Tek-Tips is the largest IT community on the Internet today!

Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!

  • Congratulations TugboatEng on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Lintel Arching effect interupted? 2

Status
Not open for further replies.

mar2805

Structural
Joined
Dec 21, 2008
Messages
375
Location
HR
HI guys.
This is my First time caclulating a amasonry structure and Ive been reading a lot of topics on this matter but few things are not clear to me.
If you look at the picture attach:
Imaginary arching triangle is intrupted by a RC floorslab (green line).
Should I still provide the traingle load from wall Dead weight or should I take full rectangular block from the lintel top side to the top side of the slab + tringular shape from the top side of the slab ?
 
Tony,
please look at the pictures attached.
Isometric view
You basicly have a beam wich is suporting an wall above it.
In picture "view" you can see the view and a section thru it.
We usualy converted the whole wall to the line load acting on a beam.
Should the wall above beam also be converted to a triangle or not?
 
 http://files.engineering.com/getfile.aspx?folder=e985a7e2-948f-4d2e-9bb2-e2e943bcc169&file=view.jpg
If that was my job, I would take the whole load of the wall above as a udl on the beam. Is it a concrete frame?
 
Mar2805:
In your last two attachments the entire brick wall load is on the spandrel beam below that particular wall, and the spandrels at each level should be designed accordingly. Keep in mind than concrete shrinks and clay masonry tends to expand with age, so you must leave a joint at the top of the brick, at each story, to tolerate this expansion/contraction activity.

Take a look at a few good masonry design texts, masonry construction handbooks or masonry association tech notes for some better understanding of the loading of lintels and the arching action of various types and ages of masonry. The 45̊ triangle is a practical representation of the masonry below an arch, which might/would fall out without support, or is loading the lintel. The exact angle is subject to a bunch of variables about the make-up of the wall, otherwise the arch is in compression and starts to support the brick. The 60̊ angle is a reasonable representation of a line where the masonry is being held up by a normal corbeling action of the wall system above and around it. And, you might think of the space btwn. the two angles almost as a real arch that has its two arcs separated by a height about twice the thickness of the wall. Both of these actions require some monolith of masonry wall beyond the extent (width) of the opening below. That is, the arch needs some structure beyond the opening to take the horiz. thrust induced by the arching action. If you cut these angled lines (the arch) by an opening above your lintel, then the arching action can’t really take place, and all bets are off. Thus, Tony’s first attachment 27FEB14, 16:16, is really only correct for a large expanse of wall with no openings or concentrated loadings in the region of the triangles. His attachment from is code book gives a pretty good general explanation of the way the loads are figured, note the “common sense...” admonition. But, in your earlier attachments I would apply the full slab above and the couple feet of masonry to the lintel, and quit the screwin around with the load, unless I needed to finesse the shear, reaction or bearing a bit at a jamb. In your sketch showing symmetrical windows above, you’ve cut into the loading triangles, so it is doubtful that arching action can take place. I would apply all the wall load btwn. the windows directly to the slab. I would allow that the inner window jamb load corbel out (45̊) down to the slab, and then apply that to the lintel. The outer window jambs pretty much get carried straight down and added to the jamb load at your large, lower opening. On your sketch with the door opening on the left, there is just no arching action possible because of the door opening. Again, if I needed to finesse things I would allow the 60̊ corbeling thinking on the right side, but all of this load must still be taken down to the lower, large opening jambs.

The thing that you have to be careful about, for all this chasing the loads around, is that you don’t lose or forget any loads in the process. That is to say... even if the lintel doesn’t carry all the loading from above it in bending, due to the arching action; the opening jambs and wall section immediately adjacent to the jambs must carry all the loading (total actual load) from above the opening, and concentrated loads above may concentrate these total loads more on one jamb than the other.
 
@tony
"Is it a concrete frame"
Yes the wall is sourronunede by EC columns and beams (beam slab action)

@dhenger
So for the situation with 2 Windows above the garage door the load should be something like this (attach)
I included the load from the slaba bove since it sits inside the 60deg triangle. The load from the walls is included inside the red polygon.
As tony mentioned reaction from the lintels above should be also included, right? Should they be coverted to unifrom load also or....? How?
 
 http://files.engineering.com/getfile.aspx?folder=d1c4a360-d72a-4bb1-8f77-ac74e336e7d4&file=gar1.jpg
@dhengr,
You've explained it better than I could, and clarified a point for me also.
Thanks.
 
Keep in mind that all the wall load needs to go somewhere. It doesn't disappear just because you drew a triangle on the elevation sketch.

When there is a door or window offeset from the opening below it serves to concentrate the loads at the opening jamb. This might include a great deal of tributary area from floors and walls above. When openings are offset at the lower level of a multi-story assembly it can be a real killer for the lintels below.

