Good morning! This is Back of the Envelope – the place to learn structural engineering in tiny bites .

In today’s article, I will talk about how to do a quick-n-dirty preliminary check to make sure your composite steel deck meets the fire rating requirement.

This is something that confuses me almost all the time… but there are some rule of thumb that could be applied for back-of-the-envelope checks.

Let’s dive in.

(Estimated read time = 1 minute and 30 seconds — I told you, tiny bites)

By the way, this is a rehash of an article I wrote in my weekly newsletter, “Back of the Envelope” — where I teach you SE-related things in 5 minutes (or less), once a week.

What is the fire rating requirement?

The first step is to obtain the structure’s required fire rating from the client/architect. This is based on the type of building and other goodies per IBC Table 601.

Spray-applied fire resistive material underside of deck?

Once you know the required rating, you should then find out if “spray-applied fire resistive materials” (SFRM) will be applied to the underside of the deck.

(Some people refer to the SFRM using the product name Monokote).

Rule of thumb

Now here comes the rule of thumb:

If SFRM will be applied to the underside of the deck, then the concrete thickness above the deck would generally be 2-1/2” thick.

And it can be either normal weight or lightweight.

You’ll achieve a 1 to 4-hour rating with most UL assemblies (the architect and/or fire protection engineer need to detail that).

(source: Verco Deck Binder)

On the other hand, if SFRM will NOT be applied to the underside of the deck, then the thickness of the concrete varies depending on the required rating.

The legacy Verco catalog had this handy table below (it’s a rough generalization of all the UL assemblies):

(source: legacy Verco catalog)

For example, the thinnest configuration to get a 1-hour rating would be using 2-1/2” lightweight concrete over 1-1/2” deck (watch out for unshored span though – topic for another email).

A very common 2-hour rating configuration that I have seen is 3-1/4” lightweight concrete over 3” deck.

And there you have it. Hope that all makes sense.

Now you should be able to come up with a preliminary concrete thickness based on the required rating.

You could then move on to figure out the other design requirements (e.g., unshored span, vertical capacity, diaphragm capacity…etc.). We'll save those for another email.

Thanks for reading and enjoy the rest of your week!

The post Rule of thumb for concrete thickness (over steel deck) to meet fire rating requirement appeared first on Structural Engineer HQ.

In today’s post, I will quickly summarize what I learned from a recent webinar called “Common Challenges in Wood Lateral System Layout.” It was presented by Terry Malone, Senior Technical Director at WoodWorks, who has decades of experience designing structures.

The webinar’s goal was to help us recognize things that we need to watch out for during schematic design (so that we can be prepared to face the challenges… or propose alternatives to the client).

Let’s dive in:

•  Terry’s thought process
•  Common challenges
•  My non-technical takeaway

(Estimated read time: 4 minutes and 10 seconds )

By the way, this is a rehash of an article I wrote in my weekly newsletter, “Back of the Envelope” — where I teach you SE-related things in 5 minutes (or less), once a week.

If you enjoyed reading it, consider subscribing at the end of the post to be one of the first to get new emails every Thursday (which I eventually rehash onto LinkedIn and Structural Engineer HQ a couple of weeks later).

Terry’s thought process

Terry has a standard process that he goes through when looking at a wood building:

1. First, using PDF markups, overlay the shear walls from floor to floor so you can see how they stack (or not). Use different colors to represent shearwalls on different floors.
2. Next, locate the diaphragm boundaries to get a sense of the load paths.
3. Based on what you see, identify potential irregularity and load path challenges, such as offset walls, large openings, cantilever diaphragms…etc.
4. Lastly, form opinions and develop possible solutions based on what you’ve identified.

Here is an example of what that pdf might end up looking like:

(Source: WoodWorks/Terry – link at the end)

In a nutshell, step back to look at the building from a global perspective before pulling out your calculator to crunch the numbers.

This will help you better understand how the building might behave or perform and help you recognize what you could potentially improve.

Common challenges

Now, here are some of the lateral design challenges that come up frequently according to Terry (or maybe common “misses” by engineers?):

1/ Shear wall horizontal (out-of-plane) offset (i.e., non-stacking wall)

If you are curious and want to open your handy ASCE 7, this is Table 12.3-1 “Horizontal Structural Irregularity Type 4.”

When this occurs, all supporting members need to be designed for overstrength.

(Or convince the client that “stacking the walls” is the way to go.)

(Source: ASCE41-17)

2/ Shear wall in-plane offset (continuous and discontinuous)

Similar to the last one but offset in-plane (“Vertical Structural Irregularity Type 4” per Table 12.3-2).

It is straightforward-ish when the shearwall boundaries come straight down like this:

(Source: Common Challenges in Wood Lateral System Layouts)

It gets tricky when the walls don’t line up nicely (see below) – in this case, the headers and the jambs would need to be designed for overstrength.

(Source: Common Challenges in Wood Lateral System Layouts)

3/ Large diaphragm openings

We should pay attention to how diaphragm forces transfer through drag and chords when we have a large opening.

But what is a “large” opening?

