design of axial members - wood

7/21/2019 Design of Axial Members - Wood

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Axial Forces and Combined

Axial and Bending Forces

Axial Forces and Combined Bending and

Axial Forces Four Types of Applications:

Axial Tension Axial Compression

Combined Bending and Tension

Combined Bending and Compression

Most common axial load members are columns

Most common combined stress member is beam-

column

Magnitude of lateral force in wall elements depends

on design load and how the wall is framed

(horizontally or vertically).


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Examples of

Axial and Axial

plus Bending

Elements

Axial Tension Members

Number of structural applications:

Trusseshave numerous axial force members;

about in tension

Chordsof shearwalls and diaphragms

Collector or drag strut members (when length of

diaphragm is greater than shearwall).

Axial members allow for direct solution in design

Parallel to grain applications avoid cross grain

tension General formula:

ft= P/An Ft


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Capacity of

tension

members is

affected by

connections


Gross area vs. Net area(projected fastener area is

subtracted - nails usually disregarded)

Avoid tight-fitting installations that require forcible

driving of the bolt

Perfect installations are difficult may have to drill

from both sides

Hole diameter can be taken to be bolt diameter

plus 1/16 in., as specified in NDS


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General formula:

ft= P/An Ft

Where Ft= Ft(CD)(CM)(Ct)(CF)(Ci)

DEAD + LIVE LOADS 44 psf

Determine required size

of truss bottom chord

Loads only applied to

top chord (convert to

point loads)

Use method of joints

truss joints assumed

pinned

Connections made with

a single row of 1/2 in.

bolts

Trusses 4 ft o.c., lumber

No. 1 SPF South MC < 19%, normal

temperatures

16 ft 12 ft 12 ft

2.5 ft

8 ft

125

44 psf

Example Size truss bottom chord


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Example Size truss bottom chord

44 psf (4 ft o.c.) = 176 plf

4(6) = 24

5

5P/2 = 528 lb

528 lb

P = 176(6) =1056 lb1056 lb

1056 lb

2,112 lb 2,112 lb

Example Size truss bottom chord cont.

Find forces by method of joints;T = 3.8k

FromTable 4Afor No.1 SPF South:Ft= 400 psi

FactorscD = 1.15, andassume cF = 1.3

DetermineFt init.= Ft(cD) (cF)= 400(1.15)(1.3) =598 psi

Find required An= P/Ft= 3,802/598 = 6.36 in.2

Consider bolt holes:

Reqd Ag= An+Ah= 6.36+1.5(1/2 + 1/16) = 7.2 in.2

Try 2 x 6: A = 8.25 in.2 > 7.2 in.2 OK

Backchecksize factorfor 2x8: cF = 1.3 - same as assumed.

44 psf (4 ft o.c.) = 176 plf

4(6) = 24

5

5P/2 = 528 lb

528 lb

P = 1056 lb 1056 lb

1056 lb

2,112 lb 2,112 lb

FBD

RA= 2,112 lb

528 lb

2,112 528 = 1,584 lb

1,584(12)/5 = 3,802 lb

F

TAC=3,802 lbA


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Columns

Columns are generally sufficiently long thatbuckling

needs to be considered

Direct design is not feasible (trial size,iteration)

General formula:

fc = P/A Fc

Where Fc= Fc(CD)(CM)(Ct)(Cp)(CF)

Cross sectional area to be used (gross or net) should

be considered along the length of member, together

withtendency of member at that point to buckle

laterally

Column Design Considerations

Size factor applies only to dimension lumber

Column stability factor CPtakes buckling into

consideration slenderness ratio is a measure ofbuckling propensity

Slenderness ratiois generally expressed in terms of

effective (unbraced, corrected for end condition)

length, divided with least radius of gyration: (le/r).

