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Chapter 7. Flexure and Axially Loaded Wood Structures

Chapter 7. Flexure and Axially Loaded Wood Structures

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118



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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119



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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Flexure and Axially Loaded Wood Structures



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



73397_C007.indd 126



 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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