Question 10.SP.2: The uniform column AB consists of an 8-ft section of structu...

Effective Length. Since the column has one end fixed and one end free, its effective length is

L_{e}=2(8 \mathrm{ft})=16 \mathrm{ft}=192 \mathrm{in}

Critical Load. Using Euler’s formula, we write

P_{\text {cr }}=\frac{\pi^{2} E I}{L_{e}^{2}}=\frac{\pi^{2}\left(29 \times 10^{6} \mathrm{psi}\right)\left(8.00 \mathrm{in}^{4}\right)}{(192 \mathrm{in} .)^{2}} \quad P_{\text {cr }}=62.1 \mathrm{kips}

a. Allowable Load and Stress. For a factor of safety of 2, we find

P_{\text {all }}=\frac{P_{\text {cr }}}{F . S .}=\frac{62.1 \mathrm{kips}}{2} \quad P_{\text {all }}=31.1 \text { kips }

and

\sigma=\frac{P_{\text {all }}}{A}=\frac{31.1 \mathrm{kips}}{3.54 \mathrm{in}^{2}} \quad \sigma=8.79 \mathrm{ksi}

b. Eccentric Load. We observe that column A B and its loading are identical to the upper half of the column of Fig. 10.19 which was used in the derivation of the secant formulas; we conclude that the formulas of Sec. 10.5 apply directly to the case considered here. Recalling that P_{\text {all }} / P_{\text {cr }}=\frac{1}{2} and using Eq. (10.31), we compute the horizontal deflection of point A:

\begin{aligned} y_{m} & =e\left[\sec \left(\frac{\pi}{2} \sqrt{\frac{P}{P_{\mathrm{cr}}}}\right)-1\right]=(0.75 \mathrm{in} .)\left[\sec \left(\frac{\pi}{2 \sqrt{2}}\right)-1\right] \\ & =(0.75 \mathrm{in} .)(2.252-1) \quad y_{m}=0.939  \mathrm{in.} \end{aligned}

The maximum normal stress is obtained from Eq. (10.35):

\begin{aligned} \sigma_{m} & =\frac{P}{A}\left[1+\frac{e c}{r^{2}} \sec \left(\frac{\pi}{2} \sqrt{\frac{P}{P_{\text {cr }}}}\right)\right] \\ & =\frac{31.1 \mathrm{kips}}{3.54 \mathrm{in}^{2}}\left[1+\frac{(0.75 \mathrm{in} .)(2 \mathrm{in} .)}{(1.50 \mathrm{in} .)^{2}} \sec \left(\frac{\pi}{2 \sqrt{2}}\right)\right] \\ & =(8.79 \mathrm{ksi})[1+0.667(2.252)] \quad \sigma_{m}=22.0  \mathrm{ksi} \end{aligned}