any engineers in the house?

[Jens]

I have a need for someone to run some figures to tell me if a steel beam is going to be big enough for a particular application.  All the engineers I know are mechanical or civil engineers.  I can give you all details as far as the construction that will make the load (not the actual load however).  I had an engineer come out and give a quote...$750!  And I told her the beam I wanted to use, where I wanted to place it, how I wanted to place it, etc.  I just wanted her to say yes it will work, or no it won't.  Wasn't expecting $750 for a couple of calcs.  I'm not saying that everyone's time isn't worth something, but maybe I should go back to school and change my line of work.  I'll send you a case of beer, some homemade soup and bread, cookies...give you building advice...I don't know.

Anybody able to help?  Just let me know what you need from me to calc.

[MountainDon]

 It's not so much the time it might take to crunch the numbers as it is the years that went into earning the diploma. Those years give the engineer the knowledge of how to do the calculations as well as the right to apply the stamp to the plans. The engineer assumes some risk by applying their stamp of approval to the plan. You are also buying that assurance with the services of a professional engineer.

[eddiescabin]

Oh, THAT knid of engineer...I thought you just needed me to pull the locomotive around to the front of the cabin!?!

[Redoverfarm]

Maybe one of these might help Jens

http://www.toolbase.org/PDF/DesignGuides/ResidentialSteelLoad_SpanTables.pdf

http://www.windyhilllogworks.com/Calcs/steelbeamclc.htm

[desimulacra]

Insurance is a big part of it also you have to be bonded at a very high rate. $750.00 seems a little high though. If you know the load and span I can make a "guess" at it.

[John_M]

If you know the load and span I can make a "guess" at it.

That is why you are paying $750.  You are getting sound advice from a qualified individual who is insured.  I'm guessing the individual above isn't going to pay your medical bills if your steel beam fails!

I do agree that the price seems high though....

[Jens]

I understand the whole thing about schooling and insurance.  It is no different than what a good builder goes through though.  How do you think most people would respond to a contractor bidding even a small job out at $175 per hour, and saying that it was going to take 8 hours to put that standard split jamb interior door in?  I did get the junior engineer though, and she was "only' $75 per hour.  Still said that it would be 3 1/2 hours to write the prescription, only 3 I think it was to make the calculations.

I don't get it.  Most good contractors are just as invested, have just as much responsibility to their client.  It can take way longer to become a good contractor, and cost way more money, than it does to go to school to be an engineer.  Oh well, uneven scales like this are the way of the world. 

wanna know the funny part?  What I want to do is give extra support to joists that have been doing their job for 90 years, and have ended up suffering about 2 inches of sag over 18'.  2x8's going 18 feet, supporting the second floor exterior wall load @ 4' in from their end.  Wasn't even going to restructure, just help support.  Its like having to get an engineer to be able to buy a bra.

[Jens]

btw, how do I actually calculate the load?  Walls are 9 ft tall, sheathing is board sheathing 3/4", siding is 4" clapboard,  second floor is 30' long, with 14' over the joists in question (30'x24' overall), roof is 3.5 in 12 pitch, 1/2" sheathing, architectural shingles.  Beginning of load is about 8' from beam end.  There will be a support at the end of each 14' beam, and a central support.

[NM_Shooter]

I don't get it.  Most good contractors are just as invested, have just as much responsibility to their client.  It can take way longer to become a good contractor, and cost way more money, than it does to go to school to be an engineer. 

Not usually though.  The time, money and effort involved in becoming a PE is huge.  Anybody with two years experience pushing a broom and $2k can become a contractor in NM.  It definitely takes a lot more discipline and mental acuity to become an engineer.  Most engineers started being serious about their careers back in middle school... making A's in math and science.  I don't know many contractors who have had to suffer and make grades through mechanics of materials, thermodynamics, differential equations, and the like.  Those sorts of classes will just piss you off and make you want to charge more money.  And after all that schooling, then you get to take more tests... the EIT first, then the PE. 

That PE stamp is worth more than its weight in gold. 

[poppy]

Where's Don_P when you need him?

Figuring the load is not necessarily so hard.  Dead weights can be estimated from knowledge of the materials used.

Live loads can be determined by codes.  Snow loads should be a known factor.

Then the question becomes how is the beam going to be supported on the ends?

Another question is are you going to jack up the floor or just add the beam to prevent deflection beyond the 2"?

How long is the beam?  A 6 or 8" wide-flange beam can take some pretty hefty loads.

