Does this negate the use of the usual sign convention generally used on this channel, where counterclockwise rotation results in positive value? At 6:41, it shows that all the FEM values are positive regardless of its position, whereas in previous lectures it is commonly known that the moment on the left side of a FEM analysis is deemed to be going in CCW, thus the positive value; and the right side to be going in CW; thus the negative value.
We have three different/independent sign conventions, used for different purposes. 1. When writing the static equilibrium equations, we choose a Cartesian coordinate system in which a direction along the x, y, and z axes, and a rotation along these axes is selected to be positive. This choice is completely arbitrary. For example, we can assume either a clockwise or a counterclockwise rotation to be positive. 2. The beam sign convention. This is associated with a pair of forces, not a single force. For example, bending moment in a beam segment is considered positive when the pair of moments acting at the ends of the segment cause it to deflect downward (concave up). In this scenario, the bending moment at the left end of the segment has to act in the clockwise direction and the moment at the right end of the segment must be in the counterclockwise direction for the beam to bend concave up. 3. When formulating a method, such as the slope-deflection method, we assume a positive direction for each member end moments. For example, we can assume counterclockwise direction to be positive for the member-end moments. This is similar to the assumption that we make when analyzing trusses. We assume all the unknown member forces to be in the tensile direction. Strictly speaking, these sign conventions could be viewed contradictory. For example, in (1) above, we have to decide whether clockwise or counterclockwise bending is positive. In (2) however, a pair of clockwise-counterclockwise moments is considered positive. But this does not matter since these conventions are independent of each other, each used for a different purpose. In this video, the moments shown at 6:41 are following sign convention 2 above. The moments labeled Ma, Mb, and Mc are not fixed-end moments, they are the actual bending moments in the beam at points A, B, and C. They are placed on the beam to suggest a concave up bending of each beam segment. Since these moments are unknown, we assume them to be in the positive direction. As for the fixed-end moments, by definition, they are the support reactions of a fixed-fixed beam. If the load is downward, then the moment reaction at the left end of the beam has to be counterclockwise and the moment reaction at the right end of the beam has to be in the clockwise direction. The video shows them in these directions. If the applied load was upward, these moments would have been placed in the opposite direction. When a beam is subjected to a downward load, causing a downward deflection, we say bending moment in the beam is positive. Meaning, the internal moments that cause the deflection are acting in the positive direction per item (2) above (clockwise at the left end and counterclockwise at the right). These moments are action (not reaction) forces. In contrast, the fixed-end moments are reaction forces, meaning they act in the opposite direction to the action forces. So, for the said beam, the reaction moment at the left end has to be placed in the counterclockwise direction whereas the reaction moment at the right end of the beam should be placed in the clockwise direction.
Adobe Illustrator for writing/drawing the lecture notes, and for animating them. MathType for writing equation. SketchUp and/or Cinema4D for creating and animating 3D models. Iclone for creating and animating the avatar. Camtasia Studio for synchronizing audio and video, and creating the final video.
The more I read your presentation, the more I got confused with the rotational sign on the support. If you look at support C , both rotation sign go downward. Whereas the pdf copy provided via the link on page 3, Figure 4 are upward. Which one is correct?
The diagram shown in Figure 4 in the lecture notes is the same diagram shown in the video @1:16. The moment directions are identical in both diagrams. Please be more specific about the source of your confusion.
The moment diagram can be represented using quadratic equation of the form: M(x) =a x^2 + b x + c Bending moment at x = 0 and x = L is zero, and the moment reaches its maximum height (say, P) at x=L/2. We can use these conditions to find the coefficients a, b, and c. Since the curve passes through the origin (0,0), we know c=0. The moment equation at x = L can be written as: M(L) = a L^2 + b L = 0. Use the fact that the maximum height P is reached at x=L/2, we can write: M(L/2) = a (L/2)(L/2) + b (L/2) = P Solving the two equations in (2) and (3), we get: a = -4P/(L^2), and b = 4P/L. Substituting these values in the moment equation, we get: M(x) = -4P/(L^2) x^2 + 4P/L x To determine the area under M(x), we can integrate M(x) from 0 to L. This integral equals 2PL/3.
In the starting segment of the video, an example of the three-moment equation is provided without giving any derivation details. The rationale behind that example is to show how a three-moment equation looks like. The rest of the video provides the necessary explanation for deriving such equations. After watching the video, you should be able to apply the general form of the three-moment equation to that starting example in order to arrive at -3PL/8.
