Load-carrying capacity of timber frame diaphragms with unidirectional support

Load-carrying capacity of timber frame diaphragms with unidirectional support Jørgen Munch-Andersen, Danish Timber Information, 2012-06-26 Introduction The rules for determining the load-carrying capacity for roof, floor and wall diaphragms is usually only dealing with cases where each panel is supported by timber frame or noggins along all edges. The conventional plastic model with a hold down at the end appears as Model A in Eurocode 5. This paper proposes a simple plastic model for cases where all supports are in one direction. It is justified for walls using tests results from Källsner (1983) who measured the racking load carrying capacities for shear walls with different support conditions and panel types. It is also shown how the model can be used for roof and floor diaphragms. Fully supported panels Figure 1 illustrates the case covered by Method A in Eurocode 5 (section 9.2.4). It is a plastic method based on the assumptions that all edges are supported by the timber frame and that all forces between panels and timber frame are parallel to the edges of the panels. Further, it is presupposed that the bottom rail can transfer the horizontal F v load to the support and that there is a hold down at the end to withstand the vertical reaction Ft = H/ B. When the load on each fastener is F f, the fastener spacing along the edges is s and the effective number of fasteners along the horizontal edges is defined as n x = b/s and along the vertical edges as n y = h/s equilibrium of a panel then requires n x F f h = n y F f b. This is seen to be fulfilled when the fastener spacing is the same along all edges. The definition of the effective number of fasteners implies that the fasteners in the corners are assumed to take the load ½ F f in both the horizontal and the vertical direction. (The fasteners in the intermediate studs are only used to prevent buckling so the fastener spacing there can be larger than s). Figure 1. Shear wall where all edges are supported, as presupposed in Eurocode 5. 1

When the lateral load carrying capacity of each fastener is R f the racking load-carrying capacity of one panel becomes R = nr = bsr / (1) ' v x f f The racking load carrying capacity of a diaphragm with several panels supported along all edges can be determined by replacing the dimensions of the panel (b and h) with the dimensions of the diaphragm (B and H) so R v = BsR / (2) f Limitations In order to prevent buckling of the panels Eurocode 5 requires for fully supported panels that a net /t < 100 where a net is the inner distance between studs and t is the thickness of the panel. Eurocode 5 also prescribes that the racking load carrying capacity for narrow panels should be reduced by the factor 1 1 2 bh / 2 1 1 2/ bh 4 bh / < 2 c = 2/ hb 2 < bh / 4 < > 1 0 bh / 4 or bh / 4 The reason is presumably that for a given rotation of a narrow panel the displacement of the short edges along the top and bottom rails becomes much larger than the displacements of the long edges along the vertical studs. This means that is not reasonable to utilize the full load carrying capacity of the fasteners in both directions at the same time. (3) Unidirectional supported panels Panels not supported along the internal vertical edges but only along intermediate horizontal rails, as shown in Figure 2, are hereinafter referred to as Unidirectional supported panels. The model in this section deals with a shear wall as in Figure 2. Figure 3 illustrates how the load is assumed to be distributed over the fasteners. The fasteners within the length e at each horizontal rail are attributed to take the vertical force. The forces will act perpendicular to the rails but forces perpendicular to the edges of the panel only occurs along part of the top and bottom rails. The vertical load taken by the fasteners at the top and Figure 2. Shear wall which internally is only supported by horizontal rails. 2

bottom rail is assumed to be only half of the load taken at the fasteners at the intermediate rails. For a configuration like in Figure 2 and 3 the effective number of fasteners to take up the vertical load along one edge becomes n y h e 1 1 e = + + a s + 1 2 s where the first term is the number of nails within distance e in the intermediate rails and the second term is half of the nails within e along the top and bottom rail (half because they are assumed only to carry half load). (4) Figure 3. Load distribution over the fasteners in an intermediate panel of a shear wall with internal horizontal rails. The horizontal load is taken by the fasteners along the top and bottom rails, but those fasteners partly used to take vertical load are assumed to take only half of the others. The effective number of fasteners to take the horizontal load along one edge is then n = ( b e)/ s x As the lever arm for the vertical forces is (b e) equilibrium requires n F h= n F ( b e) (6) x f y f Inserting (4) in (5) gives the relation n y = h/s. The value of e which ensures equilibrium can be determined from (3) when n y is substituted by h/s as ha ( ss ) 2 e = ( h a) s+ as 2 Obviously e is limited by b/2. In order to ensure that the deformation at all fasteners within e is sufficient to develop the plastic load carrying capacity R f the distance e must be limited. Based on the test results discussed below it is found that the limit should be e 0,35 b. When the limit becomes effective n y will be too small to use all n x fasteners in the horizontal direction. The load carrying capacity should therefore be calculated using the effective spacing of the fasteners hs / when e 0,35b sef = hn / y, ny = As per Eq.(4) when e= 0,35b The panel will then act as a panel with height h, width (b e) and fastener spacing s ef. The racking load carrying capacity of one panel is then (5) (7) (8) 3

