The style here is AMAZING. As far as I'm concerned, you can make difficult to understand when I was in the University and extremely easy here. Bravo, you are definitely deserve to have more and more subscribers and views. Congratulations on your content. Again BRAVO !

It is incredible that you explain the information and make it more understandable with the help of visuals, thank you a lot :)

Thank you Dr. Structure! You save my structural analysis exam at hku today!

why the internal moment at G is equal to -4KN.m in the last example? If we draw the FBD of segment GD, there will be a upward shear force and clockwise moment at G as well as a downward reaction force at D , so if we take moment at support D: -0.5(4)-M=0 so M=-2 isnt it? I'm confusedddddd please helpppppp >

@DrStructure

8 жыл бұрын

+Rachel Wong Your calculation is correct, for determining moment at G due to a unit load. But this value has to be multiplied by the magnitude of the applied load (2 kN) in order to get moment at G due to the actual (applied) load.

@rachelwong3807

8 жыл бұрын

Thankyou so muchhhhhhhh: D

@TAGDAGame

8 жыл бұрын

+Rachel Wong Thank you for asking this question and thank you doctor for your answer. :)

@marksuryaharja

6 жыл бұрын

same question i wanna ask, and thanks for the respond doctor :)

@datauser1529

6 жыл бұрын

Great, I've got the same question in mind, thank you for asking it, thanks so much Dr for your answer which removed all confusions.

Massive thank you for this wonderful explenation!

Thoroughly enjoyed the video! So informative 👌

If anyone explain the subject then it should be like this....which make us learn the subject easily without mugging... Thank you for the awesome content.....I love you❤️

You're hands down the best teacher.... You this topic so easy to understand. Can't thank you enough. God bless 🙏

@DrStructure

Жыл бұрын

Glad to be of help.

I really appreciate your work. These videos are very helpful fo me,specially to understand influence line👍,so thank you. and please upload a video on how to calculate maximum bending moment under a train of concentrated load.

From Bangladesh ,Thanks a Lot ,make more more video like this, only you can give me clear concept.

+Faizan elahi The influence line is drawn for moment at G. The question asks for the load location that causes maximum negative moment at G. Only the middle region of the diagram is negative, with the value at F being maximum. Therefore, maximum negative moment at G occurs when the load is at F. If we were interested in determining the maximum positive moment at G, then according to the diagram, the load can be placed at G since the influence line shows a maximum positive value at that point.

@CarlosPDV

7 жыл бұрын

Hi, I think the moments drawn at G at 0:43 are in the wrong direction if we are using the sign-deformation convention. I may be wrong however, because I just started learning this topic.

@DrStructure

7 жыл бұрын

The directions are consistent with our beam sign convention: bending moment causing concave up deformation is considered positive.

@CarlosPDV

7 жыл бұрын

Sorry, I didn't see the animation where the bending moment flipped with the movement of the load along the beam. Makes sense. Thank you.

@DrStructure

4 жыл бұрын

@@smitpatel5908 The moment influence line for G tells us that the positive moment at G reaches its maximum value when the unit load is either at E or at G. In this case, given the geometry of the beam, both load locations result in the same moment value, since the two positive triangles have the same height.

Your are the best. Thanks a lot....

Ur explainatation is awesome

Great work. İ know a little english but i understand this lesson. Thank you very much.

Este video me fue de muchisisisisma ayuda :) This video helped me a lot, thanks!

In the last example how come just by seeing at the diagram you can say that maximum bending moment occurs at G when the car is at F.If you apply trigonometry on the diagram bending moment at G and F must me the same.Please help me out

correct me if im wrong please ,but does the segment CD (at 7':51'') turn in that direction because the point D can't go up due to support conditions at the right of segment ?

@DrStructure

4 жыл бұрын

Yes, exactly!

Using animation Very easy to understand. Thank you very much.

You said no formula for ILD for BM but I think there is one. It is x/L * (L-x) , where L is not actually the length of the beam but it is the span from the left support to the right support of the point whose ILD we are drawing.

just to add you can always visualise a moment measuring gauge at the point of interest and imagine it to ive you readings as the load travels through the structure

for determining numbers of the diagram, we look for the max Neg bending moment only ? or positive as well? if not, how do we know which one is the actual Max magnitude since the diagram just shows qualitative info

@DrStructure

4 жыл бұрын

In an actual design scenario, we need to determine both maximum positive and negative moments in beams and columns. A positive moment puts the top fibers of the beam in compression whereas a negative moment places the bottom fibers of the member in compression. Since for design purposes, the treatment of compression vs tension parts of the beam differ, it is imperative that we know the maximum values for both positive and negative bending (and their locations) before we can safely design the member.

