APPLYING WOOD ADJUSTMENT FACTORS TO DETERMINE STRESSES
Solution
Example 6.6
A glued laminated 6¾ × 30 beam of stress class
24F-1.7E western species is simply supported
over a span of 40 ft. The governing load
combination is a uniformly distributed dead
plus live load, and the beam is laterally braced
at midspan. Determine the adjusted allowable
bending stress and the adjusted factored
bending stress.
C M  1.0, Ct  1.0,
Cc  1.0, C fu  1.0, C I  1.0
From NDS Supp. Table 5A.
Fb  2400 lbf/in 2
Ey (min)  0.69  106 lbf/in 2
From Ex. 6.4, the volume factor is
CV  0.832
The distance between lateral restraints is
lu 
40 ft
2
 20 ft
 in 
(20 ft) 12 
 ft 
lu

d
30 in
 8.0
7
For a uniformly distributed load and an
𝑙𝑙𝑢𝑢 ⁄𝑑𝑑 ratio > 7, the effective length is
obtained from Fig. 6.1 as
ASD Method
The adjusted modulus of elasticity for stability
calculations with CM = Ct = 1.0 is
le  1.63lu  3d
 in 
 (1.63)(20 ft) 12   (3)(30 in)
 ft 
= 481 in
The slenderness ratio is given by NDS Eq.
3.3-5.
RB 
 0.69  106 lbf/in 2
b2

(481 in)(30 in)
(6.75 in)2
 17.80
 50 [satisfies criteria of
NDS Sec. 3.3.3]
The critical buckling design value is
FbE 
  Ey (min)C MCt
Emin

lbf 
 0.69  106
 (1.0)(1.0)
2

in 
led

1.20Emin
2
RB


1.200.69  106
lbf 

in 2 
2
17.80
 2613 lbf/in 2
The load duration factor for occupancy live
load is
C D  1.00
The reference flexural design value
multiplied by all applicable adjustment
factors except 𝐶𝐶𝐿𝐿, 𝐶𝐶𝑉𝑉, and 𝐶𝐶𝑓𝑓𝑓𝑓 is
Fb*  FbC MCtCcC I

lbf 
 2400
 (1.0)(1.0)(1.0)(1.0)(1.0)

2
in 
The beam stability factor is given by
NDS Sec. 3.3.3 as
1.0  F

CL 
1.9
 2400 lbf/in 2
F
FbE
Fb*
2613

2400
1.0  1.09


1.9

2
 1.0  F   F
0.95
 1.9 
 1.0  1.09 2 1.09

 


1.9
0.95
lbf
 0.85
in 2  1.09
lbf
 volume factor derived in Ex. 6.4
in 2
The volume factor governs. The adjusted allowable flexural stress is
Fb  FbC DC MCtC fuCcC ICV

lbf 
 2400
 (1.0)(1.0)(1.0)(1.0)(1.0)(1.0)(0.832)

2
in 
 1997 lbf/in 2
LRFD Method
The critical buckling LRFD design value is
From Table 6.4, the time effect factor for
dead load plus occupancy live load is
  0.8
FbE 
The adjusted modulus of elasticity for LRFD
stability calculations with 𝐶𝐶𝑡𝑡 = 𝐶𝐶𝑀𝑀 = 1.0 is

1.20Emin
2
RB

lbf 
(1.2) 1.032  106

2


in

(17.80)2
  EminC MCt KF s
Emin
 3909 lbf/in 2

lbf 
 0.69  106
 (1.0)(1.0)(1.76)(0.85)

5
in 
 1.032  106 lbf/in 2
The reference flexural design value multiplied by
all applicable LRFD adjustment factors except
𝐶𝐶𝑉𝑉, 𝐶𝐶𝐿𝐿, and C𝑓𝑓𝑓𝑓 is
Fb*  FbC MCtCcC I KF b

lbf 
 2400
 (1.0)(1.0)(1.0)

in 2 
 (1.0)(2.54)(0.85)(0.8)
 4145 lbf/in
F
FbE
Fb*
2
3909

4145
The LRFD beam stability factor given by NDS Sec.
3.3.3 is
 1.0  F 2
F
1.0  F

 
CL 
 

1.9
0.95
 1.9 
1.0  0.943


1.9
 0.792  0.832

2
 1.0  0.943   0.943
1.9
0.95


[volume factor derived
from Ex. 6.4]
lbf
in 2  0.943
lbf
in 2
Note: The CL equation given is actually a modified
version from SE Structural Engineering Reference
Manual and not the equation from NDS Sec. 3.3.3.
The stability factor governs, and the adjusted factored bending stress is
Fb  FbC MCtC LC fuCcC I KF b

lbf 
 2400
 (1.0)(1.0)(0.792)(1.0)(1.0)(1.0)(2.54)(0.85)(0.8)

