Good day Dr Structure Is the next SA video going to me on Moment area theorems, or its something else? Are you going to have the videos for hanging cables anytime soon? Can this SA 59 video have any settlements, if so, how do we involve them in the calculations?
The next SA lecture is going to be on the three-moment equation/theorem. We have the analysis of cable on the list, but I am unable to give you a reasonable time frame for its release at this point. Problems involving support settlement need to be addressed using analysis techniques such as the slope-deflection method or the displacement method (see SA32, for example). In such problems the objective is to determine internal forces (shear and moment) due to a settlement. The moment-area theorem deals with calculating displacements only, it cannot be used to determine internal forces, whether due to a settlement problem or otherwise.
We draw the moment diagram, then divide each segment of the diagram by its EI. When EI is not constant, we end up with a jumb/drop in the moment diagram at the intersection of the two beam segments, since for one segment we are dividing M by EI, and in the other segment the same M is divided by a different EI.
Hi ! i have a question, in the second problem when finding Theta D at the third page of the solution is: Determine the slope at D using the first moment-area theorem. θd -θc = area under the M/EI diagram between C and D The area is the Highlighted Triangle with a Base of 2 Height of -4. According to me this area is -4 units^2 but in the solution there is an extra value (2x2/3) wich seems like an arm to me against point D. So my question is: Where does this Term come from (2x2/3), since the First Moment-Area Theorem states that the difference in slopes between two points equals to the area under them ? Thank you !
Thanks for the question. You are correct, that term should not be there. The area under the shaded diagram is -4/EI which makes theta_D = 1.5/EI. We will make the correction and update the solution file accordingly.
@@DrStructure sir do you have some kind of portal where we can send doubts.. I was solving some questions of the m-a theorem and i got stuck on a que.. I cannot send picture of the que in utube comments .. Sir can u help ?
Dear Dr. Structure, in question 2 of the practice problems, you have drawn an elastic curve. Don't you think that its curvature should change at the point of inflection which is 1.09m left of point C?
Thank you for pointing out the issue with the elastic curve. Yes, the actual curvature of the elastic curve is different than the assumed curvature given in the solution. Of importance here is to note that even an incorrect assumption about the shape of the elastic curve yields the correct value for the vertical displacement at B. Using the correct shape of the elastic curve could have added a bit of complexity to the presentation that we felt was unnecessary for the problem at hand. However upon examining the solution, we found a calculation error which now has been corrected. The solution pdf file on courses.structure.education site has been updated accordingly.
I am not sure what you mean by social aspects of engineering. Although we don't have any pdf files to share outside the video lectures, I am certain there are ample books and articles out there that could address your learning needs. You should be able to generate a list of references around such topics via internet search.
In such a case, one first needs to find the slope at A (using the second moment-area theorem). Knowing the slope at A, the first moment-area theorem can be used to determine the slope at C.
@@DrStructure sir in the pdf solution of excercise que 2 .. The load of 6 kn acts at B which is the mid point of AC sir then can we calculate the slope of c taking 1st m-a theorem between C and B taking thitaB= 0?... But in the pdf solution u calculate deflection at A then divided by 6 to get thitaC that i understood but why cannot we use the 1st m-a theorem btwn C and B ?
@@rikdevghosh7324 No, that would not work. It is true that B is at the mid point of AC, but that does not automatically make Theta_B = 0. For Theta_B to be zero, the applied load has to be symmetrical about B. In this, if we remove the load at D, making the 6 kN load at B the only load on the system, then Theta_B becomes zero. Alternatively, if we add an overhang of 2m to left of A, then apply a load of 2 kN at the free-end of that overhang, which results in having the beam subjected to symmetrical loading about B, then Theta_B would be zero.
Slope at D is not 5.33/EI, it is 0.17/EI. Slope at D is calculated using the first moment-area theorem. The moment of the M/EI diagram between C and D equals to the difference between slopes at C and D. We can calculate the moment of the area under the M/EI diagram between C and D by first calculating the area. It equals the area of the part of the diagram shown in red. The height of the triangle is -4/EI, its base is 2, therefore, the area is (-4/EI)(2)/2. Now, multiply the area by the distance from the center of the area to point D in order to determine the moment of the area about D. The center of the right triangle, measured from D, is equal to 2/3 of the base of the triangle. That gives us (2/3)(2). Multiplying the two quantities, we get: (-4/EI)(2)(2/3)(2)/2 = -5.33/EI The above value, per first moment-area theorem, is the difference between the two slopes. And since we already have calculated the slope at C, we can determine the slope at D by adding the slope at C to -5.33/EI
I am so impressed from your lectures, concepts and your soft. applications Iframe & Itruss. God bless you with lot of knowledge and bounties.
I really love this channel
Good day Dr Structure
Is the next SA video going to me on Moment area theorems, or its something else?
Are you going to have the videos for hanging cables anytime soon?
Can this SA 59 video have any settlements, if so, how do we involve them in the calculations?
The next SA lecture is going to be on the three-moment equation/theorem.
We have the analysis of cable on the list, but I am unable to give you a reasonable time frame for its release at this point.
Problems involving support settlement need to be addressed using analysis techniques such as the slope-deflection method or the displacement method (see SA32, for example). In such problems the objective is to determine internal forces (shear and moment) due to a settlement. The moment-area theorem deals with calculating displacements only, it cannot be used to determine internal forces, whether due to a settlement problem or otherwise.