شكرااااااا ع الترجمة ❤️

what if you want to draw the influence line with z-axis is positive? Because at 3:57 you show the influence line is upwards, when in my case it should be downwards.. Can i still use your method?

@DrStructure

6 жыл бұрын

Yes, you can label the axes however you wish. You can refer to the x axis as z axis, if you like. These labels are all by convention and not laws of physics. The method still works regardless of your chosen Cartesian coordinate system and how you decide to label its axes.

How do you determine the heights at the maximum moments at point E F and G for the last example?

@DrStructure

8 жыл бұрын

+Amos Mensah Generally speaking, the moment influence line is used to determine the location(s) of the applied loads that cause maximum internal moment in the beam. We then need to actually analyze the beam in order to calculate these maximum/minimum values. The pick values of the diagram cannot be determined without writing and solving the beam's equilibrium equations.

+Dr. Structure How can you say that the segment AF carries no portion of the unit load when the load is just to the right of F ? Will the hinge at F not take shear...? Well, I got that mathematically by solving equations but don't know how to get that by just examining visually... Please explain...

@DrStructure

8 жыл бұрын

+VIVEK SINGH Place the unit load to the right of F. Now, examine segment AE. Separate the segment at E from the rest of the beam. The free body diagram of the segment would consist of a shear force at E and a vertical reaction at A. Since the segment has to be in equilibrium, and there is no external load applied to it, both shear at E and reaction at A must be zero. With practice, you can draw this conclusion without actually writing the equilibrium equations. Similarly, examine segment EBF. There is no shear at E, there is a vertical reaction at B and a shear at F, but no applied force appears on the segment. So, using the same reasoning as above, we can conclude that reaction and B and shear at F are zero. Then, we are left with segment FCD only for carrying the unit load.

U really structural doctor!

@3:55 , how come the hinge is displaced upwards ? Doesn't it resists horizontal and vertical motion ?

@DrStructure

7 жыл бұрын

Keeping in mind that the hinge is fictitious, it does not resist bending moment. This means it would be forced to rotate when subjected to bending moment. Basically, the system acts like a truss, so the joint can move up or down.

U are best👍

you had said that to introduce an internal hinge and apply moment in order to calculate moment at a particular point for moving load but moment is zero for an hinge then how can we apply moment there?

@DrStructure

7 жыл бұрын

Think of a cantilever beam subjected to a distributed load. In such a beam, we know that the (internal) bending moment at the free end is zero. But does that mean we cannot apply/place an (external) bending moment (as an applied load) at the free end. No, it does not. We can place any applied load/moment anywhere on the beam including at its free end. If we did, then we can no longer argue that there is no moment at the free end, since we actually placed one there. The same idea holds true for a beam with an internal hinge. In the absence of any direct bending moment placed at the hinge, internal bending moment there would be zero. But, we are always place an external moment at the hinge, if we choose to do so.

Hello Dr. Structure, do you have solutions manual to the exercise problems given at the end of each video? We need to check whether our solutions are correct or not! Thanks a lot!

@DrStructure

2 жыл бұрын

You can find the solutions to the exercise problems, and updated lectures, in our free online course. See the video description field for the link.

Great work.. It helps me much

@MehediHasan-cq4oz

3 жыл бұрын

from Bangladesh?????

Thank you a lot ...😀😀

good day! what app did you use to make this video? thank you.

@alyssacuray500

5 жыл бұрын

i need something like this for me to be able to pass my project. help please.

In the last example, if i want to calculate the deflection under F? How to do that?

@DrStructure

6 жыл бұрын

You can use the method of virtual work (SA19-SA23). 1. Write the moment equation(s) for the entire beam under the applied load. Call it M(x). 2. Place a unit vertical load at F, then write the moment equation(s) for the beam. Call this m(x). 3. Integrate M(x)m(x)/EI over the entire beam. That would give you vertical deflection at F.

how about a video of IL for trusses?tnx

at 8:09, I understand that the CD segment rotates downward due to DEF conditions. Would CD still rotate downwards, if we did not have the DEF conditions, and instead for example, D was the connection of a cantilever beam to the wall?