R = nr = ( b e)/ s R (9) * v x f ef f The racking load-carrying capacity of a diaphragm with several panels as in Figure 2 can be determined as B * B e Rv = Rv = R f 1 (10) b s b ef The horizontal rails shall transfer the vertical load across the joint between two panels. The load is anti-symmetric as illustrated in Figure 3 which means that the moment is nil at the joint. The imum shear force in the rail occurs at the joint. In the intermediate rails equilibrium requires e V +1 b= Ff ( e b s ) or 2 e e V +1 F = f 1 s (11) 2 b where (e/s 2 + 1) is the number of fasteners within the distance e. The imum moment in the rail becomes M e e s2 V 1 1 2 b b where the term (1 s 2 /b) is an approximate but very accurate correction from uniformly distributed load to discrete loads with spacing s 2. It occurs usually at the second fastener from the joint. If the rail has to be joined within a diaphragm the best place is at the middle of a panel where the moment is nil and the shear force is V mid = V e/(b e). Experimental justification Källsner (1984) has carried out numerous tests with timber framed shear walls clad with different types of panels and with different configurations of the frame. The tests are carried out with a vertical hold down at one end of the wall so the tests reflect the assumptions for Method A in Eurocode 5. All panels were 1200 2400 mm and most tests were carried out with standing panels on a frame with continuous top and bottom rail and vertical studs spaced 600 mm apart, see type i in Figure 4. The timber was 50 mm wide and the walls were mostly 2 or 3 panels wide. The types of panels use here were 13 mm gypsum board fastened with 4 38 mm gypsum screws 13 mm bitumen impregnated soft fiberboard fastened with 1,9 35 mm round nails 9 mm gypsum board (tested at 90 % RH) fastened with 2,5 35 mm clout nail For these types of panels tests were also carried out with the panels fixed to horizontal rails in the form of 50 50 mm battens per a = 600 mm. The battens were mounted onto the frame used for type i. Both configurations with standing (type ii) and lying (type iii) panels has been investigated, see Figure 4. Note that for type ii and iii b and B is the dimensions along the rail and h and H the dimensions perpendicular to the rails. Results from type i tests are shown in Table 1. From the measured racking load carrying capacity R v,obs the lateral load-carrying capacity R f of the fasteners are found from (1) in accordance with Method A in the Eurocode. The values also agree very well with the capacities measured by Källsner directly in a testing machine. Tests with 13 mm gypsum boards were carried out with both 1, 2 and 3 panels. It is seen that the fastener load carrying capacity R f at failure is very similar for the four tests. The mean value 0,753 kn is used henceforth. The coefficient of variation is as low as 2,6%. For the 9 mm gypsum board two different fastener spacing have been used. It appears that halving the spacing does not double the failure load. The observations about fracture at failure given by Källsner indicate that the small spacing causes rupture in the panel at numerous fasteners not only at the corner fastener. This explains the apparent reduced failure load of the fastener when spaced only 50 mm. This value is not used henceforth. (12) 4

