SA35: Influence Line and Moving Load Series in Trusses
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- เผยแพร่เมื่อ 16 ต.ค. 2024
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This channel is underrated.
These are very great lectures, presented in a precise, concise and very understandable way. thank you very much!
This is extremely IMPRESSIVE. I don't know any other words to use to express my admiration. If profanity is allowed, maybe I could have used it before the word IMPRESSIVE 😂 . Bravo.
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thank you very Much Dr. structure. you're doing a good job
Absolutely mind blowing explanation!! Good to have a live load model as shown by vehicle representation.
Beautiful teaching with detail, and it's soo unfortunat that such scientific video does have a million view.
What a lecture!Exclaimed by Leon Raj PhD scholar at IIT Madras.
Thank you very much Dr. Structure. This video is really helping me to understand how the influence line works in reality. =) thank you
I do not have words for you description (hardworking) WHO to appreciate
Always bestwishes
Good to see this type of animation is used for civil engineers , as these concepts consume more brain to get deep understanding.Thank you for making such lectures.🙂😀
#StepByStepcivil
What a explanation,
If all subjects like this way no student will fail
This animation helped me in visualization, which further helped me in understanding the concept........
Thank you Dr. Structure for this video.......🙏🙏
thank you so much for your hardwork.. this is really helpful and truly you are impacting the world.
Thank u very much for such lecture please provide more lectures on structural engineering
great animation and a very impressive explanation... thank u very much sir...
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thank you very much!!
Very nice explanation, thank you very much...
Thank you very much for all very helpful videos.
Dr structure.can i you send a document for this lesson
@@Yaonestorgbeve What is the document for? Please explain the nature of the document and why you wish to send it to us.
How to make such videos?
Very good explanation thanks
This video was made using SketchUp, VideoScribe, and Camtasia Studio.
Excellent¡ I´m from Panama
beautifully explained! GREAT!
I appreciate your effort 😌
Great one,
you are wonderful teacher. thank you sooooooo much.
Sorry to say that my question is about concentrated moment acting on beams which is obviously not found in this video but discussed in one of your previous ones. In the real world problem, where do we see this concentrated moment or what causes this? Looking forward to your reply and explanation. Thanks in advance.
Various scenarios could lead to the placement of concentrated moments (as applied loads) on beams and frames. Suppose a beam has an overhang (say, 2 meters in length) subjected to a distributed load of 3 kN/m. This causes a clockwise moment of 6 kN-m about the nearest joint. For analysis purposes, we can remove and replace the overhang with a 3 kN downward load and a 6 kN-m moment at the joint that attaches the overhang to the rest of the beam. Also, architectural elements in a building could result on eccentric loading which introduce bending moments on the supporting frame. For example, depending on how it is places and secures inside the skeleton of the structure, a staircase could subject the supporting beam/frame to a bending moment.
@@DrStructure Thanks a lot for the quick reply and explanation.
I do get the 6 kN-m (from 3 kN/m x 2 m x 1m) moment at the joint, but why 3 kN downward load at the joint? Shouldn't it be 6 kN (from 3 kN/m x 2 m)? Please clarify.
Is the 2m long beam carrying 3 kN/m distributed load connected perpendicular to the other beam?
@@gnidnoeled786 Yes, yes! The downward force is (3 kN/m)(2m) = 6 kN.
@@DrStructure Thanks indeed. Would greatly appreciate if you could reply my other query?
@@gnidnoeled786 In this example, we are assuming the overhang is horizontal, it is placed on the same horizontal line as the rest of the beam. That means, if we cut the overhang close to the joint that connects it to the rest of the beam, the effect of the distributed load on the beam would be the clockwise moment of 6 kN-m and the vertical downward (shear) force of 6 kN. If the overhang is inclined at an angle (say 30 degrees counterclockwise) relative to the rest of the beam on the horizontal axis, we can still cut/separate the overhang and replace it with a bending moment and a force. But the magnitude of the moment and the force depend on the orientation of the load(s) acting on the overhang. If, for example, the distributed load of 3 kN/m is still acting perpendicular to the inclined overhang segment, the bending moment that it produces at the cut point comes out the same: (3 kN/m)(2 m)(1 m) = 6 kN-m clockwise. The magnitude of the shear force also remains the same: (3 kN/m)(2 m) = 6 kN. However, the force needs to be placed on the beam at an angle. That is, the shear force is not going to be perpendicular to the main beam, the direction of the force is going to deviate from the vertical (y) axis by 30 degrees. The 6 kN force is going to be inclined by 30 degrees from the y axis pointing downward.
excellent teaching
I don't understand how you determine how to center the loading? For example, around minute 9:40, what about the scenario of the middle axel above the peak?
When in doubt, we try all viable truck/load positions. That way we make sure the critical load pattern was not overlooked.
@9:40, if we move the truck 1 meter to the left, we get a smaller force in member AB. Why? Because the line to the left of the peak value has a sharper slope, it drops much faster than the line to the right of the peak. The height of the influence line 1 meter to the left of the peak is Sqrt(13)/36. Multiplied by 65, we get 6.5 kN. If the 65 kN force remains at the peak, the corresponding force value would be 39 kN. So, by shifting the load to the left, we are reducing the total force by: 39-6.5= 32.5 kN.
