1. No category

MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
CE-03014
DESIGN OF TIMBER STRUCTURES
WORKED-OUT EXAMPLES
B.Tech. (First Year)
Civil Engineering
Note
*
Must Know
**
Should Know
*** Could Know
1
Yangon Technological University
Department of Civil Engineering
Course No. CE 05024 + CE 06024 Design of Timber Structures I + II
Worked-out Examples
CE 05024 Design of Timber Structures I
Chapter 1
1.3 Design of Compression Members
1(1-5).* Determine an acceptable section surfaced-four-sides (S4S), for a laterallybraced double-hinged column 14 ft long, subjected to an axial load P = 40
Kips due to snow- loading conditions. The column is to be used in an exposed
structure and may be subjected to wetting for an extended period of time. Use
wood pressure- impregnated with preservatives. Consider wood for which
Fc = 1.3 ksi, F b = 1.6 ksi and E = 1800 ksi.
Solution:
Trial 1
Consider an 8 x 8 S4S section.
Actual dimensions = 7.5 x 7.5 in.
A = 56.2 in2
fc =
P
40
=
= 0.71 Ksi
A 56.2
For a double-hinged laterally-braced column, K = 1,
∴ kl =
d
14 x 12
= 22.40
7 .5
Fc = 1.15 x 0.91 x 1.3 = 1.36 Ksi
E = 1800 Ksi
∴ K = 0.671
E
1800
= 0.671
= 24.41
Fc
1.36
11 < Kl/d < K ∴Intermediate column:
 1  kl / d  ↑ 
F = Fc 1 − 
 
 3  K  
'
c
2
 1  22.4  4 
= 1.36 1 − 
 
 3  24.41  
=1.04 ksi
The ratio of actual to design compression stresses =
f c 0.71
=
= 0 .68 < 1
Fc' 1.04
Thus, the 8 x 8 S4S section is acceptable, but somewhat overdesigned.
Trial 2.
Consider 6 x 8 S4S section.
Actual dimensions = 5.5 x 7.5 in
A = 41.2 in2
fc =
40
= 0.971 Ksi
41.2
kl / d =
14 x 12
= 30.55
5 .5
K = 24.41
K < kl/d < 50 ∴Long column.
Fc' =
0 .3 E
0.3 x 1800
=
= 0.579 Ksi
2
(kl / d )
(30.55) 2
fc
=
'
c
F
0.971
= 1.68 > 1.0
0.579
∴6 x 8 S4S section is unacceptable.
∴Use 8 x 8 S4S timber
2(1-5).* Determine an acceptable section, surfaced four-sides (S4S), for a column 14
ft long subjected to the combined action of longitudinal load P = 40 Kips
with an eccentricity of 3 in at each end and a uniformly distributed transverse
load of 0.245 klf, all due to normal loading conditions ( α = 1.00). Sides way
of the column is prevented by bracing at the end supports but rotation of the
latter is unrestrained. The column is to be used in a covered structure and will
remain dry thereby requiring no downward adjustments in the allowable
stresses to increasing moisture content. Use wood for which Fc = 1.3 ksi,
Fb = 1.6 ksi and E = 1800 ksi.
3
Solution:
Trial 1.
Assume a 12 x 16 (S4S) section.
Actual dimension = 11.5 x 15.5 in.
(
A = 178.2 in , S = 460.5 in , CF = 12
2
3
)
1
9
15.5
= 0.972
narrow face = d1 = 11.5 in
long face
= d2 = 15.5 in
For a double-hinged laterally-braced, k = 1.0
Max: slenderness ratio
= kl/d1
=
14 x 12
11.5
= 14.61
E
1800
= 0.671
= 24.97 ≅ 25
Fc
1.3
K = 0.671
11 < kl/d1 < K ∴intermediate column.
 1  kl / d  4 
F = Fc 1 − 
 
