Example 1
Develop the applicable seismic forces for a one-story, box-type industrial building located in Southern California. Assume partially grouted CMU walls weighing 61 lb/ft2, a roof dead load of 9 psf, and the building is not located near (further than 9.3 miles) a seismic source. No geotechnical investigation was completed.
- Base shear coefficient, V.
- The base shear equation(s) are quite cumbersome to use, unless
on knows beforehand which equation governs.
- Recall that middle equation is for buildings medium to long fundamental T's. The left-hand equations are lower bound values. The right-hand equation is for short (stiff) T buildings.
- You can determine if its the right-hand equation quickly by
comparing the building's T to Ts:
- TS is a limiting period of vibration that is used to differentiate between stiff and flexible buildings.
- The seismically-induced forces in stiff buildings are
related to the bedrock acceleration. The forces in flexible
buildings are related more to bedrock velocity.
- Calculate T and Ts:
Zone = 4 (figure 16-2)
Z = .4 (Table 16-I)
Soil profile type = SD
NV = 1.0 (Table 16-T)
CV = .64 (Table 16-R)
Na = 1.0 (Table 16-S)
Ca = .44 (Table 16-Q)
TS = .64/(2.5(.44)) = .582 sec
Therefore, use short T base shear equation.
- Calculate V:
Zone = 4 (Figure 16-2)
Z = .4 (Table 16-I)
Na = 1.0 (Table 16-S)
Ca = .44 (Table 16-Q)
I = 1.0(Table 16-K)
R = 4.5 (Table 16-N)
- The base shear equation(s) are quite cumbersome to use, unless
on knows beforehand which equation governs.
- Story force for vertical elements, F1
- Recall that for a one story structurewhere T £
.7, the vertical story force and the base shear are equal.
F1 = V = .244W
- Recall that for a one story structurewhere T £
.7, the vertical story force and the base shear are equal.
- Story force for horizontal elements, Fp1
- In a one story masonry building with flexible diaphragms, recall
that:
Fp1 = 1.125V
Fp1 = 1.125(.244)W = .275W
- In a one story masonry building with flexible diaphragms, recall
that:
- Calculate the diaphragm design forces in the transverse loading direction
due to seismic.
- To do this, need to first determine W, the weight of the structure
that is supported by the diaphragm.
- Consider this weight (and resulting force) on a per foot basis.
A 1' strip of dead load = the mass that causes the inertial forces on a per foot basis in the diaphragm.
- Similarly in the longitudinal direction:
- It is customary to ignore wall openings in these diaphragm
force calculations, as the added accuracy is generally not warranted.
- This uniform diaphragm force is at the strength level, and has
not yet been adjusted by r and 1.4.
- In other words, Fp in the transverse direction
could be generically labeled Eh from the equation:
E = r Eh + 0
- Recall that r = 1.0 for this type of building when
- In other words, Fp in the transverse direction
could be generically labeled Eh from the equation:
- On the right-hand side of our building, a window occurs; lessening
the length of wall to resist shear to 2 - 17.5 ft segments.
- Adjusting to ASD:
transverse:
Fp = 476/1.4 = 340 lb/ft
longitudinal:
Fp = 600/1.4 = 429 lb/ft
- Adjusting to ASD:
- Comparing wind vs. seismic forces, it is apparent that seismic will govern the lateral design of the diaphragm in both directions.
- Transverse: 340 plf > 93 plf.
- Longitudinal: 429 plf > 93 plf.
- To do this, need to first determine W, the weight of the structure
that is supported by the diaphragm.
- Calculate the unit shear forces in the flexible diaphragm.
- Flexible diaphragms are like deep, thin, uniformly-loaded beams that are simply supported by the shear walls.
Consider the transverse direction:
Longitudinal force direction:
- The transverse direction is often the more critical direction for rectangular buildings due to the longer diaphragm span and the shorter shear walls.
- In summary: the diaphragm spans between the supporting shear walls, transferring the inertial affect of the perpendicular walls and itself to those walls located parallel to force direction.