@DrStructure

3 жыл бұрын

If there was a fixed support at D, and we wanted to draw the moment influence line for D, we would place a fictitious hinge at D and apply a positive moment to the hinge. That would give us exactly the same diagram as here. Here we have a real hinge at D, in the case of a fixed support at D, we end up introducing a fictitious hinge at D. These hinges behave the same way, enabling CD to rotate, of course, assuming joint C can move vertically.

thank you so much

where can I see the answers for the example problems?

@DrStructure

7 жыл бұрын

Problem 1: kzread.info/dash/bejne/daWV0bmLZq-YpMY.html Problem 2: kzread.info/dash/bejne/Z2qN0LSYmZObZ7Q.html Problem 3: kzread.info/dash/bejne/io6px6R-c7mvmKw.html

at minute 9:54, when you calculated the Moment at G, what's the "-2" , does it come from the influence line diagram?

@DrStructure

7 жыл бұрын

No, 2 does come from the influence line. It is the magnitude of the moving load (2 kN).

@yohanesingkiriwang6571

7 жыл бұрын

Dr. Structure I still don't get it.. if 2 is the moving load, then when we multiply them all, we gonna have unit of kN^2m

@DrStructure

7 жыл бұрын

Not quite. When drawing the influence line, we assume a unit load. The magnitude of the actual load then can be viewed as a multiplier. If the bending moment due to the unit load (a load having a magnitude of 1 kN) is, say 3 kN-m, then, if we replace the unit load with a load having a magnitude of 2 kN, what would be the moment value? It would be 2 times 3, or 6 kN-m. If the load is 20 kN, instead of being a unit load, then the moment would become 20 times 3, or 60 kN-m. And so on.

@yohanesingkiriwang6571

7 жыл бұрын

Awesomee.. cheers mate!

I wish you showed unit moment effect in influence lines, too

At the last example, how to get the height of maximum postive at G It is 4+4/8 ?

@DrStructure

4 жыл бұрын

Generally speaking, we do not know (cannot determine) the heights in a qualitatively drawing moment influence line. This is unlike the shear influence line in which the heights can be easily determined. We use moment influence lines for determining the load pattern that causes maximum bending moment. Once the pattern is determined, we then need to analyze the beam to determine the peak moment values.

At 9 minute 52 second ,Mg=-2(0.5)(4) ,Where is from the 2 come ? please explain

@DrStructure

7 жыл бұрын

That is the weight of the vehicle, 2 kN.

love u guys for this #civilengineeringbrotherhood #civilengineersrock #drstructureurdbest

can u explain why the beam also rotate at G ?

@DrStructure

7 жыл бұрын

Are you referring to the beam at the beginning of the video? Regardless, the technique explained for drawing influence line does not show the actual deformation of the beam; the assumed deformations are fictitious. If we place a (fictitious) hinge (which is free to rotate) at G and apply a positive pair of moments there (a counterclockwise moment and a clockwise moment), then the left segment attached to G would want to turn counterclockwise and the segment to the right of G wants to turn clockwise. This forces point G to move upward as shown in the video.

Please make video on influence line diagram of indeterminate structure

Please provide lectures for ILD for Indeterminate structures...

What a marvellous way to describe things! Which tool do you use for animation and writing?

@DrStructure

3 жыл бұрын

For this video, Camtasia Studio was used for creating the animations. The writing was created using VideoScribe.

@tanveeryounasbhatti3128

3 жыл бұрын

@@DrStructure Thank you very much for sharing. So nice of you.

@DrStructure

3 жыл бұрын

You're welcome!

@7.16 what will be the moment influence line diagram if the end F is free?

@DrStructure

6 жыл бұрын

That would make the beam unstable. We need to have a stable beam before we can analyze it.

@vikymehra5902

6 жыл бұрын

Dr. Structure upload some videos related to theory of elasticity n thanks. ☺️

In the last example where is the internal loads at Hing when you cut the beam at F ?! and thanks a lot

@DrStructure

4 жыл бұрын

It depends on the location of the unit load. If the unit load is just to the right of F, the internal shear force at the hinge would be zero. That is, when the unit load moves to the right of F, the entire load is going to be carried by segment FCD, the left part of the beam carries no load. Therefore, no shear force develops at F. However, when the unit load is to the left of F, a shear force does develop at F. The best way to see how shear force changes as a function of the position of the unit load is to draw the influence line for shear at F.

