Question 11.1: A round compression member with both ends pinned and made of...

Objective   Compute the critical buckling load for the column and the allowable load for a design factor of N = 3.

Given          L = 950 mm. Cross section is circular; D = 25 mm. Pinned ends

Column is steel; SAE 1020 cold drawn

From Appendix A–10:  s_{y}   = 441 MPa; E = 207 GPa = 207 × 10^{9}   N/m²

Analysis     Use the Method of Analyzing Columns.

Results       Step 1. Determine the end-fixity factor. For the pinned-end column, K = 1.0.

Step 2. Compute the effective length:

L_{e} =  KL = 1.0 (L) = 950   mm

Step 3. Compute the smallest value of the radius of gyration. From Appendix A–1, for any axis of a circular cross section, r = D/4. Then,

r = \frac{D}{4} = \frac{25  mm}{4} = 6.25 mm

Step 4. Compute the slenderness ratio, SR = L_{e } /r:

SR = \frac{L_{e}}{r} = \frac{950  mm}{6.25  mm} = 152

Step 5. Compute the column constant, Cc

  C_{c} = \sqrt{\frac{2 \pi^{2}E}{s_{y}}} = \sqrt{\frac{2 \pi^{2}(207 \times 10^{9}  N/m^{2})}{441 \times 10^{6}  N/m^{2}}} = 96.3

Step 6. Compare C_{c} with SR and decide if column is long or short. Then use the appropriate column formula to compute the critical buckling load. Since SR is greater than C_{c} , Euler’s formula applies:

P_{cr} = \frac{ \pi^{2} EA}{(SR)^{2}}

The area is

A= \frac{ \pi D^{2}  }{(4} = \frac{ \pi (25  mm)^{2}}{4} = 491 mm²

Then

P_{cr} = \frac{ \pi^{2} (207 \times 10^{9}  N/m^{2})(491  mm^{2})}{(152)^{2}} \times \frac{1  m²}{(10³  mm)²} = 43.4 kN

Step 7. A design factor of N = 3 is specified.

Step 8. The allowable load, P_{a} , is

P_{a} = \frac{P_{cr}}{N} = \frac{43.4  kN}{3} = 14.5 kN