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Section 6.4
Example Problem 6.2
Last Revised:
11/04/2014
The example problems presented in this section have a spreadsheet solution.
You will need this file to follow along with the presented solutions. You can
click on the following link to get the file:
Chapter 6: Excel Spreadsheet Solutions
In addition to the spreadsheet solution this problem also has a downloadable
hand solution:
BGSCMExample6_2.pdf
Given: The sections listed below are to be concentrically
loaded in compression.
Wanted: Determine the Fy for which each of the
sections will be considered slender.
a) W18x35; b) C9x13.4; c) HSS 10x4x1/8; d) HSS 7.5x0.188; e) M4x6
Solution:
The equations for lr can be solved for
Fy for given cross sections. Doing so yields the value of Fy
for which the section is slender. A similar approach can be made to
determine the value of Fy for which a section is compact or
non-compact, however, in the case of members subjected to concentric axial
compression, the only real concern is determining if the section is slender or
non-slender.
To solve this problem, the initial challenge is to determine the applicable
cases in SCM Table B4.1 (SCM pg 16.1-16). Then, the appropriate section
properties for the member are found in the section property tables or computed
from the section properties. Using the criteria from SCM Table B4.1 and
the section properties, we can then determine Fy.
a) W18x35: This section has both unstiffened and stiffened
elements.
The unstiffened element (1/2 a flange) has a tabulated width/thickness
ratio (bf/2tf) in the tables. This value is set
equal to the case 3 lr equation then
the equation is solved for Fy.
The stiffened element (the web) has a tabulated width/thickness ratio (h/tw)
in the tables. This value is set equal to the case 10
lr equation then the equation is
solved for Fy.
The value of Fy that is the least is the controlling value for
the problem. It is the value for which the member first becomes
slender.
b) C9x13.4: This member is essentially the same as for the W section.
The same SCM Table B4.1 cases apply.
In this case, the unstiffened element is the whole flange. The
width/thickness ratio for the unstiffened element is not tabulated in the
tables, so it is computed from the tabulated values for the flange width, bf,
and flange thickness, tf.
The width/thickness ratio is for the unstiffened element is not
tabulated, so it is computed from the tabulated values for "T" and tw.
c) HSS 10x4x1/8: This member only has stiffened elements.
This section has two different stiffened elements, both with tabulated
values of width/thickness ratio. Reading the verbiage for case 12
might lead you to select the wrong width/thickness ratio if you don't think
a bit about what is happening. For concentric axial loading there are
no "flanges" and "webs" in the bending sense. All the elements are
subjected to uniform compression. In this case, all the elements can
be considered to be a "flange", so you want the larger of the tabulated
values for b/t and h/t.
Once you select the appropriate width/thickness ratio, this is equated to
the correct lr equation and solved for
Fy.
d) HSS 7.5x0.188: This is a round section that has no plate members in
the cross section.
For this member, SCM Table B4.1 case 15 applies. For this case we
need the D/t ratio for the member and the equation for lr.
Note that for this case that there is NOT a square root in the equation for
lr. The equation is then solved
for Fy.
e) M4x6: This is a small "I" shaped member and is solved exactly as the
W18x35 was above.
Observations:
- Neither the W18x35 nor the HSS 10x4x1/8 make good sections for use as
columns as both are slender and will have low compressive strength values.
They are probably best used in flexure applications.
- This approach can be applied to all sections in the database to find the
value of Fy for which the member is considered to be compact,
non-compact, or slender for concentric axial compression and/or pure bending
(the values for each member a different in each case)
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