General Description of Example
The structure under consideration is a two-story planar wall with 6’ wide doorways centered horizontally on the wall. No other structural elements (slabs, beams, columns, etc.) are considered, so that the wall design can be isolated. The stories are 12’-6” high and the walls are 12” thick.
The design is in accordance with ACI 318-08, and is categorized as a special reinforced concrete structural wall per Section 22.214.171.124. Thus the provisions of Chapter 21, sections 21.9 and 21.1.3 through 21.1.7 are enforced. The coupling beam under consideration is at the 2nd level.
Figure 1 - Elevation view of wall system
Coupling beam dimensions are as follows:
The coupling beam is reinforced for flexure with a uniform mat of longitudinal bars, consisting of (5) #5 at each face of the wall.
Shear reinforcing consists of #5 stirrups at 4” along the entire span.
Figure 2 - Cross section view of coupling beam
Ag = 72” x 12” = 864 in2
As = 5 x 2 x 0.31 = 3.10 in2
1.5” clear cover top and bottom
d = 0.8 x 72” = 57.6”
A vertical point load is applied at midspan of the coupling beams at each story, 150 kips dead and 75 kips live. No other gravity loads are considered (self-weight is turned off in RAM model).
Seismic lateral loads are applied at each story such that axial force exists within the coupling beams, in order to demonstrate the effects of compression/tension on shear strength calculations.
Figure 3 - Applied dead, live, and earthquake loads
Coupling Beam Design Forces
The forces in the coupling beam can be retrieved in RAM Frame using the shear wall forces module (see Figure 4). A section is drawn at each end of the coupling beam span (inset 3” from the face) and at midspan. There is a dead, live, and earthquake load case.
Only one load combination is considered for this example:
1.2D + 0.5L + E
In this case the coupling beam is in compression.
Shear (at span ends):
Moment (positive flexure at midspan):
Figure 4 - Forces at critical points in coupling beam span as taken from RAM Frame
Check the coupling beam dimensions and shear force per section 21.9.7:
(Ln/h) = 1 < 2
> Vu = 108.40, therefore per section 126.96.36.199, diagonal reinforcing is not required and the coupling beam will be designed according to sections 21.5.2 through 21.5.4.
The design shear force Ve is calculated in accordance with section 188.8.131.52 and figure R21.5.4:
Mpr is calculated as the flexural strength of the coupling beam section assuming:
Using these assumptions for the section in Figure 2, we find:
phiMn = 590.25 k-ft (assuming 75 ksi rebar)
Mpr = phiMn/phi = 590.25/0.90 = 655.83 k-ft
Note: The flexural strengths above can be verified by reconstructing the section in a separate model with the reinforcing strength set to 75 ksi rather than 60 ksi.
Check the conditions of 184.108.40.206. If both of the following apply, then Vc must be taken as zero:
a) The earthquake-induced shear force, 2x655.83/6 = 218.61 k, represents more than half of maximum required shear strength Ve = 345.70 k. Thus, this condition applies.
b) The factored axial compressive force, Pu = 47.17 k, is less than Agf’c/20 = 216 k. Thus, this condition does not apply.
Therefore, Vc need not be taken as zero.
Concrete shear strength for members subjected to axial compression, Equation (11-4):
Vc = 2 x (1 + 0.0273) x 48,875.2 = 100.42 k
Steel shear strength per section 220.127.116.11, Equation (11-15):
Maximum steel shear strength per 18.104.22.168:
phiVn = phi(Vc + Vs) = 0.75 x (100.42 + 391.00) = 368.57 k
Figure 5 - Shear design results from RAM Concrete View/Update dialog
Maximum flexural bar spacing per section 7.6.5:
s ≤ 18” OK
Stirrup spacing limit per section 22.214.171.124:
s = 4” < 5”, OK
Minimum area of shear reinforcement per section 126.96.36.199, Equation (11-13):