@@DrStructure oh ok i was thinking that the settlement might occur when there are deflections.
Will remain posted on the new upcoming videos
how do you construct the M/EI diagram of a beam with the given EI like the 1st question of the 2 questions given at the end of the lecture ?
We draw the moment diagram, then divide each segment of the diagram by its EI. When EI is not constant, we end up with a jumb/drop in the moment diagram at the intersection of the two beam segments, since for one segment we are dividing M by EI, and in the other segment the same M is divided by a different EI.
Hi ! i have a question, in the second problem when finding Theta D at the third page of the solution is:
Determine the slope at D using the first moment-area theorem.
θd -θc = area under the M/EI diagram between C and D
The area is the Highlighted Triangle with a Base of 2 Height of -4.
According to me this area is -4 units^2 but in the solution there is an extra value (2x2/3) wich seems like an arm to me against point D.
So my question is:
Where does this Term come from (2x2/3), since the First Moment-Area Theorem states that the difference in slopes between two points equals to the area under them ?
Thank you !
Thanks for the question. You are correct, that term should not be there. The area under the shaded diagram is -4/EI which makes theta_D = 1.5/EI. We will make the correction and update the solution file accordingly.
@@DrStructure sir do you have some kind of portal where we can send doubts.. I was solving some questions of the m-a theorem and i got stuck on a que.. I cannot send picture of the que in utube comments .. Sir can u help ?
@@rikdevghosh7324 You can send email to Dr.Structure@EducativeTechnologies.net
Dear Dr. Structure, in question 2 of the practice problems, you have drawn an elastic curve. Don't you think that its curvature should change at the point of inflection which is 1.09m left of point C?
Thank you for pointing out the issue with the elastic curve. Yes, the actual curvature of the elastic curve is different than the assumed curvature given in the solution. Of importance here is to note that even an incorrect assumption about the shape of the elastic curve yields the correct value for the vertical displacement at B. Using the correct shape of the elastic curve could have added a bit of complexity to the presentation that we felt was unnecessary for the problem at hand.
However upon examining the solution, we found a calculation error which now has been corrected. The solution pdf file on courses.structure.education site has been updated accordingly.
@@DrStructure Thank you so much dear Sir. You are doing wonderful. Keep it up!
Good day Dr Structure
Do you happen to have the solutions for the exercises in video format, or its now in pdf format only?
For now, they are in pdf format only.
@@DrStructure ok thank you
well done!
Are there any videos about moment-area method in this channel?
th-cam.com/video/q1KY4-5g6E0/w-d-xo.html
Will you send me a pdf file on social aspects of engineering please. Dr. Structure
I am not sure what you mean by social aspects of engineering. Although we don't have any pdf files to share outside the video lectures, I am certain there are ample books and articles out there that could address your learning needs. You should be able to generate a list of references around such topics via internet search.
In simply supported beam where you used 2nd method if there will b a point c at the distance 1m from a point A then how will find slop at c?
In such a case, one first needs to find the slope at A (using the second moment-area theorem). Knowing the slope at A, the first moment-area theorem can be used to determine the slope at C.
@@DrStructure sir in the pdf solution of excercise que 2 .. The load of 6 kn acts at B which is the mid point of AC sir then can we calculate the slope of c taking 1st m-a theorem between C and B taking thitaB= 0?... But in the pdf solution u calculate deflection at A then divided by 6 to get thitaC that i understood but why cannot we use the 1st m-a theorem btwn C and B ?
@@rikdevghosh7324 No, that would not work. It is true that B is at the mid point of AC, but that does not automatically make Theta_B = 0. For Theta_B to be zero, the applied load has to be symmetrical about B. In this, if we remove the load at D, making the 6 kN load at B the only load on the system, then Theta_B becomes zero. Alternatively, if we add an overhang of 2m to left of A, then apply a load of 2 kN at the free-end of that overhang, which results in having the beam subjected to symmetrical loading about B, then Theta_B would be zero.
For the question 2. I don't seem to understand how the slope at d was gotten. I would appreciate any help.
Is there any specific step in the pdf solution that you would like to be explained more/better?
Yes. @drstructure the slope at D = 5.33/EI
Slope at D is not 5.33/EI, it is 0.17/EI.
Slope at D is calculated using the first moment-area theorem. The moment of the M/EI diagram between C and D equals to the difference between slopes at C and D.
We can calculate the moment of the area under the M/EI diagram between C and D by first calculating the area. It equals the area of the part of the diagram shown in red. The height of the triangle is -4/EI, its base is 2, therefore, the area is (-4/EI)(2)/2. Now, multiply the area by the distance from the center of the area to point D in order to determine the moment of the area about D. The center of the right triangle, measured from D, is equal to 2/3 of the base of the triangle. That gives us (2/3)(2). Multiplying the two quantities, we get: (-4/EI)(2)(2/3)(2)/2 = -5.33/EI
The above value, per first moment-area theorem, is the difference between the two slopes. And since we already have calculated the slope at C, we can determine the slope at D by adding the slope at C to -5.33/EI
Plz plz send me the solution of the last 2 question given in the end...i tried my level best but couldn't solved them
Please see the video description field. The links to the problem solutions are provided there.
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