2
in 
 3283 lbf/in 2
DESIGN OF WOOD DEMAND/CAPACITY RATIO
A beam is subjected to a dead load stress of
800 lbf/in2, a live load stress of 600 lbf/in2, and
a wind load stress of 500 lbf/in2. The beam is
made from lumber with a reference design
strength value of 1500 lbf/in2, and is used in a
situation where all adjustment factors except
the load duration factor are 1.0. Find the
demand/capacity ratio for all applicable IBC
load combinations.
Solution
From Table 6.1, the load duration factor for
dead load is 0.9. The allowable strength
design value for the dead load loading
condition is

lbf 
Fb  0.90Fb  (0.90) 1500


in 2 
 1350 lbf/in 2
The demand/capacity ratio for the dead load
loading condition is
fb
Fb
lbf
in 2
 600
lbf
in 2
For the dead load plus live load plus wind
load loading condition, the total applied
stress is
fb  D  0.45W  0.75L

lbf 
 (0.45) 500



in
in 2 

lbf 
 (0.75) 600


in 2 
lbf
2
 1475 lbf/in 2
The load duration factor for a wind load is
1.60, and the largest load duration factor
governs, so the allowable design value for
1350
in 2  0.593
lbf
in 2
 1500 lbf/in 2
 1400 lbf/ft2
The load duration factor for an occupancy
live load is 1.0, and the largest load duration
factor governs, so the allowable design value
for the dead load plus live load loading
condition is
 800

lbf

lbf 
Fb  1.0 fb  (1.0) 1500


in 2 
For dead load plus live load loading
condition, the total applied stress is
fb  800
800
The demand/capacity ratio for the dead load
plus live load loading condition is
fb

Fb
1400
1500
lbf
in 2  0.933
lbf
in 2
the dead load plus live load plus wind
load loading condition is

lbf 
Fb  1.60Fb  (1.60) 1500


in 2 
 2400 lbf/in 2
The demand/capacity ratio for the dead load
plus live load plus wind load loading
condition is
fb

Fb
1475
2400
lbf
in 2  0.61
lbf
in 2
DESIGN OF WOOD SHEAR
Example 6.8
A glued laminated 6¾ × 30 beam of
stress class 24F-1.7E western species
is notched and loaded as shown. The
beam has a moisture content
exceeding 16% and is subjected to
sustained temperatures between
100°F and 125°F. The governing load
combination is dead plus occupancy
live load. Determine the maximum
allowable shear force at each support
and at the hanger connection.
Solution
The reference design value for shear stress,
tabulated in NDS Supp. Table 5A, is
Fv  210 lbf/in 2
The applicable shear reduction factor for the
design of members at a notch is
From Table 6.7, the applicable temperature
factor for wet conditions between 100°F and
125°F is
Ct  0.7
From Table 6.6, the applicable wet service
factor for moisture content exceeding 16% is
C vr  0.72
ASD Method
For normal occupancy live load, the load
duration factor given in Table 6.5 is
C D  1.00
The adjusted allowable shear stress is
Fv  FvC DC MCtC vr

lbf 
 210
 (1.00)(0.875)(0.7)(0.72)

in 2 
 93 lbf/in
2
At the right support, 𝑒𝑒 < 𝑑𝑑𝑛𝑛′ and from NDS
Eq. 3.4-5, the allowable shear force is

d  d  
2
n e 
Vr  Fvb d 
 
3

 dn  
  lbf 

(2) 93
 (6.75 in)

 

in 2 

 


3






 30 in  27 in 




(12
in)
 30 in  




27 in




 11,997 ibf
C M  0.875
At the left support and from NDS Eq. 3.4-3,
the allowable shear force is
2
2
 d 
n


Vr   Fvbdn   
 3
  d 
  lbf 

(2) 93 2  (6.75 in)(27 in)  27 in 2
  in 

 
 




