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Пікірлер: 169
@user-ci2mg9fu9n8 жыл бұрын
You have saved a guy who is going for exam with nothing! Thanks!
@winthial8274 жыл бұрын
I wish I have this information to watch when I was in engineering school. Thank you for putting this info for other people to learn.
@sal1096 Жыл бұрын
I love this! takes less time in exams by drawing ILD and completing the table. Thank you!
@bendeng27685 жыл бұрын
Best Video of Influence Line ever.
@pawanbudhlakoti22836 жыл бұрын
Dr Structure. Your videos deserve 1 billion hits.
@nazeefaiman22474 жыл бұрын
Thank you so much !! Been struggling to understand this topic until I came across your video
@savagelyanimal62518 жыл бұрын
I should be paying you for my tuition fee.....Your delivery is spot on ....Appreciated... This will help in my finals today
@manikantagodi32395 жыл бұрын
the best lectures for structural design...you are really helpful...make it for concrete and steel structures also please...
@cabdimuse98565 жыл бұрын
1a: at A is one and B is zero at the end is Negative one 1 b: at B and free end is One at A is zero 2a: at A is one and the left hing is zero just like at another fixed end 3a: at A is 1 and B is zero , second hing is negative 1 at C is Zero 3b: from left first hing is zero at B is 1 At C is zero 3c: at c is 1 at B is zero
@aliahmedsententious81813 жыл бұрын
may god give you long life you really help the studentsabbat everywhere
@ryandika_putra2 жыл бұрын
the videos of your channel are better at explaining the material than my lecturer, it's sad to waste time in class
@asaddavar438 жыл бұрын
Mr. Structure you made it very easy... i request to you please upload video on influence line of frame and truss.... will you upload it ???..please reply......
@shexmurad1416 Жыл бұрын
Indeed it is very important , thank you very much
@chintanpatel25008 жыл бұрын
Lucid Explanations...Thank you Dr.Structure:-)
@FaizYah3 жыл бұрын
this is SOOOO helpful thank you very much!!
@aprilrosemarquita78606 жыл бұрын
you're so amazing!!! wooow thank you very much
@samuelhampongo72785 жыл бұрын
Thanks alot...your vidéos are REALLY helpful 👷👷👷
@fatihasouradi11333 жыл бұрын
Thank you so much for this video. Coul you plz traduce your courses to frensh. They are so helpful
@dayana65324 жыл бұрын
Thank u so much! I really understood with tour explanation!
@mohd1997m5 жыл бұрын
you are amazing!! thank you so much.
@exambuddyyt78966 жыл бұрын
Do you also have videos for finding the deflection of a beam using moment area method? Btw, your videos are very helpful! Thank you so much!!! 🙂
@kefasjanuar41654 жыл бұрын
Thanks! It means a lot to me
@rost1878 жыл бұрын
really great series of videos thank you
@dhuraiathi52916 жыл бұрын
Excellent
@anazreedzuan57176 жыл бұрын
great video
@TheMjol6 жыл бұрын
Thanks for the video! I fail to understand the concept intuitively. How can reaction force due to the unitary load be maximum anywhere else than directly on support like you mentioned in the intro? I am guessing it is due to moment generated at that point and grows as the load moves away, whereas the vertical force exerted on the support reduces as we move closer to the other support. So, does the influence line (in this case) show the change in reaction that is caused by both shear and bending moment? I hope I expressed my doubts in a clear way.
@DrStructure6 жыл бұрын
You are right. Say in the case of a simply supported beam, a reaction would be maximum only when the unit load is located at the support. However, when the beam has a long overhang, that unit force causes a large moment that needs to be countered by other reaction forces. Depending on the position of supports, then, a reaction force could be significantly larger than 1, if the unit load placed at some distance from the support causes a large moment. Yes, the influence line reflects the effect of bending moments and shear forces in the beam.
@TheMjol6 жыл бұрын
Thank you! I am still amazed that this channel answers my questions within hours. Usually I am answered within months or even never.
@DrStructure6 жыл бұрын
Thanks for the comment. Resolving questions is an important part of learning. We try our best to facilitate that.
