Chapter 7. Flexure and Axially Loaded Wood Structures
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Principles of Structural Design: Wood, Steel, and Concrete
Fbn′ is the adjusted LRFD reference design value for bending. To start
with, the reference bending design value, Fb for the appropriate species and
grade is obtained. These values are listed in Appendices B.2 through B.4
for sawn lumber and Appendices B.7 through B.9 for GLULAM and LVL.
Then the value is adjusted by multiplying the reference value by a string of
factors. The applicable adjustment factors for sawn lumber were given in
Table 6.5, and for GLULAM in Table 6.6 and for LVL in Table 6.7.
For sawn lumber, the adjusted reference bending design value is
restated as
Fbn′ = φFbλC M CtCFCrC fu CiCL K F
(7.2)
For GLULAM and SCL, the adjusted reference bending design value is
restated as
Fbn′ = φFbλC M CtCcC fu (Cv or CL )K F
Figure 7.1 Buckling
due to compression.
(7.3)
where
Fb is the tabular reference bending design value
ϕ is the resistance factor for bending = 0.85
λ is the time factor (Table 6.2)
CM is the wet-service factor
Ct is the temperature factor
CF is the size factor
Cr is the repetitive member factor
Cfu is the flat use factor
Ci is the incision factor
CL is the beam stability factor
Cc is the curvature factor
Cv is the volume factor
KF is the format conversion factor = 2.16/ϕ
Using the assessed value of Fbn′ , from Equations 7.2 or 7.3, based on the adjustment factors
known initially, the required section modulus, S is determined from Equation 7.1 and a trial section
is selected having the section modulus S higher than the computed value. In the beginning, some
section-dependent factors such as CF, Cv, and CL will not be known while the others such as λ, KF,
and ϕ will be known. The design is performed considering all possible load combinations along with
the relevant time factor. If loads are of one type only, i.e., all vertical or all horizontal, the highest
value of the combined load divided by the relevant time factor determines which combination is
critical for design.
Based on the trial section, all adjustment factors including CL are then computed and the magnitude of Fbn′ is reassessed. A revised S is obtained from Equation 7.1 and the trial section is modified,
if necessary.
Beam Stability Factor, CL
As stated earlier, the compression stress, besides causing an axial deformation, can cause a lateral
deformation if the compression zone of the beam is not braced against the lateral movement. In the
presence of the stable one-half tensile portion, the buckling in the plane of loading is prevented.
However, the movement could take place sideways (laterally), as shown in Figure 7.2.
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Flexure and Axially Loaded Wood Structures
p
b
d
(a)
(b)
(c)
Figure 7.2 Buckling of a bending member. (a) Original position of the beam. (b) Deflected position without lateral instability. (c) Compression edge buckled laterally.
The bending design described in Chapter 6 had assumed that no buckling was present and adjustments were made for other factors only. The condition of no buckling is satisfied when the bracing
requirements, as listed in Table 7.1, are met. In general, when the depth-to-breadth ratio is 2 or less,
no lateral bracings are required. When the depth-to-breadth ratio is more than 2 but does not exceed
4, the ends of the beam should be held in position by one of these methods: full-depth solid blocking, bridging, hangers, nailing, or bolting to other framing members. The stricter requirements are
stipulated to hold the compression edge in line for a depth-to-breadth ratio of higher than 4.
When the requirements of Table 7.1 are not met, the following beam stability factor has to be
applied to account for the buckling effect.
