AISC Design Guide 1 - Appendix B

[Jerehmy]

I have spreadsheets for both methods from AISC Design Guide 1 edition #2. I sent the triangular distribution method to a friend and he found an interesting situation where the anchor rod tension is negative even though e > e_kern. I for the life of me cannot figure out where the error is.

P_u = 110.55 kip
M_u = 47 kip-ft
B = 20 in
N = 16 in
N' = 14.5 in
A' = 6.5 in
f_pn = 2.224ksi (nominal concrete bearing capacity)

e = M_u/P_u = 5.102 in
e_kern = N/6 = 2.667 in

e > e_kern, thus large eccentricity moment

A = 0.5 * (3N` ± [ (3N`)² - 24(P_uA` + M_u)/(f_pnB) ]^0.5)

A = 0.5 * ((43.5 in ) ± [ (43.5 in)² - 24((110.55 kip)(6.5 in) + 47 kip-ft)/((2.224 ksi)(20in))]^0.5)

A = 39.072 in ; 4.428 in (4.428 in is obviously correct one)

T_u = f_pnAB/2 - P_u

f_pnAB/2 = (2.224 ksi)(4.428 in)(20 in)/2 = 99.5 kip

T_u = 98.5 kip - 110.6 kip = -12.1 kip


Can't figure out why. I thought it might be an error calculating A but I have triple checked it.

 

[mike20793]

Enabling strain compatibility makes the baseplate rigid. This is necessary because a 4th equation of equilibrium is required to solve the problem. The attached paper should help with this case, even though it is geared towards biaxial bending in baseplates.

 

[BAretired]

Your 'A' distance of 4.428" is wrong by inspection.

BA

 

[BadgerPE]

How did you arrive at a different "A" value BA? I used DG 1 and came up with an identical value of 4.428".

 

[BAretired]

If 4.428" is correct, the total compression is 4.428(20)2.224/2 = 98.4k which is less than Pu, so 4.428 is too small.

BA

 

[BAretired]

Pu = 110k
Mu = 47'k or 564"k
e = 564/110 = 5.13"

Effective width = (8-5.13)3 = 8.61"
Average pressure = 110/8.61(20) = 0.6388ksi
Maximum pressure = 1.28ksi
Minimum pressure = 0

Check
C = 0.6388(8.61)20 = 110k
M = 110(8 - 2.87) = 564"k = 47'k

This means the bolts are not stressed and the triangular pressure diagram extends over 8.61".

BA

 

[pete600]

I do not see the equation for 'A' in the design aid, can you provide the location?

 

[pete600]

The equation for 'A' appears to be incorrect.

A=[f+/-(f^2-4*f*B*(P*A'+M)/6)^0.5]/(f*B/3)

It looks like the equation you are using is similar but has been edited algebraicly. The algebra appears to have lead to the answer.

 

[BAretired]

pete600,

I didn't use the equation in the design guide; it's at the top of page 58. I am attaching a pdf of it. It doesn't seem to yield the correct answer, but I haven't really checked it thoroughly yet.

I found that A = 8.61" based on the fact that the eccentric load occurs at the kern of an area measuring 8.61" x 20". The pressure varies from 0 to a maximum of 1.28 ksi.



BA

 

[pete600]

BAretired,
I also derived it but I am getting the same answer as Jerehmy.

Jerehmy,
I think that the issue maybe the bearing pressure.

I think the pressure used in the equation may the allowable bearing pressure (f_allow). Can you confirm?

Should the pressure be the applied pressure and not the allowable pressure?

Applied Pressure from Moment - maximum magnitude at edge
f_m = (6*M)/(B*N^2) = 0.661 ksi

Applied Pressure from point load - evenly distributed
f_p = P/(B*N) = 0.345 ksi

Max Applied Pressure
f_max = F_m + f_p = 1.006 ksi <= 2.224 psi = f_allow

Using f_max instead of f_allow yields the following:
A = 12.224"
Tu = 161.3 kip

It is worth noting that the value of Tu is greater than if it was calculated as a simple couple between two anchors (not what I was expecting).

tension from simple couple between anchors
Tu = (N'/2 + e)*P/N' = 94.2 k

 

[BAretired]

In the example given in the OP,
P = 110.5k
fp = 2.224ksi
N = 16", B = 20"
If the volume of the stress block is equal to Pu
i.e. 2.224AB/2 = 110.5
so A = 110.5*2/2.224(20) = 4.97"
e = N/2 - A/3 = 6.34"

The base plate can handle an ultimate load of 110.5k at an eccentricity of 6.34" without relying on tension in the anchor bolts.