@taongamaiwanga98904 жыл бұрын
thankyou Doc
@Brutal_Husky9 жыл бұрын
great explanation
@attarmohammed92292 жыл бұрын
thank you very much
@kamalpm3248 жыл бұрын
very nice,practically understending
@patrisagar65483 жыл бұрын
2:42 didn't get it. Can you elaborate it a bit please. How do we know that 10 kN is compressive or tensile. Also i think we drew the influence line diagram for the reaction in the column so whatever sign ILD shows must be the sign of the reaction in the column. Please explain how to know which is which
@DrStructure3 жыл бұрын
We need to distinguish between action and reaction forces. The support reaction is the force that the support/column exerts on the beam. An action force is what the beam exerts on the column. This is basically the third Newton's Law (i.e., for every action there is an equal and opposite reaction). So, the beam is exerting a downward (compressive) force on the column, and the column reacts by applying an upward force on the beam. The upward force is the support reaction.
@muskduh2 жыл бұрын
So good thanks
@aswinjomike94669 жыл бұрын
realii gud vid , presentation ,xplanation n the tiny li'l animated head tht prevents boredom!:P i'd realli appreciate it if u cld explain this method a bit more.. this is not the method that i've studied and find it a bit difficult.. thnx anyway!
@DrStructure9 жыл бұрын
Let me know what it is about the method that you find difficult to understand either by explaining what you are confused about or by asking a few questions. I will then be able to address your concerns directly in the next lecture.
@aswinjomike94669 жыл бұрын
ma'am (judging by ur dp:P) .. im confused by the way you jus elevate or increase the ordinate of the influence line-by unit qty. jus considering the support conditions.. while i'm used to calculating the value of reaction components (moment or shear..)w.r.t the position of load along the beam.. ur method seems much easier but i'm not getting the hang of it .. plz explain..
@annejenethabellar51336 жыл бұрын
Hi dr. Structure. I just want to know how can i calculate the heights 1,2 and -4 in the influence lines for your first example. Thankyou so much. Your videos helped me a lot with my studies. I just dont know how to calculate reactions when 2 or more internal hinges are present in the beam.
@DrStructure6 жыл бұрын
Hi Anastasia, In the case of this specific example, we are not really calculating any support reactions in the conventional way (using the equilibrium equations), we are just lifting Point B by one unit (since we want to draw the reaction influence line for B), and use geometry to determine the remaining heights. Knowing that the influence line consists of straight lines, we can use similar triangles to determine the remaining height. For example, the segment of influence line between A and C can be viewed as two triangles. The smaller triangle has a base of 5 and height of 1 (the triangle between A and B). The larger triangle has a based of 10, so the height of it (say, h) must satisfy the following relationship: 1/5 = h/10 Solving for h, we get: h = 2. Similarly, the two triangles between C and E must obey the same geometric property. That is, the ratio of height to base of the triangle above the x axis must be equal to the height to base ratio of the triangle below the x axis. So we can write: 2/2 = q/4 where q is the unknown height of the second triangle. That gives us: q = 4. To see examples of how statically determine beams with internal hinges can be analyzed using equilibrium equations, review lecture: kzfaq.info/get/bejne/Z7-goq-bsNrQY4U.html There are three exercise problems at the end of the video, all involving internal hinges. Their solution is also given.
@annejenethabellar51336 жыл бұрын
Thankyou so much for such detailed answer Dr. Structure.
@gnidnoeled7869 ай бұрын
The maximum positive reaction at point B is 2 x 5 when the moving load is at point C. Is this positive force 2 x 5 reacting upward but compression on column BG?
@DrStructure9 ай бұрын
Correct!
@gnidnoeled7869 ай бұрын
@@DrStructure Thanks a lot.
@reemelshorbagy81815 жыл бұрын
شكرا على الترجمة
@ieuhfuffffff4 жыл бұрын
So, can you help me, what is program or software do you use to make this video? I just want to make theses video like you, to teach my friends, my students. It is amazing and intelligence. Thank you so much.
@DrStructure4 жыл бұрын
This particular video was done by first tracing the writings manually using a stylus on an ipad, and saving the tracing in SVG file format. SVG files were turned into animations using VideoScribe (an online tool). And, the video segments were put together, and the audio was added and sync with the video using Camtasia Studio.
