Question 6.10: For the one-story warehouse illustrated in Figure 6.12, dete......

Part 1: Determine design seismic forces on the diaphragm

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

Using the USGS Ground Motion Parameter Calculator, S_S = 1.51 and S_1 = 0.76 .

Using Tables 11.4-1 and 11.4-2, the soil-modified accelerations are S_{MS} = 1.51 and S_{M1} = 1.14 .

Design accelerations: S_{D S}={\frac{2}{3}}\times1.51=1.01 and S_{D1}={\frac{2}{3}}\times1.14=0.76

• Step 2: Determine the SDC from Flowchart 6.4.

From IBC Table 1604.5, the Occupancy Category is II.

Since S_1 > 0.75, the SDC is E for this building (11.6).

• Step 3: Use Flowchart 6.8 to determine the seismic base shear using the equivalent lateral force procedure.

1. Check if equivalent lateral force procedure can be used (see Flowchart 6.6).

a. The building has a Type 2 re-entrant corner irregularity, since 37.0 ft >
0.15×185.0 = 27.8 ft and 192.0 ft > 0.15× 256.0 = 38.4 ft (see Table 12.3-1 and Table 6.2).

b. Determine the period in the N-S direction by Eqs. 12.8-9 and 12.8-10 for concrete shear wall structures:

T_{a}=\frac{0.0019}{\sqrt{C_{W}}}\,h_{n} \,

C_{W}={\frac{100}{A_{B}}}\sum_{i=1}^{X}\left\lgroup{\frac{h_{n}}{h_{i}}}\right\rgroup^{2}\frac{A_i}{\left[1+0.83\left\lgroup\frac{h_{i}}{D_{i}}\right\rgroup^{2}\right]}

where

A_B= area of base of structure = 40,256 sq ft

h_n = average height of building = 22.67 ft

h_i = average height of shear wall i = 22.67 ft

A_i = area of shear wall i : A_1={\frac{7}{12}}\times185=107.9 sq ft, A_{2}={\frac{7}{12}}\times37 = 21.6 sq ft, A_{3}={\frac{7}{12}}\times148=86.3 sq ft

D_i = length of shear wall i: D_1 = 185 ft, D_2 = 37 ft, D_3 = 148 ft

Thus,

C_{W}={\frac{100}{40,256}}\Bigg\{\frac{107.9}{\left[1+0.83\left\lgroup\frac{22.67}{185}\right\rgroup^{2}\right]}+\frac{21.6}{\left[1+0.83\left\lgroup\frac{22.67}{37}\right\rgroup^{2}\right]}+\frac{86.3}{\left[1+0.83\left\lgroup\frac{22.67}{148}\right\rgroup^{2}\right]}\Bigg\} \,

={\frac{100}{40,256}}(106.6+16.5+84.7)=0.52 \,

T_{a}=\frac{0.0019}{\sqrt{0.52}}\times22.67=0.06 sec

c. Determine the period in the E-W direction.

C_{W}={\frac{100}{40,256}}\Bigg\{\frac{37.3}{\left[1+0.83\left\lgroup\frac{22.67}{64}\right\rgroup^{2}\right]}+\frac{112.0}{\left[1+0.83\left\lgroup\frac{22.67}{192}\right\rgroup^{2}\right]}+\frac{149.3}{\left[1+0.83\left\lgroup\frac{22.67}{256}\right\rgroup^{2}\right]}\Bigg\} \,

={\frac{100}{40,256}}(33.8+110.7+148.3)=0.73 \,

T_{a}=\frac{0.0019}{\sqrt{0.73}}\times22.67=0.05 sec

3.5\,T_{s}=3.5\times0.76/1.01=2.63 sec > period in both directions

In accordance with Table 12.6-1, the equivalent lateral force procedure is permitted to be used for this irregular structure with a Type 2 re-entrant corner irregularity and with T < 3.5T_s .