@noorhanalkhatib4341

4 жыл бұрын

@@DrStructure thanks a lot .. you are so kind and I love your channel .. keep going

How to find max bending moment at any location with internal hinge

@DrStructure

5 жыл бұрын

To find the maximum bending moment in a beam, which is different than finding the peak value in a moment influence line, we can find/write the moment equation, then set its derivative to zero. This is the same as finding the point at which shear equals zero. This can be done graphically too. Draw the shear diagram and find the location(s) at which it crosses the x axis.

at 9:55, why is there no equation for influence line? Should it be like this: Mg +2(4) + 0.5(4) = 0? And also, why did you multiply the load of the truck in the reaction: 2(0.5)(4)?

@DrStructure

6 жыл бұрын

The free-body diagram shown at the bottom of the page give us the following moment equation (written about point G). Mg + 0.5(4) = 0. Or, Mg = -(0.5)(4). But, this is the bending moment at G due to a unit load, not the actual load of the truck. Since the truck load is 2 kN, we multiply the moment due to the unit load by the load of the truck to get the actual moment value. So, Mg = - 2(0.5)(4) = -4 kN.m

@jericotemplado

6 жыл бұрын

To Hector Regala : I know that the way that the lecture is presented is a little bit confusing, however I wanna make this clear that the downward reaction at point D is as a result of the analysis using 1 unit of load which is supposedly negative 0.5 unit an so a the reaction at point C which is 1.5; they are just both multiplier to whatever is the applied load so that you can get their reactions for both points C and D. Therefore using 2KN moving load, the reaction at point D should be (- 0.5 x 2KN) = - 1KN, and then taking moment at point G : the equation should look like this>>M@g = negative reaction @D x lever arm = (-1Kn)(4M) = - 4Kn-m.

for Mg why did you do -2(0.5)(4). Where does the 2 come from?

@DrStructure

3 жыл бұрын

2 kN is the magnitude of the real (moving) load.

@joseh7612

3 жыл бұрын

@@DrStructure thanks

If moment at an internal hinge is always zero, how does the bar BC & CD bends ( at 10.59 min of the video ) ? Why don"t they resist the moment ?

@DrStructure

7 жыл бұрын

Are you referring to segment BCD in Exercise 1 at the end of the video?

@tasnimshoily1971

7 жыл бұрын

I am so sorry. There's been a typing mistake . I was referring to the figure at 6.59 min of the video.

@DrStructure

7 жыл бұрын

Okay. First, we need to keep in mind that the influence line does not represent the actual deformation of the beam, it represents bending moment. @6.59 we are drawing influence line for moment at E, not at C where the real hinge is located. The shape of the influence line (which has nothing to do with the actual deformation of the beam) tells us that if we place a unit load at the hinge (C), there would be a negative moment at E. That is all. The diagram does not tell us anything about the real deformation of the beam. It is imperative that we don't mix up the technique we are using here to draw influence line with how we draw the actual deformation of the beam. One has nothing to do with the other. But, your question suggests something more. It suggests that if bending moment at a point is zero, there should be no deflection at that point. That is not correct. To make this clear, consider a cantilever beam (fixed at one end but free at the other end). If we place a vertical load at the free end what would happen to the beam? It is going to displace downward even although bending moment is zero at that end.

The robotic voice sounds so realistic and it's so clear that it makes focusing easier

@DrStructure

3 жыл бұрын

That is a human voice.

Hello, Could you please explain the reason behind why the bending moment is always zero at a real hinge?

@DrStructure

7 жыл бұрын

Moment develops in at a point when its rotation is restrained. Think of a cantilever beam, it is fixed at one end and free at the other end. Since the beam is not going to rotate at the fixed end, a bending moment develops at that end. On the other hand, the free end is free to rotate as well as move up and down. Since the rotation of the joint is not prevented, then no bending moment develops at that end. A pin or a roller act the same way as a free end with regard to rotation. Each allows rotation, therefore no bending moment can develop at any of these joints. An internal hinge acts the same way. Such a hinge connects two beam segments whereby one can rotate relative to the other, therefore no bending moment can develop at the hinge either.

@idfk3302

7 жыл бұрын

+Dr. Structure yes thank u however when you did the moment influence lines for a hinge or roller support you did not consider moment to be always zero no matter the position of the load .. And you're saying internal hinge works the same as roller and hinge support in regards to it allowing rotation and thus having no moment reaction. This is where my confusion is. Isn't this contradicting ?