@ieuhfuffffff4 жыл бұрын
@@DrStructure Wow, so difficult. If you draw by a stylus, you draw perfectly ^^. It's very hard for me to do that. I don't know how to use SVG file too.
@DrStructure4 жыл бұрын
The type of stylus that you use is key, the slippery kind does not work that well. You need to pick the type that does not slip on the glass, the kind that has rubber/silicon tip, like this: www.amazon.com/dp/B006843RUC?tag=new-best-seller-20
@ieuhfuffffff4 жыл бұрын
@@DrStructure Thank you so much. I'm going to try it. How do you draw on Ipad? I mean What is software you use to draw and save it in SVG file?
@DrStructure4 жыл бұрын
@@ieuhfuffffff Graphic for Ipad (graphic.com)
@saronberhane8853 ай бұрын
thank you Dr structure where can i find the solution for question you did at the END
@DrStructure3 ай бұрын
The solutions for the expertise problems are available in the free online course referenced in the video description field.
@yosuftraders51284 жыл бұрын
thank u very much ... i easily understood it....
@zaidradwan40005 жыл бұрын
Thank you sir I appreciate your help
@DrStructure5 жыл бұрын
You're welcome.
@kshitijverma253 жыл бұрын
New subscriber
@itzcloudydrey6 жыл бұрын
Very helpful...
@nickoles1330 Жыл бұрын
thank you
@meghabista27896 ай бұрын
@Dr.structure how do we write linear equations to find out maximum and minimum reaction at B due to moving 1 Kn load ?i didn't understand it
@DrStructure6 ай бұрын
The qualitative approach for drawing influence lines circumvents the need for writing equations. But, if one wants to write the algebraic equation for a support reaction, one can easily write the equilibrium equations for the beam in terms of x (the distance from the left end of the beam to the position of a unit load), then solve the equations for the reaction force. The result is an equation in terms of x. Example: Consider a beam with a pin support at its left end and a roller support at a distance L from the pin support. The length of the beam is 2L. Let’s refer to the position of a unit load moving from left to right as x. We can write two equilibrium equations. The sum of the forces in the y direction must be zero. Or, Ra + Rb = 1, where Ra is the vertical reaction at the pin and Rb is the vertical reaction at the roller. The sum of the moments about the left end of the beam must be zero. Or, (L)(Rb) - (1)(x) = 0 Solving the two equations for Ra and Rb, we get: Rb = x/L Ra = 1 - x/L Where x is between 0 and 2L So, Rb is at its minimum when x = 0 and at its maximum when x = 2L, where x is the position of the moving load. Similarly, Ra is at its minimum when x = 2L (Rb = -1 at x=2L) and at its max value when x = 0.
@meghabista27896 ай бұрын
@@DrStructure thank you so much .Your content is truly amazing
@MehediHasan-ue8py Жыл бұрын
How can I match my answer to be right or wrong? Is the solution of the last second exercise available?
@DrStructure Жыл бұрын
The exercise problem solution are available in the free online course referenced in the video description field.
@emmanueltemenu75942 жыл бұрын
Where do we find the solution to the questions at the end of the video? Would be nice to compare
@DrStructure2 жыл бұрын
The solution for exercise problems are available through our (free) online course. The link is given in the video description field.
@sremix28182 жыл бұрын
This channel 💓
@nathannael61243 жыл бұрын
how do we calculate the value (such as in the 1:36 minute mark), where there are 1,2, and 4. where do these number come from?
@DrStructure3 жыл бұрын
@NATHAN NAEL For statically determinate beams, the influence line diagram consists of straight lines often forming triangles. We can use the height/base ratio property of similar triangles to determine the unknown heights. Here, we know the height of the influence line at B, it is 1. The height of a reaction influence line at the point that the reaction force is acting always equals 1. In this case, there are two triangles between A and C, a small triangle and a large triangle. We know the height of the smaller triangle, it is 1. And we know the base of the two triangles. The smaller one has a base of 5 m and the larger one has a base of 10 meters. So we can write: (height/base) of the smaller triangle = (height/base) of the larger triangle. Or, (1/5) = (h/10). Solving for h, we get: h = 2. Similarly, between C and E we have two triangles. The height of the left triangle is 2 (we just computed it). The base of the left triangle is 2, and the right triangle has a base of 4. So, we can write: (2/2) = (h/4) which gives us: h = 4.
