SA34: Influence Line in Trusses

[@Civiltechie]

I have seen many videos on this topic and left them halfway, In case of this one I understood each and every thing all along. Nice explanation.

[@FrancescoLogozzo]

Excellent dual coverage of influence lines and method of sections. Thanks!

[@HemantSingh-gg5kd]

wow this men needs appreciation 💖💖💖

[@looloolalaable]

You guys need to upload videos more. One video per month won't suffice. Awesome job. Keep it up

[@nikocelia9778]

Studying for the FE this video saved us, thank you so much!!

[@ahmadzia5896]

LOVE For Dr. Structure!

[@himachaldayal9458]

Better teacher than my lecturer 🙏

[@pankajmaurya5326]

Awesome explanation in short time... Thanx a lot dr. Structure

[@pankajkaanti1610]

Nicely explained sir, appreciable work

[@samekid273]

This is the only video that I actually understood

[@shieldonsanchez8955]

Hi, Thank you very much for this video. It has been a huge help in my preparation for my FE Exam.

[@matavalamuttej841]

Understood clearly, Thanks a lot..

[@mudharalobaidi4243]

Great Explanation !! Thank You

[@AmmarNehlawi]

Hey, thanks for the explanation it's really help! please i want to know in ( 5:00 ) how you get the angle 45? and how you get the route of 2 over 2?

[@DrStructure]

The height of the truss = the height of the right triangle = L. The width of each truss segment = the base of the right triangle = L. The makes the angle 45 degrees. Cosine (45) = sqrt (2) / 2

[@AmmarNehlawi]

what if the height does "not" equal the width .. how can i find the angle? for example the height = 6m and width = 5m , i think we should use (tan angle = height\width) that mean to find the angle ( angle = arc tan height\width ) am i right?

[@DrStructure]

Yes, arc tangent can be used to find the angle. However, tangent of an angle is the ratio of the side of the right triangle facing the angle divided by the side adjacent to the angle. In our case, if the height of the truss is 6m, then we need to find arc tan(5/6).

[@aamirabdulsalam]

@@DrStructure now i got it 45 90 45

[@Patrickorwa]

I know you got lost because you got FdjCos45 = 1/square root 2 . This is also correct since 1/sqr root 2= sqr root 2 / 2.

[@thansiyapa4721]

Thank you ssoo much!!! Dr. Structure :)

[@Leandroz15]

Awesome lesson is short time! Thanks a lot for your job.

[@venkyrce]

Wowww., So simple, I love it

[@nipurnkhatri]

Your videos are Awesome... Hates off Please, upload video about structural design (Steel & RCC)

[@SaurabhSingh-bs1cd]

Mind blowing😇

[@ecuawezzy]

Nice! Do influences lines in frames next. Please!

[@Advait_thakur]

Good explaination 👌

[@gh5030]

Thank you for this great explanation.

[@HossamKorin]

Great Vid, Don't need to subscribe for all vids but will do so anyway to promote your awesome efforts!

[@DrStructure]

Thanks for your support.

[@koketsolegoka6242]

Thank you so much ,this video was really useful💖.

[@birdsarentreal3054]

Thank you so much Doctor 💙🙏🙏

[@ismairig6746]

Amazing, thanks for the video!

[@mitch3850]

Is it possible for a member's influence line to exceed the value of 1? I'm grappling a problem where the truss is 12m long (2m segments), and only 1.5m in height. The truss is upside down like the one shown 2:52. I haven't seen any examples where this is the case.

[@DrStructure]

Is it possible for the axial force in a member exceed the applied load? Absolutely. This depends mainly on the geometry of the truss. A unit force applied far away from an end of a truss could result in large member forces near that end.

[@mitch3850]

Understood. Thank you for the quick response!

[@orbital16]

Wondering how he came up with sqrt.2/2Fdj at 5:13..thanks..

[@DrStructure]

cos (45) = sqrt(2)/2 = 0.707

[@orbital16]

@Drstructure..lots of love..thank you.

[@susmitaa248]

@@DrStructure Isn't cos (45) = 1 / sqrt(2)?

[@DrStructure]

​@@susmitaa248 Yes, it is. If we multiply the numerator and denominator of 1/sqrt(2) by sqrt(2), we get sqrt(2)/2. So the two expressions are equal to each other.

[@susmitaa248]

@@DrStructure Thanks Dr Structure :) I have an exam tomorrow on influence lines and your videos have been extremely useful!

