Figure 5.10 (a) Frame with diaphragm resisting forces. (b) Resisting forces applied as uniformly distributed in-plane loads for frame component stress analysis.
Vh = Chαs[A sinh(α x) + B cosh(α x)] (5-12) y = A cosh(α x) + B sinh(α x) + R/k (5-13)
ye = R / [ k (1 – D)] (5-14)
where:
Vh = diaphragm shear force, lbf (N) x = distance from endwall, in. (mm)
R = eave load, lbf (N)
s = frame spacing, in. (mm)
y = lateral displacement of diaphragm at a distance x from the endwall, in.
(mm)
ye = lateral displacement of the endwall, in. (mm)
k = stiffness of interior frames, lbf/in.
(N/mm)
ke = stiffness of endwall frames (or shearwalls), lbf/in. (N/mm)
Ch = horizontal shear stiffness for a width s of the diaphragm, lbf/in. (N/mm) L = Distance between endwalls, in.
(mm)
sinh = hyperbolic sine cosh = hyperbolic cosine
( k / Ch )1/2
α = s
A = ye – R/k
A ( 1 – cosh(α L)) B = sinh(α L)
ke sinh(α L) D = α Ch s (1 - cosh(α L))
5.7 Component Design
5.7.1 General. All building components must be checked to ensure that actual loads do not ex-ceed allowable design values. In this section, special attention is given to components that are involved in load transfer by diaphragm action.
5.7.2 Diaphragms. The maximum shear in a diaphragm section, Vp,i, cannot exceed the al-lowable shear strength of the section, va,i, multi-plied by the diaphragm length.
Vp,i < va,i d i (5-15) where:
Vp,i = in-plane shear force in diaphragm section i from equation 5-9 lbf (N) va,i = allowable in-plane shear strength of
diaphragm i (see Section 6.3.3), lbf/ft (N/m)
d i = slope length of diaphragm i, ft (m) Qb
Roof Gravity Loads
Ceiling Gravity Loads
s x qw w
s xqwr s xqlr
s x ql w
Qa
Qc
Roof Gravity Loads
Ceiling Gravity Loads
s x qw w
s xqwr s xqlr
s x ql w
q p,a q p,b
q p,c
(a)
(b)
5.7.3 Diaphragm Chords. In addition to shear forces, a roof/ceiling diaphragm assembly must also resist bending moment. The magnitude of this bending moment is dependent on a number of factors. For design, this bending moment is assumed to be no greater than:
Md = Vh L / 4 (5-16)
where:
Md = diaphragm bending moment, lbf-ft (N m)
Vh = maximum total shear in roof/ceiling diaphragm assembly, lbf (N)
L = distance between shearwalls, ft (m) Equation 5-16 treats the roof/ceiling assembly as a uniformly loaded beam that is simple sup-ported by two shearwalls spaced a distance L apart. Each shearwall is assumed to be sub-jected to a force that is equal to the maximum total shear in the roof/ceiling assembly, Vh. The maximum total shear in the roof/ceiling assem-bly, Vh, can be obtained from computer output (e.g. figure 5.9), or equation 5-7 or 5-12 if appli-cable. The uniform load on the roof/ceiling as-sembly (w in figure 5.11a) is set equal to 2Vh/L.
This quantity is multiplied by L2/8 to obtain Md.
The bending moment applied to a roof/ceiling diaphragm assembly is resisted by axial forces (a.k.a. chord forces) in members orientated per-pendicular to trusses/rafters. This includes roof purlins and analogous framing members in the ceiling diaphragm. For bending moment calcula-tions, these members are referred to as dia-phragm chords (figure 5.11a). Any connection in the chords, either between intermediate chord members or where they are connected to the endwalls, must be designed to resist the calcu-lated axial force.
