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
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.
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......
@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.
This is to illustrate that I understand correctly your tutorials about reaction influence lines. I did consider the influence line at the rections at A, D and F. Reactions at A, max positive reaction = 2x5 = 10kN when load is at E, max negative reaction = 1x5 = 5kN when load is at C. Reaction at D, max positive reaction = 1x5 = 5kN when load is at D. Reaction at F, maximum positive reaction = 1x5 =5kN when load is at F. May I know your comments on these answers?
The reactions at A and F are correct. The reaction at D is 3 x 5 = 15. The influence line for the reaction force at D forms a triangle with a height of 1 at D and a height of 3 at E. When D is pushed up by a unit, E rises correspondingly by three units. As a result, segment CD rotates counterclockwise (about C) , while segment EF rotates clockwise (about F).
@DrStructure For the reaction at D, I try to figure out how at E the height is 3 if at D it's only 1. Isn't it height at E is just 3/7? Thus, the reaction at D = 1x5 = 5kN when load is at D and it's 3/7x5 = 15/7kN when load is at E.
When point D is displaced upward by 1 unit, point C remains stationary but rotates due to the movement. This rotation causes point E to move upward. A right triangle forms with the base between points C and E, which is 6 meters, and the height at E, which we’ll refer to as h. The distance between points C and D is 2 meters. From the geometry of the triangle, the relationship between the displacement at D (1 unit) and the base and height of the triangle is given by: 1/2 = h/6. Solving for h: h = (1/2) * 6 = 3. The upward movement of point E results in a support reaction at D. If the load applied at E is 5 (units of force), the reaction at D can be calculated as: Reaction at D = h * Load at E = 3 * 5 = 15.
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?
+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.
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
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.
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?
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.
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.
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.
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.
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: ruclips.net/video/1YpoLjKtl3Q/видео.html There are three exercise problems at the end of the video, all involving internal hinges. Their solution is also given.
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.
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.
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.
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
@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?
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.
A statically indeterminate beam can be analyzed using the force method, the slope-deflection method, the moment distribution method, or other such techniques.
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.
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 ?
+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!
@@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 ❤️❤️❤️
Solution for Exercise Problem 1: ruclips.net/video/vZIVyIoYycA/видео.html Solution for Exercise Problem 2: ruclips.net/video/NxkXHDp18dA/видео.html Solution for Exercise Problem 3: ruclips.net/video/8JgbArDo1yo/видео.html Solution for Exercise Problem 4: ruclips.net/video/m5THLpFhWt4/видео.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??
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.
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
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!
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..
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/
+Noor Rababa`h Problem 1: ruclips.net/video/vZIVyIoYycA/видео.html Problem 2: ruclips.net/video/NxkXHDp18dA/видео.html Problem 3: ruclips.net/video/8JgbArDo1yo/видео.html Problem 4: ruclips.net/video/m5THLpFhWt4/видео.html
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.
@@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.
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.
You have saved a guy who is going for exam with nothing! Thanks!
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.
Best Video of Influence Line ever.
I love this! takes less time in exams by drawing ILD and completing the table. Thank you!
Dr Structure. Your videos deserve 1 billion hits.
Thank you so much !! Been struggling to understand this topic until I came across your video
the best lectures for structural design...you are really helpful...make it for concrete and steel structures also please...
I should be paying you for my tuition fee.....Your delivery is spot on ....Appreciated... This will help in my finals today
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
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.
may god give you long life you really help the studentsabbat everywhere
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......
Indeed it is very important , thank you very much
the videos of your channel are better at explaining the material than my lecturer, it's sad to waste time in class
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?
@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.
This is to illustrate that I understand correctly your tutorials about reaction influence lines. I did consider the influence line at the rections at A, D and F. Reactions at A, max positive reaction = 2x5 = 10kN when load is at E, max negative reaction = 1x5 = 5kN when load is at C. Reaction at D, max positive reaction = 1x5 = 5kN when load is at D. Reaction at F, maximum positive reaction = 1x5 =5kN when load is at F. May I know your comments on these answers?
The reactions at A and F are correct.
The reaction at D is 3 x 5 = 15.
The influence line for the reaction force at D forms a triangle with a height of 1 at D and a height of 3 at E. When D is pushed up by a unit, E rises correspondingly by three units. As a result, segment CD rotates counterclockwise (about C) , while segment EF rotates clockwise (about F).
@DrStructure For the reaction at D, I try to figure out how at E the height is 3 if at D it's only 1. Isn't it height at E is just 3/7? Thus, the reaction at D = 1x5 = 5kN when load is at D and it's 3/7x5 = 15/7kN when load is at E.
When point D is displaced upward by 1 unit, point C remains stationary but rotates due to the movement. This rotation causes point E to move upward. A right triangle forms with the base between points C and E, which is 6 meters, and the height at E, which we’ll refer to as h. The distance between points C and D is 2 meters.
From the geometry of the triangle, the relationship between the displacement at D (1 unit) and the base and height of the triangle is given by:
1/2 = h/6.
Solving for h:
h = (1/2) * 6 = 3.
