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Steven Fowler
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Добавлен 15 сен 2014
Statics Lecture 19 part 4 example
Moments of Inertia with composite bodies and some circular holes. Here we solve an example problem to find the moment of inertia of a composite shape involving two circular holes. We employ parallel axis theorem to solve the problem. This is the fourth part of our nineteenth lecture in an Engineering Statics course.
Просмотров: 81
Видео
Statics Lecture 19 part 3 example
Просмотров 31Месяц назад
Composite bodies moment of inertia. Here we show an example of finding the moment of inertia for a shape using both composite bodies analysis and applying parallel axis theorem.
Statics Lecture 19 part 2
Просмотров 69Месяц назад
Introduction to Parallel Axis Theorem. Here, we further explain moments of inertia and discuss using parallel axis theorem to verify the result of the example problem we solved in part 1 of lecture 19. This is continuing in our lectures for Engineering Statics.
Statics Lecture 19 part 1 example
Просмотров 53Месяц назад
Moments of Inertia. Here we introduce moments of inertia and work a very simple calculus based moment of inertia example for an Engineering Statics course.
Statics Lecture 18 part 2 example
Просмотров 35Месяц назад
Composite body centroid example. Here we apply the principle of composite bodies to find the centroid of a shape for the eighteenth lecture of an Engineering Statics course.
Statics Lecture 18 part 1
Просмотров 45Месяц назад
Composite body centroids. Here in this first part of an eighteenth lecture in an Engineering Statics course, we introduce the theory behind finding the centroid of a shape using the composite body approach.
Statics Lecture 17 part 4 bonus example
Просмотров 37Месяц назад
Centroid between two functions. Here we solve a simple example involving double integrals to find the centroid between two functions. This is the last part of our seventeenth lecture in an Engineering Statics course.
Statics Lecture 17 part 3 example
Просмотров 29Месяц назад
Double integral centroid example. Here we find the centroid of the area under a curve described by the function y = x^2. This is the third and final part of our seventeenth lecture for an Engineering Statics course.
Statics Lecture 17 part 2
Просмотров 22Месяц назад
More theory on finding centriods of objects. Here we introduce the concept of finding the centroid of a function through the use of a double integral in an Engineering Statics class.
Statics Lecture 17 part 1
Просмотров 57Месяц назад
Introduction to Center of Gravity, Center of Mass, and Centroids. In this first part of our seventeenth lecture in an Engineering Statics course, we discuss the theory behind determining the center of gravity for an object. Then we describe the relationship between center of gravity, center of mass, and centroids.
Statics Lecture 16 part 5 example
Просмотров 17Месяц назад
An example problem involving friction and a block on an inclined plane for an Engineering Statics course.
Statics Lecture 16 part 4 example
Просмотров 69Месяц назад
Friction example of a classic motor and break problem in an Engineering Statics course.
Statics Lecture 16 part 3 example
Просмотров 43Месяц назад
Simple classic friction example problem from an Engineering Statics course.
Statics Lecture 16 part 2
Просмотров 44Месяц назад
Continued theory of dry friction in an Engineering Statics course. Discussion about different scenarios encountered involving friction.
Statics Lecture 16 part 1
Просмотров 33Месяц назад
An introduction to dry friction applied to Engineering Statics
thank you soooooo much for this video he helped me so much
@geordansmith4035 you're very very welcome. I'm glad it was helpful. There's several others in my play list for my Graphics course.
I have a question regarding how the forces are drawn in the FBD. If, for FBD of CDE, the Dx force is drawn in the positive X direction, wouldn’t the Bx force in FBD of ABC, be drawn in the negative x direction? I’m thinking that if Dx is drawn in the positive direction, then if there was a FBD of member BD, Dx for that would be drawn in the negative x direction and force Bx in the positive x direction. Then Bx on member ABC would be drawn in the negative x direction. Does that make sense?
yes, you definitely could look at the diagrams and interpret in advance the direction of the forces. The disadvantage of that: you have to be sure to draw the opposite force on the other side of the pin. For me personally, its too easy to get my signs/directions mixed up. So I have learned for me personally, I always draw an unknown pin force in the positive direction on the first appearance of that force, and then I draw it negative on the next free body diagram. If I find through the algebra, it actually has a negative value, I simply plug that negative value into the equations on the other FBD. This is what helps me keep track of signs and directions. If it helps you, feel free to use it. You are welcome to use any method that works for you, so long as you can keep track of the signs/directions of the forces from one FBD to the next. Does this help?
