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7/21/2019 Design of Axial Members - Wood
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www.sgh.com
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
Axial Tension Members
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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Axial Tension Members
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
Slenderness Considerations
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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Slenderness Considerations
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
Slenderness Considerations
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
c = 0.8 for visually graded lumber
= 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
c = 0.8 for visually graded lumber
= 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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www.sgh.com
Combined Axial and Bending
Forces
Combined Bending and Tension
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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Combined Bending and Tension
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
Combined Bending and Tension
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
Example: Combined Bending and Tension
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
Example: Combined Bending and Tension
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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
Example: Combined Bending and Tension
Combined Bending and Compression
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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Combined Bending and Compression
General formula:
Where
Combined Bending and Compression
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
Example Beam-Column cont.
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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Example Beam-Column cont.
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
Example Exterior Wall Beam-Column
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
Example Exterior Wall Beam-Column
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
Example 7.7 Design Sawn Lumber Column
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
(le/d)max=kle/dy=1(120)/3.5 = 34.3; Emin=Emin=620,000psi
c = 0.8 for visually graded lumber
= 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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Example 7.7 Design Sawn Lumber Column
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
(le/d)max=kle/dy=1(120)/5.5 = 21.8; Emin=Emin=580,000psi
c = 0.8 for visually graded lumber
= 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
c = 0.8 for visually graded lumber
= 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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Example 7.15 Combined Bending and Tension
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
Example 7.15 Combined Bending and Tension
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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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
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
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
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
Example 7.20 Glulam Beam-Column
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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Example 7.20 Glulam Beam-Column
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
Example 7.20 Glulam Beam-Column
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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Example 7.20 Glulam Beam-Column
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
Example 7.20 Glulam Beam-Column
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