SA36: Analysis of a Roof Truss: Method of Joints
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- เผยแพร่เมื่อ 21 มี.ค. 2017
- This lecture is a part of our online course on introductory structural analysis. Sign up using the following URL: courses.structure.education/
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Your force at member bh is wrong .. And the rest of the members for compression and tension. Maybe u need to review the forces or ur computations..
Why is the force in member BH incorrect? can you explain? can you support your assertion?
@@DrStructure i used the equation in my scical..
@@DrStructure the principle in solving the forces using method of joint is correct but u need to review the values of the forces maybe u need to recalculate it, anyways im just telling what ive found.
BH should be 14.07kN in compression and JH should be 86.42 KN in tension
@@chinitoengineer5462 Let’s do some qualitative reasoning here to see why JH cannot be a tensile member.
Think of the truss as a hollowed beam which is subjected to a series of downward loads. Consequently, the “beam” is going to bend downward which causes compression along the top fiber and tension along the bottom fiber of the member. Translating this back into the truss terminology, we can conclude that the members along the top of the truss are going to be compressed and the members along the bottom chord of the truss are going to be stretched. This simple reasoning invalidates the assertion that JH is in tension.
I leave the calculations and validations of the force magnitudes as an exercise problem, and welcome any solid reasoning and/or proof that you can offer in support of your conclusion.
Omg, how cool is this! Thank you so much for these videos, I wish my professors were as creative as you!
Thank you so much! Finally I have found a video where it breaks down the whole picture of structural analysis, from the structure to it's simplified set up problem!!!
More like this please Dr.Structure.
I agree. We need more such presentations :)
Really fascinated by seeing the actual application of concepts. Loved the video.
This channel is golden. I wish I had this resource when i took this course at uni.
this is a best structural course on the world. thank you Dr
Thank you so much for these videos, they are very helpful to see how all those theories I learned in uni apply in real life situations.
A wonderful lecture. It reaches us how to calculate each component stress of roof trusses. With this we can design the extraordinary trusses while the safety is guaranteed.
Thank you so much Dr. Structure. These lectures are very much easy to understand and well explained. :}
Awesome! Requesting to publish a load comparison video between different truss models. That would further help viewers to chose the best truss for their roofs!
Wow! Fantastic lecture.
FANTASTIC Video!
Thank u so much for contributing to the human world. amazing amazing video.
Awesome!! Please make more & more such interesting video about structural analysis. I have one request, please make such awesome videos on "Structural Design" RCC and Steel.
Wow! Thank you so much... I love this video. Thank you for sharing.
Massive respect Dr.
u r the best dr...
Hello... Your videos are Awasome!!!
Please upload videos on RCC building design..... Means how to design reinforcement in each member.... How to evaluate loads acting on the structure etc.. make videos of A to Z for the reinforcement... Also include how structural designer design reinforcement in the member... So make videos about how to design slab (including beams)... From A 2 Z......
Very useful simply explaining, thanks a lot
thanks very helpful. im working on a roof trus design and this has helped me understand how load transfers from roof to a trus. i thought truss design was useless as i did it in yr 1 of my structural mechanics class. now after 5 years i am needing to use that method again xD
Wow this is so cool now I understand thanks a lot maam 🙂
Woooow These videos from real life structures are amazing, keep going, people I love you, they are very helpful for my students, I advice them to watch these videos, and I ' d like to ask for a similar related channel like that which is more specialized in steel structures that shows them, for example how to connect many columns above each other in multistorey frame/truss .
Thanks in advance
Thanks for your comment.
Thank you for the info.
I wish this is how they teach you structural analysis in college.
wowwwwww I love this video thank you so much
So simple and attractive, can we compare these manual results with staad pro out put ??
Thanks a lot.
