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 >
@DrStructure9 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.
@rachelwong38079 жыл бұрын
Thankyou so muchhhhhhhh: D
@TAGDAGame9 жыл бұрын
+Rachel Wong Thank you for asking this question and thank you doctor for your answer. :)
@marksuryaharja7 жыл бұрын
same question i wanna ask, and thanks for the respond doctor :)
@datauser15296 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.
@gnidnoeled7862 жыл бұрын
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 !
@tabzz03 Жыл бұрын
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.
@ahmetazrak1530 Жыл бұрын
It is incredible that you explain the information and make it more understandable with the help of visuals, thank you a lot :)
@leungtimothy28275 жыл бұрын
Thank you Dr. Structure! You save my structural analysis exam at hku today!
@DrStructure9 жыл бұрын
+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.
@CarlosPDV8 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.
@DrStructure8 жыл бұрын
The directions are consistent with our beam sign convention: bending moment causing concave up deformation is considered positive.
@CarlosPDV8 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.
@DrStructure4 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.
@anupamkushwaha48403 жыл бұрын
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❤️
@CalmInsider7 жыл бұрын
Thoroughly enjoyed the video! So informative 👌
@juhirao40756 жыл бұрын
Ur explainatation is awesome
@chaiupala2 жыл бұрын
Using animation Very easy to understand. Thank you very much.
@seymurakbarov17429 жыл бұрын
Your are the best. Thanks a lot....
@kartiksinoliya52457 жыл бұрын
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.
@Nico-qh9ew7 жыл бұрын
Massive thank you for this wonderful explenation!
@arifabdullah99527 жыл бұрын
From Bangladesh ,Thanks a Lot ,make more more video like this, only you can give me clear concept.
@aalokkafle9 жыл бұрын
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.
@ahmetkasmoglu15765 жыл бұрын
Great work. İ know a little english but i understand this lesson. Thank you very much.
@shravan45735 жыл бұрын
U really structural doctor!
@ninoenriquez24314 жыл бұрын
The robotic voice sounds so realistic and it's so clear that it makes focusing easier
@DrStructure4 жыл бұрын
That is a human voice.
@herahemel7 жыл бұрын
Great work.. It helps me much
@MehediHasan-cq4oz4 жыл бұрын
from Bangladesh?????
@a.msayem16639 жыл бұрын
love u guys for this #civilengineeringbrotherhood #civilengineersrock #drstructureurdbest
@Pankajkumar-es1ug3 жыл бұрын
U are best👍
@a.msayem16639 жыл бұрын
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
@vikymehra59026 жыл бұрын
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.
@DrStructure6 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.
@vikymehra59026 жыл бұрын
Dr. Structure plz upload solution for this question.
@DrStructure6 жыл бұрын
The links to the solutions are given in the video description field.
@vikymehra59026 жыл бұрын
Dr. Structure I want the solution for question having udl load instead of point load
@DrStructure6 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.
@veeramathy10384 жыл бұрын
Thank you a lot ...😀😀
@deepsarkar32606 жыл бұрын
Please make video on influence line diagram of indeterminate structure
@tabark25835 жыл бұрын
شكرااااااا ع الترجمة ❤️
@carlosandreszambranoespino38707 жыл бұрын
Este video me fue de muchisisisisma ayuda :) This video helped me a lot, thanks!
@onkarpatil31667 жыл бұрын
Sir pls give solution of example.. give atleast ILD diagram for check right or wrong... thank for all this sir 👍👍
@DrStructure7 жыл бұрын
Solution for Exercise Problem 1: kzbin.info/www/bejne/eaLEoIqIa7FlqsU Solution for Exercise Problem 2: kzbin.info/www/bejne/a2e8n4WVnpVobbM Solution for Exercise Problem 3: kzbin.info/www/bejne/jovYlnV7eLt8nqs
@gnidnoeled7869 ай бұрын
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?
@DrStructure9 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.
@gnidnoeled7869 ай бұрын
@@DrStructure Thank you, I now got it. 👍
@tanveeryounasbhatti31283 жыл бұрын
What a marvellous way to describe things! Which tool do you use for animation and writing?
@DrStructure3 жыл бұрын
For this video, Camtasia Studio was used for creating the animations. The writing was created using VideoScribe.
@tanveeryounasbhatti31283 жыл бұрын
@@DrStructure Thank you very much for sharing. So nice of you.
@DrStructure3 жыл бұрын
You're welcome!
@minniec22893 жыл бұрын
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?
@DrStructure3 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.
@faizanelahi32679 жыл бұрын
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
@SHUBHAMKumar-qd1pi5 жыл бұрын
When there is an anticlockwise moment why cd part moves downward pls explain
@DrStructure5 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-qd1pi5 жыл бұрын
@@DrStructure I have a question may you solve it for me
@SHUBHAMKumar-qd1pi5 жыл бұрын
I would be grateful
@DrStructure5 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
@ricardoherrera90522 жыл бұрын
Excelente
@biswaruppati89626 жыл бұрын
Please provide lectures for ILD for Indeterminate structures...
