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Quest Zero
India
Приєднався 25 кві 2020
Rayleigh Ritz Method - Simply supported beam with Point Load @ centre and Uniformly Distributed Load
Obtaining the solution for maximum deflection and general relation for deflection of a simply supported beam subjected to point load at centre and uniformly distributed load throughout the length of beam by Rayleigh Ritz Method is discussed. Trigonometrical trial function assumption is made for obtaining the solution
Переглядів: 4 586
Відео
Projection of Planes - 09
Переглядів 463 роки тому
Problem statement : A hexagonal lamina of sides 25 mm is rests on one of its sides on VP. The lamina makes 45 deg to VP and the sides on which the lamina rests makes 45 deg to HP. Draw its projections. Don't forget to Like, Share and Subscribe
Projection of Planes - 08
Переглядів 1623 роки тому
In this session the top view and front view of a pentagonal lamina resting on one of its corner on HP and side opposite to resting corner making an angle of 45 deg to VP and being 20 mm above HP is discussed. The projections are in first angle.
Projection of Planes - 07
Переглядів 2493 роки тому
In this session a pentagonal lamina is considered. It's projected front view and top view when resting on HP on one of its edges, entire lamina being inclined at 60 deg to HP and resting edge inclined at 45 deg to VP is obtained in 1st angle of projection
Projection of Planes - 06
Переглядів 3,1 тис.3 роки тому
In this session a rectangular lamina resting on VP with one of its shorter edge, seen as a square in front view and resting edge inclined at 30 deg to HP is discussed. The solution is obtained in First angle of Projection
Projection of Planes - 05
Переглядів 1043 роки тому
In this session a rectangular lamina resting on HP on one of its shorter edges and inclined at 45 deg to HP and resting edge inclined at 30 deg to VP is considered. Solution is obtained in First angle of projection
Projection of Planes - 04
Переглядів 813 роки тому
In this session, the solution to a square lamina resting on HP on one of its side and making 30 deg to HP and side resting on HP making 45 deg to VP is obtained by change of position method. All solutions are in first angle of projection.
Projection of Planes - 03
Переглядів 1003 роки тому
In this session, solution to projection of an Isosceles triangle lying on HP, seen as an equilateral triangle in Front view and edge parallel to VP being inclined at 45 deg to HP is solved. The solution in in 1st angle of projection. Exact inclination of the surface with respect to VP is 52 deg.
Projection of Planes - 02
Переглядів 913 роки тому
Engineering Graphics in First Angle of Projection Projection of Planes 02 In this session projection of an equilateral triangular lamina resting on HP on one its corner is discussed. The edge opposite to the resting corner is specified distance above HP and makes specified angle with VP.
Projection of Planes - 01
Переглядів 1143 роки тому
Engineering Graphics - Projection of Planes in 1st angle of projection In this session the basic planar lamina - equilateral triangular lamina, is considered for projection. The triangle lies on HP on one of its edge, the surface makes 60 deg to HP and the resting edge makes 60 deg to VP
Projection of Lines - 05
Переглядів 1223 роки тому
Solution to projection of lines in 1st angle of projection. Step by step construction for a given problem is explained.
Projection of Lines - 04
Переглядів 1553 роки тому
Solution to Projection of Lines in 1st angle of projection. Data pertaining to projection length of lines in front view and top view, true length of line is only provided. Construction is made to obtain solution of true and apparent angle of inclination in both front and top views. My apology for the bad audio
Projection of Lines - 03
Переглядів 1973 роки тому
Engineering Graphics/Drawing in 1st Angle of Projection, where a length of line, distance from HP and VP of one of its end, and true inclination of line with HP and VP are given. The solution of projection of given line and value of apparent length and inclinations are found.
Projection of Lines - 02
Переглядів 2373 роки тому
Projection of a line whose length and distance between the end projectors is given. The true inclination of line wrt to VP is known and solution to find the true inclination to HP is obtained. All drawings are in 1st angle of projection.
Projection of Lines - 01
Переглядів 1893 роки тому
Projection of Lines from Engineering Graphics/ Engineering Drawing in 1st angle of Projection. In this session a line inclined to both HP and VP and one end distance from both projection planes is defined. Solution to obtain the projection length and angle is solved.
