huangyhg

shear in circular concrete columns
i have two questions regarding circular concrete columns that i could use some help with:
1) how do you calculate the effective shear area in a circular concrete column? what is analagous to b x d in a rectangular column? are others just using a percentage of the gross area (85%) or is there specific code guidance provided?
2)when vc is insufficient to resist shear, can spiral stirrups be used to resist shear? if so, how is vs calculated? in my office, we generally just slip rectangular hoops into the column cage when shear reinforcing is needed. is this what others typically do?
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1.) see aci 318-05 11.3.3
2.) see aci 318-05 11.5.1.1c and 11.5.7.3
you can use the spirals to resist shear see the sections noted above.
well... now i feel a little silly. i was looking in the canadian code and neglected to check out aci.
regardless, thanks for your help.
this spawns two more questions though:
1) according to aci, you can use 0.8 x d^2 for the effective shear area. however, that's 1.0186 more than the gross area. how the heck can the effective area be more than the gross area? short of ordering the aci references, can anybody offer an explanation?
2) when using spiral reinforcement for shear with reversible loads, do you treat the spirals like regular inclined reinforcement working at an unfavorable pitch? by unfavorable, i mean a pitch oriented in the same sense as the expected shear crack?
my copy of aci 318-05 states 0.8 x d not d^2 and the commentary states "effective area can be taken as the gross area of the section or as an equivlent rectangular area".
it does say you can take the effective depth (0.8d) times the diameter of the section to get the shear, vc. this is 0.8d^2. as steve states, though, i think it is meant to give you the area of the entire section - it was just much easier to use 0.8 instead of 0.785398 (or pi/4).
huh. gross area it is then.
btw: what does aci 11.5.7.3 say? my 2002 aci only goes up to 11.5.6.9.
it is an equation for stirrups or spirals oriented at some angle, alpha, to the longitudinal axis of the
no rush -- thanks for your help. does it look like:
vs = (av x fy x(sin(alpha) + cos(alpha))*d)/s ? that's what i've got as 11.5.6.4
i actually forgot my code in the office, but that does look like it.
the approx method is to treat it as an equivalent rectangular section 0.8d deep and 2 layers of reinforcement 0.67d apart, the width is then ag/0.8d to keep the area.
when i have checked this for bending it has come in less than 5% different from the exact method.
there is an exact method for these in the istructe (uk) magazine which actually allows for the slope and shape of the helical reinforcement.

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