The process of analyzing a timber column goes like this.
In hand solution we will stop and pick a different section if the slenderness ratio (SR) is over 50 rather than doing additional work. You should be aware that in this course I will want you to solve all the values regardless if you find a failure mode early on.
Just so you know, in section 3.7.1.4 you will see that the slenderness ratio is allowed to be up to 75 during construction1. This means prior to putting any axial load on it. By the time the column is loaded the slenderness ratio must be no more than 50.
Let's work through an example to see the whole process.
Footnote 1 - See C3.7.1.4 in NDS.
Values from problem statement:
Values from Table 1B:
Given that the species of the column is Southern Pine, the grade is No.2 and its size is 4x6, what Table should we look up the Unmodified Allowable Stresses in?
(Click on the light blue down arrow to find out what table we need to use.)
Well the beam is 4" thick, so we don't use Table 4D; it is for members that are 5"×5" or larger.
The species is Southern Pine, so than means we use Table 4B of the Supplement.
On page 39 of the Supplement, under the 2" - 4" Size classification, we find that,
column.unmodified_stresses_latex
On page 38 of the Supplement there is information about CF, Cr, Cfu, and CM.
If there had been wet service conditions then we would need to find the values for CM for Fc and Emin. We would have to further check that Fc was greater than 750 psi to use the CM value for Fc.
Section 3.7.1.2 in the Timber NDS covers effective length. In it you will see that
Le=Ke×L
(If you have Thumbnail Zoom Plus or Imagus installed you can hover over the column image to get a better view of it.)
This is the same K as the one that was discussed in the optional lecture on buckling and here. In timber all connections are considered pinned. This is due to the flexibility of both the material and connections. So Ke is always equal to 1.
I am not saying that Le=L all the time because we need to account for bracing.
In the column shown to the left (similar to Figure 3F in the 2015 Timber NDS) you will see that the effective length values of the column are measured from midpoint of brace to midpoint of brace. In this case if there were no top and bottom braces it would be measured from end to end for Lx and from end to midpoint of the brace for Ly.
Only if no braces are present then Le=L for both Lx and Ly.
Also you should note that the variable for effective length in section 3.7.1.2 is le not Le. I'm using L to avoid confusion in hand written assignments where l may look too much like the number 1. That being said if I inadvertently show lower a lower case l instead of an L you will know why and what it means.
The slenderness ratio is a measure the buckling resistance of a column in one direction. You calculate it by dividing the effective length (distance between supports, if the column was bending in one direction) by the thickness (or width) of the column in that same direction.
In other words,
In Section 3.7.1.2 of the Timber NDS - Column Stability Factor CP
Le,xd=144"5.5"=26.18 Le,yb=144"3.5"=41.14 ⇐ Controls SR=41.14(Note that its the largest of these values that controls, and is the slenderness ratio.)
F∗c is found just like F′c except you don't include the Column Stability Factor (CP) in the calculation. It is used to take into account that the column could crush rather than buckle. The adjustment values we need to find F′c are listed in Table 4.3.1 of the Timber NDS. Because the problem statement did not indicate otherwise CD=1.25 and because there are "normal temperature and moisture conditions", Ct=1 and CM=1. Since the column is Southern Pine, CF=1. Finally, the column is not incised so Ci=1.
F∗c=Fc×CD×CM×Ct×CF×Ci F∗c=1400psi×1.25×1×1×1×1 F∗c=1750psiWe also need to know the allowable value for E′min because it is used in the elastic buckling stress equation, (FcE).
From Table 4.3.1 we see that we need to know CM, Ct, Ci, and CT.
column.E_min_latex
FcE is the Elastic buckling capacity of the column (with a factor of safety of 2.75 applied). It is defined in section 3.7.1.5 of the Timber NDS.
FcE=0.822×E′min(SR)2FcE=0.822×510,000psi(41.14)2FcE=247.7psiThe value for c is listed in section 3.7.1.5 of the Timber NDS and it depends on the shape and type of the material the column is made out of. For a sawn lumber column, like a 4x6, c=0.8. I'll leave it to you to read about the other possible values.
CP=1+FcE/F∗c2×c−√(1+FcE/F∗c2×c)2−FcE/F∗ccLet's find the components of CP separately.
FcEF∗c=0.1415 1+FcE/F∗c2×c=1+0.14152×0.8=0.7134 FcE/F∗cc=0.14150.8=0.1769so
CP=0.7134−√0.71342−0.1769=0.1372Because F∗c is the same as F′c without CP being applied we can just find F′c like this,
F′c=F∗c×CP
F′c=1750psi×0.1372
F′c=240.0psi
or we can find it explicitly.
column.F_c_latex
To find the actual compressive stress we divide the axial load by the column cross sectional area.
column.f_c_latex
After an analysis of the No. 2 Southern Pine 4x6, under normal temperature and moisture conditions, and with only end point support, we find that the allowable compressive stress parallel to grain compared with the actual compressive stress is,
column.column_adequacy_check_latex
(Click on the blue right arrow to see the column capacity.)
That actual capacity of the column is found by multiplying the column's cross sectional area by its allowable compressive stress.
Pallowed=F′c×A
Pallowed=240.0psi×19.25in2
Pallowed=4,621lbs
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