2015ndsexamples2015 nds examples … · 2021. 1. 7. · (nds®) for wood construction (ansi/awc...

2015 NDS Examples – Beams, Columns, and2015 NDS Examples Beams, Columns, and Beam-Columns (DES220)John “Buddy” Showalter, PE Lori Koch, PEy ,Vice President, Technology TransferAmerican Wood Council

,Manager, Educational Outreach American Wood Council

Copyright Materials

This presentation is protected by US and InternationalThis presentation is protected by US and International Copyright laws. Reproduction, distribution, display and use of the presentation without written permission of AWC is

prohibited.

© American Wood Council 2016

2

• The American Wood Council is a Registered Provider with The American I tit t f A hit t C ti i

• This course is registered with AIA CES for continuing professional education. A h it d t i l d t tInstitute of Architects Continuing

Education Systems (AIA/CES), Provider # 50111237.

As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or

• Credit(s) earned on completion of this course will be reported to AIA CES for AIA members. Certificates of

any method or manner ofhandling, using, distributing, or dealing in any material or product.

Q ti l t d t ifi t i lAIA members. Certificates of Completion for both AIA members and non-AIA members are available upon request.

• Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation.

Description

• For those seeking practical application of the provisions of the National Design Specification®provisions of the National Design Specification® (NDS®) for Wood Construction (ANSI/AWC NDS-2015) which is referenced in the 2015 International Building Code, this presentation provides several design examples including beams, columns, and structural elements under combined bending andstructural elements under combined bending and axial loading. Design provisions and equations from the 2015 NDS and reference design values from the 2015 NDS Supplement will be used to calculate2015 NDS Supplement will be used to calculate capacities for these elements under various loading conditions. Each example will include discussion of design value adjustment factors and load combinations.

Learning Objectives

1. Understand application of NDS design provisions for beams, columns and structural elements under combined bendingcolumns, and structural elements under combined bending and axial loading.

2. Be familiar with reference design values from the NDS Supplement.

3. Be familiar with design value adjustment factors from the NDS and NDS Supplement.pp

4. Be familiar with the impact of combinations of loads of different durations on design of structural wood members.

Polling QuestionWhat is your profession?

a) Architecta) Architectb) Engineerc) Code Officialc) Code Officiald) Building Designere) Othere) Other

6

General Notes

• Example Problems completed in MATHCAD• Small green lines under some variables have no impact

on our problems, these are a MATHCAD feature• PDF of MATHCAD problems has been provided, MATHCADPDF of MATHCAD problems has been provided, MATHCAD

sheets are not available for distribution• Free Design Example Publication planned for 2017• 1991 NDS Commentary

• http://www.awc.org/codes-standards/publications/nds-1991-1997standards/publications/nds 1991 1997

AMERICAN WOOD COUNCIL

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29NATIONAL DESIGN SPECIFICATION FOR WOOD CONSTRUCTION

4.3.5 Beam Stability Factor, CL

Reference bending design values, Fb, shall be mul-tiplied by the beam stability factor, CL, specified in 3.3.3.

4.3.6 Size Factor, CF

4.3.6.1 Reference bending, tension, and compres-sion parallel to grain design values for visually graded dimension lumber 2" to 4" thick shall be multiplied by the size factors specified in Tables 4A and 4B.

4.3.6.2 Where the depth of a rectangular sawn lum-ber bending member 5" or thicker exceeds 12", the ref-erence bending design values, Fb, in Table 4D shall be multiplied by the following size factor:

1 9FC (12 / d) 1.0 (4.3-1)

4.3.6.3 For beams of circular cross section with a diameter greater than 13.5", or for 12" or larger square beams loaded in the plane of the diagonal, the size fac-

tor shall be determined in accordance with 4.3.6.2 on the basis of an equivalent conventionally loaded square beam of the same cross-sectional area.

4.3.6.4 Reference bending design values for all species of 2" thick or 3" thick Decking, except Red-wood, shall be multiplied by the size factors specified in Table 4E.

4.3.7 Flat Use Factor, Cfu

When sawn lumber 2" to 4" thick is loaded on the wide face, multiplying the reference bending design value, Fb, by the flat use factors, Cfu, specified in Tables 4A, 4B, 4C, and 4F, shall be permitted.

4.3.8 Incising Factor, Ci

Reference design values shall be multiplied by the following incising factor, Ci, when dimension lumber is incised parallel to grain a maximum depth of 0.4", a maximum length of 3/8", and density of incisions up to

Table 4.3.1 Applicability of Adjustment Factors for Sawn Lumber

ASD only

ASD and LRFD LRFD only

Load

Dur

atio

n Fa

ctor

Wet

Ser

vice

Fac

tor

Tem

pera

ture

Fac

tor

Bea

m S

tabi

lity

Fact

or

Size

Fac

tor

Flat

Use

Fac

tor

Inci

sing

Fact

or

Rep

etiti

ve M

embe

r Fac

tor

Col

umn

Stab

ility

Fac

tor

Buc

klin

g St

iffne

ss F

acto

r

Bea

ring

Are

a Fa

ctor

Form

at C

onve

rsio

n Fa

ctor

Res

ista

nce

Fact

or

Tim

e Ef

fect

Fac

tor

KF

Fb' = Fb x CD CM Ct CL CF Cfu Ci Cr - - - 2.54 0.85

Ft' = Ft x CD CM Ct - CF - Ci - - - - 2.70 0.80

Fv' = Fv x CD CM Ct - - - Ci - - - - 2.88 0.75

Fc' = Fc x CD CM Ct - CF - Ci - CP - - 2.40 0.90

Fc' = Fc x - CM Ct - - - Ci - - - Cb 1.67 0.90 -

E' = E x - CM Ct - - - Ci - - - - - - -

Emin' = Emin x - CM Ct - - - Ci - - CT - 1.76 0.85 -

Internal AWC Use Only

Simply Supported Beam Capacity Check Example (ASD)

A Select Structural Douglas Fir-Larch 4X16 beam on a 20 ft span supports a hoist located at the center of thespan. Determine the maximum allowable load on the hoist (including its weight) based on bending. Assumenormal load duration. The beam is supported on a 2x4 top plate. Lateral support is provided only at the ends ofthe member and the ends are considered pinned.

Check beams capacity to resist shear stress from maximum (moment controlled) load; determine deflectionfrom maximum load and check bearing capacity.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.

Reference and Adjusted Design Values for 4x16 Select Structural DF-L (size adjusted 4x12 values)

Fb 1500 psi E 1900000 psi Emin 690000 psi (Table 4A)

Fc⊥ 625 psi Fv 180 psi

CD 1.0 CM 1.0 Ct 1.0

Cfu 1.0 Cr 1.0 Ci 1.0

CT 1.0 CF 1.0 (Table 4A 14" and wider)

E' E CM Ct Ci E'min Emin CM Ct Ci CT

E' 1900000 psi E'min 690000 psi

Member length and properties

l 20 ft b 3.5 in d 15.25 in wbearing 3.5 in

Ib d

3

12

Ag b d Sb d

2

6

Ag 53.38 in2

S 135.66 in3

I 1034 in4

Beam Stability Factor

F'b* Fb CD CM Ct CF Ci Cr F'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL and flat use factorCfu applied. The following calculations determine the beamstabilty factor CL:

F'b* 1500 psi

lu 12in

ft l lu 240 in laterally unsupported length

lu

d15.7 lu/d >7 (Table 3.3.3)

le 1.37 lu 3 d le 375 in (Table 3.3.3)

RB

le d

b2

RB 21.6 slenderness ratio for bending (3.3-5)

FbE1.20 E'min

RB2

FbE 1776 psi critical bucking design value for bending (3.7.1)

CL

1FbE

F'b*

1.9

1FbE

F'b*

1.9

2FbE

F'b*

0.95

CL 0.876

F'b F'b* Cfu CL

F'b 1313 psi F'b is adjusted bending design value with all adjustmentfactors.

Determine Maximum Moment Allowed on Beam

Maximum total moment is the adjusted bending design value F'b times the section modulus S

Mmax F'bS

12in

ft

Mmax 14849 ft·lbf

Determine Maximum Hoist Load P

Maximum hoist load P is determined from subtracting moment due to beam weight from the maximum total momentallowed on the beam and solving for hoist load P. Load P creates a moment on beam length L of PL/4. Assumedensity of beam material is 34.2 lbs/ft3 (110% of tabulated of the specific gravity G for DF-L).

ρ 34.2lbf

ft3

wbeamweight ρb

12in

ft

d

12in

ft

wbeamweight 12.677 plf

Mbeamweight

wbeamweight l( )2

8 Mbeamweight 634 ft·lbf

Mallow Mmax Mbeamweight Mallow 14215 ft·lbf

P 4Mallow

l

Result: The total allowable concentrated moment-limited midspan load (hoist plus payload) is P 2843 lbf

Check Beam's Capacity to Resist Shear from Maximum (bending controlled) Load

VP

2 V 1421 lbf

fv3 V

2 b d fv 40 psi

F'v Fv CD CM Ct Ci F'v 180 psi fv<F'v okay

Check Compressive Stress at Bearing Points

fc⊥V

b wbearing fc⊥ 116 psi

F'c⊥ Fc⊥ CM Ct Ci F'c⊥ 625 psi fc⊥< F'c⊥ okay

Note: NDS Section 4.3.12 allows Fc⊥ to be increased by Cb as specified in Section 3.10.4. That increase was

not used in this example.

Check Deflection

Total deflection is the combination of deflection from beam weight and deflection from the applied crane load.Deflection from beam weight is considered long term deflection. Deflection from crane load may be consideredshort-term.