If the roof/ceiling assembly behaves as a single beam in resisting bending moment, the maxi-mum chord force (which is located in the edge chords) can be calculated as:
Pe = Md α / b (5-17)
where:
Pe = axial force in edge chord, lbf (N) Md = diaphragm bending moment from
equation 5-16, lbf-ft (N m)
α = reduction factor dependent on chord force distribution
b = horizontal distance between edge chords, ft (m)
Figure 5.11. (a) Plan view of a diaphragm under a uniform load, w. Chord force distribu-tions when (b) moment resisted by edge chords only, (b) chord force distribution is linear, and (c) chord force distribution is linear, but diaphragm halves assumed to act independ-ently in resisting moment.
Vh
w
(a) (b) (c)
Trusses/rafters Chords
(d) Vh
L
Shearwall
The axial force in an edge chord is dependent on chord force distribution as indicated by the presence of α in equation 5-17. The current ASAE EP484 diaphragm design procedure (ASAE, 1999a) assumes that edge chords act alone in resisting bending moment (figure 5.11b). For this case, α is numerically equal to one (1). This is a conservative approach. Alter-natively, many engineers assume a linear distri-bution of chord forces as shown in figure 5.11c.
When a linear distribution is assumed, the re-duction factor α is a function of chord location. If chords are evenly spaced, then α is given as:
(n – 1 )2 α = n / 2
Σ
(n – 2 i + 1)2i =1
when n is even
(n – 1 )2 α = (n-1)/2
Σ
(n – 2 i + 1)2i =1
when n is odd
where:
α = reduction factor when chords are evenly spaced and chord forces are linearly distributed
n = number of chord rows, including the two rows of edge chords
The preceding equations were used to calculate the values given in table 5.3.
If a linear distribution of chord force is assumed (figure 5.11c), and interior chords are evenly spaced, the load in an interior chord, Pi, is given as:
Pi = 2 Pe x i / b (5-18)
where:
Pi = axial force for chord in row i, lbf (N) Pe = axial force in edge chord from
equa-tion 5-17, lbf (N)
b = horizontal distance between edge chords, ft (m)
x i = horizontal distance from center of diaphragm to chord row i.
Table 5.3. Reduction Factor, α, for Axial Force in Edge Chords
n* α n* α
2 1.000 22 0.249
3 1.000 23 0.239
4 0.900 24 0.230
5 0.800 25 0.222
6 0.714 26 0.214
7 0.643 27 0.206
8 0.583 28 0.200
9 0.533 29 0.193
10 0.491 30 0.187
11 0.455 31 0.181
12 0.423 32 0.176
13 0.396 33 0.171
14 0.371 34 0.166
15 0.350 35 0.162
16 0.335 36 0.158
17 0.314 37 0.154
18 0.298 38 0.150
19 0.284 39 0.146
20 0.271 40 0.143
21 0.260 41 0.139
* n is the number of chord rows, including the two rows of edge chords
Technical Note Chord Forces
The axial force induced in an individual chord by applied building loads is a function of many complex, interacting design variables. For this reason, designers have had to rely on simplify-ing assumptions in order to approximate chord forces.
One common assumption is that the roof/ceiling assembly acts as a large deep beam that is simply supported by two end shearwalls. This assumption is used to calculate the maximum in-plane bending moment to which a diaphragm is subjected. This assumption is conservative in that it neglects the resistance to in-plane bend-ing contributed by sidewalls. Sidewalls help re-sist (and thereby reduce) in-plane bending mo-ments in two ways. First they brace endwalls
and other shearwalls, which limits rotation of the diaphragm at these shearwalls. Second, they resist a change in eave length (and hence changes in eave chord forces) by virtue of their own in-plane shear stiffness.
Because of the influence of sidewalls, the distri-bution of in-plane bending moment will not follow that for a typical simple supported beam (i.e., zero moment at the supports, and maximum moment at midspan). For this reason, Pollock and others (1996) recommend modeling the roof/ceiling assembly as a deep beam with fixed supports.
Because of uncertainty surrounding variation in in-plane bending moment with building length, some designers will assign the maximum calcu-lated in-plane bending moment (Md from equa-tion 5-16) to every locaequa-tion along the length of the building. This is obviously a conservative approach.