@cand.20717 жыл бұрын
I wish you showed unit moment effect in influence lines, too
@yohanesingkiriwang65717 жыл бұрын
at minute 9:54, when you calculated the Moment at G, what's the "-2" , does it come from the influence line diagram?
@DrStructure7 жыл бұрын
No, 2 does come from the influence line. It is the magnitude of the moving load (2 kN).
@yohanesingkiriwang65717 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
@DrStructure7 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.
@yohanesingkiriwang65717 жыл бұрын
Awesomee.. cheers mate!
@VIVEKSINGH-bp6fu8 жыл бұрын
+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...
@DrStructure8 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.
@hectorregala77777 жыл бұрын
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)?
@DrStructure7 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
@jericotemplado6 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.
@gentleman74673 жыл бұрын
The worked out example has an error.. It would be -2KN
@DrStructure3 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.
@gentleman74673 жыл бұрын
@@DrStructure but.. We r drawing influence line... Watch again... In the animation 1kip is illustrated and calculated for 2kip...
@DrStructure3 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.
@ubaidullah-pj9mu2 жыл бұрын
At 9:55 , why is the BM equation "-2*0.5*4"? Why is it not just "-0.5*4"?
@DrStructure2 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.
@lakkojisantoshkumar26387 жыл бұрын
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?
@DrStructure7 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.
@CivilSigmaengineers6 жыл бұрын
Superbbbbbbbbbbbb
@totatoto987 жыл бұрын
thank you so much
@marshymellie97307 жыл бұрын
where can I see the answers for the example problems?
@DrStructure7 жыл бұрын
Problem 1: kzbin.info/www/bejne/eaLEoIqIa7FlqsU Problem 2: kzbin.info/www/bejne/a2e8n4WVnpVobbM Problem 3: kzbin.info/www/bejne/jovYlnV7eLt8nqs
@phiwahalala94363 жыл бұрын
Great work, but why did we multiply the moment at G by 2?
@DrStructure3 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.
@smithakm11744 жыл бұрын
In last example, what to do if we want to find positive max BM
@DrStructure4 жыл бұрын
We place the load at either E or G.
@smithakm11744 жыл бұрын
@@DrStructure Thank you so much
@DrStructure4 жыл бұрын
You're welcome.
@noorhanalkhatib43414 жыл бұрын
In the last example where is the internal loads at Hing when you cut the beam at F ?! and thanks a lot
@DrStructure4 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.
@noorhanalkhatib43414 жыл бұрын
@@DrStructure thanks a lot .. you are so kind and I love your channel .. keep going
@tasnimshoily19717 жыл бұрын
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 ?
@DrStructure7 жыл бұрын
Are you referring to segment BCD in Exercise 1 at the end of the video?
@tasnimshoily19717 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.
@DrStructure7 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.
@nickganigan2154 жыл бұрын
At the last example, how to get the height of maximum postive at G It is 4+4/8 ?
@DrStructure4 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.
@amosmensah59029 жыл бұрын
How do you determine the heights at the maximum moments at point E F and G for the last example?
@DrStructure9 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.
@sheikhali51435 жыл бұрын
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
@DrStructure5 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.
@Only1oswaldfr Жыл бұрын
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.
@sibtainali47517 жыл бұрын
@3:55 , how come the hinge is displaced upwards ? Doesn't it resists horizontal and vertical motion ?
@DrStructure7 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.
@Kai6737 жыл бұрын
In the last example, if i want to calculate the deflection under F? How to do that?
@DrStructure7 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.
@abhishekverma14875 жыл бұрын
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.
@DrStructure5 жыл бұрын
The moment at C is zero in the example @7:54. Why do you think there is a bending moment at C?
@abhishekverma14875 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).
@DrStructure5 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.
@abhishekverma14875 жыл бұрын
@@DrStructure ok got it. Thank you
@taongamaiwanga98904 жыл бұрын
cheers
@omidhassani7 жыл бұрын
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.
@DrStructure7 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
@SHUBHAMKumar-qd1pi5 жыл бұрын
How to find max bending moment at any location with internal hinge
@DrStructure5 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.
@djalelfellah894 жыл бұрын
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 ?
@DrStructure4 жыл бұрын
Yes, exactly!
@mariajonalyndesilvagangan2857 жыл бұрын
how about a video of IL for trusses?tnx
@tluanga316 жыл бұрын
why at hinge u say moment is 0 but ...when there is are 2 support moment is not zero
@DrStructure6 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.
@idfk33027 жыл бұрын
Hello, Could you please explain the reason behind why the bending moment is always zero at a real hinge?
@DrStructure7 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.