According to the recommendations by FPInnovations (Canadian non-profit wood people, link at the end), if the opening has any of the following, it is considered large:

• Opening depth > 0.15 x (Diaphragm depth)
• Opening length > 0.15 x (Diaphragm length)
• “Distance from diaphragm edge to the nearest opening edge” < 3 x max(Opening depth or opening length)
• “The diaphragm portion between opening and diaphragm edge” exceeds the maximum aspect ratio requirement (i.e., check the diaphragm aspect ratio around each side of the opening against the SDPWS requirements.)

(Source: SDPWS 2015)

If any of the criteria above is true, it is recommended that you should do a detailed analysis (the FPInnovations document has examples on how to do this. Again, link at the end).

4/ Discontinuous chord

Essentially, watch out for corners.

Terry dives into this in more detail in a different WoodWorks document called “The Analysis of Irregular Shaped Diaphragm.” Check out the link at the end to learn more.

(Source: The Analysis of Irregular Shaped Diaphragm)

5/ Cantilever diaphragms

This is allowed by code, but you must meet the SDPWS requirement for “Open Front Structures” (Section 4.2.5.2).

(Source: SDPWS 2015)

6/ Flexible vs. semi-rigid or rigid diaphragm (and global uneven stiffness)

So technically, the code allows you to idealize a diaphragm as flexible if you meet the criteria in ASCE 7 12.3.1.1. This means the analysis is relatively simpler compared to rigid or semi-rigid diaphragms.

However, according to NEHRP Seismic Design Technical Brief No 10 (link at the end), most light-framed wood diaphragms behave semi-rigidly in reality.

The implication is that if you have a building with uneven lateral stiffness on each end, you could have excessive drift on one end even though you technically designed it to meet code.

(Source: ASCE41-17)

So what to do?

You could run a back-of-the-envelope calc treating it like a rigid diaphragm (or cantilevered beam). Based on that, estimate the drift, then stiffen up the lateral system based on educated guess (ahem, I mean engineering judgment).

7/ Combined lateral system

Lastly, according to ASCE 7 12.2.3.3, if you combine different lateral systems (e.g., wood shear wall + moment frame), you should use the lowest R for the direction under consideration.

Don’t need to follow that if you meet all of the exceptions though:

• Risk Category I or II
• Two stories or less
• Light-frame construction or flexible diaphragm

Non-technical Takeaways

Well, that was a lot of technical info. Here are some of my non-technical thoughts:

1. If it feels like you are overthinking things, you are probably not. 90% of the time, we are not designing simple rectangular boxes, and stuff could get pretty complicated, so don’t feel bad. You are not the only one.
2. Being able to identify these challenges early on in the design means you could potentially get the client to change some of them by offering alternatives.
3. And if you are very early (e.g., proposal phase), you might be able to even bake that into your fee, knowing that more effort will be required later down the road.

That is all – thanks for reading!

If you enjoyed reading this, check out the rest of my articles on Back of the Envelope and subscribe maybe?

And be sure to leave a reply to let me know what you think!
(It helps motivate me to keep on writing )

Resources

1. “The Analysis of Irregular Shaped Diaphragm” (link to pdf)
2. “Seismic Design of Wood Light-Frame Structure Diaphragm Systems” (NEHRP Tech Brief) (link)
3. WoodWorks Events Archive with link to pdf (link)
4. WoodWorks Free project support (link)
5. FPInnovations/WoodWorks recommendation for large opening (link)

The post Webinar Recap: “Common Challenges in Wood Lateral System Layout” appeared first on Structural Engineer HQ.

This is a rehash of an article I wrote in my weekly newsletter, “Back of the Envelope” — where I teach you SE-related things in 5 minutes (or less), once a week.

If you enjoyed reading it, consider subscribing at the end of the post to be one of the first to get new emails every Thursday (which I eventually rehash onto SEHQ & LinkedIn a couple of weeks later).

Have you ever set a pencil down vertically on the table to pretend it's a skyscraper, only to then shake that table violently (or blow on it) to knock it down? (you monster)

A standard #2 pencil has a diameter of 0.28 inches, and a length of 7.48 inches — making your skyscraper's aspect ratio equal to about 1:27.

Turns out that they've actually built one of these pencil towers…

(Today's estimated read time = 4 minutes and 30 seconds)

When I first saw that picture (top of this page) of 111 West 57th Street (aka Steinway Tower), I thought it was perhaps Photoshopped.

Except… it's real life, not fantasy.

With a width of 59 ft and a height of 1,428 ft, its ratio is about 1:24 – making it the slenderest tower ever built (so far).

You probably have a lot of questions, such as:

1/ What is the building for?

Super-high-end residential; we are talking about $8M-$66M per unit.

2/ What does the view look like from up there?

(Source: streeteasy.com)

3/ Who designed it?

Arch: SHOP.

Struct: WSP – formerly Parsons Brinckerhoff if you are wondering.

4/ How did they make the structure work?

Thick, high strength concrete shear walls on east & west + core; lots of 70-foot long rock anchors; plus a tuned mass damper at the top.

5/ What's the overturning and uplift?

You nerd.