For rectangular columns, slenderness ratio can be

expressed as le/d (radius of gyration is a direct

function of the width)

For non-rectangular columns, r 12 can be substituted

for d


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Column Design Considerations

Behavior of wood columns is the result of

interaction between two modes of failure:

buckling and crushing

For use in design, theEuler critical buckling stress

is expressed in NDS as:

Crushingis measured by compressive design value

multiplied byall factors except CP:

F*c= Fc(CD)(CM)(Ct)(Ci)(CF)

Column Stability Factor

Takes into consideration Euler critical stress and

crushing strength:

where buckling and crushing interaction factor

for columns c:

= 0.8 for sawn lumber= 0.85 for timber poles and piles

= 0.9 for SCL columns


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Slenderness can really affect column strength

Slenderness Considerations

By inspection the slenderness ratio about the y axis is

larger and therefore critical


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In this case, (l/d)xmay govern


Conditions can exist under which the column may

buckle about the strong axis

In Stud walls, weak axis is

braced, but strong is not


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End conditions are important

Theeffective length eused in slenderness ratiosis

theoretically the unbraced length of a pinned-end

column

For other column-end conditions, the effective

length is taken as the distance between inflection

points (point of reverse curvature and zeromoment)

For purposes of column analysis, the inflection

point is considered as pinned end

Effectivelength

factor: ke


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Sidesway means

that top of column

is relatively free to

displace laterally:

In braced

situations or in

shearwalls,

sidesway is

prevented)

In rigid frames,

system is

flexible, and

sidesway can

occur


Effective length = effective length factor x unbraced

length:Le= Kex L


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16 ft 12 ft 12 ft

125

44 psf

17 psf

750 plf

330 plf

8 ft

45 psf72 plf

7 ft

55 psf

96 plf

96 plf

Example: Interior Bearing Wall Capacity

No bending although on

exterior walls there could

be wind loads

Continuous lateral support

provided in y direction

Need to consider bearing

capacity of top and bottom

plates

Load is D + L, disregard self

weight of wall See if Standard grade Hem-

Fir, 2 x 4 studs at 16 in. o.c.

are adequate

No special temperature

and moisture requirements

Total load: 17(12) + 45(12) + 72 = 816 lb/ft

Total wall height, including top and bottom plates, is 8 ft

Total load per stud, assuming 16 in. o.c.:

P =816 lb/ft (16/12) =1,088 lbs From Table 4A, for Standard grade Hem-Fir:

Fc= 1300 psi, Fc= 405 psi, Emin= 440,000 psi

Example: Interior Bearing Wall Capacity


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Example Continued

CF= 1.0 for compression standard grade,

A = 5.25 in2; le= 8(12) 3(1.5) = 91.5 in.

(le/d)x= 1(91.5)/3.5 = 31.3 (about y axis studs are braced)

Emin = Emin = 440,000psi

c = 0.8 for visually graded lumber

= 369 psi; Fc* = 1300(1.0) = 1,300 psi

= 0.265 ;

Fc= 1,300(.265) = 344 psi;

P = Fc A = 344(5.25) = 1,806 lb > 1,088 lb OK

Fc= Fc= 405 psi > 344 psi; so column capacity governs

x

x

P = 1,088 lb

Fc= 1300 psi,Fc= 405 psi,

Emin= 440,000 psi

16 ft 12 ft 12 ft

3

1

5

125

44 psf

17 psf

750 plf

330 plf

8 ft

96 plf

72 plf

72 plf

7 ft

45 psf

55 psf

96 plf

96 plf

BASEMENT COLUMN

Column height 7 ft

From before, load on beam

due to D+L: w = 1,583 plf

2

4

Example Design of a Column

W

Trib. L = 12 ft

Trib. length for column = 12 ft

Column load:

PD+L = 12(1,583) = 18,996 lb


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Example Design Sawn Lumber Column

Design column, No. 1 Douglas Fir - Larch

Bracing the same for buckling about x and y axes

(non existent)

Dead and live load apply

Temperature, moisture, incision do not apply

P = 19k D + L

L

=

7

Try 4x6 (dimension lumber):

Example Design Sawn Lumber ColumnFrom Table 4A for No 1 DFL:

Fc= 1500 psi, Emin= 620,000 psi; A = 19.25 in2, CF= 1.1

(le/d)max=kle/dy=1(84)/3.5 = 24; Emin = Emin = 620,000psi


= 885 psi; Fc* = 1500(1.0)(1.1) = 1,650 psi

= 0.46 ;

Fc= 1650(.46) = 757 psi

Pallow.= Fc A = .757(19.25) = 14.57k < 19k NG P = 19k D + L

L

=

7


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Trial 2: Try 6x6 (P&T); From Table4Dfor No 1 DFL:

Fc= 1000 psi, Emin= 580,000 psi, CF= 1.0, A = 30.25 in2

(le/d)max=kle/dy=1(84)/5.5 = 15.3; Emin=Emin=580,000psi


= 2037 psi; Fc* = 1000(1.0)(1.0) = 1,000 psi

= 0.87 ;

Fc= 1000(.87) = 870 psi

Allow. P = Fc A = .87(30.25) = 26.3 k > 19 kOK,

Use 6x6 column, No. 1 DF-L

P = 19k D + L

L

=

7

Example Design Sawn Lumber Column

Example: Axial Capacity of Glulam

Determine if a 3-1/8x6 axial-

load glulam combination 2 DF

works for the previous

example.

MC less than 16%, normal

temp.

Loads are D+L assumed to

govern

Different unbraced lengths

about x and y axes (due to

installed kickers).

5

5

3

1

5


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Example: Axial Capacity of Glulam

A = 18.75 in2, and from Table 5B: Fc= 1,950 psi;

Exmin = 830,000psi

Eymin= 830,000psi

(le/d)x=1(7)(12)/6 =14; Governs

(le/d)y=1(3.5)(12)/3.125 =13.44

c = 0.9 for glulam; Emin= Emin

= 3,481 psi; Fc* = 1,950(1.0) = 1,950 psi

= 0.91;

Fc= Fc* (CP) = 1,950(0.91) = 1,767 psi

P = Fc A = 1.767(18.75) = 33.1 k > 19 k OK

In this case, same Emin properties

apply for x and y axes, but could

be different in glulam.

Built-Up Columns

Constructed from several parallel wood members

that arenailed or bolted togetherto function as a

composite column NDS Section 15.2

Connectingfasteners do not fully transfer the shear

between the various pieces, and capacity of a built-

up column is less than that of a solid sawn or glulam

column of same size and grade

Capacity of an equivalent solid column is reduced by

anadjustment factor Kf(only applies to the columnslenderness ratio for the axis parallel to the weak

axis of individual laminations)


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Built-Up Columns

Questions?


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Combined Axial and Bending

Forces


Effects ofsimultaneousbending and tensile stresses

must be considered

On one face, axial stressesaddto bending stresses, on

other face theycanceleach other

Therefore, capacity of member is either governed by

combined tension criterion or by net compression

criterion


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Interaction equation:

Because actual bending stress fbis a bending tensile

stress, F*bis used (does not include CL)

If combined stress is compressive, bending analysis

must be performed (CLis included but Cvis not):

Fb** = Fb CLapplies, but CVdoes not


Designer should be cautious about net compressive

stress:maximum bending compressive stress may occur

with or without tensile stress to cancel it.

Truss bottom chords are good example (tensile force

may or may not be present)

Conservative approach is toignore reduction in bending

stress due to tension:

Fb** = Fb CLapplies, but CVdoes not


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Combined Biaxial Bending and Tension

Axial tension and bending tension:

Net Compressive stress

Example: Combined Bending and Tension

Truss example similar to

earlier example

additional uniform load

applied to bottom chord

Use Select Structural DFL

with MC


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44 psf (4 ft o.c.) = 176 plf

4(6) = 24

5

5

P/2=528+408 = 936 lb

6(176) = 1056 lb

1056 lb

RA= 2,928 lb

2,928 lb

W = 17 psf (4 ft o.c.) = 68 plf

P = 1056 lb

P = 68(12)= 816 lb

936 lb


Trial and error procedure try a nominal 2x8:

Ag=10.88 in.2; S=13.14 in.3; An=1.5(7.25 9/16) = 10.03 in

2

Fb= 1500 psi, Ft= 1000 psi; cF=1.2 for tension & bending

Check axial tension at An: ft= T/An= 4.78/10.03 = 476 psi

Ft= Ft(cD)(cF) = 1000(1.15)(1.2) = 1380 psi > 476 psi OK

Check Bending: M=wl2/8= 14,688 in-lb; fb= M/S= 1118 psi

Fb= Fb(cD)(cF) = 1,500(1.0)(1.2) = 1800 psi > 1118 psi OK

RA= 2,928 lb

936 lb

2,928 938 = 1,990 l b

1,990(12)/5 = 4,776 lb

F

TAC= 4,776 lbA

176 plf

24

5

5

P/2 = 936 lb

1056 lb

1056 lb

RA= 2,928lb

2,928lb

W = 68 plf

P = 10 56 lb

P = 816 lb

936 lb

FBDW = 68 plf

12

4,776 k



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Combined Stresses:

Fb= 1500 psi, Ft= 1000 psi;

cF= 1.2 for tension & bending

Ag =10.88 in.2; S = 13.14 in.3;

ft=T/Ag= 4,776/10.88 = 439 psi

F*b= Fb(cD)(cF) = 1500(1.15)(1.2) = 2,070 psi

ft/Ft+ fbx/F*bx= 439/1380+1118/2070 = 0.86 < 1 OK

(D+S governs; typically would need to check D alone too)

Net bending compressive stress: automatically ok, since we already

checked it.

Also, since combined stress index is0.86, we theoretically could find

a smaller size that would work (likely wont).

In general, CSI values of 1.02 or 1.03 could be acceptable.

W = 68 plf

12

4,776 k



NDS interaction formula considers column buckling,

lateral torsional buckling of beams, and beam-column

interaction In a beam-column, additional bending stress is created

due to P- effect

To analyze combined stresses, following convention

used:

Column buckling isgoverned by whichever the

larger slenderness ratio is(Fcincludes CP)

When bending momentabout x axis, the value of

FCEfor use in amplification factor is to bebased on

(le/d)x


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General formula:

Where


Effect of column buckling, lateral torsional buckling,

and P-D effect is demonstrated on the following

interaction diagrams:


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Biaxial Bending and Compression

General formula:

where:

Truss T.C. Example: Beam-Column

Determine required size of truss top chord

D + S loads, effects of roof slope already

accounted for

Connections made with a single row of in.

bolts

Trusses 4 ft o.c., lumber is No. 1 Dense

Southern Pine

MC < 19%, normal temperatures, top chord

laterally bracedL =6 f t

5

RA= 2,928 lb

936 lb

2,928 938 = 1,990 lb

1,990(12)/5 = 4,776 lb

F

TAC= 4,776 lbA

5,174 lb


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Example Beam-Column cont.

Try 2x8:A = 10.88 in.2; S = 13.14 in.3 CF= 1 for both C & BFc= 1,800 psi, Fb= 1,650 psi, Emin= 620,000 psi

Axial check: fc= P/A = 5174/10.88 = 476 psi;

Stability: (le/d)x=kle/dx=1(6.5)(12)/7.25 = 10.75;

Emin= Emin = 620,000 psi; c = 0.8 for visually graded lumber

= 4,410 psi; Fc* = 1,800(1.15)(1.0) = 2,070 psi

= 0.88; Fc= 2,070(.88) = 1,822 psi

Fc= 1,822 > fc= 476 OK

x

x

L =6 f t

5


From before:Fc= 1,800 psi, Fb=1,650 psi, Emin= 620,000 psi

2x8: A = 10.88 in.2; S = 13.14 in.3 CF= 1 for both C & B

Net section compression check (not really needed here):An=1.5(7.25 (9/16)) = 10.03 in

2

fc= P/A = 5,174/10.03 = 516 psi

Fc= Fc* = 1,650(1.15)(1.0) = 1,898 psi > 516OK

Bending Check:

M = wl2/8 = 176(6)2/8 = 792 ft-lb; fb= M/S = 723 psi

Fb= 1800(1.15)(1.0) = 2,070psi > 723 psiOK

L =6 f t

5


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From before:Fc= 1,800 psi, Fb=1,650 psi, Emin= 620,000 psi

2x8: A = 10.88 in.2; S = 13.14 in.3 CF= 1 for both C & B

Combined Stress check :

CDbased on shortest duration of load in combination

FcEx= FcE= 4,410 psi (coincidence that le/dx= le/dmax)

fb= 849 psi; Fbx= 2,070 psi; fc= 476 psi ; Fc= 1,822 psi(476/1822)2 + (1/(1 476/4410))(723/2070) = 0.46 < 1 OK

Can use 2 x 8 No. 1 SP, but

perhaps another, weaker

grade or a smaller size works.