2. Use Flowchart 6.8 to determine the lateral seismic forces in the N-S and E-W directions.

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

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

For a building frame system with intermediate precast concrete shear walls (system B9), R = 5. Note that the average building height of 22.67 ft is less than the 45 ft height limit for this system assigned to SDC E (see footnote k in Table 12.2-1).

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

For Occupancy Category II, I = 1.0.

d. Determine the period of the structure T.

It was determined above that T_a = 0.06 sec in the N-S direction and T_a = 0.05 sec in the E-W direction.

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

For San Francisco, CA, T_L = 12 sec > T_a in both directions.

f. Determine seismic response coefficient C_s .

The value of C_s from Eq. 12.8-2 is:

C_{s}={\frac{S_{D S}}{R/I}}={\frac{1.01}{5/1.0}}=0.20

Also, C_s must not be less than the larger of 0.044S_{DS} I = 0.013 and 0.044 (Eq. 12.8-5). or the value obtained by Eq. 12.8-6 (governs), since S_1 > 0.6 :

C_{s}={\frac{0.5\,S_{1}}{\left\lgroup{\frac{R}{I}}\right\rgroup}}={\frac{0.5\times0.76}{\left\lgroup{\frac{5}{1}}\right\rgroup}}=0.08

Thus, the value of C_s from Eq. 12.8-2 governs.

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

The effective weight W is equal to the weight of the roof framing plus the weight of the walls tributary to the roof: ^{19}

Weight of roof framing = 0.015× 40,256 = 604 kips

Weight of walls =\,\frac{7}{12}\times0.15\times\frac{22.67}{2}\times\left[2(256+185)\right]=875 kips

W = 604 + 875 = 1,479 kips

h. Determine seismic base shear V.

Seismic base shear is determined by Eq. 12.8-1 and is the same in both the NS and E-W directions, since the governing C_s is independent of the period:

V = C_sW = 0.20×1,479 = 296 kips

3. Determine the design seismic forces in the diaphragm in both directions by Eq. 12.10-1.

Diaphragm design force F_{px}=\frac{\sum^{n}_{i=x}\,F_i}{\sum^{n}_{i=x}\,w_i}w_{px}

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

Since this is a one-story building, Eq. 12.10-1 reduces to F_{px} = 0.20w_{px}

Minimum F_{p x}=0.2S_{D S}I w_{p x}=0.2\times1.01\times1.0\times w_{p x}=0.20w_{p x}

Maximum F_{p x}=0.4S_{D S}I w_{p x}=0.40w_{p x}

The wood sheathing is permitted to be idealized as a flexible diaphragm in accordance with 12.3.1.1. Seismic forces are computed from the tributary weight of the roof and the walls oriented perpendicular to the direction of analysis.^{20}

\underline{\mathrm{N-S \,\,direction}}

Uniform diaphragm loads w_{N1} and w_{N2} are computed as follows (see Figure 6.13):

w_{{N1}}=0.20\times15\times185+0.20\times87.5\times2\times{\frac{22.67}{2}}=952 plf

w_{N2}=0.20\times15\times148+0.20\times87.5\times2\times{\frac{22.67}{2}}=841 plf

Also shown in Figure 6.13 is the shear diagram for the diaphragm.

\underline{\mathrm{E-W \,\,direction}}

Uniform diaphragm loads w_{E1} and w_{E2} are computed as follows (see Figure 6.14):

W_{E1}=0.20\times15\times64+0.20\times87.5\times2\times{\frac{22.67}{2}}=589 plf

W_{E2}=0.20\times15\times256+0.20\times87.5\times2\times{\frac{22.67}{2}}=1,165 plf

Also shown in Figure 6.14 is the shear diagram for the diaphragm.

