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.

Thank you! I am still amazed that this channel answers my questions within hours. Usually I am answered within months or even never.

Thanks for the comment. Resolving questions is an important part of learning. We try our best to facilitate that.

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.

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.

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..

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.

Thankyou so much for such detailed answer Dr. Structure.

Correct!

@@DrStructure Thanks a lot.

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.

@@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.

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

@@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?

@@ieuhfuffffff Graphic for Ipad (graphic.com)

The solutions for the expertise problems are available in the free online course referenced in the video description field.

You're welcome.

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.

@@DrStructure thank you so much .Your content is truly amazing

The exercise problem solution are available in the free online course referenced in the video description field.

The solution for exercise problems are available through our (free) online course. The link is given in the video description field.

@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.

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.

Thank you!

+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.

+Dr. Structure Thank you!

thanks alot dr.structure

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

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??

You are welcome. No, we don't have lectures on strength of materials yet.

Great explanation.....Very noble work. May God bless u. Please make more videos, entirely covering the structural analysis....Amazing

+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.

+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!

What is the answer of question 3 in the exirces for reaction influence line...

kzfaq.info/get/bejne/brCXlaSjqdWVqaM.html

Can you please tell me how to write static determinant equation for beam to find out maximun and mimimum reaction at B?

See the video description field for the links.

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.

@@DrStructure Thanks Doc! Can you do a video where we do an analysis of an indeterminate beam?

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.

The solutions, and updated lectures, can be found in the free online course referenced in the video description field.

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.

@@DrStructure ok I understood Thanks a lot, You deserve a great Appreciation ❤️❤️❤️❤️❤️❤️❤️

@@DrStructure I am wondering that you posted a video 7 years ago and now replied for a question in minutes, it is so amazing❤️❤️

Thanks for the note. We do our best to be responsive in supporting students.

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.

@@DrStructure Thanks for replying!!

The solution to the exercise problems are provided in the (free) course referenced in the video description field.

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).

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

@@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.

A statically indeterminate beam can be analyzed using the force method, the slope-deflection method, the moment distribution method, or other such techniques.

The exercise problem solutions, and updated lectures, are provided in the (free) online course referenced in the video description field.

You can find the solution for the exercise problems, and updated lectures in the free online course referenced in the video description field.

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

+Abdullah Ehsan Prium Here is the direct youtube link for Problem 4: kzfaq.info/get/bejne/o5uEe6-hq867pGg.html

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

More like structural engineering

The links are given in the video description field.

@@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 ❤️❤️❤️

Thanks for the feedback.

Links to solutions activate toward the end of the video. Starting at 10:15, click on a problem to access its solution video.

The solutions, and updated lectures, are available in the (free) online course referenced in the video description field.

See the video description field for the links.

I am not sure what you mean by "what is point G?" It is the support point at the base of the column.

My bad.. sorry I did not see point G at the bottom of the column before. Cheers

Not for this lecture, not yet. We are working on creating a pdf version of each video lecture, but that will take some time.

We can use similar triangles to determine the value at B. (Length BD)/(Value at B) = (Length CD)/1

Dr. Structure thank u so much! !!!! i barelu figure it out! your the best! i love ur videos! nice and simple!

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.

The links are provided in the video description field.

+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

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/

If you post your specific question on this topic here, we will give you feedback accordingly.

​@@DrStructure the difference between constructing an influence line and constructing a shear or moment diagram.

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.

@@DrStructureThanks.

Is that a technical question or a philosophical one? If technical, please elaborate.

@@DrStructure I mean. is there any mathematical justification for this method.

​@@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...

​@@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.