2
1+ α
1+ α α
CL =
−
−
1.9
1.9 0.95
(7.4)
Table 7.1
Bracing Requirements for Lateral Stability
Depth/Breadth Ratioa
≤2
>2 but ≤4
>4 but ≤5
>5 but ≤6
>6 but ≤7
Combined bending and
compression
≤1
>1
Bracing Requirements
Sawn lumber
No lateral bracing required
The ends are to be held in position, as by full-depth solid blocking, bridging, hangers, nailing,
or bolting to other framing members, or other acceptable means
The compression edge is to be held in line for its entire length to prevent lateral displacement,
as by sheathing or subflooring, and the ends at points of bearing are to be held in position to
prevent rotation and/or lateral displacement
Bridging, full-depth solid blocking, or diagonal cross bracing is to be installed at intervals not
exceeding 8 ft, the compression edge is to be held in line for its entire length to prevent lateral
displacement, as by sheathing or subflooring, and the ends at points of bearing are to be held
in position to prevent rotation and/or lateral displacement
Both edges of a member are to be held in line for their entire length, and the ends at points of
bearing are to be held in position to prevent rotation and/or lateral displacement
The depth/breadth ratio may be as much as 5 if one edge is held firmly in line. If under all load
conditions, the unbraced edge is in tension, the depth/breadth ratio may be as much as 6
Glued laminated timber
No lateral bracing required
The compression edge is supported throughout its length to prevent lateral displacement, and
the ends at point of bearing are laterally supported to prevent rotation
Nominal dimensions.
a
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Principles of Structural Design: Wood, Steel, and Concrete
where
α=
FbEn
Fbn′*
(7.5)
where
F bn
′∗ is the reference bending design value adjusted for all factors except Cv, Cfu, and CL
FbEn is the Euler-based LRFD critical buckling stress for bending
FbEn =
1.2 E ′y min (n )*
RB2
(7.6)
*
where
′ n ) is the adjusted nominal stability modulus of elasticity
Emin(
R B is the slenderness ratio for bending
RB =
Le d
≤ 50
b2
(7.7)
where Le is the effective unbraced length, discussed in the “Effective Unbraced Length” section
below.
When R B exceeds 50 in Equation 7.7, the beam dimensions should be revised to limit the slenderness ratio to 50.
Effective Unbraced Length
The effective unbraced length is a function of several factors such as the type of span (simple, cantilever, continuous), the type of loading (uniform, variable, concentrated loads), the unbraced length,
Lu which is the distance between the points of lateral supports, and the size of the beam.
For a simple one span or cantilever beam, the following values can be conservatively used for
the effective length.
For
For 7 ≤
For
Lu
< 7, Le = 2.06 Lu
d
Lu
≤ 14.3, Le = 1.63Lu + 3d
d
Lu
> 14.3, Le = 1.84 Lu
d
(7.8)
(7.9)
(7.10)
Example 7.1
A 5-½ in. × 24 in. glued laminated timber beam is used for a roof system having a span of 32 ft,
which is braced only at the ends. GLULAM consists of the Douglas Fir 24F-1.8E. Determine the
beam stability factor. Use the dead and live conditions only.
* Use Y-axis.
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121
Solution
1. Reference design values
Fb = 2400 psi
E = 1.8 × 106 psi
Ey (min) = 0.83 × 106 psi
2. Adjusted design values
Fbn* = λFbφKF
= (0.8)( 2400)(φ)
( 2.16) = 4147 psi or 4.15 ksi
φ
Emin(
′ n) = Ey (min)KF
1 .5
= 1.25 × 106 psi or 1.2
25 × 103 ksi
= 0.83 × 106 (φ)
φ
(
)
3. Effective unbraced length
Lu 32 × 12
=
= 16 > 14.3
d
24
From Equation 7.10
Le = 184
. Lu = 184
. (32) = 58.88 ft
4. From Equation 7.7
Led
b2
RB =
(58.88 × 12)(14)
(5.5)2
=
= 23.68 < 50
5. FbEn =
1.2Emin
′ ( n)
RB2
=
OK
(
1.2 1.25 × 103
( 23.68)
2
) = 2.675 ksi
2.675
FbEn
=
= 0.645
*
4.15
Fbn
6.α =
1+ α
1+ α
α
−
−
7. Cl =
1.9
1.9
0.95
2
2
=
1.645
1.645
0.645
−
−
1 .9
1.9
9
0.95
= 0 .6
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Principles of Structural Design: Wood, Steel, and Concrete
Shear Criteria
A transverse loading applied to a beam results in vertical shear stresses in any transverse (vertical)
section of a beam. Due to the complimentary property of shear, an associated longitudinal shear
stress acts along the longitudinal plane (horizontal face) of a beam element. In any mechanics of
materials text, it can be seen that the longitudinal shear stress distribution across the cross section
is given by
fv =
VQ
I
(7.11)
where
f v is the shear stress at any plane across the cross section
V is the shear force along the beam at the location of the cross section
Q is the moment of the area above the plane where stress is desired to the top or bottom edge of
the section. Moment is taken at neutral axis
I is the moment of inertia along the neutral axis
Equation 7.11 also applies for the transverse shear stress at any plane of the cross section as well
because the transverse and the longitudinal shear stresses are complimentary, numerically equal
and opposite in sign.