[ak_rob]

I'll give it a shot being more of a steel guy than a wood guy. If you can come up with a weight (building ,snow load, everything in it)  and a span most good steel yard or online  have a beam chart to spec.how big a beam for how much span and deflection. For reference I'm doing a 14x18 cabin with 100 lbs snow load 4x6 beams 7lbs beam I thing don't have the numbers here. When I did the cal. One beam would support the hold cabin. I went with two to hard to balance one beam.
Looking at your numbers you only have two 7' spans with good footing its not going to take much of a beam.

[ak_rob]

Something like this it's 4'+/- spacing. The single beam would support the cabin. This is old photo before cross bracing.

[Don_P]

Hey Poppy, been drawing and redrawing a house, sized several beams this morning. It will go through an engineer before its over. An engineer's fee? It takes an awful expensive lawyer to blame their error on me  .

First thing out of the gate... if this is helping out a grandfathered structure and if you are not changing occupancy class or the structure then you are proposing a piece of trim.

Generally you can throw 10 psf at the wall so 90 lbs/lineal foot of wall. It looks like you are in 15 psf snow, the roof is generally 10psf dead load so 25 psf total on the roof. I'd be happy to show you how to calculate tributary load on the beam if you'd like, a good skill to pick up and practice. It'll take a dimensioned sketch, you can do it in paint.

[John Raabe]

My engineer usually charges $300 to $400 for the whole house. This includes checking the beams, tracing all the loads, designing shear walls and any modifications to the foundation, tie downs, etc. He marks up the plans and provides calc sheets for the inspector. If the plans need to be wet stamped I make the changes to the drawings, and take them back for the stamps.

When I have needed something simple like a few steel beams I do not leave the drawings but sit down and work with the engineer at the desk. Call it a walk-in review. Then it is harder to charge more than the 1/2 hour that it actually took to do the work. I write a check right then so there is no billing cycle and he is happy to do this again next time.

You might find if you ask for a quick consult and will bring your checkbook or cash that you will get the help you want. If you don't need documents for the inspector and just need advice, tell the engineer that up front.

[Jens]

"You might find if you ask for a quick consult and will bring your checkbook or cash that you will get the help you want. If you don't need documents for the inspector and just need advice, tell the engineer that up front."

I did.  Maybe that's why it was only billed out for the junior engineer at $75 per hour, instead of the head engineer at $175 per hour. 

NM Shooter, notice that I said "good" contractors.  Those aren't the guys who just simply have a tool belt and an old pickup truck.  The investment that I speak of that most have, cannot be learned in any classroom.  Monetarily speaking, there are many contractors who also have huge amounts of money invested in there business.  Here in TN, to be able to build a $500,000 house, you have to have personal assets of $100,000 ($50k liquid, $50k secured), just to be able to have your license.  Many of the good contractors that I have known, also have bachelors degrees (usually in business), yet people still wouldn't pay any of them the same rates as a doctor or engineer.  The $1m insurance and bonding that my partner and I had in New Hampshire cost over $10k for the year...more than the malpractice policy of a friend who is a therapist ($6000 per year). 

Its just the way it is.  I am almost to the point of accepting it in my life, that some people just plain make more money than others.  Why?  Because when you need a doctor, a lawyer, or an engineer, there are a lot fewer to choose from than when it comes to builders.  Good builders are still just as scarce.  Plus, when you need a doctor, lawyer, or engineer, it is because insurance is paying for it, your freedom is on the line, or you are cornered into it by a building department (driven by society's stupidity over the years) that makes it so that there is no other way.  Oh well.

don-p.  in a 4' section of wall, there is about 250 pounds or so (sheathing, siding, framing, insulation, drywall).  Of course, the wall is only 4" wide, so I would assume that the 62.5 lbs per foot would be greater because of this point loading.  But then again, the load is not strictly on one point, but is spread out over the length of the plate.  Does that mean that it is less than 62.5? 

What is the difference between dead and live loads (dead is always there, live is people/furniture, and stuff right?), and which would this be considered? 

The joists are on 19" centers, and are true 2x8, heart pine.  They span 18 feet across the living room.  4 feet from the first floor exterior wall, is where the second floor wall sits.  The spans will be 14' each.  They will meet in the center over a post that will be continued to a foundation footer in the crawlspace.  I will join them with a steel plate, and 1/2" bolts.  The beams on hand are 6.5" wide, and 8" tall, with a thickness of approx 3/8".  I believe that it is 25lbs per foot.  There will be a bit of jacking up under the joists, and the plan right now is to attach the joists to a wood plate on top of the beams.  Preferable to this (although considerably more work), is to cut the joists, and pocket the beam into the ceiling/floor, hanging the joists on top mounted hangers (or pad out the web, and use standard hangers). 