Δbeam_weight5

384 E I

wbeamweight

12

ft

in l 12

in

ft

4

Δbeam_weight 0.023 in

Δcrane_loadP

48 E Il 12

in

ft

3

Δcrane_load 0.417 in

Δtotal Δbeam_weight Δcrane_load Δtotal 0.44 in

Calculate Span/Deflection Ratio

12 lin

ft

Δtotal546 L/Δtotal


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3.2 Bending Members – General

3.2.1 Span of Bending Members

For simple, continuous and cantilevered bending members, the span shall be taken as the distance from face to face of supports, plus ½ the required bearing length at each end.

3.2.2 Lateral Distribution of Concentrated Load

Lateral distribution of concentrated loads from a critically loaded bending member to adjacent parallel bending members by flooring or other cross members shall be permitted to be calculated when determining design bending moment and vertical shear force (see 15.1).

3.2.3 Notches

3.2.3.1 Bending members shall not be notched ex-cept as permitted by 4.4.3, 5.4.5, 7.4.4, and 8.4.1. A gradual taper cut from the reduced depth of the member to the full depth of the member in lieu of a square-cornered notch reduces stress concentrations.

3.2.3.2 The stiffness of a bending member, as de-termined from its cross section, is practically unaffected by a notch with the following dimensions:

notch depth (1/6) (beam depth) notch length (1/3) (beam depth) 3.2.3.3 See 3.4.3 for effect of notches on shear

strength.

3.3 Bending Members – Flexure

3.3.1 Strength in Bending

The actual bending stress or moment shall not ex-ceed the adjusted bending design value.

3.3.2 Flexural Design Equations

3.3.2.1 The actual bending stress induced by a bending moment, M, is calculated as follows:

IbMc Mf

S (3.3-1)

For a rectangular bending member of breadth, b, and depth, d, this becomes:

b 2

M 6MfS bd

(3.3-2)

3.3.2.2 For solid rectangular bending members with the neutral axis perpendicular to depth at center:

I3

4bd moment of inertia, in.12

(3.3-3)

I 23bdS section modulus, in.

c 6 (3.3-4)


3.3.3.1 When the depth of a bending member does not exceed its breadth, d b, no lateral support is re-quired and CL = 1.0.

3.3.3.2 When rectangular sawn lumber bending members are laterally supported in accordance with 4.4.1, CL = 1.0.

3.3.3.3 When the compression edge of a bending member is supported throughout its length to prevent lateral displacement, and the ends at points of bearing have lateral support to prevent rotation, CL = 1.0.

3.3.3.4 Where the depth of a bending member ex-ceeds its breadth, d > b, lateral support shall be provid-ed at points of bearing to prevent rotation. When such lateral support is provided at points of bearing, but no additional lateral support is provided throughout the length of the bending member, the unsupported length,

u, is the distance between such points of end bearing, or the length of a cantilever. When a bending member is provided with lateral support to prevent rotation at intermediate points as well as at the ends, the unsup-ported length, u, is the distance between such points of intermediate lateral support.

3.3.3.5 The effective span length, e, for single span or cantilever bending members shall be determined in accordance with Table 3.3.3.



16 DESIGN PROVISIONS AND EQUATIONS

Table 3.3.3 Effective Length, e, for Bending Members

Cantilever1 where u/d < 7 where u

Uniformly distributed load e=1.33 u e=0.90 u + 3d

Concentrated load at unsupported end e=1.87 u e=1.44 u + 3d

Single Span Beam1,2 where u/d < 7 where u

Uniformly distributed load e=2.06 u e=1.63 u + 3d

Concentrated load at center with no inter-mediate lateral support

e=1.80 u e=1.37 u + 3d

Concentrated load at center with lateral support at center

e=1.11 u

Two equal concentrated loads at 1/3 points with lateral support at 1/3 points

e=1.68 u

Three equal concentrated loads at 1/4 points with lateral support at 1/4 points

e=1.54 u

Four equal concentrated loads at 1/5 points with lateral support at 1/5 points

e=1.68 u

Five equal concentrated loads at 1/6 points with lateral support at 1/6 points

e=1.73 u

Six equal concentrated loads at 1/7 points with lateral support at 1/7 points

e=1.78 u

Seven or more equal concentrated loads, evenly spaced, with lateral support at points of load application

e=1.84 u

Equal end moments e=1.84 u


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3.3.3.6 The slenderness ratio, RB, for bending members shall be calculated as follows:

eB 2

dRb

ed (3.3-5)

3.3.3.7 The slenderness ratio for bending members, RB, shall not exceed 50.

3.3.3.8 The beam stability factor shall be calculated as follows:

2* * *bE b bE b bE b

L

1 F F 1 F F F FC

1.9 1.9 0.95 (3.3-6)

where:

Fb* = reference bending design value multiplied by all applicable adjustment factors except Cfu, CV, and CL (see 2.3), psi

3.3.3.9 See Appendix D for background infor-

mation concerning beam stability calculations and Ap-pendix F for information concerning coefficient of vari-ation in modulus of elasticity (COVE).

3.3.3.10 Members subjected to flexure about both principal axes (biaxial bending) shall be designed in accordance with 3.9.2.

3.4 Bending Members – Shear

3.4.1 Strength in Shear Parallel to Grain (Horizontal Shear)

3.4.1.1 The actual shear stress parallel to grain or shear force at any cross section of the bending member shall not exceed the adjusted shear design value. A check of the strength of wood bending members in shear perpendicular to grain is not required.

3.4.1.2 The shear design procedures specified here-in for calculating fv at or near points of vertical support are limited to solid flexural members such as sawn lumber, structural glued laminated timber, structural composite lumber, or mechanically laminated timber beams. Shear design at supports for built-up compo-nents containing load-bearing connections at or near points of support, such as between the web and chord of a truss, shall be based on test or other techniques.

3.4.2 Shear Design Equations

The actual shear stress parallel to grain induced in a sawn lumber, structural glued laminated timber, struc-tural composite lumber, or timber pole or pile bending member shall be calculated as follows:

IvVQf

b (3.4-1)

For a rectangular bending member of breadth, b, and depth, d, this becomes:

v3Vf2bd

(3.4-2)

3.4.3 Shear Design

3.4.3.1 When calculating the shear force, V, in bending members:

(a) For beams supported by full bearing on one surface and loads applied to the opposite sur-face, uniformly distributed loads within a dis-tance from supports equal to the depth of the bending member, d, shall be permitted to be ig-nored. For beams supported by full bearing on one surface and loads applied to the opposite surface, concentrated loads within a distance, d, from supports shall be permitted to be multi-plied by x/d where x is the distance from the beam support face to the load (see Figure 3C).

Figure 3C Shear at Supports

minbE 2

B

1.20 EF

R


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(b) The largest single moving load shall be placed at a distance from the support equal to the depth of the bending member, keeping other loads in their normal relation and neglecting any load within a distance from a support equal to the depth of the bending member. This con-dition shall be checked at each support.

(c) With two or more moving loads of about equal weight and in proximity, loads shall be placed in the position that produces the highest shear force, V, neglecting any load within a distance from a support equal to the depth of the bend-ing member.

3.4.3.2 For notched bending members, shear force, V, shall be determined by principles of engineering me-chanics (except those given in 3.4.3.1).

(a) For bending members with rectangular cross section and notched on the tension face (see 3.2.3), the adjusted design shear, Vr', shall be calculated as follows:

2n

r v nd2V F bd

3 d (3.4-3)

where:

d = depth of unnotched bending member, in.

dn = depth of member remaining at a notch measured perpendicular to length of mem-ber, in.

Fv' = adjusted shear design value parallel to grain, psi

(b) For bending members with circular cross sec-tion and notched on the tension face (see 3.2.3), the adjusted design shear, Vr', shall be calculat-ed as follows:

2n

r v nd2V F A

3 d (3.4-4)

where:

An = cross-sectional area of notched member, in2

(c) For bending members with other than rectangu-lar or circular cross section and notched on the tension face (see 3.2.3), the adjusted design shear, Vr', shall be based on conventional engi-neering analysis of stress concentrations at notches.

(d) A gradual change in cross section compared with a square notch decreases the actual shear

stress parallel to grain nearly to that computed for an unnotched bending member with a depth of dn.

(e) When a bending member is notched on the compression face at the end as shown in Figure 3D, the adjusted design shear, Vr', shall be cal-culated as follows:

nr v

n

d d2V F b d e3 d

(3.4-5)

where:

e = the distance the notch extends from the inner edge of the support and must be less than or equal to the depth remaining at the notch, e

dn. If e > dn, dn shall be used to calculate fv using Equation 3.4-2, in.

dn = depth of member remaining at a notch meet-ing the provisions of 3.2.3, measured per-pendicular to length of member. If the end of the beam is beveled, as shown by the dashed line in Figure 3D, dn is measured from the in-ner edge of the support, in.

Figure 3D Bending Member End-Notched on Compression Face

3.4.3.3 When connections in bending members are

fastened with split ring connectors, shear plate connect-ors, bolts, or lag screws (including beams supported by such fasteners or other cases as shown in Figures 3E and 3I) the shear force, V, shall be determined by prin-ciples of engineering mechanics (except those given in 3.4.3.1).

(a) Where the connection is less than five times the depth, 5d, of the member from its end, the ad-justed design shear, Vr', shall be calculated as follows:

2e

r v ed2V F bd

3 d (3.4-6)



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where: for split ring or shear plate connections:

de = depth of member, less the distance from the unloaded edge of the member to the nearest edge of the nearest split ring or shear plate connector (see Figure 3E), in.

for bolt or lag screw connections:

de = depth of member, less the distance from the unloaded edge of the member to the center of the nearest bolt or lag screw (see Figure 3E), in.

(b) Where the connection is at least five times the depth, 5d, of the member from its end, the ad-justed design shear, Vr', shall be calculated as follows:

r v e2V F bd3

(3.4-7)

(c) Where concealed hangers are used, the adjusted design shear, Vr', shall be calculated based on the provisions in 3.4.3.2 for notched bending members.

