Buttress Wall Relative Rigidity

[HMROSS]

I am using the method of relative rigities to distribute horizontal forces due to a rigid roof diaphram for a single story structure. The structure is comprised of rectangular boxes all of which are open on one end and closed on the other (earth covered-into hillside). I am looking at placing buttresses on both sides of the open-ends to move the center-of-rigidity towards the center-of-mass and reduce torsional effects.

Being that all the other wall rigidities are based on superposition of their solid pier(s)relative deflections, what is the best assumptions to make regarding a 45 degree sloped buttress wall from some height h. Examples (cantilever action):
A) h=8';L=8';h/L=1.0; d=0.7, R=1.4.
B) h=4';L=8';h/L=0.5; d=0.2; R=5.0
C) h=8';L=4';h/L=2.0; d=3.8; R=0.26

My objective is to model the rigidity as close as possible to the other relative rigidties. Noting that the method I am employing gives somewhat bigger overall values of rigidty than more advanced methods.

I am chosing to assume example B at this time and would like some feedback--shoot away at my ass---umption please.

 

[tincan]

Woould this be similar to the end counterfort of a retaining wall? might not be getting the correc picture.

tincan

 

[HMROSS]

To tincan,

You can think of it as a counterfort only there is no retaining wall but there is a bearing/shear wall normal to the seismic force. The open side of the box has no lateral resisting elements other than the wall normal (as a retaining is) to the seismic force. In-plane with what would be an interior shear wall, where large openings are now proposed, but outside the building, I am proposing the counterfort to resist seismic forces and reduce torsion. These would be like wings. They happen to fit this particular building plan since they will also act to somewhat enclose a courtyard space--its a U shape building-a plan irregularity. Instead of an internal shear wall at the end of the legs of the U, I want to move the shear resisting element to the ouside of the building. Hope this helps.

 

[tincan]

Whats the height, width, and number of units? The open end of these units sounds almost like a multi-cell reinforced concrete box. Also, whats the depth of bury.
Tincan.

 

[HMROSS]

tincan, thanks for the interest. Here is more information. There are 5 cells, cells share a common wall. I will be comparing costs for cast-in-place and prestressed rein. concr. roof systems. The cell width is 16-feet, interior space. My goal is to design with rein. cmu under the UBC-97 no-inspection cluase for the working stress design method (1/2 stresses)--we are building the structure ourselves with friends and relatives--what else is there to due. If need be, I will jump to the special inspection criteria.

The roof slopes at 3% from north to south, north being the low point and against the hill which slopes at approx. 15%. The height of the wall against the hill is 8'8", the heigth the south end of the n-s walls varies with length. The height of the walls in the west-east direction are constant. A minimum of 1-foot (max 3') of earthen cover in combination with a drainage, water barrier, and insulation layers over the structural roof system. The earthen cover will tie back into the hill, portins of the west and east outer walls will be earthen bermed. With: W=wall no.; L=length,ft; n-w= north or east wall heigth, ft; s-e = south or east wall end heigth,ft; here is the wall info:

Wall L,ft n or e, h, ft
(North-South Walls)
W L N-W S-E
1 50'8" 8'8" 10.19
2 50.67 8.67 10.19
3 32 8.67 9.63
4 50'8" 8.67 10.19
5 50'8" 8.67 10.19
6 16' 8.67 8.69
(WEST-EAST WALLS)
AA 16' 8.67 8.67
AB 16' 9.75 9.75
AC 16' 10.11 10.11
BC 16' 8.67 8.67
BB 16' 9.53 9.53
CA 16' 8.67 8.67
CB 16' 9.43 9.43
DA 16' 8.67 8.67
DB 16' 10.11 10.11
EA 40.64 8.67 8.67
EB 40.64 9.17 9.17

There are five modules-A,B,C,D,&E. Modules A,B,C,&D bearing walls 1,2,3,4,&5 create the "U" plan-courtyard to the south, with module E--the garage--creating a leg from the north side of the "U". The "U" plan is open to the south. Hope this helps. I am currently trying to keep walls to 8-inch cmu.