@DrStructure

7 жыл бұрын

Good question. When a pin or a roller is located at an interior point, when it is not at the end of the beam, it cannot freely rotate. Why? because the beam segment to the right of the roller wants it to rotate in a certain way, and the beam segment to the left of the roller wants the joint to rotate differently. This "competition" between the two segments prevents the free rotation of the joint. Consequently, a bending moment develops at the joint. If the roller/pin was at the end of the beam. however, there would be no "competition" between segments as there is only one beam segment attached to the joint. The segment dictates the rotation of the joint. That is, the joint would be free to rotate as dictated by the beam segment. As a result, no bending moment develops at the joint. An internal hinge acts differently than a pin or a roller. To form a hinge, we cut the rigid beam at the point and reconnect the two segments using a frictionless (truss) pin where relative rotation between the segments is permitted. That is quite different than the behavior of an interior roller/pin. In the latter scenario, the beam remains rigid over the pin/roller support whereas in the former scenario the beam acts more like a truss joint. An internal hinge often is associated with a beam splice where we connect two beam segments using gusset plates and bolts. In reality such a splice carries some moment, but for design purposes we often assume the (hinge) connection is frictionless and carries no bending moment.

@idfk3302

7 жыл бұрын

+Dr. Structure Bless u for taking the time to answer us students in need. that was very helpful thank you. 🤗

How about the influence line for the truss members force under moving live load. Let say a vehicle with 3 points load. How to calculate the other points when putting one on the joint? By using force result or similar triangles? I could send the question's drawing if I had your email address.

@DrStructure

7 жыл бұрын

I think I understand your question. However, if you need to clarify the question with a drawing, feel free to send email to: Dr.Structure@EducativeTechnologies.net Suppose we have three concentrated loads in series with a distance of 2 m between each pair of adjacent loads. The middle load has a magnitude of 2P and each exterior load has a magnitude of P. The moment influence line for our truss tells us that the middle load of 2P has to be placed at joint B. Then, one of the exterior loads would have to be placed 2 m to the left of B and the other exterior load would be placed 2 m to the right of B. Suppose the truss geometry is such that there is a joint (labeled A) 6 m to the left of B, and a joint (labeled C) 6 m to the right of B. This means we have a load of P placed at an interior point on member AB, and a load of P placed at an interior point on member BC. Since in trusses we assume loads are placed only at the joints of the structure, we need to distributed the load on AB to its end joints, and do the same for the load on BC. Since the load (P) on AB is placed 4 m from end A, and 2 m from end B, then 2/3 of P goes to B and 1/3 of it goes to A. (This is a linear proportional load distribution, just assume the truss member is a simply supported beam subjected to a concentrated load, then calculate the vertical support reactions.) Similarly, for member BC, we place 2/3 of P at B and 1/3 of it at C. This means we need to analyze the truss under the following loading case: P/3 at A P/3 at C 2P + 2(2/3)P = 10P/3 at B

Great work, but why did we multiply the moment at G by 2?

@DrStructure

3 жыл бұрын

2 kN is weight of the vehicle. We need to multiply the calculated moment due to the unit load by the weight of the vehicle to get the real moment value.

Excelente

How come a roller have a downward reaction?

@DrStructure

7 жыл бұрын

Reaction at a roller could be upward or downward. If the roller is placed horizontally, then it can only move along the x axis. Such a roller, by definition, cannot move up or down in the y direction.

@itachiuchiha6155

7 жыл бұрын

I think reactions are the resisting force of the supports. Just for example a roller, if you apply force acting downward on a roller the resisting force on it is upward, that's why roller support only have upward reactions because no resisting force on it when you applied upward force.

At 9:55 , why is the BM equation "-2*0.5*4"? Why is it not just "-0.5*4"?

@DrStructure

2 жыл бұрын

-0.5*4 is the moment due to a unit load. Since we want to determine the bending moment due to a moving load of 2 kN, we need to multiply -0.5*4 by 2.

what is the difference between roller..hindge and triangle shape support

@DrStructure

6 жыл бұрын

I covered the difference between roller and hinge in my response to your other message. The triangular support is called a pin. It behaves almost like a roller. A roller, as the name suggests, can move/roll on the surface that it is resting on. Pin is a roller that cannot roll. Look up images of pin and roller online, you should be able to easily find pictures of bridges that use these support types.

@tluanga31

6 жыл бұрын

Dr. Structure thanks anyway iam still in the process

Where can I check the answers for the examples at the end of video?

@DrStructure

2 жыл бұрын

The solutions for the exercise problems can be found in the (free) online course referenced in the video description field.