 3  K  
'
c
 1  14.61  4 
= 1.3 1 − 
 
 3  25  
= 1.25 Ksi
le = 1.92 lw
= 1.92 x 14
= 26.88 ft
C s = le d
b
2
=
26.88 x 12 x 15.5
= 6.15 < 10
(11.5) 2
∴No correction due to lateral stability is required.
Fb' = C F Fb = 0.972 x 1.6 = 1.56 ksi
fc =
P
40
=
= 0.224 ksi
A 178.2
fb =
M
S
4
M=
1  0.245 
2
x
 x (14 x 12)
8  12 
= 72.0 K-in
72
= 0.156 ksi
460.5
fb =
( kl d − 11)
J=
kl
2
K − 11
d2
=
14 x 12
= 10.84
15.5
∴J = 0
Using interaction equation;
f c f b + f c ( 6e d ) (1 + 0.25 J )
+
Fc'
Fb' − J f c
 6 x 3
0.156 + 0.224 
 (1 + 0)
0.224
 15.5 
=
+
1.25
1.56 − 0
= 0.446 < 1.0
The 12 x 16 S4S section is acceptable, but somewhat big.
A closer section may be obtained using:
A2 = 0.446 x 178.2 = 79.5 in2
S2 = 0.446 x 460.5 = 205.4 in3
To use values somewhat larger than A2 and S2 , various S4S sections such as
8 x 14, 10 x 12, 8 x 16 and others may be selected.
Trial 2.
Select 8 x 14 S4S section.
Actual dimension = 7.5 x 13.5 in
1
A = 101.2 in2 , S = 227.8 in3 and CF = (12
kl/d =
14 x 12
= 22.4 , K = 25
7 .5
11 < kl/d < K; ∴ Intermediate column.
 1  22.4  4 
∴ F = 1.3 1 − 
  = 1.02 Ksi
 3  25  
'
c
13.5
) 9 = 0.987
5
le = 26.88 ft
CS =
26.88 x 12 x 13.5
= 8.80 < 10
7.5 2
∴ Fb' = C F Fb = 0.987 x 1.6 = 1.58 Ksi
fc =
40
= 0.395 Ksi
101.2
fb =
72
= 0. 316 Ksi
227.8
14 x 12
− 11
13
.
5
J =
= 0.10
25 − 11
Substituting in the interaction equation:
 6 x 3
0.316 + 0.395 
 ( 1 + 0.25 x 0.1)
0.395
 13.5 
+
= 0.943 < 1, O.K .
1.02
1.58 − 0.10 x 0.395
The section is acceptable.
∴Use an 8 x 14 S4S timber
3.(1-5).*Determine an acceptable section for a double- hinged braced column
subjected to P = 40 K, e = 3 in, w = 0.245 K/ft, and made out of wood for
which Fc = 1.3 ksi, Fb = 1.6 ksi and E = 1800 ksi. Consider that wind loading
conditions apply ( α = 1.33) and the column is braced at midheight by wood
girts attached to the wide face of the section.
Solution:
Trial 1.
Assume a 6 x 12 S4S section.
Actual dimensions = 5.5 x 11.5 in
A = 63.2 in2 , S = 121.2 in3 , CF = 1.0
For a double-hinged column braced also at midheight about its weak axis of
bending, K = ½ .
1
x 14 x 12
kl = 2
= 15.27
d1
5.5
14 x 12
At the strong axis of bending, kl d =
= 14.61 < 15.27
2
11.5
6
∴ max: slenderness ratio = 15.27
E
= 0.671 1800 (1.33 x 1.3) = 21.65
α Fc
K = 0.671
11 < kl/d < K ∴Intermediate column.
4

 kl  
 1  
'
Fc = α Fc 1 −  d  
 3K  
  

  

