Wind Loads - Lateral Force Example

CENE 437: The Class: Loads: Lateral Loads: Wind Loads: Lateral Force Example

Wind Loads

Example

Develop the applicable wind forces for a one-story, box-type industrial complex located in Southern California, in a suburban area with roof D = 12 psf and 16" o.c. partially grouted CMU walls.

Design pressure for primary LFRS:

For this example, the pressure coefficients are the same regardless of load direction.

P = C_e C_q q_s I_w

C_e Þ Exposure B Þ	{	.62	(0 - 15')	(UBC '94 Table 16-G)
		.67	(20')

C_qÞ	Method 1 (Table 16-H)	Þ	{	.8 windward	Assuming » flat roof and closed building.
				-.5 leeward
				-.7 roof

The resulting pressure diagram for wind loading in the transverse direction:

Diaphragm Loading:
- Showing transverse loading here.
- Consider a 1' width of wall that is simply supported at foundation and roof.
- Examine the normal wall load effect on the supporting diaphragm:
  - Since the diaphragm must support both, simultaneously, the inward and outward pressures, model these effects as a resultant pressure.
- The roof diaphragm must resist 125.4 lb/ft (or, in other words, each foot of wall produces a force of 125 lbs on the diaphragm.)
Diaphragm to shear walls for flexible diaphragms:
- Showing only the transverse loading.
- We will only work with flexible diaphragms in this class.
Diaphragm Plan View:

In this example, the shear wall at each end must resist a resultant load of 7837.5 lbs.
Upon development of the shear wall loads, examine shear wall drift and overturning.
1. Shear wall drift:
  - L/h = 66/17 = 3.88, drift dominated by shear deflection.
  - Therefore during wall design, ensure that D » 1.2 P h / G A £ .0025 h
2. Overturning (OM)
  - Since this is a shear wall building, every shear wall should be checked for OM.
  - Check all shear walls, in lieu of an overall building OM check (In other words, if the shear walls are adequate then the overall building is adequate. But, remember that overall building stability does not ensure that all shear walls are ok).
    
    RM = Resisting Moment
    RM = 87120 (33) = 2,874,960 ft-lbs
    OM = Overturning Moment
    OM = 7837.5(17) = 133,240 ft-lbs
  - If the shear wall supported a significant % of the roof framing, then a portion of roof uplift should be included in OM. Likewise, a portion of the roof dead load should be included in RM.
  - 2 / 3 RM = 1,916,640 > > OM
  - No special anchorage required for the shear wall, which is typical for masonry walls that support wood roof systems. (This is not, however, true for all wood buildings with wood shear walls. Very often, these lightweight walls will need special hold down anchors to resist the overturning potential.)
Lateral forces normal to the wall.
- In addition to acting as shear walls, these walls must be designed to resist forces acting perpendicular to them.
- These perpendicular forces do not come from the primary LFRS analysis as shown in step 1. Develop these forces by considering localized, higher loads taken on portions of the building as reflected in C_q.
- A careful reading of '97 UBC Table 16-H indicates two load cases here.
  - One for all structures and the other for enclosed and unenclosed structures.
  - For outward (negative) forces, use C_q/C_e based on mean roof heights.
  - For inward (positive) forces, use C_q/C_e based on the actual height of the element.
Repeat wind load development and distribution for wind in longitudinal direction.

Send Email to Deb Larson at Debra.Larson@nau.edu

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