Hi, is there a way of solving for the fixed end moment of various type of loading? i know you gave the values for a u.d.l and point load but will like to know how to derive those instead of lifting values from textbooks. Thank you
Dr Structure i recall there used to be some formula for finding the Centroid of triangle from either end . Like if the perpendicular drawn divided base as "a" and "b" then when we talked about location from "b" end we said directly something like (b+L)/3. where L=a+b ,, i don't remember exactly but i need your help what was it actually. THANKS
I am not sure if this is what you are looking for, but… Suppose the right triangle has corner points (a,0), (b,0), and (b,h) where h is the height of the triangle at the far end. Then, the geometric center of the area can be written as x = (a + b + b)/3 = (a + 2b)/3 and y = (0 + 0 + h)/3 = h/3. That is, the coordinates of the centroid in the average of the coordinates of the three corner points.
@@DrStructure i apologise that i said centroid rather i was asking about centre of Area of triangle and thus its location from either of horizontal vertical . More precisely i wanted short formula of ( X bar ). Say we have a triangle ABC and A&C lying on horizontal line while B is vertical vertex . If i draw perpendicular from B on AC intersecting AC at point m (say) .if Am='a' and Cm= 'b' then if if write X coordinate of center of Area of Triangle ABC from horizontal vertex A Is it ( a+L)/3. ??? ..... Where L=a+b(length of horizontal base of ∆) Ans similarly from C =(b+L)/3. ???
The example at the beginning of the lecture is meant to show the process without any derivation. The derivation for the three-moment equation is given afterward. According to the three-moment equation (see 5:50), the middle term coefficient is 2 times the sum of the (L/I) for each segment. In this case, since L is the same for both segments and they both have the same moment of inertia (I), the coefficient for the middle term simplifies to 2( 1 + 1) = 4.
@@tusharkumarkumawat8741 A negative sign on the rhs of the equation is linked to the sign convention that we use for bending moment. In this presentation, we've assumed a bending moment that causing a concave up deflection of the beam segment to be positive. If we assume the alternative, that is, if we assume bending moment is positive when the beam deflection is concave down, the rhs of the equation would become positive.
If I understand your question correctly, you are asking about the software that we've used to create the talking 3D character. It was done using Reallusion Character Creator and iClone software tools.
The three-moment equation is for calculating member forces in an indeterminate beam only. Once the structure has been analyzed and the support reactions have been calculated, techniques such as the conjugate beam or the virtual work can be used to calculate deflections.
There are two equations @12:20: 32 Mb + 8 Mc = -750 and 8 Mb + 36 Mc = -4800. We need to solve them simultaneously for Mb and Mc. If we do, we get: Mb = 10.48 kN and Mc = -135.66 kN.
At the beginning of the lecture the three-moment equation for a beam is given as a way of letting you know what such an equation looks like, without showing how it was obtained/derived. The derivation for the general form of the three-moment equation follows that example. After reviewing the entire lecture, go back to that initial example and see if you can derive that three-moment equation.
The FEM version of the equation is very handy. Thanks!!!
you are doing a great job. Thank you very much
Does this negate the use of the usual sign convention generally used on this channel, where counterclockwise rotation results in positive value?
At 6:41, it shows that all the FEM values are positive regardless of its position, whereas in previous lectures it is commonly known that the moment on the left side of a FEM analysis is deemed to be going in CCW, thus the positive value; and the right side to be going in CW; thus the negative value.
6:20 to be precise, there it shows that all FEM values are positive regardless of its rotation.
We have three different/independent sign conventions, used for different purposes.
1. When writing the static equilibrium equations, we choose a Cartesian coordinate system in which a direction along the x, y, and z axes, and a rotation along these axes is selected to be positive. This choice is completely arbitrary. For example, we can assume either a clockwise or a counterclockwise rotation to be positive.
2. The beam sign convention. This is associated with a pair of forces, not a single force. For example, bending moment in a beam segment is considered positive when the pair of moments acting at the ends of the segment cause it to deflect downward (concave up). In this scenario, the bending moment at the left end of the segment has to act in the clockwise direction and the moment at the right end of the segment must be in the counterclockwise direction for the beam to bend concave up.
3. When formulating a method, such as the slope-deflection method, we assume a positive direction for each member end moments. For example, we can assume counterclockwise direction to be positive for the member-end moments. This is similar to the assumption that we make when analyzing trusses. We assume all the unknown member forces to be in the tensile direction.
