Question 11.2: Determine the critical load on a steel column having a squar...

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

Given          L = 300 mm. Cross section is square; each side is b = 12 mm.

One pinned end; one fixed end; column is steel; SAE 1040 hot rolled

From Appendix A–10: s_{y} = 414 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 fixed-pinned-end column, K = 0.80 is a practical value (Figure 11–4).

Step 2. Compute the effective length:

L_{e}  =  KL = 0.80(L)= 0.80(300  mm) = 240 mm

Step 3. Compute the smallest value of the radius of gyration. From Appendix A–1, for a square cross section, r b=/ \sqrt{1} 2. Then,

r =   \frac{b}{ \sqrt{12}} = \frac{12  mm}{ \sqrt{12}} = 3.46 mm

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

SR =   \frac{L_{e}}{ r} = \frac{KL}{r}=\frac{(0.8)(300  mm)}{ 3.46  mm} = 69.4

Step 5. Normally, we would compute the value of the column constant, C_{c} . But, in this case, let’s use Figure 11–6. For a steel having a yield strength of 414 MPa, C_{c} = 96, approximately.

Step 6. Compare C_{c} with SR and decide if the column is long or short. Then use the appropriate column formula to compute the critical buckling load. Since SR is less than C_{c} , the Johnson formula, Equation (11–6), should be used:

P_{cr} = As_{y}[1- \frac{s_{y}(SR)^{2}}{4 \pi^{2}E} ]   (11–6)

The area of the square cross section is

A = b² = (12 mm)² = 144 mm²

Then,

P_{cr} =\left\lgroup144  mm^{2}\right\rgroup \left\lgroup\frac{414  N}{mm^{2}}\right\rgroup [1- \frac{(414 \times 10^{6}  N/m^{2})(69.4)^{2}}{4 \pi^{2}(207 \times10^{9}  N/m^{2})} ]

P_{cr} = 45.1  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{45.1  kN}{3} = 15.0 kN