Question: Determine the column allowable compressive force capacity, Pa for the given 3 x 8 S 4 S Douglas Fir sawn lumber column with an Fc

Determine the column allowable compressive force capacity, Pa for the given 3 x 8 S4S Douglas Fir
sawn lumber column with an Fc =1000 psi and E =1.6 x 106 psi. Mid-height bracing is provided for
the weak axis of buckling. Assume CM =1.0, Ct =1.0, CF =1.0. Assume a 7-day load duration. An
illustration of this column is found below. (40 points total) Use Table A1-a on pg.567 for the column
dimensional properties.
[Hint: See Example problem 9.8 on slide 23 in the lecture PowerPoint presentation titled Chapter 9
REV wood only found in Module 22.]
2.1. Find the column cross-
cross-sectional
area, A (in2).(2
points) Make sure
you look up the
true dimensions
in Table A1-a in
the Appendix on
pg.567 of the
textbook.
2.2. Find the weak
axis effective
length, Le,y (ft).
The weak axis is
the one that
makes the stud
depth wider than it
is deep. For this cross-section that would be the direction that is braced at mid-height and is
marked by a) Weak axis in the sketch. (2 points)
2.3. Using the smaller dimension of the stud cross-section calculate the weak axis slenderness
ratio, Le,y /dy . Dont forget to convert Le,y to inches. (2 points)
2.4. Find the strong axis effective length, Le,x (ft). The strong axis is the one that makes the stud
depth deeper than it is wide. For this cross-section that would be the direction that is not
braced at mid-height and is marked by b) Strong axis in the sketch. (2 points)
2.5. Using the larger dimension of the stud cross-section calculate the strong axis slenderness
ratio, Le,x /dx . Dont forget to convert Le,x to inches. (2 points)
2.6. Determine which slenderness ratio, L/d
governs. The largest one governs. (2
points)
2.7. Using Figure 9.23 to the right determine
the appropriate load duration factor, CD .
(2 points)
2.8. The equation for allowable stress in a
wood column is:
FC = F*C Cp however
F*C = FC CD CM Ct CF
Calculate F*C using the information you
were given above and the load duration
factor, CD from part 2.7. We will calculate
Fc later once we have Cp .(4 points)
3x8 S4S
Pa
2.9. Calculate the Euler critical buckling stress for columns, FcE . The value of KcE =0.3 for sawn
lumber.
FcE =(KcE E)/(L/d)2 for each direction.
2.9.1. For the weak axis use Le,y /dy from part 2.3 above. (4 points)
2.9.2. For the strong axis use Le,x /dx from part 2.5 above. (4 points) You must show your work
to get credit however if you did part 2.9.2 correctly the buckling stress limit for the
strong axis, FcE,x should be 891.8 psi.
This time the smaller one will govern the columns buckling capacity. This should be the same
direction that governed in the answer to part 2.6 above.
2.10. Now calculate the ratio FcE /F*C for only the direction that governed in both part 2.6(and 2.9).
Use the smaller FcE from part 2.9 and the F *C from 2.8. Round the number to two digits (the
nearest 100th ).(2 points)
2.11. Using Table 9.3 on page 478 of your textbook (see partial Table below) and the ratio from part
2.10 determine the appropriate column stability factor, Cp .(4 points) Note: If you cannot find
your ratio from part 2.10 above in the table below then you have made a miscalculation.
2.12. The columns allowable buckling compressive stress, FC = Cp F*C . Calculate the allowable
buckling compressive stress, F C .(4 points)
2.13. Using the area, A calculated in part 2.1 above calculate the column allowable load Pa = FC A.(4
points)
Determine the column allowable compressive force

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Civil Engineering Questions!