  30 in 
3




 9153 lbf
The hanger connection is less than 5𝑑𝑑 from
the end of the beam, and from NDS Eq. 3.4-6,
the allowable shear force is
2
2
 d 
e


Vr   Fvbde   
 3
  d 
  lbf 

(2) 93
 (6.75 in)(27 in)


2
  in 2 
  27 in 

  30 in 

3








 9153 lbf
LRFD Method
From Table 6.4, the applicable time effect factor
for dead load plus occupancy live load in
  0.8
From Table 6.3, the applicable resistance
factor for shear is
The adjusted factored shear stress is
Fv  FvC MCtC vr KF v

lbf 
 210
 (0.875)(0.7)(0.72)
2

in
 (2.88)(0.8)(0.75)
  
From Table 6.2, the applicable format
conversion factor for shear is
 160 lbf/in 2
KF  2.88
At the left support and from NDS Eq. 3.4-3,
the allowable shear force is
2
2
 d 
Vr   Fvbdn   n 
 3
  d 


(2) 160 lbf  (6.75 in)(27 in)
2

 
  27 in 
2

in

 
 

 30 in 
3




At the right support, 𝑒𝑒 < 𝑑𝑑𝑛𝑛, and from NDS
Eq. 3.4-5, the allowable shear force is

d  d  
2


n e 


Vr  Fvb d 
 
3
 dn  



(2) 160 lbf  (6.75 in)



 
in 2 

 


3






 30 in  27 in 




(12 in)
 30 in  





27 in


 15,747 lbf
 20,640 lbf
The hanger connection is less than 5d from the
end of the beam, and from NDS Eq. 3.4-6, the
allowable shear force is
2
2
 d 
e
Vr   Fvbde   
 3
  d 


(2) 160 lbf  (6.75 in)(27 in)


 27 in 2
 

2

 
in 
 




  30 in 
3




 15,747 lbf
DESIGN OF WOOD AXIAL + BENDING
Example 6.9
The select structural 2 × 6 Douglas fir-larch
top chord of a truss is loaded with the
service level loads shown in the illustration.
The governing load combination consists of
dead plus occupancy live load, and the
moisture content exceeds 19%. The chord is
laterally braced at midlength about the
weak axis, and the self-weight of the chord
and bracing members may be neglected.
Determine whether the member is
adequate.
Solution
The reference design values for compression
and modulus of elasticity are tabulated in
NDS Supp. Table 4A and are
Fc  1700 lbf/in 2
Emin  0.69  106 lbf/in 2
CT  1.0
Ct  1.0
The applicable adjustment factors for
compression and modulus of elasticity are as
follows.
C M  wet service factor from Table 6.6
= 0.8 [compression member]
= 0.9 [modulus of elasticity]
C F  size factor from NDS Supp. Table 4A
=1.1
[compression member]
C i  1.0
From the illustration, the slenderness ratio
about the strong axis is
 in 
(1.0)(8 ft) 12 
 ft 
Kel1

5.5 in
d1
 17.46
From the illustration, the slenderness ratio
about the weak axis is
 in 
(1.0)(4 ft) 12 
 ft 
Kel2

d2
1.5 in
 32.00
[governs]
ASD Method
The reference compression design value
multiplied by all applicable adjustment
factors except CP is given by
The adjusted modulus of elasticity for
stability calculations is

lbf 
  EminC M  0.69  106
Emin
(0.9)

in 2 
6
 0.62  10 lbf/in
Fc*  FcC DC MC F

lbf 
 1700
 (1.00)(0.8)(1.1)

in 2 
2
 1496 lbf/in 2
The load duration factor for dead load plus
occupancy live load from Table 6.5 is
C D  1.00
The critical buckling design value is
FcE 2 

0.822Emin
2
l 
 e 2 
 d2 
 498 lbf/in

The ratio of FcE 2 to Fc* is

lbf 
(0.822) 062  106


in 2 
F 
2

 in 
(4 ft) 12 
 ft 

 1.5 in





2
FcE 2
Fc*
498

1496
lbf
in 2
lbf
in 2
 0.333
The column parameter is obtained from
NDS Sec. 3.7.1.5 as
c  0.8 [for sawn lumber]
The column stability factor is specified by NDS
Sec. 3.7.1 as
CP
1.0  F 


