Timber Design Tip Qc 1s1415 p1

8/10/2019 Timber Design Tip Qc 1s1415 p1

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TIMBER DESIGNTIP-QC 13 September 2014Lecturer: Richard S Regidor

Visually Graded Lumber

Defects are allowed to become larger and more frequentas the grade drops. Commonly considered defects:

1.Location, size and placement of knots2.Slope of grain3.manufacturing defects (splits from drying4.Wane5.Warp

Machine stress-grading

Timber loaded about minor axis (on flat) Small loads applied - timber is loaded at much less than design strength Intention is to find poor pieces by measuring stiffness High speed operation - timber in the machine for typically 1 second

Proof-grading

Timber loaded about major axis (on edge) High load applied - timber is loaded at loads near the design strength Intention is to find poor pieces by breaking them


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Low speed process - timber may take 15 to 20 seconds to pass through themachine

FLEXURAL MEMBERS

Fb=FbCDCMCtCLCFCVCfuCrCcCf

Applicable adjustment factors

Load Duration factor (CD)

If a member is fully stressed to the maximum allowable stress, continuous orcumulative, for more than 10 years, CD= 0.90

Period CD

for seven days duration, as for roof loads 1.25

for earthquake 1.33

for wind (connections and fasteners) 1.33

for wind ( members only) 1.6

for impact 2


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This is not applicable to compression perpendicular to grain.

Size factor adjustment (CF)

Applicable to flexural members with a thickness exceeding 125 mm and depth

more than 300 mm.

1 9300

FC

d

=

Where:CF= size factorD = depth of beam, mm

If the beam is circular in cross-section, with D> 340 mm, convert the section toan equivalent conventionally loaded square beam with the same cross sectionalarea.


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Slenderness Factor and Flexural stress (CSand Fb)

2eS

l dC b=

Where:

Cs= slenderness factorle= effective length of the beam, mmd = depth of the beam, mmb = breadth of the beam, mm

If Cs10, full value of Fbmaybe used.


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Effective lengths of beams ( maybe multiplied by ( )0.85 2.55 ul d+ ), exceptthe highlighted values

Single-span beampoint load at center ---------------- 1.61lu

uniformly loaded ------------------- 1.92luequal end moments -------------- 1.84lU

Cantilever beamsLoad concentrated at free end --- 1.69lU

Load uniformly distributed -------- 1.06lUPoint load plus uniform load ------ 1.69lu

Single span/CantileverOther loads -------------------------- 1.92lU


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Allowable bending stress (Fb),

if 10 < CSCk

4

1' 13

sb b

k

CF FC

=

Where:

0.811k

b

EC

F=

E = modulus of elasticityFb= allowable unit stress for extreme fiber in

bending ( basic value)


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if Ck< CS50

2

0.438'b

s

EF

C=

If Cs> 50, not allowed by Code.

Form Factor Adjustment (Cf)

For circular beams: Cf= 1.18

Square (with diagonal vertical): Cf= 1.414

Lumber I beams and box beams

2

2

1430.81 1 1

88f g

dC C

d

+= +

+


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Where:

Cf= form factorCg= support factor = p2(6-8p+3p2)(1-q) + qP = ratio of depth of compression flange to full

depth of beamq = ratio of thickness of web(s) to the full width

of beam.

** Cumulative with the form factor adjustment except for lumber I beams and box

beams.

Wet Service Factor, CM

When the moisture content of structural members under service conditions will

exceed 19% for an extended period of time, design values should be multiplied bythe appropriate wet-service factor.


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Design Value CMFb 0.85Ft 1.0Fv 0.97

FC 0.67Fc 0.80E 0.90

Temperature Factor, Ct

Reduction of the design values for wood may be necessary for memberssubjected to elevated temperatures for repeated or prolonged periods.

Design Values038T C

0 038 52C T C< 0 052 66C T C<

Ftand E 1.0 0.90 0.90Fb, FV, Fcand Fc

Dry 1.0 0.80 0.7Wet 1.0 0.70 0.5


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Volume Factor, CV---For glued-laminated beams only

Flat-Use Factor, Cfu

When the load is applied to the wide face ( flatwise) dimension of lumber,

design values should be multiplied by the appropriate flat-use factor.

Thickness

Width, mm 50 and 75 mm 100 mm

50 and 75 1

100 1.1 1125 1.1 1.05

150 1.15 1.05

200 1.15 1.05

250 and wider 1.2 1.1


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Beam Stability Factor, CL

( ) ( ) 2

* * *1 1

1.9 1.9 0.95

bE b bE b bE bL

F F F F F FC

+ + =

Where:Fb*= design value for bending multiplied by all applicable adjustment

factors, except Cfu,Cvand CL

2

'bE

bEB

K E

F R=

KbE= 0.438 ( for visually graded lumber and machine rated lumber)= 0.609 ( for products with a coefficient of variation of 0.11 or

less)E = design modulus of elasticity multiplied by applicable adjustment

factors.


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Bearing Area Factor, Cb

For compression perpendicular to grain where the bearing is less than 150 mmlong and at least 75 mm from the end of a member.

9.525bb

b

LCL

+=

Where: Lb= bearing length in millimeters, measured parallel to grain.

Repetitive-Member Factor, Cr

Design values for bending Fbmay be increased when three or more membersare connected so that they act as a unit. The design value for bending of dimension

lumber 50 to 100 mm thick may be multiplied by the repetitive-member factor Cr=1.15.