Just kidding… well, we are structural engineers, so we should run some quick numbers for funsies. In today's issue, we will do a back of the envelope calc on 111-West-57-St for the following:

• Dead load
• Seismic load
• Wind load
• Overturning & uplift

Let's do this.

—

Side note: By the way, obviously they've done tons of serious analyses with this thing (wind tunnel tests, 3d dynamic models etc.); but hey, we are only running calcs with our trusty TI-83 and HP33s, so be easy on me here.

—

Quick-n-dirty dead load

(source: 111w57.com)

Area: The width of the tower is constant at 59 feet, but the length sort of tapers back slightly as it gets taller. I scaled off some plans and got 76′. (So rough floor plate = 4484 sqft).

Slab: I have no idea how thick the slab is, but some videos mentioned 15′ floor-to-floor and 13′ ceiling. So maybe 8″ slab sounds about right (Slab weight = 100 psf).

Wall: 36″ thick at the lower levels and 16″ at the top, so average 26″. Guestimate some wall length (288′) x 15′ tall, divide by area, we get (Wall weight = 313 psf).

Terracotta: Estimate 15 psf but smeared into the floor plate (Cladding = 10 psf).

Miscellaneous: (Misc = 10 psf).

Tune mass damper: 800 ton (TMD = 1600 kip)

(Source: New York Times)

Steel structure: This is the unoccupied structure at the pinnacle. Guessing roughly 10 floors at 30 psf, with an average floor area of 2,000 sqft. (Pinnacle = 900 kip).

All in all, DL = 4484 sqft x 84 floors x (100+313+10+10)/1000 + 1600 + 900 = 165,000 kip.

Put that into perspective in terms of compression on the concrete walls: 165,000 kip / (36 in x 70 ft of wall x 2 x 12) = 2,728 psi

Nice. Seems right.

Quick-n-dirty seismic load

Grabbing some essentials from ATC:

• Sds = 0.304
• Sd1 = 0.096

Make some may-or-may-not-be-true assumptions:

• Risk category = III (not just II because the tower is for the super-rich. Jk? I am guessing it should be higher because it could cause serious damage to its surroundings if it falls down.)
• I = 1.25
• R = 5 (special reinforced concrete shear wall)
• h = 1428’

Not going to bore you with all of the math here but if we follow along with ASCE 7 Chapter 12:

• Cs (Eq 12.8-2) = 0.076
• Cs (Eq 12.8-3) = 0.005
• Cs (Eq 12.8-5) = 0.017

And if I put ASCE 7 words into a formula that you and I intuitively understand (from years of Excel-ing):

Cs = max(min(0.076,0.005),0.017) = 0.017

(I'm definitely not used to seeing this since I am in California… did I make a mistake?)

Seismic base shear is then 165000 x 0.017 = 2,759 kip

(For fun: 2759 / (70’ x 2) = 19.7 k/ft per shear wall.)

Not bad at all. So seismic probably not an issue compared to wind. Let’s see…

Quick-n-dirty wind load

Back to good o’ ATC:

V = 125mph

Make some not-so-accurate-but-probably-ok-enough assumptions:

• Kd = 0.85
• Exposure = C (I have no clue – feels like things are different high up there)
• Kzt = 1.0
• Ke = 1.0
• G = 0.85

Chugging along with ASCE 7 Chapter 26:

• Kz (@100’) = 1.27
• Kz (@500’) = 1.77
• Kz (@1000’) = 2.06
• Kz (@1400’) = 2.21
• qz (@100’) = 43 psf
• qz (@500’) = 60 psf
• qz (@1000’) = 70 psf
• qz (@1400’) = 75 psf

(Phew… almost there)

I am probably butchering this last part following ASCE 7 Chapter 27 but remember, this is just for fun – don't put me in timeout.

• GCp (windward) = 0.8
• GCp (leeward) = -0.5
• GCp (net) = 1.3
• p_net (@100’) = 56 psf
• p_net (@500’) = 79 psf
• p_net (@1000’) = 91 psf
• p_net (@1400’) = 98 psf

And the moment of truth:

Wind base shear = (56×100+79×400+91×500+98×400)x76’ = 9,264 kip

Hey look at that, at least three times higher than seismic.

Alright, I am running out of time (and steam) so going to bump this up a notch to “super-quick-n-dirty.

Super-quick-n-dirty overturning and uplift

OT = 9264 * 1400 / 2 = 6,484,800 kip-ft

T = 6484800 / 58’ = 111,807 kip (uplift from wind)

Net uplift = 111807 – 0.9 x (165000×0.5)  = 37,557 kip

So many digits.

I saw a video that says there are about 200 x 3″ diameter rock anchors. (Here is the video – it's long but pretty good. Watch it at 2x speed for extra fun).

So let's say half of them take the tension.

Tension per anchor = 37557/100 = 376 kip

Stress per anchor = 376 / 7 in^2 = 54 ksi

Not too bad – seems reasonable.

(original image source: Youtube)

And… I am out. (why did I do that to myself?)

I suppose the point is that anyone, including you, could run a back-of-the-envelope calc to see if things make sense-ish.

