Timber Column

8/13/2019 Timber Column

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Example 1GIVEN: Nominal 4x6 Douglas Fir-Larch No. 2 grade columns are used asformwork to carry temporary construction loading of 125 PSF (includingdead loads) as shown below. The frame is subject to rain and wetconditions. The columns are laterally braced at the top and bottom only.The 4x6 is considered to be Dimension Lumber.REQUIRED:

1) What is the maximum allowable load, Pallowfor each of thecolumns?

2) Are the columns acceptable?

Step 1 Determine Fc*:

Fc* = Fc(CD)(CM)

Fc= 1350 PSI from Table in Lecture 29 notes

CD= 1.25 for construction loads (see Lecture 29 notes)

CM= 0.80 since MC > 19%

Fc* = 1350 PSI(1.25)(0.80)= 1350 PSI

Lecture 30 - Page 3 of 8


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Step 2 Determine maximum slenderness ratio (SR)max:

(SR)strong axis="5.5

)/"12)("0'10( ft

= 21.8

(SR)max= 34.3

(SR)weak axis="5.3

)/"12)("0'10( ft

= 34.3

Step 3 Determine FcE:

FcE=max

2)(

))((

SR

EKcE

KcE= 0.3 since it is sawn lumber

E = 1,600,000 PSI (see Table from Lecture 29 notes)

FcE= 2)3.34(

)000,600,1)(3.0( PSI

= 408 PSI

Step 4 Determine column stability factor Cp:

Cp=c

FF

c

FF

c

FF

c

cE

c

cE

c

Ec )*

(

2

)*

(1

2

)*

(1

2

+

+

where 30.01350

408

*==

PSI

PSI

F

F

c

cE

c = 0.8 since column is sawn lumber

Cp=8.0

3.0

)8.0(2

)3.0(1

)8.0(2

)3.0(12

+

+

= 0.81 0.53

= 0.28



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Step 5 Determine allowable axial load, Pallow:

Pallow= (Fc*)(Cp)(A)

= (1350 PSI)(0.28)(3.5 x 5.5)

Pallow= 7277 lbs.

Step 6 Determine actual load on each column, Pactual:

Pactual= (Trib. Area)(PSF)

= (14-0 x 9-0)(125 PSF)

Pactual= 3938 lbs. < 7277 lbs. column is acceptable

Design of Wood Columns:

The direct design (selection) of wood columns is complicated because ofthe number of unknowns. Therefore, column design is typically one ofTRIAL and ERROR.

Design aids such as pre-engineered graphs and spreadsheets are oftenused for the selection of column members as shown in the table below:

Col. Section Unbraced Length about weak axis (feet)Nom.Size:

Area(in2)

8 10 12 14 16 18 20 22

4x4 12.25 7.3 4.9 3.5 2.6

4x6 19.25 11.4 7.8 5.5 4.1

6x6 30.25 24.8 20.9 16.9 13.4 10.7 8.7 7.2 6.5

6x8 41.25 33.9 28.5 23.1 18.3 14.6 11.9 9.8 8.9

8x8 56.25 51.5 48.1 43.5 38.0 32.3 27.4 23.1 19.7

8x10 71.25 65.3 61.0 55.1 48.1 41.0 34.7 29.3 24.9

10x10 90.25 85.9 83.0 79.0 73.6 67.0 60.0 52.9 46.4

10x12 109.25 104.0 100.0 95.6 89.1 81.2 72.6 64.0 56.1Al lowable axial l oad, Pallow, (kips) for Douglas Fir-Larch No. 1 grade CD= 1.0



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Example 2GIVEN: An exterior wall is built using Southern Pine No. 2 grade 2x6studs at 16 o.c. with thick plywood sheathing nailed to the studs. Thewalls are to be designed to carry snow as well as live and dead loads.REQUIRED: What is the maximum permissible load on the wall in termsof pounds per linear foot of wall?

Step 1 Determine Fc*:

Fc* = Fc(CD)(CM)

Fc= 1650 PSI from Table in Lecture 29 notes

CD= 1.15 for snow loads (see Lecture 29 notes)

CM= 1.0 since MC < 19%

Fc* = 1650 PSI(1.15)(1.0)= 1898 PSI



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Step 2 Determine maximum slenderness ratio (SR)max:

(SR)weak axis= Negligible since it iscontinuously laterally braced by theplywood sheathing

(SR)strong axis="5.5

)/"12)("0'9( ft

= 19.6

(SR)max= 19.6

Step 3 Determine FcE:

FcE=max

2)(

))((

SR

EKcE

KcE= 0.3 since it is sawn lumber

E = 1,600,000 PSI (see Table from Lecture 29 notes)

FcE= 2)6.19(

)000,600,1)(3.0( PSI

= 1249 PSI

Step 4 Determine column stability factor Cp:

Cp=c

FF

c

FF

c

FF

c

cE

c

cE

c

Ec )*

(

2

)*

(1

2

)*

(1

2

+

+

where 66.01898

1249

*==

PSI

PSI

F

F

c

cE

c = 0.8 since stud is sawn lumber

Cp=8.0

66.0

)8.0(2

)66.0(1

)8.0(2

)66.0(12

+

+

= 0.54



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Step 5 Determine allowable axial load, Pallow:

Pallow= (Fc*)(Cp)(A)

= (1898 PSI)(0.54)(1.5 x 5.5)

Pallow= 8456 lbs. per stud

Al lowable wall load = 6342"16

"128456 =

pounds per linear foot


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