The walls I am considering would be walls AC.L, for Left of wall AC.; AC.R, for right of wall AC, so on and so forth; walls AC.L, AC.R, DB.L, DB.R, 6.L, & 6.R. Each of these principle walls have very large openings and offer very little with respect to lateral resistance. This cuases an "relatively" large eccentricity. By placing the support walls in-plan with these walls but outside the building I can satisfy lateral resistance and reduce eccentriciies. The walls will be partially hidden with earthen berming.

I just have to define a reasonable deflection/rigidity for the buttress type walls. this nust be "relative" to the way rigidity values have been created for the main walls.

 

[tincan]

Wall at Face of Hill
1 2 3 4 5 6
-----------------------------
' ' ' ' ' '
' ' ' ' ' '
' ' ' ' '----'
' ' ' ' '
' ' ' ' ' cave drawing
' '-----' ' ' dont understand
' ' ' ' ' the letters
' ' ' ' '
' ' ' ' '
' ' ' ' '
'-----' '-----'-----'

Is this your layout? Whats your location? CMU's might not work on the exterior wall for a moist situation. You might think about reinforced concrete for those walls in contact with soil. CMU walls should be ok for the interior. I would think that the walls (except for the open ends) could be designed as a basement wall. As noted, the open ends will require work.

let me know on the above, best

 

[HMROSS]

The letters represent Modules and N-S walls; all *A walls are against the hillside. Thus, Wall AB, is south of Wall AA and Wall AC is south of Wall AB. I did not give you the x,y cooridnates, just the heights. Here is a cave(BEAR) drawing.

:-AA-:-BA-:-CA-:-DA-:---EA---:
1 2 3 4 5 6
: : : : :---EB---:
: : :-CB-: :
: :-BB-: : :
:-AB-: : :
: : * : :
:-AC-: :-DB-:

* = COURTYARD

Walls AA, BA, CA, & DA have the same rigidty(R), Wall EA is solid but is longer; Walls AB has about 80% of AA's R, Wall AC is essentially not there and has less than 1% of AA's R, Wall BB has about 25% of AA's R, Wall CB has about 13% of AA's R, Wall DB has about 25% of AA's R, and Wall 6 is essentially not there with less than .01% of Wall 5's R--which is the highest of all the walls.

I can work with the eccentricities but I am looking at the wing walls to the left(L) and right(R) of Walls 1,2,4,5, EA, and EB at Walls AC,DB, and 6 to reduce torsional effects; hence walls designated as AC.L, AC.R, DB.L and DB.R, 6.L, and 6.R. It is my own wall designations, sorry if they are confusing. The longest walls, Walls 1,2,4,5 are 50.67'. Wall 2 is 32', Wall EA and EB are 40.67'. all other walls are 16'.

R
:-AA-:-BA-:-CA-:-DA-:---EA---:
1 2 3 4 5 6
: : : : :---EB---:
: : :-CB-: : L
: :-BB-: : :
:-AB-: : :
: : * : :
L-:-AC-:-R L-:-DB-:-R

 

[tincan]

Sorry in taking so long to answer. I had a car to get ready for a show this week end. Heres my thoughts on the subject.
The relative rigidity is based on equal deflections of the various units resisting the moment/shear acting on the units, D1 = D2 = D3........The assumed deflection is D=VL^3/12EI. Taking equal D's, the deflection is proportional to 1/I. I is dependent upon the depth of the wall/column/ resisting the shear/moment/deflection. The tapered sections have a variable I. Taking an average I at the 3/8 point from the bottom,

h = 8 L = 8 LI = 5 I~5^3 = 125
h = 4 L = 8 LI = 5 I~5^3 = 125
h = 8 L = 4 LI = 2.5 I~2.5^3 = 15.3

Without knowing the full particulars, I think I would go with the 8 x 8.