There are, of course, wind tunnel tests to make the wind load more accurate. Also, in most cases for these towers, drift and occupancy comfort probably govern more than anything else.

But user “MIStructE_IRE” at Eng-Tips said it best:

“…they didn't have ETABS or spreadsheets when they built the empire state! I believe if you can't explain a fancy computer model using a couple of quick calcs and some free body diagrams – then either there's something wrong or the engineer hasn't a clue what they're doing!”

That's all for now. Hope you liked this, and I'll see you next week.

The post Skinniest Tower, Ever appeared first on Structural Engineer HQ.

In today's post, I’ll talk about something I learned recently related to “embedded posts and poles.”

• Quick overview: “nonconstrained” & “constrained”
• What exactly is “rigid floor or pavement”?
• Another option for “chain-link fence”

(Estimated reading time = 3 minutes and 10 seconds)

Overview

You’ve probably seen these two goodies more than a few times throughout your career:

(Source: UpCodes)

They are equations for “embedded posts and poles” from IBC section 1807.3.

We use them to design pole footings for miscellaneous things that cantilever off the ground (e.g., fence posts and flag poles etc.)

The first equation is for “nonconstrained.” Basically, the top of the footing is surrounded by dirt, so it’s free to move horizontally. (By the way, “nonconstrained” is not an actual word in the dictionary, so I am getting the red squiggly line in Word. Why is it not called “unconstrained”… sorry I digress.)

The second equation is for, you guessed it, “constrained.” The code defines it: “lateral constraint is provided at the ground surface, such as by a rigid floor or pavement.”

Using the constrained equation usually allows the footing embedment depth to be shorter than using the nonconstrained equation (potentially saving you or the owner money.)

“Rigid floor or pavement”

The question is then, what exactly is “rigid floor or pavement”?

Some jurisdictions consider it constrained if the top of footing (or post) is surrounded by a concrete slab.

For example, if the post is in the middle of a slab-on-grade, you are good to go.

On the other hand, certain jurisdictions require the top of the footing to be doweled into the slab. Or you can have bars welded to the steel post with enough length to develop the bars into the slab.

Either way, the code is perhaps being intentionally vague so it’s up to interpretation.

Which got me thinking.

These equations have been around for a long, long time. Where did they come from?

Apparently… from tons of research throughout the years (just skim through some of these titles):

(Source: American Society of Agricultural and Biological Engineers)

After some digging, I found this document called ANSI/ASAE EP486.1 Shallow Post Foundation Design, published by the American Society of Agricultural and Biological Engineers (ASABE).

Say what? Agricultural?

Yeah I imagine they probably build tons of posts out there.

My guess is that ASABE studied the researches and compiled their findings into a detailed document. Then the code writers studied the ASABE document and simplified it down to two easy-to-follow equations so we can all enjoy them (if you happen to know the real history – let me know!)

Anyhow. Back to what is considered “constrained.”

Interestingly… check this out.

ANSI/ASAE EP486.1 actually gives you a formula with a nice picture and everything:

(Source: ANSI/ASAE EP486.1)

(In case you are wondering, “table 1” referenced here is the same as IBC Table 1806.2 Presumptive Load-Bearing Values.)

How cool is that?

Essentially, the report considers the constraint resistance to be based on friction + lateral pressure. Sounds reasonable.

If we do a fun run:

• Say you have 5ft x 5ft x 4” concrete slab (Wd = 1250 lbf)
• S = 150 lbf/ft^3
• k_cs = 0.25

The resulting “lateral resistance of constraint” = 400 lbf.

Not bad. And most likely adequate for things you are using the pole foundation for.

Your miles may vary depending on the jurisdiction, but it’s food for thought.

(By the way, for you die-hard enginerds, there is another related document by ASABE that you can dig into: ASABE Meeting Presentation.)

Chain-link fence

Now, suppose you are just using the footings for a chain-link fence. In that case, there is another goodie you can potentially utilize: ASTM 567-14a(2019) Standard Practice for Installation of Chain-Link Fence.

The standard uses a prescriptive method for the footing, which states (just skim through it):

Let me unwrap that for you.

First, the posts have to be spaced at 10’o.c. or less.

Then,

• If posts are 4” or smaller, footing diameter = 4 x post diameter
• If posts are larger than 4″, footing diameter = 3 x post diameter

Now let’s say H = fence height (ft):

• If H is 4’ or less, then footing depth = 24”
• If H is between 4′ and 20′, then footing depth = 24″ + 3″ x (H – 4′)

In other words, if you have the post size, spacing, and fence height – boom! You get footing size.

You can also go the detailed route and run ASCE 7 wind load + pole footing per 1807.3 — but if you are in a hurry and need to get something out quickly to the client, this could be helpful.

(Again, mileage may vary depending on the jurisdiction.)

Alright, and that is all for this week. Hopefully it’s helpful – thanks for reading!

The post Another take on the good ol’ “embedded posts and poles” appeared first on Structural Engineer HQ.

In today's post, I’ll talk about something I learned recently related to “embedded posts and poles.”

