Question 6.4: For the seven-story office building in Example 6.3, determin......

Use Flowchart 6.6 to determine the permitted analytical procedure.

1. Determine the SDC from Flowchart 6.4.

It was determined in Step 2 of Example 6.3 that the SDC is D.

2. Determine S_{DS} and S_{D1} from Flowchart 6.2.

The design accelerations were determined in Step 1, item 4 of Example 6.3:
S_{DS} = 0.90 and S_{D1} = 0.41.

3. Determine T_S .

It was determined in Step 2, item 4 of Example 6.3 that T_{S}=S_{D1}/\,S_{D S}=0.46 sec.

4. Determine the fundamental period of the building T.

The periods T were determined in Step 2, item 4 of Example 6.3 as 0.61 sec in the N-S direction and 1.1 sec in the E-W direction.

5. Check if T\lt 3.5T_{S}.

3.5T_S = 3.5× 0.46 = 1.6 sec, which is greater than the periods in both directions.

6. Determine if the structure is regular or not.

a. Determine if the structure has any horizontal structural irregularities in accordance with Table 12.3-1.

i. Torsional irregularity

In accordance with Table 12.3-1, Type 1a torsional irregularity and Type 1b extreme torsional irregularity for rigid or semirigid diaphragms exist where the ratio of the maximum story drift at one end of a structure to the average story drifts at two ends of the structure exceeds 1.2 and 1.4, respectively. The story drifts are to be determined using code-prescribed forces, including accidental torsion. In this example, the floor and roof are metal deck with concrete, which is considered to be a rigid diaphragm (12.3.1.2).

At this point in the analysis, it is obviously not evident which method is
required to be used to determine the prescribed seismic forces. In lieu of using a more complicated higher order analysis, the equivalent lateral force procedure may be used to determine the lateral seismic forces. These forces are applied to the building and the subsequent analysis yields the story drifts Δ, which are used in determining whether Type 1a or 1b torsional irregularity exists. The results from the equivalent lateral force procedure will be needed if it is subsequently determined that a modal analysis is required (see 12.9.4).

Use Flowchart 6.8 to determine the lateral seismic forces from the equivalent lateral force procedure:

a) The design accelerations and the SDC have been determined in Example 6.3.

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

In the N-S direction, special steel concentrically braced frames are required, since the building is assigned to SDC D (system B3 in Table 12.2-1). In this case, R = 6. Note that height of the building, which is 96 ft, is less than the limiting height of 160 ft for this type of system in SDC D.^{11}

In the E-W direction, special steel moment frames are required (system C1 in Table 12.2-1). In this case, R = 8, and there is no height limit.

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 in Step 2, item 4 of Example 6.3 that the approximate period of the structure T_a, which is permitted to be used in the equivalent lateral force procedure, is 0.61 sec in the N-S direction and 1.1 sec in the E-W direction.

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

For Memphis, TN, T_L = 12 sec, which is greater than the periods in both directions.

f) Determine seismic response coefficients C_s in both directions.

• N-S direction:

The seismic response coefficient C_s is determined by Eq. 12.8-3:

C_{S}={\frac{S_{D1}}{T(R/I)}}={\frac{0.41}{0.61(6/1.0)}}=0.11

The value of C_s need not exceed that from Eq. 12.8-2:

C_{S}={\frac{S_{D S}}{R/I}}={\frac{0.90}{6/1.0}}=0.15

Also, C_s must not be less than the larger of 0.044S_{DS} I = 0.04 (governs) and 0.01 (Eq. 12.8-5).

Thus, the value of C_s from Eq. 12.8-3 governs in the N-S direction.

• E-W direction:

C_{s}={\frac{S_{D1}}{T(R/I)}}={\frac{0.41}{1.1(8/1.0)}}=0.05

The value of C_s need not exceed that from Eq. 12.8-2:

C_{S}={\frac{S_{D S}}{R/I}}={\frac{0.90}{8/1.0}}=0.11

Also, C_s must not be less than the larger of 0.044S_{DS} I = 0.04 (governs) and 0.01 (Eq. 12.8-5).

Thus, the value of C_s from Eq. 12.8-3 governs in the E-W direction.

