engineering with wood tension & compression

Engineering with WoodTension & 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

square, rectangular, circular Built-up column

mechanically laminated, nailed, bolted

Glued laminated column Studs

Column Failure Modes

Crushing – short

Crushing and Buckling - intermediate

Buckling - long

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

Effective Column Length, l When end fixity conditions are

known:

elelel

el

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 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

Adjustment Factors

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

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

Stress Check

pcc

c

cc

CFF

APfFf

'

'

/

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

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

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

Column Capacity

'

22

'

25.305.5

44758.0770

cc

c

Ff

inA

psiF

klbPAfA

Pf

Ff

c

c

cc

5.13520,1344725.30

'

Tension

tt

t

FfAPf

/

Adjustment Factors

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

Bending and Axial Compression

0.1

11 2122

'

2

11'

1

2

'

bEbcEcb

b

cEcb

b

c

c

FfFfF

f

FfF

f

F

f

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

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

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

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

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

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

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

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)

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)