• Quick overview: “nonconstrained” & “constrained”
• What exactly is “rigid floor or pavement”?
• Another option for “chain-link fence”

(Estimated reading time = 3 minutes and 10 seconds)

Good morning. Andy here from Back of the Envelope — the place to be for all your p-delta soul-crushing info.

Just kidding. No but seriously. Over the weekend, as I was working on a RISA Floor/3d model for a looming deadline, I repeatedly ran into the dreaded “P-delta diverging” error. It drove me crazy.

So today, I am going to cover:

• What is P-delta
• Why it is important
• What is “P-delta divergence”
• Why it happens and how to resolve it

(Estimated reading time = 4 minutes and 30 seconds)

What is P-delta

Quick refresher.

Say you have a column.

“P” is the axial load applied to the column.

“Delta” is the lateral displacement on one end of the column (caused by whatever reason. E.g., earthquake load or miscellaneous small displacement that just happens in a complex 3d model).

P-delta effect is essentially the extra (aka secondary) moment and shear due to this displacement.

Here is a nice picture from RISA explaining it:

Why is P-delta important?

Because if we don’t account for that secondary moment in the design, we could be under-designing the column.

For example, if your DC ratio for axial load is already high, a little extra moment can put your column into the overstress territory, which is obviously not good.

What is P-delta divergence?

As you can see from the picture above, the secondary shear will cause more lateral displacement, which means… more p-delta effect.

In other words:

Lateral displacement -> secondary shear -> more lateral displacement -> and so on

In practical terms, if we were to do this by hand (don’t):

1/ Apply the load and apply a displacement

2/ Determine the secondary shear caused by p-delta

3/ Apply that shear and determine the new displacement and new secondary shear

4/ Rinse and repeat until the new displacement is so small that it won’t make any more difference — aka the solution converges.

RISA does all this for you in the backend.

But it gets tricky (and soul-crushing) when the solution cannot converge. In other words, the “new displacement” in step 3 is not getting any smaller on each iteration – hence the “P-delta divergence” error.

Why does it happen

Because you screwed up, and your model is broken.

Jk, but actually, yes the model is broken-ish. But how?

If you think about it: What could cause the new displacement to get larger at each iteration?

One reason could be that the column is not restrained by other members. So one end of the column is either moving or rotating out whack — aka “instability.”

Another reason could be because the applied load is causing the member to buckle (remember Euler’s formula from school?). So a solution literally cannot be derived because the column is unstable.

How to resolve it

“How do I get rid of this error asap so I can move on with the rest of my life!” I asked myself.

RISA has a few helpful info to help you debug this (link here and here and here).

They created all of these because, apparently, it is a very common issue.
(“I know I’m not the only one,” sings Sam Smith when he found out that other people were also having p-delta divergence issues).

Long story short, there are 3 steps:

1/ Run the load combination without p-delta and see what happens to the deflected shape (difficulty = easy-ish)

This helped me find my first issue.

The top of my columns did not have rotation stiffness since it’s pinned all around.

The result = whacky deflected shape.

(See the long lines at the bottom? Those are some beams deflecting miles into the center of earth.)

Easy fix — go back to RISA Floor and change the top of columns to “fixed”. This will hopefully resolve the entire issue.

If not, move on to step 2.

2/ Turn on p-delta, but run load combo with only a fraction of the load (difficulty = medium)

In other words, to debug: Create a self-weight dead-load-only combo. Change the load factor to something small like 0.1 to see if the divergence error goes away.

If it does, then it means there is a buckling issue somewhere.

You can potentially find this by looking at the deflected shape in the plan view.

For example, here is a top view of a truss buckling out-of-plane:

3/ Look at the coordinate of the error and hypothesize (difficulty = sucks)

Sometimes the lateral displacement is so tiny that the deflected shape from steps 1 or 2 tells you literally nothing.

And because you know nothing at this point, Jon Snow, the only thing you could really do is locate the coordinate (RISA tells you where it is) and just “look around it” to see if you can figure out what’s wrong.

Luckily for me, after fidgeting for hours with this, I was able to see that one of my columns in the area was not restrained in one of the directions.

Adding a beam where my arrow is showing resolved the issue. And the soul has been uncrushed, for now.

And that’s all – thanks for reading! 

Hope this is helpful (at least maybe someday).

If p-delta divergence ever occurs to you… good luck, but never give up, never surrender. You got this.

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The post Site Walls and Earthquake appeared first on Structural Engineer HQ.

Steel is actually one of my favorite structural materials. For some reason, the design outlined by AISC just seems much more straightforward compared to other materials (e.g. concrete). Perhaps I just had a really good professor in college… (thanks Professor Uang!)

Anyways, if you are planning to take the PE or the SE but have very limited knowledge or experience with structural steel design, my goal is to be able to teach you 70% of what you need to know (for gravity design) with the least amount of effort.

I intend to cover these topics below in a number of posts (I'll start from the top down and if people find them useful, I'll continue. If not, perhaps I'll move on to talk about something else instead):

• Design strengths
• Load combinations
• Shapes and materials
• Flexural design
• Shear design
• Compression design
• Tension design
• Bolted connections
• Welded connections
• Composite beam design

…so hopefully by the end, you will know a thing or two about steel!