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

The member sizes and superimposed dead loads are given in Figure 6.7 and the effective weights at each floor level are given in Tables 6.10 and 6.11. The total weight W is the summation of the effective dead loads at each level.

h) Determine seismic base shear V.

Seismic base shear is determined by Eq. 12.8-1 in both the N-S and E-W directions.

• N-S direction: V=C_{s}W=0.11\times9,960=1,096 kips

• E-W direction: V=C_{s}W=0.05\times9,960=498 kips

i) Determine exponent related to structure period k in both directions.

Since 0.5 sec < T < 2.5 sec in both directions, k is determined as follows:

• N-S direction: k = 0.75 + 0.5T = 1.06

• E-W direction: k = 0.75 + 0.5T = 1.30

j) Determine lateral seismic force F_x at each level x.

Lateral forces F_x are determined by Eqs. 12.8-11 and 12.8-12.

A summary of the lateral forces F_x and the story shears V_x are given in Table 6.10 and 6.11 for the N-S and E-W directions, respectively.

Three-dimensional analyses were performed independently in the N-S and EW directions for the seismic forces in Table 6.10 and 6.11 using a commercial computer program. In the model, rigid diaphragms were assigned at each floor level. In accordance with 12.8.4.2, the center of mass was displaced each way from its actual location a distance equal to 5 percent of the building dimension perpendicular to the applied forces to account for accidental torsion in seismic design.

A summary of the elastic displacements δ_{xe} at each end of the building in both the N-S and E-W directions due to the code-prescribed forces in Tables 6.10 and 6.11 is given in Table 6.12 at all floor levels.

According to Table 12.3-1, a torsional irregularity occurs where maximum story drift at one of the structure is greater than 1.2 times the average of the story drifts at the two ends of the structure. The average story drift Δ_{avg} and the ratio of the maximum story drift to the average story drift Δ_{max} / Δ_{avg} are also provided in Table 6.12.

For example, at the first story in the N-S direction:

Δ_{1} = 0.46 in.

Δ_{2} = 0.18 in.

\Delta_{aνg}=\frac{0.46+0.18}{2}=0.32 in.

{\frac{\Delta_{\mathrm{max}}}{\Delta_{aνg}}}={\frac{0.46}{0.32}}=1.44~\gt 1.4

Therefore, a Type 1b extreme torsional irregularity exists at the first story in the N-S direction.

According to 12.8.4.3, where torsional irregularity exists at floor level x, the accidental torsional moments M_{ta} must be increased by the torsional amplification factor A_x given by Eq. 12.8-14:

A_{x}=\left\lgroup{\frac{\delta_{\mathrm{{max}}}}{1.2\delta_{aνg}}}\right\rgroup^{2}

At the first story in the N-S direction:

A_1=\left[\frac{0.46}{1.2\left\lgroup\frac{0.46+0.18}{2}\right\rgroup}\right]^{2}=1.44\gt 1.0

Therefore, the accidental torsional moment at the first story is:^{12}

(M_{t a})_{1}=A_{1}F_{1}e

\qquad =1.44\times71\times(0.05\times180)=920 ft – kips

ii. Re-entrant corner irregularity

By inspection, this irregularity does not exist.

iii. Diaphragm discontinuity irregularity

This irregularity does not exist in this building when opening sizes for typical elevators and stairs are present.

iv. Out-of-plane offsets irregularity

In the first story, the seismic force-resisting system consists of special steel concentrically braced frames along column lines 1 and 7. Above the first floor, there is a 30-ft offset of the braced frames, which occur along column lines 2 and 6.

Therefore, a Type 4 out-of-plane offset irregularity exists.

Note that the forces from the braced frames along column lines 2 and 6 must be transferred through the structure to the braced frames along column lines 1 and 7, respectively.

v. Nonparallel systems irregularity

This discontinuity does not exist, since all of the braced frames and momentresisting frames are parallel to a major orthogonal axis of the building.

b. Determine if the structure has any vertical structural irregularities in accordance with Table 12.3-2.

i. Stiffness–Soft Story Irregularity

A soft story is defined in Table 12.3-2 based on the relative lateral stiffness of stories in a building. In general, it is not practical to determine story stiffness. Instead, this type of irregularity can be evaluated using drift ratios due to the code-prescribed lateral forces.^{13} A soft story exists when one of the following conditions are satisfied:

Soft story irregularity: 0.7\frac{\delta_{1e}}{h_{1}}\gt \frac{\delta_{2e}-\delta_{1e}}{h_{2}} or

0.8\frac{\delta_{1e}}{h_{1}}\gt \frac{1}{3}\bigg[\frac{\delta_{2e}-\delta_{1e}}{h_{2}}+\frac{\delta_{3e}-\delta_{2e}}{h_{3}}+\frac{\delta_{4e}-\delta_{3e}}{h_{4}}\bigg]

Extreme soft story irregularity: 0.6\frac{\delta_{1e}}{h_{1}}\gt \frac{\delta_{2e}-\delta_{1e}}{h_{2}} or

0.7\frac{\delta_{1e}}{h_{1}}\gt \frac{1}{3}\bigg[\frac{\delta_{2e}-\delta_{1e}}{h_{2}}+\frac{\delta_{3e}-\delta_{2e}}{h_{3}}+\frac{\delta_{4e}-\delta_{3e}}{h_{4}}\bigg]

Check for a soft story in the first story:
In the N-S direction, the displacements of the center of mass at the first, second, third and fourth floors are δ_{1e} = 0.31 in., δ_{2e} = 0.51 in., δ_{3e} = 0.69 in., and δ_{4e} = 0.91 in.

0.7\,\frac{\delta_{1e}}{h_{1}}=0.7\,\frac{0.31}{18\times12}=0.0010\lt \frac{\delta_{2e}-\delta_{1e}}{h_{2}}=\frac{0.51-0.31}{13\times12}=0.0013 \,

0.8\,\frac{\delta_{1e}}{h_{1}}=0.7\,\frac{0.31}{18\times12}=0.0011\lt \frac{1}{3}\biggl[\frac{\delta_{2e}-\delta_{1e}}{h_{2}}+\frac{\delta_{3e}-\delta_{2e}}{h_{3}}+\frac{\delta_{4e}-\delta_{3e}}{h_{4}}\biggr] \,

\qquad\qquad\quad=\,\frac{1}{3}\Bigg[\frac{0.51-0.31}{13\times12}+\frac{0.69-0.51}{13\times12}+\frac{0.91-0.69}{13\times12}\Bigg]=0.0013

In the E-W direction, the displacements of the center of mass at the first, second, third and fourth floors are δ_{1e} = 0.61 in., δ_{2e} = 1.49 in., δ_{3e} = 2.68 in., and δ_{4e} = 3.75 in.

0.7\,\frac{\delta_{1e}}{h_{1}}=0.7\,\frac{0.61}{18\times12}=0.0020\lt \frac{\delta_{2e}-\delta_{1e}}{h_{2}}=\frac{0.49-0.61}{13\times12}=0.0056 \,

0.8\frac{\delta_{1e}}{h_{1}}=0.8\frac{0.61}{18\times12}=0.0023\lt \frac{1}{3}\left[\frac{\delta_{2e}-\delta_{1e}}{h_{2}}+\frac{\delta_{3e}-\delta_{2e}}{h_{3}}+\frac{\delta_{4e}-\delta_{3e}}{h_{4}}\right] \,

\qquad\qquad\quad={\frac{1}{3}}\biggl[{\frac{1.49-0.61}{13\times12}}+{\frac{2.68-1.49}{13\times12}}+{\frac{3.75-2.68}{13\times12}}\biggr]=0.0067

Therefore, a soft story irregularity does not exist in the first story.

ii. Weight (mass) irregularity.

Check the weight ratio of the first and second stories: 2,037/1,381 = 1.48 < 1.50. Thus, this irregularity is not present.

iii. Vertical geometric irregularity.

A vertical geometric irregularity is considered to exist where the horizontal dimension of the seismic force-resisting system in any story is 1.3 times that in an adjacent story.

In this case, the setbacks at the first floor level must be investigated for the moment-resisting frames along column lines A and E:

Width of floor 1/Width of floor 2 = 180/120 = 1.5 > 1.3

Thus, a Type 3 vertical geometric irregularity exists.

iv. In-plane discontinuity in vertical lateral force-resisting element irregularity.

There are no in-plane offsets of this type, so this irregularity does not exist.

v. Discontinuity in lateral strength-weak story irregularity.

