Engineering with Wood
Tension & Compression
Presenters:
David W. Boehm, P.E.
Gary Sweeny, P.E.
The information presented in this seminar is based on the
knowledge and experience of the engineering staff at
Engineering Ventures. Background and support information
comes from various sources including but not limited to:




NDS
IBC
BOCA
Simplified Design for Wind and Earthquake Forces, Ambrose
& Vergun
 The project files at Engineering Ventures
All designs of structures must be prepared under the direct
supervision of a registered Professional Engineer.
Wood Compression & Tension
Members
(ALLOWABLE STRESS DESIGN)
Wood Compression and Tension
Members







Definitions
Parameters of design
Design procedure – Axial compression
Bending and Axial compression
Bending and Axial Tension
Sample Problems
Questions
What are compression members?




Structural members whose primary loads
are axial compression
Length is several times greater than its
least dimension
Columns and studs
Some truss members
What Are Tension Members?




Structural members whose primary loads
are axial tension
Some truss members
Rafter collar ties
Connections are critical
Types of wood columns

Simple solid column


Built-up column



square, rectangular, circular
mechanically laminated, nailed, bolted
Glued laminated column
Studs
Column Failure Modes

Crushing – short

Crushing and Buckling - intermediate

Buckling - long
Slenderness Ratio
Slenderness ratio:
Ke l r
Kel  le
NDS  l e d
NDS=National Design Specification
The larger the slenderness ratio, the greater the instability of the column
Effective Column Length, ll e

When end fixity conditions are known:
le
Simple Solid Column
l l1& l2 = distances
between points of lateral
support
d1& d2 = cross-sectional
dimensions
Slenderness Ratio

Simple solid columns:


< 50
Except during construction < 75
A large slenderness ratio indicates a
greater instability and tendency to buckle
under lower axial load
Design of Wood Columns



fc ≤ F’c (Allowable Stress Design)
fc = P/A, Actual compressive stress
= load divided by area
F’c = Allowable compressive stress
Design of Wood Columns
Determination of Allowable Stress, F’c

Compressive stress parallel to grain
adjustment factors:






Load duration
Wet service
Temperature
Size
Incising
Column stability
NDS table values
Adjustment Factors
Column Stability Factor, Cp

Where:
FcE  K cE E ' le d 
2
KcE is defined by the Code (NDS) for the particular type of wood selected
Modulus of Elasticity is adjusted by the following factors:
CM, Ct, Ci, CT
Column Stability Factor, Cp


Compression members supported throughout
the length: Cp = 1.0
C is given in the Code (NDS) for the type of
column selected
F
1   cE * 
F c
Cp  

2c
2
  FcE
  FcE
1


*
 
F *c  
F
c

 
2c
c




c is given in the Code (NDS) for the type of
column selected.
F*c is Fc multiplied by all of the adjustment
factors except Cp
Stress Check
fc  F
'
c
fc  P / A
'

c
F


c
F
Cp
Example #1 – Column Design
Example: Find the capacity of a 6x6 (Nominal) wood column.
Given:
Height of Column = 12’-0”
End conditions are pinned top & bottom
Wood species & Grade = Spruce-Pine-Fir No. 1
Visually graded by NLGA
Interior, dry conditions, normal use for floor load support (DL &LL)
Tabulated Properties:
(from NDS Tables)
Factors: Compression members (from NDS)
E= 1,300,000 psi
Fc= 700 psi
CM=1.0 (moisture)
CD=1.0 (duration)
Ct=1.0 (Temp) Temperature
CF=1.1 (Size)(Table)
Ci=1.0 (Incising)
Cp=TBD
Slenderness Ratio
l = 12’-0”
For columns, slenderness
le= Kel
Ratio = le1/d1 or le2/d2
whichever is larger
Pin-pin: Ke = 1.0
le = 1.0 (12)= 12.0
le (12)12

 26.2  50
Slenderness Ratio =
d
5.5
Cp – Stability Factor
F
1   cE * 
F c
Cp  

2c
2
  FcE
  FcE
1



*
*
 
F c 
F
c

 
2c
c




= 0.58
FcE  K cE E ' le d   568.1
2
C = 0.8 For sawn lumber
KcE= 0.3 For Visually Graded Lumber
Fc* = Fc x CD CM Ct CF Ci
E’ = 1,300,000 psi
= 700 (1)(1)(1)(1.1)(1)
= 770 psi
l e/d = 26.2
Column Capacity
Fc  7700.58  447 psi
'
A  5.52  30.25in 2
f c  Fc
'
f c  Fc
'
fc  P
A
Af c  P
30.25447   13,520 lb  13.5k
Tension

