For the timber column shown in the two figures below, E 12 GPa and the allowable stress for compression parallel to the grain is 9 MPa. Using Code 4, Table 11-1, to find the allowable stress for a centric load, use the allowable stress method to determine the maximum load P that can be applied to the column if - (a) The column is eccentrically loaded as shown on the left. (b) The column is eccentrically loaded as shown on the right.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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4
Code No. Source
3
TABLE 11-1 Some Representative Column Codes for Centric Loading
2
1
с
b
b
a
Material
Structural steel
with a yield
point σy
2014-T6 (Alclad)
Aluminum alloy
6061-T6
Aluminum alloy
Timber with a
rectangular
cross section
bxd where d <b
Compression-Block and/or Intermediate-Range
Formulas and Limitations
(L/r is the effective ratio L'/r)
11-4 EMPIRICAL COLUMN FORMULAS-CENTRIC LOADING 667
0 ≤ ≤ Cc
L
L
12 = = = 55
11 ≤
r
VI
L
9.566
12
L≤9.5
Y
IA
L
</sk
11
Jall =
C²---
Jall
1
=
FS
217² E
σy
5
3 (L/
L/
FS = 1 + 1 (²4) - (²4/7)
3 8 Cc
8 Cc
1
28 ksi
193 MPa
Fall - [30.7 -0.23 (-)] ksi
= [212 1.585
()]
all 19 ksi
—
= 131 MPa
2 C
()))
Ơall = F.*
Fall 20.2 0.126
= [20.2-
-
-
= [139-0
Fall Fo
139 - 0.868
6 (²)] k
F. [1 - 1 (1/4)]
3
k = 0.671VE/F.
())
MPa
ksi
MPa
Slender Range
L z. Cc
Gall
L
Fall
L
Jall
≥ 55
=
ks
O all
=
T²E
1.92(L/r)²
54,000
(L/r)²
-
≥ 66
372(10³)
(L/r)²
L
d
51,000
(L/r)²
ksi
≤50
351(10³)
(L/r)²
0.30E
(L/d)²
MIPa
ksi
MPa
Transcribed Image Text:4 Code No. Source 3 TABLE 11-1 Some Representative Column Codes for Centric Loading 2 1 с b b a Material Structural steel with a yield point σy 2014-T6 (Alclad) Aluminum alloy 6061-T6 Aluminum alloy Timber with a rectangular cross section bxd where d <b Compression-Block and/or Intermediate-Range Formulas and Limitations (L/r is the effective ratio L'/r) 11-4 EMPIRICAL COLUMN FORMULAS-CENTRIC LOADING 667 0 ≤ ≤ Cc L L 12 = = = 55 11 ≤ r VI L 9.566 12 L≤9.5 Y IA L </sk 11 Jall = C²--- Jall 1 = FS 217² E σy 5 3 (L/ L/ FS = 1 + 1 (²4) - (²4/7) 3 8 Cc 8 Cc 1 28 ksi 193 MPa Fall - [30.7 -0.23 (-)] ksi = [212 1.585 ()] all 19 ksi — = 131 MPa 2 C ())) Ơall = F.* Fall 20.2 0.126 = [20.2- - - = [139-0 Fall Fo 139 - 0.868 6 (²)] k F. [1 - 1 (1/4)] 3 k = 0.671VE/F. ()) MPa ksi MPa Slender Range L z. Cc Gall L Fall L Jall ≥ 55 = ks O all = T²E 1.92(L/r)² 54,000 (L/r)² - ≥ 66 372(10³) (L/r)² L d 51,000 (L/r)² ksi ≤50 351(10³) (L/r)² 0.30E (L/d)² MIPa ksi MPa
- For the timber column shown in the two figures below, E = 12 GPa and the allowable
stress for compression parallel to the grain is 9 MPa. Using Code 4, Table 11-1, to find
the allowable stress for a centric load, use the allowable stress method to determine
the maximum load P that can be applied to the column if
(a) The column is eccentrically loaded as shown on the left.
(b) The column is eccentrically loaded as shown on the right.
2m
150 mm
P
(a)
Y
X
100 mm
2 m
150 mm
(b)
P
Y
100 mm
Transcribed Image Text:- For the timber column shown in the two figures below, E = 12 GPa and the allowable stress for compression parallel to the grain is 9 MPa. Using Code 4, Table 11-1, to find the allowable stress for a centric load, use the allowable stress method to determine the maximum load P that can be applied to the column if (a) The column is eccentrically loaded as shown on the left. (b) The column is eccentrically loaded as shown on the right. 2m 150 mm P (a) Y X 100 mm 2 m 150 mm (b) P Y 100 mm
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