Figure 3E Effective Depth, de, of Members at Connections

3.5 Bending Members – Deflection

3.5.1 Deflection Calculations

If deflection is a factor in design, it shall be calcu-lated by standard methods of engineering mechanics considering bending deflections and, when applicable, shear deflections. Consideration for shear deflection is required when the reference modulus of elasticity has not been adjusted to include the effects of shear deflec-tion (see Appendix F).

3.5.2 Long-Term Loading

Where total deflection under long-term loading must be limited, increasing member size is one way to

provide extra stiffness to allow for this time dependent deformation (see Appendix F). Total deflection, T, shall be calculated as follows:

T = Kcr LT + ST (3.5-1)

where:

Kcr = time dependent deformation (creep) factor

= 1.5 for seasoned lumber, structural glued laminated timber, prefabricated wood I-joists, or structural composite lumber used in dry service conditions as defined in 4.1.4, 5.1.4, 7.1.4, and 8.1.4, respectively.


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= 2.0 for structural glued laminated timber used in wet service conditions as defined in 5.1.4.

= 2.0 for wood structural panels used in dry service conditions as defined in 9.1.4.

= 2.0 for unseasoned lumber or for seasoned lumber used in wet service conditions as de-fined in 4.1.4.

= 2.0 for cross-laminated timber used in dry service conditions as defined in 10.1.5.

LT = immediate deflection due to the long-term component of the design load, in.

ST = deflection due to the short-term or normal component of the design load, in.

3.6 Compression Members – General

3.6.1 Terminology

For purposes of this Specification, the term “col-umn” refers to all types of compression members, in-cluding members forming part of trusses or other struc-tural components.

3.6.2 Column Classifications

3.6.2.1 Simple Solid Wood Columns. Simple col-umns consist of a single piece or of pieces properly glued together to form a single member (see Figure 3F).

3.6.2.2 Spaced Columns, Connector Joined. Spaced columns are formed of two or more individual members with their longitudinal axes parallel, separated at the ends and middle points of their length by blocking and joined at the ends by split ring or shear plate connectors capable of developing the required shear resistance (see 15.2).

3.6.2.3 Built-Up Columns. Individual laminations of mechanically laminated built-up columns shall be designed in accordance with 3.6.3 and 3.7, except that nailed or bolted built-up columns shall be designed in accordance with 15.3.

3.6.3 Strength in Compression Parallel to Grain

The actual compression stress or force parallel to grain shall not exceed the adjusted compression design value. Calculations of fc shall be based on the net sec-tion area (see 3.1.2) where the reduced section occurs in the critical part of the column length that is most sub-ject to potential buckling. Where the reduced section does not occur in the critical part of the column length that is most subject to potential buckling, calculations of fc shall be based on gross section area. In addition, fc based on net section area shall not exceed the reference

compression design value parallel to grain multiplied by all applicable adjustment factors except the column stability factor, CP.

Figure 3F Simple Solid Column

3.6.4 Compression Members Bearing End to End

For end grain bearing of wood on wood, and on metal plates or strips see 3.10.

3.6.5 Eccentric Loading or Combined Stresses

For compression members subject to eccentric loading or combined flexure and axial loading, see 3.9 and 15.4.



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and 2

c b1

cE2 bE

f f1.0

F F (3.9-4)

where: for either uniaxial edgewise bending or biaxial bending

and for uniaxial flatwise bending or biaxial bending

and

minb1 bE 2

B

1.20 Ef F

(R ) for biaxial bending

fb1 = actual edgewise bending stress (bending load applied to narrow face of member) , psi

fb2 = actual flatwise bending stress (bending load applied to wide face of member) , psi

d1 = wide face dimension (see Figure 3H), in.

d2 = narrow face dimension (see Figure 3H), in.

Effective column lengths, e1 and e2, shall be de-termined in accordance with 3.7.1.2. Fc', FcE1, and FcE2 shall be determined in accordance with 2.3 and 3.7. Fb1', Fb2', and FbE shall be determined in accordance with 2.3 and 3.3.3.

3.9.3 Eccentric Compression Loading

See 15.4 for members subjected to combined bend-ing and axial compression due to eccentric loading, or eccentric loading in combination with other loads.

Figure 3H Combined Bending and Axial Compression

3.10 Design for Bearing

3.10.1 Bearing Parallel to Grain

3.10.1.1 The actual compressive bearing stress par-allel to grain shall be based on the net bearing area and shall not exceed the reference compression design value parallel to grain multiplied by all applicable adjustment factors except the column stability factor, CP.

3.10.1.2 Fc*, the reference compression design val-

ues parallel to grain multiplied by all applicable ad-justment factors except the column stability factor, ap-plies to end-to-end bearing of compression members provided there is adequate lateral support and the end cuts are accurately squared and parallel.

3.10.1.3 When fc > (0.75)(Fc*) bearing shall be on a

metal plate or strap, or on other equivalently durable, rigid, homogeneous material with sufficient stiffness to distribute the applied load. Where a rigid insert is re-quired for end-to-end bearing of compression members,

it shall be equivalent to 20-gage metal plate or better, inserted with a snug fit between abutting ends.

3.10.2 Bearing Perpendicular to Grain

The actual compression stress perpendicular to grain shall be based on the net bearing area and shall not exceed the adjusted compression design value per-pendicular to grain, fc Fc '. When calculating bearing area at the ends of bending members, no allowance shall be made for the fact that as the member bends, pressure upon the inner edge of the bearing is greater than at the member end.

minc cE1 2

e1 1

0.822 Ef F

( / d )2e1 1/ d )e1 1

minc cE2 2

e2 2

0.822 Ef F

( / d )2e2 2/ d )e2 2


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3.10.3 Bearing at an Angle to Grain

The adjusted bearing design value at an angle to grain (see Figure 3I and Appendix J) shall be calculated as follows:

*c c

* 2 2c c

F FF

F sin F cos (3.10-1)

where:

= angle between direction of load and direction of grain (longitudinal axis of member), de-grees

3.10.4 Bearing Area Factor, Cb

Reference compression design values perpendicular to grain, Fc , apply to bearings of any length at the ends of a member, and to all bearings 6" or more in length at any other location. For bearings less than 6" in length and not nearer than 3" to the end of a member, the ref-erence compression design value perpendicular to grain, Fc , shall be permitted to be multiplied by the following bearing area factor, Cb:

bb

b

0.375C b 0.37

b

(3.10-2)

where:

b = bearing length measured parallel to grain, in.

Equation 3.10-2 gives the following bearing area factors, Cb, for the indicated bearing length on such small areas as plates and washers:

Table 3.10.4 Bearing Area Factors, Cb

b 0.5" 1" 1.5" 2" 3" 4" 6" or moreCb 1.75 1.38 1.25 1.19 1.13 1.10 1.00

For round bearing areas such as washers, the bear-

ing length, b, shall be equal to the diameter.

Figure 3 Bearing at an Angle to Grain



STR

UC

TU

RA

L GLU

ED

LAM

INATED

TIM

BER


5

5.3.4 Temperature Factor, Ct

When structural members will experience sus-tained exposure to elevated temperatures up to 150 F (see Appendix C), reference design values shall be multiplied by the temperature factors, Ct, specified in 2.3.3.


Reference bending design values, Fb, shall be mul-tiplied by the beam stability factor, CL, specified in 3.3.3. The beam stability factor, CL, shall not apply simultaneously with the volume factor, CV, for struc-tural glued laminated timber bending members (see 5.3.6). Therefore, the lesser of these adjustment factors shall apply.

5.3.6 Volume Factor, CV

When structural glued laminated timber members are loaded in bending about the x-x axis, the reference bending design values, Fbx

+, and Fbx-, shall be multi-

plied by the following volume factor:

(5.3-1)

where:

Table 5.3.1 Applicability of Adjustment Factors for Structural Glued Laminated Timber

ASD only

ASD and LRFD LRFD only

Load

Dur

atio

n Fa

ctor

Wet

Ser

vice

Fac

tor

Tem

pera

ture

Fac

tor

Bea

m S

tabi

lity

Fact

or 1

Vol

ume

Fact

or 1

Flat

Use

Fac

tor

Cur

vatu

re F

acto

r

Stre

ss In

tera

ctio

n Fa

ctor

Shea

r Red

uctio

n Fa

ctor

Col

umn

Stab

ility

Fac

tor

Bea

ring

Are

a Fa

ctor

Form

at C

onve

rsio

n Fa

ctor

Res

ista

nce

Fact

or

Tim

e Ef

fect

Fac

tor

KF

Fb' = Fb x CD CM Ct CL CV Cfu Cc CI - - - 2.54 0.85

Ft' = Ft x CD CM Ct - - - - - - - - 2.70 0.80

Fv' = Fv x CD CM Ct - - - - - Cvr - - 2.88 0.75

Frt' = Frt x CD CM Ct - - - - - - - - 2.88 0.75

Fc' = Fc x CD CM Ct - - - - - - CP - 2.40 0.90

Fc' = Fc x - CM Ct - - - - - - - Cb 1.67 0.90 -

E' = E x - CM Ct - - - - - - - - - - -

Emin' = Emin x - CM Ct - - - - - - - - 1.76 0.85 -


Glulam Beam Design Example (ASD) - Bending, Shear, Bearing, and Deflection

Design a simple roof supporting beam spanning 32 ft, with 5000 lb loads (1000 lb DL + 4000 lb SL) applied bypurlins at 8 ft on center (at 1/4 points plus he ends). Member has lateral supports at the ends andcompression edge supports at the purlins locations. Beam supports are 6 inches long. Assume dry serviceconditions. Temperature is less than 100o degreees but occasionally may reach 150o. Use 24F-1.8Estructural glued laminated Southern Pine timber.


