1997 UBC Earthquake Design
Example 2
- Continue with the previous example, Example
1, and examine the following:
- Shear wall design force.
- Lateral forces normal to the wall.
- Shear wall overturning.
- Shear wall drift.
- Shear wall forces
- Consider seismic loading in the transverse direction.
- From example 1, step 5, the diaphragm force (at ASD level)
supported by one of the 50' endwall was determined to be 17000
lbs.
- Recall, however, that this was developed from Fpx
based upon a R = 4.0, and not Fx with a R = 4.5.
Since the shear wall is a vertical element of the LFRS, it is
permissible to reduce this reaction force by 89%; accounting
for this difference between 4.0 and 4.5.

- Calculate the top half of the wall's inertia force assuming
no openings in the wall:
- Total unit shear, applied at midheight:
v` = (15111 + 5582) / 50 = 414 lb/ft.
- Since this building is located in seismic zone 4, increase
v by 50 %:
v` = 414 (1.5) = 621 lb/ft
- Similarly, for seismic movement in the longitudinal direction:

- The above unit shear shear wall forces were developed for the
50' end wall that had no openings. What happens when the wall has
openings, like th 15' on on the other end wall?
- The lateral load must be carried by the effective wall segments
known as shear panels if wood walls or piers if concrete or
masonry walls.
- Different procedures are used to distribute the horizontal
diaphragm reaction to the effective wall segments, depending
if wood or masonry walls.
- In shear panels, the unit shear is the same in every panel
due to the assumption that the panel force is inversely
proportional to the panel length.
- In piers, the pier force in inversely proportional to
pier rigidity. The unit shear in wider pier will be greater
than the unit shear in a narrow pier.
- In this example, however, the endwall piers are the same with
the same rigidities. The unit shear in each 17.5' pier will
be:
- Lateral forces normal to walls:
- It is assumed that the walls perpendicular to the ground motion
span vertically between the roof diaphragm and foundation.

-
Force on main wall using '97 UBC Eqn 32-2:

Also note: Fp = .587Wp is greater than .7(.44)
(1.0)Wp = .308Wp and less than 4.0(.44)(1.0)Wp
= 1.76 Wp according to '97 UBC 1632.2
-
Force on cantilvered parapet:
-
Anchorage design force at diaphragm:
- Recall that if in seismic zones 3 or 4, ap for the main wall
is increased to 1.5. This increases the main wall force by 50%
for use in anchorage force calculation.

- This exceeds the code minimum requirement of 420/1.4 = 300
lb/ft (at ASD level).
- Therefore, provide an anchorage system capable of resisting
520.8 lb/ft.
- Overturning check on shearwalls:
- A lower factor of safety is permitted in seismic design vs. that
used in wind because of the transient and reversing nature of the
seismic forces.
- The factor of safety is accounted for by reducing the resisting
moment to dead load by .9.
- Consider overturning of the short walls due to seismic forces
acting in the transverse building direction.

- Story drift, D:
- According to '97 UBC 1630.10.2, the maximum in elastic displacement
DM, should not exceed .025h
for structures with T < .7 seconds.
- A serviceability consideration.
- DM is a strength level
inelastic displacement due to a design earthquake.
- DM = .7RDS
where R is from Tabe 16-N and DS
is static story drift due to strength-level forces.
- Consider the following calculation for the 50' shear wall:

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