[@Haiderali-dp3yz]

Waooo Thanku #Doctor

[@SA-wb4qb]

Amazing, thank you a lot guys

[@Kreighzey]

amazing vid keep it up

[@debangadutta6517]

can you please provide ILD, SF and BM (also maxm) at specific location and maxm absolute BM for rolling load on SSB or girder....

[@frankieeog1625]

everything good but are changing the sign conventions for the moments at a?

[@lamillado]

Hello. Why do diagonal members in trusses have positive and negative influnce line?

[@DrStructure]

Depending on the location of the load, a diagonal member could be subjected to either a compressive (negative) force or a tensile (positive) force. So, as the load moves across the truss (as the load location changes), the direction of the axial force in the member could change from positive to negative. This positive/negative force direction is reflected in the influence line for the member.

[@lamillado]

@@DrStructure thank you very much!! Just one more question. Since trusses have slender members in capable of compression that cause buckling (dangerous), what do we do in actual to prevent this failure in trusses? I hope you also answers this. Will really appreciate it. Thank you!!

[@DrStructure]

We can control (prevent) buckling of the member by decreasing its unbraced length (by bracing it) and/or increasing its radius of gyration (a cross sectional property).

[@lamillado]

@@DrStructure thank you so much!! ❤

[@sidiskieeldani71]

Thanks you sir its very understandable 😘

[@keithcoless2140]

THANK YOU SO MUCH

[@AdanEntertainment]

appreciable work tnx

[@konstantinoskritos3082]

In analyzing trusses we said that loads are applied in the joints, because truss members are only under axial forces. Then how is it that a concentrated force is acting in the middle of a truss member, and what happens in between? Is this idealization valid (of connecting points in the influence line diagram with staights) or is there a different model dealing with that case?

[@DrStructure]

The moving load does not act directly on the truss. The load acts on the bridge deck which then is transferred to the cross beams supporting the deck before finding its way to the truss joints. As the unit load travels on the deck, it is distributed proportionally to the two cross beams supporting the deck segment. For example, when the load is at the midpoint of a deck segment, half the load goes to the left support beam and half the load goes to the right support beam. In turn, these 1/2 loads are transferred to two distinct truss joints. When we are drawing the influence line, there is no need to show these intermediate steps (since the diagram is a line and we only need its end points to draw it), we just analyze the truss when the unit load is applied directly to the joints.

[@queenop3493]

best vdo

[@venkyrce]

Sir can you do the process for top and bottom chord members,

[@camposkarlryanl.6368]

THANKSS!

[@supriyakhedkar7175]

Sir, which software were used to make this vedio

[@DrStructure]

For this particular video, we used videoscribe (an online tool) to create the (text/writing) animations. The additional (motion) animations were done using Camtasia Studio.

[@sicasica211]

sorry may I ask if the truss member receive vertical load e.g. G with 30N downwards will there have shear diagram being drawn?

[@DrStructure]

No, truss members carry axial loads only. There is no shear force or bending moment in a truss member.

[@sicasica211]

@@DrStructure ok thanks

[@ecuawezzy]

If the load is at the top chord then the I.L should be -1. Right!

[@DrStructure]

It depends on the member that we are drawing the influence line for. The vertical members would be in tension if the load moves along the bottom chord. They go under compression if the unit load moves along the top chord. So, the influence line for a vertical member needs to be drawn below the x-axis when the unit load is on the top, and above the x-axis when the unit load is at the bottom. However, some of the diagonal members may not switch sign (from tension to compression, or vice versa) regardless of the (top/bottom) position of the unit load. When in doubt, it is best not to make any assumptions and analyze the member to make sure the right diagram is drawn.

[@tandindorji9575]

Why is the component JF*Cos45 not used in the equilibrium equation? Shouldnt the equation be like 1/5-JD*Cos45-JF*Cos45=0,When unit load is place at B.

[@DrStructure]

No, the force in JF does not appear in the equilibrium equation because there is no force labeled JF on the free-body diagram. That force appears only if we cut member JF. But since the member has been left intact on that particular free-body diagram, its internal force does not play a role in the equilibrium equations.

[@tandindorji9575]

@@DrStructure Thank you so much for the information.

[@DrStructure]

You're welcome.

[@neltan3842]

Is there a software that can be used for calculating influence line for truss?

[@DrStructure]

Robot Structural Analysis by Autodesk is one, there might be other, less expensive tools out there as well.

[@FlabeCoStructuralEngineers]

Hello sir,I want to design lenticular truss and its shape is complicated. So is it possible to draw the ILD of these type of trusses?

