Question 11.5: A steel wide-flange column of HE 320A shape (Fig. 11-29a) is...
A steel wide-flange column of HE 320A shape (Fig. 11-29a) is pin supported at the ends and has a length of 7.5 m. The column supports a centrally applied load P_{1}=1800 \mathrm{~kN} and an eccentrically applied load P_{2}=200 \mathrm{~kN} (Fig. 11-29b). Bending takes place about axis 1–1 of the cross section, and the eccentric load acts on axis 2–2 at a distance of 400 mm from the centroid C.
(a) Using the secant formula, and assuming E = 210 GPa , calculate the maximum compressive stress in the column.
(b) If the yield stress for the steel is \sigma_{Y}=300 \mathrm{~MPa} , what is the factor of safety with respect to yielding?
(a) Maximum compressive stress. The two loads P_{1} and P_{2} acting as shown in Fig. 11-29b are statically equivalent to a single load P = 2000kN acting with an eccentricity e = 40 mm (Fig. 11-29c). Since the column is now loaded by a single force P having an eccentricity e, we can use the secant formula to find the maximum stress.
The required properties of the HE 320A wide-flange shape are obtained from Table E-1 in Appendix E:
A=124.4 \mathrm{~cm}^{2} \quad r=13.58 \mathrm{~cm} \quad c=\frac{310 \mathrm{~mm}}{2}=155 \mathrm{~mm}TABLE E-1
Properties of European Wide-Flange Beams
| Designation | Mass per meter |
Area of section |
Depth of section |
Width of section |
Thickness | Strong axis 1-1 | Weak axis 2-2 | |||||
| G | A | h | b | t_{w} | t_{t} | I_{1} | S_{1} | r_{1} | I_{2} | S_{2} | r_{2} | |
| Kg/m | {cm}^{2} | mm | mm | mm | mm | {cm}^{4} | {cm}^{3} | cm | {cm}^{4} | {cm}^{3} | cm | |
| HE 1000 B HE 900 B HE 700 B HE 650 b HE 600 B |
314 291 241 225 212 |
400 371.3 306.4 286.3 270 |
1000 900 700 650 600 |
300 300 300 300 300 |
19 18.5 17 16 15.5 |
36 35 32 31 30 |
644700 494100 256900 210600 171000 |
12890 10980 7340 6480 5701 |
40.15 36.48 28.96 27.12 25.17 |
16280 15820 14440 13980 13530 |
1085 1054 962.7 932.3 902 |
6.38 6.53 6.87 6.99 7.08 |
| HE 550 B HE 600 A HE 450 B HE 550 A HE 360 B HE 450 A |
199 178 171 166 142 140 |
254.1 226.5 218 211.8 180.6 178 |
550 590 450 540 360 440 |
300 300 300 300 300 300 |
15 13 14 12.5 12.5 11.5 |
29 25 26 24 22.5 21 |
136700 141200 79890 111900 43190 63720 |
4971 4787 3551 4146 2400 2896 |
23.2 24.97 19.14 22.99 15.46 18.92 |
13080 11270 11720 10820 10140 9465 |
871.8 751.4 781.4 721.3 676.1 631 |
7.17 7.05 7.33 7.15 7.49 7.29 |
| HE 340 B HE 320 B HE 360 A HE 340 A |
134 127 112 105 |
170.9 161.3 142.8 133.5 |
340 320 350 330 |
300 300 300 300 |
12 11.5 10 9.5 |
21.5 20.5 17.5 16.5 |
36660 30820 33090 27690 |
2156 1926 1891 1678 |
14.65 13.82 15.22 14.4 |
9690 9239 7887 7436 |
646 615.9 525.8 495.7 |
7.53 7.57 7.43 7.46 |
| HE 320 A HE 260 B HE 240 B HE 280 A HE 220 B HE 260 A HE 240 A |
97.6 93 83.2 76.4 71.5 68.2 60.3 |
124.4 118.4 106 97.26 91.04 86.82 76.84 |
310 260 240 270 220 250 230 |
300 260 240 280 220 260 240 |
9 10 10 8 9.5 7.5 7.5 |
15.5 17.5 17 13 16 12.5 12 |
22930 14920 11260 13670 8091 10450 7763 |
1479 1148 938.3 1013 735.5 836.4 675.1 |
13.58 11.22 10.31 11.86 9.43 10.97 10.05 |
6985 5135 3923 4763 2843 3668 2769 |
465.7 395 326.9 340.2 258.5 282.1 230.7 |
7.49 6.58 6.08 7 5.59 6.5 6 |
| HE 180 B HE 160 B HE 140 B HE 120 B HE 140 A |