The upward movement of point E results in a support reaction at D. If the load applied at E is 5 (units of force), the reaction at D can be calculated as:
Reaction at D = h * Load at E = 3 * 5 = 15.
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?
+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
Thanks! It means a lot to me
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
thank you Dr structure where can i find the solution for question you did at the END
The solutions for the expertise problems are available in the free online course referenced in the video description field.
@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
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 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?
Correct!
@@DrStructure Thanks a lot.
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.
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.
you are amazing!! thank you so much.
you're so amazing!!! wooow thank you very much
Thank you so much for this video. Coul you plz traduce your courses to frensh. They are so helpful
Where can I find solutions to the problems given at the end?
The solutions, and updated lectures, can be found in the free online course referenced in the video description field.
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!!! 🙂
Lucid Explanations...Thank you Dr.Structure:-)
Thanks alot...your vidéos are REALLY helpful 👷👷👷
Thank u so much! I really understood with tour explanation!
How can I match my answer to be right or wrong? Is the solution of the last second exercise available?
The exercise problem solution are available in the free online course referenced in the video description field.
If we have fixed support at one end and roller at the mid, so how would be influence line for roller at mid??
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.
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.
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:
ruclips.net/video/1YpoLjKtl3Q/видео.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.
this is SOOOO helpful thank you very much!!
How would you compute the reactions if there is a uniform load?
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!!
شكرا على الترجمة
how do you find the value of the peak? in other words the magnitude of the influence line where the worst moment is experienced
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.
Will you please upload influence line diagram for statically indeterminate beams.
Where do we find the solution to the questions at the end of the video? Would be nice to compare
The solution for exercise problems are available through our (free) online course. The link is given in the video description field.
you have the solutions for the ending ?
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
Where do you give solutions to the questions given at last??
See the video description field for the links.
@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?
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!
how do we know the value of B at 7:17
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!
Where can we get the solutions to the questions you've given
The solution to the exercise problems are provided in the (free) course referenced in the video description field.
how to calculate reaction of continuous beam , where no hinge are provide?
A statically indeterminate beam can be analyzed using the force method, the slope-deflection method, the moment distribution method, or other such techniques.
thankyou Doc
really great series of videos thank you
Thanks. Can I know what software you using for drawings
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.
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 ?
+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...
ruclips.net/video/8JgbArDo1yo/видео.html
Can you please tell me how to write static determinant equation for beam to find out maximun and mimimum reaction at B?
Dr, at minute 2:48, what's the point G in your beam or column drawn? Thanks
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
mr. structure please provide influence line dig. for 2hing & 3 hing arch.
Sir , where is solutions of ending questions for cross check
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.
Excellent
Extremely helpful videos. But where are the answers to the questions you gave at last?
Solution for Exercise Problem 1: ruclips.net/video/vZIVyIoYycA/видео.html
Solution for Exercise Problem 2: ruclips.net/video/NxkXHDp18dA/видео.html
Solution for Exercise Problem 3: ruclips.net/video/8JgbArDo1yo/видео.html
Solution for Exercise Problem 4: ruclips.net/video/m5THLpFhWt4/видео.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
This channel 💓
great video
Pls provide solutions at the end of each given exercise.
The exercise problem solutions, and updated lectures, are provided in the (free) online course referenced in the video description field.
i couldn't solve the problem 4, its not even in ur website... please give the solution
+Abdullah Ehsan Prium Here is the direct youtube link for Problem 4:
ruclips.net/video/m5THLpFhWt4/видео.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
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.
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)
So good thanks
very nice,practically understending
great explanation
How can i get exercises answer??
The links are provided in the video description field.
where are the last questions' solve???
Links to solutions activate toward the end of the video. Starting at 10:15, click on a problem to access its solution video.
Thank you sir I appreciate your help
You're welcome.
thank u very much ... i easily understood it....
Thanks
Do you have pdfs???
Not for this lecture, not yet. We are working on creating a pdf version of each video lecture, but that will take some time.
Very helpful...
where are the solutions?
The solutions, and updated lectures, are available in the (free) online course referenced in the video description field.
Where are the answers for given questions
See the video description field for the links.
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!
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..
really nice videos
thank you very much
Hi guys. Do you have a video which explains how to draw an influence diagram for distributed and concentrated load..thanks
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/
please give us the solutions for problem 2&3
+Noor Rababa`h
Problem 1: ruclips.net/video/vZIVyIoYycA/видео.html
Problem 2: ruclips.net/video/NxkXHDp18dA/видео.html
Problem 3: ruclips.net/video/8JgbArDo1yo/видео.html
Problem 4: ruclips.net/video/m5THLpFhWt4/видео.html
Thanks mam.
Actualy Very nice
What are u guys major in? Civil Engineering?
More like structural engineering
New subscriber
Thank you
hello Dr. Structure . please help me .
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.
Cheers!
Very nice
THis is good Reccomeded. Im not a bot
interesting, but why?
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.
I did four problems and it’s all wrong,and I have final tomorrow...
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.
thank you very much
thank you