## New AISC spec: compressive capacity for members with slender elements

## New AISC spec: compressive capacity for members with slender elements

(OP)

I’ve been studying the new AISC Spec, and I’ve developed a bit of confusion and uncertainty about how to apply the provisions in Section E7, which concern members with slender elements in their cross-section. I would like to start a thread on this topic. I apologize for the very very long post, its as concise as I could make it.

To start, I’d like to make some observations on the definition of “slender elements”. The new spec points to Table B4.1 for the definitions(Table B5.1 in 1989 ASD.) This table is somewhat different from the one in the 1989 ASD. (I am not familiar with the LRFD, so I cannot comment on that one.) The new table has more different “cases”, and for example, I found that under the old code, uniform compression on webs was not an element in its own right, but fit under the listing “all other uniformly compressed stiffenened elements. I also note that the dimension parameter referenced for webs has changed from h/tw with h = clear distance between flanges(1989 ASD), to h/tw with h = clear distance between flanges less the fillet or corner radius (new spec.). Not only that but a few of the limiting values have changed: flexure in flanges of rolled I shapes was 95/sqrt(Fy), now it is 1.0*sqrt(E/Fy) = 170/sqrt(Fy). Of particular interest to me here is the limiting value for non-compact elements, webs under uniform compression, case 10 in Table B4.1. The limit is h/tw = T/tw = 1.49*sqrt(E/Fy). (this is numerically equal to the old limit from ASD of 254/sqrt(Fy).

When I go on to try and apply section E7 in the new code, I immediately find an anomaly. Under Paragraph 2(a) the code defined how to calculate be, the effective width of the slender element. It defines be for those elements with T/tw = 1.49*sqrt(E/f). This is different from the limit above, from Table B4.1. (Note the denominator under the radical = Fy in table B4.1, while the denominator here is f.) f is defined below as being based on Fcr as calculated with Q=1.00. For a compression member, Fcr is the critical buckling load for either buckling about the X axis or buckling about the Y axis. There are apparently two different values. The anomaly is that f is different from Fy. As far as I can tell, for all rolled W shapes, f < Fy. This means then, that there are some W shapes with slender webs (eg W14X38, W14X34 & lighter) for which the T/tw is still less than the limiting value of 1.49*sqrt(E/f), so the spec does not define be. There is a gap. I’m not certain, but I think that most or all of the W shapes with slender webs will fall into this category. So how do you calculate the value of be for these shapes? I need this to be able to calculate Qa for the slender web shapes.

When I asked this question to AISC, I got two rather vague replies. First: yes that anomaly is there, and the answer might be to delete the limit from paragraph E7 2(a). This would eliminate that gap. I later got another response that said if you had a case where T/tw was less than the limit 1.49*sqrt(E/f) then you should use Qa = 1.00. This seemed counter-intuitive to me: I expected that any time you had a T/tw less than the limit in Table B4.1, you had Qa = 1.00, and any time you had T/tw greater than that limit, then Qa was < 1.00. Apparently this is not the case, for some cases with T/Tw > Table B4.1 limit, Qa still was 1.00. Furthermore, look at the Eq. E7-17. The caveat on the right side, <b makes it seem like even if T/tw is large enough to apply Eq. E7-17, there might be some cases where the calculated value > T, and you must use the limiting max value of T. All of this seems to me to make the limit indicated in Table B4.1 somewhat abastract. Does anyone here understand what that limit really means?

Now, I’ve proceded assuming that the limit b/t>=1.49*sqrt(E/f) does apply and tried to calculate compressive capacity of the following member:

W14X34 Material A992 Fy = 50 ksi Fu = 65 ksi

Lx = 12’ Ly = 12’

This member has a non-slender flange based on Table B4.1.

This member has slender web based on Table B4.1.

From the new code, I find that Qs = 1.00 for non-slender flanges.

For slender webs, I need to do the following: run two parallel calculations as follows:

X-axis buckling: calculate

[KL/r]x = 24.7

Fex = 469.148 ksi from Eq. E3-4

Fcrx = 47.819 ksi (note less than Fy) from Eq. E7-2 with Q = 1 governs

T/tw = 11.625/.285 = 40.789

1.49*sqrt(E/Fy) = 35.884 so web is slender by Table B4.1

1.49*sqrt(E/Fcrx) = 36.693 so I must calculate be

be = 1.92*t*sqrt(E/Fcrx)*[1 – 0.34/(TD/tw)*sqrt(E/Fcrx)] = 10.709 in (note be < T = 11.625 in).

Then, Aeff = A – T*tw + be*tw = 9.739 in^2

Qa = Aeff/A = 9.829/10 = .9739

Qs = 1.00

Q = Qs*Qa = 1.00 * .9829 = .9739

Now using Eq. 7-2 or 7-3, Eq.7-2 governs and Fcrx = 46.625 ksi.