@asifurrahmanmahin43116 жыл бұрын
@Dr. Structure Is it important or necessary to do the calculation before drawing the IL diagram in the exam or in the practical life? Or can we apply Muller Breslau's principle directly?
@DrStructure6 жыл бұрын
What kind of calculations? If they are necessary for drawing the influence line, then yes. Otherwise, we can draw the influence line directly and then use it as a visual tool it to better see the load locations that result in maximum reaction/shear/moment in the structure.
@asifurrahmanmahin43116 жыл бұрын
Thank you!
@Khanguyen-uu4vl8 жыл бұрын
really nice videos
@noorhanalkhatib43414 жыл бұрын
Thanks
@omidhassani8 жыл бұрын
Cars, vehicles, trucks, ... mostly are considered as 2 or more point loads, but not 1! How to calculate the maximum reactions when a car with 3 point loads?
@DrStructure8 жыл бұрын
+Omid Hassani We always draw the influence line for a point using a single unit load regardless of the actual number of axles of the vehicle. Once the influence line is drawn, we then need to determine the location of the concentrated load series (due to the multiple axle vehicle) that results in maximum shear at the point of interest (say, Point A). Example: Suppose we have three concentrated loads in series having magnitudes P1, P2 and P3. Say, the distance between P1 and P2 is d1 and the distance between P2 and P3 is d2. Also, say our shear influence lines has two peak values (either maximum or minimum). Let's refer to the location of the first peak value as L1 and the location of the second peak value as L2. Place P1 at L1. This means P2 is going to be to the right of P1 by d1, and P3 is to the right of P2 by d2. We know the height of the influence line at L1. Using simply geometry (similar triangles), we can determine the height of the influence line under P2 and P3. Now, shear at A under this load pattern is: (height of influence line at L1)(P1) + (height of influence line at L1+d1)(P2) + (height of influence line at L1+d1+d2)(P3). This equation gives us a number which represents the shear value at A for this particular location of the load series. Now, move the load series so that P2 is at L1. This means P1 is to the left of L1 by d1 and P3 is to the right of L1 by d2. Again, using simple geometry we can figure out the height of shear diagram under P1 and P3. Then, calculate shear at A for this location of the load series using an equation similar to the previous step. Next, move the load series so that P3 is at L1. Then, calculate shear at A due to this load series location in a similar manner. Compare the three shear values to determine the load series location that results in maximum shear at A when the vehicle is around L1. Repeat the same calculations for the case that the vehicle (load series) is around L2. Then, compare the two results to determine absolute maximum/minimum shear at A due to the moving vehicle load.
@omidhassani8 жыл бұрын
+Dr. Structure Thank you!
@roshn82746 жыл бұрын
thanks alot dr.structure
@Kai6736 жыл бұрын
Extremely helpful videos. But where are the answers to the questions you gave at last?
@DrStructure6 жыл бұрын
Solution for Exercise Problem 1: kzfaq.info/get/bejne/rMB5idx61L_dk3U.html Solution for Exercise Problem 2: kzfaq.info/get/bejne/hN6bi6t11ZeclHU.html Solution for Exercise Problem 3: kzfaq.info/get/bejne/brCXlaSjqdWVqaM.html Solution for Exercise Problem 4: kzfaq.info/get/bejne/o5uEe6-hq867pGg.html
@Kai6736 жыл бұрын
Thank you so much. Your videos are helping me to prepare for Gate 2018 examination. Do you have lectures on Deflection topic from Strength of Material??
@DrStructure6 жыл бұрын
You are welcome. No, we don't have lectures on strength of materials yet.
@prof.dadapatilcivilenginee93866 жыл бұрын
Great explanation.....Very noble work. May God bless u. Please make more videos, entirely covering the structural analysis....Amazing
@FRSHBL8 жыл бұрын
Actualy Very nice
@declanconnor50428 жыл бұрын
Hi, i'm really struggling to grasp/visualise this concept.. I don't understand why the maximum reaction load at B occurs when the unit load is at E. Is there anything else that i could do which would help me understand it better ?
