99999999% You don't design a bearing as a mechanical engineer. You choose your bearing based on your engineering calculation. Unless of course, you are working in SKF, NTN or other bearing manufacturers.

Harmonics typically related to vibration. Something, say "A", will vibrate erratically under a certain condition. When "A" vibrate erratically, it will likely to fail due to excessive deformation. This "certain condition" could be related to wind flow (for lighting pole), machine rotation (for a shaft) or other factors. Let's say you design a lighting pole equipped with a small windmill on top, in location X. You want to design the whole system (i.e. design the pole) to avoid them having harmonics (excessive vibration) when it installed with local wind condition at location X. (difficult, wind always varied and difficult to predict) You also want to design the pole to avoid them having harmonics with the windmill operational rotation. (Easier to predict)

Not quite understand what you trying to say. But, I take a shot here. Let's say you want to select screw jack for lifting purpose. Typically, the manufacturer of the screw jack has catalog that define the safe limit to lift and max limit. You could start from contacting these manufacturers. Usually though, forklift or other lifting device uses (telescopic) hydraulic cylinders. One manufacturer: http://www.princehyd.com/Products/Hydraulic-Cylinders/Telescopic