[@DrStructure]

An influence line can be drawn for any type of truss. If you can analyze the truss, you can draw its member influence lines. Of course, here we are assuming the path of load for the truss is a straight line which is always the case for bridges. If such a path does not exist, the influence may not serve any useful purpose.

[@dopaumingodonquixote1144]

Will you pls tell me how you assume EJ is a zero force member at if the force is at any joint other than E. Your reply will be much appreciated.

[@DrStructure]

Consider the free-body diagram of joint E when no load is being applied to the joint. Three forces would be acting at the join: the force in member DE, the force in EF, and the force in EJ. The first two forces are in the horizontal (x) direction, and the last one is in the vertical (y) direction. For the joint to be in the state of equilibrium the sum of the forces in the x and y directions must be zero. Assuming Fef, Fde and Fej are the three member forces, the equilibrium equations can be written as: Fef + Fde = 0 Fej = 0. As you can see, from the second equation we can conclude that EJ is a zero-force member.

[@mattxyj]

Amazing Video! This really helped! I just want to ask if the reference used for this was Kassimali's. Thank you.

[@DrStructure]

Thanks for the feedback. To answer your question, both Kassimali and Hibbeler textbooks are good references, and offer excellent presentations of various structural analysis topics. However, none of our lectures are based entirely on textbook materials.

[@mattxyj]

@@DrStructure Oh I see. Thank you again.

[@nikhilchoudhary2943]

sir where's answers of exercise problems?

[@DrStructure]

You should be able to see the links at the end of the video, though they may not be visible on platforms other than desktop. Here are the links: Exercise 1: kzbin.info/www/bejne/anTMeJyOqLiXY8k Exercise 2: kzbin.info/www/bejne/a6CVfKF-ar99rKc

[@nikhilchoudhary2943]

Dr. Structure thankyou sir

[@DrStructure]

You are welcome!

[@hellsing3062]

@@DrStructure thank you so much! It really helped!!

[@manishjangid7981]

Shouldn't be the ordinate at D for Influence line for Force in member DJ be {3*Sqrt(2)/5} ?

[@DrStructure]

No. (sqrt(2)/2) Fdj = 3/5

[@hugokung6991]

Grade Safeguarded!

[@skychahal1518]

Thank you soooo muchh

[@hamedabdelazim7717]

fantastic

[@ericfranzen]

7:24 - Where did the -1 come from? Cryptic!

[@DrStructure]

That is the unit moving load.

[@veronicaadelamonzonpando1969]

The underlines(shading) are wrong ,it think they must be vertical

[@DrStructure]

I am not certain what you are referring to, feel free to elaborate.

[@debangadutta6517]

it would be of great help

[@dstutz54]

What if the selected member is horizontal. Say I-H

[@DrStructure]

Place the unit load at B and calculate the support reactions. Then, cut the truss vertically thru members IH, IC, and CD, diving the structure into two parts. Take the left part, and write the moment equilibrium equation at point C. The only unknown force in that equation is the internal force in member IH. Solve the equation for the unknown force. Place the unit load at C and repeat the above steps. Place the unit load at D and repeat the steps. For each unit load position, we get a value for the force in IH. Plot them to get the influence line for the member.

[@riathakur5472]

I almost didn’t see the man speaking 😂

[@PavanAK74]

please upload about moment distribution method

[@Thomasdekkar]

owsum x 10^100

[@randomx9864]

thanks ™

[@rupeshprajapati8288]

can you please provide the answers for exercise problems sir!!!

[@DrStructure]

The links are given in the project description above.

[@rupeshprajapati8288]

Dr. Structure i couldnt find it....can you provide the link sir!!!!

[@DrStructure]

Exercise 1: kzbin.info/www/bejne/anTMeJyOqLiXY8k Exercise 2: kzbin.info/www/bejne/a6CVfKF-ar99rKc

[@rupeshprajapati8288]

Dr. Structure thank you sir!!!!!!

[@catobsessed379]

can you please upload a video on influence line - (indeterminate structure) !!!!

[@amritashgautam2335]

❤️

[@TheSterlingArcher16]

It's an absolute joke the FE expects you to perform problems this difficult or more in less than 3 minutes.

[@jonnyfull317]

@Archer They want you in the DANGER ZONE

[@jbraddy2795]

this voice is so deep it gives me anxiety...

[@damienchand3503]

Bring back the lady voice

[@elenamarievillamin2706]

The lecture is supposedly good but Watching at midnight is not advisable because of the animated man talking beside it. Gosh i stopped watching because it creeps me out

[@athan6025]

😂😂😂😂😂 the 2nd take away exercise is not funny!!