Type i: Supported 4 sides Type ii. Horizontal support, standing panels Figure 4. Definition of wall types and geometry. Type iii. Horizontal support, lying panels Table 1. Results for type i walls (Källsner, 1984). The lateral load-carrying capacity R f of the fasteners are back calculated according to Eq. (1). Panel type 13 mm gypsum boards Soft fiber 9 mm moist gypsum boards board Test no 1.1 1.2 2 3 13 U1 U2 No of panels 1 1 2 3 3 2 2 B, mm 1200 1200 2400 3600 3600 2400 2400 H, mm 2400 2400 2400 2400 2400 2400 2400 s, mm 200 200 200 200 75 100 50 R v,obs, kn 4,59 4,52 8,71 13,84 16,71 9,72 17,98 R f, kn 0,765 0,753 0,726 0,769 0,348 0,405 0,375 Results for wall type ii and iii are given in Table 2. The fastener load carrying capacity R f is taken from Table 1. From the last row in Table 2 it is seen that the model predicts the measured racking load-carrying capacity quite well, except for Test no U6 in the last column. In that test the panels is reported to bulge severely and the bending stress σ m in the batten is very high. Further, the load V at the end of the batten is so high that it would require more than one fastener to withstand it. It is not specified by Källsner how this connection is made, but it is most likely by only one fastener. For practical design the cases represented by Test U6 would be ruled out because the horizontal rail could not be proven to be strong enough, so Test U6 is disregarded in the evaluation of the model. Figure 5 shows the observed versus the estimated values of R v for the other tests. The slope 0,975 indicates that the model on average overestimates R v by 2,5 %. The coefficient of variation of R v,est /R v,obs is only 6,4 %. These small values describes the model uncertainty and can easily be covered by a model modification factor of 0,9 applied to the estimated load carrying capacity. There is no sign of the model error being dependent on the type of panel, the orientation of the panels (type ii or type iii) or the fastener type. It also works for different ratios of fastener spacing in the outer and in the intermediate rails, as long as the load carrying capacity of the rails and their fastening at the end is sufficient. The three types of panels used have all a quite small stiffness and some bulging has been reported by Källsner. This might be assumed to decrease the load capacity. Using a stiffer panel like plywood is expected to reduce the bulging so the higher capacity of the fasteners in such materials can safely be used, especially when the unsupported edges are fitted with tongue and groove. 5

Table 2. The observed and estimated racking load carrying capacity for wall type ii and iii. Wall type ii iii Panel type (Equation no.) 13 mm gypsum Soft fiber board 9 mm moist gypsum boards Soft fiber board 9 mm moist gypsum boards Test no. 24 22 U3 U4 23 U5 U6 No of panels 3 2 2 2 2 2 2 R f, kn 0,753 0,348 0,405 0,405 0,348 0,405 0,405 H, mm 2400 2400 2400 2400 2400 2400 2400 B, mm 3600 2400 2400 2400 2400 2400 2400 h, mm 1200 1200 1200 1200 2400 2400 2400 b, mm 2400 2400 2400 2400 1200 1200 1200 a, mm 600 600 600 600 600 600 600 s, mm 200 75 100 50 75 100 50 s 2, mm 300 150 100 50 150 100 50 R v,obs, kn 6,96 4,12 5,37 10,00 7,03 8,05 10,67 e, mm (7) 533 840 500 550 700 500 550 e 0,35b, mm 420 420 420 420 700 500 550 n y (4) 10,3 18,0 20,8 37,6 16,0 12,0 24,0 s ef, mm (8) 233 133 115 64 75 100 50 R v,est, kn (10) 7,56 4,07 5,48 9,90 7,89 7,70 14,99 V, kn (11) 1,18 0,86 1,37 2,47 1,40 1,92 3,75 M, knm (12) 0,120 0,103 0,171 0,324 0,325 0,365 0,778 σ m, MPa * 5,8 4,9 8,2 15,5 15,6 17,5 37,3 R v,est /R v,obs 1,09 0,99 1,02 0,99 1,12 0,96 1,40 σ m is the bending stress in the 50 x 50 mm rails from the moment M. 10 9 R_v,obs, kn 8 7 6 5 4 4 5 6 7 8 9 10 R_v,est, kn Figure 5. The observed versus the estimated load carrying capacity for the tests in Table 2 excluding Test U6. The slope of the best fit straight line through the origin is 0,975. 6