My original plan, was to do this cut and hang method using a doubled LVL, 1.75/11 or so.  The lumber yards guy told me that it would need 5 pieces of LVL though.  Plan B, was to replace all of the joists with 2x12's.  That would've taken the bounce and sag out of the floor, but left it with a too-heavy load IMO.  Plan C entailed sistering 2x10's to the old joists, lowering the ceiling a bit (to make it straight), and straightening out the floor above.  The steel beam turned into the plan, when my clients dad (who is definitely a good, and intelligent man) took her to the salvage yard to see what the beams would cost.  When they returned, they told me that they had bought them.  They made up my mind for me, which is fine I guess.  Now I just have to figure out how to make it work, and get them in place (safely).

If I were to pocket the beam in, I am sure that the inspector would raise his eyebrows at it.  If I just add it under the joists (which have been holding up for 90 years BTW), I have a feeling it won't be such a problem.  Even still, I'd like to know it is big enough (I have a feeling it is at least 30% bigger than necessary).  I will try to add a drawing soon.

[Jens]

drawings





let me know if that is clear as mud.  just a quickie drawing, don't know if you need other info.

[Jens]

that's better.  can actually see it.  BTW, hold down the control key, and scroll mouse wheel to zoom on Windows 7

[davidj]

The difference isn't between a good contractor and a good engineer - both have spent a long time to get where they are and have a huge range of skills and experience.  The difference is between a bad contractor and a bad engineer - the bad engineer has still spent 4 years and $100K+ getting a degree, the same again under close supervision, and then passed some well-controlled professional exams.  After all of this, even the bad ones shouldn't be too incompetent.  When it comes to bad contractors, all bets are off!

[NM_Shooter]

Know what?  I betcha the world's best contractor charges way, way, way more for his time than the world's best PE 

[RainDog]

The difference isn't between a good contractor and a good engineer - both have spent a long time to get where they are and have a huge range of skills and experience.  The difference is between a bad contractor and a bad engineer - the bad engineer has still spent 4 years and $100K+ getting a degree, the same again under close supervision, and then passed some well-controlled professional exams.  After all of this, even the bad ones shouldn't be too incompetent.  When it comes to bad contractors, all bets are off!

Ooh, excellent insight. I like that! Never would have occurred to me to think of it in those terms.

[Don_P]

in a 4' section of wall, there is about 250 pounds or so (sheathing, siding, framing, insulation, drywall).  Of course, the wall is only 4" wide, so I would assume that the 62.5 lbs per foot would be greater because of this point loading.  But then again, the load is not strictly on one point, but is spread out over the length of the plate.  Does that mean that it is less than 62.5?

My wife has accused me of starting a conversation in the middle and working out towards both ends  I'm assuming someone has determined that the wall weighs 250 lbs per 4 lineal feet. or 62.5 lbs/lineal foot or 7 pounds per square foot...sounds reasonable.

What is the difference between dead and live loads (dead is always there, live is people/furniture, and stuff right?), and which would this be considered?

Dead loads are the permanent loads, the materials of which the building is constructed. Dead loads can have some special effect such as inducing creep, sag, over time. Dead load contributes to some responses, the resistance to uplift and overturning in wind. It can be disregarded in live load deflection calculations. Live loads are the shorter term loads of occupancy and environment. The occupants furniture and debris is considered a 10 year live load.  In the case of wood, it can handle short duration high loads well so the design values for wood reflect an adjustment for short duration live loads...snow, max load for 2 months 15%, construction load 1 week 25%, wind/earthquake 10 minutes 60%, impact 200%

I don't understand the 4' between the I beam and the 1st floor wall, is that 4' exposed to the outdoors on the 2nd floor?

Lets look at the 14' left hand I beam span in the drawing. The floor is spanning 4' from foundation wall to the beam, so 2' is bearing on the foundation and 2' is bearing on the beam. On the other side of the beam ~7' is bearing on the beam and 7' is bearing on some other support. Combined, we have a tributary width of 9' bearing on the beam. Multiply that by the 14'span and we have a tributary load area of 126 square feet. A floor is designed with 10psf dead and typically 40psf live load, 126sfx50psf=6300lbs. This is uniformly distributed by the joists along the length of the beam.