 1  15.27  4 
= (1.33 x 1.3) 1 − 
 
 3  21.65  
= 1.59 ksi
Unsupported length of beam, lu = 14 /2 = 7 ft.
le = 1.92 x 7 = 13.44 ft
C S = le d
b
2
=
13.44 x 12 x 115
= 7.83 < 10
5.5 2
∴No correction due to lateral stability is required.
∴ Fc' = C F x α x Fb = 1.0x 1.33 x 1.6 = 2.13 Ksi
fc =
fb =
P
40
=
= 0.633 Ksi
A
63.2
M
72
=
= 0.594 Ksi
S 121.2
 kl − 11
 d


2
 14.61 − 11
J=
=
= 0.34
K − 11
21.65 − 11
Using interaction equation,
 6e 
f b + f c   (1 + 0.25J)
fc
d 
+
'
Fc
Fb' − Jf c
 6 x 3
0.594 + 0.633
 (1 + 0.15 x 0.34)
0.633
 11 .5 
+
= 1.27 > 1 , N.G.
1.59
2.13 − 0.34
The 6 x 12 S4S section is not acceptable.
7
Trial 2.
Consider using a section having:
A = 1.27 x 63.2 = 80.3 in2 , S = 1.27 x 121.2 = 153.9 in3
Take a 6 x 14 S4S section
A = 74.2 in2 , S = 167.1 in3 ., CF = 0.987
1
x 14 x 12
kl = 2
= 15.27
d1
5.5
kl
=
d2
14 x 12
= 12.44
13.5
K = 21.65
11 < kl/d1 < K ∴Intermediate column.
 1  15.27  4 
F = 1.33 x 1.3 1 − 
  = 1.59 Ksi
 3  21.65  
'
c
le = 1.92 x 7 = 13.44 ft.
CS =
13.44 x 12 x 13.5
= 8.48 < 10
5.5 2
fc =
40
= 0.539 Ksi
74.2
fb =
72
= 0.431 Ksi
167.1
J=
12.44 − 11
= 0.14
21.65 − 11
Using interaction equation,
 6 x 3
0.431 + 0.539 
 (1 + 0.25 x 0.14)
0.539
 13.5 
+
= 0.919 < 1, O.K .
1.59
2.10 − 0.14 x 0.539
The section is acceptable.
∴ Use a 6 x 14 S4S section for the column.
4(1.5).* Determine three acceptable S4S sections, for a column 19 ft long subjected to
the combined action of longitudinal load P = 40 Kips with an eccentricity of
3 in., and a midspan moment M – 72 K- in. due to transverse loading. Assume
α = 1.0 and such wood that Fc = 1.3 ksi, Fb = 1.6 ksi and E = 1800 ksi. The
8
column is to be used in a covered structure, hence, no adjustments in the
allowable stresses are needed.
Solution:
Trial 1.
Assume a 10 x 14 S4S section
Actual dimension = 9.5 x 13.5 in
A = 128.25 in2 , S = 288.56 in3 ,. CF = 0.987
d1 = 9.5 in, d2 = 13.5 in
Assume double-hinged, laterally-braced col:, K = 1.0
kl
=
d1
19 x 12
= 24
9 .5
K = 0.671
1800
= 24.97 ≅ 25
1.3
11 < kl/d1 < 25 ∴Intermediate col:  1  24  4 
F = 1.3 1 −    = 0.932 Ksi
 3  25  
'
c
le = 1.92 x 19 = 36.48 ft
CS =
36.48 x 12 x 13.5
= 8.1 < 10
9.5 2
No correction due to lateral stability is required.
Fb' = C F Fb = 0.987 x 1.6 = 1.58 ksi
19 x 12
− 11
13
.
5
J=
= 0.421
25 − 11
fc =
40
= 0.312 Ksi
128.25
fb =
72
= 0.25 Ksi
288.56
Using interaction equation:
fc
+
Fc'
6e
) (1 + 0 .25 J )
d
Fb' − J f c
f b + fc (
9
=
0.312
+
0.932
6x3
) (1 + 0.25 x 0.421)
13.5
= 0.335 + 0.49 = 0.825 < 1.0
1.58 − 0.421 x 0.312
0.25 + 0.312 (
∴The 10 x 14 S4S section is acceptable.
Trial 2.
Assume a 8x18 S4S section.
Actual dimension = 7.5 x 17.5 in
A = 131.25 in2 , S = 382.81 in. 3 , CF = 0.959
d1 = 7.5 in, d2 = 17.5 in.
kl
=
d1
K
19 x 12
= 30.4
7 .5
= 25
K < kl/d < 50 ∴ Long column.
∴ Fc' =
0.3 E
0.3 x 1800
=
= 0.584 Ksi
2
(30.4) 2
( kl d )
le = 1.92 x 19 = 36.48 ft
CS =
36.48 x 12 x 17.5
= 11.67
7.5 2
CK =
3E
3 x 1800
=
= 25.98 ≈ 26
5Fb
5 x 1.6
10 < CS < CK ∴Intermediate beam.
 1C
F = α Fb 1 −  S
 3  C K
'
b



