Design of sawn timber columns and compressive members
Design requirements.
-
The slenderness
ratio Le/d for solid column shall not exceed 50 for service load
and shall not exceed 75 for construction. Le=Ke´L
is effective length of column, Ke is slenderness ratio, L is unsupported
length of column. For
rectangular section, Le/d shall be evaluated in both directions.
-
Maximum
compressive stress, fc must not exceed allowable stress parallel to grain,
F’c = Fc*CD*CM*Ct*CF*Cp
Where
Fc
is allowable bending stress in NDS supplement.
CD is load duration factor, (see
beam design)
CM is wet service factor, (use
when moisture of timber is higher than 19%)
Ct
is temperature factor, (when
timber is used in temperature higher than 150°F)
CF is size factor, (apply
only to visually graded sawn lumber members, and to round timber bending
members, not apply simultaneously with Cv for glued laminated timber)
Cp is column stability factor (see
below)
Slenderness ratio,
Ke
Column stability
factor Cp
According
to NDS 3.7.1, column stability factor shall be determined as
-
Fully supported
laterally throughout its length, Cp=1.
-
Otherwise, Cp
shall be calculated as
Fc*=Compressive
design value in NDS tables multiplied by all other adjustment factor except Cp,
FcE=
KcE´E’/(Le/d)2,
KcE=0.3
for visually graded lumber and machine evaluated lumber, (note: KcE=0.418
for machine stress rated lumber and glued laminated timber),
C=0.8
for sawn lumber, (note: c = 0.85 for round timber piles and 0.9 for glued
laminated timber).
Design procedure
for timber column and compressive members
-
Select timber
species and section.
-
Calculate
slenderness ratio for both axes, Lex/dx, Ley/dy,
where Lex=Lx*Kex, Ley=Ly*Key.
Kex and Key, are slenderness ratios in x and y
direction. Lx and Ly are unsupported length in x and y
direction.
-
Determine
maximum compressive stress, f”c=P/A. P is column axial
load.
A is cross section area.
-
Determine
allowable compressive stress, Fc*
Fc*
= Fc´CD´CM´Ct´CF
Where
Fc
is allowable bending stress in NDS supplement.
CD
is duration factor,
CM is wet service factor, (use
when moisture of timber is higher than 19%)
Ct
is temperature factor, (when
timber is used in temperature higher than 150°F)
CF is size factor, (apply
only to visually graded sawn lumber members, and to round timber bending
members, not apply simultaneously with Cv for glued laminated timber)
-
Calculate
elastic modulus
E’=E´CM´Ct
Where E is modulus of elasticity in NDS supplement
-
Calculate FcE=
KcE*E’/(Le/d)2
-
Calculate Cp
-
Calculate
allowable compressive stress,
F”c
= Fc*´Cp
Example
4: Design of sawn timber column:
Design
data:
Floor
area supported by column: A = 80 ft2
Unsupported
length of column, L = 10 ft
Hinge
support at top and bottom of column
Design
load:
Floor
live load: WL = 30 psf
Floor
dead load: WD = 10 psf
Superimposed
dead load: WSD = 5 psf
Timber:
Southern pine, moisture less than 19%, used in normal room temperature.
Solution:
1.
Select southern pine, 4"x4" stud grade, d = 3.5 in
Actual
cross section: Ac = 12.25 in2.
Allowable
compressive stress parallel to grain: Fc = 975 psi
2.
Calculate slenderness ratio: Ke = 1, Le =Ke´L
= 10 ft, Le/d = 34 < 50
3.
Calculate compressive stress with load duration factor
Load
duration factor for dead load: CD = 0.9
Load
duration factors for live load: CD = 1.0
(Use 1 per NDS)
Calculate
Design load: P = [WD + WSD+ WL]´A
= 3600 lb
Column
compressive stress, fc=P/Ac = 293.8 psi
4.
Calculate allowable stress without Cp.
CM=1,
Ct=1, Cf=1
Fc*
= Fc´CM´Ct´CF
= 975 psi
5.
Calculate elasticity modulus
E’=E´CM´Ct
= 1.4´106
psi
6.
Calculate FcE
KcE=0.3
FcE= KcE*E’/(Le/d)2=
357.3 psi
7.
Calculate Cp
c = 0.8
Cp = 0.333
8.
Calculate allowable compressive stress
F”c
= Fc*´Cp
= 324.8 psi
>
fc= 293.8 psi
O.K.
Example
5: Design of timber load bearing stud wall
Design
data:
Tributary
width of floor supported by wall: B = 20 ft2.
Unsupported
height of stud wall, L = 10 ft
Hinge
support at top and bottom of stud wall
Design
load:
Floor
live load: WL = 30 psf
Floor
dead load: WD = 10 psf
Superimposed
dead load: WSD = 5 psf
Timber:
Southern pine, moisture less than 19%, used in normal room temperature.
Solution:
1.
Select southern pine, 2"x4" stud grade at 16" O.C. d1
= 3.5 in, d2 = 1.5 in, s = 16 in
Actual
cross section: Ac = 5.25 in2.
Allowable
compressive stress parallel to grain: Fc = 975 psi
2.
Calculate slenderness ratio:
Ke = 1, Lex = Ke´L = 10 ft, Lex/d1=34 < 50
Provide
blocking at mid-height in d2 direction
Ley=Ke´(L/2)
= 5 ft, Ley/d2=40 < 50 Govern
3.
Calculate compressive stress with load duration factor
Load
duration factor for dead load: CD = 0.9
Load
duration factors for live load: CD = 1.0
(Use 1 per NDS 2001)
Calculate
Design load: P = [WD + WSD+ WL]´B´s
= 1200 lb
Column
compressive stress, fc=P/Ac = 228.5 psi
4.
Calculate allowable stress without Cp
CM=1,
Ct=1, Cf=1
Fc*
= Fc´CD´CM´Ct´CF
= 975 psi
5.
Calculate elasticity modulus
E’=E´CM´Ct
= 1.4´106
psi
6.
Calculate FcE
KcE=0.3
FcE= KcE*E’/(Le/d)2=
262.5 psi
7.
Calculate Cp
c = 0.8
Cp = 0.252
8.
Calculate allowable compressive stress
F”c
= Fc*´Cp
= 246 psi
>
fc= 228.5 psi
O.K.
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