# Bridge Column Overstrength Plastic Demand question



## bassplayer45 (May 10, 2013)

Alright, i apologize for the length of this in advance. Working in Indiana and obtaining my SE, the lateral portion of the test is very tricky since i am basically teaching myself everything seismic related, since we basically don't do it. I have been making up example problems and running them in RC-Pier, then doing them by hand. I am currently working on a column design in zone 3 and i would like some guidance.

I have developed a simple 1 round column model with a pier and pile cap. After running through the extreme event 1 load case, i arrived at the following. Pu = 629.1 kips Mu (dead) = 206.25 ft-k, Mu (eq) = 4350 ft-k

From 3.10.9.4, the design forces shall be the lesser of the provisions in 3.10.9.4.2 and 3.10.9.4.3.

For my column design, i adhere to 3.10.9.3. With an R factor of 3.0, i performed the following calculation. (If i did this correctly, i satisfy the first provision)

Pu = 629.1 kip (dead)

Mu = 206.25 + (4350 / 3) = 1656 ft-k

Per 3.10.9.4.3, i must now check inelastic hinging forces. From 3.10.9.4.3b, it says for single column bents that are concrete, determine the overstrength moment resistance. Using a resistance factor of 1.3 for the moment, use the axial load from the extreme event 1 load case.

My first question is: Do i multiply my Mu by 1.3? 1656 * 1.3? or 4350 * 1.3? Or do i multiply my column moment resistance Mn by 1.3?

Second question: This basically would result with me having a Pu = 626.1 with my new Mu. If these are not lower than the calculated forces above (reduced, or not reduced?), then my original forces govern for the column design?

Third and final question: Once i calculate my overstrength moment, i would divide this by the distance between top of column and bottom of column to get my maximum shear force to resist, correct? Without reduction by R?

I appreciate the help or any links to resources.


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## Layman (May 13, 2013)

See my replies after each of your question below

"My first question is: Do i multiply my Mu by 1.3? 1656 * 1.3? or 4350 * 1.3? Or do i multiply my column moment resistance Mn by 1.3? "

Neither of these two. Multiply by 1.3 the Mn (nominal strength) of the column you just designed with method using R. (See first paragraph of AASHTO 3.10.9.4.3a General)

"Second question: This basically would result with me having a Pu = 626.1 with my new Mu. If these are not lower than the calculated forces above (reduced, or not reduced?), then my original forces govern for the column design? "

Correct.

"Third and final question: Once i calculate my overstrength moment, i would divide this by the distance between top of column and bottom of column to get my maximum shear force to resist, correct? Without reduction by R? "

Or divide the overstrenght by half of the distance between top and bottom hinges, depending on end conditions.This gives you the design force already. No R factors need to be applied. (AASHTO Third paragraph of 3.10.7.1)


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## Layman (May 13, 2013)

More comments on question 2.

Axial force P almost never governs a column design of any aspects, but we know it has a great impact on shear design as concrete may be allowed to contribute more shear resistance. So, using a greater Pu of the results from using the R and plastic hinge mothds may not lead a conservative shear design (spiral steel spacing or size).

These are my understanding. I never designed a bridge in real world.


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## bassplayer45 (May 13, 2013)

Appreciate the help! I had been reading that paragraph for awhile and was almost positive we multiplied Mn by 1.3, just wasn't 100% sure.

I have been using RC-Pier for awhile and it has been giving me some screwy answers with the ouput when it involves seismic. I was getting a design moment of 1450 (EQ / response) + 206.25 = 1656, but it keeps wanting to design for the original 4556.3, still trying to figure out where it is getting its numbers.

Again, thanks for the help. I will finish this example problem up and see where it takes me!


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## bassplayer45 (May 13, 2013)

Sorry, forgot to ask this last question related to my previous post.

I obtained a nominal moment strength of 2136 ft-k for my column from the seismic design. After multiplying by 1.3, i get 2136 * 1.3 = 2778 ft-k. NOW, is this compared to the 1656 ft-k from above which was reduced? Or is it compared to the 4350 ft-k which was unreduced? Based on the code, i interpret it as the reduced moment. Thus, 2778 is greater than 1656, so my moment design strength is governed by the original seismic design forces.


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## Layman (May 14, 2013)

Compare the overstrength Moment to the reduced value, 1656 ft-k, based on 3.10.9.4.1. This comparsion is not needed as 1.3Mn&gt;Mn, and Mn was designed to be greater than Mu in your preliminary design, 1656. So the second dot of 3.10.9.4.3d, Column and Pile Bent Design Forces, simply states to use "modified design moments determined for extreme Event Limit State Load Combination I", which in this case is 1656.


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## McEngr (May 14, 2013)

I think it's important to ask yourself why you are performing the design in the first place. Do you understand the philosophy of ductile concrete seismic design? Why would they apply 1.3Mn? You may want to get a grasp of capacity design concepts and then tackle these problems. PCA Notes and Alan Williams has good literature on these subjects.


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## bassplayer45 (May 15, 2013)

Thanks Layman. McEngr, appreciate the input and resources. Ive slowly been reading up more on the subject. It has definetly been an undertaking so far


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