@DrStructure8 жыл бұрын
+Declan Connor The underlying relationship between the position of the unit load and the reaction force is a mathematical one. It might be beneficial to write the equation(s) that define the relationship so that you can see/examine how the reaction force changes as the load moves across the beam. Since the beam is statically determinate, you should be able to come up with the equation(s) with rather ease. They are: x: position of the unit load Rb: reaction at B 0 < x < 10 x - 5 Rb = 0 10 < x < 16 12 - x - Rb = 0 16 < x < 21 84 - 4x + 5Rb = 0 Since these are linear equations, to determine their max/min values we simply need to evaluate them at their boundaries. So, in the first interval Rb=0 at x=0, and Rb=2 at x=10. For the second interval we get: Rb=2 at x=10 and Rb=-4 at x=16. For the third interval, we get: Rb=-4 at x=16 and Rb=0 at x=21. So, maximum positive reaction at B is 2 (when x=10), and maximum negative reaction is 4 when x=16. By the way, if we graph these equations, we get the influence line for reaction at B.
@declanconnor50428 жыл бұрын
+Dr. Structure thanks very much for your very detailed an thorough explanation. I get it😎. Fantastic video's. No doubt I will use many more and there will be more questions to follow!
@hakanyalciner18 жыл бұрын
What is the answer of question 3 in the exirces for reaction influence line...
@DrStructure8 жыл бұрын
kzfaq.info/get/bejne/brCXlaSjqdWVqaM.html
@meghabista27896 ай бұрын
Can you please tell me how to write static determinant equation for beam to find out maximun and mimimum reaction at B?
@vishalkesharwani8654 жыл бұрын
Where do you give solutions to the questions given at last??
@DrStructure4 жыл бұрын
See the video description field for the links.
@ubaidullah-pj9mu2 жыл бұрын
how do you find the value of the peak? in other words the magnitude of the influence line where the worst moment is experienced
@DrStructure2 жыл бұрын
For moment influence lines, peak values cannot be determined directly from the diagram. You need to actually analyze the beam in order to determine the peak moment value.
@ubaidullah-pj9mu2 жыл бұрын
@@DrStructure Thanks Doc! Can you do a video where we do an analysis of an indeterminate beam?
@DrStructure2 жыл бұрын
We have a few lectures on the analysis of indeterminate beams/frames/trusses. You can find the updated versions of these lectures in our free online course. See the video description field for the link to the course.
@mariamios7427 Жыл бұрын
Where can I find solutions to the problems given at the end?
@DrStructure Жыл бұрын
The solutions, and updated lectures, can be found in the free online course referenced in the video description field.
@saeedhijab76552 жыл бұрын
If we have fixed support at one end and roller at the mid, so how would be influence line for roller at mid??
@DrStructure2 жыл бұрын
The technique described in this lecture applies to statically determinate beams only. The beam you are describing is statically indeterminate. For such a beam, the influence diagram is not a line. Rather, it is a curve.
@saeedhijab76552 жыл бұрын
@@DrStructure ok I understood Thanks a lot, You deserve a great Appreciation ❤️❤️❤️❤️❤️❤️❤️
@saeedhijab76552 жыл бұрын
@@DrStructure I am wondering that you posted a video 7 years ago and now replied for a question in minutes, it is so amazing❤️❤️
@DrStructure2 жыл бұрын
Thanks for the note. We do our best to be responsive in supporting students.
@ankuskumar50166 жыл бұрын
Will you please upload influence line diagram for statically indeterminate beams.
@FranCoVids5 жыл бұрын
How would you compute the reactions if there is a uniform load?
@DrStructure5 жыл бұрын
We determine the area under the influence line for the segment of the beam subjected to the distributed load, then multiply that area by the distributed load intensity to get the support reaction due to the load.
@FranCoVids5 жыл бұрын
@@DrStructure Thanks for replying!!
@gazikaushik4905 жыл бұрын
Thanks mam.
@gideongitumbi76223 жыл бұрын
Where can we get the solutions to the questions you've given
@DrStructure3 жыл бұрын
The solution to the exercise problems are provided in the (free) course referenced in the video description field.