For a rectangular cross section that is usually the case with wood beams, the shear stress distribution by above relation is parabolic with the following maximum value at the center.
fv max = Fvn′ =
3 Vu
2 A
(7.12)
In terms of Vu, the basic equation for shear design of beam is
Vu = 2 / 3Fvn′ A
(7.13)
where
Vu is the maximum shear force due to factored load on beam
Fvn′ is the adjusted reference shear design value
A is the area of the beam
The NDS permits that the maximum shear force, Vu, might be taken to be the shear force at a distance equal to the depth of the beam from the support. However, usually Vu is usually taken to be
the maximum shear force from the diagram, which is at the support for a simple span.
For sawn lumber, the adjusted reference shear design value is
Fvn′ = φFv λC M CtCi K F
(7.14)
For GLULAM and SCL, the adjusted reference shear design value is
Fvn′ = φFv λC M Ct K F
(7.15)
where
Fv is the tabular reference shear design value
ϕ is the resistance factor for shear = 0.75
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Flexure and Axially Loaded Wood Structures
λ is the time factor (see the “Time Effect Factor, λ” section)
CM is the wet-service factor
Ct is the temperature factor
Ci is the incision factor
KF is the format conversion factor = 2.16/ϕ
Deflection Criteria
The actual maximum deflection for the beam is calculated using the service load (not the factored
load). The maximum deflection is a function of the type of loading, type of beam span, moment of
inertia of the section, and modulus of elasticity. For a uniformly loaded simple span member, the
maximum deflection at mid-span is
δ=
5wL4
384 E ′I
(7.16)
where
w is the uniform combined service load per unit length
L is the span of beam
E′ is the adjusted modulus of elasticity
E ′ = EC M Ct Ci
E is the reference modulus of elasticity
I is the moment of inertia along neutral axis
The actual maximum deflection should be less than or equal to the allowable deflections, Δ. Often
a check is made for live load alone as well as for the total load. Thus,
Max. δ L ≤ allow. ∆ L
(7.17)
Max. δ TL ≤ allow. ∆ TL
(7.18)
The allowable deflections are given in Table 7.2.
Table 7.2
Recommended Deflection Criteria
Classification
Roof beams
No ceiling
Without plaster ceiling
With plaster ceiling
Floor beamsa
Highway bridge stringers
Railway bridge stringers
Live or Applied Load Only
Dead Load Plus Applied Load
Span/180
Span/240
Span/360
Span/360
Span/300
Span/300–Span/400
Span/120
Span/180
Span/240
Span/240
Source: American Institute of Timber Association, Timber Construction Manual, 5th edn., John
Wiley & Sons, New York, 2005.
a Additional limitations are used where increased floor stiffness or reduction of vibrations is desired.
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Principles of Structural Design: Wood, Steel, and Concrete
When the above criteria are not satisfied, a new beam size is determined using the allowable
deflection as a guide and computing the desired moment of inertia on that basis.
Example 7.2
Design the roof rafters spanning 16 ft and spaced 16 in. on center (OC). The plywood roof sheathing prevents the local buckling. The dead load is 12 psf and the roof live load is 20 psf. Use
Douglas Fir-Larch #1 wood.
