The RAM Structural SystemTM, V8i
Release 13.0
RAM ConcreteTM
Shear Wall Design Verification
Example
Bentley Systems, Inc.
2744 Loker Avenue West, Suite 103
Carlsbad, CA 92010
Telephone: (760) 431-3610
Toll Free: (800) 726-7789
RAM Concrete Shear Wall Design Verification Example
Table of Contents
1
2
3
4
5
6
Overview..................................................................................... 3
Wall Design Forces used by RAM Concrete .............................. 5
Design of Walls for Shear......................................................... 12
Design of Walls for Axial-Flexural Forces............................... 14
Special Boundary Element Extents........................................... 16
Special Boundary Tie Design ................................................... 19
1
Overview
This document is intended to provide technical verification for the analysis and design results within the
RAM Concrete Shear Wall design module for a sample RAM Structural System model. This example
utilizes RAM Structural System v8i, r13.0, which is available for download on Bentley Select as of
December 16th, 2008. The model used in this document is available upon request by emailing Josh Taylor
at Josh.Taylor@bentley.com.
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 does not
intend to exhaustively report the design results for the entire model.
Figure 1 – RAM Frame 3D view showing wall and floor slab mesh
Figure 2 – Model 3D view showing only walls
Model Properties
Model Overview:
 (4) full stories plus (1) partial story (62”-0” total height)
 195’ long by 75’ wide footprint
 9” concrete flat slab floors
 12” + 18” interior concrete shear walls
 Perimeter special moment frames: 20” wide by 24” deep beams with 20” square columns
Design Criteria:
 IBC 2006
o Bearing Wall System – Special Reinforced Concrete Shear Walls
o Occupancy Category “I”
o Site Class “C”
 ACI 318-05
o Special Reinforced Concrete Walls & Special Moment Resisting Frames
Floor Loads:
 Office: 25 psf superimposed dead, 60 psf reducible live
 Storage: 5 psf superimposed dead, 150 psf storage
 Roof: 20 psf superimposed dead, 20 psf roof
Wind Loads:
 Exposure: B
 Basic Wind Speed: 90 mph
 Topographical Factor, Kzt = 1.0
 Directionality Factor, Kd = 0.85
Seismic Loads:
 Equivalent Lateral Force Procedure
 Response Modification Coefficient, R = 5.0
 Displacement Amplification Factor, Cd = 5.0
 Importance Factor, I = 1.0
 Ss = 1.00, S1 = 0.30
Frame Analysis Assumptions:
 All floors treated as semi-rigid diaphragms with maximum mesh node spacing of 4 ft
 P-Delta effects considered with scale factor of 1.0
 Out-of-plane wall stiffness ignored
2
Wall Design Forces used by RAM Concrete
Verify Section Cut SC2H:133.
Figure 3 shows a screen shot of the View/Update dialog for Wall Design Group 2. The currently selected
Section Cut is SC2H:133 (at the base of the wall) and the tabular data in the Results page reflect this
Section Cut, as does the cross-section view at the bottom center. The worst case interaction value for
axial/flexural design is produced by Load Combination 260, 0.9D – 1.4E6, as shown directly above the
tabular data sheet. Vertical reinforcement within wall boundaries is colored red in the 3D view box (upper
left). Confinement ties are shown in boundary regions in the cross section sketch.
Figure 3 – RAM Concrete Shear Wall View/Update, Axial/Flexural results
