Question 6.11: For the one-story retail building illustrated in Figure 6.17......

• Step 1: Determine the seismic ground motion values from Flowchart 6.2.

Using the USGS Ground Motion Parameter Calculator, S_S = 1.47 and S_1 = 0.50 .

Using Tables 11.4-1 and 11.4-2, the soil-modified accelerations are S_{MS} = 1.47 and S_{M1} = 0.65 .

Design accelerations: S_{D S}={\frac{2}{3}}\times1.47=0.98 and S_{D1}={\frac{2}{3}}\times0.65=0.43

• Step 2: Determine if the simplified method of 12.14 can be used for this building.

The simplified method is permitted to be used if the following 12 limitations are met:

1. The structure shall qualify for Occupancy I or II in accordance with Table 1-1.

From Table 1-1, the Occupancy Category is II. O.K.

2. The Site Class shall not be E or F.

The Site Class is C in accordance with the design data. O.K.

3. The structure shall not exceed three stories in height.

The structure is one story. O.K.

4. The seismic force-resisting system shall be either a bearing wall system or building frame system in accordance with Table 12.14-1.

The seismic force-resisting system is a bearing wall system. O.K.

5. The structure has at least two lines of lateral resistance in each of the two major axis directions.

Masonry shear walls are provided along two lines at the perimeter in both directions. O.K.

6. At least one line of resistance shall be provided on each side of the center of mass in each direction.

The center of mass is approximately located at the geometric center of the building and walls are provided on all four sides at the perimeter. O.K.

7. For structures with flexible diaphragms, overhangs beyond the outside line of shear walls or braced frames shall satisfy: a ≤ d / 5 .

The diaphragm in this building is rigid, so this limitation is not applicable.

8. For buildings with diaphragms that are not flexible, the distance between the center of rigidity and the center of mass parallel to each major axis shall not exceed 15 percent of the greatest width of the diaphragm parallel to that axis.

Assume that the center of mass is at the geometric center of this building.^{21} This limitation is satisfied with respect to the east-west direction, since the center of rigidity and center of mass are on the same line due to the symmetrical layout of the walls on the east and west elevations.

The center of rigidity must be located in the north-south direction, since the north and south walls are not identical. By inspection, the center of rigidity is located closer to the south wall, since the stiffness of that wall is greater than the stiffness of the north wall.

To locate the center of rigidity, the stiffnesses of the north and south walls must be determined. Assuming that the piers are fixed at the top and bottom ends, the stiffnesses (or rigidities) of the walls and piers can be determined by the following:

Total displacement of pier or wall i: δ_i = δ_{fi} + δ_{νi}

δ_{fi} = displacement due to bending =\frac{\displaystyle{\left\lgroup\frac{h_{i}}{\ell_{i}}\right\rgroup^{3}}}{Et} \,

δ_{νi} = displacement due to shear =\frac{\displaystyle{3\left\lgroup\frac{h_{i}}{\ell_{i}}\right\rgroup}}{Et} \,

where h_i = height of pier or wall
\qquad \quad \ell_i = length of pier or wall
\qquad \quad t = thickness of pier or wall
\qquad \quad E = modulus of elasticity of pier or wall

Stiffness of pier or wall

k_{i}={\frac{1}{δ_{i}}}

In lieu of a more rigorous analysis, the stiffness of a wall with openings is determined as follows: first, the deflection of the wall is obtained as though it were a solid wall with no openings. Next, the deflection of a solid strip of wall that contains the openings is subtracted from the total deflection. Finally, the deflection of each pier surrounded by the openings is added back.

Table 6.20 contains a summary of the stiffness calculations for the north wall.

Similar calculations for the south wall are given in Table 6.21. The pier designations are provided in Figure 6.18.

North wall stiffness = 0.635Et

South wall stiffness = 0.818Et

East and west wall stiffness = 1.65Et .

The location of the center of rigidity in the north-south direction can be determined from the following equation:

\overline{{{y}}}_{r}=\frac{\sum\limits{k_{i}y_{i}}}{\sum{k_{i}}}

where y_i is the distance from a reference point to wall i.

Using the centerline of the south wall as the reference line (see Figure 6.19),

{\overline{{y}}}_{r}={\frac{{\mathrm{0.635}}E t\left\lgroup60-{\frac{7.625}{12}}\right\rgroup}{{\mathrm{0.635}}E t+{\mathrm{0.818}}E t}}=25.9 ft

e_{1}={\left\lgroup30-{\frac{7.625}{2\times12}}\right\rgroup}-25.9=3.8 ft

0.15× 60 = 9.0 ft > 3.8 ft O.K.

In addition, Eqs. 12.14-2A and 12.14-2B must be satisfied:

\sum\limits_{i=1}^{m}k_{1i}d_{1i}^{2}+\sum\limits_{j=1}^{n}k_{2j}d_{2j}^{2}\geq2.5\left\lgroup0.05+\frac{e_{1}}{b_{1}}\right\rgroup b_{1}^{2}\sum\limits_{i=1}^{m}k_{1i} \,

\sum\limits_{i=1}^{m}k_{1i}d_{1i}^{2}+\sum\limits_{j=1}^{n}k_{2j}d_{2j}^{2}\geq2.5\left\lgroup0.05+\frac{e_{2}}{b_{2}}\right\rgroup b_{2}^{2}\sum\limits_{i=1}^{m}k_{2i} \,

where the notation is defined in Figure 12.14-1 and 12.14.1.1.