@shiqing2990

2 жыл бұрын

@@DrStructure I signed up using the link provided in description but still can’t view the answers

@DrStructure

2 жыл бұрын

Login to the course, navigate to the influence lines section, then select moment influence line lecture/video as if you are going to watch it. Two pdf files are listed under the video frame: lecture notes and solution to exercise problems. Open this latter file to view the solutions.

why there is no influence line for normal forces in beams ?

@DrStructure

5 жыл бұрын

If the beam carries an axial force, then one can draw the influence line for it. When an axial force is present in the member (like in frame members), it is often constant, it does not change value across the length of the member. So, the influence line for it would be a a straight line (a rectangle). Such a diagram is not really helpful in determining max/min values. Put it differently, we don't need the influence line to tell us where the axial force becomes max positive or max negative; it is max positive (or negative) throughout the length of the member.

@amrabdullah1995

5 жыл бұрын

@@DrStructure thank you

When there is an anticlockwise moment why cd part moves downward pls explain

@DrStructure

5 жыл бұрын

In the example where CD is the segment between two hinges, there is an anticlockwise moment at D. This means the moment wants to turn the segment in the anticlockwise direction. For CD to turn that way, end C needs to move down relative to D.

@SHUBHAMKumar-qd1pi

5 жыл бұрын

@@DrStructure I have a question may you solve it for me

@SHUBHAMKumar-qd1pi

5 жыл бұрын

I would be grateful

@DrStructure

5 жыл бұрын

We would be glad to try and address questions pertaining to these lectures. Please feel free to post them here, or if diagrams/drawings are involved, email them to: Dr.Structure@EducativeTechnologies.net

Thanks

Considering segment GD to obtain the internal bending moment at G, the equation is Mg = -2 x 0.5 x 4 = -4 as shown in this video. In this equation, where this factor -2 came from? Considering segment FG to solve for moment at G, the equation is Mg = (1.5 x 4) - (1 x 8) = -2. Thus, internal bending moment considering segment GD = -4 and considering segment FG = -2. Why are values of the internal bending moment at point G obtained not the same, considering my solution to get bending moment at G using segment FG is correct?

@DrStructure

7 ай бұрын

The bending moment at G due to a unit load is -0.5 x 4 = -2. Since the weight of the moving vehicle is 2 kN, we need to multiply -2 by 2 kN to obtain the bending moment due to the vehicle’s weight.

@gnidnoeled786

7 ай бұрын

​@@DrStructure Thank you, I now got it. 👍

how do we get the number for the influence line?

@DrStructure

8 жыл бұрын

+legion Generally, we cannot determine the peak values of a moment influence line directly from the diagram. The diagram is used only for determining the load location at which bending moment at a given point (say, B) is maximum. Once that location is determined, you need to place the load there and actually analyze the beam in order to determine the bending moment value at B.

@jasonRhawt

8 жыл бұрын

+Dr. Structure u have any videos on this would like to watch it

@DrStructure

8 жыл бұрын

+legion The exercise problems at the end of the video demonstrate the process of calculating maximum/minimum moment values.

Why is the moment influence line at 8:07 like that? :( Why did it went downward instead of rotating upward at D? :(((

@DrStructure

7 жыл бұрын

For the beam to rotate upward at D, D must be able to move up. Examine segment DEF. For D to move up, E has to move up, or it must rotate. But it can do neither. E cannot move up since there is a roller support at the joint. E cannot rotate either since the pin at F prevents the right end of the segment from moving. So, segment DEF cannot move at all, it remains stationary. This means, under the applied moment, CD has to rotate counterclockwise as shown in the diagram.

@jssplrs

7 жыл бұрын

Dr. Structure Thank you!! Really helped a lot. 😄

hey. thanks for the video. do you have the link solution for the problems? thaanks. you're awesome

@DrStructure

7 жыл бұрын

Problem 1: kzread.info/dash/bejne/daWV0bmLZq-YpMY.html Problem 2: kzread.info/dash/bejne/Z2qN0LSYmZObZ7Q.html Problem 3: kzread.info/dash/bejne/io6px6R-c7mvmKw.html

where does the 2 come from when solving the internal moment?

@DrStructure

3 жыл бұрын

That is the magnitude of the truck load.

@l1a2g3e4r5

3 жыл бұрын

@@DrStructure oh ok I see where it comes from, but why do you multiply it by the moment you get from the end reaction?

@DrStructure

3 жыл бұрын

The computations for the example at the end of the lecture were done using a unit load. So, we need to multiply the results by the actual load magnitude in order to get the correct answer. Alternatively, we can replace the unit load with the actual load to begin with, then go through the computations. Either way, we arrive at the same answer.

