sir your lectures are so amazing and so knowledgable and in depth that i love the most these are the best. i request you to please make lectures on Design of Reinforced Concrete Structures thanks
Your work is fantastic, in-depth and helpful for students . Thanks for that . Could you please make videos on 2 hinged arches of ist order and 3rd order indeterminacy
Awesome and incredible channel. All can easily understand about structure related math. Hope you make more and full vedio about structure. Best wishes for your success
Your all lectures is so amazing to realize. For all sfd and bmd of three hinged arch is same as curve even for single point load or triangular load??Thanks.
Different loads result in different arch free-body diagrams, therefore, the force equations (and their graphs) would be different than the example given in the lecture.
Dr. Structure, First of all i appreciate ur work. I want some videos on the " How to draw influence line diagram in INDETERMINATE structure for reaction, shear and moment".
Each type of arch has its advantages and disadvantages. For example, a 3-hinged arch is not as rigid as the other types of arches. Consequently, it is less sensitive to support settlement. That is, while a hinge-less (more rigid) arch would undergo additional stresses, if a support settles due to, say, weak soil, the ability of the hinges in a 3-hinged arch to rotate would prevent the development of those additional stresses in the two arch legs.
It very amazing work.thank you a lot I'm a teaching assistant and I want to make some lectures for my student by the same way ,but I don't know how?,so I hope you help me . How I do it or what're the softwares which can help me too. .
Thanks for the note. We use several software tools for creating the lectures. The lecture notes are created/written in Adobe Illustrator, and animated using an Illustrator script. Additional animation, when needed, is done using Adobe Animate. Character animation is done using Iclone. Then the entire video is put together and video and audio is synchronized using Camtasia Studio.
If we label the axes along H and R as x and y, and label the axes along N and V as x’ and y’, then we can see that H has a component along x’ and y’. And, R has a component along x’ and y’. Note that the x’-y’ coordinate system is obtained by rotating the x-y system counterclockwise by angle theta. The component of H along x’ is H cos theta. The component of H along y’ is -H sin theta. The component of R along x’ is R sine theta. And the y’ component of R is R cos theta. Then the total force (due to H and R) along x’ is H cos theta + R sin theta. And the total force along y’ is R cos theta - H sine theta.
No, a flat arch (arch at the bottom, flat on the top) has a variable cross-section which would result in a change in the distributed load pattern and the geometric properties of the system. These changes are not consistent with the assumptions used for analyzing regular (circular or parabolic) arches with a constant cross-section.
There are not that many textbooks written exclusively on arches. I can suggest Theory of Arched Structures by Igor Karnovsky. But I am not certain that is what you are looking for. You can find the table of contents of the book and some of its pages on Amazon. Take a look and see if that suits you. You may also be able to find some older textbooks on the subject. As for youtube or website, I am not aware of any that covers the topic in a comprehensive way. It seems you are looking for a substantive online course on the topic. As far as I can tell, none exist. If you come across one, let us know.
You can use wolfram alpha (www.wolframalpha.com) to perform the integration. In the input field type: Derivative of (198-1.92*x)/sqrt(1+(0.8-0.032*x)^2 Then click on the equal sign on the right side of the input field. Set the resulting equation equal zero and simplify the expression. You should get: 1875 - 150 x.
There are a few online resources that could provide the information you are looking for, or guide your in the right direction. Here is one: www.concretecentre.com/
H has a component along the N-axis and a component along the V-axis. Since H is a horizontal force that makes angle theta with the N-axis, its projection onto the axis is: H cos(theta). H also has a projection along the V-axis, it is H sin(theta). This projection force, however, is going to be in the direction opposite to what is shown on the diagram for the V-axis. The arrow for the axis is pointing downward, meaning we are assuming downward to be positive alone the V-axis. Since the projection of H onto the V-axis is going to be upward, we attach a negative sign to it. That is, we write the projection of H onto the V-axis as -H sin(theta). The vertical force R also has a projection along the N and V axes. R makes angle theta with the V-axis. Therefore, its component along the V-axis is: R cos(theta), and the projection of R onto N-axis is R sin(theta). Note that these projections are acting in the positive directions of the N and V axis, hence no negative sign needs to be used to label either of these components. Since H and R each has a component along the N and V axes, the total force along N and V is the sum of the two components. For N we get H cos(theta) + R sin(theta). For V, we get -H sin(theta) + R cos(theta).
Unpopular opinion: The good thing about this channel is that we can also get the pdf version of notes. Thank you soo much Dr. Structure.
Amazing work. Its quite easy to understand when someone explains with this depth. Keep updating us !!!
Man i was searching for mobile tutor and it is here.... really good👌👌🥰
Thanks for clarifying the concept.
sir
your lectures are so amazing and so knowledgable and in depth that i love the most these are the best.
i request you to please make lectures on Design of Reinforced Concrete Structures
thanks
Your work is fantastic, in-depth and helpful for students . Thanks for that . Could you please make videos on 2 hinged arches of ist order and 3rd order indeterminacy
Awesome and incredible channel. All can easily understand about structure related math. Hope you make more and full vedio about structure. Best wishes for your success
Thank you for your feedback.
Amazing work
Concise and clear, thanks.
Your all lectures is so amazing to realize. For all sfd and bmd of three hinged arch is same as curve even for single point load or triangular load??Thanks.
Different loads result in different arch free-body diagrams, therefore, the force equations (and their graphs) would be different than the example given in the lecture.
Dr. Structure thanks for the reply
You're welcome!
