Engineering with Wood Tension & Compression Presenters: David W. Boehm, P.E. Gary Sweeny, P.E....

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

llK

rlK

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

APf

Ff

'

'

/

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) Temperature

CF=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= Kel

Pin-pin: Ke = 1.0

le = 1.0 (12)= 12.0

Slenderness Ratio =

502.265.5

12)12(

d

le

For columns,

slenderness

Ratio = le1/d1 or

le2/d2

whichever is larger

Cp – Stability Factor

KcE= 0.3 For Visually Graded Lumber

E’ = 1,300,000 psi

l 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 lumber

Fc* = 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

klb

PAfA

Pf

Ff

c

c

cc

5.13520,1344725.30

'

Tension

tt

t

Ff

APf

/

Adjustment Factors

Bending and Axial Tension

0.1' *

b

b

t

t

F

f

F

f0.1**

b

tb

F

ffand

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)

okd

l

d

e 505.185.5

102

"5.5

1

1

1

okd

l

d

e 33.55.1

8

"5.1

2

2

2

Therefore 18.5 governs

Allowable Compressive Stress

1227

1455

2

'

1

*

dl

EKF

psiF

e

cEcE

c

5.18

000,400,1

3.0'

dl

psiE

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 Stresses

psiA

Pfc 3.363

25.8

3000

psiS

M

I

Mcfb

570

1

ftwindpsfW #40"0'220

356.7

6

5.55.1

6

22

inbh

S

psiF

psiF

psiF

cE

b

c

2.1227

2093

917

1

1

'

2.1227

5.18

000,400,13.022

11

'

1

dl

EKF

e

cEcE

lbft

wlM 360

8

5.840

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

sod

landl u

u ,7"8 "5.16

06.2

ue ll

ok

b

dlrRatiosSlendernes eB 5035.6

5.1

5.55.16

22

(Assume blocked)