4. Determine connection forces between the diaphragm and the shear walls.

Since the building has a Type 2 horizontal irregularity, the diaphragm connection design forces must be increased by 25 percent in accordance with 12.3.3.4. Thus, F_{px} = 1.25× 0.20w_{px} = 0.25w_{px} . This force increase applies to the row of diaphragm nailing that transfers the above diaphragm shears directly to the shear walls (and to the collectors) and to the bolts between the ledger beams and the shear walls.

Part\,\, 2: \,\,Determine \,\,design \,\,seismic\,\, forces \,\,on\,\, the\,\, steel collector \,\,beam \,\,in \,\,the\,\, N-S \,\,direction

From the diaphragm shear diagram in Figure 6.13, the maximum collector load is equal to 30.5(148/185) + 80.7 = 105.1 kips tension or compression.

The uniform axial load can be approximated by dividing the maximum load by the length of the collector: 105,100/148 = 710 plf.

The uniform axial load can be used to determine the axial force in any of the beams at any point along their length. For example, at the midspan of the northernmost collector beam, the axial force is equal to 710 × (148 − 37 / 2) /1,000 = 92 kips tension or compression. In accordance with 12.3.3.4, this force must be increased by 25 percent unless the collector is designed for the load combinations with overstrength factor of 12.4.3.2 (see 12.10.2.1).

The collector beams are subsequently designed for the combined effects of gravity and axial loads due to the design seismic forces.

Part \,\,3: \,\,Determine\,\, out-of-plane\,\, design \,\,seismic\,\, forces \,\,on\,\, the precast\,\, concrete\,\, wall\,\, panels
1. Solid wall panels

According to 12.11.1, structural walls shall be designed for a force normal to the surface equal to 0.4S_{DS}I times the weight of the wall. The minimum normal force is 10 percent of the weight of the wall.

For a solid precast concrete wall panel:

Weight W_p = (7 /12) × 0.15× 22.67 = 2.0 kips/ft

0.1× 2.0 = 0.2 kips/ft

0.4×1.01×1.0 × 2.0 = 0.8 kips/ft (governs)

This force governs over the three minimum anchorage forces specified in 12.11.2 as well.

Distributed load = 0.8/22.67 = 0.04 kips/ft/ft width of wall

This uniformly distributed load is applied normal to the wall in either direction.

Anchorage of the precast walls to the flexible diaphragms must develop the out-ofplane force given by Eq. 12.11-1:

F_p = 0.8S_{DS} IW_p = 1.6 kips/ft

Note that the 25 percent increase in the design force for diaphragm connections is not applied to out-of-plane wall anchorage force to the diaphragm (12.3.3.4).

2. Wall panels with openings

A typical wall panel on the east or west faces of the building is shown in Figure 6.15.

In lieu of a more rigorous analysis, the pier width between the two openings is commonly defined as a design strip. The total weight used in determining the out-ofplane design seismic force is taken as the weight of the design strip plus the weight of the wall tributary to the design strip above each adjacent opening (see Figure 6.16):

W_{p{1}}=\frac{7}{12}\times150\times4=350 plf

W_{p{2}}=\frac{7}{12}\times150\times6=525 plf

W_{p{3}}=\frac{7}{12}\times150\times1.5=131 plf

The out-of-plane seismic forces are determined by 12.11.1:

F_{P{1}}=0.4\,S_{D S}I W_{P{1}}=0.4\times1.01\times1.0\times350=141.4 plf from 0 ft to 20 ft

F_{P2}=0.4S_{D S}I W_{P1}=0.4\times1.01\times1.0\times525=212.1 plf from 14 ft to 20 ft

F_{P3}=0.4S_{D S}I W_{P1}=0.4\times1.01\times1.0\times131=52.9 plf from 7 ft to 20 ft

The wall is designed for the combination of axial load from the gravity forces and bending and shear from the out-of-plane seismic forces.

^{19} The weight of the walls is conservatively calculated assuming that there are no openings in the walls.
^{20} Walls parallel to the direction of the seismic forces are typically not considered in the tributary weight, since these walls do not obtain support from the diaphragm in the direction of the seismic force.