2c

2
 1.0  F    F 
 2c 
c
1.0  0.333


(2)(0.8)
 1.0  0.333 2 0.333

 
 (2)(0.8) 
0.8
 0.31
The allowable compression design value
parallel to grain is
Fc  FcC MC FC P

lbf 
 1700
(0.8)(1.1)(0.31)

in 2 
 464 lbf/in 2
Applying IBC Eq. 16-9, the factored load is
P D L
 1000 lbf + 1800 lbf
= 2800 lbf
The actual compression stress on the
chord is given by
fc 
2800 lbf
P

A (1.5 in)(5.5 in)
 339 lbf/in 2
 Fc
The chord is adequate.
DESIGN OF WOOD AXIAL + BENDING
Example 6.10
The select structural 2 × 6 Douglas fir-larch
top chord of a truss is loaded with the
service level loads shown in the
illustration.
The
governing
load
combination consists of dead plus live
load, and the moisture content exceeds
19%. The chord is laterally braced at
midlength about the weak axis, and the
self-weight of the chord and bracing
members may be neglected. Determine
whether the member is adequate.
Solution
The distance between lateral restraints is
The reference design values for bending and
modulus of elasticity are tabulated in NDS
Supp. Table 4A, and they are
Fb  1500 lbf/in 2
Emin  0.69  106 lbf/in 2
C fu  1.0, C i  1.0,
C r  1.0, Ct  1.0
lu 
8 ft
2
 4 ft
From Fig. 6.1, for a concentrated load at
midspan and with lateral restraint at
midspan, the effective length for flexure is
le  1.11lu
 in 
 (1.11)(4 ft) 12 
 ft 
 53.28 in
The slenderness ratio for flexure is given by
NDS Sec. 3.3.3 as
RB 
led1
d22

(53.28 in)(5.5 in)
(1.5 in)2
 11.41
 50
 satisfies criteria of 


 NDS Sec. 3.3.3 
From Table 6.6, the applicable wet service
factor for flexure is
C M  0.85
From NDS Supp. Table 4A, the applicable size
factor for flexure is
C F  1.3
ASD Method
From Table 6.5, the load duration factor for
dead load plus occupancy live load is
The critical buckling design value for
flexure is
C D  1.00
From Ex. 6.9, the adjusted modulus of
elasticity for stability calculations is
  EminC M
Emin
FbE 


1.20Emin
2
RB

lbf 
(1.20) 0.62  106


in 2 
(11.41)2
 5715 lbf/in 2

lbf 
 0.69  106
 (0.9)

2
in 
 0.62  106 lbf/in 2
The reference flexural design value
multiplied by all applicable adjustment
factors except CL is
Fb*
 FbC MC F

lbf 
 1500
 (0.85)(1.3)
2

in 
 1657 lbf/in 2
F
FbE
Fb*
5715

1657
lbf
The beam stability factor is given by NDS
Sec. 3.3.3 as
 1.0  F 2
1.0  F
  F
 
CL 

 1.9 
1.9
0.95
1.0  3.45


1.9

2
 1.0  3.45   3.45


1.9
0.95
 0.98
in 2  3.45
lbf
in 2
The allowable flexural design value for load
applied to the narrow face is
Fb1  FbC MC LC F

lbf 
 1500
 (0.85)(0.98)(1.3)

in 2 
 1626 lbf/in 2
Applying IBC Eq. 16-9, the factored vertical
load is
W  1.0L  (1.0)(200 lbf)  200 lbf
The actual edgewise bending stress is
fb 1 

WL
4S
 in 
(200 lbf)(8 ft) 12 
 ft 
(4)(7.56 in 3 )
 635 lbf/in 2
From Ex. 6.9,
Kel1
d1
FcE 1 
 17.46

0.822Emin
l 2
 e 1 
 d 
 1

lbf 
(0.822) 0.62  106

2


in

Fc  464 lbf/in 2
fc  339 lbf/in 2

2
(8 ft) 12 in 


 ft 

 5.5 in





The critical buckling design value, in the
plane of bending, for load applied to the
narrow face is
 1672 lbf/in 2
The moment magnification factor for axial
compression and flexure with load applied to
the narrow face is
C m 3  1.0 
fc
The interaction equation for bending load
applied to the narrow face of the member
and concentric axial compression load is
given in NDS Sec. 3.9.2 as
 f 2
 c   fb 1  1.0
 Fc 
Fb1C m 3
FcE 1
339
 1.0 
1672
lbf
in 2
lbf
in 2
 0.797
The left side of the expression is
635 lbf

