Mechanical Engineering with Dr. Sanei

8:54

Part B: Chat GPT-Genie’s limitations

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7:25

Part A: ChatGPT- Genie' Capabilities

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5:14

Thermoset vs Thermoplastic Torch Experiment

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3:09

Experiment showing conservation of angular momentum

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9:19

How to use Mendeley Reference Manager Software

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18:15

Helical Gears: Geometry and force analysis- Direction of tangential, radial and axial loads

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10:07

Spur Gear Force Analysis: Tangential and Radial Load

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20:31

Gear Train: Regular Vs Reverted Gear Train

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22:38

Gear Ratio: Velocity Ratio, Torque Ratio

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9:14

Problem 13-101: Kinetics of a particle example solved using polar coordinate system

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12:01

Problem 13-98: Kinetics of a particle example using polar coordinate

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10:59

Problem 13-95: Kinetics of a particle using polar coordinate system,

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14:34

Problem 13-90: Kinetics of a particle using polar coordinate: A boy sliding down the spiral slide

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7:47

Kinetics of a particle

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12:55

Stress concentration-Static vs Fatigue(Kt vs Kf), Open vs closed hole, bending vs torsion(Kt vs Kts)

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6:24

Difference Between Different Academic Titles- Assistant Professor vs Lecturer

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4:20

The Difference between Shear Stress Equations: Tau=VQ/IT vs Tau= V/A

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1:01

3D printing of a threaded elbow with ABS

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10:00

Lab 6-Material Treatment: Effect of cold working and process annealing on material properties

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5:19

Statics Practice Problem 9-74: Center of gravity of a three wheeler using composite body formulation

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4:44

Statics Practice Problem 9-65: Centroid of a composite body by subtracting the missing parts

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5:10

Statics Practice Problem: Finding the centroid of a composite body

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2:43

Statics Practice Problem 9-60; Finding the centroid of a composite body

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6:04

How to find the centroid of a composite body

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4:50

Statics practice Problem 9-37: centroid of a circular sector

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5:21

Statics Practice Problem 9-28: Centroid of an area bounded by two curves

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4:36

Statics Practice Problem 9-11: Centroid of an area defind by a curve

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3:53

Statics Practice Problem 9-40: Finding the centroid of a volume (paraboloid)

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4:13

Example of center of gravity of homogenous rod

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Пікірлер

@ADITISHASTRI-i5h 10 күн бұрын

great vid

@engineeringvision 10 күн бұрын

Thank you! I am glad you found it helpful. Which university are you visiting from?

@selvamraj11 13 күн бұрын

In the previous video, you used a fixed beam with UDL to calculate the equivalent load for an evenly distributed load. My question is, how does this work for determining the equivalent load for the second element of this problem?

@engineeringvision 12 күн бұрын

You put it in the matrix form with all elements, all you need to know is what reactions you have at each support, if it is fixed you will have a moment and for pinned supports, you set zero for the moment value

@selvamraj11 13 күн бұрын

Thanks for video, Why is the equivalent load calculated on the fixed beam case always? Why not in the simply supported beam?

@engineeringvision 13 күн бұрын

That is the harder case as there are moments at the two end involved make the system statically undermined, the simply supported case is an easier system where reaction forces can be determined using equilibrium equations

@user-zo4uf9gb4e 16 күн бұрын

Do professors gets extra pay when they help phd student? Or get pay for writing papers?

@engineeringvision 14 күн бұрын

Great questions! Not normally, that is part of the job! Unless you have an accepted grant, then you can pay yourself summer salary or supplement your academic year salary up to 20%.

@urpapi6969 22 күн бұрын

thank you for sharing your knowledge

@engineeringvision 21 күн бұрын

You are welcome! I am glad you found it helpful

@CalebPiatt-zc1fv 23 күн бұрын

thank you so much

@engineeringvision 23 күн бұрын

You're welcome! I am glad you found it helpful

@tori3522 Ай бұрын

Thank you so much! I was stuck on this problem and you were the only one able to help me.

@engineeringvision Ай бұрын

You are welcome

@soheilgh1845 Ай бұрын

I have a question about compensating the toe region. A continuation of the linear part should be generated in order to find the corrected zero strain point. Am I right? after that, should stress start from zero too? after this procedure, at the beginning, before the linear part, still there will be an inclined part. Is it right?

@engineeringvision Ай бұрын

At the zero strain, the force is not necessarily zero due to tightening the grip! Stress-strain curve is not a good indication of initial conditions and it is hard to find the force at zero strain unless you read the machine recording at the time of experimentation! Extrapolation of stress strain curve will give you inaccurate beginning point. You can shift the stress-strain curve to give you more reasonable stress and strain values but do not extend it beyond its limit of recording

@kalosalbertonoleordinola2228 Ай бұрын

Dear, from what book do you get those exercises? Greetings from Peru

@engineeringvision Ай бұрын

It is Hibbler book

@user-hg5jv9nq9f 2 ай бұрын

Thank you so much ❤

@engineeringvision 2 ай бұрын

You're welcome 😊

@patrickchomba-ii3ft 2 ай бұрын

Thanks very much for this it has really helped me to understand this ❤❤❤

@engineeringvision 2 ай бұрын

You are welcome! I am glad you found it helpful

@sinakhokhumalo5485 2 ай бұрын

Sir why is it negative when it is moving the beam clockwise?