Another major assumption that a designer must make involves the distribution of chord forces across a building. Three different chord force distributions are shown in figure 5.11b, 5.11c, and 5.11d. Whether or not edge chords resist virtually all of the in-plane bending moment (fig-ure 5.11b), or a linear distribution of axial forces exists in chords between edge chords (figure 5.11c) is a question that is at the heart of ongo-ing research. In reality, the distribution of chord forces lies somewhere in between these two extremes, exactly where being dependent on specifics of the design and on the magnitude of the applied load (Note: at higher load levels, load distributions change due to geometric and material nonlinearities). Presently, there is very little research data to support one specific de-sign procedure/assumption. The most extensive investigation of chord forces was by Niu and Gebremedhin (1997) who strain gauged purlins in a full-scale building and in a diaphragm test assembly. The data collected in this study does not strongly support any particular hypotheses regarding chord force distribution. The only other research of significance to chord force dis-tribution was the comprehensive finite element analyses of diaphragm assemblies by Wright (1992) and Williams (1999). Both of these re-searchers found that in-plane bending
moment in their models was resisted almost en-tirely by the edge purlins. Bohnhoff and others (1999) showed that as the shear stiffness of cladding is increased, interior purlins get more involved in resisting in-plane bending moments.
Chord force distribution has also been shown to depend on the degree of interaction between individual diaphragms. Figure 5.11d illustrates the distribution of chord forces when there is no interaction between individual diaphragms on both sides of a ridge. Note that interaction be-tween individual diaphragms on opposites sides of a ridge is highly dependent on: (1) the spac-ing between ridge purlins, and (2) the rigidity of the ridge cap and other elements joining the two diaphragms.
5.7.4 Shearwalls. End and intermediate shear-walls must have sufficient strength to transmit forces from roof and ceiling diaphragms to the foundation system. In equation form:
va > Vs / (W – DT) (5-19) openings in the shearwall, ft (m) The allowable shear capacity of end and inter-mediate shearwalls, va, is obtained from vali-dated structural models, or from tests as out-lined in ASAE EP558 (see Section 6.5). The total force in the shear wall, Vs, is obtained from computer output (e.g. figure 5.8), or equation 5-7 or equation 5-12 if applicable.
The total width of door and window openings, DT, generally varies with height as shown in fig-ure 5.12. At locations where DT is the greatest (section b-b in figure 5.12) additional reinforcing may be required to ensure that the allowable shear stress is not exceeded.
The structural framing over a door or window opening will act as a drag strut transferring
shear across the opening. The header over the opening shall be designed to carry the force in tension and/or compression across the opening.
Figure 5.12. Shearwall showing variations in opening width, DT, with height.
Shearwall strength can easily be increased when the applied load exceeds shearwall capac-ity. For example, the density of stitch screws can be increased and additional fasteners can be added in panel flats (on both sides of each ma-jor rib is the most effective). If only one side of the wall has been sheathed, add wood paneling or metal cladding to the other side. Metal diago-nal braces can also be added beneath any wood paneling or corrugated metal siding.
5.7.5 Shearwall Connections. Connections that fasten (1) roof and ceiling diaphragms to a shearwall, and (2) shearwalls to the foundation system, must be designed to carry the appropri-ate amount of shear load. The design of these connections may be proved by tests of a typical connection detail or by an appropriate calcula-tion method.
At end shearwalls it is not uncommon to use the truss top chord to transfer load from roof clad-ding to endwall cladclad-ding. Sidewall steel is fas-tened directly to the truss chord, as is the roof steel when purlins are inset. In buildings with top-running purlins, roof cladding can not be fastened directly to the truss. In such cases, blocking equal in depth to the purlins is placed between the purlins and fastened to the truss.
Roof cladding is then attached directly to this blocking.
5.7.6 Shearwall Overturning. Diaphragm load-ing produces overturnload-ing moment in shearwalls.
This moment induces vertical forces in shear-wall-to-foundation connections that must be added to vertical forces resulting from tributary loads. In the case of embedded posts, increases in uplift forces may require an increase in em-bedment depth, and increases in downward force may require an increase in footing size (see Chapter 8).