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Stresses at base of Shear Wall (capacity check)

Stresses at base of Shear Wall (capacity check)

Stresses at base of Shear Wall (capacity check)

(OP)
Hi,

Any comments are welcome. I have a tall building and need to check concrete stress at the base of its shear walls. The FE software I am using gives stresses sigma x, y (orthogonal in plane) principal stresses and von misses. The question is what to compare against the compressive capacity of concrete. From a quick search i did, i didn't find any specific answer for that. Is it reasonable to use von misses for concrete (combine 3 stresses into one equivalent stress and compare that to the compressive strength of concrete)?

Thanks.

RE: Stresses at base of Shear Wall (capacity check)

Being aerospace i havn't done much in the way of concrete!!
However, von Mises is a yield based interaction and concrete doesn't have any appreciable elasic deformation. For brittle (i would have thought it fell into this category) materials i would use Min-Principal for the direct stresses, and also have a look at the shear.
The compressive allowable of the concrete would have been obtained from direct loads anyhow.

 

RE: Stresses at base of Shear Wall (capacity check)

IMHO, concrete by itself is brittle and fails in shear (maybe a more knowledgeable civil guy could comment).

but isn't most concrete reinforced with rebar ?  which will clearly fail in a different mode. (no?)

but don't the structures codes define allowable stresses, or required sections, for the required loads ?

RE: Stresses at base of Shear Wall (capacity check)

You should be looking at the max. principal tensile and compressive stresses (check that steel will fully take the tensile stresses and that the compressive stresses are OK with respect to the max. compressive strength of the concrete) and at the max. shearing stress for shear capacity......Since Von Mises is a shear based value it should NOT be compared against max. compressive strength....

Ed.R.

RE: Stresses at base of Shear Wall (capacity check)

The wall will have reinforcrmrnt in vertical and horizontal directions only. The principal stresses will invariably be in directions which are different from reinforcement directions. This creates a problem to work out how much of the rebar is effective in resisting principal stresses. So it will be simpler to check the 3 orthogonal stresses to check the adequacy of the section.
Axial tensile stress is checked against vertical rebar capacity only.
Axial compressive stress is checked against concrete compressive strength plus vertical rebar compressive strength.
Shear is checked against concrete shear stress and steel tensile strength in horizontal rebar.

Methods of checking these stresses are given in Codes of Practice like ACI 318 or BS 8110.
 

RE: Stresses at base of Shear Wall (capacity check)

(OP)
Thanks for the comments guys,

normm: I agree with that, the "problem" though is that in some cases, compressive component of stress exceeds compressive capacity of concrete (locally-especially at the corners of the base)-this is at least what the software gives (not necessarily true since there's no problem with existing buildings). However, when needed to check for a case that you believe you might have problem, it is not easy, at least for me, to judge with confidence that everything is OK.

 

RE: Stresses at base of Shear Wall (capacity check)

GioC

It does not matter if the compressive stress output exceeds the allowable stress for concrete by a reasonable amount. Because the compressive resistance comes not only from concrete but also from steel rebar which is embedded in concrete. Unfortunately, I am based in UK and the method of measurement of concrete strenghth and allowable stress Code values will be different from yours. But if you treat the corner as a 1 meter wide reinforced concrete wall, you will be OK. Let us assume the wall is 400 mm thick, and the extreme stres is 20 N/mm^2.

Hence the compressive load on the 1m length of wall is ( 20 x 400 x 1000 )/1000 = 8000 kN.

The allowable stress in concrete alone is 0.4.fcu which for grade 40 concrete is 16 N/mm2.
Assume we put in Asc area of steel. and fy is the yield stress in steel. Then all we have to make sure for satisfactory design is :
((16 X 400 x 1000)/1000 + Asc. fy) greter than 8000 kN.

I hope this helps.

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