Question 6.12: Determine the seismic base shear for the nonbuilding illustr......
Determine the seismic base shear for the nonbuilding illustrated in Figure 6.20 using (1) 2L4x4x1/2 braces and (2) 2L4x4x1/4 braces, given the design data below.
| DESIGN DATA | |
| Location: | Phoenix, AZ (Latitude: 33.42°, Longitude: -112.05°) |
| Soil classification: | Site Class D |
| Structural system: | Ordinary steel concentrically braced frame |
Determine the seismic base shear from Flowchart 6.11.
This nonbuilding structure is similar to buildings and the appropriate design requirements from Chapter 15 are used to determine the seismic base shear.
1. Determine S_{DS} , S_{D1} and the SDC from Flowchart 6.4.
Using the USGS Ground Motion Parameter Calculator, S_S = 0.18 and S_1 = 0.06 .
Using Tables 11.4-1 and 11.4-2, the soil-modified accelerations are S_{MS} = 0.28 and S_{M1} = 0.15 .
Design accelerations: S_{D S}={\frac{2}{3}}\times0.28=0.19 and S_{D1}={\frac{2}{3}}\times0.15=0.10
From IBC Table 1604.5, the Occupancy Category is I, assuming that the contents of the storage bin are not hazardous and that the structure represents a low hazard to human life in the event of failure.
From Table 11.6-1, for 0.167 < S_{DS} < 0.33 , the SDC is B.
From Table 11.6-2, for 0.067 < S_{D1} < 0.133 , the SDC is B.
Therefore, the SDC is B for this nonbuilding structure.
2. Determine the importance factor I in accordance with 15.4.1.1.
Based on Occupancy Category I, the importance factor I is equal to 1.0 from Table 11.5-1.
3. Determine the period T in accordance with 15.4.4.^{22}
In lieu of a more rigorous analysis, Eq. 15.4-6 is used to determine the period T:
T=2\pi\sqrt{\frac{\sum\limits_{i=1}^{n}w_{i}\delta_{i}^{2}}{g\sum\limits_{i=1}^{n}f_{i}\delta_{i}}}where \delta_i are the elastic deflections due to the forces f_i, which represent any lateral force distribution in accordance with the principles of structural mechanics.
For this one-story nonbuilding structure, this equation reduces to
T=2\pi{\sqrt{\frac{w}{g k}}}where k is the lateral stiffness of the structure.
The stiffness can be obtained by applying a unit horizontal load to the top of the frame. This load does not produce any forces in the columns.
Assuming that the elastic shortening of the beams is negligible, only the braces in a given direction contribute to the stiffness of the frame.
From statics, the force in one of the four braces due to a horizontal load of 1 applied to the top of the frame is equal to 0.5592. The horizontal deflection δ due to this unit load can be obtained from the following equation from the virtual work method:
\delta=\sum u^{2}L/\,A Ewhere u = force in a brace due to the virtual (unit) load = 0.5592
\qquad \,\,\, L = length of a brace = {\sqrt{6^{2}+12^{2}}}=13.4 ft = 161 in.
\qquad \,\,\, A = area of a 2L4x4x1/2 brace = 7.49 sq in.
\qquad \,\,\, E = modulus of elasticity = 29,000 ksi
Thus,
\delta=\frac{4\times0.5592^{2}\times161}{7.49\times29,000}=0.0009 in.
The stiffness k={\frac{1}{\delta}}=1,079 kips/in.
Therefore, the period T is
T=2\pi\sqrt{\frac{30}{386\times1,079}}=0.05 sec^{23}
4. Determine the base shear V.
Since the period is less than 0.06 sec, use Eq. 15.4-5 to determine V:
V=0.30\,S_{D S}WI=0.30\times0.19\times30\times1.0=1.7 kips
The calculations are similar to those in Part 1, except the stiffness and the period of the structure are different due to the use of lighter braces.
\delta=\frac{4\times0.5592^{2}\times161}{3.87\times29,000}=0.0018 in.
Stiffness k={\frac{1}{\delta}}=556 kips/in.
T=2\pi{\sqrt{\frac{30}{386\times556}}}=0.07 sec > 0.06 sec
Therefore, the base shear V can be determined by the equivalent lateral force procedure (15.1.3).
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{0.19}{1.5/1.0}}=0.13where the seismic response coefficient R = 1.5 from Table 15.4-1 for an ordinary steel concentrically braced frame with unlimited height, which is permitted to be designed by AISC 360, Specification for
Structural Steel Buildings (i.e., without any special seismic detailing).
Also, C_s must not be less than the larger of 0.044S_{DS} I = 0.008 and 0.01 (governs) (Eq. 12.8-5).
Thus, the value of C_s from Eq. 12.8-2 governs.
V=C_{s}W=0.13\times30=3.9 kips
^{22} The approximate fundamental period equations in 12.8.2.1 are not permitted to be used to determine the period of a nonbuilding structure (15.4.4).
^{23} The weight of the steel framing is negligible.