Oops…disregard my question. I should have watched the entire video.
no worries at all. glad you got squared away. Let me know if you have other questions.
Regarding the FBD of BC, when summing the forces in the X direction, wouldn’t the equation be BX + 800lb - Cx = 0?
I think the k component of the unit vector BC should be 3/7.
Expert catch Katrina! Keep up the great work!
Would Tcb not be 52.5 lbs? *Sampling Equilibrium of forces in the Z axis* (6/7)Tcb = 45 lbs Tcb = 45(7/6) lbs *Tcb= 52.5 lbs*
Peyton, Katrina has us on the right track.... when I constructed the unit vector up at the very top, somehow 6/14k incorrectly became 6/7k. This should be 3/7k. Therefore 45=3/7 Tcb then Tcb = 105
Great video. Thanks for the insight.
Glad you enjoyed it!
Thanks
you are most welcome!
Amazing video thank you for sharing this information
Video isnt synced with voice :/
Thanks for pointing that out! I made some cuts and edits to the to the video, but something caused the sound to get cut shorter than the video. I will work to fix it.
Ok, I think it is fixed now. Again, thanks for letting me know!! Good luck on your finals!
Amazing video thank you
Amazing video thank you
Amazing video thank you
Amazing video thank you for sharing
Thank you for providing this video. I was able to grasp normal surfaces easier after watching it.
Hola Steven Fowler, pimer comentario¡¡, dame un saludito :3, te gustaria dejarme un comentario? :3,
I am a recent graduate from a university majoring in mechanical engineering. I want you to help us and solve questions about engineering fundamentals in mechanical fe exams such as materials. Mechanics, thermodynamic fluids, resistance to mechanical design materials, etc.
Hi, I am not really set up to do review for an FE exam. You are more than welcome to all of the videos on my channel! My entire summer thermodynamics course is here. Also, if you search my channel for "review" you will find my review videos for Thermodynamics and Fluid Mechanics. Hopefully those help. If you need more, email me, and we can try to find you some resources.
@@professorstevenfowler can I communicate with you? Facebook what's up anything
bookerystore.com/downloads/fe-electrical-and-computer-practice-problems/
does Gibbs equation only work for ideal gases?
Gibbs Equation in its raw differential form will work for any substance, but the versions of Gibbs equations where we have used Ideal Gas Law to simplify into an algebraic expression can only be used for ideal gasses
I thought an easier way to find T_L was to use the Q_L we found then put that into Q_H=Q_L+W_in then use (Q_H/Q_L)=(T_H/T_L). I figured algebra was much easier to do it that way and got the same answer.
yes, this will work IFF you have a reversible situation
this review serves as a great formula sheet for the final exam! Thanks for posting , from UBC School of Engineering!!
Hey I'm really glad I could help. I wanted something positive to come out of the challenges of this year! Let us know how your final went!
When calculating the boundary work out, you have it as c/3 but then only multiplied by the full c. Was that an accident or did the 1/3 go somewhere that I didn't catch?
Yes you are exactly right! that's a good catch. The final answer should be divided by 3.
at the beginning of the video you show the previous example. How did you get .175 for your c_v avg?
That's because our average temperature is 350 F. At 300 F cv is 0.174 and at 400 F cv is 0.176. So at 350 F we would have cv = 0.175. This comes from Table A-2E (b) in our text. Table A-2 goes on for three pages (a, b, and c). This value comes from the second page (b). Does this help?
Steven Fowler yes that helps, thank you
you used v at 20* instead of 30*, I got v=0.04582
yes you are absolutely right!
thank you professor!
You're very welcome!
I got 94214 for the sum of the moment. 6399 + 87816 = 94215 . I need extra points for this :)
I'll take a look, no promises though :-)
I see what happened, the alarm went off before I typed the second 9 so it was 87816 + 639. Good catch!
Instead of dealing with the fractions of Th and Tl and COP.. can we use the fraction relationship (Qh/Ql) = (Th/Tl)?
If you have a reversible cycle and you use absolute temperatures, yes!