4

 1  11.67  4 
 = 1.6 1 − 
  = 1.578 Ksi

 3  26  
For long column, J = 1.0
fc =
40
= 0.305 Ksi
131.25
fb =
72
= 0.188 Ksi
382.81
Using interaction equation.
 6 x 3
0.188 + 0.305 
 (1 + 0.25)
0.305
 17.5 
+
= 0.522 + 0.456 = 0.978 < 1.0
0.584
1.578 − 1.0 x 0.305
∴ The 8 x 18 S4S section is acceptable.
10
Trial 3.
Assume a 12 x 12 S4S section.
Actual dimension = 11.5 x 11.5 in.
= 132.25 in2 , S = 253.48 in4 , CF = 1.0
A
kl = 19 x 12 = 19.826
d
11.5
K
= 25
11 < kl/d < K ∴Intermediate column.
 1  19.83  4 
F = 1.3 1 − 
  = 1.128 Ksi
 3  25  
'
c
le = 36.48 ft
CS =
36.48 x 12 x 11.5
= 6.17 < 10
11.5 2
∴ No correction for lateral stability is required.
Fb' = C F Fb = 1.0 x 1.6 = 1 .6 Ksi
J=
19.83 − 11
= 0 .631
25 − 11
fc =
40
= 0.302 Ksi
132.25
fb =
72
= 0.284 Ksi
253.48
 6 x 3
0.284 + 0.302 
 (1 + 0.25 x 0.631)
0.302
 11.5 
+
= 0.268 + 0.589 = 0.858 < 1.0
1.128
1.6 − 0.631 x 0.302
∴ The 12 x 12 S4S section is acceptable.
5.(1-5).*Determine three acceptable S4S sections, for a column 10 ft long subjected to
the combined action of longitudinal load P = 40 kips with an eccentricity of 3
in, and a mid-span moment M = 72 K-in due to transverse loading. Assume
α = 1.0 and such wood that Fc = 1.3 ksi, Fb = 1.6 ksi and E = 1800 ksi. The
column is to be used in a covered structure, hence, no adjustments in the
allowable stresses are needed.
11
Solution:
Trial 1.
Assume a 6 x 18 S4S section.
Actual dimension = 5.5 x 17.5 in
A = 96.25 in2 , S = 280.73 in3 , CF = 0.959
Assume K = 1.0
kl
=
d1
10 x 12
= 21.82
5 .5
K = 0.671
1800
= 24.97 ≈ 25
1.3
11 < kl/d < K ∴ Intermediate column.
 1  21.82  4 
F = 1.3 1 − 
  = 1.048 Ksi
 3  25  
'
c
le = 1.92 x 10 = 19.2 ft
CS =
19.2 x 12 x 17.5
= 11.545
5.5 2
CK =
3E
=
5Fb
3 x 1800
= 25.98 ≈ 26
5 x 1.6
10 < CS < CK ∴Intermediate beam.
 1  11.545  4 
F = 1.6 1 − 
  = 1.579 Ksi
 3  26  
'
b
10 x 12
− 11
For Int: Col:, J = 17.5
=0
25 − 11
fc =
40
= 0.416 Ksi
96.25
fb =
72
= 0.256 Ksi
280.73
Using Interaction equation:
 6 x 3
0.256 + 0.416 
 (1 + 0)
0.416
 17.5 
+
= 0.397 + 0.433 = 0.83 < 1.0 O.K.
1.048
1.579 − 0
∴ The 6 x 18 S4S section is acceptable.
12
Trial 2.
Assume a 8 x 14 S4S section.
Actual dimension – 7.5 x 13.5 in.
A = 101.25 in2 , S = 227.81 in3 , CF = 0.987
kl = 10 x 12 = 16
d
7 .5
K = 0.671
1800
= 24.97 ≈ 25
1.3
11 < kl/d < 25 ∴ Intermediate column.
 1  16  4 