This type of irregularity exists where a story lateral strength is less than 80 percent of that in the story above. The story strength is considered to be the total strength of all seismic-resisting elements that share the story shear for the direction under consideration.

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

Determine whether a weak story exists in the first story in the E-W direction. In this case, the story strength is equal to the sum of the column shears in the moment-resisting frames in that story when the member moment capacity is developed by lateral loading. It is assumed in this example that the same column and beam sections are used in the moment-resisting frames in the first and second stories.

Assume the following nominal flexural strengths:^{14}

Columns: M_{nc} = 550 ft-kips

Beams: M_{nb} = 525 ft-kips

• First story shear strength

Corner columns A1/E1 and A7/E7 are checked for strong column-weak
beam considerations:

2M_{nc} = 2 × 550 = 1,100 ft-kips > M_{nb} = 525 ft-kips

The maximum shear force that can develop in each exterior column is
based on the moment capacity of the beam (525/2 = 263 ft-kips), since it is less than the moment capacity of the column (550 ft-kips) at the top of the column:^{15}

V_1=V_{7}={\frac{263+550}{18}}=45 kips

Interior columns A2 through A6/E2 through E6 are checked for strong column-weak beam considerations:

2M_{nc} = 2 × 550 = 1,100 ft-kips > 2M_{nb} =1,050 ft-kips

The maximum shear force that can develop in each interior column is based on the moment capacity of the beam (525 ft-kips), since it is less than the moment capacity of the column (550 ft-kips) at the top of the column:

V_{2}=V_{3}=V_{4}=V_{5}={\frac{525+550}{18}}=60 kips

Total first story strength = 2(V_{1}+V_{2}+V_{3}+V_{4}+V_{5}+V_{6}+V_{7})=780 kips

• Second story shear strength

V_1=V_{7}={\frac{263+263}{13}}=41 kips

V _2 = V_3=V_4=V_5=V_6=\frac{525+525}{13}=81   kips

Total second story strength \begin{array}{c}{{=~2(V_{1}+V_{2}+V_{3}+V_{4}+V_{5}+V_{6}+V_{7})=}}\end{array} 974 kips

780 kips > 0.80 × 974 = 779 kips

Therefore, a weak story irregularity does not exist in the first story in the E-W direction.

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

Assuming the same beam, column and brace sizes in the first and second floors, the shear strengths of these floors are essentially the same, and no weak story irregularity exists in the N-S direction.

In summary, the building is not regular and has the following irregularities: horizontal Type 1b extreme torsional irregularity, horizontal Type 4 out-of-plane offsets irregularity and vertical Type 3 vertical geometric irregularity.

7. Determine the permitted analytical procedure from Table 12.6-1.

The structure is irregular with T < 3.5T_S. If the structure had only a Type 4 out-ofplane offsets irregularity, the equivalent lateral force procedure could be used to analyze the structure. However, since a Type 1b extreme torsional irregularity and a Type 3 vertical geometric irregularity also exist, the equivalent lateral force procedure is not permitted; either a modal response spectrum analysis (12.9) or a seismic response history procedure (Chapter 16) must be utilized.^{16}

^{11} The increased building height limit of 12.2.5.4 is not considered.
^{12} It is assumed that the center of mass and center of rigidity are at the same location in this building.
^{13} Story displacements based on the code-prescribed lateral forces can be used to evaluate soft stories when the story heights are equal.
^{14} The assumed nominal flexural strengths of the columns and beams are based on preliminary member sizes and are provided for illustration purposes only.
^{15} At the bottom of the column, it is assumed that the full moment capacity of the column can be developed.
^{16} See the reference sections in Tables 12.3-1 and 12.3-2 that must be satisfied for these types of irregularities in structures assigned to SDC D. Columns B2, C2, D2, B6, C6 and D6 must be designed to resist the load combinations with overstrength factor of 12.4.3.2, since they support the discontinued braced frames along column lines 2 and 6 (12.3.3.2). Design forces shall be increased 25 percent for connections of diaphragms to vertical elements and to collectors and for connection of collectors to the vertical elements (12.3.3.4). Members that are not part of the seismic force-resisting system must satisfy the deformational compatibility requirements of 12.12.4.