Oh my gosh thank you!
good explanation
Dr. Structure you is the best structural analysis professor to me
for that, i ask if there are any videos about finite element method and plates .membrane analysis
Mahmoud Taha I think they upload it before .. Check out the videos on their channel and if your re using the web you can easily serach in the channel using words
thanks alot for you, I check the channel and not found anything
You are correct, we have not yet produced lectures on advanced structural analysis topics. We will get to them eventually, after the basic topics are covered thoroughly.
thanks a lot, I will waiting for your awesome explanation
Awesome.
it is very useful .
well explain
Dr. Structure, can i have a pdf copy on how you derived the formula of the reactions of the indeterminate beams on your lecture SA36?thanks
Unfortunately, we don't yet have a pdf for this lecture. We are working on creating pdfs for all the videos. Hopefully we can get them out sooner rather than later.
Nice presentation. Thank you. Why did not consider the angle force acting in purlins to truss?
Sure, other forces such as the weight of purlins as well as own weight of the truss are in play here. In this lecture, the focus is on the roof load only. In the actual design process, we often analyze the structure under each load type, then use the principle of superposition to determine the combined load effects on the members.
Impressive. Which software did you use for the animation part for the building?
The building was modeled and animated using SketchUp.
dear admin.
do we neet to put purlin at the joint of the truss? and so. we will get axial tension and compression force.
The purlins are there, directly supporting the roof. Yes, they may carry axial force depending on the geometry of the structure and the load pattern. For example, if wind load is present in the direction parallel to the purlins, significant axial compresssive force develops in those members.
The analysis of the purlins however is not a part of the 2D truss analysis since they are not in the plane of the truss.
Awesome video. At around 4:50 you say "If we analyzed this statically indeterminate beam, we would get ..." Was this done in another video? What method was used to do it if not? Thanks
Since that is an indeterminate beam, either the slope-deflection method or the displacement method would work. No, we don't have the video for that analysis. But you can find videos on several indeterminate analysis techniques on this channel.
What about including the weight of the truss and the ceiling etc. attached to it at the bottom chord? How do we show the load arrows in the truss diagram?
The weight of each truss member can be distributed equally between its end joints. In this case, there is no ceiling, there is only a roof. If a ceiling exists and it is supported along the bottom chord of the truss, then its weight can be distributed to the truss joint along the bottom chord, similar to how the roof load is distributed among the joints along the top chord.
how did you get the reactions of the statically indeterminate beams?
We can use the slope-deflection method or the matrix displacement method to analyze such indeterminate beams in order to determine their reactions.
Wch application and pen u hv used
May I know what type of truss is this? I can’t found it from any book.
This is called a double fan truss.
after the actual application, we rely on the computer to calculate the structure?
Yes, in most cases we use software to analyze structures. But, the main responsibility of structural engineers is not to learn how to use software tools, rather it is to use their knowledge of how structures behave, and the principles of design to ensure the safety and reliability of the system. No software system can do that on our behalf, at least not yet. :)
this video is awesome
Do we really need all these zero force members when we are going to design a portal frame in real life? Is it necessary to keep these members?
Sometimes, for economic reasons we may want to use prefabricated trusses. In such cases, depending on how the truss is being integrated with the rest of the structural system, we may end up with zero-force members for some loading cases.
Also, although some of our truss members carry no force under the roof (gravity) load, but they could end up carrying axial force under the wind (horizontal ) load. For example, AG and GH would end-up carrying an axial force, if a horizontal (wind) load is present at G. So, this "zero-force" labeling varies from load case to load case for a given truss.
Further, if a truss has long compression members, say like member HG, then for stability reasons (to prevent member buckling) we may need to introduce additional members (like member BI) which for the sake of static analysis can be considered zero-force members.
Thanks a lot for reply. Would you please continue another video. What type of truss (K truss, Pratt truss, Warren truss) can we use to minimize cost for same load applied (38.5 kN on one side)? What shape of steel (I beam, W beam, H beam and channel) and size (web, flange, thickness) can we use to minimize cost for same load applied?
How did you calculate the beam support reactions?
The long beam is a continuous one, meaning that we need to employ an analysis technique for statically indeterminate beams to calculate its support reactions.
hi dr. do we always used pin and roller support? because in the actual its usually a fixed support both side. does it has a big difference if we used fixed support?
please advice.
thank you.
Yes, all sorts of connections are used specially in steel structures. If a support is fixed, a bending moment develops at the joint which could make a big difference in design calculations. Here, the truss is assumed to be supported by a pin and a roller in order to make it statically determinate. Otherwise, we could not have analyzed it using the Method of Joints. We would have had to use a more advanced technique to analyze the truss.