2
 339 lbf 

in 2  
in 2
 0.535  0.490



lbf
 464 lbf 
1626
 (0.797)

2

2



in
in 
 1.025
 1.0
The chord is adequate.
DESIGN OF WOOD POST AXIAL + BENDING
A short post is subject to an axial wind load
and a compressive dead load, as shown. The
post is made from select structural 4×10
Douglas fir-larch members (actual
dimensions: 3.5 in × 9.25 in) and is fixed at the
base and free at the top. The post is used such
that all adjustment factors except the load
duration factor and column stability factor are
1.0. Does the post satisfy the NDS interaction
equations for combined axial compression
and flexure?
Determine the applied stresses.
fc 
The reference design values for select
structural Douglas fir-larch members are
20,000 lbf
P

A (3.5 in)(9.25 in)
Fb  1500 lbf/ft2
Fc  1700 lbf/ft2
 618 lbf/in 2
Emin  690,000 lbf/ft2
25,000 in-lbf
M
fb 

1
S
  (3.5 in)(3.5 in)(9.25 in)
 6 
Find the effective length. The post is a
cantilever from the ground, so K = 2.1 in.
le  Kl  (2.1 in)(30 in) = 63 in
 1323 lbf/in 2
Find the compressive design value of the
member. The load duration factor for a dead
load plus wind load loading condition is 1.60.
Find the critical buckling design value. The
shortest dimension of the cross section will
buckle, so d = 3.5 in.
Fc*  FeC DC M

lbf 
 1700
(1.60)(1.0)

in 2 
 2720 lbf/in 2
FcE 

0.822Emin
l 2
 e 
 d 
 

lbf 
(0.822) 690,000

2


in

 1751 lbf/in 2
 63 in 2




 3.5 in 
The ratio of the compressive design value
and the critical buckling design value is
F 
FcE
Fc*
1751

2720
CP
lbf
1.0  F 


2c
 1.0  F  2 F 

 
 2c 
c
1.0  0.644


(2)(0.8)
2
in  0.644
lbf
 1.0  0.644 2 0.644

 
 (2)(0.8) 
0.8
 0.526
in 2
The allowable compression design
value is
Find the column stability factor. Select
structural Douglas fir-larch is a sawn lumber,
so c = 0.8.

lbf 
F  Fc*C P  2720
 (0.526)

in 2 
 1431 lbf/in 2
Find the moment magnification factor.
C m 1  1.0 
fc
FcE 1
618
 1.0 
Check the interaction equation. There is no biaxial bending,
so disregard the biaxial bending ratio.
 f 2
 c   fb 1  fb 2  1.00
 F  
Fb1C m 1 Fb2C m 2
 c
2
f 
fb 1
 c  
Fb1C m 1
 Fc 
1751
lbf
in 2  0.647
lbf
in 2

2
lbf
 617 lbf 
1323
2 

in  
in 2



 1431 lbf 
1500 lbf  (0.647)
(1.6)



in 2 
in 2 
 1.04 [ 1.00, not acceptable]
DESIGN OF WOOD CONNECTION WITH BOLTS
Example 6.12
A bolted connection in tension consists of a single row of
eight ¾ in diameter bolts in two select structural 2 × 6
Douglas fir-larch members in single shear. The governing
load combination consists of dead plus occupancy live load,
and the moisture content exceeds 19%. The bolt spacing and
end distance are 4 in. Determine the capacity of the
connection.
Solution
The ¾ in diameter bolt reference design
value for single shear is tabulated in NDS
Table 12A as
The specified minimum end distance for
the full bolt design value is specified in
NDS Sec. 12.5.1 as
a p  7D
Z  720 lbf
Ct  1.0
 (7)(0.75 in)
As  Am  8.25 in 2
= 5.25 in
The applicable adjustment factors for the bolts are as follows.
C M  0.7 [wet service factor from NDS Table 11.3.3]
C g  0.71 [group action factor from NDS Table 11.3.6A]
C 

actual end distance
specified minimum end distance
4 in
5.25 in
 0.76 [geometry factor from NDS Sec. 12.5.1]
ASD Method
From Table 6.5,
The allowable capacity for eight bolts is
C D  1.00
From Ex. 6.11, the allowable tension capacity
of the members is
T  FtAn