Let's get started with design strengths (this may be pretty basic to most of you out there, but we have to start somewhere).

Required Strength vs. Nominal Strength vs. Allowable Strength vs. Design Strength???

If you are new to “strength” design, all of these different terminologies may sound confusing to you. Let me just clear that up by using a simple diagram.

We'll start from the top. Basically, you have the demand, which is the required strength. This can be either ASD () or LRFD () depending on which load combination you are using (we'll talk about ASD & LRFD load combinations in another post).

In the manual, the subscript for required strength is “a”. Personally I find that slightly confusing so when I do my calculations, I like to write “req” as the subscript instead. Like this: .

Next, you have the nominal capacity (or nominal strength) which is the “unfactored” capacity. Meaning you basically calculate this out using the formulas in the manual for axial (), moment (), or shear () without applying the resistance factor or safety factor (more on this below).

ASD

From the nominal capacity, if you divide it by Ω, you end up with the allowable strength (e.g. . I like to call this ).

Ω is the safety factor which is always greater than or equal to one.

Note that for for seismic design, there is a thing called the overstrength factor which uses the symbol . This has nothing to do with the safety factor Ω that we use here, so make sure you don't get those two confused.

So now you got your demand and capacity, you can check to see if your design is “OK” or “No Good (NG)” (e.g. if , OK, otherwise NG).

LRFD

Similarly, you do the same thing for LRFD.

From the nominal capacity, you would multiply by to obtain the design strength (e.g. ).

 is the resistance factor and it's always less than or equal to one.

From there you check your demand vs capacity (e.g. if , OK, otherwise NG).

The End

From time to time, in the PE or the SE exam, there may be questions that specifically use the word “nominal” and then asks you to calculate some stuff out. By knowing the difference between these 4 different “strength” terminologies, you should have no problem understanding what you have to do.

Hopefully you find this helpful. Next up: load combinations.

Stay tuned.

The post Steel: Required Strength vs. Nominal Strength vs. Allowable Strength vs. Design Strength appeared first on Structural Engineer HQ.

Introduction

If you have studied or have done some wood design, you certainly have came across the “beam stability factor”.

It looks something like this ((NDS Equation 3.3–6):

Now, I am not going to go into details about all the variables; I mainly just wanted to talk about the final C_L factor.

As you already know, once you obtained F_bE and F_b^*, calculating C_L is merely plug-and-chug. It is very straightforward yet tedious…

I can’t remember the number of times that I accidentally plugged in the wrong numbers or made a silly mistake when punching the numbers into a calculator. The result, as you guessed it, is usually not pretty (depending on how early I find out about the mistake).

So I came up with a simpler way to avoid doing all of these tedious “plug-and-chug”.

Introducing – Tabulated Table For Beam Stability Factor:

Wood CL Tabulation

1 file(s) 291.59 KB

Download

This table not only helped me avoid making arithmetical errors, it also saved me quite some times during the SE exam (which you know, every second counts!) And before I knew it, life was good again (ok that might be exaggerating a little but you get the idea).

How To Use The Tabulated Table for Beam Stability Factor

Since C_L only depends on [F_bE / F_b^*], I figured if I have a table that lists out all of the possible [F_bE / F_b^*] and its corresponding C_L, I’ll never have to use a calculator to “plug-and-chug” again.

So the way to use it is very simple:

1. First, determine the [F_bE / F_b^*] ratio up to 2 or 3 decimal points.
2. Then, find your C_L.

Example 5.3 (SERM)

Let’s test this out using Example 5.3 of the Structural Engineering Reference Manual (SERM).

1. [F_bE / F_b^*] = 2.83 from the example.
2. 2.83 falls in between 2.54 and 3.42; therefore, use C_L = 0.97.

Calculate C_L Using Tabulated Table (Example 5.3)

And we are done! (Although the actual C_L is 0.974, a difference of 0.004 will not make much difference to your design 99% of the time.)

Example 5.4 (SERM)

Let us do another one using Example 5.4.

1. [F_bE / F_b^*] = 1.06.
2. For this part, we first match 1.0 to the first column then we find 0.06 in the top row. The result is C_L = 0.840. Done!

Calculate C_L Using Tabulated Table (Example 5.4)

Imagine doing this simple two-step process using the table versus hand calculating the following using a non-graphical calculator…

See the difference? I hope so because it sure helped me quite a bit.

Next: Tabulated Table for Column Stability Factor

Similar to the Beam Stability Factor, Column Stability Factor also requires you do the same thing – tedious plug-n-chug (NDS Equation 3.7-1):

In this case, besides [F_cE / F_c^*], there is one more variable, c, which can be 0.8, 0.85, or 0.9 depending on the type of lumber.

No problem, I’ll just create 3 more tables.

Example 5.7 (SERM)

We’ll test this out using again, the example from SERM. From example 5.7:

1. [F_cE / F_c^*] = 0.333 and c = 0.8.
2. 0.33 corresponds to C_P = 0.304… and done.

Calculate C_P Using Tabulated Table (Example 5.4)

Again, by using the table, you avoided having to do this during the exam:

Thoughts?