Your included load needs to go to the centerline of the windows above, not just the lintel's edge as shown in mar2805's sketch. And if it's a multi-story building ALL the load from above needs a path to the ground. It does not magically disappear when it encounters window or door lintel.

If you think in terms of stiffness and load path you can work these out for yourself (this is what an engineer is, and does). If you think in terms of rules of thumb and shortcuts you won't have a feel for what's happening and you might end up with a grossly unconservative design.
 
Mar2805:
I’ll say it again..., Tony’s first sketch and the first diag. on his code copy really only applies in a solid (continuous) masonry wall. You can’t get the arching or corbeling action the 45̊ or 60̊ triangles contemplate when you cut the imaginary arch rib by an opening like your door (on the left) or your two windows. That 3rd fl. (roof) slab is carried by the window lintels to their jambs and by the red polygon (width btwn. the windows) directly down to the large, lower lintel. I would not count on any arching effect btwn. the windows. The loads on the outer window jambs can start to distribute (corbel) out into the wall beyond, but mostly they go right down into the large, lower jambs as a concentrated loading. At least that’s they way I’d design those lower jambs and a few feet adjacent to them. Some vert. rebar right at the jamb and maybe some extra cells grouted immediately adjacent. And, some of that vert. rebar might be carried right up to the 3rd fl. bond beam if the window openings permitted it. Your red polygon is mostly the masonry DL, plus the slab loading btwn. the windows, but within that polygon you also have the inner window jambs, and they would distribute out below the window sills at 45̊ like the ‘point load from partition’ as is shown on the second page of Tony’s code copy. I think you should remove the 60̊ triangle from your last sketch. You could grout and reinforce the masonry from the large lintel to the 2nd fl. slab and you would have a good deep lintel, tied into the bond beam under the 2nd fl.
 
Guys Im sorry but any chance that you might skecth what you said?
Its much easier to understand it.
YOu got me confusing...

@JLNJ
Can you see the attachement. Two situation:
1) Windows aligned vertical
2.) Windows not aligned verticaly

YOu say that the 2 situation is very dangeruous if you floolow "by the book" 45 and 60 deg load paths.
Can you please explain and post a sketech of your comment.
Thanx.
 
 http://files.engineering.com/getfile.aspx?folder=72b69835-041c-43d2-88b7-3920de94df2a&file=window.jpg
Ok.
Plase just simple yes or no.
Sketch attached.
Load on the lintel is coming from walls Dead weight (indicated in red polygon) + line load from the slab (indicated in yellow)
Correct?

If yes one can determine the lintels support reaction, bending moment....

Can someone please now sketch the load thats transfered ONLY from this window lintel to the large garage lintel.
 
 http://files.engineering.com/getfile.aspx?folder=653e1f56-f213-4841-a960-7abebbbab862&file=garf_copy.jpg
errrr....guys you wrote half a page on "How to"...so can you please just answer my question from my last post "5 Mar 14 7:53 "
Thanx
 
Hi!
I found an article here that explains how to deal with concetrated loads.
It seems that you disperse the load to a width of 4x the wall thickness and then divide this lenght with the concetrated load in order to get a liner load acting on the lintel below.
See link on page 2.
If I use this on my previous post were I got a support lintel reaction of 20 kN and use this force as the attacking force on the lintel below I get an value of 33.33kN/m. See sketch attached.
Is this correct?
Come on guys I need some advice :-(

 
Considering onlyyour 20KN reaction from the lintel,I would have thought that it would have dispersed downward at 45 deg, as the red lines, as far as the lower slab. What happens to that load below the slab would depend on its stiffness. Conservatively, you could assume it continues down at 45 deg until it reaches the lintel.
 
 http://files.engineering.com/getfile.aspx?folder=b11ea8ad-c8e6-40ed-b8fe-141e8d93eb38&file=Scan0026.jpg
Are these CMU walls supporting the slab/concrete beam or does the concrete beam/slab support the block or is this like a brick veneer?
Is this new design or analysis of existing?
The southeast brick institute has a publication on brick lintels which you could add some continuous longitudinal joint reinforcement to the courses just above the lintel.
Are we talking about behaviour or design? The brick or block especially if placed in a stack bond will not only arch but act as a deep beam. This is behaviour but you need to be careful when using this information for design as you do not want excessive deflection.

I would think that the 60deg would be more appropriate considering there is masonry only on one side of the point load (reaction), but this is only a guess. Tony may know better. But before we get too carried away, I suppose no matter how you slice it these are approximations.