Dr. Structure ... If the supports could be fixed supports on both sides...what method could we use to analyse the structure...??
That would make the structure indeterminate, slope-deflection method, force method or matrix displacement method can be used to analyze such structures. Here however, given that we have a truss structure, using a fixed support (capable of carrying bending moment) does not make much sense since no such force generally develops in trusses.
I have a question dr. Structure ...we have analyse the roof truss by considering only the incoming load from the roof cover and transmit it to the supporting members of the structure as 38.5 Kn but what about the self weight of the surpotting members such as purlins and the principal rafter don’t they contribute in adding some weight to the supports...?? I’ll be delighted to get the feedback
Yes, weight of the various structural elements need to be considered in the design process. Here, we are not actually designing the system, we are just illustrating the analysis under a common load, and not all the loads. If this was a real design project, then we would have considered the other members' own weight as well as the various live loads such as snow, wind,...
Very well...would you please please make a video on how to design a steel roof truss by considering other member’s own weight so that we can understand how get the actual load transmitted to the supporting columns...i and others with the same problem we’ll be very thankful
I personally understand preparation of these videos takes a while and i understand ...for the sake of time will you please link me up with a clear information on how to calculate dead load,live and wind load of the steel roof truss to the supporting members and by then i’ll be able to continue wth my design ...please help Even if you can take information interms of pictures of your work or anyhow i’ll be very glad ...loshylarrystylz@gmail.com thats my gmail...thank You dr. Structure in advance
A simple way to get the truss members' own weight into the calculations is to determine the weight of each member, then add them all up to get the total weight of the truss. Divide that total weight by the length of the truss to get a uniformly distributed load which can then be placed as a gravity load on the truss. For analysis purposes, this requires converting the distributed load to joint loads to be placed either along the top chord or bottom chord of the truss. For example, if the bottom truss chord consists of four joints, then the middle joints each takes W/3 portion of the total load (W being the total wight of the truss), and the two end joints each takes W/6 portion of the load.
If the truss rests on columns at its ends, then each column would take half of the weight of the truss. That is, each column would take W/2 compression force from the truss.
Dr. Structure ...So im gonna need to find the weight of each member in my truss and add them up with the incoming load from the roof cover...and divide the overall load by teo because my truss rest on two supporting columns...iz dat right if the structure iz symetrical??
Thanks for great presentation , but you refer to lecture 5 that I could not find.
Lecture SA04 is being referenced at the beginning of this lecture. Below is the link to that lecture as well as lecture SA05. Generally, you can search for a particular lecture on youtube using keywords like this: "Dr. Structure SA04". Alternatively, visit our website: Lab101.Space and look under Content Library.
SA04: th-cam.com/video/Evsjp0zKeGw/w-d-xo.html
SA04U (an updated version of the lecture): th-cam.com/video/wxAsjywBbMg/w-d-xo.html
SA05: th-cam.com/video/LAb_ezELQ_Y/w-d-xo.html
how did you determine the span to depth ratio. in many cases it ranges from 12 -24
In principle, that is a design consideration often decided by the architect based on the underlying architectural design criteria and the cost of construction.
Dr structure...i have a question after obtaining the member forces of the truss structure like the above which method am i using to design members such as Rafter and struts...??
The actual design of structural members is governed by design coded and standards which vary from region/country to region/country. For example, in the U.S., the American Society of Civil Engineers, American Concrete Institute and American Institute of Steel Construction publish codes and standards that are heavily used in the design of concrete and steel structures.
Im having sloping parallel truss...can i analyse it by using this method of joint??i tried to analyse it by using this method but the member segments of parallel rafters from one joint to another keeps rising...i started from the left support of the reaction and it keeps rising as it goes to the top...so i js wonder where m i wrong??
I need to see the truss you are trying to analyze before I can answer your question.
How come the reactions from the beams transferred to the truss is directing downward? Shouldn't be those reactions must be normal to the truss joints?