Solution
A. Loads
16
× 1 = 1.333 ft 2 /ft
12
1. Tributary area/ft =
2. Loads per feet
wD = 12 × 1.333 = 16 lb/ft
wL = 20 × 1.333 = 26.66 lb/ft
3. Loads combination
w u = 1.2w D + 1.6w L
= 1.2 (16) + 1.6 ( 26.66) = 61.86 lb/ft
4. Maximum BM
w uL2 (61.86 )(16 )
=
= 1974.52 ft lb or 23.75 in.-k
8
8
2
Mu =
5. Maximum shear
Vu =
w uL (61.86)(16)
=
= 494.9 lb
2
2
B. Reference design values (Douglas Fir-Larch #1, 2 in. and wider)
1. Fb = 1000 psi
2. Fv = 180 psi
3. E = 1.7 × 106 psi
4. Emin = 0.62 ×106 psi
C. Preliminary design
1. Initially adjusted bending design value
Fbn
′ (estimated) = φλFbKFCr
= φ (0.8)(1000)( 2.16 /φ)(1.15) = 1987 psi
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2. Sreqd =
( 23.75 × 1000) = 11.95 in.3
Mu
=
Fbn
1987
′ (estimated)
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Flexure and Axially Loaded Wood Structures
3. Try 2 in. × 8 in. S = 13.14 in.3
A = 10.88 in.2
I = 47.63 in.4
D. Revised design
1. Adjusted reference design values
Reference Design
Values (psi)
λ
CF
Cr
1000
0.8
1.2
1.15
2.16/ϕ
2384.6
Fvn′
180
0.8
2.16/ϕ
311
E′
1.7 × 10
0.62 × 106
—
—
—
Fbn′
a
6
Emin(
′ n)
a
F(′)n (psi)
KF
1.5/ϕ
1.7 × 106
0.93 × 106
Without the CL factor.
2. Beam stability factor CL = 1.0
E. Check for bending strength
Bending capacity = Fbn
′ S in.-k
=
( 2384.6)(13.14) = 31.33 in.-k > 23.75 in.k
1000
OK
F. Check for shear strength
2A
2
= 311 × 10.88 = 2255 lb > 494.5 lb
Shear capacity = Fvn′
3
3
OK
G. Check for deflection
1. Deflection is checked for service load, w = 16 + 26.66 = 42.66 lb/ft
2.δ =
3. Allowable deflection
5 wL4
5 ( 42.66)(16) (12)
= 0.78 in.
=
384 E ′I
384 1.7 × 106 ( 47.63)
4
∆=
(
3
)
L
16 × 12
=
= 1.07 in. > 0.78 in.
180
180
OK.
Example 7.3
A structural glued laminated timber is used as a beam to support a roof system. The tributary
width of the beam is 16 ft. The beam span is 32 ft. The floor dead load is 15 psf and the live load is
40 psf. Use Douglas Fir GLULAM 24F-1.8E. The beam is braced only at the supports.