@sounhournim164311 ай бұрын
THis is good Reccomeded. Im not a bot
@ghassan38704 жыл бұрын
Thanks. Can I know what software you using for drawings
@DrStructure4 жыл бұрын
For this particular video, we manually traced the writings/drawing using an ipad and stylus, saved the tracings in svg file format, then animated the svg files using VideoScribe (an online whiteboard animation tool).
@ghassan38704 жыл бұрын
Dr. Structure is there is any software you can recommend it for making the drawings easier because I'm working on structure lectures and make drawings with visio and it's taking long time . Thanks for reply
@DrStructure4 жыл бұрын
@@ghassan3870 If you just want to draw, without any sort of animation, you may want to look into Adobe Illustrator. That is what we have been using for the most recent videos. You might find other similar, yet less expensive, drawing tools that could fit your needs.
@shrimadbhagavadgita16177 жыл бұрын
how to calculate reaction of continuous beam , where no hinge are provide?
@DrStructure7 жыл бұрын
A statically indeterminate beam can be analyzed using the force method, the slope-deflection method, the moment distribution method, or other such techniques.
@neerajrana83297 жыл бұрын
mr. structure please provide influence line dig. for 2hing & 3 hing arch.
@jonathandewilde21126 жыл бұрын
Cheers!
@nisreensalameh18878 жыл бұрын
Thank you
@andrewanim54322 жыл бұрын
Pls provide solutions at the end of each given exercise.
@DrStructure2 жыл бұрын
The exercise problem solutions, and updated lectures, are provided in the (free) online course referenced in the video description field.
@somdutta17688 жыл бұрын
Very nice
@dougwoolery97212 жыл бұрын
you have the solutions for the ending ?
@DrStructure2 жыл бұрын
You can find the solution for the exercise problems, and updated lectures in the free online course referenced in the video description field.
@DrStructure2 жыл бұрын
To balance joint B (showing a moment of -0.4), we need a positive moment of 0.4, to be distributed to between the two columns using the distribution factors: 3/5 and 2/5 as follows: 0.4(3/5) = 0.24 0.4(2/5) = 0.16
@abdullahehsanprium19478 жыл бұрын
i couldn't solve the problem 4, its not even in ur website... please give the solution
@DrStructure8 жыл бұрын
+Abdullah Ehsan Prium Here is the direct youtube link for Problem 4: kzfaq.info/get/bejne/o5uEe6-hq867pGg.html
@abdullahehsanprium19478 жыл бұрын
Dr Structure, you are the awesomest, i tried problem 4 and only because of you I succeeded, I just needed the solution to see if i am correct
@zy46634 жыл бұрын
What are u guys major in? Civil Engineering?
@DrStructure4 жыл бұрын
More like structural engineering
@AmitYadav-mv9ez4 жыл бұрын
Sir , where is solutions of ending questions for cross check
@DrStructure4 жыл бұрын
The links are given in the video description field.
@AmitYadav-mv9ez4 жыл бұрын
@@DrStructure thank you So much Sir.. I have doubt for last 1-2 months for simplest method.. but now all of doubt clear ❤️❤️... Your teaching methods, Concept are unique and great.. Once again Thank You ❤️❤️❤️
@DrStructure4 жыл бұрын
Thanks for the feedback.
@TabassumMalihaHassan8 жыл бұрын
where are the last questions' solve???
@DrStructure8 жыл бұрын
Links to solutions activate toward the end of the video. Starting at 10:15, click on a problem to access its solution video.
@joecheung7968 Жыл бұрын
where are the solutions?
@DrStructure Жыл бұрын
The solutions, and updated lectures, are available in the (free) online course referenced in the video description field.
@mmahesh42904 жыл бұрын
Where are the answers for given questions
@DrStructure4 жыл бұрын
See the video description field for the links.
@yohanesingkiriwang65717 жыл бұрын
Dr, at minute 2:48, what's the point G in your beam or column drawn? Thanks
@DrStructure7 жыл бұрын
I am not sure what you mean by "what is point G?" It is the support point at the base of the column.
@yohanesingkiriwang65717 жыл бұрын
My bad.. sorry I did not see point G at the bottom of the column before. Cheers
@samuelhampongo72785 жыл бұрын
Do you have pdfs???