51.2 42.6 33.7 26.7 24.7 |
65.25 54.25 42.96 34.01 31.42 |
180 160 140 120 133 |
180 160 140 120 140 |
8.5 8 7 6.5 5.5 |
14 13 12 11 8.5 |
3831 2492 1509 864.4 1033 |
425.7 311.5 215.6 144.1 155.4 |
7.66 6.78 5.93 5.04 5.73 |
1363 889.2 549.7 317.5 389.3 |
151.4 111.2 78.52 52.92 55.62 |
4.57 4.05 3.58 3.06 3.52 |
| HE 100 B HE 100 A |
20.4 16.7 |
26.4 21.24 |
100 96 |
100 100 |
6 5 |
10 8 |
449.5 349.2 |
89.91 72.76 |
4.16 4.06 |
167.3 133.8 |
33.45 26.76 |
2.53 2.51 |
| Note: Axes 1-1 and 2-2 are principal centroidal axes. | ||||||||||||
The required terms in the secant formula of Eq. (11-67) \sigma_{\max }=\frac{P}{A}\left[1+\frac{e c}{r^{2}} \sec \left(\frac{L}{2 r} \sqrt{\frac{P}{E A}}\right)\right] are calculated as follows:
\begin{aligned}\frac{P}{A} &=\frac{2000 \mathrm{kN}}{124.4 \mathrm{~cm}^{2}}=160.77 \mathrm{~MPa} \\\frac{e c}{r^{2}} &=\frac{(40 \mathrm{~mm})(155 \mathrm{~mm})}{(13.58 \mathrm{~cm})^{2}}=0.336 \\\frac{L}{r} &=\frac{7.5 \mathrm{~m}}{13.58 \mathrm{~cm}}=55.23 \\\frac{P}{E A} &=\frac{2000 \mathrm{~kN}}{(210 \mathrm{~GPa})\left(124.4 \mathrm{~cm}^{2}\right)}=765.6 \times 10^{-6}\end{aligned}Substituting these values into the secant formula, we get
\begin{aligned}\sigma_{\max } &=\frac{P}{A}\left[1+\frac{\mathrm{ec}}{r^{2}} \sec \left(\frac{L}{2 r}\sqrt{\frac{P}{E A}}\right)\right] \\&=(160.77 \mathrm{~MPa})(1+0.466)=235.6 \mathrm{~MPa}\end{aligned}This compressive stress occurs at midheight of the column on the concave side (the right-hand side in Fig. 11-29b).
(b) Factor of safety with respect to yielding. To find the factor of safety, we need to determine the value of the load P, acting at the eccentricity e, that will produce a maximum stress equal to the yield stress \sigma_{Y} = MPa Since this value of the load is just sufficient to produce initial yielding of the material, we will denote it as P_{Y}.
Note that we cannot determine P_{Y} by multiplying the load P (equal to 2000 kN) by the ratio \sigma_{Y} /\sigma_{max} . The reason is that we are dealing with a non-linear relationship between load and stress. Instead, we must substitute \sigma_{max} = \sigma_{Y} = 300 MPa in the secant formula and then solve for the corre-sponding load P, which becomes P_{Y}. In other words, we must find the value of P_{Y} that satisfies the following equation:
\sigma_{Y}=\frac{P_{Y}}{A}\left[1+\frac{e c}{r^{2}} \sec \left(\frac{L}{2 r} \sqrt{\frac{P_{Y}}{E A}}\right)\right] (11-70)
Substituting numerical values, we obtain
300 \mathrm{~MPa}=\frac{P_{Y}}{124.4 \mathrm{~cm}^{2}}\left[1+0.336 \mathrm{sec}\left(\frac{55.23}{2} \sqrt{\frac{P_{Y}}{(210 \mathrm{~GPa})\left(124.4 \mathrm{~cm}^{2}\right)}}\right)\right]or 3732 \mathrm{~kN}=P_{Y}\left[1+0.336 \sec \left(5.403 \times 10^{-4} \sqrt{P_{Y}}\right)\right]
in which P_{Y} has units of kN. Solving this equation numerically, we get
P_{Y}=2473 \mathrm{~kN}This load will produce yielding of the material (in compression) at the cross section of maximum bending moment.
Since the actual load is P = 2000 kN , the factor of safety against yielding is
This example illustrates two of the many ways in which the secant formula may be used. Other types of analysis are illustrated in the problems at the end of the chapter.