The wall adds 62.5 lbs per lineal foot, 62.5x14=875 lbs. This is also uniformly distributed along the beam. If you hang a chain from the beam and use that to pull the engine of your truck, then we would have a point load concentrated at some place on the beam.

The beam weighs 24plf X 14lf= 336 lbs

So far I have 6300+875+336 lbs=7511 lbs uniformly distributed along the beam span.

Now we need to know what is bearing on the wall above. Is the roof center bearing or does it span from a far wall at some distance away and on the wall over the beam? Or is there a floor above the wall

The joists are on 19" centers, and are true 2x8, heart pine.  They span 18 feet across the living room.  4 feet from the first floor exterior wall, is where the second floor wall sits.  The spans will be 14' each.  They will meet in the center over a post that will be continued to a foundation footer in the crawlspace.  I will join them with a steel plate, and 1/2" bolts.  The beams on hand are 6.5" wide, and 8" tall, with a thickness of approx 3/8".  I believe that it is 25lbs per foot.  There will be a bit of jacking up under the joists, and the plan right now is to attach the joists to a wood plate on top of the beams.  Preferable to this (although considerably more work), is to cut the joists, and pocket the beam into the ceiling/floor, hanging the joists on top mounted hangers (or pad out the web, and use standard hangers). 

My original plan, was to do this cut and hang method using a doubled LVL, 1.75/11 or so.  The lumber yards guy told me that it would need 5 pieces of LVL though.  Plan B, was to replace all of the joists with 2x12's.  That would've taken the bounce and sag out of the floor

A W8x24 is an 8" tall 6.5" widex24lbs per foot beam, there is a W8x28 with similar dimensions but I'll be conservative. Strength of materials... Its probably A36, structural steel with a yield strength of 36,000 psi, divide that by 1.67 to get a safe working stress. So allowable extreme fiberstress (Fb) is 21,500 psi in bending. I've used the lower allowable design stress, the higher value is 24ksi. The southern pine joists by comparison have an Fb of about 1200 psi. In stifness the steel has a modulus of elasticity (E) of 29,000,000psi, the pine 1,600,000.

From what I understand thus far I can do this, IF the load is evenly distributed along its length...a "uniformly distributed load" as opposed to some form of "point load" (a concentrated load in one area). A W8x24 spanning 14' can carry 18,000 lbs with about 1/2" deflection. Max allowable deflection is L/360 for a floor...168" (14')/360= .466"... ~1/2". We still have about 11kips unused capacity but I haven't figured any loads from above the wall. The AISC manual lists the allowable uniform load on that beam as 24,000 lbs at 14' span, it would deflect about 5/8" under that load.

The simple stuff should be in a carpenters bag of tricks. The public probably hasn't noticed the changes to code that have occured just during my career. The old codebooks allowed me to design simple beams and girders as long as my math was according to accepted engineering practices. It was a realization that good design not embossed paper was the goal. People tend to fulfill our expectations, there has been some good marketing. 

You will need the services of an engineer throughout your career, become friends with one. The semi retired are smart and often want to help you learn, a younger one can be with you through life and learn with you. The little jobs are sometimes just a conversation then. I've had a beam checked for blueberries. They were good blueberries though 

[Jens]

"I don't understand the 4' between the I beam and the 1st floor wall, is that 4' exposed to the outdoors on the 2nd floor? "

The second floor exterior walls, do not line up with the first floor.  It is a 720 square foot space, framed up in the middle of a 2080 square foot space.  The exterior walls of the second floor, have roof that goes from them to the exterior walls on the first floor.  Think of it as being a 24x30 foot two story, with a wraparound porch that has been completely enclosed (but the railing of the porch is the actual exterior walls of the first floor.

"The AISC manual lists the allowable uniform load on that beam as 24,000 lbs at 14' span, it would deflect about 5/8" under that load."

this is point load, yes?  Surely an evenly distributed load cannot exert these same stresses if, "So allowable extreme fiberstress (Fb) is 21,500 psi in bending", which I assume means that it would take a point load of 21,500 psi to bend it.  Am I correct in this thought process?  Some of these joists do not even have any bearing on the 1st floor plate (they float above it), because they are held up by adjacent joists and the bottom wall plate of the wall above (one 20d nail).  I have one section that I had to replace the top plates (about 6 ft worth), and I didn't even have to jack up on them to do so. 

BTW, I do have friends who are engineers.  Unfortunately, none of them are in structural.  Chemical engineers, mechanical engineers, genetic engineers, nuclear engineers, biological engineers...no structural.  I think that it is actually one of the less common engineering fields.  I even have long conversations with one acquaintance who is a quantum physicist!  No structural engineers though.