L =6 f t

5

Beam Columns with Transverse and

Eccentric Loads

In most cases with square-cut column ends, effects of

eccentricity are ignored by designers

Minimum recommended eccentricity to consider is 1 in.

or one tenth of column width.

No code requirement to design for minimum eccentricload left to thejudgment of engineer


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Example Exterior Wall Beam-Column Check column in

exterior wall that

supports gravity and

wind loads

Consider two L.C.:L.C.1: D+.75(L)+.75(S)

L.C. 2: D + W(there are others that may govern:D+L; D+S; D+.75W+.75L+.75 S)

Try 4x4 SPF#2 column,disregard self weight.

8 ft

8 ft

125

Large double windows

on either side

D + L = 7 +10 = 17 psf

D+S =14+30 = 44 psf

W

=20psf

6 6

85

1

2

1260 lb

360lb

4 x 4

5.524

Example Exterior Wall Beam-Column

Gravity Loads:

PD= 14(6)(12)+7(6)(6)

PD = 1260 lb

PL= 10(6)(6) = 360 lb

PS= 30(6)(12) = 2160 lb

PT= PD+.75PL+.75PS =

= 1260+.75(360+2160)

PT = 3,150 lb

8 ft

8 ft

125

TA 2 Attic

TA 1 Roof

PS = 2160 lb

20 psf


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L.C. 1, Gravity loads: PD+ 0.75 (PL+S) = 3.15 k; cD= 1.15;

Neglect column fixity @ ends, assume k=1:

(le/d)x= 1(8)(12)/3.5 = 27. 4

(le/d)y= 1(5)(12)/3.5 = 17.14; 27.4 governs

Emin= 510,000 psi, c = 0.8 for sawn lumber

= 558 psi; Fc* = 1150(1.15) = 1,322 psi

= 0.38;Fc= 1,322(0.38) = 502 psi

fc = P/A=3,150/12.25 = 257 psi OK

OK

From Table 4A:Fc = 1150 psi;

Fb = 875 psi;

Emin= 510,000 psi

A =12.25 in2;

S = 7.15 in3

xx


8

5

1

2

PT= 3.15k

840lb

20 psf

L.C. 2: (D + W)

Axial alone (dead load): PD= 1260 lb

(le/d)max= (le/d)x= 27.4 from beforeEmin= 510,000 psi, c = 0.8 for sawn lumber

= 558 psi; Fc* = 1150(1.6) = 1,840 psi

1.6 taken throughout combined stress check

= 0.28;

Fc= 1840(0.28) = 517 psi

fc = P/A = 1260/12.25 = 103 psi

fc/Fc=0.2

xx

Example Exterior Wall Beam-ColumnFc = 1150 psi;

Fb = 875 psi;

Emin= 510,000 psi

A =12.25 in2;

S = 7.15 in3

8

5

1

2

PD= 1.26k

840lb

20 psf


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L.C. 2 (D +W):Bending alone (wind)

Wind on header: w1= 20(3) = 60 lb/ft

Wind on sill: w2= 20(3.5) = 70 lb/ft

Reaction on column from headers: P1= 60(6) = 360 lb

Reaction on column from sills: P2= 70(6) = 420 lb

Mx= 720(12) = 8640 in-k; f b= M/S = 8,640/7.15 = 1208 psi

Fb=Fb(cD)(cL)

Load & shear & momentWall Framing and Tributary heights


1260 lb

.5

3

1

3.5

P=1.26k

5

P=1.26k P=1.26k

P=1.26k

8

P1=.36k

P2=.42k

2

1

420

360

60

420

720

Fc = 1150 psi;Fb = 875 psi;

Emin= 510,000 psi

A =12.25 in2;

S = 7.15 in3

L.C. 2Bending alone (wind) - continued:

Find cL: Lu= 5 ft (window height); Lu/d=60/3.5=17.1

14.3 17.1: Le

= 1.84Lu

= 110.4 (Table 3.3.3)** formula applies conservatively for all cases

= 5.6; = 19,402; Fb*= 875(1.6) = 1400 psi

= 0.996;

Fb= 1400(.996) = 1,395 psi

Stress ratio:

fb/Fb= 1208/1395 =0.866

Example Exterior Wall Beam-ColumnFc = 1150 psi;

Fb = 875 psi;

