9.1 Introduction
9.1.1 General. Structurally efficient post-frame buildings utilize the roof as a diaphragm to resist horizontal wind forces. This chapter presents an example of diaphragm design following the five steps outlined in Section 5.1.4.
9.1.2 Building Specifications. Table 9.1 lists design parameters for the example building.
Table 9.1. Example Building Specifications Width (truss length) 36 ft Length (along ridge) 60 ft Height at post bearing 12 ft
Roof slope 4/12 (18.43°)
Bay spacing 10 ft
Number of frames
(including end walls) 7
Post embedment depth 4 ft
Post grade & species No.2 S. Pine Post size Nom. 6- by 6-in.
Roof snow load 30 psf
Roof dead load 5 psf
Concrete slab? Yes
Ceiling? No
9.1.3 Wind Loads. It is assumed that the ex-ample building is located in a jurisdiction that has adopted the 1994 Uniform Building Code.
Design wind loads calculated according to this code are presented in Table 9.2
Table 9.2. Wind Loads
Wind speed 80 mph
Exposure category B
Windward wall, qww 8.13 psf
Leeward wall, qlw -5.08 psf *
Windward roof, qwr 3.05 psf
Leeward roof, qlr -7.12 psf *
* Negative loads act away from the surface in question. Positive loads act toward the sur-face in question.
9.2 Step 1: Modeling
9.2.1 General. The structural model for this ex-ample building follows that in Section 5.2. The frames are numbered from one to seven begin-ning on the left end. That portion of the roof dia-phragm between each frame is broken into two discrete segments labeled 1a, 1,b, …6a, 6b.
See Figure 9.1.
Figure 9.1. Identification of frame elements and roof diaphragm segments.
9.3 Step 2: Stiffness Properties
9.3.1 Frame Stiffness, k. One reliable way to determine frame stiffness is to use a plane-frame analysis program such as the PPSA pro-gram mentioned in Section 5.3.2. In this exam-ple, all posts will be considered fixed at the grade line and pin connected to trusses (figure 5.5). Consequently, the stiffness of each em-bedded post can be calculated using equation 5-3 which is given as:
kp = 3 E I / Hp 3
For the nominal 6- by 6-inch No. 2 Southern Pine posts:
E = 1.2 x 106 lbf/in.2 (No adjustment for wet conditions is necessary for Southern Pine timbers. It is gener-ally required for laminated posts.) I = 76.26 in.4
Hp = 144 in.
Thus, kp = 91.9 lbf/in.
1a
1b 2a
2b 3a
3b 4a
4b
1 2 3 4 5 6 7
5a
5b 6a
6b
Frame stiffness, k, is obtained by summing indi-vidual post stiffness values (equation 5-2). This summation yields:
k = 184 lbf/in.
9.3.2 Diaphragm Stiffness, Ch. The diaphragm assembly used in this example is Test Assembly 11 in Table 6.1. Its properties are summarized in Table 9.3.
Table 9.3. Diaphragm Properties
Metal thickness 29 gage
Assembly width, 3 x a 36 ft
Assembly length, b 12 ft
Allowable shear strength, va 107 lbf/ft Effective in-plane shear
stiffness, c 3700 lbf/in.
Effective shear modulus, G 3700 lbf/in.
In-plane shear stiffness for a single diaphragm section is calculated using equation 6-9, which is given as.
G bh
cp = s cos(θ)
Substitution of appropriate values yields:
(3700 lbf/in.)(18 ft) cp =
(10 ft)(cos(18.43)) cp = 7020 lbf/in.
The horizontal shear stiffness, ch, of a single diaphragm section is calculated using equation 6-10 which is given as:
ch = cp cos2(θ)
Substitution of appropriate values yields:
ch = (7020 lbf/in.) cos2(18.43°) = 6320 lbf/in.
Total horizontal shear stiffness of a diaphragm element, Ch, is found by summing the stiffness values of the two sections that comprise each diaphragm element (see equation 5-4).
Ch = 6320 + 6320 = 12,640 lbf/in.
9.3.3 Shearwall Stiffness, ke. There are no large doors in the endwalls of the example build-ing. Lacking a specific tested endwall assembly, the 12 ft high endwalls will be assumed to have the same shear stiffness as an 8 ft section of the roof diaphragm; that is, ke will be set equal to Ch or 12,640 lbf/in.
9.4 Step 3: Eave Loads
9.4.1 Windward Roof Pressures. As noted in Section 9.1.3, this example uses wind loads cal-culated in accordance with the 1994 UBC. Pres-sure coefficients (from UBC table 16-H) for windward roof slopes between 2/12 and 9/12 are -0.9 (outward) and 0.3 (inward). It is impor-tant to recognize the significant impact that wind direction (inward or outward) has on calculated eave loads. The 3.05 psf design windward roof pressure listed in table 9.2 was calculated using the 0.3 pressure coefficient. When combined with the negative pressure of 7.12 psf on the leeward roof, the net lateral roof pressure is 10.17 psf. If the –0.9 pressure coefficient would have been used, the net lateral roof pressure would have been –2.03 psf.
9.4.2 Fixity Factors, f. Based on the assump-tion of a post fixed at the groundline (see Sec-tion 9.3.1), a fixity factor of 3/8 is appropriate for this example.
9.4.3 Eave Load, R. Since this example uses symmetrical base restraint and frame geometry, equation 5-6 may be used.
R = s [hr (qwr – qlr) + hw f (qww – qlw)]
where:
hr = (36 ft /2) (4/12) = 6 ft hw = 12 ft
s = 10 ft f = 0.375
or
R = 10 ft [6 ft (3.05 psf + 7.12 psf) + 12 ft (.375)(8.13 psf + 5.08 psf)]
R = 1205 lbf
For later calculations, it is convenient to calcu-late R in terms of its components – roof, wind-ward wall and leewind-ward wall.