Applications In the following the use of the model for shear walls, flooring and roofing is illustrated. It is assumed that the fastener spacing is similar along all rails so s 2 = s. It is quite advantageous for the load carrying capacity and the equations become much simpler. The estimated load carrying capacity R v should be reduced by the model modification factor 0,9. Similarly, V and M ought to be increased by 1/0,9 because the model uncertainty might mean that the distribution of the load on the rails are not quite as assumed. This is not included in the examples in the following. Simplification for s 2 = s Equations (4) becomes n y h e = + 1 a s Equation (7) reduces to e= a s 0,35b (7a) and (8) to s ef a = es / + 1 Finally (8a) into (10) gives an equation independent of s ef, so only e needs to be determined from (7a) B e Rv R e = f + 1 1 a s b (4a) (8a) (10a) By means of (10a) with the capacities R v and R f replaced by the forces F v and F f (11) is reduced to a V = F a t B = H (11a) where the latter term is obvious because H/a is the effective number of rails, each contributing by V to the reaction F t (the top and bottom rail contributes by ½ V ). Limitations In order to prevent buckling of unidirectional supported panels the requirement to the thickness in Eurocode 5 should be tightened to a net /t < 50. This conforms with that the 9 mm gypsum panels in Table 2, which has the ratio 550/9 ~ 60 were reported to bulge severely in some cases, so they ought to be excluded from the design rule. The reduction of the racking load carrying capacity for narrow panels in Eurocode 5 represented by the factor c in (3) should for unidirectional supported panels be applied to e. This will ensure the same minimum displacement for a narrow panel as for a full panel as used in the tests. (7a) therefore becomes ca ( s) e = min c0,35b (7b) End panels If the horizontal rails are supported at the ends such that the load V can be transferred to a structure behind the end panels (first and last panel) will act like an intermediate panel as described above. Another and usually quite simple possibility is to fix the outer vertical edge of the end 7

panel to a vertical stud as shown in Figure 6. This solves the problem that the load V might be so high that it is difficult to transfer it from the rail to another structure by a simple connection. Figure 6. Unidirectional supported shear wall with vertical studs at the ends. This reduces the need for fastening the ends of the rails to the structure behind the rails. Example 1: Shear wall A wall consisting of four panels as in Figure 7 is considered. The input parameters are: R f = 0,25 kn. H = 2400 mm B = 4800 mm h = 1200 mm b = 2400 mm a = 600 mm s = 100 mm Figure 7. Shear wall with panel length parallel to the support. 8

The necessary calculations become: Eq. (3): b/h = 2 => c = 1 Eq. (7b): e = 1 (600 100) = 500 mm 4800 500 500 Eq. (10a): R v = 0, 25 + 1 1 = 9,50 kn 600 100 2400 If the load is F v = 8 kn the rails should withstand 600 Eq. (11a): V = 8 = 1,00 knand 4800 500 500 100 Eq. (12): M = 1, 00 1 1 = 190 knmm = 0,19 knm 2 2400 2400 Example 2: Roof diaphragm (wind bracing) When for example plywood is used on roofs the panels will usually be supported only by the rafters as shown in Figure 8. Plywood panels 600 2400 mm are placed across the rafters.this diaphragm can be used to bring the wind load q on the upper part of the gable wall down to the head beam on the façade. The line load, which here is assumed to be triangular, can be replaced by a concentrated load at the centre of gravity for the distributed load. It is seen that the configuration in principle is the same as in Figure 2, except that the rails are turned 90 and the external load is acting perpendicular to the support and therefore becomes F t. The distance from the gravity centre to the head beam becomes B and the length of the roof becomes H. The load carrying capacity should therefore be determined as H Rt = Rv (13) B where R v is determined from (10a). q Ft s h H=l B b a Ft Rafter Head beam Figure 8. Roof slab with panel length across the supporting rafters. The slab supports the gable wall for wind load. 9