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Principles of Structural Design: Wood, Steel, and Concrete
Solution
A. Loads
1. Tributary area/ft = 16 × 1 = 16 ft 2/ft
2. Loads per ft
wD = 15 × 16 = 240 lb/ft
wL = 40 × 16 = 640 lb/ft
3. Design load, w u = 1.2w D + 1.6w L
= 1.2 ( 240) + 1.6 (640) = 1312 lb/ft or 1.31k/ft
4. Design bending moment
w uL2 (1.31)(32)
=
= 167.88 ft-k or 2012.16 in.-k
8
8
2
Mu =
5. Design shear
Vu =
w uL 1.31(32)
=
= 20.96 k
2
2
B. Reference design values
Fb = 2400 psi
Fv = 265 psi
E = 1.8 × 106 psi
Ey(min) = 0.83 × 106 psi
C. Preliminary design
1. Initially adjusted bending reference design value
= (0.8)(2400)ϕ(2.16/ϕ) = 4147 psi or 4.15 ksi
2012.16
2. Sreqd =
= 484.86 in.3
4.15
Try 5 1/2 in. × 24 in. S = 528 in.3
A = 132 in.2
I = 6336 in.4
D. Revised adjusted design values
Type
Reference Design
Values (psi)
λ
KF
F(′)n (psi)
Fbn′ a
2400
0.8
2.16/ϕ
4147.2 (Fba)
Fvn′
265
0.8
2.16/ϕ
457.9
E′
Emin(
′ n)
1.8 × 106
0.83 × 106
—
—
—
1.5/ϕ
1.8 × 106
1.25 × 106
E. Volume factor, Cv
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5.125
Cv =
b
1/ 10
12
d
1/ 10
21
L
5.125
=
5.5
1/ 10
12
24
1/ 10
1/ 10
21
32
1/ 10
= 0.89
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Flexure and Axially Loaded Wood Structures
F. Beam stability factor, CL
From Example 7.1, CL = 0.60
Since CL < Cv, use the CL factor
G. Bending capacity
1. Fbn
′ = (4147.2) (0.6) = 2488.32 psi or 2.488 ksi
2. Moment capacity = Fbn
′S
= 2.488(528)
= 1313.7 in.-k < 2012.16(Mu) NG
A revised section should be selected and steps E, F, and G should be repeated.
H. Check for shear strength*
2A
2
= 457.9 × 132 = 40295 lb or 40.3 k > 20.29 k
Shear capacity = Fvn′
3
3
OK
I. Check for deflection
1. Deflection checked for service load w = 240 + 640 = 880 lb/ft
4
3
5 wL4
5 (880)(32) (12)
2. δ =
=
= 1.82 in.
384 EI
384 1.8 × 106 (6336)
(
)
3. Permissible deflection
∆=
L
32 × 12
=
= 2.13 in. > 1.82 in.
180
180
OK
Bearing at Supports
The bearing perpendicular to the grains occurs at the supports or wherever a load bearing member
rests onto the beam, as shown in Figure 7.3. The relation for bearing design is
Pu = FC′ ⊥ n A
(7.19)
Pu
Bearing
area
R
Figure 7.3 Bearing perpendicular to grain.
* Based on the original section.
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Principles of Structural Design: Wood, Steel, and Concrete
The adjusted compressive design value perpendicular to grain is obtained by multiplying the reference design value by the adjustment factors. Including these factors, Equation 7.19 becomes
For sawn lumber
Pu = φFC ⊥ λC M ClCiCb K F A
(7.20)
Pu = φFC ⊥ λC M CtCb K F A
(7.21)
For GLULAM
where
Pu is the reaction at the bearing surface due to factored load on the beam
FC ⊥ is the reference compressive design value perpendicular to grain
FC′ ⊥ n is the adjusted compressive design value perpendicular to grain
ϕ is the resistance factor for compression = 0.9
λ is the time effect factor (see the “Time Effect Factor, λ” section in Chapter 6)
CM is the wet-service factor
Ct is the temperature factor
Ci is the incision factor
Cb is the bearing area factor as discussed below
KF is the format conversion factor for bearing = 1.875/ϕ
A is the area of bearing surface
Bearing Area Factor, Cb
The bearing area factor is applied only to a specific case when the bearing length lb is less than 6 in.
and also the distance from the end of the beam to the start of the contact area is larger than 3 in.,
as shown in Figure 7.4. The factor is not applied to the bearing surface at the end of a beam that
may be of any length or where the bearing length is 6 in. or more at any other location than the end.
This factor accounts for the additional wood fibers that could resist the bearing load. It increases the
bearing length by 3/8 in. Thus,
Cb =
lb + 3 / 8
lb
(7.22)
where lb is the bearing length is the contact length parallel to the grain.
Less than
6 in.
lb +
3
in.
8
lb
More than 3 in.
End of beam
Figure 7.4 Bearing area factor.
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