Emin= 510,000 psi

A =12.25 in2;

S = 7.15 in3


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L.C. 2Combined Stresses D+W:

(le/d)bending= (le/d)x= 27.4

= 558; From before:

fc= 103 psi; fc/Fc= 0.2; fb/Fb= 0.866

Amplification factor: 1/(1-fc/FcE) = 1/(1-103/558) = 1.226;

= (0.2)2 + 1.226(0.866) = 1.1 NG

NOTE:4x6 would work for sure, but more effective to use better grade

NOTE:D+L; D+S; D+.75W+.75L+.75 S should also be checked

NOTE: Shear, deflection, and bearing perpendicular to grain need checking too

also need to be checked.

xx

Example Exterior Wall Beam-ColumnFc = 1150 psi;Fb = 875 psi;

Emin= 510,000 psi

A =12.25 in2;

S = 7.15 in3

QUESTIONS?


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Example 7.2 Size truss bottom chord

Determine required size of truss bottom chord Loads only applied to top chord

Loads assumed to be applied to T.C. joints

Use method of joints truss joints assumed pinned

D + S loads, effects of roof slope already accounted for

Connections made with a single row of in. bolts

Trusses 4 ft o.c., lumber No. 1 SPF South

MC < 19%, normal temperatures

Example 7.2 Size truss bottom chord

Find forces by method of joints; T = 3.96k

Factors cD = 1.15, and assume cF = 1.3

Determine Ft= Ft(cD) (cF) = 400 (1.15)(1.3) = 598 psi

Find required An= P/Ft= 3960/598 = 6.62 in.2 Consider bolt holes:

Reqd Ag= An+Ah= 6.62+1.5(3/4 + 1/16) = 7.84 in.2

Try 2 x 6: A = 8.25 in.2 > 7.84 in.2 OK

Backcheck size factor cF = 1.3 - same as assumed.

FBD


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Example 7.7 Design Sawn Lumber Column

Design column, No. 1 Douglas Fir - Larch Bracing the same for buckling about x and y axes

Dead and roof live load apply

Temperature, moisture, incision do not apply


Try 4x6 (dimension lumber); From Table 4A for No 1 DFL:

Fc= 1500 psi, Emin= 620,000 psi, CF= 1.1, A = 19.25 in2



= 433 psi; Fc* = 1500(1.25)(1.1) = 2062 psi

S = 0.2 ;

Fc= 2062(.2) = 412 psi

Allow. P = Fc A = .412(19.25) = 7.94 k < 15 kNG


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Trial 2: 6x6 (P&T); From Table 4D for No 1 DFL:Fc= 1000 psi, Emin= 580,000 psi, CF= 1.0, A = 30.25 in

2



= 1003 psi; Fc* = 1000(1.25)(1.0) = 1250 psi

S = 0.611 ;

Fc= 1250(.611) = 764 psi

Allow. P = Fc A = .764(30.25) = 23.1 k > 15 k

OK, use 6x6 column, No. 1 DF-L

Example 7.9 Axial Capacity of Glulam

Determine axial compression capacity of 6-3/4 x 11

24F-1.7E Southern Pine glulam

MC will exceed 16% (consider treatment), normal temp

Loads are D+S assumed to govern Different unbraced lengths about x and y axes,

different material properties for x and y axes

A = 74.25 in2


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Example 7.9 Axial Capacity of Glulam

Table 5A: Fc= 1000 psi (CM= 0.73),Exmin= 880,000 psi (CM= 0.833)

Eymin= 670,000 psi (CM= 0.833)

(le/d)x=1(22)(12)/11 = 24; Exmin= 860,000(.833) = 733000 psi

(le/d)y=1(11)(12)/6.75 = 19.6; Eymin= 670,000(.83) = 558000 psi

c = 0.9 for glulam

= 1046 psi; Fc* = 1000(1.15)(0.73) = 840 psi

=1200 psi; Fc* is the same;

cPx= .832; cPy= .867; Fc= 1000(1.15)(.73)(.832) = 699 psi

P = Fc A = .699(74.25) = 51.9 k

y

y

y

x

x

Example 7.11 Capacity of a Bearing Wall

No bending although on exterior walls there could be

wind loads

Continuous lateral support provided in x direction

Need to consider bearing capacity of top and bottom plates Load is D + S

Lumber is Standard grade Hem-Fir, 2 x 4 studs

No special temperature and moisture requirements


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Example 7.11 Capacity of a Bearing Wall

Table 4A: Fc= 1300 psi, Fc= 405 psi, Emin= 440,000 psi,CF= 1.0 for compression, A = 5.25 in

2

le= 9.5(12) 3(1.5) = 109.5 in.