The input parameters are: R f = 0,5 kn H=l = 5600 mm B = 2400 mm h = 2400 mm b = 600 mm a = 800 mm s = 75 mm The necessary calculations become: Eq. (3): b/h = ¼ => c = ½ Eq. (7b): e = ½ 0,35 600 = 105 mm 2400 105 105 Eq. (10a): R v = 0,5 + 1 1 = 2,97 kn 800 75 600 5600 Eq. (13): R t = 2,97 = 6,93kN 2400 If the load is F t = 6 kn the rafters should withstand 800 Eq. (11a): V = 6 = 0,86 knand 5600 105 105 75 Eq. (12): M = 0,86 1 1 = 46 knmm = 0,046 knm 2 600 600 Each rafter must be fastened to the head beam for V. Some blocking will be needed to prevent torsion. The moment in the rafters is seen to be negligible. The reactions F v should be transferred to the gable walls. When the panels are staggered some panels will be smaller than the full panels. There is no reason reduce the capacity due to the smaller panels because value of c is higher for the panels with reduced h and therefore has a higher load-carrying capacity per length. It is probable that the staggering causes other mechanisms, e.g. contact forces and bending, to contribute to the load-carrying capacity for narrow panels. Using these effects requires another model and experimental verification. Example 3: Floor diaphragm Timber panels laid across floor beams can enable a timber floor to transfer wind loads on the facades to transverse shear walls. Figure 9 shows the same structure as Figure 8, but the wind load is represented by the line load q. The floor can be regarded as two virtual unidirectional supported shear walls rotated 90 when cut into halves. The line load is converted to two concentrated forces acting at distance H = l/4 from the ends. The edge beam shall be able to transfer F t between the two virtual walls. The input parameters for each virtual wall are: R f = 0,5 kn H = l/4 = 1400 mm B = 3600 mm h = 2400 mm b = 600 mm a = 800 mm s = 75 mm 10

Edge beam Ft s a B l h H=l/4 b Ft H q Figure 9. Floor diaphragm which supports the facade for wind load. The necessary calculations become: Eq. (3): b/h = ¼ => c = ½ Eq. (7b): e = ½ 0,35 600 = 105 mm 3600 105 105 Eq. (10a): R v = 0,5 + 1 1 = 4, 45 kn 800 75 600 If the line load is q = 1,4 kn/m the concentrated load on each virtual wall is F v = ½ 1,4 5,6 = 3,92 kn. The floor beams should withstand 800 Eq. (11a): V = 3,92 = 0,87 kn and 3600 105 105 75 Eq. (12): M = 0,87 1 1 = 47 knmm = 0,047 knm 2 600 600 The edge beam shall transfer 1400 Eq. (13): F t = 3,92 = 1,52 kn 3600 F t could also be obtained directly as Ft = ql / B= 1,4 5,6 /3,6= 1,52kN 1 2 1 2 8 8 Conclusions A model is developed to estimate the racking load carrying capacity of timber framed diaphragms when the panels are only supported in one direction Unidirectional supported panels. The model is plastic (as the model for fully supported panels in Eurocode 5) and is based the load distribution on the fasteners shown in Figure 10. The load along the unsupported edges is supposed to be transferred by the nails within the distance e from the edge. It is important that the shear force V can be sustained by the supports and transferred to a structure behind at the ends. 11

Figure 10. The assumed load distribution on the fasteners for a unidirectional supported diaphragm. The model agrees very well with a comprehensive test series carried out by Källsner (1984) when the load carrying capacity of the fasteners is determined from tests with fully supported panels. The model becomes quite simple when the fastener spacing is the same in all supports (s = s 2 ). The racking load carrying capacity R v can for non-narrow panels be estimated as B e e Rv R = f + 1 1 a s b where: e= a s 0,35b R f is the load carrying capacity of the fasteners s is the fastener spacing B is the width of the diaphragm, along the supports b is the width of each panel, along the supports a is the distance between the supports (c-c) The shear force becomes V = a B The model can be used for both shear walls, roofs and floors sheathed with panels. In order to compensate for minor model uncertainties it is recommended to reduce the racking load carrying capacity by 10 %. The tests concerns rather soft panels but using a stiffer panel like plywood is expected to reduce the bulging so the higher capacity of the fasteners in such materials can safely be used. Tongue and groove along the unsupported edges will further reduce the bulging. It is also recommended to reduce the capacity of narrow panels (width to length ratio smaller than 1:2) similarly to Eurocode 5. This reduction compensates for the quite different translation at the fasteners in the length and width direction of the panel. References Källsner, B: Panels as wind-bracing elements in timber-framed walls (in Swedish). TräteknikCentrum, Stockholm, 1984. (Träteknik Rapport Nr 56). Eurocode 5: Design of timber structures - Part 1-1: General and rules for buildings. CEN 2004. 12