That’s it for now. Do you find this helpful? Let me know in the comments below.

Thank you for reading and good luck studying!

The post Wood Design Shortcut – Beam Stability Factor (and Column Stability Factor) appeared first on Structural Engineer HQ.

While studying the ASCE 7 Chapter 11 & 12, do you frequently get lost in its”wall of text“?

No problem, we are here to help!

Ian has created and shared three very valuable flow charts (you can download them below). By following along while studying, you will gain a much better understanding of  many of the seismic design requirements and calculations described in the code.

If you are not already familiar with these concepts, I highly recommend that you study them thoroughly in preparation for the SE exam!

Files	Description
Seismic Ground Motion Values Flow Chart (Flow Chart 1) 1 file(s) 326.11 KB Download Seismic Design Category Flow Chart (Flowchart 2) 1 file(s) 328.85 KB Download Equivalent Lateral Force Procedure (Flowchart 3) 1 file(s) 402.45 KB Download Created by Ian Riley	The first part of these flow charts outlines the appropriate procedures for determining “seismic ground motion values” and “seismic design category” which goes along very well with my previous post. The last part of the flow charts goes through the “equivalent lateral force procedure” step-by-step. It is very useful for making sure that you don't miss all the little details in the code.

PS: Thanks to David A. Fanella, Ph.D., S.E., P.E, for his original publications in cenews.com.

The post Three “Seismic Design Requirements” Flow Charts: The Quickest & Easiest Way To Learn The Design Values In “ASCE 7 Chapter 11 & 12” appeared first on Structural Engineer HQ.

Since I've been practicing structural engineering in California, I've kind of taken my experience with seismic design for granted.

Not only it’s something that we do nearly every single day, it’s also required in order to get the PE license (CA has an additional 4-hour seismic portion on top of the regular NCEES 8-hour exam for the PE exam).

I understand that a lot of people in “non-earthquake areas” are not used to seismic design; since it will be on the SE exam, my goal is to help you as much as possible by covering a number of important topics.

If you practice structural engineering in states that rarely deal with earthquakes or if you are fairly new to the world of lateral design, I hope you find these articles helpful in your preparation for the SE exam.

In this post, I will start with the explanation of “ASCE 7–05 Chapter 11: Seismic Design Criteria”.

Note that this chapter is very closely related to Section 1613 of IBC 2009. Parts of them are pretty much identical. To me, ASCE 7–05 is slightly easier to read so once you get the hang of it, finding and understanding the related sections in IBC should be cake.

Description – Variables of “Seismic Design Criteria”

Every lateral design problem usually starts with the variables described in Chapter 11.

It’s important to understand what these variables are and how to obtain/calculate them so that you can start on the right track.

This chapter of ASCE 7 is the very basic of determining the required seismic demand. So if you get this part wrong, the rest of the design is essentially incorrect.

Now, since this part is so important, what is the best way to understand them?

My recommendation: Think of “reading the code” like “solving a problem”. There are “givens” and there are “things to be determined”. That’s how I am going to break down these variables.

“Given” (4 Variables)

1. S_S

By ASCE 7–05 definition, this is the “mapped MCE, 5 percent damped, spectral response acceleration parameter at short periods as defined in Section 11.4.1”. (IBC 2009 defines short periods as 0.2 seconds).

If you have time, you can dig in to different seismic related books and figure out what this is all about. But as far as the exam goes, this number is usually given.

If not given, you’ll have to look at Figure 1613.5(1) to 1613.5(14) of the IBC. It’s good know what they look like although I’ve never had to use them.

One thing you need to pay attention to when looking at the figures is the title. If it mentions “0.2 sec” then you are looking at the S_S. Otherwise, it’ll say “1.0 sec” which is the S₁ value that I’ll talk about next.

If you need to determine this number outside of the exam settings (like in your actual actual job), USGS has a handy tool that helps you determine the values for you. You just have to enter site latitude and longitude which you can obtain from Google Maps by right clicking on a location and select “what’s here”.

2. S₁

S₁ is pretty similar to S_S except that it’s for longer period.

By ASCE 7–05 definition, this is the “mapped MCE, 5 percent damped, spectral response acceleration parameter at a period of 1 s as defined in Section 11.4.1”.

Everything I said above for S_S applies here also so I am not going to repeat it.

Moving on.

3. Site Class

I've came up with three possibilities of how this can be given in a problem.

Possibility 1: Given Directly

On the exam, the problem may just flat out tell you what the site class is.

In actual practice, most of time, this is determined by the soil engineer.

Possibility 2: Given Indirectly

The exam might also give you bunch of variables like “soil shear wave velocity” or “standard penetration resistance”. In that case, you just need to look at IBC 2009 Table 1613.5.2 and match the numbers to the correct class.

Occasionally, you may need to do some calculations if the problem gives you properties of different soil layers. I am not going to go into details in this post but basically you’ll have read and understand IBC 2009 Section 1613.5.5 and 1613.5.5.1 which has all the formulas and explanations.

Possibility 3: Unknown

Sometimes the problem might not give you any clues at all regarding site class.

