wide beam shear one-way shear

huangyhg

wide beam shear/ one-way shear
the attached document is an example of how i check wide beam/one-way shear. as shown the column is centered on a square slab on grade so its straight forward check.
i now have a situation where a 30"x30" column is on the corner of the slab. (from the north face of the column to edge of slab is 0'-9",from south face of column to edge of slab is 30'-0", from east face of column to edge of slab is 6'-3", and from west face of column to edge of slab is 1'-3".) i take d/2= 13.25" which is greater than north face of column to edge of slab.
i am not sure what to take the bw as? i would assume that the south face would have to take the full vertical load from the column since d/2 is greater than north face of column to edge of slab, but since the west face is pretty close to edge of slab the bw value is not clear to me?
without providing shear ties i would have to increase the depth to the foundation to 54" to support a load of 500k from the column. thats taking the bw as 108" (6'-3"+ 30" +1'-3")  i appreciate any feedback!  
your vu is not pu/2, it is the sum of the bearing pressure on the side of the critical shear plane opposite of the column (which in your case would be less than pu/2).
kbvt is right.  this is why aci allows you to use vu at a distance d from the face of the column.  your one way shear will less than pu/2, more often than not, much less.
can you upload the word file as the 2003 version for those who have not upgraded to 2007 yet?
clansman
if a builder has built a house for a man and has not made his work sound, and the house which he has built has fallen down and so caused the death of the householder, that builder shall be put to death." code of hammurabi, c.2040 b.c.fff">
just to make sure i understand.
1. so if the distance from face of column to edge of concrete is less than d the shear check is neglected on the faces of the columns that fall less than d from the edge?
2. even if the column is located at the corner of the slab assum the pu is distrubuted evenly throughout the entire slab?
this is how i am interpreting it neglecting slab dl. if the column is located in the corner of say 10'x 10' slab and take d = 36" (column = 24"x24")(pu =500k)[qu = pu/(10'x10') = 5ksf] then vu = [10'-0"*(10'-0"-36"-24")*qu] = 250k??  
clansman
the attached word file is as follows:
for a 10' x 10' x 30" slab on grade with column centered on slab the one-way shear strength is:
ultimate vertical load from column:  pu=500k
?vc= ?2λ√(f'c )*bw*d
clear cover = 3", rebar = #8's, ?=0.75, bw=120", λ=1, f'c=4000psi, d = (30"-3.5") = 26.5"
?vc=301.7k
since the column is centered on slab assume half of the vertical load from the column to be resisted by one way shear (ie.  pu/2≤?vc=).

1. yes
2. no, you have to use a triangular (or trapezoidal) pressure distrubution, as appropriate.
for non-concentric applied column loads, convert the load to a concentric axial load with an applied moment (two moments in your case as you are eccentric in two directions).  m1 = p1*e1, m2 = p1*e2.  determine your pressure distribution below the footing (should be trapezoidal in this case).  sum the bearing pressure outside on the opposite side of the critical shear plane (just as you would for the concentric case).  this is your vu.
stucturaleit,kbvt
thanks for your replys!
would either of you have any refferences/examples to a case simialiar to this? i have several sources but each one has examples of columns placed directly in center of slabs. by the sounds of it i have been overly concervative over the years.
bowles text has a couple examples in chapter 8.