∴ F = 1.3 1 −    = 1.227 Ksi
 3  25  
'
c
le = 1.92 x 10 = 19.2 ft
CS =
19.2 x 12 x 13.5
= 7.436 < 10.0
7.5 2
∴ Fb' = 1.6 x 0.987 = 1.579 Ksi
10 x 12
− 11
For Intermediate col:; J = 13.5
=0
25 − 11
fc =
40
= 0.395 Ksi
101.25
fb =
72
= 0.316 Ksi
227.81
Using Interaction equation.
 6 x 3
0.316 + 0.395 
 (1 + 0)
0.395
 13.5 
+
= 0.322 + 0.534 = 0.856 < 1.0 O.K.
1.227
1.579 − 0
∴ The 8 x 14 S4S section is acceptable.
Trial 2.
Assume a 10 x 12 S4S section.
Actual dimension = 9.5 x 11.5 in.
A = 109.24 in2 , S = 209.4 in3 , CF = 1.0
kl = 10 x 12 = 12.63
d
9 .5
K
= 25
13
11 < kl/d < 25 ∴ Intermediate column.
 1  12.63  4 
F = 1.3 1 − 
  = 1.272 Ksi
 3  25  
'
c
le = 10 x 1.92 = 19.2 ft
CS =
19.2 x 12 x 11.5
= 5.418 < 10
9.5 2
∴ No correction for lateral stability is required.
∴ Fb' = 1.0 x 1.6 = 1.6 Ksi
10 x 12
− 11
J = 11.5
=0
25 − 11
fc =
40
= 0.366 Ksi
109.25
fb =
72
= 0.344 Ksi
209.4
Using Interaction equation:
 6 x 3
0.344 + 0.366 
 (1 + 0)
0.366
 11.5 
+
= 0.288 + 0.573 = 0.861 < 1.0 O.K.
1.272
1 .6 − 0
∴ The 10 x 12 S4S section is acceptable.
6(1-5)** Design a spaced column as a braced double-hinged member for which the
effective length is the actual length. Consider using glued-laminated wood
for which Fc = 2.2 ksi, Fb = 2.2 ksi for load parallel to wide face of
laminations, Fb = 2.6 ksi for load perpendicular to wide face of laminations
and E = 2000 ksi, P = 104 Kips and length = 24.8 ft. Assume snow- loading
condition.
Solution:
Assume two laminated leaves may be used.
The min: width of the leaves: L b = 80
14
b=
297
= 3.71 in
80
1
The practical min: width = 5 in
8
But it is convenient to use at least a 6
3
in width to avoid penalization in
4
allowable stress due to large slenderness.
Then L b = 297 6.75 = 44
Assume condition “ a “
K = 1.06
E
2000
= 1.06
= 29.8
α Fc
1.15 x 2.2
K < L/b < 80 ∴Long column.
Fc' =
0.75E 0.75 x 2000
=
= 0.775 Ksi
( 44) 2
(L b ) 2
The required cross sectional area is:
A=
P
104
=
= 134 in 2
'
Fc 0.775
Use two 6.75 x 12 in leaves ( A = 162 in2 )
Check as a solid column
In lieu of an exact analysis the load may be used considered as having a
min: eccentricity of 1 in or 1.0 d, whichever is greatest.
Using leaves as deep as 12 in. the min: eccentricity is 1.2 in.
The section properties of two 6.75 x 12 in leaves are:
A = 2 x 6.75 x 12 = 162 in2
 6.75x12 2 
 = 324 in 3
S = 2 x 
6