And Pcrx = 46.625*10 = 466.25 k

In parallel, for y axis buckling

[KL/r]y = 94.118

Fey = 32.311 ksi from Eq. E3-4

Fcry =26.163 (note less than Fy) from Eq. E7-2 with Q = 1 governs

T/tw = 11.625/.285 = 40.789

1.49*sqrt(E/Fy) = 35.884 so web is slender by Table B4.1

1.49*sqrt(E/Fcry) = 49.607 so be = T

So now:

Aeff = A – T*tw + T*tw = 10.00 in^2

A = 10 in^2

Qa = Aeff/A = 1.00

Qs = 1.00

Q = Qs*Qa = 1.00 * .1.00 = 1.00

Now using Eq. 7-2 or 7-3, Eq.7-2 governs and Fcry = 26.163 ksi.

And Pcry = 26.163*10 = 261.63 k

So, y axis buckling governs because it has the smallest Pcr value, and Qa = 1.00, Qs = 1.00 and Q = 1.00.

Also Pcr = Pcry = 261.63 k.

Note that the smallest Qa was for X axis buckling but that did not yield the lowest Pcr. So it does not govern.

My confusion is at two locations: the slender limit of 1.49*sqrt(E/f), because it doesn’t match the limit in table B4.1; and have I calculated Aeff correctly?

Any comments, advice or corrections on my analysis here? I would really appreciate any feedback.

Regards,

chichuck

To start, I’d like to make some observations on the definition of “slender elements”. The new spec points to Table B4.1 for the definitions(Table B5.1 in 1989 ASD.) This table is somewhat different from the one in the 1989 ASD. (I am not familiar with the LRFD, so I cannot comment on that one.) The new table has more different “cases”, and for example, I found that under the old code, uniform compression on webs was not an element in its own right, but fit under the listing “all other uniformly compressed stiffenened elements. I also note that the dimension parameter referenced for webs has changed from h/tw with h = clear distance between flanges(1989 ASD), to h/tw with h = clear distance between flanges less the fillet or corner radius (new spec.). Not only that but a few of the limiting values have changed: flexure in flanges of rolled I shapes was 95/sqrt(Fy), now it is 1.0*sqrt(E/Fy) = 170/sqrt(Fy). Of particular interest to me here is the limiting value for non-compact elements, webs under uniform compression, case 10 in Table B4.1. The limit is h/tw = T/tw = 1.49*sqrt(E/Fy). (this is numerically equal to the old limit from ASD of 254/sqrt(Fy).

When I go on to try and apply section E7 in the new code, I immediately find an anomaly. Under Paragraph 2(a) the code defined how to calculate be, the effective width of the slender element. It defines be for those elements with T/tw = 1.49*sqrt(E/f). This is different from the limit above, from Table B4.1. (Note the denominator under the radical = Fy in table B4.1, while the denominator here is f.) f is defined below as being based on Fcr as calculated with Q=1.00. For a compression member, Fcr is the critical buckling load for either buckling about the X axis or buckling about the Y axis. There are apparently two different values. The anomaly is that f is different from Fy. As far as I can tell, for all rolled W shapes, f < Fy. This means then, that there are some W shapes with slender webs (eg W14X38, W14X34 & lighter) for which the T/tw is still less than the limiting value of 1.49*sqrt(E/f), so the spec does not define be. There is a gap. I’m not certain, but I think that most or all of the W shapes with slender webs will fall into this category. So how do you calculate the value of be for these shapes? I need this to be able to calculate Qa for the slender web shapes.

When I asked this question to AISC, I got two rather vague replies. First: yes that anomaly is there, and the answer might be to delete the limit from paragraph E7 2(a). This would eliminate that gap. I later got another response that said if you had a case where T/tw was less than the limit 1.49*sqrt(E/f) then you should use Qa = 1.00. This seemed counter-intuitive to me: I expected that any time you had a T/tw less than the limit in Table B4.1, you had Qa = 1.00, and any time you had T/tw greater than that limit, then Qa was < 1.00. Apparently this is not the case, for some cases with T/Tw > Table B4.1 limit, Qa still was 1.00. Furthermore, look at the Eq. E7-17. The caveat on the right side, <b makes it seem like even if T/tw is large enough to apply Eq. E7-17, there might be some cases where the calculated value > T, and you must use the limiting max value of T. All of this seems to me to make the limit indicated in Table B4.1 somewhat abastract. Does anyone here understand what that limit really means?

Now, I’ve proceded assuming that the limit b/t>=1.49*sqrt(E/f) does apply and tried to calculate compressive capacity of the following member:

W14X34 Material A992 Fy = 50 ksi Fu = 65 ksi

Lx = 12’ Ly = 12’

This member has a non-slender flange based on Table B4.1.