Strictly speaking, these sign conventions could be viewed contradictory. For example, in (1) above, we have to decide whether clockwise or counterclockwise bending is positive. In (2) however, a pair of clockwise-counterclockwise moments is considered positive. But this does not matter since these conventions are independent of each other, each used for a different purpose.
In this video, the moments shown at 6:41 are following sign convention 2 above. The moments labeled Ma, Mb, and Mc are not fixed-end moments, they are the actual bending moments in the beam at points A, B, and C. They are placed on the beam to suggest a concave up bending of each beam segment. Since these moments are unknown, we assume them to be in the positive direction.
As for the fixed-end moments, by definition, they are the support reactions of a fixed-fixed beam. If the load is downward, then the moment reaction at the left end of the beam has to be counterclockwise and the moment reaction at the right end of the beam has to be in the clockwise direction. The video shows them in these directions. If the applied load was upward, these moments would have been placed in the opposite direction.
When a beam is subjected to a downward load, causing a downward deflection, we say bending moment in the beam is positive. Meaning, the internal moments that cause the deflection are acting in the positive direction per item (2) above (clockwise at the left end and counterclockwise at the right). These moments are action (not reaction) forces. In contrast, the fixed-end moments are reaction forces, meaning they act in the opposite direction to the action forces. So, for the said beam, the reaction moment at the left end has to be placed in the counterclockwise direction whereas the reaction moment at the right end of the beam should be placed in the clockwise direction.
Thanks for this video.I get it now
What are the software's required to make these kind of animated video lecturer.. please tell sir/madam.
Adobe Illustrator for writing/drawing the lecture notes, and for animating them.
MathType for writing equation.
SketchUp and/or Cinema4D for creating and animating 3D models.
Iclone for creating and animating the avatar.
Camtasia Studio for synchronizing audio and video, and creating the final video.
The more I read your presentation, the more I got confused with the rotational sign on the support. If you look at support C , both rotation sign go downward. Whereas the pdf copy provided via the link on page 3, Figure 4 are upward. Which one is correct?
The diagram shown in Figure 4 in the lecture notes is the same diagram shown in the video @1:16. The moment directions are identical in both diagrams.
Please be more specific about the source of your confusion.
the formulation for Aab is 2/3 x base x height, from where do you get those formula? can you explain ?
The moment diagram can be represented using quadratic equation of the form:
M(x) =a x^2 + b x + c
Bending moment at x = 0 and x = L is zero, and the moment reaches its maximum height (say, P) at x=L/2. We can use these conditions to find the coefficients a, b, and c.
Since the curve passes through the origin (0,0), we know c=0.
The moment equation at x = L can be written as: M(L) = a L^2 + b L = 0.
Use the fact that the maximum height P is reached at x=L/2, we can write:
M(L/2) = a (L/2)(L/2) + b (L/2) = P
Solving the two equations in (2) and (3), we get: a = -4P/(L^2), and b = 4P/L.
Substituting these values in the moment equation, we get:
M(x) = -4P/(L^2) x^2 + 4P/L x
To determine the area under M(x), we can integrate M(x) from 0 to L.
This integral equals 2PL/3.
Thx a lot, you're doing great@@DrStructure
Ma'am will u please illustrate from where this -3pl/8 have came???
In the starting segment of the video, an example of the three-moment equation is provided without giving any derivation details. The rationale behind that example is to show how a three-moment equation looks like. The rest of the video provides the necessary explanation for deriving such equations. After watching the video, you should be able to apply the general form of the three-moment equation to that starting example in order to arrive at -3PL/8.
so touching for an excellent video
Thank you very much 🥳
Hi, is there a way of solving for the fixed end moment of various type of loading? i know you gave the values for a u.d.l and point load but will like to know how to derive those instead of lifting values from textbooks.
Thank you
Take a look at this lecture:
ua-cam.com/video/Z_jxaKQirC0/v-deo.html
Dr Structure i recall there used to be some formula for finding the Centroid of triangle from either end .
Like if the perpendicular drawn divided base as "a" and "b" then when we talked about location from "b" end we said directly something like (b+L)/3. where L=a+b ,, i don't remember exactly but i need your help what was it actually. THANKS
I am not sure if this is what you are looking for, but… Suppose the right triangle has corner points (a,0), (b,0), and (b,h) where h is the height of the triangle at the far end. Then, the geometric center of the area can be written as x = (a + b + b)/3 = (a + 2b)/3 and y = (0 + 0 + h)/3 = h/3. That is, the coordinates of the centroid in the average of the coordinates of the three corner points.