EIT
 
Yes, the attachment you've posted 5 Mar 14 7:53 shows the correct solution for the top left window. I don't know if I'd bother taking it to this level, as it would be easier to simply carry everything down to the lower lintel which is actually above the top left window. That said, you are absolutely able to have arching above the slab where the opening and insufficiently deep masonry occurs below.

Note that beyond all the academic discussion, brick needs to be in at least fair to good condition, properly maintained (ie: 1/3 depth repointed with matching mortar), in order to arch. Where in doubt, carry loads straight down without 45 or 60 degree triangles. You will then be conservative and avoid a common source of failure.
 
Mar2805:
You haven’t bothered to show us details of the wall construction, or told us what type of mat’l. the wall or lintel is made of. You haven’t bothered to put any dimensions on your sketches or the loads. These are all important bits of info. if you are wanting a meaningful discussion, they do influence the discussion and a proper design. And, an experienced engineer will draw conclusions and first impressions from that more complete info. fairly quickly, and without it he is almost blind, just guessing what you are really doing. Without it, your question and sketches seem kinda hypothetical and textbookish, so it is tough to give anything other than general answers. You have gotten some good info. from a number of the posters here, and you should read it a couple times for its full meaning. It seems that you did finally do a little literature digging on your own, I think someone suggested that too, and that is a really important thing to do if you want to become a good engineer and retain what you’re learning. Your latest sketch looks about right. And, if you look back, I said about the same thing in talking about the jambs and an adjacent couple of feet of wall taking all the loads from above. There’s that 60̊ again, and give or take a few degrees it doesn’t make much difference in the final analysis. Don’t undersize the lintel and don’t forget, for all the screwin around with the loads, the two jambs plus a couple feet of adjacent wall must carry all the loads from above; don’t chase the loads around and lose them, and then not account for them, that’s dangerous. Learn to think like the loads, how will they act, and take the most direct path to the foundation; look at the structure and follow this path all the way to the ftg.. Remember compatibility (re: strain and various element movement), if an element starts to be overloaded (a lintel, a jamb) can, or will, above elements (w.r.t. the lintel), or adjacent elements (the jambs) pick up some of that load, without any serious repercussions. If you do cut that load path (your windows or door vs. arching) or you can’t draw a clean load path, then you better think damn hard about how to provide a proper load path around the void. I don’t think I’ve ever seen a properly designed structure fail because it was overloaded by 5-10%, but I’ve seen plenty of failures because a proper load path wasn’t provided or because of poor detailing. The load didn’t know you shot a bull, it tried to go where it knew it should, and you didn’t provide a path or detail which would take that loading.
 
Thanx for advices.
I attached an .pdf drawing of the problem.
You have some dimensions in there as well as mine "load chategorisation" on the garage door lintel.
Wall self weight, slab support reactions, lintel reactions.
I wonder if the 2nd floor lintel raction have any influence on the garage door lintel.
I think that their influence is lost due to fact that the wall self weight, at some point going down from the small window lintel, becomes bigger then the window lintels disperesed reaction force.
Something like Boussinesq formula for stresess in soild due to surface loads...
I now it suonds funny but its kinnda similiar situation...
What do you think
 
 http://files.engineering.com/getfile.aspx?folder=3a579c07-2562-458f-ad6e-3fbfba59fe1a&file=sketch.pdf
That looks correct - BUT why would you ever want to be so very precise? There is no justification for spreading the load in the brick "just because you can".

Apply arching ABOVE a lintel.
Apply loads straight down from reactions.
Ensure you continue all loads down to foundation.

These are safe, time tested methodologies that won't fail you.
 
Mar2805:
For about the eighth time, you are chasing the friggin loads around and losing half of them along the way. You have an arching type triangular load right over the window lintels for lintel design and lintel reactions, but you don’t indicate any way that the rest of the loads over that window (above the arching triangle and added to the lintel reactions) get concentrated at the window jambs due to the arching action. That extra (lost) load includes masonry and slab loads above the triangle area on both sides of the window center line. And, you have the masonry weight btwn. the windows to add to the red slab loads. Don’t worry real much about finessing Boussinesq’s work until you have a better grasp of basic loads and load paths. CELinOttawa and I are saying essentially the same thing.
 
I STRONGLY urge you to listen to dhengr. The more you over-complicate this, the more opportunity you ggive yourself to screw up. Relax, a firm and deep understanding of load paths and all the finess in the world will come with time. Slow down, and account for all loads as *simply* and *reliably* as you can.

The Cross method works not because it tells you some magically "correct" state of stress in a structure, but because it ensures that there is a possible state of stress at which the structure will be strong enough.

Worry about these three basics:

- Strength
- Stiffness
- Stability

Forget the rest; You'll pick it up along the way whether you want to or not. Right now you're just cruzing for a brusing.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor

Back
Top