Thank you to everyone for your help here. 

[Don_P]

Hmm, I've confused you.

Let's look first at the AISC table. It is for uniformly distributed loads and is giving maximum allowable load in bending. The max safe load is 24,000 lbs uniformly distributed along the length of the beam. If I hang the entire load from the middle of the beam it will have a much greater bending moment than if the load is evenly spread along the beam. I can show you how to calculate the bending moment for each condition but here's the short version, the beam has half the capacity if point loaded at midspan. You could only hang 12,000 lbs safely from the middle of the beam. Either condition, 24 kips uniformly distributed or 12 kips point loaded at midspan, produces the same 42,000 lb-ft bending moment. Another table in the manual lists beams and spans by maximum allowable bending moment, at 14' span this beam is listed for ...42,000 lb-ft. A 24,000 pound evenly distributed weight or a 12,000 pound gorilla hanging from the middle is a 42,000 lb-ft load on a 14' beam.

You've worked on cars. The beam is a torque wrench handle and the moment is the torque setting. The max allowable moment in the beam is the same as the maximum torque setting on the torque wrench, go over that and you risk permanently bending the handle.


The Fb number is the strength of the material and does not indicate the load capacity of the beam but gives the bending strength of A36 steel in pounds per square inch. Fb, E, Fv, etc are design values giving the strength in bending, the stiffness, and the maximum shear the material itself can safely handle, the material properties.  Since in this post I've used the steel manual numbers I'll use their higher Fb number to continue, that was Fb 24,000 psi instead of 21,500.
Getting our moment units into inches... 42,000 lb-ft x 12...maximum bending moment=504,000 in-lbs.
Divide that by the maximum allowable fiberstress in bending (Fb) 24,000 pound/square inch
Section Modulus required: 504,000/24,000 =21"³

This gives me the "section modulus" of the needed beam. Think about what just happened, I took the bending force and divided it by the strength of the desired material. I now have a size that I'm searching for, an A36 steel beam with a section modulus of 21. For an I beam determining section modulus is mind numbing, but they have a table, a W8x24 has a section modulus of 20.9

Soo, Fb is not the load capacity it is the unit strength of the material. If we wanted to use yellow pine for that same load, I would divide the the moment by the design value for that material, Fb 1200psi
504,000/1200=420. so to carry the same load but with a pine beam I need a section modulus of 420"³. That would take a 6x22, 8x20, 10x18,12x16 or 14x14 yellow pine timber to do the same work as our 8" I beam...same load, different strength of materials. Section modulus is describing the geometry of the beam, the required tradeoff of width to depth to maintain the same strength.

I've wandered from determining the load to sizing the beam but it is all steps along a common path, is this making sense?

The roof is quite steep rising 9' in 4' of run, does it then gambrel from the top of the second floor wall?
Has the roof also sagged 2"?

I'm not sure if you've worked with wood that has taken a creep set. The deflection will not jack out in a day. I'd use screwjack posts under the beam and sneak them up a crank or three at a time over as long a period as it takes to slowly nurse them back to something approaching flat.

[Jens]

So what you are saying is (I am not trying to be a smartass here, BTW), an 8" tall, 6.5" wide beam, is not enough to give some extra support to my joists?  Remember what I said earlier too, about the joists not all actually even being loaded.  The bottom plate of the wall actually seems to be acting like a mini-header. 

It is not a gambrel roof, all the roofs are 3-12.  This is the house.  Notice how the second floor, and first floor walls do not line up with eachother.

[Don_P]

 I've given the capacity of the beam at 14' span both in full allowable safe load, 24,000 lbs, and its capacity at code L/360 deflection, 18,000 lbs.

We are still trying to determine the load, the photo really helps.
I've identified thus far 7,511 lbs of load, the floor, wall and self weight of the beam. I've drawn my assumptions so far. The loads we've identified so far are in blue, the new beam is in red. There is alot of capacity left at this point but we haven't identified all the loads.



What is the width of the 2nd floor? Can you draw in all other load bearing supports and distances for the second floor? Is there a bearing under the ridge or does the upper roof span from wall to wall?

Without the I beam the wall in blue is delivering a point load to the span of the 1st floor joists.

The joists not being evenly loaded means that some are taking the lions share of the load and others are coasting, "Load goes to stiffness". If they are rough sawn thick and thin then some shimming over the beam will be neccessary as you bring it up to try to even things out.