也可以把竖向力分解成垂直于杆件和平行于杆件的分力
Are those downward forces/loads at the bottom chord (not in this problem) are loads due to ceiling loads? Where/How to include the dead load of the truss?
You can calculate the weight of each truss member (area * length * specific weight of the material), then place half of the weight at each end node of the member as a point gravity load.
Yes, generally speaking, the loads along the bottom chord of a roof truss come from the ceiling.
Why y dont find the reactions on other beams? Since we have different loads on other beams we have to determine the reactions right?
We have 5.5 kn and 11 kn
Each side of the roof is supported by three beams. We do calculate the load for all three beams (@6:00). Two of the beam have the same tributary area, hence, they carry the same load. The other beam, at the top of the roof, having a smaller tributary area, carries a smaller load.
how did we get 1,58m and the other differences between 2 member ?
Not sure what distance you are referring to. If you are referencing the distance between the roof beams/purlins, they are obtained from the given dimensions of the system. The side of each roof panel is about 9.18 m. The beams/purlins divide this length into 5 equal distances, hence each segment ends up having a width of 1.84 m.
Is it possible to have trusses instead of beams in the roof ?
Yes, roofs in houses (like the one discussed in this lecture) are often designed using trusses.
Nice video... What softwares are used for making this video???
Several software tools were used for this video. The 3D model and its animation was done in SketchUp. The text and graphic animation was done by manually tracing static pages on an ipad and then using VideoScribe to turn the tracings into video segments. The entire video was then put together using Camtasia Studio.
Thanks for the kind Answers
Thankyou for your honest response, may god bless you.
What about the weight of the truss itself? How do we compute and include them as dead load?
We can calculate the total weight of each truss member, then place half of it at each end joint of the member as a downward (gravity) load. A member’s weight can be calculated by first multiplying the specific weight of the material (steel, wood, ...) by the cross-sectional area of the member to get weight per unit length. Then multiply that by the member’s length to get the total weight.
@@DrStructure Thanks a lot.
You are welcome!
hi dr.. can you include windload calculation?
Yes, we will include it in a future lecture.
ขอบคุณครับ
But if we put purlins instead of beams, how do we transfer the weight of the roof and the purlins to the truss? And how do we transfer the calculated weight of the truss to the column. Thank you in advance.
For analysis purposes, purlins are beams. The roof sits on top of the purlins which in turn rest on the roof truss.
The truss has a vertical support reaction at each end. The reaction force is the force/weight that the truss transfers to the column/wall.
@@DrStructure so i need to calculate the weight of the roof cover and the purlins and then add that to the self weight of the truss to get the vertical reaction on both ends of the truss?
@@miguelcarloperez7770 For dead load analysis, yes.
can anyone explain to me Tan 0.96m/ 1.8m in 9:42 minutes what i know the length of y axes is 0.6m how you got the difference which is 0.36m i will be so much thankful
Say, the angle that the inclined (roof) member makes with horizontal axis is alpha.
tangent (alpha) = 9/1.8 where 9 is half of the span length and 1.8 is the height of the inclined member.
At H, the overall height can be written as (0.6 + h). We can use tangent (alpha) to determine h.
Note the small right triangle that has GH as its hypotenuse. The base of that triangle is 1.8 and its height is h. And, the angle that the hypotenuse makes with horizontal axis is alpha. So, we can write:
tangent (alpha) = 9/1.8 = 1.8/h
Solving the above equality for h, we get: h = 0.36
Then, the overall height at H becomes: (0.6 + h) = (0.6 + 0.36) = 0.96
what type of truss is this?
I mean the specific name for this type of a truss and what is the column height that should be provided for such type of trusses ....
It has the fan truss configuration. The height of the supporting columns is a design issue, there is no standard height. This type of truss is commonly seen in industrial buildings. Such buildings typically have a height of about 40 feet (12 m).
Dr. Structure it's called fan truss?
does it come under the king post truss classification?