 6.75 x 12 3 
 = 1944 in 4
I = 2 x 
12


kL = 297 = 24.8
d 12
K = 0.671
E
2000
= 0.671
= 18.9
α Fc
1.15 x 2.2
15
K < kL d < 50 ∴ long column
J = 1.0
Fc' =
0.3 E
0.3 x 2000
=
= 0.976 Ksi
2
( 24.8) 2
(KL d)
le = 1.84 lu , where lu is one half the unsupported length because of the
midspan spacer between column leaves.
l e = 1.84 x ( 297 2 ) = 273 in
Cs =
le d
=
b2
273 x 12
= 8.5
6.75 2
Cs = 10 and CF = 1.0
∴ Fb' = α Fb = 1.15 x 2.6 = 2.99 Ksi
Using interaction formula:
P
A+
Fc'
104
(P A)  bed  (1 + 0.25 J)
162 +
0.976
 
Fb' − J ( P A)
≤ 1.0
(104162)  6 x121.2  (1 + 0.25 )
2.99 − (104
)
= 0.863 < 1.0
162
∴ The spaced column with two 6.75 x 12 in leaves is acceptable.
7(1-5).**Design a solid column as a braced double- hinged member for which the
effective length is the actual length. Consider using glued- laminated wood for
which Fc = 2.2 ksi, Fb = 2.2 ksi for load parallel to wide face of laminations, Fb
= 2.6 ksi for load perpendicular to wide face of laminations and E = 2000 ksi,
P = 104 Kips and length = 24.8 ft. Assume snow-loading condition.
Solution
Assume 10¾ x 18 in. glued laminated section.
A = 194 in2 , S = 581 in3 , I = 5225 in4 , CF = 0.956
16
KL = 24.8 x 12 = 27.68
d
1075
2000
= 18.9
1.15 x 2.2
K = 0.671
K < KL d < 50 ∴ long column
Fc' =
0.3 E
(Kld )
2
=
0.3 x 2000
= 0.783 Ksi
( 27.68) 2
le = 1.84 lu
= 1.84 x 24.8 x 12 = 547.58 in.
547.6 x 18
= 9.24 < 10
10.75 2
Cs =
∴ Fc' = 1.15 x 2.6 x 0.956 = 2.858 Ksi
Using min: specified eccentricity = 1.2 in.
fc =
P 104
=
= 0.536 Ksi
A 194
For long column, J = 1.0
Using interaction equation:
 6e 
f c   (1 + 0.25 J )
fc
d 
+
≤ 1 .0
'
Fc
Fb' − J f c
 6 x 1 .2 
0.536 
 (1 + 0.25 )
0.536
 18 
+
= 0.856 + 0.115 = 0.8 < 1.0
0.783
2.858 − 0.536
∴10¾ x 18 in. glued laminated section is acceptable.
8(1-5).***The interior column of a building supports the box-girder at its continuous
end. The column is subjected to a load equal to P = 41.22 Kips, which is
transmitted to it by direct compression between column and box-beam. The
length of the column is 21 ft from finished floor elevation to bottom of boxgirder.
Column
may
be
considered
braced
against
relative
lateral
displacements of the ends and a solid timber section, S4S wood having
allowable stresses as follows: Fb = 1.65 ksi, Fc = 1.05 ksi, and E = 1600 ksi.
Assuming that it is hinged at both ends about both axes and that it is subjected
17
to end moments created by a specified minimum eccentricity of the load.
Assume snow- loading conditions.