The kern of the plate is at N/6 = 2.67" from the center of plate.

The actual eccentricity, e = 47*12/110.5 = 5.1"

Conclusion:
The kern of the base plate is not where the AISC method starts to apply. The proper location is at an eccentricity of N/2 - A/3 when A is calculated assuming the compression block carries the total load by itself without help from the anchor bolts. If the eccentricity exceeds that, the AISC method works properly.

BA

 

[pete600]

@BAretired:
the value 'fpn' given was defined as follows:

fpn = 2.224ksi [highlight #EDD400](nominal concrete bearing capacity)[/highlight]

I do not think the 'bearing capacity' is appropriate to use.

What if the column overturns prior to inducing the maximum bearing pressure of 2.224 psi?

I think the bearing capacity is a design constraint, the applied pressure cannot exceed the bearing capacity, but the actual load analysis should be based on the applied loads.

 

[BAretired]

the value 'fpn' given was defined as follows:
fpn = 2.224ksi (nominal concrete bearing capacity)

I do not think the 'bearing capacity' is appropriate to use.

I tend to agree. The bearing capacity is probably too low, but that makes the procedure conservative which is okay.

What if the column overturns prior to inducing the maximum bearing pressure of 2.224 psi?

That is not possible with four anchor bolts present. Failure cannot occur until two bolts are stressed to their yield point and the ultimate compressive stress is reached under the opposite end of the base plate.

I think the bearing capacity is a design constraint, the applied pressure cannot exceed the bearing capacity, but the actual load analysis should be based on the applied loads.

I agree; applied loads for ASD, factored loads for LRFD. In my earlier analysis I found the maximum pressure f_p to be 1.28ksi, less than 60% of f_pn.


BA

 

[pete600]

That is not possible with four anchor bolts present. Failure cannot occur until two bolts are stressed to their yield point and the ultimate compressive stress is reached under the opposite end of the base plate

I think it is unlikely that the anchors and the concrete bearing will reach failure at the same time. The anchors may or may not fail first. The anchors if very weak could fail before the bearing strength is reached.

The equation for e_kern is derived using the 'Applied loads'. The reason the tension is coming out negative is because the stress used in the original question is not the applied stress but the bearing capacity. See quote below.

I sent the triangular distribution method to a friend and he found an interesting situation where the anchor rod tension is negative even though e > ekern. I for the life of me cannot figure out where the error is.

 

[BAretired]

The anchors may not reach yield at the same time as the maximum concrete bearing but the anchor bolts are capable of straining while maintaining yield stress. Failure is deemed to occur when maximum concrete bearing occurs.

The reason the tension is coming out negative is explained in the attached pdf file.

BA

 

[pete600]

The anchors may not reach yield at the same time as the maximum concrete bearing but the anchor bolts are capable of straining while maintaining yield stress. Failure is deemed to occur when maximum concrete bearing occurs.

I do not think this is always true. What if there is anchor pullout or breakout?

What if there was not an applied moment? Is the bearing pressure the concrete bearing capacity or is it P/A?

Applied stress:
σ_a = P/A; not f_allow (axial load)
σ_b = M/S; not f_allow (flex Ural load)

If there is combined loading
σ_a +/- σ_b = P/A +/- M/S

E_kern is defined as the location the load acts (e = M/P) that causes the magnitude of the stress distribution at the bolt side of the base plate to be zero

σ_a - σ_b = P/A - M/S = 0

Therefore P/A = M/S

P/(N*B) = P*e/(B*N^2/6)

Solving for e
e=e_kern=N/6

The triangular stress on the concrete used must be caused by 'Applied Loads', using the bearing capacity will lead to ubconservatice tension loads in the anchors.

I am only questioning, because I want to understand why I am incorrect, if I am.

 

[KootK]

@Jerehmy: any chance you'd want to post this MathCAD sheet for review? Of course, I'll understand if there are intellectual property issues.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[BAretired]

I do not think this is always true. What if there is anchor pullout or breakout?

There should not be anchor pullout or breakout if the anchorage develops the yield strength of the anchor bolts.

What if there was not an applied moment? Is the bearing pressure the concrete bearing capacity or is it P/A?

