Question 11.5: A steel wide-flange column of a HE 320A shape (Fig. 11-29a) ......

Use a four-step problem-solving approach.
Part (a): Maximum compressive stress.
1. Conceptualize: The two loads P_1 ~ and ~ P_2 acting as shown in Fig. 11-29b are statically equivalent to a single load P = 2000 kN 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, use the secant formula to find the maximum stress.
The required properties of the HE 320A wide-flange shape are obtained from Table F-1 in Appendix F:
A = 124.4 ~ cm^2 \quad r = 13.58 ~ cm \quad c = \frac{310~mm}{2} = 155~mm
2. Categorize: The required terms in the secant formula of Eq. (11-67) are calculated as
\quad\quad\quad\quad {\frac{P}{A}}={\frac{2000\,\mathrm{kN}}{124.4\,\mathrm{cm^2}}}=160.77\,\mathrm{MP}{\mathrm{a}} \\ \quad\quad\quad\quad {\frac{e c}{r^{2}}}={\frac{(40\operatorname*{mm})(155\operatorname*{mm})}{(13\cdot58\operatorname*{cm})^{2}}}=0.336 \\ \quad\quad\quad\quad {\frac{L}{r}}={\frac{(7.5m)}{13.58\,\mathrm{cm}}}= 55.23 \\ \quad\quad\quad\quad {\frac{P}{E A}}={\frac{2000\,\mathrm{kN}}{(210{\mathrm{GPa}})(124.4\,\mathrm{cm}^{2})}}=765.6\times10^{-6}
3. Analyze: Substitute these values into the secant formula to get
\quad\quad\quad\quad\sigma_{\mathrm{max}}\ \ ={\frac{P}{A\!}}\!\left[1+{\frac{e c}{r^{2}}}sec {\Bigg\lgroup}{\frac{L}{2r}}{\sqrt{\frac{P}{E A}}}{\Bigg\rgroup}\right]  \\  \quad\quad\quad\quad = (160.77 ~MPa)(1 + 0.466) = 235.6 ~MPa
4. Finalize: This compressive stress occurs at mid-height of the column on the concave side (the right-hand side in Fig. 11-29b).
Part (b): Factor of safety with respect to yielding.
1. Conceptualize: To find the factor of safety, 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 = 300 ~MPa. Since this value of the load is just sufficient to produce initial yielding of the material, denote it as P_Y.
2. Categorize: Note that force P_Y cannot be determined by multiplying the load P (equal to 2000 kN) by the ratio \sigma_Y /\sigma_{max}. The reason is that there is a nonlinear relationship between load and stress. Instead, substitute \sigma_{max} = \sigma_Y = 300~ MPa in the secant formula and then solve for the corresponding load P, which becomes P_Y. In other words, find the value of P_Y that satisfies
\quad\quad\quad\quad \sigma_{\mathrm{Y}}={\frac{P_{\mathrm{Y}}}{A}}{\Bigg[}{1+{\frac{e c}{r^{2}}}}{\mathsf{s e c}}{{\Bigg\lgroup}}{\frac{L}{2r}}{\sqrt{\frac{P_{\mathrm{Y}}}{E A}}}{\Bigg\rgroup}{\Bigg]}\quad\quad(11-70)
3. Analyze: Substitute numerical values to obtain
\quad 300\;M P{a}=\frac{P_{Y}}{124.4\;\mathrm{cm}^{2}}\bigg[1+0.336\,\mathrm{sec}{\Bigg\lgroup}\frac{55.23}{2}\sqrt{\frac{P_Y}{{(210~\mathrm{GP}a)(124.4~\mathrm{cm}^{2})}}}{\Bigg\rgroup}\bigg]
so
\quad\quad\quad\quad 3732\,\mathrm{KN}={P}_{Y}[1+0.336\,s e c\,(5.403\times10^{-4}\sqrt{P_{Y}}\,)]
in which P_Y has units of kN. Solving this equation numerically gives
\quad\quad\quad\quad P_Y = 2473 ~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
\quad\quad\quad\quad n = \frac{P_Y}{P} = \frac{2473 ~KN}{2000~ kN} = 1.236
4. Finalize: 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.