Question 11.4: A steel wide-flange column of W 14 × 82 shape (Fig. 11-28a) ...
A steel wide-flange column of W 14 × 82 shape (Fig. 11-28a) is pin-supported at the ends and has a length of 25 ft. The column supports a centrally applied load P_{1} = 320 k and an eccentrically applied load P_{2} = 40 k (Fig. 11-28b). Bending takes place about axis 1-1 of the cross section, and the eccentric load acts on axis 2-2 at a distance of 13.5 in. from the centroid C.
(a) Using the secant formula, and assuming E = 30,000 ksi, calculate the maximum compressive stress in the column.
(b) If the yield stress for the steel is \sigma_{Y} = 42 ksi, 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. 11-28b are statically equivalent to a single load P = 360 k acting with an eccentricity e = 1.5 in. (Fig. 11-28c). 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 W 14 × 82 wide-flange shape are obtained from Table E-1 in Appendix E:
A = 24.1 in.² r = 6.05 in. c = \frac{14.31 in.}{2} = 7.155 in.
The required terms in the secant formula (Eq. 11-59) are calculated as follows:
\frac{P}{A}=\frac{360 k}{24.1 in.^{2}}=14.94 ksi
\frac{ec}{r^{2}}=\frac{(1.5 in.)(7.155 in.)}{(6.05 in.)^{2}} =0.2932
\frac{L}{r}=\frac{(25 ft)(12 in./ft)}{6.05 in.} =49.59
\frac{P}{EA}=\frac{360 k}{}=497.9\times 10^{-6}
Substituting these values into the secant formula, we get
\sigma_{\max}=\frac{P}{A}\left[1+\frac{ec}{r^{2}}\sec\left(\frac{L}{2r}\sqrt{\frac{P}{EA}}\right)\right] (11-59)
= (14.94 ksi)(1 + 0.345) = 20.1 ksi
This compressive stress occurs at midheight of the column on the concave side (the right-hand side in Fig. 11-28b).
(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} = 42 ksi. 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 360 k) by the ratio \sigma_{Y}/\sigma_{\max}. The reason is that we are dealing with a nonlinear relationship between load and stress. Instead, we must substitute \sigma_{\max}=\sigma_{Y} = 42 ksi in the secant formula and then solve for the corresponding 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{ec}{r^{2}}\sec\left(\frac{L}{2r}\sqrt{\frac{P_{Y}}{EA}}\right)\right] (11-62)
Substituting numerical values, we obtain
42 ksi=\frac{P_{Y}}{24.1 in.^{2}}\left[1+0.2939 \sec\left(\frac{49.59}{2}\sqrt{\frac{P_{Y}}{}}\right)\right]
or
1012 k=P_{Y}\left[1+0.2939\sec \left(0.02916\sqrt{P_{Y}}\right)\right]
in which P_{Y} has units of kips. Solving this equation numerically, we get
P_{Y} = 716 k
This load will produce yielding of the material (in compression) at the cross section of maximum bending moment.
Since the actual load is P = 360 k, the factor of safety against yielding is
n=\frac{P_{Y}}{P}=\frac{716 k}{360 k}=1.99
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.