Yes, there is such a thing as a fan truss. It is more like a fink truss than a king post one. King post has a vertical member at the center of truss, something that is missing in our truss. Of course, we can modify the fan configuration by adding a vertical member at the center of the truss making it related to the king post.
thank you soo much
would you plz discuss the method of moment distribution in the future
Yes, moment distribution is the topic of the next two or three lectures.
thx a lot you are amazing teacher ^_^
how do I know the load of the roof in my house ?! ... I can't even give an approximate number to it
It depends on the type (material) of the roof and its dimensions. For example, for a concrete roof, one can use the specific weight of concrete (~150 lb/cubic foot) as a starting point. But, for actual design purposes you always need to study the governing local design codes for such information.
+Dr. Structure Thank you 😘😁😀
Hope you can teach us concrete and steel structures.
Yes, we are moving in that direction, slowly but surely. Once we are done with the analysis topics, we’ll
get into design issues.
what if C-purlins are used instead of beams?
As far as this analysis is concerned, the shape of the cross-section of the purlins, or the other structural members, has no bearing on the results.
👍 👌 👍
Hi Dr. Structure and everyone. Can anyone explain why HJ carries 86KN and BC carries 91KN when the whole roof's weight is only about 77KN? Thanks in advance for the answer.
It is not uncommon for the applied loads to produce bigger internal forces in the structure. The equilibrium equations can show us the mathematical necessity for this phenomenon. For example, consider a truss joint connecting two members each making a 10 degrees angle with the horizontal axis. If the joint is subjected to a downward force of 50 kN, since the sum of the forces in the vertical direction must be zero, we can write: 2F sin(10) = 50 where F is the axial force in each truss member. Solving the above equation for F, we get: 144 kN, a significantly bigger force than what is being applied at the joint.
i wonder who is the awesome man/woman behind these videos
good day to you good sir, i have 2 question below, would appreicate if you can help me with it
1. 4:52 video time, can you show us how you determine the reaction at the beam load to be 1.4w, 4.1w, 3.5w 3.5w, 4.1w, 1.4w ?
2. 6:45 video time, if i were to lock both end using bolts and nuts to column ?
@ 4:52 we have a statically indeterminate beam. To see how the reaction forces are determined, we need to actually analyze the beam. But since the lecture is not about the analysis of indeterminate beams, the process was not explained. You can use iFrame (See Lecture IT02) to quickly analyze the beam and validate the results, although, when using iFrame, you should use a numeric value for the load instead of w.
Yes, the ends of the truss are physically attached to the columns using either bolts/nuts or weld. Either way, the same member forces result. A welded connection makes the joint more rigid than a bolted connection, but for our analysis purposes here in this case, the difference between the two types of connections is insignificant.
p.s. Like your screen name :)
@@DrStructure Your work is TRULY AMAZINGLY VERY AWESOME, that's how i really felt, i am not sue how many people you have unconciously help, but i hope the Lord is with you and he bless you and bless you again
sorry i have some additional question ( sorry i am not a structure engineer, most of my structural knowledge comes from your work
1. do you have a lecture to determine load of indeterminate beam ?
2. if the ends of bolt are welded, wouldn't it become "fixed" with moment ? that would cause the column to take moment as well too right ?
3. If we lock using bolt and nuts it will be sondier as "pinned" and if we weld it will be consider as "fixed" ?
4. Now we assume 1 side is pinned and the other end is roller, but in actual both side of truss are supported by pinned ( bolts and nuts ), in that case how will it affect the support reaction ? will they be bigger/smaller ?
PS : i have one more question on free body diagram, it's a simple frame ( 4 column and 4 beam )
something like below
magnumpiering.com/wp-content/uploads/2017/12/Moment-Frame-Details-100x100.jpg
and a tank on top of it, problem is really the wind, i have no idea how the wind force will go into the column and beam, because the wind will act in the center of the tank and it's a vertical surface but my beam is horizontal, i have some PDF which are very clear ( isometric drawing ) on the example, if you are comfortable with it may i send the question to you ?
also base on the picture of the frame ( simple 4 leg column and 4 beam ), is that a fixed support or pinned support if i bolts and nuts my column to the ground ?
once again, thank you very much
@@DOTA2MAJISTRATE Thanks for your feedback.