(le/d)x= 1(109.5)/3.5 = 31.3; Emin = Emin = 440,000psi


= 370 psi; Fc* = 1300(1.15) = 1495 psi

= 0.233 ;

Fc= 1495(.233) = 348 psi;

P = Fc A = 348(5.25) = 1829 lb ; max w = 1829/1.33 = 1371 lb/ft

Fc= Fc= 405 psi > 348 psi; column capacity

governs

x

x

Example 7.15 Combined Bending and Tension

Truss example similar to truss in example 7.2

additional uniform load applied to bottom chord

Use No.1 and Better Hem-Fir with MC


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Trial and error procedure try a nominal 2x8 first: Fb= 1100 psi, Ft= 725 psi; cF= 1.2 for tension & bending

Ag=10.875 in.2; S=13.14 in.3; An=1.5(7.25 (13/16)=9.66 in

2

Check axial tension at An: ft= T/An= 4.44/9.66 = 460 psi

Ft= Ft(cD)(cF) = 725 (1.15)(1.2) = 1000 psi > 460 psi OK

Check Bending: M = wl2/8=10,800 in-lb; fb= M/S = 822 psi

Fb= Fb(cD)(cF) = 1100(0.9)(1.2) = 1188 psi > 822 psi OK


Combined Stresses:

Fb= 1100 psi, Ft= 725 psi; cF= 1.2 for tension & bending

Ag=10.875 in.2; S=13.14 in.3; ft=T/Ag=4440/10.875= 408psi

F*b= Fb(cD)(cF) = 1100(1.15)(1.2) = 1518 psift/Ft+ fbx/F*bx= 408/1000+822/1518 = 0.95 < 1 OK

(D+S governs, but D alone is close)

Net bending compressive stress: automatically ok, since

we already checked it.

Also, since combined stress index is 0.95, we are likely

not going to find a smaller size that would work.

In general, CSI values of 1.02 or 1.03 could be

acceptable.


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Example 7.18 Beam-Column

Determine required size of truss top chord (example 7.15)

D + S loads, effects of roof slope already accounted for

Connections made with a single row of in. bolts

Trusses 4 ft o.c., lumber is No. 1 Southern Pine

MC < 19%, normal temperatures, top chord laterally braced

Example 7.18 Beam-Column Try 2x8: A = 10.875 in.2; S = 13.14 in.3 CF= 1 for both C & B

Fc= 1650 psi, Fb= 1500 psi, Emin= 620,000 psi

Axial: Stability check: fc= P/A = 4960/10.875 = 456 psi

(le/d)x=kle/dx=1(8.39)(12)/7.25 = 13.9;Emin=Emin= 620,000psi c = 0.8 for visually graded lumber

= 2638 psi; Fc* = 1650(1.15)(1.0) = 1898 psi

S = 0.791; Fc= 1898(.791)=1501 psi

1501 > 456 OK

x

x


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Try 2x8: A = 10.875 in.2; S = 13.14 in.3 CF= 1 for both C & B


Net section check:

An=1.5(7.25 (13/16)) = 9.66 in2

fc= P/A = 4960/9.66 = 514 psi

Fc= Fc* = 1650(1.15)(1.0) = 1898 psi > 514 OK

Bending Check:

M = wl2

/8 = 0.176(7.5)2

/8 = 1.24 ft-k; fb= M/S = 1130 psiFb= 1500(1.15)(1.0) = 1725 psi > 1130 psi OK


Try 2x8: A = 10.875 in.2; S = 13.14 in.3 CF= 1 for both C & B


Combined Stress check :

CDbased on shortest duration of load in combination

FcEx= FcE= 2638 psi (coincidence that le/dx= le/dmax)

Fb= 1130 psi; Fbx= 1725 psi; fc= 456 psi; ; Fc= 1501 psi

(456/1501)2 + (1/(1 456/2638)(1130/1725) = 0.884 < 1 OK

Use 2 x 8 No. 1 SP


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Example 7.20 Glulam Beam-Column

