Building design example of a special concentrically braced frame This problem requires the determination of the seismic loads based on the location of the building and design of a steel special concentric-braced frame system based on AISC seismic provisions to resist seismic lateral loading. In

Question

Building design example of a special concentrically braced frame
This problem requires the determination of the seismic loads based on the location of the building and design of a steel special concentric-braced frame system based on AISC seismic provisions to resist seismic lateral loading. In order to put it all together, the following information must be obtained for analysis. The office building shown in Figure 6.2 is in SDCD (Figure 6.3).
• [latex]S_{DS} = 1.0, S_{D1} =0.42[/latex]
• The floors are assumed to be rigid diaphragms
• The redundancy ρ = 1.0: two bays of seismic force resisting elements are provided on each side of the building and at the perimeter. Therefore, a ρ = 1.0 requirement is met
• For a SCBF system, the response modification factor/coefficient R = 6 and over-strength factor [latex]Ω_0 = 2.0[/latex]

Accepted Answer

Typical gravity loading is as follows:
• [latex]D_{ ext{floor}}[/latex] = 85 psf
• [latex]D_{ ext{roof}}[/latex] = 70 psf
• Curtain wall [latex]=200^{\# / ft}[/latex] along the building perimeter at levels 2–4, and [latex]120^{\#/ft}[/latex] at the roof
• [latex]L_{ ext{floor}}[/latex] = 80 psf
• S = 20 psf
• [latex]T_L[/latex] = 8 s
The following steps are required:
1. Determine the period of structure
2. Determine the seismic response coefficient
3. Determine the seismic weight of structure
4. Determine the seismic base shear
5. Vertical distribution of seismic forces to each level
6. Determine the seismic design story shear
7. Horizontal distribution to vertical elements of seismic-force-resisting system (rigid diaphragm analysis)
8. Design SCBF Brace BR-1
9. Design SCBF Column CL-1
10. Design SCBF Beam BM-1
Step 1: Determine the period of the structure T.
[latex]T_a=C_t h_h^x [/latex] (2.10) (ASCE 7 Equation 12.8–7)
From table ASCE 7 T 12.8–2
Structure type: SCBF
[latex]C_t[/latex] = 0.02
x = 0.75
The height of the building, [latex]h_h[/latex] = 51 ft
Then the approximate fundamental period:
[latex]T_a=(0.02)(51)^{0.75}=0.382 [/latex] s
Step 2: Determine the seismic response coefficient, [latex]C_s[/latex].
[latex]C_s=\frac{S_{DS}}{(R/I_e)}[/latex] (2.12) (ASCE 7 Equation 12.8–8)
An office building is assigned to risk category II, which corresponds to a seismic
importance factor [latex]I_e = 1.0[/latex] , from ASCE T 1.5-2.
Then,
[latex]C_s=\frac{S_{DS}}{(R/I_e)}=\frac{1.0}{(6/1.0)}=0.167[/latex]
and [latex]C_s[/latex] does not need to be greater than
[latex]C_s=\frac{S_{D1}}{T(R/I_e)}=\frac{0.42}{0.382(6/1.0)}=0.183[/latex] for T[latex]\leq T_L[/latex][latex]C_s=\frac{S_{D1}T_L}{T^2(R/I_e)}=\frac{0.42(8)}{(0.382)^2 (6/1.0)}=0.436[/latex] for T[latex]\gt T_L[/latex]
However, [latex]C_s[/latex] should not be less than
[latex]C_s=0.044S_{DS}I_e[/latex] (2.13) (ASCE 7 Equation 12.8–5)
= 0.044 (1.0) (1.0) = 0.044
Therefore, [latex]C_s[/latex] = 0.167
Step 3: Determine the seismic weight as tabulated in Table 6.1.

Table 6.1 Tabulation of seismic weight for each oor of the building
Level	Area (Ft²)	DL (ksf )	Curtain wall weight (k)	Total [latex]W_i[/latex] (k)
Roof	96 × 120 = 11.520	0.07	0.120 × 432′ = 51.84	858.24
4th	11,520	0.085	0.200 × 432 = 86.4	1065.6
3rd	11,520	0.85	86.4	1065.5
2nd	11,520	0.85	86.4	1065.6
				w = 4055.0

Step 4: Determine the seismic base shear.
[latex]V = C_s W[/latex]

[latex]V= 0.167 (4055)=677^k[/latex]
Step 5: Determine the vertical distribution of seismic force to each level.
[latex]F_x=C_{vx}v[/latex] (3.23) (ASCE 7 Equation 12.8–11)
[latex]C_{vx}=\frac{w_x h_x^k}{\sum_{i=1}^{n} {w_i h_i^k}}[/latex] (3.24) (ASCE 7 Equation 12.8–12)
For T = 0.382s < 0.5s, k = 1.0
The vertical distribution of seismic forces is tabulated in Table 6.2 (Figure 6.4).

Table 6.2 Tabulation of seismic forces for each oor of the building
Level x	[latex]H_x[/latex] (Ft)	[latex]W_x[/latex](kips)	[latex]w_x h_x[/latex] ( k −ft)	[latex]C_{vx}=\frac{w_x h_x}{\sum_{i=1}^{n} {w_i h_i}}[/latex]	[latex]F_x=C_{vx}V[/latex] (kips) [latex]V=677^k[/latex]
Roof	51	858.24	43,770.24	0.3365	227.81
4th	39	1065.6	41,535.00	0.3194	216.23
3rd	27	1065.6	28,771.2	0.2212	149.75
2nd	15	1065.6	15,984.0	0.1229	83.2
			∑130,060.44	1.000	677.0

Step 6: Determine the seismic design story shear, see Figure 6.5.
Step 7: Determine the horizontal distribution of seismic forces to the vertical elements of seismic-force-resisting system (rigid diaphragm analysis).
1. The center of mass and center of rigidity are coincidental for this building’s structural systems’ configuration.
2. For simplicity, ignore the effects from torsional forces due to accidental eccentricity.
3. Assume that seismic forces at each level are horizontally distributed evenly to the braced frames on grid lines ① and ⑤.
Then, the story force at each braced frame is equal to the story force divided by the number of frames (Figure 6.6; Table 6.3).

Table 6.3 Story forces at each braced frame at each level
Level	[latex]F_x[/latex] (k)
Roof	[latex]F_{xR}=227.18/4=57.0[/latex]
Fourth	[latex]F_{x4}=216.23/4=54.0[/latex]
Third	[latex]F_{x3}=149.75/4=37.4[/latex]
Second	[latex]F_{x2}=83.2/4=20.8[/latex]

Question 6.1: Building design example of a special concentrically braced f...