(Ans: 10 x 10 S4S section, α = 0.92)
Solution:
Consider a section 10 x 10 in., S4S.
Actual dimension = 9.5 x 9.5 in.
A = 90.25 in2 , S = 142.9 in3 , I = 678.75 in4 , CF = 1.0.
It is hinged at both ends about both axes, i.e. K1 = K2 = 1.0, and that it is
subjected to end moments created by an eccentricity of the load given by the
maximum of 1 in. or 0.1 times the side of the column.
∴ The min: specified eccentricity, e = 1 in.
Kl = 21 x 12 = 26.53
d
9 .5
1600
= 24.4 < 26.53 ∴ longcolumn
1.15 x 1.05
K = 0.671
∴ Fc' =
0.3 x 1600
= 0.682 Ksi
( 26.53) 2
le = 1.84 lu = 1.84 x 21 x 12 = 464 in.
C S=
le d
b
2
=
464 x 9.5
= 6.99 < 10 and C F = 1.0
9 .5 2
∴ F = α Fb = 1.15 x 1.65 = 1.90 Ksi
'
b
fc =
P 41.22
=
= 0.457 Ksi
A 90.25
For long column, J = 1.0
Using interaction equation :
f c f c ( 6e d ) (1 + 0.25 J )
+
≤ 1.0
Fc'
Fb' − Jf c
 6 x 1
0.457 
 (1 + 0.25)
0.457
 9 .5 
+
= 0.67 + 0.25 = 0.92 < 1.0
0.682
1.9 − 0.457
∴ 10 x 10 S4S Section is acceptable
18
9(1-5).***The interior column of a building is subjected to a load equal to P = 41.22
Kips. The length of the column is 21 ft from finished floor elevation to bottom
of box- girder. Design the column assuming that it is hinged to the foundation
pedestal, but fully fixed to the supported beam. Consider as a solid timber
section, S4S wood having allowable stresses as follows: Fb = 1.65 ksi,
Fc = 1.05 ksi and E = 1600 ksi. Consider two cases: (a) full lateral bracing, and
(b) no lateral bracing. Assume normal loading conditions.
Solution:
Case (a) Full lateral bracing.
P = 41.22 Kips
Fb = 1.65 ksi
L = 21 ft
Fc = 1.05 ksi
E = 1600 ksi
Assume 10 x 10 S4S section.
Actual dimension = 9.5 x 9.5 in.
A = 90.25 in2 , S = 142.9 in3 , I = 678.75 in4 , CF = 1.0
It is hinged to the foundation pedestal, but fully fixed to the supported beam,
K = 0.7
∴ Kl =
d
0.7 x 21 x 12
= 18.57
9 .5
Min: specified eccentricity, e = 1 in
K = 0.671
1600
= 26.19
1.05
11 < kl d < K ∴ Intermedia tecolumn .
 1  18.57  4 
F = 1.05 1 − 
  = 0.962 Ksi
 3  26.19  
'
c
For full lateral bracing, lu = 0 ∴le = 0
Fb' = ∝ C F Fb = 1.65 Ksi
d 2 − 11 18.57 − 11
J=
=
= 0.498
K − 11
26.19 − 11
kl
fc =
P 41.22
=
= 0.457 Ksi
A 90.25
Using interaction equation:
19
6e
f c f c ( d ) (1 + 0.25 J)
+
≤ 1 .0
Fc'
Fb' − J f c
 6 x 1 .0 
0.457 
 (1 + 0.25 x 0.498)
0.457
 9 .5 
+
0.962
1.65 − 0.498 x 0.457
= 0.475 + 0.228
= 0.703 < 1.0
∴ 10 x 10 S4S section is acceptable.
Case (b) No lateral bracing.
Assume 10 x 10 in S4S section
A = 90.25 in2 , S = 142.9 in3 , I = 678.75 in4
