The Force Transfer Around Openings (FTAO) Calculator, developed by APA – The Engineered Wood Association, helps designers add structural strength and lateral force resistance to buildings with multiple openings. The ratio of sheathing to windows in large buildings has tipped in favor of the windows, reducing the surface area available for wood structural panel sheathing that would otherwise provide the lateral load resistance needed for the building. As trends toward larger buildings with bigger and more numerous windows continue to grow, designers will need a tool they can count on to help them easily calculate the shear walls necessary to resist the lateral forces. The FTAO approach is one of three recommended by the American Wood Council (AWC) in their publication Special Design Provisions for Wind & Seismic (SDPWS) for designing wood shear walls to resist lateral forces. The other two options include the Individual Full-Height Wall Segments, a more “traditional” approach, and the Perforated Shear Walls, which is an empirical design method based on the percentage of full-height wall segments adjacent to openings. The FTAO method, also referred to as the Force Transfer Shear Walls method, allows the utilization of the full-wall geometry, including sheathed areas above and below openings. Historically, the FTAO method has been considered a last resort in the design industry, due in large part to the fact that an FTAO analysis is more comprehensive, and therefore assumed to be more time consuming than the other two design options. The application of the FTAO method also varies greatly within the industry, as the code simply states that the design must be based on rational analysis. With no clear directive in the design codes, there have been multiple design rationale techniques developed (e.g., Drag-Strut Analogy, Cantilever Beam, Diekmann), causing some dispute among design professionals as to which technique is the most precise. These discrepancies are what led APA to perform full-scale wall tests on FTAO design methodology, intended to prove that the application of FTAO can be more practical and simultaneously provide the most design flexibility. This article will discuss the benefits of 20 • RCI Interface October 2018 Figure 1 – Illustration of aspect ratio (h/b) comparison for all three shear wall methods. Photo credit: iStock.com/JohnnyH5 FTAO for wood-framed structures, illustrate the testing performed on wood structural panel-sheathed wood-framed shear walls for the enhancement of FTAO methodology, and provide a design example for a wood-framed shear wall with multiple openings and asymmetric wall piers. BENEFITS OF FTAO WOOD SHEAR WALLS There are many benefits to the FTAO method, such as utilizing the entire shear wall geometry to resist the applied load, potentially reducing the number of hold-downs, and reducing the base plate shear anchorage. But the greatest benefit of FTAO over other shear wall design methods lies within the definition of the wall pier aspect ratio. In the segmented and perforated methods, the aspect ratio of the contributing wall piers is defined as the full height of the wall system divided by the length of the wall pier. (Figure 1, Parts A and B). When utilizing the FTAO method, the aspect ratio of the wall pier is defined as the height of the opening divided by the length of the adjacent wall pier (Figure 1, Part C). For example, given the wall illustrations in Figure 1, where the full height (h) is equal to 10 ft. and the opening is 4 ft. tall, the minimum wall pier length (b) allowed to be included in the design at the 2:1 aspect ratio for each method would be: segmented and perforated are equal to (10/2) or 5 ft., and FTAO is equal to (4/2) or 2 ft. This benefit not only allows for narrower wall segments but also helps increase the overall height of the wall system where other methods might be limited. Using wood structural panel sheathing adds benefits. The governing aspect ratio limitation in most applicable building codes for wood structural panels is 2:1 with a maximum of 3.5:1 (with the application of an adjustment factor). Although the use of other building materials is permitted, per SDPWS Section 4.3.4, the aspect ratios either cannot exceed 2:1, or a load reduction factor is applied when the aspect ratio surpasses 1:1. TESTING FTAO DESIGN METHODOLOGY FTAO has become more popular among design professionals as a viable shear wall analysis for wood-framed walls. The majority of the published FTAO design examples, however, are limited to the wall containing a single opening; and, until recently, the recognized FTAO methods were tested in a limited number of installations. In 2009, a joint research project was conducted by APA, the University of British Columbia (UBC), and the USDA Forest Products Laboratory to examine the internal forces generated in wood structural panel-sheathed wood-framed shear walls during a lateral event. The test results, in conjunction with analytical computer-based models from UBC (12 full-scale, 8- x 12-ft. wall configurations) were used to develop an enhanced FTAO design methodology and evaluate the accuracy of the calculated forces in the walls using historical FTAO methods. The initial testing determined the Diekmann rational analysis as the most accurate simplified method to estimate the forces in the shear wall. Using this approach as the basis, APA expanded the methodology to incorporate multiple openings and asymmetric piers. This methodology is based on the following key design assumptions: 1.The unit shear is equivalent aboveand below the openings. 2.The corner forces are based on theunit shear above and below theopenings and only the wall piersadjacent to that unique opening. 3.The tributary length of the openingis the basis for calculating the shearto each wall pier. 4.The shear to each wall pier is thetotal shear divided by the lengthof the wall, multiplied by the sumof the pier length, plus the tributary width of the adjacent openings,divided by the pier length: v1 =(V/L)*((L1 + T1)/L1). 5.The unit shear in the corner zonesis equal to subtracting the cornerforces from the panel resistance, R.R is equal to the shear of the piermultiplied by pier length: Va1 =(v1L1 – F1)/L1. 6.The design is checked by summingthe shears along each vertical line.The first and last lines sum to thehold-down force, and the interiorlines sum to zero (Figure 2). Testing data were also used to analyze the overall deflection of the FTAO wall systems, which verified that the sheathing below the openings aided in resisting the overall deflection of the wall. To attain YOU CALL ME RAINHYDROTECH CALLS ME OPPORTUNITY2.3889×10.indd 18/1/17 5:29 PM October 2018 RCI Interface • 21 the most accurate fastener deformation variable, the basis of the testing was completed using the four-term deflection equation, 2015 International Building Code (IBC) Equation 23-2: Δ = 8vh3/EAb + vh/Gt + 0.75hen + da h/b Note that the three-term deflection equation provided in SDPWS (Equation 4.3-1) may also be used, but the deflection calculations must be consistent throughout the design. The wall deflection assumption (Figure 3) is that the total deflection of the FTAO shear wall is equivalent to the average of the deflection of each wall pier in both the positive and negative directions. The wall pier heights also vary depending on the deflection direction and amount of sheathing below the openings. For example, in Figure 3, positive deflection of wall pier 1 (δ1+) is determined using the height measured from the bottom of the opening to the top of the wall due to the resistance of the wall sheathing below the opening. Negative deflection of wall pier 1 (δ1-) is determined using the full wall height. Δ = average (δ1+, δ2+, δ3+, δ1-, δ2-, δ3-) FTAO DESIGN EXAMPLE Historically, FTAO design examples have been completed showing symmetric wall pier widths and a single opening in the wall system. Figure 4 illustrates the benefits of FTAO for walls with multiple openings and asymmetric wall pier widths. Given a 20-ft.-long wall that is 10 ft. tall with a 4000-pound shear force applied to the top of the wall, use the FTAO method to calculate hold-down forces, required strap forces, and required wall sheathing capacity. 