Dr. Structure, First of all i appreciate ur work. I want some videos on the " How to draw influence line diagram in INDETERMINATE structure for reaction, shear and moment".
*God of structure*
Fabulous 🎁
Very very helpfull,I want to know what the concept and pylosophy the 3 hinged arch,what the reason we suppose hing in midspan?
Each type of arch has its advantages and disadvantages. For example, a 3-hinged arch is not as rigid as the other types of arches. Consequently, it is less sensitive to support settlement. That is, while a hinge-less (more rigid) arch would undergo additional stresses, if a support settles due to, say, weak soil, the ability of the hinges in a 3-hinged arch to rotate would prevent the development of those additional stresses in the two arch legs.
from jordan ,thank you
great channel and great videos❤
It very amazing work.thank you a lot
I'm a teaching assistant and I want to make some lectures for my student by the same way ,but I don't know how?,so I hope you help me .
How I do it or what're the softwares which can help me too. .
This is my email:
islam.engineer91@gmail.com
Thanks for the note. We use several software tools for creating the lectures. The lecture notes are created/written in Adobe Illustrator, and animated using an Illustrator script. Additional animation, when needed, is done using Adobe Animate. Character animation is done using Iclone. Then the entire video is put together and video and audio is synchronized using Camtasia Studio.
Thank you Dr. Structure 🤗😍
I am confused with the way you obtain the N and V In terms of H ,R, Sine and cosine
If we label the axes along H and R as x and y, and label the axes along N and V as x’ and y’, then we can see that H has a component along x’ and y’.
And, R has a component along x’ and y’.
Note that the x’-y’ coordinate system is obtained by rotating the x-y system counterclockwise by angle theta.
The component of H along x’ is H cos theta. The component of H along y’ is -H sin theta.
The component of R along x’ is R sine theta. And the y’ component of R is R cos theta.
Then the total force (due to H and R) along x’ is H cos theta + R sin theta. And the total force along y’ is R cos theta - H sine theta.
Thanks sir
Will you please upload two hinged arch.
For a flat arch does the same method applies? as thrust would be extremely high. Thanks
No, a flat arch (arch at the bottom, flat on the top) has a variable cross-section which would result in a change in the distributed load pattern and the geometric properties of the system. These changes are not consistent with the assumptions used for analyzing regular (circular or parabolic) arches with a constant cross-section.
thank you ….
can you explain me..why thrust daigram is parabolic where as shear daigram is linear??
The shear diagram, when drawn, looks linear, but it is not actually a linear function. The function is given @9:48.
@@DrStructure I want complete information about 3 hinged arches ..pls suggest you tube channel or website or a text book ..plss!!
There are not that many textbooks written exclusively on arches. I can suggest Theory of Arched Structures by Igor Karnovsky. But I am not certain that is what you are looking for. You can find the table of contents of the book and some of its pages on Amazon. Take a look and see if that suits you. You may also be able to find some older textbooks on the subject.
As for youtube or website, I am not aware of any that covers the topic in a comprehensive way. It seems you are looking for a substantive online course on the topic. As far as I can tell, none exist. If you come across one, let us know.
waiting for fem
Nice 1
Can you explain how to get the result 1875-150x=0
That is the simplified expression for dN/dx = 0. Are you getting a different value for the expression?
Oh okay @@DrStructure, maybe I still don't understand, but if you don't mind, can you share the long method to find dN/dx=0
You can use wolfram alpha (www.wolframalpha.com) to perform the integration.
In the input field type:
Derivative of (198-1.92*x)/sqrt(1+(0.8-0.032*x)^2
Then click on the equal sign on the right side of the input field.
Set the resulting equation equal zero and simplify the expression. You should get: 1875 - 150 x.
I want some information and latest issues related to light weight concrete. Plzzz help me out
There are a few online resources that could provide the information you are looking for, or guide your in the right direction. Here is one: www.concretecentre.com/
Dr. Structure thank you sir plzzzz suggest me 2 more sites so that I can work on my project.
how to get 62.5?
The equation for f(x) is (4 h x/ L^2)(L - x). When L = 50 and h = 10, the equation simplified to (50 x - x^2)/62.5
At 6:23 how N and V have derived.
H has a component along the N-axis and a component along the V-axis. Since H is a horizontal force that makes angle theta with the N-axis, its projection onto the axis is: H cos(theta). H also has a projection along the V-axis, it is H sin(theta). This projection force, however, is going to be in the direction opposite to what is shown on the diagram for the V-axis. The arrow for the axis is pointing downward, meaning we are assuming downward to be positive alone the V-axis. Since the projection of H onto the V-axis is going to be upward, we attach a negative sign to it. That is, we write the projection of H onto the V-axis as -H sin(theta).
The vertical force R also has a projection along the N and V axes. R makes angle theta with the V-axis. Therefore, its component along the V-axis is: R cos(theta), and the projection of R onto N-axis is R sin(theta). Note that these projections are acting in the positive directions of the N and V axis, hence no negative sign needs to be used to label either of these components.
Since H and R each has a component along the N and V axes, the total force along N and V is the sum of the two components. For N we get H cos(theta) + R sin(theta). For V, we get -H sin(theta) + R cos(theta).
Thank you dr Structure, you are an amazing teacher.
@@stanzinnorboo7083 You're welcome, and thank you!
Sorry for again disturbing you, i want to ask , will you please make a video on two hinge arch and arch with different support level?😀
@@stanzinnorboo7083 Eventually we will cover those topics, the when of it however is not known at this point.