This member has slender web based on Table B4.1.

From the new code, I find that Qs = 1.00 for non-slender flanges.

For slender webs, I need to do the following: run two parallel calculations as follows:

X-axis buckling: calculate

[KL/r]x = 24.7

Fex = 469.148 ksi from Eq. E3-4

Fcrx = 47.819 ksi (note less than Fy) from Eq. E7-2 with Q = 1 governs

T/tw = 11.625/.285 = 40.789

1.49*sqrt(E/Fy) = 35.884 so web is slender by Table B4.1

1.49*sqrt(E/Fcrx) = 36.693 so I must calculate be

be = 1.92*t*sqrt(E/Fcrx)*[1 – 0.34/(TD/tw)*sqrt(E/Fcrx)] = 10.709 in (note be < T = 11.625 in).

Then, Aeff = A – T*tw + be*tw = 9.739 in^2

Qa = Aeff/A = 9.829/10 = .9739

Qs = 1.00

Q = Qs*Qa = 1.00 * .9829 = .9739

Now using Eq. 7-2 or 7-3, Eq.7-2 governs and Fcrx = 46.625 ksi.

And Pcrx = 46.625*10 = 466.25 k

In parallel, for y axis buckling

[KL/r]y = 94.118

Fey = 32.311 ksi from Eq. E3-4

Fcry =26.163 (note less than Fy) from Eq. E7-2 with Q = 1 governs

T/tw = 11.625/.285 = 40.789

1.49*sqrt(E/Fy) = 35.884 so web is slender by Table B4.1

1.49*sqrt(E/Fcry) = 49.607 so be = T

So now:

Aeff = A – T*tw + T*tw = 10.00 in^2

A = 10 in^2

Qa = Aeff/A = 1.00

Qs = 1.00

Q = Qs*Qa = 1.00 * .1.00 = 1.00

Now using Eq. 7-2 or 7-3, Eq.7-2 governs and Fcry = 26.163 ksi.

And Pcry = 26.163*10 = 261.63 k

So, y axis buckling governs because it has the smallest Pcr value, and Qa = 1.00, Qs = 1.00 and Q = 1.00.

Also Pcr = Pcry = 261.63 k.

Note that the smallest Qa was for X axis buckling but that did not yield the lowest Pcr. So it does not govern.

My confusion is at two locations: the slender limit of 1.49*sqrt(E/f), because it doesn’t match the limit in table B4.1; and have I calculated Aeff correctly?

Any comments, advice or corrections on my analysis here? I would really appreciate any feedback.

Regards,

chichuck

## RE: New AISC spec: compressive capacity for members with slender elements

When you're evaluating the Table B4.1 limit, you do not yet know the possible applied stress level at which lateral buckling is predicted to occur. It's just a way to point you to the correct Ch E section. If you evaluate the equation with Fy and get that it's nonslender, then you know it's nonslender regardless of unbraced length. Think of it as "if Table B4.1 limit indicates nonslender, then you know your section is nonslender, but if it is not satisfied, you still MIGHT have a section that doesn't require reduction (not that it's nonslender--but doesn't require reduction for local buckling)." In the E7 calculations, you have a better idea of the possible stress, so f is in there instead of Fy. This f=Fcr might be very small, far below a stress that would cause web local buckling.

As an extreme example, to illustrate the idea, consider the effect of local buckling for your W14x34 if it's 8-ft long. Local buckling will play a big role in this case because the section can develop a very large stress before some other (besides web local) buckling occurs. If your W14x34 column is near KL/r=200, then it can never get anywhere near Fy, so f=Fcr is used. If it's really long, it will buckling in some other mode long before local bucklng has an effect.

I admire anyone who can really get through your post, BTW. It would take me at least an hour to really study it.

DBD

## RE: New AISC spec: compressive capacity for members with slender elements

Your explanation and example make it very clear what the reasons for the two different "limits" on slenderness in Table B4.1 and Section E7. Thank you. And a red star for you as well.

As for the length, I was striving to provide a complete example and explanation, so I would not have to put everyone through the pain of repeated cycles of ".. you need to provide more information." This section of the spec gives rise to a fairly lengthy calculation.

Regards,

chichuck

## RE: New AISC spec: compressive capacity for members with slender elements

I've put up a website, and I posted the mathcad file i've been working on with this calculation in it. The worksheet includes tensile or compressive, major axis bending, minor axis bending and appropriate interaction equations from the new AISC code.

All are invited to check it out, and I would be grateful for any feedback you might offer. Perhaps a live calculation sheet is easier to work with than a long-winded post.

the address is http://home.mn.rr.com/gbbadger/

Regards,

chichuck