The term E6 in the controlling load combination 0.9D – 1.4E6 represents the load case EQ_IBC06_X_+E_0.3Y_+E_F from the RAM Frame analysis. The applied story forces for this load case are shown in the
RAM Frame Loads and Applied Forces report below.
Figure 4 – RAM Frame Loads and Applied Forces report
Load case E6 coincides with the full X-Direction seismic story force applied coincidentally with 30% of
the Y-direction seismic story force, with both loads applied at an eccentricity that generates a positive Zaxis moment. This is illustrated in Figure 5.
0.3FY
Fx
Y
X
Figure 5 – Orientation of applied loads with respect to mass center for seismic load case E6
The design forces at the Section Cut SC2H:133 can be verified, among other means, by forming a Wall
Group in RAM Frame of the same extents as the Section Cut. The resultant forces at the base of the Wall
Group for each load case or load combination can then be processed.
Figure 6 – RAM Frame Wall Group numbers
The forces at the base of RAM Frame Wall Group 2 are listed in Table 1.
Table 1 –Forces at base of RAM Frame Wall Group 2
Load Condition
Dead
X_+E_-0.3Y_+E_F
0.9D - 1.4E6
P (kips)
810.34
-394.38
1281.44
Mxx (k-ft)
142.76
3625.83
-4947.68
Myy (k-ft)
961.85
20855.04
-28331.39
Note: The RAM Frame Wall Group forces listed above are taken at the wall base, and thus the moments
will be slightly larger in magnitude than the corresponding moments at Section Cut SC2H:133, which is
located 6 in above the base.
Figure 7 through Figure 12 show RAM Frame screen shots for Wall Group 2 forces for each load case for
axial loads and moments.
Figure 7 – Wall axial forces for Dead Load case from RAM Frame
Figure 8 – Wall major moments for Dead Load case from RAM Frame
Figure 9 – Wall minor moments for Dead Load case from RAM Frame
Figure 10 – Wall axial forces for seismic load case from RAM Frame
Figure 11 – Wall major moments for seismic load case from RAM Frame
Figure 12 – Wall minor moments for seismic load case from RAM Frame
Design shear forces can be verified in a similar manner. For Section Cut Segment SC2H:133A, the
controlling load combination is 224, 0.9D + 1.4E6. For Section Cut Segment SC2H:133B, the controlling
load combination is 157, 1.2D + 1.4E11. The design shear forces are summarized in Table 2. The term E11
represents the load case EQ_IBC06_-0.3X_+E_Y_+E_F from RAM Frame.
Table 2 – Section Cut Segment Forces
Section Cut Segment
SC2H:133A
Load Condition
Dead
-0.3X_+E_Y_+E_F
0.9D + 1.4E6
V (kips)
0.56
583.18
816.86
SC2H:133B
Dead
-0.3X_+E_Y_+E_F
1.2D + 1.4E11
6.54
334.12
475.62
3
Design of Walls for Shear
Verify Section Cut Segment SC2H:133A.
Figure 13 – RAM Concrete Shear Wall View/Update – shear design results
The shear strength calculation is per ACI 318-05, Sections 11.10.6 (as specified in the design criteria
setting) and 11.10.9. The relevant parameters are summarized in Table 3:
Table 3 – Summary of parameters used in shear strength equations
Parameter
f’c
h
Lw
d
Nu
Vu
Mu
Value
8,000 psi
18.0 in
20.0 ft
16.0 ft
177.18 kips
816.86 kips
18849.63 k-ft
Equation (11-29):
Comments
Wall Thickness
Axial force acting on Section Cut SC2H:133 for LC 0.9D + 1.4E6
Shear acting Section Cut Segment SC2H:133A for LC 0.9D + 1.4E6
Moment acting Section Cut Segment SC2H:133A for LC 0.9D + 1.4E6
Vc  3.3 f c' hd 
Nud
177.18  16.0
 3.3 8,000  18.0"16.0'12 / 1000 
 1055.51
4l w
4  20.0
Equation (11-30):