The 70 kN force, when shifted to the left by 1 meter gives us an additional 7 kN force. It causes an increase in the force value from 35 kN to 42 kN. And the force in member AB increases by 6 kN due to the shifting of the 60 kN to the left. So, the total increase in axial force in member AB is 7+6 = 13 kN. But the decrease in the force is 32.5 kN. So the net change is going to be negative. Meaning the proposed loading scenario is not more critical that the original one.
Axial force acting on member AB per the calculation are 86 kN in tension and 145 kN in compression. In the real world design of that AB, are we just going to use the bigger of the two axial forces, which is the compressive axial force 145 kN?
No, the design requirements for compression members are not the same as the requirements for tension members. When a member undergoes both tension and compression, it needs to be designed such that both sets of requirements are satisfied.
@@DrStructure A thousand thanks.
see if there is some traffic and all vehicles stop simultaneously would the bridge be able to undergo such load too ? why is that we dont see any bridges that has stagnant load on in
The (multiple) truck load magnitudes and patterns given in the design code(s) have been determined experimentally; they ensure the safety of the bridge. The provisions in the code specify the need for having multiple trucks being present on the bridge at one time. So it is not like we are designing the bridge for one truck load at a time. Furthermore, depending on the design, some bridges do have and post weight limits that if exceeded could result in structural damage or faliure.
Excellent
When in doubt . overbuild haha jk,
What grade and size of beam is utilized ,factoring in the cross braces over length and height . Always calculate a trucks load assuming it’s loaded . Adding over half of the trucks weight into the formula , I was on a job where the engineers forgot to factor in the weight of the fluid where new structural was installed to support piping in an oil and gas facility resulting in the structural buckling under load. Causing a lot of rework. Not to mention the job coming to a halt . Not good.
This is so well done I love it.
In 7:38 where does 5/6 at hinge and 1/6 from roller came from?
Did you mean to write @3:38 instead of @7:38?
If yes, the support reactions at the pin and roller supports can be determined using the static equilibrium equations.
Writing the moment equilibrium equation about the left end of the truss, we get:
(1)(3 m) - (Rr)(18 m) = 0 where Rr is the support reaction at the right support. Solving for Rr, we get Rr = 1/6.
Then, summing the forces in the y-direction, we get: Rl + Rr = 1 where Rl is the reaction force at the left support. Since Rr = 1/6, Rl = 1 - 1/6 = 5/6.
what software did you use to model the truck across that briddge?
The truck, and the bridge were modeled using SketchUp.
Dr. Structure thank you for letting me know! I’m doing a design project for school over this same stuff and wanted a visual representation. I hope it’s fairly easy to use
very useful vedio. thank you so much❤😍👌
please upload for 3 hined and 2 hinged arches cable
Ty dr. Structural... usefull
awesome work. thankyou so much
Dr.Strucrure Sir/Madam! You're using which softwares in your presentation?
The bridge 3D modeling and animation was done using SketchUp. The content of the lecture was prepared in Illustrator then manually traced on an Ipad resulting in SVG files. VideoScribe was then used to generate movie (avi) files from the SVGs. Last, the entire lecture was put together, voice and additional animations were added using Camtasia Studio.
For the first exercise
NCD=-35 KN
NGH=43.33 KN
NHD=-25 KN
thank you so much, please whitch soft do you use to do this animation whit bridge ?
The animation is done using SketchUp software.
thank you
Are these 3 concentrated loads (i.e. 65, 70 and 60) P/2 at each axel load?
Yes, the given load values are the half of the total load being exerted on the bridge by the truck.
@@DrStructure Thanks.
Can you please tell me what voice software you are using in this video ?
That is a natural human voice, not a computer generated one.
👌
Excellent Video .... Keep It Up :)
Most amazing video❤❤❤
can it be applied in buildings?
Why?
Influence lines can be constructed for any structural system including building frames. We can use influence lines to calculate max/min internal forces in the frame members due to the live loads present in the building.
Will you please upload influence line diagram for statically indeterminate beams.
Gracias
great videos
good way to learn
Does temperature have an effect on load capacity? Example , Such as cold temps in Canada
Significant temperature fluctuations could induce stress in members. So yes, it could effect the system. Consult your local design code to see it that is something one needs to consider when designing a structural system.
please provide a video on moving loads series for beams..i would be very thankful to you!
Awesome!!
Thank you very much for your lectures!! Would you please upload lectures on moment distribution method ??
sir if truck front wheel is at 10m from left support then force in member AB is 148kN
Please show your calculation details.
Thank you very much
Please cover the truss analysis due to UDL
Masterpiece
thanks
its good way to relative problems with naturaltes
will you please make a lecture on arches (2-3 hinged both)
Plzz upload videos on influence line diagram for arches
interesting thank you very much !!
Thanks a lot ..... tqtq
Thank you!!!
th-cam.com/video/FnbfP-UgPws/w-d-xo.html
Check out the link for more vdo on influence line diagram
Great
Thanks!
thanks so much
Thank youuuuuu
hi there, how to make a donation?
Thanks for your note. You may do so online here: lab101.space/Sponsor.asp
OK, GOT IT. THANKS AGAIN FOR THESE AMAZING VIDEOS
love u guys
We want a video on a Girder floor
Thank you😁
please upload thrust in arches
For the second exercise
NAB=214.06 KN
Es más sencillo a como mi prf me explico gracias
thank you very much!!!!