K = 0.7
Kl = 0.7 x 21 x 12 = 18.57
d
9 .5
K = 26.19
11 < Kl d < K ∴ Intermedia tecolumn .
 1  18.57  4 
F = 1.05 1 − 
  = 0.962 Ksi
 3  26.19  
'
c
le = 1.84
CS =
lu = 1.84 x 21 x 12 = 464 in
le d
=
b2
464 x 9.5
= 6.99 < 10 and C F = 1.0
9.5 2
Fb' = α C F Fb = 1.65 Ksi
P 41.22
fc = =
= 0.457 Ksi
A 90.25
18.57 − 11
J=
= 0.498
26.19 − 11
Using interaction equation:
6e
f c f c ( d ) (1 + 0.25 J )
+
≤ 1.0
Fc'
Fb' − J f c
6 x1
0.457 (
) (1 + 0.25 x 0.498)
0.457
9.5
+
0.962
1.65 − 0.498 x 0.457
= 0.475 + 0.228
= 0.703 < 1.0
20
∴ 10 x 10 S4S section is acceptable.
10(1-5).***The interior column of a building is subjected to a load equal to P = 41.22
Kips. The length of the column is 21 ft from finished floor elevation to bottom
of box-girder. Design the column assuming that it is rotationally fixed to both
the foundation pedestal and the supported girder. Design for (a) full lateral
bracing and (b) no lateral bracing. Consider as a solid timber section, S4S
wood having allowable stresses as follows: F b = 1.65 ksi, Fc = 1.05 ksi and E =
1600 ksi. Assume snow- loading conditions.
Solution:
Case (a) Full lateral bracing.
Consider a 8 x 10 in S4S section
Actual dimension = 75 x 9.5 in
A = 71.25 in2 , S = 112.81 in3 , I = 535.86 in4 , CF = 1.0
It is rotationally fixed to both the foundation pedestal and the supported girder,
K = 0.5.
Min: specified eccentricity, e = 1 in.
∴ Kl =
d
0.5 x 21 x 12
= 16.8
7 .5
K = 0.671
E
1600
= 0.671
= 24.43 > 16.8
αFc
1.15x1.05
11 < kl d < K ∴ Intermedia tecolumn .
 1  16.8  4 
F = 1.15 x 1.05 1 − 
  = 1.117 Ksi
 3  24.4  
'
c
For full lateral bracing, lu = 0
∴ Fb' = 1.15 x 1.65 = 1.9 Ksi
P 41.22
=
= 0.579 Ksi
A 71.25
kl − 11 0.5 x 21 x 12 − 11
d2
9.5
J=
=
= 0.169
K − 11
24.4 − 11
fc =
Using interaction equation:
21
0.579
+
1.117
6 x1
) (1 + 0.25 x 0.169)
9 .5
1.9 − 0.169 x 0.579
0.579 (
= 0.518 + 0.211 = 0.73 < 1.0
∴ 8 x 10 in S4S section is acceptable.
Case (b) No lateral bracing
Consider 8 x 10 in S4S section
A = 71.25 in2 , S = 112.81 in3 , I = 535.86 in4 , CF = 1.0
K = 0.5
Min: specified eccentricity, e = 1 in.
∴ Kl =
d
0.5 x 21 x 12
= 16.8
7 .5
K – 24.43
11 < ∴ Kl d < K ∴ Intermedia te column .
Fc' = 1.117 Ksi
For no lateral bracing, le = 1.84 lu
le = 1.84 x 21 x 12 = 464 in
464 x 9.5
= 8.85 < 10
7.5 2
∴ Fb' = α C F Fb = 1.9 Ksi
CS =
J = 0.169
Fc = 0.579 ksi
Using interaction equation:
0.579
+
1.117
6 x1
) (1 + 0.25 x 0.169)
9 .5
= 0.73 < 1.0
1.9 − 0.169 x 0.579
0.579 (
∴ 8 x 10 in S4S section is acceptable.