N 
l w 1.25 f c'  0.2 u  

lw h  
Vc  0.6 f c'  
hd
M u lw





Vu
2


240.0  111.80  8.203

Vc  53.66 
  18.0  16.0  12  819.80
276.91  120.0

Equation (11-30) controls
Horizontal reinforcing is (2) curtains of #5@12” o.c. Thus,
As = 2 x 0.307 in2 = 0.614 in2
Vs 
Av f y d
s

0.614  60  16.0'12" /'
 589.44
12.0
Vn = 819.80 + 589.44 = 1409.24 kips
Equation 11.10.3 limits Vn to:
Vn  10 f c' hd  10 8,000  18.0"16.0'12 / 1000  3091.14
Thus,
Vn = 0.75 x 1409.24 = 1056.93 kips
4
Design of Walls for Axial-Flexural Forces
Verify Section SC2H:133.
Pn,max  0.800.85 f c' Ag  Ast   f y Ast 

 = 0.65
Ag = 20.5’x12x18” + 8.25’x12x12” = 5616 in2
As = 175.0 in2 (see table below)
The reinforcing zones in each wall panel are summarized in Table 4.
Table 4 – Reinforcing zones in each wall panel
Section Cut Segment
SC2H:133A
Wall Panel
WP 51
WP 51
WP 51
Reinf Zone
Zone 1
Zone 2
Zone 3
Bars
(3) #9@4” o.c.
(2) #9@12” o.c.
(3) #9@4” o.c.
Steel Area (in2)
66.0
16.0
42.0
SC2H:133B
WP 54
Zone 4
(3) #9@6” o.c.
51.0
175.0
Pn ,max  0.80  0.65  0.85  8.0  5616.0  175.0  60.0  175.0  24,699.4
kips
The worst case axial-flexural interaction is produced by load combination 260, 0.900 D - 1.400 E6, which
produces the following design forces on the section:
Pu = 1281.44 kips
Globally oriented moments,
MXX = -4836.35 k-ft MYY = -27,929.55 k-ft
Locally oriented moments,
Mmaj = -27,929.55 k-ft Mmin = 4836.35 k-ft

M u  M 2 maj  M 2 min  28,345.19 k-ft
 = tan-1(Mmaj/Mmin) = 170.17 CCW
Figure 14 – RAM Concrete Shear Wall View/Update – axial/flexural design results
5
Special Boundary Element Extents
Verify Section SC2H:133 under load combination 0.9D – 1.4E6.
From ACI 318-05, Section 21.7.6.2, compression zones shall be reinforced with special boundary elements
where:
lw
c
600
u
ACI Equation (21-8)
hw
RAM Concrete calculates u as the average horizontal displacement at the top of the Wall Design Group
minus the average horizontal displacement at the Section Cut under consideration. The average horizontal
displacement at the top of Wall Design Group 2 can be verified from the RAM Frame nodal displacements
report. Figure 15 shows the node numbers at the top of Wall Design Group 2. Table 5 lists the
displacements at these nodes for load combination 0.9D – 1.4E6.
Figure 15 – RAM Frame node numbers at top of wall group
Table 5 – Nodal deflections from RAM Frame for controlling seismic load combination
Node Number
2
3
5
X (in)
-0.8988
-1.0565
-0.9011
Y (in)
0.0094
0.0010
0.0947
Xavg = (-0.8988 - 1.0565 - 0.9011)/3 = -0.9521 in
Yavg = (0.0094 + 0.0010 + 0.0947)/3 = 0.0350 in
e 
 0.95212  0.03502
 0.9527 in
u = Cde = 5.0(0.9527 in) = 4.7637 in
lw = 21.86 ft (projected length of section in direction of resultant load)
hw = 62.0 ft (overall height of Wall Design Group 2)
u/hw = 4.7637/(62x12) = 0.00640 < 0.007, thus use u/hw = 0.007 in ACI equation (21-8):
c
21.86  12
 62.46 in = 5.205 ft
600  0.007
The required length of the boundary is specified in ACI 318-05, Section 21.7.6.4, and is equal to the lesser
of:
c – 0.1lw = 5.63 ft – 0.1(21.68 ft) = 3.45 ft
c/2 = 5.63/2 = 2.82 ft
Thus the required boundary length is 3.45 ft, or 3’-5.4”.
Figure 16 – Boundary element evaluation for load combination 0.9D – 1.4E6
In this example, boundary regions have been assigned to the Wall Panels 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 as solved for above. In Figure 16, each of the
boundary regions is tied off as shown in the cross section sketch at the bottom. 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 shot. 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 Tie/Link Design sheet provides the transverse reinforcement calculations for the each of the boundary
zones that intersect the selected Section Cut.
6
Special Boundary Tie Design
The design of transverse reinforcement in confinement zones is per ACI 318-05, Section 21.4.4.
Equation (21-4),
0.09 sbc f c'
Ash 
f yt
s = 3 x 0.196 in2 = 0.589 in2
b = 18 – 2 x (1.5 + 0.50/2) = 14.5 in
Rearranging and solving for s,
s
f yt Ash
0.09bc f
'
c

60.0  0.589
 3.385 in
0.09  14.50  8.0
The spacing of transverse reinforcing shall also conform to ACI 318-05, Section 21.4.4.2, which states the
spacing of transverse reinforcing shall not exceed the smallest of:
a) ¼ of the minimum member dimension = 18.00/4 = 4.5 in
b) 6 x the diameter of the long bar = 6 x 1.128 = 6.77 in
Therefore, the controlling maximum tie spacing for reinforcing zone 1 is 3.385 in.
Figure 17 – Boundary zone tie design for Section Cut