With no moment, the actual bearing pressure is P/A and the factored bearing pressure is P_u/A

Applied stress:
σ_a = P/A; not f_allow (axial load)
σ_b = M/S; not f_allow (flex Ural load)

If there is combined loading
σ_a +/- σ_b = P/A +/- M/S

E_kern is defined as the location the load acts (e = M/P) that causes the magnitude of the stress distribution at the bolt side of the base plate to be zero

σ_a - σ_b = P/A - M/S = 0

Therefore P/A = M/S

P/(N*B) = P*e/(B*N^2/6)

Solving for e
e=e_kern=N/6

The above is true when the bearing stress is covering the entire area of plate, but that is not the range that we are considering in this discussion.

The triangular stress on the concrete used must be caused by 'Applied Loads', using the bearing capacity will lead to unconservative tension loads in the anchors.

That is true, but it is the same principle we use in determining the required area of reinforcement in a concrete beam. The difference is that in a beam, we use a Whitney stress block whereas in the AISC guide, a triangular stress block is used. It is certainly a point which could be argued as noted in Section B.1 Introduction to Appendix B.

I am only questioning, because I want to understand why I am incorrect, if I am.

As well you should. You are not necessarily incorrect and you are entitled to use more conservative assumptions in determining the tension force in the anchor bolts if you wish.

My only goal was to explain why the tensile force in the OP's example turned out negative, so I did not quarrel with the basic assumptions made by AISC. In actual fact, there may be several contentious points in the proposed method.

BA

 

[Jerehmy]

Kootk - I didn't want to post the spreadsheet lest an error somewhere cause people to dismiss the entire post, even if the error has nothing to do with the anchor tensile force calculation. I'll post anyway. Link


If Appendix B does not work for all situations, they should say so. Also, to quote the design guide "B.4.2 Design Procedure for a Large Moment Base - When the effective eccentricity is large (greater than ekern), there is a tensile force in the anchor rods due to the moment." If this isn't always the case, they should have a corollary to the equations.


The reason it doesn't work is because of B, it's too wide. Maybe it's because the equations they use for kern distance are elastic equations and the design equations are plastic/failure equations? Seems the ekern only works with square plates, which is fine but they should change the verbiage in the Appendix to indicate this is so or change the equation for when eccentricity counts as large v small moment.

Just for reference, it doesn't count as large eccentricity per the new design method which has:

e_crit = N/2 - P/(2*qmax) = 6.76in > e = 5.108in

 

[BAretired]

If Appendix B does not work for all situations, they should say so. Also, to quote the design guide "B.4.2 Design Procedure for a Large Moment Base - When the effective eccentricity is large (greater than ekern), there is a tensile force in the anchor rods due to the moment." If this isn't always the case, they should have a corollary to the equations.

Agreed. I think they overlooked that fact.

The reason it doesn't work is because of B, it's too wide. Maybe it's because the equations they use for kern distance are elastic equations and the design equations are plastic/failure equations? Seems the ekern only works with square plates, which is fine but they should change the verbiage in the Appendix to indicate this is so or change the equation for when eccentricity counts as large v small moment.

I agree that B = 20" is too wide, but that is not the reason it doesn't work. It doesn't work because e < e_crit = N/2 - 2P_u/3B.f_pn = 8 - 110.5/(20*2.224) = 6.34" which I showed in a pdf attachment earlier.

The actual e = 47*12/110.55 = 5.10" which is less than 6.34". If the anchor bolts are considered a reaction, it stands to reason they would be in compression when the load falls between them and the centroid of concrete compression. The force T would be 110.55*1.24/(6.5 + 6.34) = 10.7k compression while the force C would be 110.55*(6.5 + 5.1)/12.84 = 99.9k and C + T = P as expected.

Just for reference, it doesn't count as large eccentricity per the new design method which has:

ecrit = N/2 - P/(2*qmax) = 6.76in > e = 5.108in

Pete600 would object to the "new" design method even more than the AISC suggestion, but it appears more in keeping with the usual concept of limit design.

BA

 

[pete600]

Pete600 would object to the "new" design method even more than the AISC suggestion, but it appears more in keeping with the usual concept of limit design.

I am unsure of the spirit of this comment, but I do not object to either design method.

The original question was why is the anchor in compression, when e_kern says it should not be.

I still believe my original conclusion is correct. Using the triangular design method noted (not the only the acceptable method) the bearing stress should be the applied bearing stress not the allowable bearing stress.

I was unable to open the MathCad spreadsheet.
See MathCAD spreadsheet in link.baseplate