@engineeringvision 2 ай бұрын

That’s our agreed convention to call a clockwise moment negative and counter clockwise moment positive, but in general if you use the right hand rule, a clock wise motion is towards negative Z which is the axes perpendicular to the page and counter clock wise would be towards positive z

@dimuswrld4706 3 ай бұрын

How did you get 300N/m cause after using similar triangles that is supposed to be 200N/m

@khiziiikhan 3 ай бұрын

Thanks sir

@engineeringvision 3 ай бұрын

You are welcome! I am glad you found it helpful

@rojherbryanguevaracerda4362 3 ай бұрын

What is the name of Book?

@kareemehab5353 3 ай бұрын

Dynamics-Hibbeler-14th-edition-book

@engineeringvision 3 ай бұрын

It’s Hibbler

@TheBigFatVladimir 3 ай бұрын

Thank yoU!

@engineeringvision 3 ай бұрын

You're welcome!

@randomplayer2387 3 ай бұрын

Excuse me, I'd like to ask, what do we do if the length of point C from either point A or B is unknown?

@engineeringvision 3 ай бұрын

Then you need to have other information available

@theb4i83 3 ай бұрын

isnt L 0.6-0.3cos45?

@yerda_music 3 ай бұрын

wouldn't the final angle be 120 since the measurement is taken from outside of the triangle

@user-ku5jk9le1t 3 ай бұрын

Sir how to learn fluid mechanics in easy way any book ?

@garrettshelton5788 3 ай бұрын

reminds me of that tragedy

@engineeringvision 3 ай бұрын

What tragedy? Did you fail statics?

@user-ku5jk9le1t 3 ай бұрын

Which book

@engineeringvision 3 ай бұрын

Hibbler

@meeranchand2970 3 ай бұрын

Ax =0 thats seems untrue

@engineeringvision 3 ай бұрын

In practice, there are forces, but in theory if only external vertical forces are applied and in absence of external horizontal forces, there is no reaction forces to oppose any forces

@kvkfncdmndjd406 4 ай бұрын

Very nice example. Could you tell me from which book is it?

@engineeringvision 4 ай бұрын

Hibbler

@Tuba4life1 4 ай бұрын

How did you find Fea and Feb, you never finished those equations

@ingridlimae Ай бұрын

if you analyse the joint A, you can notice an Y force (as well as on joint B). then, you do: Mo= 6 times YB + 900 times 4 + 600 times 8. YB= -1400 N ΣFY= YA+YB= 0 so YA= 1400N afterwards, we can find the angle of the ABE triangle: 4/3 (half of the base)= 1.333333 1.333= tg 53.13 if Fea= Ya times sin 53.13 then Fea equals 1750N you can find Feb easily after that.

@jerryrichardson2661 4 ай бұрын

Nice work

@engineeringvision 4 ай бұрын

Thanks

@jerryrichardson2661 4 ай бұрын

Good stuff

@engineeringvision 4 ай бұрын

Glad you enjoyed

@sadranajafi2529 4 ай бұрын

Tnx a lot dr ❤️

@engineeringvision 4 ай бұрын

You are welcome

@mariarodriguezhernandez4066 4 ай бұрын

why is 2/3?

@engineeringvision 4 ай бұрын

The centroid of a triangle is 2/3 from the edge of the triangle or 1/3 from the base! It can be proven but normally memorized as a rule

@sarvangmehta3383 5 ай бұрын

Can you provide matlab file?

@engineeringvision 4 ай бұрын

I don’t think I can attach them here. Which school do you go to?

@Amanda-bc5xh 5 ай бұрын

How did you figure out the 100 for the triangle when you cut the beam?

@godfreymutale4876 2 ай бұрын

Using similarities of triangles, before it was cut the bigger triangle had 200N/m, after it was cut the smaller triangle was exactly half of the bigger one, finding the ratio(i.e 200:x=6:3) , x= 100. Note that 6 and 3 are just the base distances of a bigger and smaller triangles respectively.

@emmasimz5438 2 ай бұрын

Thank you@@godfreymutale4876

@namkhanhphung2545 5 ай бұрын

how do you determine the distance from Fr1 and Fr2 from b ? i dont really get this part

@engineeringvision 5 ай бұрын

Good question! For a rectangle, it is in the middle and for a triangular load, the resultant force is acting 1/3 of the base from the right angle of the triangle.