\sum\limits_{i=1}^{m}k_{1i}d_{1i}^{2}+\sum\limits_{j=1}^{n}k_{2j}d_{2j}^{2}=(0.635E t\times33.5^{2})+(0.818E t\times25.9^{2})+\left(2\times1.65E t\times19.9^{2}\right) \,

\qquad\qquad \quad =2,568.2E t \,

2.5\left\lgroup0.05+\frac{{e}_{1}}{b_{1}}\right\rgroup b_{1}^2\sum\limits_{i=1}^{m}k_{1i}=2.5\left\lgroup0.05+\frac{3.8}{60}\right\rgroup\times60^{2}\times(0.635Et+0.818E t) \,

\qquad\qquad \quad=1,482.1E t\lt 2,568.2E t O.K.

2.5\left\lgroup0.05+\frac{{e}_{2}}{b_{2}}\right\rgroup b_{2}^2\sum\limits_{j=1}^{m}k_{2j}=2.5\left\lgroup0.05+\frac{0}{40.5}\right\rgroup\times40.5^{2}\times(2\times01.65Et) \,

\qquad\qquad \quad=676.6E t\lt 2,568.2E t O.K.

Thus, all conditions of the eighth limitation are satisfied.

Note that Eqs. 12.14-2A and 12.14-2B need not checked where the following three conditions are met:

1. The arrangement of walls is symmetric about each major axis.

2. The distance between the two most separated wall lines is at least 90 percent of the structure dimension perpendicular to that axis direction.

3. The stiffness along each of the lines of resistance considered in item 2 above is at least 33 percent of the total stiffness in that direction.

In this example, only the second and third conditions are met.

9. Lines of resistance of the lateral force-resisting shall be oriented at angles of no more than 15 degrees from alignment with the major orthogonal axes of the building.

The shear walls in both directions are parallel to the major axes. O.K.

10. The simplified design procedure shall be used for each major orthogonal horizontal axis direction of the building.

The simplified design procedure is used in both directions (see Step 3). O.K.

11. System irregularities caused by in-plane or out-of-plane offsets of lateral forceresisting elements shall not be permitted.

This building does not have any irregularities. O.K.

12. The lateral load resistance of any story shall not be less than 80 percent of the story above.

Since this is a one-story building, this limitation is not applicable. O.K.

Since all 12 limitations of 12.14.1.1 are satisfied, the simplified procedure may be used.

• Step 3: Determine the seismic base shear from Flowchart 6.9.

1. Determine S_S , S_{DS} and the SDC from Flowchart 6.4.

From Step 1, S_S = 1.47 and S_{DS} = 0.98 .

According to 11.6, the SDC is permitted to be determined from Table 11.6-1 alone where the simplified design procedure is used.

For S_{DS} > 0.50 and Occupancy Category II, the SDC is D.

2. Determine the response modification factor R from Table 12.14-1.

For SDC D, a bearing wall system with special reinforced masonry shear walls is required (system A7). For this system, R = 5.

3. Determine the effective seismic weight W in accordance with 12.14.8.1.

Conservatively assume that the masonry walls are fully grouted and neglect any wall openings. Also assume a 10 psf superimposed dead load on the roof.

Weight of masonry walls tributary to roof diaphragm

=0.081\times{\frac{12}{2}}\times2(60.0+40.5)=98 kips

Weight of roof slab ={\frac{8}{12}}\times0.150\times60\times40.5=243 kips

Superimposed dead load = 0.010 × 60 × 40.5 = 24 kips

W = 98 + 243 + 24 = 365 kips

4. Determine base shear V by Eq. 12.14-11.

V={\frac{F S_{D S}}{R}}\,W={\frac{1\times0.98\times365}{5}}=72 kips

where F = 1 for a one-story building (see 12.14.8.1).

Since this a one-story building, story shear V_x = V .

5. Distribute story shear to the shear walls.

Since the building has a rigid diaphragm, the design story shear is distributed to the shear walls based on the relative stiffness of the walls, including the effects from the torsional moment M_t resulting from eccentricity between the locations of the center of mass and the center of rigidity. Note that the simplified procedure does not require accidental torsion (12.14.8.3.2.1).

For lateral forces in the N-S direction, there is no torsional moment, since there is no eccentricity between the center of mass and center of rigidity in that direction.

The east and west walls have the same stiffness, so each wall must resist 72/2 = 36 kips.

For lateral forces in the E-W direction, the torsional moment is equal to 72× 3.8 = 274 ft-kips.

The total lateral force to be resisted by the north and south shear walls can be determined from the following equation:

V_{1i}=\frac{k_{1i}}{\sum\limits k_{1i}}\,V_{x}+\frac{d_{\mathrm{1}i}k_{\mathrm{1}i}}{\sum\limits_{i=1}^{m}k_{\mathrm{1}i}d_{\mathrm{1}i}^{2}+\sum\limits_{j=1}^{n}k_{2j}d^{2}_{2j}}M_t

For the north shear wall:

V_{11}=\frac{0.635E t}{0.635E t+0.818E t}\times72+\frac{33.5\times0.635E t}{2,568.2E t}\times274 \,

\quad =(0.437\times72)+(0.0083\times274) \,

\quad =31.5+2.3=33.8 kips

For the south shear wall:

V_{12}=\frac{0.818E t}{0.635E t+0.818E t}\times72-\frac{25.9\times0.818E t}{2,568.2E t}\times274 \,

\quad =(0.563\times72)-(0.0082\times274) \,

\quad =40.5-2.3=38.2 kips

The east and west shear walls are subjected to a shear force due to the torsional moment for lateral forces in the east-west direction, but that force is less than the 36 kip force that is required for lateral forces in the N-S direction.

^{21} The exact location of the center of mass should be computed where it is anticipated to be offset from the geometric center of the building.