If that’s the case, you can assume it’s site class D per section 1613.5.2 which states: “When the soil properties are not known in sufficient detail to determine the site class, Site Class D shall be used…”.

Note that if this occurs in the afternoon portion (written part), make sure you cite the code section!

4. Occupancy Category

This is usually pretty straightforward.

The exam problem will usually give you some kind of building usage description; you will have to match it to the description listed In IBC 2009 Table 1604.5.

Note that Occupancy Category II is for “Buildings and other structures except those listed in Occupancy Category I, III, and IV”; therefore, you should familiarize yourself with this table so that you have a good idea what kind of buildings might not belong to I, III, and IV.

“Determine” (6 Variables)

1. Importance Factor, I (I call it “I_E”, “_E” for earthquake)

Once you know what the building “Occupancy Category” is, use ASCE 7–05 Table 11.5–1 to obtain the importance factor.

Side note: Personally I like to call it “I_E” in my calcs just to distinguish it from the importance factor for wind load (which I call “I_W“).

2. Short-period Site Coefficient, F_a

This is determined based on Site Class and S_S using ASCE 7–05 Table 11.4–1.

However, note that the numbers given in the table for S_S are in increment of 0.25. So what happens if you have a S_S somewhere in between (e.g. S_S = 0.85)?

In that case, you’ll have to interpolate, which can be tedious.

To save time, I created a little handy chart for S_S up to a few decimal points. That way I only need to look at the chart to find the corresponding F_a rather than doing the cumbersome calculation myself.

3. Long-period Site Coefficient, F_v

Pretty much the same as F_a except it’s based on S₁ using ASCE 7–05 Table 11.4–2. The tabulation for different S_S is also included in the above file.

4. S_DS

Once you you have F_a, you can calculate S_MS = F_aS_s

From S_MS, you can then determine S_DS = (2/3) S_MS

5. S_D1

Similarly, S_M1 = F_vS₁

And S_D1 = (2/3) S_M1

6. Seismic Design Category, SDC

Now that you have determined S_DS and S_D1, you can determine the Seismic Design Category by looking at Table 11.6–1 and Table 11.6–2 whichever governs (i.e. SDC “D” governs over SDC “A”).

Note that this only applies if S₁ < 0.75 per ASCE 7–05 Section 11.6.

If S₁ is greater than or equal to 0.75, you basically has two possible outcomes depending on what your occupancy category (OC) is. If your OC is I, II, or III then the SDC is “E”; if your OC is IV, then you have SDC “F”.

Other Topics (3 items)

So far we have discussed the most important parts of Chapter 11. There are a few other things in this chapter that I didn't mention because I don’t encounter them too often. However, these are things that you will come across from time-to-time. I’ll just briefly go over them so that you are aware of their existence.

Design for SDC A (ASCE 7–05 Section 11.7)

I've never had to use this section myself either in real practice or in the exam. However, since the exam can cover pretty much anything they want (as long it's referenced in the spec), it’s still important to at least have some idea of what this is.

I won't go over it here since I believe it will be easier for you to just read the section. If you are familiar with seismic design concepts for higher SDCs, this section will be a light read. If not, I suggest that you come back to read this section in the ASCE7 after you have completed your study – it will make more sense.

Design Response Spectrum (ASCE 7–05 Section 11.4.5)

Sometimes you may be asked to calculate the “design spectral response acceleration, S_a” which is based on a couple of variables that you have already determined above (i.e. S_D1 and S_DS) with the exception of T_L and T. All the other terms mentioned in this section is derived from these 4 variables which you can see in this chart below:

Seismic Design Response Spectrum

• “T_L” is the long-period transition period which is determined from looking at figures 22–15 to 22–20.

• “T” is the fundamental period of the structure. It is kind of like the variable “X” in an X-Y chart whereas the “S_a” is the “Y”.

In other words, the spectrum chart describes: “What is the S_a for a building with fundamental period T in a location that has certain S_D1, S_DS, and T_L?”

General, Definition, and Notation (ASCE 7–05 Section 11.1, 11.2, and 11.3)

Finally, it is prudent to look at the “General” section and be aware of the different “exceptions”. You may or may not run into them but having a mental image of what they are will help you if they come up during the exam.

Also, “occasionally,” you’ll run into terms that you are not exactly sure what they mean. One of the first places to look at will be the “definition and notation sections”. I would just skim through them very quickly so that you have an idea of what's there.

Recap & Thank You!

To recap, these are the things in ASCE 7–05 Chapter 11 that you should know by now:

“Given” variables:

• S_s
• S₁
• Site Class
• Occupancy Category

“To be determined” variables:

• Importance Factor, I (I_E)
• F_a
• F_v
• S_DS
• S_D1
• Seismic Design Category (SDC)

Other topics:

• Design for SDC A Response spectrum
• General, Definition, and Notations

That’s it for now. If you find this helpful (or not), please let me know in the comments below. Thank you for reading!

(PS: Ian has created a handy flowchart that is related to this section. I will post them soon – stay tuned!)

The post 13 Things You Need to Know About “Seismic Design Criteria” (ASCE 7 Chapter 11) appeared first on Structural Engineer HQ.