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6.5 EXAMPLES OF DESIGN
OF COLUMNS ACCORDING TO THE ASD CODE
Example 1
The
diagonally braced frame shown in Fig. 6.5 should be designed for a lateral
earthquake load of 31 Kips. Three different openings are required as shown
in the figure. The diagonal must fit within the thickness of a 4-in.-thick
wall. Also, the width of the diagonal in the plane of the frame must not
exceed 4 in. Design a rolled section for the diagonal from the ASD manual,
using A36 steel with yield stress of 36 ksi.
Solution
The structure is statically determinate. The
diagonal must be designed as a compression member.
The
requirement of the problem will be satisfied best by selecting a tube. Try
TUBE 4x4 with wall thickness t = 0.5 in., A = 6.36 in.2, and rmin = 1.39 in.
Compression
capacity of the section =
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O.K.
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Check local flange buckling (column four of Table
5.1):
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O.K.
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USE TUBE 4x4 with t = 0.5 in.
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weight
= 21.63 lb/ft
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Example 2
The cross
section of a built-up column is shown in Fig. 6.6. It consists of two L8x8x1 angles and is made of
A36 steel with yield stress of 36 ksi. Using the ASD code, find the maximum
allowable axial load for this column for a length of 15 ft. Assume that
there exist ideal hinged conditions at the column ends and that the angles
are properly connected so that they may not buckle individually.
Solution
Properties of L8x8x1: A1 = area of cross section = 15 in.2,
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rz = 1.56 in.
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c = 2.37 in.
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w = 51 lb/ft
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We must
first calculate the moments of inertia of the built-up section with respect
to x and y (axes of symmetry of the built-up section). From the geometry
we can write (Fig. 6.6)
The
moment of inertia of one angle about the z-axis shown in Fig. 6.6 is
Thus,
Using the relation
for moment of inertia of a triangle about its base (Fig. 6.7), we have
From Eq.
(6.14):
Note: Moments of inertia of the built-up section
could also be computed by using rotation of axes and the parallel-axis
theorem.
Example 3
A column is built of two tubes 16x8 and one angle L8x8x1 available in a shop.
In order to utilize the material efficiently a designer has proposed the
arrangement shown in Fig 6.8. The sections are made of A36 steel with yield
stress of 36 ksi. Using the ASD code, determine the maximum allowable load
for this column for a length of 14 ft. Assume ideal hinged conditions at
the column ends. The sections are properly connected so that they may not
buckle individually. Thickness of the tubes is t = 0.5 in.
Solution
For L8x8x1:
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Ix1 = Iy1 = 89 in.4
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A = 15.0 in.2
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rz = 1.56 in.
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Iz = 15(1.56)2
= 36.50 in.4
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For a
tube 16x8 (t = 0.5 in.):
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A = 22.4 in.2
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C1 = centroid of the L8x8x1
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C: centroid of the
built-up section
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For one tube (Fig.
6.9):
Moment of
inertia of the built-up section about the weak axis (x-axis):
Cross-sectional
area of the built-up section:
From Eq. (6.14):
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Fa = 20.17 ksi
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P = AFa
= (59.8)(20.17) = 1206.2 K
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