Reference and Adjusted Design Values for 24F-1.8E structural glued laminated softwood timber

Fbx+ 2400 psi Fbx- 1450 psi (Table 5A)

Fc⊥x 650 psi Fvx 265 psi

Ft 1100 psi Fc 1600 psi

Ex 1800000 psi Exmin 950000 psi

CD 1.15 CM 1.0 Ct 1.0 T> 150 only occasionally (Table 5.3.1)

Cfu 1.0 Cc 1.0 Cb 1.0 CI 1.0 Cvr 1.0

E'x Ex CM Ct E'xmin Exmin CM Ct

E'x 1800000 psi E'xmin 950000 psi


l 32 ft b 5 in d 30.25 in initial iteration

Note: Beam length designated as lower case l instead of upper case L used in the Specification nomenclature

Ag b d Sxxb d

2

6 Ixx

b d3

12 lsupport 6 in

Ag 151.3 in2

Sxx 762.6 in3

Ixx 11534 in4

Beam Stability Factor

Fbx+* Fbx+ CD CM Ct Cc CI F'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL flat use factor Cfuand volume factor CV applied. Fbx+* 2760 psi

lu 12in

ft 8 ft lu 96 in laterally unsupported length

le 1.54 lu le 148 in (Table 3.3.3)

RB

le d

b2

RB 13.375 slenderness ratio for bending (3.3-5)

FbE

1.20 E'xmin

RB2


CL

1FbE

Fbx+*

1.9

1FbE

Fbx+*

1.9

2FbE

Fbx+*

0.95

CL 0.965

Volume Factor

x 20 Southern Pine

CV min

1

21 ft

l

1

x12 in

d

1

x

5.125 in

b

1

x

CV 0.936 CL and Cv shall not apply simulatenously (5.3.5). CV isless than CL. CV controls

Adjusted Bending Design Value

F'b Fbx+* minCL

CV

Cfu

F'b 2584 psi F'b is adjusted bending design value with all adjustmentfactors.

Assume Beam Weight and Determine Section Modulus Required to Resist Bending

Maximum total moment is the adjusted bending design value F'b times the section modulus S

wbeamweight 40lbf

ft P 5000 lbf

Mest Pl

2

wbeamweight l2

8

12

in

ft Mest 1021440 in·lbf

Sreqd

Mest

F'b

Sreqd 395 in3

Try a 5 X 22 member

b2 5 in d2 22 in Beam dimension for trial section. Subscripts used denotesecond iteration

Ag2 b2 d2 Sxx2

b2 d22

6

Ixx2

b2 d23

12

Ag2 110 in2

Sxx2 403.3 in3

Ixx2 4437 in4

RB2

le d2

b22

RB2 11.406 slenderness ratio for bending (3.3-5)

FbE2

1.20 E'xmin

RB22

FbE2 8763 psi critical bucking design value for bending (3.7.1)

CL2

1FbE2

Fbx+*

1.9

1FbE2

Fbx+*

1.9

2FbE2

Fbx+*

0.95

CL2 0.978

CV2 min

1

21 ft

l

1

x12 in

d2

1

x

5.125 in

b2

1

x

CV2 0.951 CL and Cv shall not apply simulatenously (5.3.5). CV isless than CL. CV controls

Adjusted Bending Design Value

F'b2 Fbx+* minCL2

CV2

Cfu

F'b2 2625 psi F'b2 is adjusted bending design value with all adjustmentfactors for the trial section considered.

wbeamweight2 30lbf

ft beamweight for 5 X 22 beam

M2 Pl

2

wbeamweight2 l2

8

12

in

ft

M2 1006080 in·lbf

Sreqd2

M2

F'b2

Sreqd2 383 in3

383 in3 < 403 in3

Okay

Shear Parallel to Grain

The two 5000 pound purlin loads at the ends of the beam are within "d" of the supports and can be ignored for shear(3.4.3.1(a)). Shear determined from remaining purlin loads

Vpurlins3 P

2

Vpurlins 7500 lbf

Vbeamweight2 wbeamweight2l

2d2

1

2lsupport

Vbeamweight2 417.5 lbf

Vtotal Vpurlins Vbeamweight2

Vtotal 7918 lbf

Allowable Shear Parallel to Grain f v

Cvr 1.0 (5.3.10)

F'v Fvx CD CM Ct Cvr

F'v 305 psi

fv3

2

Vtotal

b2 d2

fv 108 psi fv < F'v Applied stress parallel to grain less than allowable Ok

Compression Perpindicular to Grain

At Bearing Ends

The bearing ends of the beam transmit all the purling loads so the two 5000 pound purlin loads at the ends of thebeam are included in the bearing load calculations

Rpurlins1

25 P

Rpurlins 12500 lbf

Rbeamweight21

2wbeamweight2 l 2

lsupport

2

Rbeamweight2 487.5 lbf

Rtotal Rpurlins Rbeamweight2

Rtotal 12988 lbf

F'c⊥x Fc⊥x CM Ct

F'c⊥x 650 psi

fc⊥

Rtotal

b2 lsupport

fc⊥ 433 psi Applied compressive stress is less than compression designvalue. OK

At Purlins

Purlins are supported by saddle style hangers that transfer commpressive loads to the top of the beam.Deternine the area of the hangers required to support each purlin without creating compressive stresses greaterthan the compression perpindicular to grain design value

AhangerP

F'c⊥x

Ahanger 7.69 in2

Assuming that 50% of the beam width (2-1/4 in) is available to support the purlin hanger, the width of the hanger canbe calculated as follows:

whanger

Ahanger

0.50 b2

whanger 3.08 in

A 3-1/8 wide purlin hanger that extends 50% (2-1/2 inches) across the beam is adequate. Note: The compressivedesign value F'cp can be increased by the bearing area factor Cb (5.3.12). For 3 inch bearing the factor is:

lb 3 in Cb

lb 0.375 in

lb Cb 1.125

1

Cbwhanger 2.735 in

Using the bearing factor Cb confirms that a 3 inch wide hanger that extends 2-1/2 inches across the beam wouldbe adequate.

At this stage of the calculations, the span of the beam can be reviewed. The 32 foot span was based on thecenter to center distance between supports. The length of the span used in design is the face to face distanceplus 1/2 of the required bearing length at the ends (3.2.1).

In example, the face to face distance is 32 ft minus 6 inches or 31.5 feet. At the end of the beam. the requiredbearing distance is 12,998 lbs/(5 inches * 650 psi) or 4 inches. At the interior end,half the purlin load is assumedto be transferred to the beam end. Required length in bearing is (2500 lbs + 7500 lbs + 488 lbs)/5 inches * 650psi) or 3.25 inches. The two required bearing lengths and the face to face distance produces a span of 31.5 ft +1/2 (4.00/12) + 1/2 (3.25/12) or 31.8 feet which 99.4% of the center to center span. The shorter span reducesmoments and bending stresses by 1.25%. The reduction is considered insufficient to allow the uses of the nextsmaller beam.

Deflection

The specification does not include specific deflection limits for roofs. The IBC limits deflection to L/180. In someapplications, deflections may be critical and the designed may wish to limit deflections.

Dead load deflection is usually calculated to determine the desired camber of the beam. The recommended camber isusually 150% of the dead load deflection. Deflection for the 5000 lb concentrated loads and the beam weight is:

Δpurlin

19 P 12 lin

ft

3

384 E'x Ixx2 Δbeamweight

5wbeamweight2

12in

ft

l 12in

ft

4

384E'x Ixx2

Δpurlin 1.754 in Δbeamweight 0.089 in

Δtotal Δpurlin Δbeamweight

Δtotal 1.843 in

Length/Deflection Ratio

l 12in

ft

Δtotal208 L/Δ >180. The length deflection ratio satisfies IBC criteria

Polling QuestionThe NDS provides restrictions on allowable deflection limits.

a) Trueb) False)

8


DES

IGN P

RO

VIS

IONS

AND

EQ

UATIO

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3

3.6.6 Column Bracing

Column bracing shall be installed where necessary to resist wind or other lateral forces (see Appendix A).

3.6.7 Lateral Support of Arches, Studs, and Compression Chords of Trusses

Guidelines for providing lateral support and deter-mining e/d in arches, studs, and compression chords of trusses are specified in Appendix A.11.

3.7 Solid Columns

3.7.1 Column Stability Factor, CP

3.7.1.1 When a compression member is supported throughout its length to prevent lateral displacement in all directions, CP = 1.0.

3.7.1.2 The effective column length, e, for a solid column shall be determined in accordance with princi-ples of engineering mechanics. One method for deter-mining effective column length, when end-fixity condi-tions are known, is to multiply actual column length by the appropriate effective length factor specified in Ap-pendix G, e = (Ke)( ).

3.7.1.3 For solid columns with rectangular cross section, the slenderness ratio, e/d, shall be taken as the larger of the ratios e1/d1 or e2/d2 (see Figure 3F) where each ratio has been adjusted by the appropriate buck-ling length coefficient, Ke, from Appendix G.

3.7.1.4 The slenderness ratio for solid columns, e/d, shall not exceed 50, except that during construc-

tion e/d shall not exceed 75. 3.7.1.5 The column stability factor shall be calcu-

lated as follows: 2* * *

cE c cE c cE cP

1 F F 1 F F F FC

2c 2c c (3.7-1)

where:

Fc* = reference compression design value parallel to grain multiplied by all applicable adjust-ment factors except CP (see 2.3), psi

c = 0.8 for sawn lumber

c = 0.85 for round timber poles and piles

c = 0.9 for structural glued laminated timber, structural composite lumber, and cross-laminated timber

3.7.1.6 For especially severe service conditions and/or extraordinary hazard, use of lower adjusted de-sign values may be necessary. See Appendix H for background information concerning column stability calculations and Appendix F for information concern-ing coefficient of variation in modulus of elasticity (COVE).

