huangyhg

shear-friction in a concrete wall
if you read section 11.7.1 of aci 318, you will find it severely restricts the use of shear friction. you are correct that if you design the wall as pinned, you will get a crack and shear friction will apply. but practically the same shear is acting at some distance (an inch, a foot?) away from the crack and by definition shear friction does not apply there.
i would design the wall as fixed on three sides. use a reference such as "moments and reactions for rectangular plates" from the u.s. department of the interior, water resources technical publication, engineering monograph no. 27. unfortunatly this publication is out of print, but maybe someone in your office has a copy. i suspect risa is giving you some kind of unrealistic loads at the discontinuity. using the moments and reactions from the reference, bite the bullet and make the wall thick enough to carry the shear (using formula 11-3 of aci 318) at the base and sides. note that you can design for a distance of "d" away per section 11.1.3.1. i've designed a lot of these walls and shear usually controls the thickness.
a good reference for design of basement walls is "rectangular concrete tanks", revised fifth edition by javeed a. munshi, which can be purchased through the porland cement association. the purpose of the publication is as a design aid for fluid containing structures, but it works well for basement walls. there are charts for different types of support conditions, such as the fixed ends-pinned bottom-free top condition you describe. these charts provide moment, shear and deflection design coeficients for different locations at the wall.
rceng - what you can also consider is extending your model to include the perpendicular walls - extend them back some distance and fix those ends. by pinning the corners you neglect the added rotational stiffness of the intersecting walls. by adding fixity at the corners, you over estimate the fixity provided by the intersecting walls and get higher negative moments than really apply.
be sure to include lateral forces on the added walls as well.
as far as shear friction goes - i wouldn't use it in this case. usually, we check for phi x vc against the qx qy forces that risa provides and ensure adequate thickness to resist the forces. aci's equation - vc = 2 x sqrt (f'c) x b x d is quite conservative - especially with phi = 0.85.
rceng (visitor)27 aug 02 22:17
thanks everyone for your comments. i need some way to check up on risa. "rectangular concrete tanks" suggested by kramer is on the way. i re
here's what i've done on a recent study of a swimming pool wall:
1. model in risa-3d using full thickness of the wall (i.e. you are using igross).
2. calculate the cracking moment of the wall section ( a one foot section) using the aci equation mcr = fr x ig / y where fr = 7.5 x sqrt(f'c) and y is half your wall thickness.
3. run the model and look for elements that have values of (mx + mxy) or (my + mxy) greater than your mcr. if so, then this "portion" of the wall cracks and therefore - edit those specific elements to be thinner - thus modeling a cracked wall. the thinner thickness that you choose could simply be icracked. usually using about 0.15 x ig will suffice but you can be more detailed if you want.
4. re-run the model again - again search for elements which exceed mcr and size them down.
5. repeat step 4 until the model converges. thus, you will acually see the progression of the cracking in the wall. for my pool, i followed this procedure and the resulting "crack" was exactly the crack that existed in the pool wall that prompted my study in the first place.
this will/should re-distribute your q forces. note that after thinning up some elements you will get strange results in those thin elements. this takes some interpretation but i think it will do what you need.

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