Overview
This document is intended to provide technical verification for the analysis and design results within the RAM Concrete Shear Wall design module. This example utilizes RAM Structural System v8i, SELECTseries4, Release 14.04.02.00, which is available for download on Bentley SELECT.
The scope of this document is limited to providing sufficient technical background so that the user may reproduce the design results and calculations performed in the design module. This document is not intended to exhaustively report the design results for the entire model, or provide a tutorial on how to use the applications.
The RAM model referred to in this document can be downloaded at the following link:
After downloading the file, change the extension from *.123 to *.rss.
Model Properties & Assumptions
- (4) 12'-6" stories
- 12" thick x 20'-0" long concrete wall stack
- f'c = 4,000 psi, fy = 60 ksi
- Walls reinforced with (2) layers of #6@12" o.c. vertically, and (2) layers of #5@12" o.c. horizontally
- 10 klf dead load, 10 klf unreducible live load applied across full extent of wall at each story
- Static lateral seismic story forces applied as shown in diagram below
- Walls treated as Special Reinforced Concrete Shear Walls per ACI 318, Chapter 21
- Maximum mesh node spacing of 4' used in RAM Frame
- P-D effects are neglected
- Membrane modifier of 0.35 used for walls
Design for Shear
The shear strength calculation is per ACI 318-08, Section 11.9.6 (as specified in the RAM Concrete design criteria setting) and 11.9.9. The relevant parameters are summarized below:
f'c = 4,000 psi
h = 12"
Lw = 20'
d = 0.8*20' = 16'
Nu = 656 kips (combo 0.82D + E1)
Vu = 250 kips (combo 0.82D + E1)
Mu = 9,333 k-ft (combo 0.82D + E1)
Equation (11-27)
Vc = 3.3l*sqrt(f'c)hd + Nud/(4Lw)
= 3.3*1.0*sqrt(4,000)*12*(16*12)/1000 + 656*16/(4*20)
= 480.9 + 131.2
= 612.1 kips
Equation (11-28)
Vc = [ 0.6l*sqrt(f'c) + Lw[1.25*sqrt(f'c) + 0.2Nu/(Lwh)] / (Mu/Vu - Lw/2) ]hd
= [ 0.6*1.0*sqrt(4,000) + 20*12*[1.25*sqrt(4,000) + 0.2*656,000/(20*12*12)] / (9,333*12/250 - 20*12/2) ]*12*16*12
= [ 37.9 + 240*[79.1 + 45.6] / 328.0 ]*2304/1000
= 297.5 kips
Thus, Equation (11-28) controls
Horizontal reinforcing is (2) curtains of #5@12" o.c. Thus,
As = 2*0.307 in2 = 0.614 in2
Vs = Avfyd/s = 0.614*60*16*12/12 = 589.4 kips
Vn = 297.5 + 589.4 = 886.9 kips
Equation 11.9.3 limits Vn to:
Vn <= 10sqrt(f'c)hd = 10*sqrt(4,000)*12*16*12/1000 = 1457.2 kips (does not control)
Thus,
fVn = 0.75*886.9 = 665.2 kips
Since this system is considered a Special Reinforced Concrete Shear Wall per ACI 318, the shear strength per Section 21.9.4.1 is also evaluated,
Equation (21-7)
Vn = Acv(acl*sqrt(f'c) + rtfy)
hw/Lw = 12.5/20 = 0.625 < 1.5, thus ac = 3.0
Acv = 0.8*20*12*12 = 2304 in2
rt = 0.62/(12*12) = 0.004306
Vn = 2304.0*(3.0*1.0*sqrt(4,000) + 0.004306*60,000) = 1032.4 kips
fVn = 0.75*1032.4 = 774.3 kips
Therefore, Equation 11.9.3 controls
fVn = 665.2 kips
Comparing to shear results in the design report for the bottom-most section cut:
Design for Axial/Flexural Forces
Equation (10-2)
fPn,max = 0.80f[0.85f'c(Ag - Ast) + fyAst]
f = 0.65
Ag = 20*12*12 = 2,880 in2
Ast = 2 layers x 21 rows x 0.44 in2 = 18.5 in2
fPn,max = 0.80 x 0.65 x [0.85 x 4 x (2,880 - 18.5) + 60 x 18.5] = 5,636 kips
We can deduce the value of fPn,max calculated by the program by dividing the value of Pu (which has zero moment) by the interaction value:
1119.95 / 0.199 = 5,629 kips
Special Boundary Element Design
From ACI 318-08, Section 21.9.6.1, compression zones shall be reinforced with special boundary elements where:
c >= Lw / (600du/hw) ACI Equation (21-8)
du is the design displacement. ACI 318-08 elaborates on the meaning of this quantity in Section R21.9.6.2.
The value of displacement used in the calculation of du is taken from the RAM Frame Nodal Displacements report for the worst case combined load (accessible in RAM Frame Load Combinations mode).
From the report shown above, load combination 4 provides the worst case elastic lateral displacement de = 0.770".
du = Cdde = 5.0(0.770") = 3.85"
Also,
Lw = 12' x 12"/' = 240"
hw = 4 x 12.5' x 12"/' = 600" (overall height of wall)
du/hw = 3.85"/600" = 0.006417 < 0.007, thus use du/hw = 0.007 in ACI equation (21-8):
climit = 240/(600*0.007) = 57.14" = 4.76'
From the shear wall design results, the neutral axis distance from the extreme compression fiber for load combination 4, 1.28D + 0.5L - E1, is
c = 5.09' > climit
Thus a special boundary element is required. The required length of the boundary is specified in ACI 318-08, Section 21.9.6.4, and is equal to the larger of:
c - 0.1Lw = 5.09 - 0.1*20 = 3.10'
and
c/2 = 5.09/2 = 2.55'
Thus the required boundary length is 3.09', or 3'-1-3/16"
In this example, boundary regions have been assigned to the wall panel using the Assign -> Manual Reinforcement command. They have been laid out so that the resulting boundary length exceeds the minimum required length for each load combination. The point of maximum compression within the section cut for the selected load combination is denoted with a black dot. The required boundary length is then dimensioned from that point as shown in the screen capture above. If any reinforcing zones not designated as boundaries lie within this region, a design failure will be issued on the Design Warnings tab. In the scenario above, a boundary has been assigned so that the requirement is fulfilled.
The design of ties in confinement zones is per ACI 318-08, Section 21.6.4.
Equation (21-5)
Ash = 0.09sbcf'c/fyt
Per the code design criteria settings, a #4 tie is used,
Ash = 2 x 0.196 in2 = 0.392 in2
b = 12 - 2 x (0.75 + 0.50/2) = 10.0"
Rearranging and solving for s,
s = 60.0 x 0.392 / (0.09 x 10.0 x 4) = 6.53"
The spacing of transverse reinforcing shall also conform to ACI 318-08, Section 21.6.4.3, which states that the spacing of transverse reinforcing shall not exceed the smallest of:
a) 1/4 of the minimum member dimension = 12/4 = 3"
b) 6 x the diameter of the long bar = 6 x 0.75" = 4.5"
Therefore the controlling maximum tie spacing is 3"