3.7.2 Tapered Columns

For design of a column with rectangular cross sec-tion, tapered at one or both ends, the representative di-mension, d, for each face of the column shall be derived as follows:

minmin max min

max

dd d (d d ) a 0.15 1d

(3.7-2)

where:

d = representative dimension for tapered column, in.

dmin = the minimum dimension for that face of the column, in.

dmax = the maximum dimension for that face of the column, in.

Support Conditions Large end fixed, small end unsupported a = 0.70

or simply supported Small end fixed, large end unsupported a = 0.30

or simply supported Both ends simply supported:

Tapered toward one end a = 0.50 Tapered toward both ends a = 0.70

For all other support conditions:

min max mind d (d d )(1 / 3) (3.7-3)

mincE 2

e

0.822 EF

/ d 2


LKoch

Highlight

LKoch

Highlight



Calculations of fc and CP shall be based on the rep-resentative dimension, d. In addition, fc at any cross section in the tapered column shall not exceed the ref-erence compression design value parallel to grain mul-tiplied by all applicable adjustment factors except the column stability factor, CP.

3.7.3 Round Columns

The design of a column of round cross section shall be based on the design calculations for a square column of the same cross-sectional area and having the same degree of taper. Reference design values and special design provisions for round timber poles and piles are provided in Chapter 6.

3.8 Tension Members

3.8.1 Tension Parallel to Grain

The actual tension stress or force parallel to grain shall be based on the net section area (see 3.1.2) and shall not exceed the adjusted tension design value.

3.8.2 Tension Perpendicular to Grain

Designs that induce tension stress perpendicular to grain shall be avoided whenever possible (see Refer-ences 16 and 19). When tension stress perpendicular to grain cannot be avoided, mechanical reinforcement suf-ficient to resist all such stresses shall be considered (see References 52 and 53 for additional information).

3.9 Combined Bending and Axial Loading

3.9.1 Bending and Axial Tension

Members subjected to a combination of bending and axial tension (see Figure 3G) shall be so propor-tioned that:

t b*bt

f f 1.0FF

(3.9-1)

and

b t**b

f f 1.0F

(3.9-2)

where:

Fb* = reference bending design value multiplied by all applicable adjustment factors except CL, psi

Fb** = reference bending design value multiplied by all applicable adjustment factors except CV, psi

Figure 3G Combined Bending and Axial Tension

3.9.2 Bending and Axial Compression

Members subjected to a combination of bending about one or both principal axes and axial compression (see Figure 3H) shall be so proportioned that:

2

c b1

c b1 c cE1

b22

b2 c cE2 b1 bE

f f

F F 1 f F

f 1.0F 1 f F f F

(3.9-3)


Compression Member Analysis Example (ASD)

A No 2 Spruce Pine Fir 2X6 interior bearing stud, 91.5 inches long, sheathed on both sides with gypsum board,carries dead load and snow load from the roof. Determine CP and the allowable compressive stress Fc' for thestud. Assume studs are placed 16" on center and top and bottom plates are of same grade and species.Determine axial loads controlled by buckling and allowed by bearing.

Reference and Adjusted Design Values for No. 2 SPF 2x6

Fb 875 psi Emin 510000 psi (Table 4A)

Fc 1150 psi Fc⊥ 335 psi

CD 1.15 CM 1.0 Ct 1.0 CF 1.1 Ci 1.0 CT 1.0

E'min Emin CM Ct Ci CT E'min 510000 psi

F'c⊥ Fc⊥ CM Ct Ci F'c⊥ 335 psi


L 91.5 in b 1.5 in d 5.5 in

Column Stability FactorFc* Fc CD CM Ct CF Ci Fc* is adjusted bending design value with all adjustment

factors except the column stability factor CP ,Fc* 1455 psi

le2 0 le1 L Effective lengths of compression member in planes of lateralsupport. Strong axis buckling controls

le2

b0

le1

d16.636 <50 OK (3.7.1.4)

FcE0.822 E'min

le1

d

2 FcE 1515 psi critical bucking design value for compression

members (3.7.1.5)

c 0.8 sawn lumber (3.7.1.5)

CP

1FcE

Fc*

2 c

1FcE

Fc*

2 c

2FcE

Fc*

c Column Stability Factor (3.7-1)

CP 0.705

F'c Fc* CP

F'c 1025 psi F'c is adjusted compression design value with all adjustmentfactors.

Determine Axial Loads Controlled by Buckling and BearingPBuckling b d F'c PBuckling 8458 lbf

PBearing b d F'c⊥ PBearing 2764 lbf

Note: Bearing area factor (Cb) can be used to increase the bearing controlled load on interior studs. The bearingfactor for the 1-1/2 bearing length measured parallel to grain is 1.25 (Equation 3.10-2 and Table 3.10.4)

Cb 1.25

PBearingIncreased b d F'c⊥ Cb PBearingIncreased 3455 lbf

Compression Members and Column Stability Calculation (ASD)

Compare the bearing strengths of a 4x4 and 6x6 post being used for an interior column (onlycarrying gravity loads - D+L). Both members are No. 2 Southern Pine and have a length of 10feet. Both ends are assumed to be pinned (Ke=1.0 - NDS 3.7.1.2). Assume all members areloaded concentrically.

Reference and Adjusted Design Values - 4x4 PostFc 1450psi

(NDS Supplement Table4B)E 1400000psi

Emin 510000psi

CF 1.0 Size factor (NDS Supplement Table 4B)

CM 1.0 Moisture factor (NDS Supplement Table 4B)

Ct 1.0 Temperature factor (NDS Table2.3.3)

Ci 1.0 Incising factor (NDS Table4.3.8)

CT 1.0 Buckling Stiffness factor (NDS4.4.2)

CD 1.0 Load Duration factor (NDS Table2.3.2)

E' E CM Ct Ci E' 1.4 106

psi

E'min Emin CM Ci Ct CT E'min 5.1 105

psi

d 3.5in Actual member dimensions = 3.5" x 3.5"

Area 12.25in2

Ke 1.0

Length 120in

Le Ke Length

Le 120 in

Le

d34.286 Needs to be less than 50 (NDS

3.7.1.3)

Column Stability Factor Calculation

FcE0.822 E'min( )

Le

d

2 FcE 356.628 psi (NDS 3.7.1)

Fc* Fc CD CM Ct CF Ci Fc* 1.45 103

psi

(NDS3.7.1)csawn 0.8

CP

1FcE

Fc*

2csawn

1FcE

Fc*

2csawn

2FcE

Fc*

csawn

CP 0.232

Axial Capacity

F'c Fc CD CM Ct CF Ci CP

F'c 336.316 psi

P F'c Area

P 4119.867 lbf

Reference and Adjusted Design Values - 6x6 Post Note: "2" subscript indicates 6x6;differentiates between 6x6 and4x4 properties and calculationsFc2 525psi

(NDS Supplement Table 4D)E2 1200000psi

Emin2 440000psi

CF2 1.0 Size factor (NDS Supplement Table4D)

CM2 1.0 Moisture factor (NDS Supplement Table 4D)

Ct2 1.0 Temperature factor (NDS Table 2.3.3)

Ci2 1.0 Incising factor (NDS Table 4.3.8)

CT2 1.0 Buckling Stiffness factor (NDS 4.4.2)

CD2 1.0 Load Duration factor (NDS Table 2.3.2)

E'2 E2 CM2 Ct2 Ci2 E'2 1.2 106

psi

E'min2 Emin2 CM2 Ci2 Ct2 CT2 E'min2 4.4 105

psi

d2 5.5inActual member dimensions = 5.5" x 5.5"

Area2 30.25in2

Ke2 1.0

Length2 120in

Le2 Ke2 Length2

Le2 120 in

Le2

d221.818

Needs to be less than 50 (NDS 3.7.1.3)

Column Stability Factor Calculation

FcE2

0.822 E'min2 Le2

d2

2 FcE2 759.779 psi (NDS 3.7.1)

Fc2* Fc2 CD2 CM2 Ct2 CF2 Ci2 Fc2* 525 psi

(NDS 3.7.1)csawn2 0.8

CP2

1FcE2

Fc2*

2csawn2

1FcE2

Fc2*

2csawn2

2FcE2

Fc2*

csawn2

CP2 0.801

Axial Capacity

F'c2 Fc2 CD2 CM2 Ct2 CF2 Ci2 CP2

F'c2 420.648 psi

P2 F'c2 Area2

P2 12724.605 lbf

4x4 Post Capacity = 4120 lbs6x6 Post Capacity = 12725 lbs

Note: for eccentrically loaded columns, see NDS Chapter 15

Polling QuestionThe Column Stability Factor (CP) equation is the same for solid sawn and glulam members.

a) Trueb) False)

9



Calculations of fc and CP shall be based on the rep-resentative dimension, d. In addition, fc at any cross section in the tapered column shall not exceed the ref-erence compression design value parallel to grain mul-tiplied by all applicable adjustment factors except the column stability factor, CP.

3.7.3 Round Columns

The design of a column of round cross section shall be based on the design calculations for a square column of the same cross-sectional area and having the same degree of taper. Reference design values and special design provisions for round timber poles and piles are provided in Chapter 6.

3.8 Tension Members

3.8.1 Tension Parallel to Grain

The actual tension stress or force parallel to grain shall be based on the net section area (see 3.1.2) and shall not exceed the adjusted tension design value.

3.8.2 Tension Perpendicular to Grain

Designs that induce tension stress perpendicular to grain shall be avoided whenever possible (see Refer-ences 16 and 19). When tension stress perpendicular to grain cannot be avoided, mechanical reinforcement suf-ficient to resist all such stresses shall be considered (see References 52 and 53 for additional information).