Adjustment Factors
ft  P / A
ft  Ft
Bending and Axial Tension
ft
fb
 *  1.0
Ft ' Fb
and
fb  ft
 1.0
**
Fb
Where:
Fb* = tabulated bending design value multiplied by all applicable adjustment
factors except CL
Fb** = tabulated bending design value multiplied by all applicable adjustment
factors except Cv
Bending and Axial Compression
2
 f 
f b1
fb2
c

 

 1.0
 F ' c  F 'b1 1   f F  F 'b 2 1   f F    f F 2
c
cE1
c
cE 2
b1
bE


Bending and Axial Compression
fc
Compression Zone- Axial Compression
Compression Zone- Bending, X-X
fb x-x
fb y-y
Combined Max. Stress
Compression Zone- Bending, Y-Y
Max. Compression Zone- Combined
Example #2
Exterior Wall Stud Design
Problem: Is a 2x6 @ 24” oc wall stud pattern adequate for a one story exterior
wall?
Given:
Height of stud = 8’–6”
Assume Pin-Pin End conditions and exterior face is braced by wall
sheathing
Wood species/grade = Spruce-Pine-Fir No.1/No.2
Visually graded by NLGA rules
Interior, Dry conditions, normal use
Subject to wind & roof loads (snow)
Loads: 20 psf wind; 3.0k axial compression
Exterior Wall Stud Design
Solution: Combined bending & axial compression
Wood properties: (from NDS Tables)
Fc = 1150 psi
E = 1,400,000 psi
CD=1.15
Cm=1.0
Ct=1.0
Cf=1.1
Ci=1.0
Cp=TBD
Fc* = 1150 (1.15)(1)(1)(1.1)(1.0)= 1455 psi
Slenderness Ratio
for Compression Calculation
le  K el For Pin-Pin, Ke = 1.0
le1  1.0102"  102"
le2  8" (Sheathed/nailed)
d1  5.5"
d 2  1.5"
le1
l e2
102

 18.5  50 ok
d1 5.5
8

 5.33 ok
d 2 1.5
Therefore 18.5 governs
Allowable Compressive Stress
F
1   cE * 
F c
Cp  

2c
Fc  1455 psi
*
FcE1 
'
2
  FcE
  FcE
1



*
 
F c 
F *c



2
c
c




K cE E
 1227
2
le d 
K cE  0.3 table 
E '  1,400,000 psi
le d  18.5
c = 0.8 for sawn lumber (table)
Cp = 0.63
F’c = Fc(CDCMCtCFCiCp)
F’c = 1150(1.15)(1)(1)(1.1)(1)(.63)
F’c = 1455(.63) = 917 psi
Allowable Bending Stress
Fb = 875 psi
F’b1 = FbCDCMCtCLCFCfuCi Cr
= 875(1.6)(1.0)(1.0)(1.0)(1.3)(1.0)(1.0)(1.15)
= 2093 psi
F’b2 = ignore since no load in “b2” direction
CD = 1.6 (wind)
CM = 1.0
Ct = 1.0
CL = 1.0
CF = 1.3 (Table 4A)
Cfu = 1.0
Ci = 1.0
Cr = 1.15 (repetitive)
Combined Stresses
P 3000

 363.3 psi
A 8.25
fc 
K cE E '
0.31,400,000 
FcE1 

2
18.52
le1 d1 
 1227.2
Fc  917 psi
'
Fb1  2093 psi
FcE1  1227.2 psi
W  20 psf wind 2'0"  40 # ft
 
wl 2 40 8.52
M

 360 ft lb
8
8
 4,320 in  lb
bh 2 1.55.52
S

 7.56 in 3
6
6
Mc M
f b1 

I
S
 570 psi
Combined Stress Index
2
 
f
f b1
fb2
CSI   c  

 1.0
 '
'
'
2








F
F
1

f
F
F
1

f
F

f
F
b1
c
cE1
b2
c
cE 2
b1
bE
 c


2
 363.6 

 
 917 
570
  363.6 

20931  
  1227.2 
 0.54  1.0 ok
0
(Check deflection and shear)
Effective Length
for Bending Calculation
l  102"
d
b
lu  8" and

5.5
 3.67  5
C L  1.0
1.5
lu
 7, so
d
Slenderness Ratio  rB 
le  2.06lu (Assume blocked)
 16.5"
le d
b2

16.55.5
1.52
 6.35  50 ok