Coupling Beam Verification - ACI 318-08 with Seismic (RAM Concrete)

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 21.1.1.7. 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 2^nd level.

Figure 1 - Elevation view of wall system

Materials

• f’_c = 5,000 psi
• f_y = 60 ksi

Dimensions

Coupling beam dimensions are as follows:

• h = 72”
• b_w = 12”
• L_n = 72”

Reinforcing

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

A_g = 72” x 12” = 864 in²

A_s = 5 x 2 x 0.31 = 3.10 in²

1.5” clear cover top and bottom

d = 0.8 x 72” = 57.6”

Applied Loads

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.

Axial

• P_DL = 25.79 k
• P_LL = 12.89 k
• P_E  =  84.56 k (compression)

Shear (at span ends):

• V_DL = 74.76 k
• V_LL = 37.38 k
• V_E  =   0.00 k

Moment (positive flexure at midspan):

• M_DL = 163.13 k-ft
• M_LL =   81.57 k-ft
• M_E  =     8.16 k-ft

Combined loads

• P_u = 1.2x(-25.79) + 0.5x(-12.89) + 84.56 = 47.17 k (compression)
• V_u = 108.40 k
• M_u = 244.70 k-ft (positive flexure at midspan)

Figure 4 - Forces at critical points in coupling beam span as taken from RAM Frame

Shear Design

Check the coupling beam dimensions and shear force per section 21.9.7:

(L_n/h) = 1 < 2

> V_u = 108.40, therefore per section 21.9.7.3, 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 V_e is calculated in accordance with section 21.5.4.1 and figure R21.5.4:

M_pr is calculated as the flexural strength of the coupling beam section assuming:

• phi = 1.0
• Tensile strength of reinforcing = 1.25f_y
• Axial load acting in the section is neglected

Using these assumptions for the section in Figure 2, we find:

phiM_n = 590.25 k-ft (assuming 75 ksi rebar)

M_pr = phiM_n/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.

Thus,

Check the conditions of 21.5.4.2. If both of the following apply, then V_c 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 V_e = 345.70 k. Thus, this condition applies.

b)      The factored axial compressive force, P_u = 47.17 k, is less than A_gf’_c/20 = 216 k. Thus, this condition does not apply.

Therefore, V_c need not be taken as zero.

Concrete shear strength for members subjected to axial compression, Equation (11-4):

V_c = 2 x (1 + 0.0273) x 48,875.2 = 100.42 k

Steel shear strength per section 11.4.7.2, Equation (11-15):

Maximum steel shear strength per 11.4.7.9:

Controls

Thus,

phiV_n = phi(V_c + V_s) = 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 21.5.3.2:

• s ≤ d/4 = 57.6”/4 = 14.4”
• s ≤ 8d_b,min = 8 x 0.625 = 5.0” (controls)
• s ≤ 24d_b,shear = 24 x 0.625 = 15.0”
• s ≤ 12”

s = 4” < 5”, OK

Minimum area of shear reinforcement per section 11.4.6.3, Equation (11-13):

OK