@l1a2g3e4r5

3 жыл бұрын

@@DrStructure thank you so much!

Sorry, would you mind to show the answers at the end problem?

@DrStructure

8 жыл бұрын

+Pichmean veasna KZread annotation links at the end of the video point to the solution videos.

cheers

At 6:14 you said moment at hinge always zero than in next 2 examples 7:54 why the moment at hinge is not zero(i.e. at c). Pls anyone explain.

@DrStructure

4 жыл бұрын

The moment at C is zero in the example @7:54. Why do you think there is a bending moment at C?

@abhishekverma1487

4 жыл бұрын

@@DrStructurebecause in ILD there is an ordinate at c that means moment is not zero. The ILD at 6:14 in that you made a straight line from A to C because of hinge(zero moment).

@DrStructure

4 жыл бұрын

@@abhishekverma1487 That influence line is for bending moment at D. So, the height of the diagram at C indicates the moment at D when a unit load is at C. The height is not saying anything about the moment at C.

@abhishekverma1487

4 жыл бұрын

@@DrStructure ok got it. Thank you

great

In last example, what to do if we want to find positive max BM

@DrStructure

4 жыл бұрын

We place the load at either E or G.

@smithakm1174

4 жыл бұрын

@@DrStructure Thank you so much

@DrStructure

4 жыл бұрын

You're welcome.

If A is hinged and B has a roller support, also C is a free end. B is 15m away from A and C is 21m from A. Find max +/- bending moment at D and E for a moving udl 1kn/m and 3m long, where D is 12m from A and E is 17m from A.

@DrStructure

6 жыл бұрын

Point E is not going to experience any positive moment, it would only go under negative moment. Why? because the point is in segment BC which is an overhang, like a cantilever beam. Bending moment anywhere in BC is going to be negative, if the load is placed on BC, otherwise it would be zero. Then, it stands to reason that maximum (negative) moment, at any point in segment BC takes place when the load is placed at the far right end of the beam (at the free end). As for segment AB, moment in the segment (at any point) would be negative when the load is placed on BC. So, maximum moment at any point in AB (including point D), occurs when the load is placed at the far right end of the beam, similar to the previous case. Moment in AB would be positive if the load is placed on segment AB. The moment influence line for the beam (for D) is going to be triangular in shape with the vertex at D above the beam (forming a positive triangular area above the beam). Given the location of D, the left side of the triangle is going to have a shallower (smaller) slope than the right side of the triangle. Therefore, starting from D, the area under the triangle 3m (length of the load) to the left the point is going to be larger than the area under the triangle 3m to the right of D. This means, maximum positive moment at D develops when the right end of the distributed load is at D and the left end of the load is 3m to the left of D.

@vikymehra5902

6 жыл бұрын

Dr. Structure plz upload solution for this question.

@DrStructure

6 жыл бұрын

The links to the solutions are given in the video description field.

@vikymehra5902

6 жыл бұрын

Dr. Structure I want the solution for question having udl load instead of point load

@DrStructure

6 жыл бұрын

The solution that I describe is for a uniformly distributed load, of length 3m. To find maximum positive moment at D, please the distribute load on the beam such that the right end of it is at D. Then, analyze the beam and calculate bending moment at D. That would be the maximum positive moment at D. Then, place the distribute load such that its right end is at the free end of the beam. Analyze the beam again and determine bending moment at D. That would be the maximum negative moment at D. Using the same load position, determine bending moment at E. That would be the maximum negative moment at E. Maximum positive moment at E is zero.

Superbbbbbbbbbbbb

I can’t figure out where the 1.5 and 0.5 came from

@DrStructure

Жыл бұрын

You mean the support reactions for the beam segment FD? We get them by applying the static equilibrium equations to the beam segment. The segment is subjected to a downward force of 1 at F, there is a vertical reaction at C and a vertical reaction at D. Writing and solving the equilibrium equations for the segment will give us 1.5 for the reaction at C and -0.5 for the reaction at D.

Nice

Can you explain the question in the video at 7:47 min ???

@DrStructure

4 жыл бұрын

Think of a beam segment/bar. If it is supported by one pin/roller only, the bar has the capacity to rotate about that point. That is, if we push the bar up or down at either of its ends, it is going to rotate around the support point, very much like a seesaw. But, if the bar is resting on two pin/rollers, then the seesaw action cannot take place, the bar loses its ability to rotate. And that is what we have here. Segment DEF rests on two supports, therefore, it cannot rotate. Segment ABC rests on one support only, hence it rotates.