Check glulam column that supports gravity and wind loads L.C. 1: D +LR ; L.C. 2: D + W

Axial-load glulam combination: 5-1/8x7-1/2; 2 DF glulam


L.C. 1: Gravity loads: D + LR = 5+4 = 9k; cD= 1.25;

(D alone should be checked as well)

neglect column fixity @ ends, assume k=1:

(le/d)x= 1(16)(12)/7.5 = 25.6(le/d)y= 1(8)(12)/5.125 = 18.7 < 25.6

Exmin= Eymin= Emin= 830,000 psi, c = 0.9 for glulam

= 1041 psi; Fc* = 1950(1.25) = 2438 psi

= 0.4; Fc= 2438(0.4) = 975 psi

fc = P/A = 9000/38.4 = 234 psi;

OK

Table 5B:

Fc=1950 psi;

Fbx=1700 psi;

Exmin= 830,000psi

Eymin = 830,000psi

A = 38.4 in2;

S = 48 in3

xx


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L.C. 2: Could be (D +W) or (D + 0.75(W + Lr))(predetermined that D+W governs)

Axial (dead load):

(le/d)max= (le/d)x= 25.6 from before

Emin= 830,000 psi, c = 0.9 for glulam

= 1041 psi; Fc* = 1950(1.6) = 3120 psi

1.6taken throughout combined stress check

= 0.319; Fc= 3120(0.319) = 995 psi

fc = P/A = 5000/38.4 = 130 psi

fc/Fc= 0.131

Table 5B:Fc=1950 psi;

Fbx=1700 psi;

Exmin= 830,000psi

Eymin = 830,000psi

A = 38.4 in2;

S = 48 in3

xx


L.C. 2(D +W): Bending (wind)

Wind on header: w1

= 22.2(6.5) = 144 lb/ft

Wind on sill: w2= 22.2(5.5) = 122 lb/ft

Reaction on column from headers: P1= 144(12) = 1728 lb

Reaction on column from sills: P2= 122(12) = 1463 lb

Mx= 87.8 in-k; f b= M/S = 87.8/48 = 1830 psi

Fb=Fb(cD)(cL) or Fb=Fb(cD)(cV); determine which governs

Load & shear & moment

Table 5B:

Fc=1950 psi;

Fbx=1700 psi;

Exmin= 830,000psi

Eymin = 830,000psi

A = 38.4 in2;

S = 48 in3

Wall Framing and Tributary heights


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L.C. 2Bending (wind):

FindcL: lu= 8 ft (window height); lu/d=96/7.5=12.8.

7 12.8 14.3: le= 1.63lu+3d = 179 (Table 3.3.3)*

* formula applies conservatively for all cases

= 7.15; = 19,483; Fb*= 1700(1.6)=2720 psi

= 0.993

FindcV: (21/L)1/x(12/d)1/x(5.125/b)1/x x for DF glulam = 10

Cv= (21/16).1(12/7.5).1(5.125/5.125).1 =

Cv= 1.007 > 1 : Cv= 1, CLgoverns:

Fb= 1700(1.6)(.993) = 2700 psi

Stress ratio: fb/Fb= 1830/2700 = 0.678

Table 5B:

Fc=1950 psi;

Fbx=1700 psi;

Exmin= 830,000psi

Eymin = 830,000psi

A = 38.4 in2;

S = 48 in3


L.C. 2Combined Stresses:

(le/d)bending= (le/d)x= 25.6 coincidence, could differ

= 1041; From before:

fc= 130 psi; fc/Fc= 0.131; fb/Fb= 0.678

Amplification factor: 1/(1-fc/FcE) = 1/(1-130/1041) = 1.14;

= (0.131)2 + 1.14(0.678) = 0.79OK

Note: if D + 0.75(W + LR) was used, the interaction valuewould be considerably less.

NOTE: shear, deflection, and bearing perpendicular to

grain also need to be checked

Table 5B:

Fc=1950 psi;

Fbx=1700 psi;

Exmin= 830,000psi

Eymin = 830,000psi

A = 38.4 in2;

S = 48 in3

xx

design of axial members - wood

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