2 2 • RC I I n t e r f a c e Oc t o b e r 2 0 1 8 Figure 2 –Vertical shear line illustration. Figure 3 – FTAO deflection illustration. ISSUE SUBJECT SUBMISSION DEADLINE January 2019 Environmental Issues October 15, 2018 February 2019 Building envelope (misc.) November 15, 2018 March 2019 Fenestration December 14, 2018 April 2019 Codes and standards January 15, 2019 May/June 2019 Convention review February 15, 2019 July 2019 Forensics April 15, 2019 Publish in RCI Interface RCI Interface journal is seeking submissions for the following issues. Optimum article size is 2000 to 3000 words, containing five to ten graphics. Articles may serve commercial interests but should not promote specific products. Articles on subjects that do not fit any given theme may be submitted at any time. Submit articles or questions to Executive Editor Kristen Ammerman at 800-828-1902 or kammerman@rci-online.org. Begin by calculating the hold-down forces at each end of the shear wall: 1.H = (V*h)/L = ((4000 lb.)(10 ft.))/(20 ft.) = 2000 lb.The selected hold-downs must have a minimum capacity of 2000 pounds. To attain the corner forces, both the unit shear and boundary forces above and below the openings must be determined. 2. Unit shear above and below openings: va = vb = H/(ha + hb) = 2000 lb./(2 ft. + 3 ft.) = 400 plf 3.Total boundary force above and below the openings: O1 = (va)(LO1) = (400 plf)(4 ft.) = 1600 lb. O2 = (va)(LO2) = (400 plf)(6 ft.) = 2400 lb. Calculate the corner forces at each opening to determine the maximum strap force required: 4.F1 = ((O1)(L1))/(L1 + L2) = ((1600 lb.)(2 ft.))/(2 ft. + 4.5 ft.) = 492 lb. F2 = ((O1)(L2))/(L1 + L2) = ((1600 lb.)(4.5 ft.))/(2 ft. + 4.5 ft.) = 1108 lb. F3 = ((O2)(L2))/(L2+L3) = ((2400 lb.)(4.5 ft.))/(4.5 ft. + 3.5 ft.) = 1350 lb. F4 = ((O2)(L3))/(L2 + L3) = ((2400 lb.)(3.5 ft.))/(4.5 ft. + 3.5 ft.) = 1050 lb.The selected strapping to be placed above and below the openings must have a minimum tension capacity of 1350 pounds. ORDINARY ROOFS WASTE MEHYDROTECH ROOFS LEVERAGE MY POTENTIAL2.3889×10.indd 28/1/17 5:29 PM October 2018 RCI Interface • 23 Figure 4 – Design example illustration. To calculate the unit shear beside the openings and to determine our wall sheathing requirement, the tributary width of each internal shear line must be determined. 5.T1 = ((L1)(LO1))/(L1 + L2) = ((2 ft.)(4 ft.))/(2 ft. + 4.5 ft.) = 1.23 ft. T2 = ((L2)(LO1))/(L1 + L2) = ((4.5 ft.)(4 ft.))/(2 ft. + 4.5 ft.) = 2.77 ft. T3 = ((L2)(LO2))/(L2 + L3) = ((4.5 ft.)(6 ft.))/(4.5 ft. + 3.5 ft.) = 3.38 ft. T4 = ((L3)(LO2))/(L2 + L3) = ((3.5 ft.)(6 ft.))/(4.5 ft. + 3.5 ft.) = 2.63 ft. Using the tributary width to the internal shear lines, the maximum unit shear beside the openings can be calculated to determine the required wall sheathing capacity. 6.V1 = ((V/L)(L1 + T1))/L1 = ((4000 lb./20 ft.)(2 ft. + 1.23 ft.))/(2 ft.) = 323 plf V2 = ((V/L)(T2 + L2 + T3)/L2 = ((4000 lb./20 ft.)(2.77 ft. + 4.5 ft. + 3.38 ft.))/(4.5 ft.) = 473 plf V3 = ((V/L)(T4 + L3)/L3 = ((4000 lb./20 ft.)(2.63 ft. + 3.5 ft.))/(3.5 ft.) = 350 plfThe selected wall sheathing must have a minimum shear capacity of 473 plf. Note that as a calculation verification, the sum of the unit shears multiplied by the length of the adjacent wall pier should be equal to the initial shear force: (V1)(L1) + (V2)(L2) + (V3)(L3) = V (323 plf)(2 ft.) + (473 plf)(4.5 ft.) + (350 plf)(3.5 ft.) = 4000 lbs. This checks out. To verify the calculations, the final design assumption must be met. The sum of the shear-line forces at the outside lines must be equal to the hold-down force, and the sum of each interior shear line must be equal to zero. To perform this verification, the corner force resistance and corner zone unit shears must be determined. The corner force resistance, R, for each pier is determined by multiplying the unit shear in each wall pier by the corresponding wall pier length: 7.R1 = (V1)(L1) = (323 plf)(2 ft.) = 646 lb. R2 = (V2)(L2) = (473 plf)(4.5 ft.) = 2,129 lb. R3 = (V3)(L3) = (350 plf)(3.5 ft.) = 1225 lbs. To calculate the corner force unit shear in each wall pier, first the resultant shear force in the corner zone must be determined, then divided by the corresponding wall pier length: 8.R1-F1 = 646 lbs. – 492 lbs. = 154 lbs. R2-F2-F3 = 2129 lbs. – 1108 lbs. – 1350 lbs. = -329 lbs. R3-F4 = 1225 lbs. – 1050 lbs. = 175 lbs. THE GARDEN ROOF® ASSEMBLY INTRODUCED OVER 20 YEARS AGO, PROVIDING:stormwater management solutionsreduceretaindelayextended roof longevityadditional usable spacefull assembly warranty Learn more today at hydrotechusa.rainhydrotechusa.com/power-of-rainHELPING YOU HARNESS THE POWER OF RAIN™© 2017 Garden Roof is a registered trademark of American Hydrotech, Inc.Harness the Power of Rain is a trademark of American Hydrotech, Inc.2.3889×10.indd 38/1/17 5:29 PM October 2018 RCI Interface • 25 APA has automated the calculation process in the previous example with the creation of an FTAO spreadsheet (https://www.apawood.org/designerscircle-ftao). The spreadsheet is based on the tested methodology and provides the engineer with the required hold-down forces, tension strap forces, and wall sheathing capacity, and automatically completes the design check in step 9. The spreadsheet also provides shear wall deflection calculations for the three- and four-term deflection equation options. CONCLUSION With the introduction of a new method for estimating the overall wall deflection and tested verification for the accuracy of FTAO at shear walls, designers can be confident in having another valuable tool to apply when designing the lateral resisting system for wood buildings. The FTAO method helps expand the boundaries of building design with wood-framed walls, utilizing the full shear wall geometry to help increase the design flexibility and potentially reduce the economic impact of the lateral resisting system. 9.vc1 = (R1-F1)/L1 = (154 lb.)/2 ft. = 77 plf vc2 = (R2-F2-F3)/L2 = (-329 lb.)/4.5 ft. = -73 plf vc3 = (R3 – F4)/L3 = (175 lb.)/3.5 ft. = 50 plf Performing a design check by summing the forces at each shear line (Figure 5), will complete the analysis: 10. Line 1: (va1)(ha + hb) + (V1)(ho) = H? (77 plf)(2 ft. + 3 ft.) + (323 plf)(5 ft.) = 2,000 lb. Line 2: (va)(ha + hb) – (va1)(ha + hb) – (V1)(ho)= 0? (400 plf)(2 ft. + 3 ft.) – (77 plf)(2 ft. + 3 ft.) – (323 plf)(5 ft.) = 0 Line 3: (va2)(ha + hb) + (V2)(ho) – (va)(ha + hb)= 0? (-73 plf)(2 ft. + 3 ft.) + (473 plf)(5 ft.) – (400 plf)(2 ft. + 3 ft.) = 0 Line 4 = Line 3 Line 5: (va)(ha + hb) – (va3)(ha + hb) – (V3)(ho)= 0? (400 plf)(2 ft. + 3 ft.) – (50 plf)(2 ft. + 3 ft.) – (350 plf)(5 ft.) = 0 Line 6: (va3)(ha + hb) + (V3)(ho) = H? (50 plf)(2 ft. + 3 ft.) + (350 plf)(5 ft.) = 2000 lbs. Using the four-term deflection equation (Figure 3), overall wall deflection is calculated by determining the average of the positive and negative deflection values for each pier (Figure 4). Assuming a maximum hold-down capacity of 2500 pounds, nail slip of 0.125 in., and an APA-rated sheathing 15/32-in. performance category with 8d nails at 4 in. on center: 11. Δ = average (δ1+,δ2+,δ3+,δ1-,δ2-,δ3-) = average (1.393,0.692,0.508,0.708,0.692,0.963) (in) = 0.826 in. 26 • RCI Interface October 2018 Jared Hensley holds a BS in architectural engineering with an emphasis in structural design from the University of Wyoming. He is a registered professional engineer in Colorado, Wyoming, and Washington. Hisdesign and construction management experience includes structural design for new and renovated buildings incorporating steel, masonry, timber, and concrete, as well as acting as a consulting engineer on structural design and rough framing inspections in the greater Denver area. Hensley joined APA in the spring of 2015 and covers the Pacific Northwest territory. Jared Hensley Figure 5 – Wall force diagram illustration. CORRECTION A table on page 26 of the August 2018 issue of RCI Interface issue, in the article, “Mechanical Fasteners for Nailable Roof Decks,” contained an error. Metal Auger B is NOT suitable for use with lightweight insulating concrete.
