Download - Engineering with Wood Tension & Compression
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Engineering with WoodTension & Compression
Presenters: David W. Boehm, P.E.
Gary Sweeny, P.E.
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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.
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Wood Compression & Tension Members
(ALLOWABLE STRESS DESIGN)
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Wood Compression and Tension Members
Definitions Parameters of design Design procedure – Axial
compression Bending and Axial compression Bending and Axial Tension Sample Problems Questions
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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
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What Are Tension Members? Structural members whose primary
loads are axial tension Some truss members Rafter collar ties Connections are critical
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Types of wood columns Simple solid column
square, rectangular, circular Built-up column
mechanically laminated, nailed, bolted
Glued laminated column Studs
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Column Failure Modes
Crushing – short
Crushing and Buckling - intermediate
Buckling - long
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Slenderness Ratio
dlNDS e
ee
e
llKrlK
Slenderness ratio:
The larger the slenderness ratio, the greater the instability of the column
NDS=National Design Specification
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Effective Column Length, l When end fixity conditions are
known:
elelel
el
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Simple Solid Column
l l1& l2 = distances between points of lateral support
d1& d2 = cross-sectional dimensions
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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
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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
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Design of Wood ColumnsDetermination of Allowable Stress, F’c
Compressive stress parallel to grain adjustment factors: Load duration Wet service Temperature Size Incising Column stability
NDS table values
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Adjustment Factors
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Column Stability Factor, Cp
Where:
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
2' dlEKF ecEcE
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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
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
cF
F
cF
F
cF
F
C c
cEc
cE
c
cE
p
*
2
**
2
1
2
1
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Stress Check
pcc
c
cc
CFF
APfFf
'
'
/
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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)
Factors: Compression members (from NDS)CD=1.0 (duration)CM=1.0 (moisture)Ct=1.0 (Temp) TemperatureCF=1.1 (Size)(Table)Ci=1.0 (Incising)Cp=TBD
Example #1 – Column Design
Tabulated Properties:(from NDS Tables)
E= 1,300,000 psiFc= 700 psi
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Slenderness Ratio
l = 12’-0”le= KelPin-pin: Ke = 1.0le = 1.0 (12)= 12.0
Slenderness Ratio =
502.265.512)12(
dle
For columns, slendernessRatio = le1/d1 or le2/d2
whichever is larger
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Cp – Stability Factor
KcE= 0.3 For Visually Graded LumberE’ = 1,300,000 psil e/d = 26.2
cF
F
cF
F
cF
F
C c
cEc
cE
c
cE
p
*
2
**
2
1
2
1
1.568' 2 dlEKF ecEcE
C = 0.8 For sawn lumberFc* = Fc x CD CM Ct CF Ci
= 700 (1)(1)(1)(1.1)(1) = 770 psi
= 0.58
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Column Capacity
'
22
'
25.305.5
44758.0770
cc
c
Ff
inA
psiF
klbPAfA
Pf
Ff
c
c
cc
5.13520,1344725.30
'
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Tension
tt
t
FfAPf
/
Adjustment Factors
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Bending and Axial Tension
0.1' *
b
b
t
t
Ff
Ff 0.1**
b
tb
Fff
and
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
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Bending and Axial Compression
0.1
11 2122
'
2
11'
1
2
'
bEbcEcb
b
cEcb
b
c
c
FfFfF
f
FfF
f
F
f
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fc Compression Zone- Axial Compression
fb x-x Compression Zone- Bending, X-X
fb y-y Compression Zone- Bending, Y-Y
Combined Max. Stress Max. Compression Zone- Combined
Bending and Axial Compression
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Example #2Exterior 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
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Exterior Wall Stud Design
Solution: Combined bending & axial compression
Wood properties: (from NDS Tables)
CD=1.15Fc = 1150 psi Cm=1.0E = 1,400,000 psi Ct=1.0
Cf=1.1Ci=1.0Cp=TBD
Fc* = 1150 (1.15)(1)(1)(1.1)(1.0)= 1455 psi
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Slenderness Ratio for Compression Calculation
lKl ee
"102"1020.11
el "82el
For Pin-Pin, Ke = 1.0
(Sheathed/nailed)
okdld
e 505.185.5
102
"5.5
1
1
1
okdld
e 33.55.18
"5.1
2
2
2
Therefore 18.5 governs
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Allowable Compressive Stress
1227
1455
2
'
1
*
dlEKF
psiF
e
cEcE
c
5.18000,400,1
3.0'
dlpsiE
tableK
e
cE
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
cF
F
cF
F
cF
F
C c
cEc
cE
c
cE
p
*
2
**
2
1
2
1
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Allowable Bending Stress
Fb = 875 psi
F’b1
= FbCDCMCtCLCFCfuCi Cr CD = 1.6 (wind)
CM = 1.0 = 875(1.6)(1.0)(1.0)(1.0)(1.3)(1.0)(1.0)(1.15) Ct = 1.0
CL = 1.0 = 2093 psi CF = 1.3 (Table 4A)
Cfu= 1.0 Ci = 1.0 Cr = 1.15
(repetitive)
F’b2 = ignore since no load in “b2” direction
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Combined Stressespsi
APfc 3.363
25.83000
psiSM
IMcfb
570
1
ftwindpsfW #40"0'220
356.7
65.55.1
6
22
inbhS
psiFpsiFpsiF
cE
b
c
2.12272093917
1
1
'
2.12275.18000,400,13.022
11
'
1
dlEKF
e
cEcE
lbftwlM 36085.840
8
22
lbin 320,4
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Combined Stress Index
0.1
11 212
'2
2
1'1
1
2
'
bEbcEcb
b
cEcb
b
c
c
FfFfF
f
FfF
f
F
fCSI
ok0.154.0
0
2.1227
6.36312093
570
917
6.3632
(Check deflection and shear)
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Effective Length for Bending Calculation
"102l 567.35.1
5.5
b
d 0.1LC
sodlandl u
u ,7"8
"5.1606.2
ue ll
okb
dlrRatiosSlendernes eB 5035.6
5.1
5.55.1622
(Assume blocked)