1 ) Yes, we have several lectures on the analysis of statically indeterminate beams. You can see the list of our lectures and a link to each on our website: lab101.space/Course-StructuralAnalysis.asp The slope-deflection method, the force method, the moment distribution method, and the displacement method all deal with the analysis of indeterminate structures (beams and frames).
2) Correct, if the ends are welded, the connection is considered fixed, and that could cause the development of significant bending moment in the columns, depending on the location and direction of the loads. But as far as the truss itself is concerned, a fixed vs pin support does not make much of a difference in the member forces.
3) Correct, that is generally true. Although we can also design bolted connections that resist moment, act more or less as a fixed support.
4) We can actually design a connection as a roller by having the bottom chord of the truss simply rest on the column sandwiching a friction pad. A fraction pad does provide some level of resistance against slippage of the beam, but unlike a pin (bolted connection), it would allow horizontal movement. Generally, We don’t see that type of connection in residential buildings, but they can be found in modern bridges and other such structures where shrinkage and expansion of the beam due to temperature change could be an issue. A roller (friction pad) allows the member to expand without causing additional stress. If both sides of the truss are actually pinned, and if there is a horizontal force being applied to the structure, horizontal reaction forces develop at both supports, and the compression member(s) along the bottom chord of the truss would end up carrying more stress/force, as the pins prevent the members from expanding. In contrast, when one of the supports is a roller, the member(s) can expand/roll resulting in less stress. So, the difference between the two scenarios, would be the amount of displacement and stress that takes places in the members, specially along the bottom chord of the truss (assuming a similar truss configuration as shown in the video).
5) That is a rather involved problem. Wind effects the tank which effects the frame. Usually, we want to use three-dimensional analysis to accurately model the behavior of the system and figure out the forces and stresses that result. However, to do a quick back-of-envelope analysis, assuming the tank is rigidly connected to the frame. We can replace the tank with four vertical columns, each rising from the middle of the beam that the tank is resting on. Then, we can distribute the wind load among those columns. For example, we can argue that if wind is in North-South direction, then the two columns in the direction take most of the wind load, so we are going to distribute the wind load between those two columns equally and ignore the other columns. If however, wind blows in East-West direction, we assume the other two columns take most of the load. This way, we can get rid of the tank, and place the load on the frame by adding four fictitious columns to the system. This revised system is much easier to analyze, although we still need to do some more work to transfer the load from the vertical column to the actual frame.
is this text to speech engine or human voice?
Human voice.
At 9:35 the height of 0.96m is wrong, if you calculate the height correctly it would be 1050mm
Please show how you arrive at 1050mm for the height in question.
Hii what if both the end are hinged , there will be lateral load on column and that will produce huge moment in column..
But if say column is hinged at bottom then the moment will not generate and get transferred to concrete pedestal via non uniform bearing pressure..
Hii sr please send your whatsapp no.
If the truss is attached to the columns using non-rigid connections (i.e., pin and roller as was assumed in this lecture), then the columns being pin-connected at their base makes the frame unstable.
If however we make the beam-column connection rigid, at the their base, the columns can be pin-connected: either both pins, or a pin and a roller. In that case, the equilibrium equations tell us how the applied lateral load transfers to the base of the columns. In the case of a simply supported frame subjected to a lateral load, bending moment reaches its maximum value at the joint connecting the beam to the pin-based column.
If both end of truss are hinged then there will be large horizontal force component acting in the connected column how to transfer that much large moment to the soil please tell
If the truss is connected to the columns using two hinges (one on each side of the frame), then the frame (consisting of two columns and the truss) needs to be fixed at the base of at least one of the columns, and either fixed or pinned at the base of the other column. Otherwise, the system becomes unstable.
To properly transfer the forces and moments at the base of the frame to the soil, the frame needs to rest on a properly designed foundation. For example, one can design the right size footing that can effectively transfer the axial force, the shear force, and the bending moment from the column to the soil.
Hii sr please send a way to contact you i will send you image of my problem...
My question to you is if both end are hinged of truss then due to equilbrium at truss ends a large horizontal force will be generated..
And finally that force will act on the columns producing large moment if base of column is fixed...
If base of column is not fixed it is pinned then?? Like in case of base plate
A bit too technical for me.
spanish :,v please