3.9 Combined Bending and Axial Loading

3.9.1 Bending and Axial Tension

Members subjected to a combination of bending and axial tension (see Figure 3G) shall be so propor-tioned that:

t b*bt

f f 1.0FF

(3.9-1)

and

b t**b

f f 1.0F

(3.9-2)

where:

Fb* = reference bending design value multiplied by all applicable adjustment factors except CL, psi

Fb** = reference bending design value multiplied by all applicable adjustment factors except CV, psi

Figure 3G Combined Bending and Axial Tension

3.9.2 Bending and Axial Compression

Members subjected to a combination of bending about one or both principal axes and axial compression (see Figure 3H) shall be so proportioned that:

2

c b1

c b1 c cE1

b22

b2 c cE2 b1 bE

f f

F F 1 f F

f 1.0F 1 f F f F

(3.9-3)



DES

IGN P

RO

VIS

IONS

AND

EQ

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3

and 2

c b1

cE2 bE

f f1.0

F F (3.9-4)

where: for either uniaxial edgewise bending or biaxial bending

and for uniaxial flatwise bending or biaxial bending

and

minb1 bE 2

B

1.20 Ef F

(R ) for biaxial bending

fb1 = actual edgewise bending stress (bending load applied to narrow face of member) , psi

fb2 = actual flatwise bending stress (bending load applied to wide face of member) , psi

d1 = wide face dimension (see Figure 3H), in.

d2 = narrow face dimension (see Figure 3H), in.

Effective column lengths, e1 and e2, shall be de-termined in accordance with 3.7.1.2. Fc', FcE1, and FcE2 shall be determined in accordance with 2.3 and 3.7. Fb1', Fb2', and FbE shall be determined in accordance with 2.3 and 3.3.3.

3.9.3 Eccentric Compression Loading

See 15.4 for members subjected to combined bend-ing and axial compression due to eccentric loading, or eccentric loading in combination with other loads.

Figure 3H Combined Bending and Axial Compression

3.10 Design for Bearing

3.10.1 Bearing Parallel to Grain

3.10.1.1 The actual compressive bearing stress par-allel to grain shall be based on the net bearing area and shall not exceed the reference compression design value parallel to grain multiplied by all applicable adjustment factors except the column stability factor, CP.

3.10.1.2 Fc*, the reference compression design val-

ues parallel to grain multiplied by all applicable ad-justment factors except the column stability factor, ap-plies to end-to-end bearing of compression members provided there is adequate lateral support and the end cuts are accurately squared and parallel.

3.10.1.3 When fc > (0.75)(Fc*) bearing shall be on a

metal plate or strap, or on other equivalently durable, rigid, homogeneous material with sufficient stiffness to distribute the applied load. Where a rigid insert is re-quired for end-to-end bearing of compression members,

it shall be equivalent to 20-gage metal plate or better, inserted with a snug fit between abutting ends.

3.10.2 Bearing Perpendicular to Grain

The actual compression stress perpendicular to grain shall be based on the net bearing area and shall not exceed the adjusted compression design value per-pendicular to grain, fc Fc '. When calculating bearing area at the ends of bending members, no allowance shall be made for the fact that as the member bends, pressure upon the inner edge of the bearing is greater than at the member end.

minc cE1 2

e1 1

0.822 Ef F

( / d )2e1 1/ d )e1 1

minc cE2 2

e2 2

0.822 Ef F

( / d )2e2 2/ d )e2 2


Combined Bending and Axial Loading of a Truss Chord Member (ASD)

A No. 2 Hem-Fir 2x8 is considered for use as the bottom chord of a 24-ft roof truss (12 ft between panel points).The chord will be subject to a uniform dead load of 8 psf as well as tension forces (assuming pinnedconnections) of 880 lb from roof wind loads (WL), 880 lb from roof live (RLL) and 1420 lb from dead loads (DL).Trusses are to be spaced 4 ft. Framing will have a 19% (max) moisture content. Check the adequacy of thebottom chord member for bending and tension for the appropriate load cases.

Note: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 forrequirements on load combinations.

Reference and Adjusted Design Values for No. 2 Hem-Fir 2x8

Fb 850 psi E 1200000psi Emin 440000psi (Table 4A)

Ft 525 psi CM 1.0 Ct 1.0 CF 1.0 (Table 4.3.1)

Cfu 1.0 Ci 1.0 Cr 1.0

CV 1.0 CT 1.0

E' E CM Ct Ci E'min Emin CM Ct Ci CT

E' 1200000psi E'min 440000psi


l 12 ft b 1.5 in d 7.25 in

Ag b d Sb d

2

6

Ag 10.875in2

S 13.141in3

Applied Loads

wD 8lbf

ft2

wtrib 4 ft Twind 880 lbf TLive 880 lbf TDead 1420 lbf

Load Case 1: DL + RLL + WL

CD 1.6 Appendix B Section B.2 (non-mandatory)

Tension

Ft' Ft CD CM Ct CF Ci Adjusted tensile design value for tensile stresses for shortduration loads (2.3.1, and 4.3.1)

Ft' 840 psi

T1 Twind TLive TDead Subscripts refer to Load Case

T1 3180 lbf

ft1

T1

Ag Tensile stress in bottom chord

ft1 292 psi Ft' 840 psi Actual tensile stress is less than adjusted tensile designvalue. OK (3.8.1)

BendingF'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL ,flat use factor Cfuand volume factor CV applied. The following calculationsdetermine the beam stabilty factor CL:

F'b* Fb CD CM Ct CF Ci Cr

F'b* 1360 psi

Determine Beam Stability Factor CL (3.3.3)

lu 12in


lu

d19.9 lu/d >7 (Table 3.3.3)

le 1.63 lu 3 d le 256.5 in (Table 3.3.3)

RB

le d

b2

RB 28.75 RB < 50 OK (3.3.3.7)

FbE1.20 E'min

RB2

FbE 639 psi (3.3.3.6)

CL

1FbE

F'b*

1.9

1FbE

F'b*

1.9

2FbE

F'b*

0.95 (3.3-6)

CL 0.451 Resulting beam stability factor CL.

F'b F'b* CL CV Cfu F'b is the fully adjusted bending design value with alladjustment factors including the beam stability factor CLand flat use factor applied F'b 614 psi

F'b**1

CVF'b F'b** is adjusted bending design value with all adjustment

factors except the volume factor CV applied. F'b** 614 psi

Bending resulting from dead loadMmax

wD wtrib l2 12in

ft

8

Mmax 6912 in·lbf

fb

Mmax

S fb 526 psi

fb 526 psi F'b 614 psi Ok. Actual bending stress fb does not exceed adjustedbending design value F'b

Combined Bending and Axial Tension

ft1

Ft'

fb

F'b* 0.73 <1.0. Ok (3.9-1)

fb ft1

F'b**0.381 <1.0 ok (3.9-2)

Load Case 2: DL+RLL

CD 1.25 Appendix B Section B.2 (non-mandatory). Roof Live Load isa construction load.

Tension

Ft' Ft CD CM Ct CF Ci Adjusted tensile design value for tensile stresses for shortduration loads (2.3.1, and 4.3.1)Ft' 656.25 psi

T2 TLive TDead

T2 2300 lbf

ft2

T2

Ag

ft2 211 psi Ft' 656 psi


F'b* 1063 psi

CL

1FbE

F'b*

1.9

1FbE

F'b*

1.9

2FbE

F'b*

0.95

CL 0.565

F'b F'b* CL CV Cfu

F'b 600 psi

F'b**1

CVF'b

F'b** 600 psi

fb 526 psi F'b 600 psi


ft2

Ft'

fb

F'b* 0.82 <1.0. ok

fb ft2

F'b**0.52 <1.0 ok

Load Case 3: DL only

CD 0.9

Tension

Ft' Ft CD CF CM Ci Ct

Ft' 472.5 psi

T3 TDead

T3 1420 lbf

ft3

T3

Ag

ft3 131 psi Ft' 472.5 psi

Bending


F'b* 765 psi

CL

1FbE

F'b*

1.9

1FbE

F'b*

1.9

2FbE

F'b*

0.95

CL 0.734

F'b F'b* CL CV Cfu

F'b 561 psi

F'b**1

CVF'b

F'b** 561 psi

fb 526 psi F'b 561 psi


ft3

Ft'

fb

F'b* 0.96 <1.0. ok

fb ft3

F'b**0.7 <1.0 ok

Results: No 2 Southern Pine 2 x 8 satifies NDS Criteria for combined bending and axial tension.

Polling QuestionWhen calculating Fb*, the reference bending design value is multiplied by all applicable adjustment factors except which of the following?

a) Load duration, CD

b) Wet service, CM

c) Temperature, Ctt

d) Beam stability, CL

e) None of the above)

10


SAW

N LU

MB

ER

4


4.4 Special Design Considerations

4.4.1 Stability of Bending Members

4.4.1.1 Sawn lumber bending members shall be de-signed in accordance with the lateral stability calcula-tions in 3.3.3 or shall meet the lateral support require-ments in 4.4.1.2 and 4.4.1.3.

4.4.1.2 As an alternative to 4.4.1.1, rectangular sawn lumber beams, rafters, joists, or other bending members, shall be designed in accordance with the fol-lowing provisions to provide restraint against rotation or lateral displacement. If the depth to breadth, d/b, based on nominal dimensions is:

(a) d/b 2; no lateral support shall be required. (b) 2 < d/b 4; the ends shall be held in position,

as by full depth solid blocking, bridging, hang-ers, nailing, or bolting to other framing mem-bers, or other acceptable means.

(c) 4 < d/b 5; the compression edge of the mem-ber shall be held in line for its entire length to prevent lateral displacement, as by adequate sheathing or subflooring, and ends at point of bearing shall be held in position to prevent ro-tation and/or lateral displacement.