Galerkin Method - Steps and Bar subjected to Uniformly Distributed Load
Переглядів 9 тис.3 роки тому
Galerkin Method - Steps and Bar subjected to Uniformly Distributed Load
Rayleigh Ritz Method - Cantilever Beam subjected to Uniformly Distributed Load
Переглядів 2,8 тис.3 роки тому
Rayleigh Ritz Method - Cantilever Beam subjected to Uniformly Distributed Load
Rayleigh Ritz Method - Cantilever Beam subjected to point load at free end
Переглядів 7 тис.3 роки тому
Rayleigh Ritz Method - Cantilever Beam subjected to point load at free end
Rayleigh Ritz Method - Simply supported beam with uniformly distributed load
Переглядів 4,1 тис.3 роки тому
Rayleigh Ritz Method - Simply supported beam with uniformly distributed load
Rayleigh Ritz Method - Simply Supported beam using Trigonometric & Polynomial Approximation
Переглядів 4,3 тис.3 роки тому
Rayleigh Ritz Method - Simply Supported beam using Trigonometric & Polynomial Approximation
Rayleigh Ritz Method - Bar with change in cross-sectional area
Переглядів 2,2 тис.3 роки тому
Rayleigh Ritz Method - Bar with change in cross-sectional area
Rayleigh Ritz Method - Numerical on Bars with Uniformly Distributed Load
Переглядів 2,6 тис.3 роки тому
Rayleigh Ritz Method - Numerical on Bars with Uniformly Distributed Load
Rayleigh Ritz Method - Steps and Numerical on Bars with Point Load
Переглядів 4,3 тис.3 роки тому
Rayleigh Ritz Method - Steps and Numerical on Bars with Point Load
Block Diagram Reduction (Control System Engineering) using GNU OCTAVE/MATLAB - Example 1
Переглядів 2,6 тис.3 роки тому
Block Diagram Reduction (Control System Engineering) using GNU OCTAVE/MATLAB - Example 1
Reduction of Blocks Connected in Series, Parallel and Feedback Loop using GNU OCTAVE/MATLAB
Переглядів 4023 роки тому
Reduction of Blocks Connected in Series, Parallel and Feedback Loop using GNU OCTAVE/MATLAB
Creating Transfer Functions in GNU OCTAVE/MATLAB
Переглядів 7 тис.3 роки тому
Creating Transfer Functions in GNU OCTAVE/MATLAB
Installation of GNU Octave in Windows 10
Переглядів 9 тис.4 роки тому
Installation of GNU Octave in Windows 10
Surely I am not alone in the struggle to find online, any singular bit of somewhat helpful reference material, for the widely used "Advanced Mechanics of Materials" - Cook & Young. Found your video an hour ago, and was compelled enough to leave a comment. Thank you very much!
Bless you
Excellent vdeo
Please explain in imperial.
How to calaculate support reactions in this question?
Thank you bro very helpul
Well explained
very nice explained sir.................
Ths sir pls tell engineering Finite element method course
Sir,I have a doubt. In the last step when we have to find the stress. We have to find it at the midpoint of bar,i.e x=1,which gives the answer 0. However, you have substituted,x=0. Kindly clarify
Generally we find stress at support, as it will be maximum there, hence at x=0 is considered
Neatly Done ✅👍👍
thank u
Clarísimo!!!! aunque esté en otro idioma
Thank you Mr.Quest
Thank you, Sir
Sir what is the trial displacement function for a fixed beam with udl and fixed beam with point load ??
Sir, how do we solve same problem using linear polynomial a0+a1x, as in this case using both boundary conditions coefficients a0 and a1 both will become zero
Assumption of polynomial degree should be such that boundary conditions are satisfied. 1 degree polynomial as mentioned by you will not be sufficient. Hence, solution cannot be obtained
Thanks Brother
Nice class sir🤩💫Hopefully waiting for Finite Element Analysis new topics🤍
The last step where you find the constant in the denominator, what did you plug in for Potential Energy (Pi) and why?
Good work thank you sir
sir. I think u frget to take -ve sign in substiting the value of u1 for PE functional which leads to inverse results at the end..could you plz notice and clarrify the same
Hi, thanks a lot for noticing. You are right, the negative sign is missed after the flipping the page over. Will try including the correction in description. My apologies for the mistake
@@questzero4205 thanks lot for your timely response and you are doing a great job.. keep going sir..
What is " P " here ? weight or what ? Please reply
Hello, "P" is an external Load/Force applied at a point on the free end of beam
@@questzero4205 Thank you
Yes
How did you get the approximate function or was it given?
They are normally given. Trigonometric assumptions are generally taken for beams
We take a trial solution..check with B.C's & declare it as approxumate solution
We can even take y= a1sin( πx/l )+ a2sin(3πx/l) this will give deflection almost same to the sum of deflection of point load and udl. Thanks for the video sir.
Home work
Can u pls do SS beam with both point load at centre nd udl through out the section using R-R method.
Thank you for Watching! Next video will be based on your request.
ua-cam.com/video/izAE2zsUQEo/v-deo.html
Thank you sir for immediate response
Thank you sir
Dear Sir, on behalf of myself and my wife I would like to thank you for this helpful video. You saved us a lot of time and hassle. With kind regards from Tel Aviv, Israel.
TYSM SIR
what are the uses of these transfer functions? can you do an example? thanks!
does downloading in octave in windows need to install image package or by default it is installed in it?? please reply me
It should be there by default. If not, install them from Octave Forge
Very informative lec sir ...
Good video sir superb Want guidance to run program
Very well explained...thank you
Very useful vedio for those who are using Google classroom thank you sir 🎀❤️