Frame action on NCEES bride question

[jc2015]

NCEES practice exam question 903,

the seismic shear in each column is 810 kips, half the total base shear to the pier.  The moment in each column is then V*H/2 = 8100 kip-ft.  

the axial loads I would think are computed by simple frame action and be V*H/(distance between columns). = 1620*23/22=1693 kips. The book solution subtracts the moment of 8100 from 1620*23 but isn't this double counting moments? The 810 kips shear in each column is created by the 1620 kip total shear so you can't apply those at the same time. 

I must be missing something here 

thanks

 

[David Connor SE]

Because the columns are fixed at the top and bottom, the axial load is Ph/2L. On the test I would solve that as (1620 kips)*(23')/(2)*(22) = +/- 847 kips. However, they are saying that inflection point in the column is at +10'-0" instead of +11.5'. So the moments at the base are less, but the axial is more. The 847 kip axial load assumes the column inflection point at 11.5' above the footing instead of 10'-0", so the base moment would 810k * 11.5' = 9315 k-ft, instead of 8100 k-ft. Would they count that as incorrect, I don't know.

 

[bassplayer45]

Personally (from experience), I have done the simple method that you listed above and got an acceptable score.

 

[doodie96]

I think the axial force is made up of two components, the first is due to to overturning moment (or moment transfer of the base shear from the center of the superstructure to the top of the column). This could be compressive and would be 1620 (total force on bent) x 23' / 22'. The second component will be from the frame action of the shear on the bent and would be tensile and would equal 810 (the force in one column only) x 2 x (20/2) (half the height) / 22. Note that this is an approximation. The difference between the two is the net axial force. I believe you can use the portal method to solve the frame action component with the same result. This is explained in the Illinois seismic design guide.