2001 ASD Supplements - unprotected PDF - American Wood Council

SW-20OTHER CONSIDERATIONS5.1 Drag Struts/CollectorsAs defined in Section 2.2, the load path for a boxtypestructure is from the diaphragm into the shear wallsrunning parallel to the direction of the load (i.e., the diaphragmloads the shear walls that support it). Because thediaphragm acts like a long, deep beam, it loads each ofthe supporting shear walls evenly along the length of thewalls. The problem lies with the fact that seldom is eachshear wall solid throughout its full length. Typically a wallcontains windows and doors.The traditional model used to analyze shear walls onlyrecognizes full height wall segments as shear wall segments.This means that at locations with windows or doors,a structural element is needed to distribute the diaphragmshear over the top of the opening and into the full heightsegments adjacent to it. This element is called a drag strut.In residential construction, the double top-plates existingin most stud walls will serve as a drag strut. It maybe necessary to detail the double top plate such that nosplices occur in critical zones. Or, it may be necessary tospecify the use of a tension strap at butt joints to transferthese forces.The maximum force seen by drag struts is generallyequal to the diaphragm design shear in the direction ofthe shear wall multiplied by the distance between the shearwall segments.Drag struts are also used to tie together different partsof an irregularly shaped building.To simplify design, irregularly shaped buildings (suchas “L” or “T” shaped) are typically divided into simplerectangles. When the structure is “reassembled” after theindividual designs have been completed, drag struts areused to provide the necessary continuity between theseindividual segments to insure that the building will act asa whole.The following figures and generalized equations providemethods to calculate the drag strut forces.Figure 5.1 Shear Wall Drag StrutVLL 1L OL 2Figure 5.2 Shear Wall Special CaseDrag StrutVElevationVUnit shear above opening =L= aUnit shear below opening = =LMax. force in drag strut = Lv ovVvba1LUnit shear above opening = V LUnit shear below opening == v aVL−LMax. force in drag strut = greater ofVL LL 0 1 vaLoL1=( L−L ) ( L−L )000=v bL OElevationL 1orVL LL 0 2 vaLoL2=( L−L ) ( L−L )00APA – The Engineered Wood Association