Sir pls give solution of example.. give atleast ILD diagram for check right or wrong... thank for all this sir 👍👍

@DrStructure

6 жыл бұрын

Solution for Exercise Problem 1: kzread.info/dash/bejne/daWV0bmLZq-YpMY.html Solution for Exercise Problem 2: kzread.info/dash/bejne/Z2qN0LSYmZObZ7Q.html Solution for Exercise Problem 3: kzread.info/dash/bejne/io6px6R-c7mvmKw.html

why at hinge u say moment is 0 but ...when there is are 2 support moment is not zero

@DrStructure

6 жыл бұрын

Take a long piece of balsa wood, cut it at mid-point, then overlap the two pieces at the cut point and push a pin through the two pieces connecting them together again. That represents a hinge connection. It allows one segment to rotate relative to the other. That is, if you held one of the balsa wood segments in your hand firmly, you can make the other piece rotate easily around the pin. The pin (hinge) cannot resist rotation, hence no bending moment develops at the hinge. Now take a similar piece of balsa wood, but don't cut it at the midpoint. Instead, let it rest on a ball (roller) at the midpoint. We can still view the piece of wood as having two (connected) pieces, the segment to the left of the ball and the segment to the right of the ball. Now, hold on to the left segment, then push down the right segment. The connection between the pieces at the roller behaves differently this time (since there is no pin/hinge) connecting the two, they are rigidly connected. In this case, no free rotation takes place at the support. Although there is a rotation, but the material/connection resists free rotation. This resistance to free rotation is due to the presence of bending moment in the member at the roller.

Plzzz mam make a vedios on Muller bresula principle mthod

Can you provide the examples answer Please?

@DrStructure

5 жыл бұрын

See the video description field for the links.

Solution????

@DrStructure

2 жыл бұрын

The solution for the exercise problems are provided in the free online course referenced in the video description field.

The worked out example has an error.. It would be -2KN

@DrStructure

3 жыл бұрын

Your assertion would have been correct if the weight of the vehicle was 1 kN. But, the vehicle weighs 2 kN. Therefore, the correct bending moment value is -4 kN-m.

@gentleman7467

3 жыл бұрын

@@DrStructure but.. We r drawing influence line... Watch again... In the animation 1kip is illustrated and calculated for 2kip...

@DrStructure

3 жыл бұрын

Not sure what you are asserting. The moment influence line was drawn and used to determine the location of the load that would cause the maximum moment at point G. The load location was determined to be point F. Consequently, a unit load was placed at F and the resulting bending moment at G was determined to be 2 kN-m (the product of 0.5 and 4). Now, since the actual truck load is 2 kN, we need to multiply the moment due the unit load by 2. Hence, the bending moment at G due to the truck load equals 2(2) = 4 kN-m.

where is the answer of the exercise? pls help

@DrStructure

2 жыл бұрын

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

Where are the solutions

@DrStructure

7 ай бұрын

The solution for the exercise problems can be found in the free online course referenced in the video description field.

Why are you taken moment as positive (as explained in the starting) instead of negative as the moment direction you shown causes hogging effect, and as you better knew that hogging moment is taken as -ve while sagging moment is taken as +ve... So the moment should be -ve instead of +ve which you have taken..that's it

@DrStructure

6 жыл бұрын

You are correct in stating that bending moment in beams is positive when it causes the beam segment to bend in the concave up direction. Indeed, that is the sign convention used in the lecture. Please point to the location in the video that has made you think the sign convention is incorrect.

@manishgautam1318

6 жыл бұрын

Although it occurs in each example in the video, for instance at 2:57, bending moment direction drawn at C tends to cause concavity downward i.e. hogging in the beam...that bending moment should be -ve.. So why are you saying like 'apply a positive bending moment to the hinge'.

@DrStructure

6 жыл бұрын

We might be using different sign conventions in describing positive moment. As this is a matter of convention, it is of no significance for all practical purposes. A commonly used sign convention for beams, the one that I have used in these lectures, assumes that positive bending moment causes compression along the top fiber and tension along the bottom fiber of the beam. This happens when the beam bends concave up. What would be the direction of the pair of moments, placed at the ends of the beam segment, that cause such a deformation? The one at the left end of the segment should be in the clockwise direction and the one acting at the right end of the segment in the counterclockwise direction. That is what we refer to as positive moment in these lectures.