A long, slender column ABC is pin-supported at the ends and compressed by an axial load P (Fig. 11-15). Lateral support is provided at the midpoint B in the plane of the figure. However, lateral support perpendicular to the plane of the figure is provided only at the ends. The column is constructed

Question

A long, slender column ABC is pin-supported at the ends and compressed by an axial load P (Fig. 11-15). Lateral support is provided at the midpoint B in the plane of the figure. However, lateral support perpendicular to the plane of the figure is provided only at the ends.

The column is constructed of a steel I-beam section S 200 × 34 having modulus of elasticity E = 200GPa and proportional limit [latex]\sigma_{pl} = 300 MPa[/latex]. The total length of the column is L = 8m.

Determine the allowable load [latex]P_{allow}[/latex] using a factor of safety n = 2.5 with respect to Euler buckling of the column.

Accepted Answer

Use a four-step problem-solving approach.
1. Conceptualize: Because of the manner in which it is supported, this column may buckle in either of the two principal planes of bending. As one possibility, it may buckle in the plane of the figure, in which case the distance between lateral supports is L/2 = 4m and bending occurs about axis 2-2 (see Fig. 11-9c for the mode shape of buckling).

As a second possibility, the column may buckle perpendicular to the plane of the figure with bending about axis 1–1. Because the only lateral support in this direction is at the ends, the distance between lateral supports is L = 8m (see Fig. 11-9b for the mode shape of buckling).

Column properties : From Table F-2(b) obtain the following moments of inertia and cross-sectional area for a S 200 × 34 column:

[latex]I_{1}=26.9 imes 10^{6} mm ^{4} \quad I_{2}=1.78 imes 10^{6} mm ^{4} \quad A=4360 mm ^{2}[/latex]

2. Categorize:

Critical loads: If the column buckles in the plane of the figure, the critical load is

[latex]P_{ cr }=\frac{\pi^{2} E I_{2}}{(L / 2)^{2}}=\frac{4 \pi^{2} E I_{2}}{L^{2}}[/latex]

3. Analyze: Substitute numerical values to obtain

[latex]P_{ cr }=\frac{4 \pi^{2} E I_{2}}{L^{2}}=\frac{4 \pi^{2}(200 GPa )\left(1.78 imes 10^{6} mm ^{4} ight)}{(8 m )^{2}}=220 kN[/latex]

If the column buckles perpendicular to the plane of the figure, the critical load is

[latex]P_{ cr }=\frac{\pi^{2} E I_{1}}{L^{2}}=\frac{\pi^{2}(200 GPa )\left(26.9 imes 10^{6} mm ^{4} ight)}{(8 m )^{2}}=830 kN[/latex]

Therefore, the critical load for the column (the smaller of the two preceding values) is

[latex]P_{ cr }=220 kN[/latex]

and buckling occurs in the plane of the figure.

Critical stresses: Since the calculations for the critical loads are valid only if the material follows Hooke’s law, verify that the critical stresses do not exceed the proportional limit of the material. For the larger critical load, the critical stress is

[latex]\sigma_{ cr }=\frac{P_{ cr }}{A}=\frac{830 kN }{4360 mm ^{2}}=190.4 MPa[/latex]

Since this stress is less than the proportional limit [latex](\sigma_{pl} = 300 MPa)[/latex], both critical-load calculations are satisfactory.

4. Finalize:
Allowable load: The allowable axial load for the column, based on Euler buckling, is

[latex]P_{ ext {allow }}=\frac{P_{ cr }}{n}=\frac{220 kN }{2.5}=88 kN[/latex]

in which n = 2.5 is the desired factor of safety.

Table F-2(b)
Properties of I-Beam Sections (S Shapes)—SI Units (Abridged List)
Designation	Mass per Meter	Area	Depth	Web Thickness	Flange		Axis 1–1			Axis 2–2
	Mass per Meter	Area	Depth	Web Thickness	Width	Thickness	I	S	r	I	S	r
	kg	mm²	mm	mm	mm	mm	[latex] imes 10^{6} mm^{4}[/latex]	[latex] imes 10^{3} mm^{3}[/latex]	mm	[latex] imes 10^{6} mm^{4}[/latex]	[latex] imes 10^{3} mm^{3}[/latex]	mm
S 610 × 149	149	18900	610	18.9	184	22.1	991	3260	229	19.7	215	32.3
S 610 × 119	119	15200	610	12.7	178	22.1	874	2870	241	17.5	197	34.0
S 510 × 143	143	18200	516	20.3	183	23.4	695	2700	196	20.8	228	33.8
S 510 × 112	112	14200	508	16.1	162	20.2	533	2100	194	12.3	152	29.5
S 460 × 104	104	13200	457	18.1	159	17.6	384	1690	170	10.0	126	27.4
S 460 × 81.4	81.4	10300	457	11.7	152	17.6	333	1460	180	8.62	113	29.0
S 380 × 74	74.0	9480	381	14.0	143	15.8	202	1060	146	6.49	90.6	26.2
S 380 × 64	64.0	8130	381	10.4	140	15.8	186	973	151	5.95	85.0	26.9
S 310 × 74	74.0	9420	305	17.4	139	16.7	126	829	116	6.49	93.2	26.2
S 310 × 52	52.0	6580	305	10.9	129	13.8	94.9	624	120	4.10	63.6	24.9
S 250 × 52	52.0	6650	254	15.1	125	12.5	61.2	482	96.0	3.45	55.1	22.8
S 250 × 37.8	37.8	4810	254	7.90	118	12.5	51.2	403	103	2.80	47.4	24.1
S 200 × 34	34.0	4360	203	11.2	106	10.8	26.9	265	78.5	1.78	33.6	20.2
S 200 × 27.4	27.4	3480	203	6.88	102	10.8	23.9	236	82.8	1.54	30.2	21.0
S 150 × 25.7	25.7	3260	152	11.8	90.7	9.12	10.9	143	57.9	0.953	21.0	17.1
S 150 × 18.6	18.6	2360	152	5.89	84.6	9.12	9.16	120	62.2	0.749	17.7	17.8
S 100 × 14.1	14.1	1800	102	8.28	71.1	7.44	2.81	55.4	39.6	0.369	10.4	14.3
S 100 × 11.5	11.5	1460	102	4.90	67.6	7.44	2.52	49.7	41.7	0.311	9.21	14.6

Question 11.2: A long, slender column ABC is pin-supported at the ends and ...

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