Example 1:

Calculate the allowable bending stress for a 50 mm x 75 mm floor joist spaced @0.40 m on-center. The joists are assumed to be simply supported over a span of 3


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meters. Bridging are provided at every 0.6 m. The tabulated properties of the joistare: E = 7310 MPa and Fb= 16.5 MPa.

Solution:Fb= CrCDFb

Cr= 1.15 ; CD= 0.90

Fb= (1.15)(0.90)(16.5) = 17.08 MPa > 16.5 MPa, use Fb =Fb.

Example 2:Calculate the allowable bending stress for a visually graded wooden girder with adimension of 200 mm x 400 mm, supporting a series of floor joists. The beam has asimple span of 6 meters. The properties of the member are: E = 5940 MPa and Fb= 18 MPa.


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Solution:

Adjust due to size, d > 300 mm

1 9300

0.97400FC

= =

0.90DC =

' 0.97(0.9)(18) 15.71b F D b

F C C F MPa= = =

Check slenderness

600015 17

400

ul

d= = <

2 2

(1.92)(6000)(400)10.73 10

(200)

eS

l dC

b= = = >


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59400.811 0.811 13.49(21.47)

k S

b

EC CF

= = = >

2 4

1 1 10.73' 1 1 18 15.603 3 13.49

sb b

k

CF F MPaC

= = =

use Fb= 15.60 MPa.

Modulus of Elasticity Adjustment

To be applied only when:

1.the deflections are critical to the stability of structures or components.2.the member is exposed to varying temperature and relative humidity under

sustained loading conditions.


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Coefficient of Variation (CV)

CV= 0.25 ( for visually graded sawn lumber)CV= 0.11 ( for machine stress-rated sawn

lumber)

Modified E = Etable*(1-CV)

HORIZONTAL SHEAR

Fv=FvCDCMCtCH

Actual horizontal shear:

Solid sections,

3

2v

Vf

bd=


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Notched sections,3

2 ' 'v

V df

bd d

=

( at support tension side)

2 '3 '

Vd dV F b d e

d =

(at support, compression side)

d = total depthd = depth of beam at notch.

e = distance notch extends inside theinner edge of support.

** The shear for the notch on the compression side shall be further limited to thevalue determined for a beam of depth d if e exceeds d.


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Compression perpendicular to grain

' 0.73C C

F F

=

( for bearings of any length at the end of the beam and to all bearings 150mm)

0.375' b

C C

b

lF F

l

+=

( for bearings of less than 150 mm AND not nearer than 75 mm to the end of a

member),

lb = length of bearing, mm

COLUMN DESIGN

L/d 50

( ) ( ) ( )2

* * *

*1 1

'2 ' 2 ' '

CE C CE C CE C

C C

F F F F F F F F

c c c

+ + =


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Where:c = 0.8, for sawn lumber= 0.85 for round timber piles

( )2'CE

CEe

K EF

l d=

FC*= adjusted compression design valueKCE= 0.3, for visually graded lumber

= 0.418, for prodcts such as machine

stress-rated sawn lumber.E = modified modulus of elasticity = (1-Cv)E

Example 3:

Calculate the axial load capacity of a 200 mm x 300 mm wooden post. The post hasan unsupported length of 4 meters and is assumed to be pin-connected at theends. the properties of wood (80% stress grade, machine-stress rated) are E =12000 MPa and FC= 16 MPa (tabulated values), Fb=19 MPa(allowable). The post isexposed to variable temperature and relative humidity.


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Solution:

FC*= CD*FC= 0.90(16) = 14.40 MPa

CV= 0.11

E = (1 0.11)E = 0.89 (12000) = 10680 MPa

KCE= 0.418 ( machine stress-rated sawn lumber)

1 400016

250

el x

d= =

( )( )( )2 2

0.418) 10680' 17.44

(16)CE

CE

e

K EF MPal d

= = =

*

17.441.21

14.40

CE

C

F

F

= =


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c = 0.80 ( for sawn lumber)

( )2

1 1.21 1 1.21 1.21' 14.40 10.84

2(0.8) 2(0.8) 0.8C

F MPa

+ + = =

10.84(250)(300) / 1000 813capP kN= =

COMBINED AXIAL AND BENDING LOADS

AXIAL PLUS TENSION

* 1t b

t b

f f

F F

+

and

** 1b t

b

f f

F


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Fb*= CDCMCtCFCrCi (Fb)

Fb**= CDCMCtCLCFCrCi(Fb)

AXIAL PLUS COMPRESSION

1' '

c b

c b c

f f

F F Jf +

Where:

11

11

el d

J

K

=

, and

0.671C

EK

F=

0 1J


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Example 4:Calculate the axial load capacity of the previous problem if a moment of 12 kN-macts about the major axis of the beam.

Solution:

From the previous problem FC*=14.4 MPa

(1000)0.017

(200)(300)c

Pf P= = , P in kN

2 2

6 6(12000000)3.84

(300)(250)b

Mf MPa

bd= = =


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400016

250

120000.671 18.38

16

16 110.68

18.38 11

0.017 3.84110.84 19 (0.68)(0.017 )

458.86

el

d

k

J

P

P

P kN

= =

= =

= =

+ =

=

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Timber Design Tip Qc 1s1415 p1