@@DrStructure i apologise that i said centroid rather i was asking about centre of Area of triangle and thus its location from either of horizontal vertical . More precisely i wanted short formula of ( X bar ).
Say we have a triangle ABC and A&C lying on horizontal line while
B is vertical vertex . If i draw perpendicular from B on AC intersecting AC at point m (say) .if Am='a' and Cm= 'b' then if if write X coordinate of center of Area of Triangle ABC from horizontal vertex A
Is it ( a+L)/3. ??? ..... Where L=a+b(length of horizontal base of ∆)
Ans similarly from C =(b+L)/3. ???
Yes, your equations are correct.
Can anyone please explain why is it 4 Mb at 0:39
The example at the beginning of the lecture is meant to show the process without any derivation. The derivation for the three-moment equation is given afterward. According to the three-moment equation (see 5:50), the middle term coefficient is 2 times the sum of the (L/I) for each segment. In this case, since L is the same for both segments and they both have the same moment of inertia (I), the coefficient for the middle term simplifies to 2( 1 + 1) = 4.
what if the two ends are fixed in a 3-span beam? will there be 4 equations and 4 unknowns?
Yes, in that case we will end up with a beam with five spans (the end spans being fictitious) resulting in four unknown moments and four equations.
Where you get 4MB? . Thanks😊
The equation is derived in the lecture, please watch it in its entirety.
Some write the RHS of the equation as (6a1.x1/L1) + (6a2.x2/L2) then why are putting minus sign before them?
Not quite following your question, please elaborate. +/- signs generally depend on the assumed sign convention and the rules of algebra.
@@DrStructure he mean to say that some have +terms in rhs while some have in - sign. I also have same doubt plz clear
@@tusharkumarkumawat8741 A negative sign on the rhs of the equation is linked to the sign convention that we use for bending moment. In this presentation, we've assumed a bending moment that causing a concave up deflection of the beam segment to be positive. If we assume the alternative, that is, if we assume bending moment is positive when the beam deflection is concave down, the rhs of the equation would become positive.
@@DrStructure thanx for replying and clearing up the doubt
@@tusharkumarkumawat8741 You're welcome.
Hola Buenas tardes; me podrían dejar un comentario de donde editan a las animaciones que salen en el video (las personas que hablan ) .Muchas gracias
If I understand your question correctly, you are asking about the software that we've used to create the talking 3D character. It was done using Reallusion Character Creator and iClone software tools.
where are the solutions for the exercise problems?
The links are listed in the video description field.
thanks!!!!
I have tried to solve one problem from book with the first method... I am not getting the correct solution??
Feel free to email your solution to Dr.Structure@EducativeTechnologies.net, if you would like us to review your work.
Thank u sir/madam surely I will reply you back tomorrow with the solution.
Sir make a video on theorem of least work/strain energy method👍👍👍
i can't see the pdf file. :(
You should be able to access the pdf file here::
lab101.space/pdf/lectures/SA60.pdf
Thank you, Dr. Structure
Thank you Soo much..
But , i would like to know how we shall find the deflection using this equation..?😊😊🙏🙏
The three-moment equation is for calculating member forces in an indeterminate beam only. Once the structure has been analyzed and the support reactions have been calculated, techniques such as the conjugate beam or the virtual work can be used to calculate deflections.
Dear Dr. Structure:)
kindly post all videos related to three moment equation ASAP.
Thanks in advance.
i dont understand how
32Mb + 8Mc = -750
Mb = 10.48 kN
There are two equations @12:20:
32 Mb + 8 Mc = -750 and 8 Mb + 36 Mc = -4800.
We need to solve them simultaneously for Mb and Mc. If we do, we get: Mb = 10.48 kN and Mc = -135.66 kN.
Thinks my dear but why m at b equal 4m
At the beginning of the lecture the three-moment equation for a beam is given as a way of letting you know what such an equation looks like, without showing how it was obtained/derived. The derivation for the general form of the three-moment equation follows that example. After reviewing the entire lecture, go back to that initial example and see if you can derive that three-moment equation.
@@DrStructure thinks my dear to explain this your great thinks again
@@DrStructure my dear can add explain with Arabic language in all UA-cam thinks
We are adding various subtitles to the videos including Arabic. It just takes some time to get all of them done.
@@DrStructure thinks my dear to your answer