Problem 10.51

Since the easiest way is to compute using the secant formula for each diameter, I'll list the method for the calculations first and then a spreadsheet table with the results:

The critical force is given by

The effective length (8 ft) is twice the actual length (4 ft). The moment of inertia is:

The radius of gyration r is computed using

Finally, the maximum normal stress is given by the secant formula:

When setting up the spreadsheet, I named the variables that don't change (those with the arrows pointing at their values). When a variable is given a name in Excel (for example, I named "e" using "ee"), formulas may access that value using the name instead of the row-column reference. Furthermore, such access is absolute, so the dollar sign need not be used (as, for example, in $A$10).

e ==>	0.375	in
P ==>	12	kips
L_e ==>	96	in
F_all ==>	15	ksi
E ==>	2.90E+07	psi
Note that r is the radius of gyration of the area, not the radius of the rod
dia (in)	I (in⁴)	P_cr (kip)	Area (in²)	r² (in)	ec/r^2	F_max (ksi)
1.5	0.249	7.717741	1.767	0.1406	2.000	-29.116
1.625	0.342	10.630	2.074	0.165	1.846	-103.227
1.75	0.460	14.298	2.405	0.1914	1.714	70.088
1.875	0.607	18.842	2.761	0.2197	1.600	26.637
2	0.785	24.392	3.142	0.25	1.500	16.495
2.125	1.001	31.086	3.547	0.2822	1.412	11.908
2.25	1.258	39.071	3.976	0.3164	1.333	9.263
2.375	1.562	48.504	4.430	0.3525	1.263	7.528
2.5	1.917	59.550	4.909	0.3906	1.200	6.297
2.625	2.331	72.384	5.412	0.4307	1.143	5.376
2.75	2.807	87.188	5.940	0.4727	1.091	4.660
2.875	3.354	104.154	6.492	0.5166	1.043	4.088
3	3.976	123.484	7.069	0.5625	1.000	3.621

Since the stress of 11.908 ksi is the largest that may be used (the allowable stress is 15 ksi), the rod diameter to be chosen is in.