NCEES Practice Exam Question 902

[jc2015]

Bridge Spread Footing Question 902. 

In the answer section I'm curious as to why the effective bearing length is not used to determine the required footing dimension.  AASHTO 10.6.1.3 requires the B and L dimensions be reduced by 2*e for eccentric loadings.  Is it because the column is located in the center of the footing?  But because there is a moment the equivalent vertical load is eccentric.

 

[bassplayer45]

Let me look at this and get an answer to you. I am pretty sure I know the answer, just want to verify to be sure

 

[bassplayer45]

You are correct that it needs to be reduced by 2 * e for, but the trick here, is that for the Extreme Event limit state, it throws you to section 11.6.5, to check bearing distribution, which then sends you to 11.6.3

In 11.6.3, it calculates the uniform bearing resistance at the strength limits states for soil and rock spread footings using equations 11.6.3.2-2 and 11.6.3.2-3. This take into account the effective area based on eccentricity, so you don't have to do it twice essentially. If you were to use 10.6.1.3, you would essentially derive the equations in 11.6.3.2. Just remember that for foundations on soil, it wants the eccentricity to be in the middle 2/3's of the footing, middle 9/10 for rock. 902 is on rock, so it technically should be in the middle 9/10ths per the code, but I have never been marked down before for going middle 2/3's everywhere.

Hope this helps. Just remember, method over the madness for the test. Following the right path and getting the right answer are two different things. They want you to follow the right path