(d) 5 < d/b 6; bridging, full depth solid blocking or diagonal cross bracing shall be installed at intervals not exceeding 8 feet, the compression edge of the member shall be held in line as by adequate sheathing or subflooring, and the ends at points of bearing shall be held in position to prevent rotation and/or lateral displacement.

(e) 6 < d/b 7; both edges of the member shall be held in line for their entire length and ends at points of bearing shall be held in position to prevent rotation and/or lateral displacement.

4.4.1.3 If a bending member is subjected to both flexure and axial compression, the depth to breadth ra-tio shall be no more than 5 to 1 if one edge is firmly held in line. If under all combinations of load, the un-braced edge of the member is in tension, the depth to breadth ratio shall be no more than 6 to 1.

4.4.2 Wood Trusses

4.4.2.1 Increased chord stiffness relative to axial loads where a 2" x 4" or smaller sawn lumber truss compression chord is subjected to combined flexure and axial compression under dry service condition and has 3/8" or thicker plywood sheathing nailed to the nar-row face of the chord in accordance with code required roof sheathing fastener schedules (see References 32, 33, and 34), shall be permitted to be accounted for by multiplying the reference modulus of elasticity design value for beam and column stability, Emin, by the buck-ling stiffness factor, CT, in column stability calculations (see 3.7 and Appendix H). When e < 96", CT shall be calculated as follows:

M eT

T

KC 1K E

e (4.4-1)

where:

e = effective column length of truss compression chord (see 3.7), in.

KM = 2300 for wood seasoned to 19% moisture content or less at the time of plywood at-tachment.

= 1200 for unseasoned or partially seasoned wood at the time of plywood attachment.

KT = 1 – 1.645(COVE)

= 0.59 for visually graded lumber

= 0.75 for machine evaluated lumber (MEL)

= 0.82 for products with COVE 0.11 (see Appendix F.2)

When e > 96", CT shall be calculated based on e = 96".

4.4.2.2 For additional information concerning metal plate connected wood trusses see Reference 9.



APPEND

IX

A


the provisions of Appendix A.11 are followed or when the roof joists rest on and are securely fastened to the top chords of the trusses and are covered with wood sheathing. Where sheathing other than wood is applied, top chord diagonal lateral bracing should be installed.

(b) In all cases, vertical sway bracing should be in-stalled in each third or fourth bay at intervals of approximately 35 feet measured parallel to trusses. Also, bottom chord lateral bracing should be installed in the same bays as the ver-tical sway bracing, where practical, and should extend from side wall to side wall. In addition, struts should be installed between bottom chords at the same truss panels as vertical sway bracing and should extend continuously from end wall to end wall. If the roof construction does not provide proper top chord strut action, separate additional members should be provid-ed.

A.11 Lateral Support of Arches, Compression Chords of Trusses and Studs

A.11.1 When roof joists or purlins are used be-tween arches or compression chords, or when roof joists or purlins are placed on top of an arch or com-pression chord, and are securely fastened to the arch or compression chord, the largest value of e/d, calculated using the depth of the arch or compression chord or calculated using the breadth (least dimension) of the arch or compression chord between points of intermit-tent lateral support, shall be used. The roof joists or purlins should be placed to account for shrinkage (for example by placing the upper edges of unseasoned joists approximately 5% of the joist depth above the tops of the arch or chord), but also placed low enough to provide adequate lateral support.

A.11.2 When planks are placed on top of an arch or compression chord, and securely fastened to the arch or compression chord, or when sheathing is nailed proper-ly to the top chord of trussed rafters, the depth rather than the breadth of the arch, compression chord, or trussed rafter shall be permitted to be used as the least dimension in determining e/d.

A.11.3 When stud walls in light frame construction are adequately sheathed on at least one side, the depth, rather than breadth of the stud, shall be permitted to be taken as the least dimension in calculating the e/d ra-tio. The sheathing shall be shown by experience to pro-vide lateral support and shall be adequately fastened.


g e

A.11.3 When stud walls in light frame construction gare adequately sheathed on at least one side, the depth,q y , p ,rather than breadth of the stud, shall be permitted to be , ptaken as the least dimension in calculating the e/d r/ a-g etio. The sheathing shall be shown by experience to pro-g y p pvide lateral support and shall be adequately fastened.

Combined Bending and Axial Compression Example (ASD)

No. 1 Southern Pine 2x6 beam-columns are being designed to carry an axial compressive load of 840 lb (snow)and 560 lb (dead) plus a 25 psf wind load on their narrow face. Columns are 9 ft long and spaced 4 feet o.c..Their ends are held in position and lateral support is provided along the entire narrow face.

Check the adequacy of the beam-column for bending and compression for the appropriate load cases.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations. The column being considered is not subjected to especially severe service conditions or extraordinary hazard

Reference and Adjusted Design Values for No. 1 Southern Pine 2x6

Fb 1350 psi E 1600000 psi Emin 580000 psi (Table 4B)

Fc 1550 psi CM 1.0 Ct 1.0 CF 1.0 (Table 4.3.1)

Cfu 1.15 Ci 1.0 Cr 1.0 flat use factors are for weak andstrong axis bending

CV 1.0 CT 1.0 c 0.8 factor "c" in column stability factor CPequation for sawn lumber. (3.7.1)

E'min Emin CM Ct Ci CT

E'min 580000 psi


l 9 ft b 1.5 in d 5.5 in

Ag b d Sxb d

2

6 Sy

d b2

6

Ag 8.25 in2

Sx 7.562 in3

Sy 2.062 in3

Applied Loads

wstrong 25lbf

ft2

wtrib 4 ft Psnow 840 lbf PDead 560 lbf

wstrong wtrib 100lbf

ft

wweak 0 Load applied to weak axis of beam column.In this example, no load is applied.

Load Case 1: DL + SL + WL


Compression

Fc* Fc CD CM Ct CF Ci Fc* is reference compressive design value adjusted with alladjustment factors except the column stability factor CP,. Thefollowing calculations determine the column stabilty factorCP:

Fc* 2480 psi

P1 Psnow PDead Subscripts refer to Load Case

P1 1400 lbf

fc1

P1

Ag

fc1 170 psi Compressive stress in beam column

Determine Effective Lengths and Critical Buckling Design Values

Ke 1.0 Buckling length coefficient Ke for rotation free/translation fixed(pinned/pinned) column (Appendix G Table G1)

d1 d l1 l le1 Ke l1 12in

ft

beam dimensions, laterally unsupported lengths (Figure3F) and effective column lengths (3.7.1) for buckling ineach direction. Subscript 1 is strong (but unsupported)axis; Subscript 2 is the weak but laterally supported axis. d2 b l2

7

12ft le2 Ke l2 12

in

ft

le1 108 in le2 7 in Effective lengths in each axis. Assume drywall is connected tothe studs at 7 inches o.c. (Note: This can also be assumed toequal 0 in this case, but would create calculation problems withMathCAD; the 7 in effective length will not control in our caseand will have no impact on calculations.)le max

le1

le2

le 108 in

controlling effective length

FcE10.822 E'min

le1

d1

2 FcE1 1236 psi

critical bucking compression design values for compressionmember in planes of lateral support (3.7.1)

FcE20.822 E'min

le2

d2

2 FcE2 21892 psi

RB

le d

b2

RB 16 slenderness ratio for bending (3.3-5)

FbE1.20 E'min

RB2


Determine Column Stability Factor Cp

FcE minFcE1

FcE2

FcE 1236 psi critical bucking design value for compression member

CP

1FcE

Fc*

2 c

1FcE

Fc*

2 c

2FcE

Fc*

c

CP 0.433 Column Stability Factor (3.7-1)

Fc' Fc* CP Fc' 1073 psi Adjusted compressive design value

fc1 170 psi Fc' 1073 psi Ok. Actual compressive stress fc does not exceedadjusted compressive design value F'c

Bending

CL 1.0 Depth to breadth (d/b) ratio (6/2 = 3.0). 2<d/b<4 End restraints for the beam-column satisfy 4.4.1.2 (b)

F'b1 Fb CD CM CL Ct 1.0( ) CF Ci Cr F'b is adjusted bending design value with all adjustmentfactors. F'b is calculated for both strong and weak axisbending. The flat use factor for strong axis bending is 1.0;the flat use factor allowed for weak axis bending is per NDS4.3.7.

F'b1 2160 psi

F'b2 Fb CD CM CL Cfu Ct CF Ci Cr

F'b2 2484 psi

Bending stress in strong direction resulting from windload applied to narrow face of beam-columnMmax1

wstrong wtrib l2 12in

ft

8

Mmax1 12150 in·lbf

fb1

Mmax1

Sx fb1 1607 psi

Bending stress in weak direction resulting from windload applied to wide face of beam-column. In thisexample, no load is applied to the wide face of thebeam-column.

