Question 6.7: Determine the diaphragm design forces for the three-story he......

In order to determine the diaphragm design forces in accordance with 12.10.1.1, the design seismic forces must be determined at each floor level.

Assuming that the building is regular, the equivalent lateral force procedure can be used to determine the design seismic forces (see Table 12.6-1).

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

1. Determine the mapped accelerations S_S and S_1 .

In lieu of using Figures 22-1 and 22-2, the mapped accelerations are determined by inputting the latitude and longitude of the site into the USGS Ground Motion Parameter Calculator. The output is as follows: S_S = 0.58 and S_1 = 0.17 .

2. Determine the site class of the soil.

The site class of the soil is given in the design data as Site Class C.

3. Determine soil-modified accelerations S_{MS} and S_{M1}.

Site coefficients F_a and F_ν are determined from Tables 11.4-1 and 11.4-2, respectively:

For Site Class C and 0.5 < S_S < 0.75 : F_a =1.17 from linear interpolation

For Site Class C and 0.1< S_1 < 0.2 : F_ν =1.63 from linear interpolation
Thus,

S_{M S}=1.17\times0.58=0.68 \,

4. Determine design accelerations S_{DS} and S_{D1} .

From Eqs. 11.4-3 and 11.4-4:

S_{D S}=\frac{2}{3}\times0.68=0.45 \,

• Step 2: Determine the SDC from Flowchart 6.4.
1. Determine if the building can be assigned to SDC A in accordance with 11.4.1.

Since S_{S} = 0.58 > 0.15 and S_{1} = 0.17 > 0.04 , the building cannot be automatically assigned to SDC A.

2. Determine the Occupancy Category from IBC Table 1604.5.

For a health care facility, the Occupancy Category is III.

3. Since S_{1} < 0.75 , the building is not assigned to SDC E or F.

4. Check if all four conditions of 11.6 are satisfied.

• Check if the approximate period T_a is less than 0.8 T_S .

From Eq. 12.8-7 for a concrete moment-resisting frame:

T_{a}=C_{t}h_{n}^{x}=0.016(25.0)^{0.9}=0.29 sec

where C_t and x are given in Table 12.8-2.

T_{S}=S_{D1}/\,S_{D S}=0.19/0.45=0.42 sec

0.29 sec < 0.8× 0.42 = 0.34 sec

• The fundamental period used to calculate the design drift is taken as 0.29 sec, which is less than T_S = 0.42 sec.

• Equation 12.8-2 will be used to determine the seismic response coefficient C_s .

• Since the roof and the floors are cast-in-place concrete, the diaphragms are considered to be rigid.

Since all four conditions are satisfied, the SDC can be determined by Table 11.6-1 alone (11.6).

5. Determine the SDC from Table 11.6-1.

From Table 11.6-1, with 0.33 < S_{DS} = 0.45 < 0.50 and Occupancy Category III, the SDC is C.^{17}

• Step 3: Determine the design seismic forces of the equivalent lateral force procedure from Flowchart 6.8.

1. The design accelerations and the SDC have been determined in Steps 1 and 2 above.

2. Determine the response modification coefficient R from Table 12.2-1.

The moment-resisting frames in this building must be intermediate reinforced concrete moment frames, since the building is assigned to SDC C (system C6 in Table 12.2-1). In this case, R = 5. There is no height limit for this system in SDC C.

3. Determine the importance factor I from Table 11.5-1.

For Occupancy Category III, I = 1.25.

4. Determine the period of the structure T.

It was determined in Step 2, item 4 above that the approximate period of the structure T_a, which is permitted to be used in the equivalent lateral force procedure, is equal to 0.29 sec.

5. Determine long-period transition period T_L from Figure 22-15.

For St. Louis, MO, T_L = 12 sec > T_a = 0.29 sec

6. Determine seismic response coefficient C_s .

The value of C_s must be determined by Eq. 12.8-2 (see Step 2, item 4):

7. Determine effective seismic weight W in accordance with 12.7.2.

The effective weights at each floor level are given in Table 6.17. The total weight W is the summation of the effective dead loads at each level.

8. Determine seismic base shear V.

Seismic base shear is determined by Eq. 12.8-1:

9. Determine exponent related to structure period k.

Since T < 0.5 sec, k = 1.0.

10. Determine lateral seismic force F_x at each level x.

F_x is determined by Eqs. 12.8-11 and 12.8-12. A summary of the lateral forces F_x and the story shears V_x is given in Table 6.17.

• Step 4: Determine the diaphragm design seismic forces using Eq. 12.10-1.

Diaphragm design force F_{px}=\frac{\sum_{i=x}^{n}F_{i}}{\sum_{i=x}^{n}w_{i}}w_{px}

where w_i = weight tributary to level i and w_{px} = weight tributary to the diaphragm at level x .

Minimum F_{p x}=0.2S_{D S}I w_{p x}=0.2\times0.45\times1.25\times w_{p x}=0.1125w_{px}

Minimum F_{p x}=0.4S_{D S}I w_{p x}=0.2250w_{px}

Assuming that the exterior walls are primarily glass, which weigh significantly less than the diaphragm weight at each level, the weight that is tributary to each diaphragm is identical to the weight of the structure at that level (i.e., w_{px} = w_{x}).

A summary of the diaphragm forces is given in Table 6.18.

^{17} If Table 11.6-2 were also used, the SDC would also be C.