# Wood problem needing NDS

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### #1 Construction PE

Construction PE

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Posted 20 October 2011 - 06:07 PM

I posted the problem below under the forum Construction section and have gotten no replies... wanted to see if anyone in Structures can help.
I am looking for an end restraint coefficient C=2.1 referenced in the answer as coming from NDS but can't find where.
Thanks in advance for any help!

Lindeburg Practice Problems book

Problem 45.3 (11th ed)
A 4in x 4 in (nominal size) timber post is used to support a sign. The post's modulus of elasticity is 1.5x10^6 lbf/in^2. One end of the post is embedded in a deep concrete base. The other end supports a sign 9 ft above the ground. Neglect torsion and wind effects. What is the Euler load sign weight that will cause failure by buckling?
a- 2700 lbf
b- 3700lbf
c- 4000lbf
d- 5500lbf

1) I = bh^3/12= (3.5)(3.5^3)/12= 12.51 in^4 (finished lumber size)

2) k = (I/A) ^ 1/2
k= (12.51/(3.5"x 3.5"))^1/2= 1.01 inches

3) The end restraint coefficient is C= 2.1 (NDS) <--- this line is quoted directly from Lindeburg's problem book. Does anyone know where to find this C=2.1 figure in NDS? (I obviously don't know how to use this reference) Any body have any pointers???

4) The slenderness ratio is:
L'/k= CL/k= (2.1)(9'x12)/1.01 = 224.6

There is note here: L/k is well above 100. (Note that most timber codes limit L/k to 50, so this would not be a permitted application).

5) Fe= (pi)^2 E I/ (L^2) = [(pi^2) (1.5 x 10^6 lbf/in^2) ( 12.51 in^4)] / [((2.1)(9'x12"))^2] = 3673 lbf

Answer is therefore b.

Anybody? End-restraint coefficient C = 2.1 (from NDS). This is a timber post.

I glanced to through the tables and a couple spots in NDS, but I had no luck.

### #2 darius

darius

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Posted 20 October 2011 - 06:31 PM

appendix G
table G1
buckling length coeff
you can find it on the steel book as well
aisc XIII

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