Mmax2

wweak wtrib l2 12in

ft

8

Mmax2 0 in·lbf

fb2

Mmax2

Sy fb2 0 psi

Ok. Actual bending stresses fb1 and fb2 do not exceedadjusted bending design value F'b

fb1 1607 psi fb2 0 F'b1 2160 psi

Combined Bending and Compression

fc1

Fc'

2fb1

F'b1 1fc1

FcE1

fb2

F'b2 1fc1

FcE2

fb1

FbE

2

0.89 < 1.0 ok (3.9-3)

fc1

FcE2

fb1

FbE

2

0.379 < 1.0 ok (3.9-4)

fc1 170 psi FcE1 1236 psi Actual compression stress is less than critical bucking designvalues for both weak and strong axis buckling. Ok (NDS 3.9-2)

fc1 170 psi FcE2 21892 psi

fb1 1607 psi FbE 2636 psi Actual bending stress is less than critical buckling design value.Ok (NDS 3.9-2)

Load Case 2: DL+SL Load duration is not included in E' calculation. It is important to evaluatemultiple combinations such that all load duration effects are considered.CD 1.15

Compression

Fc* Fc CD CM Ct CF Ci P2 Psnow PDead fc2

P2

Ag

Fc* 1782 psi P2 1400 lbf fc2 170 psi


CP

1FcE

Fc*

2 c

1FcE

Fc*

2 c

2FcE

Fc*

c

CP 0.555

Fc' Fc* CP

Fc' 990 psi



CD 0.90

Compression

Fc* Fc CD CM Ct CF Ci P3 PDead fc3

P3

Ag

Fc* 1395 psi P3 560 lbf fc3 68 psi


CP

1FcE

Fc*

2 c

1FcE

Fc*

2 c

2FcE

Fc*

c

CP 0.648

Fc' Fc* CP

Fc' 904 psi


Polling QuestionWhich of the following adjustment factors is NOT applicable for dimension lumber modulus of elasticity design values?

a) Load duration, CD

b) Wet service, CM

c) Temperature, Ctt

d) Incising, Ci

e) None of the above)

11

Combined Bending and Axial Compression Example (ASD)

A No. 2 Southern Pine 2x4 oriented flatwise is being considered for use as a member within the top chord of aparallel chord gable end truss. The member is 3 ft long (between panel points) and will be subjected to axialcompression forces of 300 lb (DL) and 600 lb (SL), concentrated loads of 50 lb (DL) and 100 lb (SL) at themidpoint of the member on its wide face and 120 lb (WL) of the midpoint of the member on its narrow face.Lateral support is provided only at the ends of the member and the ends are considered pinned.

Check the adequacy of the beam-column for bending and compression for the appropriate load cases.


Reference and Adjusted Design Values for No. 2 Southern Pine 2x4

Fb 1100 psi E 1400000 psi Emin 510000 psi (Table 4B)

Fc 1450 psi CM 1.0 Ct 1.0 CF 1.0 (Table 4.3.1)

Cfu 1.10 Ci 1.0 Cr 1.0 flat use factors are for weak axisbending (3.9.2)

CV 1.0 c 0.8 factor "c" in column stability factor CPequation for sawn lumber. (3.7.1)E'min Emin CM Ct Ci

E'min 510000 psi


l 3 ft b 1.5 in d 3.5 in

Ag b d Sxb d

2

6 Sy

d b2

6

Ag 5.25 in2

Sx 3.062 in3

Sy 1.312 in3

Applied Loads

Axial Weak (y) Axis Stong (x) Axis

Applied loads. Subscripts depict loadtype (DL-dead load, SL-snow load andWL-wind load) and application in relationto the member (applied to strong or weakaxis).

DLAxial 300 lbf SLWeak 100 lbf WLStrong 120 lbf

SLAxial 600 lbf DLWeak 50 lbf

Load Case 1: DL + SL + WL Subscripts refer to Load Case


Compression

Fc* Fc CD CM Ct CF Ci Fc* is reference compressive design value adjusted with alladjustment factors except the column stability factor CP.

Fc* 2320 psi

P1 DLAxial SLAxial

P1 900 lbf

fc1

P1

Ag fc1 171 psi Compressive stress in beam column

Determine Effective Lengths and Critical Buckling Design Values

Ke 1.0 Buckling length coefficient Ke for rotation free/translation fixed(pinned/pinned) column (Appendix G Table G1)

d1 d l1 l le1 Ke l1 12in

ft

beam dimensions, laterally unsupported lengths (Figure3F) and effective column lengths (3.7.1) for buckling ineach direction. Subscript 1 is strong (but unsupported)axis; Subscript 2 is the weak but laterally supported axis. d2 b l2 l le2 Ke l2 12

in

ft

le1 36 in le2 36 in Effective lengths in each axis

le/d is less than 50 for each axis (3.7.1.4) le1

d110

le2

d224

critical bucking compression design values for compressionmember in planes of lateral support (3.7.1) FcE1

0.822 E'min

le1

d1

2 FcE1 3963 psi

FcE20.822 E'min

le2

d2

2 FcE2 728 psi


FcE minFcE1

FcE2

FcE 728 psi critical bucking design value for compression member

CP

1FcE

Fc*

2 c

1FcE

Fc*

2 c

2FcE

Fc*

c

CP 0.29 Column Stability Factor (3.7-1)

Determine Adjusted Compressive Design Value F'c

Fc' Fc* CP Fc' 673 psi Adjusted compressive design value

fc1 171 psi Fc' 673 psi Ok. Actual compressive stress fc does not exceedadjusted compressive design value F'c (NDS 3.6.3)

Narrow Face (Strong Axis) Bending - (load parallel to wide face)

F'b* Fb CD CM Ct CF Ci Cr F'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL and flat use factorCfu applied. F'b* 1760 psi

Determine Strong Axis Beam Stability Factor CL

lu 12in


lu

d10.3 lu/d >7 (Table 3.3.3)

le 1.37 lu 3 d le 59.8 in (Table 3.3.3)

RB

le d

b2

RB 9.6 slenderness ratio for bending (3.3-5) RB < 50 (NDS 3.3.3.7)

FbE1.20 E'min

RB2


CL

1FbE

F'b*

1.9

1FbE

F'b*

1.9

2FbE

F'b*

0.95

CL 0.982 Resulting beam stability factor CL. As at alternative, 4.4.1.2 (b)allows CL = 1.0 for d/b = (4/2) when the ends are held inposition by full depth solid blocking, bridging, hangers, nailing,bolting or other suitable means.

F'b1 Fb CD CM CL Ct CF Ci Cr F'b1 is adjusted edgewise bending design value with alladjustment factors.

F'b1 1729 psi

Determine Bending Load and Resulting Bending Stress

Bending stress in strong direction resulting from windload applied to narrow face of beam-columnM1Strong

WLStrong l 12in

ft

4

M1Strong 1080 in·lbf

fb1

M1Strong

Sx

fb1 353 psi F'b1 1729 psi Ok. Actual bending stress fb does not exceed adjustededgewise compressive design value F'b1

Wide Face (Weak Axis) Bending - (load parallel to narrow face)

CL 1.0 Since d<b (2 in < 4 in) CL=1.0

Flat use factor Cfu applies for weak axis bending.Cfu 1.1

F'b2 Fb CD CM CL Ct Cfu CF Ci Cr F'b2 is adjusted flatwise bending design value with alladjustment factors.

F'b2 1936 psi

Determine Bending Load and Resulting Bending Stress - Weak Axis Bending

Bending stress in strong direction resulting from windload applied to narrow face of beam-columnM1Weak

DLWeak SLWeak l 12in

ft

4

M1Weak 1350 in·lbf

fb2

M1Weak

Sy

fb2 1029 psi F'b2 1936 psi Ok. Actual bending stress fb does not exceed adjustedbending design value F'b (NDS 3.1.1)


fc1

Fc'

2fb1

F'b1 1fc1

FcE1

fb2

F'b2 1fc1

FcE2

fb1

FbE

2

0.98 < 1.0 ok (3.9-3)

fc1

FcE2

fb1

FbE

2

0.24 < 1.0 ok (3.9-4)

fc1 171 psi FcE1 3963 psi Actual compression stress is less than critical bucking designvalues for both weak and strong axis buckling. Ok


fb1 353 psi FbE 6577 psi Actual bending stress is less than critical buckling design value. Ok

Load Case 2: DL+SL

CD 1.15

P2 DLAxial SLAxial P2 900 lbf

fc2

P1

Ag fc2 171 psi

Compression

Fc* Fc CD CM Ct CF Ci

Fc* 1667 psi fc2 171 psi


CP

1FcE

Fc*

2 c

1FcE

Fc*

2 c

2FcE

Fc*

c

CP 0.387

Fc' Fc* CP

Fc' 646 psi


Weak Axis Bending

F'b2 Fb CD CM CL Ct Cfu CF Ci Cr

F'b2 1392 psi


DLWeak SLWeak l 12in

ft

4

M2Weak 1350 in·lbf

fb2

M2Weak

Sy

fb2 1029 psi F'b2 1392 psi Ok. Actual bending stress fb does not exceed adjustedcompressive design value F'b

Strong Axis Bending

Wind load is zero. No strong axis bending (appropriate terms in following equationsset to zero)


fc2

Fc'

2fb1 0

F'b1 1fc2

FcE1

fb2

F'b2 1fc2

FcE2

fb1

FbE

2

1.04 < 1.0 ok (3.9-3)

fc2

FcE2

fb1 0

FbE

2

0.24 < 1.0 ok (3.9.-4)



fb1 353 psi FbE 6577 psi Actual bending stress is less than critical buckling design value. Ok


CD 0.90

Compression

Fc* Fc CD CM Ct CF Ci P3 DLAxial fc3

P3

Ag

Fc* 1305 psi P3 300 lbf fc3 57 psi


CP

1FcE

Fc*

2 c

1FcE

Fc*

2 c

2FcE

Fc*

c

CP 0.473

Fc' Fc* CP

Fc' 617 psi


Weak Axis Bending

F'b2 Fb CD CM CL Ct Cfu CF Ci Cr

F'b2 1089 psi


DLWeak l 12in

ft

4

M3Weak 450 in·lbf

fb2

M3Weak

Sy

fb2 343 psi F'b2 1089 psi Ok. Actual bending stress fb does not exceed adjustedcompressive design value F'b

Strong Axis Bending

Wind load is zero. No strong axis bending


fc3

Fc'

2fb1 0

F'b1 1fc3

FcE1

fb2

F'b2 1fc3

FcE2

fb1

FbE

2

0.35 < 1.0 ok (3.9-3)

fc3

FcE2

fb1 0

FbE

2

0.08 < 1.0 ok (3.9-4)



fb1 353 psi FbE 6577 psi Actual bending stress is less than critical buckling design value.Ok

2015ndsexamples2015 nds examples … · 2021. 1. 7. · (nds®) for wood construction (ansi/awc...

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