@DrStructure5 жыл бұрын
Not for this lecture, not yet. We are working on creating a pdf version of each video lecture, but that will take some time.
@israeldmx20037 жыл бұрын
how do we know the value of B at 7:17
@DrStructure7 жыл бұрын
We can use similar triangles to determine the value at B. (Length BD)/(Value at B) = (Length CD)/1
@israeldmx20037 жыл бұрын
Dr. Structure thank u so much! !!!! i barelu figure it out! your the best! i love ur videos! nice and simple!
@ariely76013 жыл бұрын
I did four problems and it’s all wrong,and I have final tomorrow...
@DrStructure3 жыл бұрын
Feel free to share your erroneous solutions for feedback. Although our feedback is not going to be of any help for your exam tomorrow, but it could help you better understand the use of the concept beyond your class/exam.
@zihadkhan31203 жыл бұрын
How can i get exercises answer??
@DrStructure3 жыл бұрын
The links are provided in the video description field.
@noorrababah88308 жыл бұрын
please give us the solutions for problem 2&3
@DrStructure8 жыл бұрын
+Noor Rababa`h Problem 1: kzfaq.info/get/bejne/rMB5idx61L_dk3U.html Problem 2: kzfaq.info/get/bejne/hN6bi6t11ZeclHU.html Problem 3: kzfaq.info/get/bejne/brCXlaSjqdWVqaM.html Problem 4: kzfaq.info/get/bejne/o5uEe6-hq867pGg.html
@orbital164 жыл бұрын
Hi guys. Do you have a video which explains how to draw an influence diagram for distributed and concentrated load..thanks
@DrStructure4 жыл бұрын
By definition, an influence line/diagram shows the effect of a single moving load on the beam. There is no such thing as an influence line for a distributed load. Although we can such an influence line to assess the impact of distributed loads on beams. For example, see the videos listed here: lab101.space/modules/sm2/
@thetmyatko14244 жыл бұрын
hello Dr. Structure . please help me .
@DrStructure4 жыл бұрын
If you post your specific question on this topic here, we will give you feedback accordingly.
@thetmyatko14244 жыл бұрын
@@DrStructure the difference between constructing an influence line and constructing a shear or moment diagram.
@DrStructure4 жыл бұрын
They are completely different. An influence line (diagram) represents the value of reaction/shear/moment at a specific point in the beam. On the other hand, the shear and moment diagrams show the shear and moment values for all the points in the beam. We are dealing with two somewhat different concepts, therefore, the processes for constructing the two types of diagrams are unrelated.
@thetmyatko14244 жыл бұрын
@@DrStructureThanks.
@ineverlickyoghurtlid39035 жыл бұрын
interesting, but why?
@DrStructure5 жыл бұрын
Is that a technical question or a philosophical one? If technical, please elaborate.
@ineverlickyoghurtlid39035 жыл бұрын
@@DrStructure I mean. is there any mathematical justification for this method.
@DrStructure5 жыл бұрын
@@ineverlickyoghurtlid3903 Yes, the justification for this can be seen mathematically. In essence, all we need to do is to come up for an algebraic equation for the reaction in terms of x (the position of the unit load). The graph of that equation is the influence line. For example, for a simply supported beam, the reaction at the left end of the beam, as a function of x, can be written as: R = 1 - x/L where L is the length of the beam and x is the position of the unit load measured from the left end. The graph of this equation takes the shape of a right triangle with the height of 1 at the left end (where x = 0). Conceptually, we can view this graph as a horizontal line (i.e., the beam) being pushed up at the left end by one unit. Knowing the linear nature of the equation and knowing how it looks like, we don't really need to explicitly write it, and graph it. We just follow the conceptual/qualitative approach of graphing it, by pushing the beam up at the reaction point by a unit...
@ineverlickyoghurtlid39035 жыл бұрын
@@DrStructure Thank you for your reply. I think we can go a step further. As the functions of support reaction are linear, we can determine one segment of the function graph with two points. one is at the point, y equals one when the point is right below the load and carries all the load, one is at an adjacent support point, which is zero when the adjacent support point carries all the load. for more complex cases, they are the combination of such segments.