Question 11.4: A steel wide-flange column of W 14 × 82 shape (Fig. 11-28a) ......

(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:

σ_{max}=\frac{P}{A}\left[1+\frac{ec}{r^2}sec\left(\frac{L}{2r}\sqrt{\left(\frac{P}{EA}\right)}\right)\right]                     (11-59)

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\quad \quad \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}{(30,000\ ksi)(24.1\ in.^2)}=497.9×10^{-6}

Substituting these values into the secant formula, we get

σ_{max}=\frac{P}{A}\left[1+\frac{ec}{r^2}sec\left(\frac{L}{2r}\sqrt{\frac{P}{EA}}\right)\right]

= (14.94 ksi)(1 + 0.345) = 20.1 ksi

This 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 σ_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 σ_y/σ_{max}. The reason is that we are dealing with a nonlinear relationship between load and stress. Instead, we must substitute σ_{max} = σ_{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:

σ_{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}{(30,000\ ksi)(24.1\ in.^2)}}\right)\right]

or                  1012 k = P_y[1 + 0.2939 sec(0.02916\sqrt{P_y})]

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_v}{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.