@idfk33027 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 ?
@DrStructure7 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.
@idfk33027 жыл бұрын
+Dr. Structure Bless u for taking the time to answer us students in need. that was very helpful thank you. 🤗
@amrabdullah19955 жыл бұрын
why there is no influence line for normal forces in beams ?
@DrStructure5 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.
@amrabdullah19955 жыл бұрын
@@DrStructure thank you
@arifabdullah99527 жыл бұрын
At 9 minute 52 second ,Mg=-2(0.5)(4) ,Where is from the 2 come ? please explain
@DrStructure7 жыл бұрын
That is the weight of the vehicle, 2 kN.
@pichmeanveasnatann18578 жыл бұрын
Sorry, would you mind to show the answers at the end problem?
@DrStructure8 жыл бұрын
+Pichmean veasna KZbin annotation links at the end of the video point to the solution videos.
@emadziedan2 жыл бұрын
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!
@DrStructure2 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.
@jssplrs7 жыл бұрын
Why is the moment influence line at 8:07 like that? :( Why did it went downward instead of rotating upward at D? :(((
@DrStructure7 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.
@jssplrs7 жыл бұрын
Dr. Structure Thank you!! Really helped a lot. 😄
@manishgautam13186 жыл бұрын
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
@DrStructure6 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.
@manishgautam13186 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'.
@DrStructure6 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.
@nishi92327 жыл бұрын
Plzzz mam make a vedios on Muller bresula principle mthod
@portisikz8 жыл бұрын
hey. thanks for the video. do you have the link solution for the problems? thaanks. you're awesome
@DrStructure8 жыл бұрын
Problem 1: kzbin.info/www/bejne/eaLEoIqIa7FlqsU Problem 2: kzbin.info/www/bejne/a2e8n4WVnpVobbM Problem 3: kzbin.info/www/bejne/jovYlnV7eLt8nqs
@aiaalsaadi86136 жыл бұрын
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?
@DrStructure6 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.
@tluanga316 жыл бұрын
what is the difference between roller..hindge and triangle shape support
@DrStructure6 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.
@tluanga316 жыл бұрын
Dr. Structure thanks anyway iam still in the process
@balajigr28744 жыл бұрын
Can you explain the question in the video at 7:47 min ???
@DrStructure4 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.
@auliamuhardiarifin84277 жыл бұрын
can u explain why the beam also rotate at G ?
@DrStructure7 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.
@borhanfa89705 жыл бұрын
Can you provide the examples answer Please?
@DrStructure5 жыл бұрын
See the video description field for the links.
@joseh76123 жыл бұрын
for Mg why did you do -2(0.5)(4). Where does the 2 come from?
@DrStructure3 жыл бұрын
2 kN is the magnitude of the real (moving) load.
@joseh76123 жыл бұрын
@@DrStructure thanks
@jacoblo35672 жыл бұрын
where is the answer of the exercise? pls help
@DrStructure2 жыл бұрын
The solution to the exercise problems are provided in the (free) online course referenced in the video description field.
@FRSHBL6 жыл бұрын
Nice
@alyssacuray5006 жыл бұрын
good day! what app did you use to make this video? thank you.
@alyssacuray5006 жыл бұрын
i need something like this for me to be able to pass my project. help please.
@vikymehra59026 жыл бұрын
@7.16 what will be the moment influence line diagram if the end F is free?
@DrStructure6 жыл бұрын
That would make the beam unstable. We need to have a stable beam before we can analyze it.
@vikymehra59026 жыл бұрын
Dr. Structure upload some videos related to theory of elasticity n thanks. ☺️
@shiqing29902 жыл бұрын
Where can I check the answers for the examples at the end of video?
@DrStructure2 жыл бұрын
The solutions for the exercise problems can be found in the (free) online course referenced in the video description field.
@shiqing29902 жыл бұрын
@@DrStructure I signed up using the link provided in description but still can’t view the answers
@DrStructure2 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.
@itachiuchiha61558 жыл бұрын
How come a roller have a downward reaction?
@DrStructure8 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.
@itachiuchiha61558 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.
@accessuploads78346 жыл бұрын
Thanks
@juthinalex31068 жыл бұрын
great
@alfredjrtovar14999 ай бұрын
Where are the solutions
@DrStructure9 ай бұрын
The solution for the exercise problems can be found in the free online course referenced in the video description field.
@jasonRhawt9 жыл бұрын
how do we get the number for the influence line?
@DrStructure9 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.
@jasonRhawt9 жыл бұрын
+Dr. Structure u have any videos on this would like to watch it
@DrStructure9 жыл бұрын
+legion The exercise problems at the end of the video demonstrate the process of calculating maximum/minimum moment values.
@coolbro15022 жыл бұрын
Solution????
@DrStructure2 жыл бұрын
The solution for the exercise problems are provided in the free online course referenced in the video description field.