1 
 FtA  D  b 

16  



3
lbf  
1 
in (1.5 in)
 1300
8.25 in 2   in 


 4

16 
in 2  
 9140 lbf
T  nZC MC gC C DCt
 (8)(720 lbf)(0.7)(0.71)
 (0.76)(1.00)(1.0)
= 2176 lbf
[governs]
Example 6.16
DESIGN OF WOOD SHEAR RINGS
The Douglas fir-larch select structural
members shown in the illustration are
connected with 2 5/8 in shear plate
connectors. The governing load
combination consists of dead plus
occupancy live loads. The connector
spacing and end distances are as
shown. Determine the capacity of the
connection.
Solution
The specific gravity of Douglas fir-larch is
obtained from Table 12.3.3A as G = 0.5. From
NDS Table 13A, the species group is B. The
reference 2 5/8 in shear plate design value for
the 2½ in thick main member of group B
species with a connector on two faces is
tabulated in NDS Table 13.2B as
Pmain  2860 lbf
The reference 2 5/8 in shear plate design
value for a 1½ in thick side member of
group B species with a connector on one
face is tabulated in NDS Table 13.2B as
Pside  2670 lbf
CM  1.0, Ct  1.0
As  (2)(8.5 in 2 )  16.5 in 2
Am  18.13 in 
As
Am
The specified minimum spacing for the full
shear plate design value is given in NDS
Table 13.3 as
s  6.75 in
The applicable adjustment factors for the
bolts are as follows.
C g  group action factor from NDS
Table 11.3.6B
= 0.98
[governs]
 0.91
C   geometry factor from NDS
Sec. 13.3.2.1 for a spacing of 6 in
= 0.5 +
 0.885
(0.5)(6 in  3.5 in)
6.75 in  3.5 in
ASD Method
From Table 6.5, the load duration factor is
C D  1.0
The allowable design value for four shear
plates is
T  nPsideC gC C DC MCt
 (4)(2670 lbf)(0.98)(0.885)(1.0)(1.0)(1.0)
 9263 lbf
LRFD Method
From Table 6.4, the time effect factor for
occupancy live load is
The format conversion factor from NDS Table
11.3.1 is
The strength level design value for four shear
plates is
T  nPsideC gC C MCt KF 
 (4)(2670 lbf)(0.98)(0.885)
 (1.0)(1.0)(3.32)(0.8)(0.65)
 15,991 lbf
The resistance factor given by NDS Table
11.3.1 is
DESIGN OF WOOD NAIL CONNECTIONS
Example 6.19
Determine the lateral design value for the
3 in long 10d common wire nail in the
toe-nailed connection shown. Loading
applied to the connection is due to wind
load, and all members are Douglas firlarch.
As specified in NDS Sec. 12.1.5, toe nails are
driven at an angle of 30 degrees from the
face of the member, with the point of
penetration one-third the length of the nail
from the member end. In accordance with
NDS Comm. Sec. C12.1.5, the side member
thickness is taken to be equal to this end
distance and
ts 
C M  1.0
Ct  1.0
Z  118 lbf
L 3 in

 1 in
3
3
The applicable adjustment factors for the
nail are as follows.
•
The nominal design value for single shear is
tabulated in NDS Table 12N as
The penetration of the nail into the
main member, in accordance with NDS
Comm. Sec. C12.1.6, is taken as the
vertically projected length of the nail in
the member and
p  L cos 30 
L
3
 (3 in)(0.866) 
 1.60 in
3 in
3
• This is greater than 10D, and from NDS
Table 12N, the penetration depth factor is
Cd  1.0
• Ctn = toe-nail factor from NDS Sec. 12.54
Ctn  0.83
ASD Method
From Table 6.5, the load duration factor is
C D  1.60
The allowable lateral design value for the nail is
Z   ZC DCdCtnC MCt
 (118 lbf)(1.60)(1.0)(0.83)(1.0)(1.0)
= 157 lbf
LRFD Method
From Table 6.4, the time effect factor for
wind load is
  1.0
The format conversion factor given in NDS
Table 11.3.1 is
KF  3.32
The resistance factor given in NDS Table
11.3.1 is
  0.65
The strength level lateral design value for the
nail is
T  ZC MCtCdCtn KF 
 (118 lbf)(1.0)(1.0)(1.0)
 (0.83)(3.32)(1.0)(0.65)
= 211 lbf