RR = 10(6)(3.05 + 7.12) = 610.2 lbf RW = 10(12)(.375)(8.13) = 365.8 lbf RL = 10(12)(.375)(-5.08) = -228.6 lbf RR + RW – RL = 1205 lbf
9.5 Step 4: Load Distribution
9.5.1 Introduction. For this example problem, diaphragm shear stiffness, Ch, frame stiffness, k, endwall stiffness, ke, and eave load, R, are all constant. Consequently, in addition to analysis methods such as DAFI, load distribution can be determined using the ANSI/ASAE EP484.2 ta-bles (Section 5.6.3) and the simple beam anal-ogy equations (Section 5.5.6). For comparison purposes, all three methods are demonstrated here (Sections 9.5.2 – 9.5.4). The information obtained from these analyses is then used to determine the maximum diaphragm shear force (Section 9.5.6) and maximum post forces (Sec-tion 9.5.7).
9.5.2 ANSI/ASAE EP484.2 Tables. In this de-sign manual, the ANSI/ASAE EP484.2 tables are tables 5.1 and 5.2. Table 5.1 contains shear force modifiers or “mS” values. The product of this modifier and eave load, R, is the maximum shear force in the diaphragm, Vh. Table 5.2 con-tains sidesway restraining force modifiers or
“mD” values. The product of this modifier and eave load, R, is referred to as the horizontal re-straining force, Q, which is the amount of eave load transferred away from the center post-frame(s) by the diaphragm.
Use of tables 5.1 and 5.2 requires two ratio: ke/k and Ch/k. For this example analysis, both ratios are equal to 69 (12640/184). Using linear inter-polation, the mS value from table 5.1 is 2.77, and the mD value from table 5.2 is 0.90.
The maximum diaphragm shear force, Vh, which occurs adjacent to each endwall, is given as:
Vh = mS R = 2.77(1205 lbf) = 3340 lbf The horizontal restraining force, Q, that must be applied to the center post frame (i.e., frame 4 in figure 9.1) is given as:
Q = mD R = 0.90(1205 lbf) = 1085 lbf The difference between eave load, R, and the horizontal restraining force, Q, is the amount of the eave load that is transferred by the center post-frame to the foundation.
R – Q = 120 lbf
The eave deflection, ∆, for a post-frame with stiffness, k, subjected to an eave load, R, and horizontal restraining force, Q, is given as:
∆ = (R – Q) / k
Eave deflection for the center post-frame is given as:
∆ = (1205 lbf – 1085 lbf) / 184 lbf/in.
∆ = 0.652 in.
9.5.3 Simple Beam Analogy Equations. As previously noted, the simple beam analogy equations for diaphragm shear force, Vh, and diaphragm displacement, y, can be used when R, k, ke and Ch are constant. These two equa-tions are given in Section 5.6.6 as:
Vh = Ch α s [A sinh(α x) + B cosh(α x)]
y = A cosh(α x) + B sinh(α x) + R/k
Input parameters and calculated equation con-stants for the simple beam analogy equations have been compiled for this example analysis in Table 9.4.
Maximum diaphragm shear is calculated by set-ting x = 0 in., or:
Vh = 12,640 lbf/in.(1.0054x10-3 in.-1) •(120 in.)[-6.286 in.(0) + 2.181 in.(1)]
Vh = 3326 lbf
Maximum diaphragm displacement is calculated by setting x = L/2 = 360 in. , or:
y = -6.286 in.( 1.0662) + 2.181 in.( 0.3699) + 1205 lbf/(184 lbf/in.)
y = 0.6535 in.
Table 9.4. Parameters for Simple Beam Anal-ogy Equations
R 1205 lbf
s 120 in.
L 720 in.
ke 12,640 lbf/in.
k 184 lbf/in.
R/ k 6.549 in.
Ch 12,640 lbf/in.
α 1.0054x10-3 in.-1 *
α L 0.7239
cosh(α L) 1.2737
sinh(α L) 0.7888
D -23.890 *
ye 0.2631 in. *
A -6.286 in. *
B 2.181 in. *
cosh(0) 1
sinh(0) 0
cosh(α 360 in.) 1.0662
sinh(α 360 in.) 0.3699
* Equations for calculation of these values are given in Section 5.6.6.
The force transferred to the foundation by the center frame (frame 4) is equal to the product of eave displacement, y, and frame stiffness, k, or:
y k = 0.6535 in. (184 lbf/in.) = 120.2 lbf The horizontal restraining force, Q, for the frame 4 is equal to the difference between the eave load, R, and the 120.2 lbf, or
Q = 1205 lbf – 120.2 lbf = 1084.8 lbf Note that ye in table 9.4 is the eave displace-ment of the endwall.
9.5.4 DAFI. As previously mentioned, DAFI is a computer program specifically written for deter-mining load distribution between diaphragms and frames. DAFI can be downloaded free from the NFBA web site (www.postframe.org).
The maximum shear force in the diaphragm. Vh, is numerically equal to the load resisted by the endwall frame. In figure 9.2, this value is given as 3353.2 lbf. Note that this value is more pre-cise than the 3340 lbf value calculated from the mS values in table 5.1 because the values in table 5.1 are only given to three significant fig-ures. It is important to note that the shear load
FRAME FRAME APPLIED HORIZONTAL LOAD RESISTED FRACTION OF