Column AB consists of a W10 × 39 rolled-steel shape made of a grade of steel for which

Question

Column [latex]A B[/latex] consists of a [latex]W 10 imes 39[/latex] rolled-steel shape made of a grade of steel for which [latex]\sigma_{Y}=36 \mathrm{ksi}[/latex] and [latex]E=29 imes 10^{6} \mathrm{psi}[/latex]. Determine the allowable centric load [latex]\mathbf{P}(a)[/latex] if the effective length of the column is [latex]24 \mathrm{ft}[/latex] in all directions, [latex](b)[/latex] if bracing is provided to prevent the movement of the midpoint [latex]C[/latex] in the [latex]x z[/latex] plane. (Assume that the movement of point [latex]C[/latex] in the [latex]y z[/latex] plane is not affected by the bracing.)

Accepted Answer

We first compute the value of the slenderness ratio from Eq. 10.41 corresponding to the given yield strength [latex]\sigma_{Y}=36 \mathrm{ksi}[/latex].

[latex]\frac{L}{r}=4.71 \sqrt{\frac{29 imes 10^{6}}{36 imes 10^{3}}}=133.7[/latex]

a. Effective Length [latex]=\mathbf{2 4} \mathrm{ft}[/latex]. Since [latex]r_{y}

[latex]\frac{L}{r_{y}}=\frac{(24 imes 12) \mathrm{in.}}{1.98 \mathrm{in.}}=\frac{288 \mathrm{in} .}{1.98 \mathrm{in.}}=145.5[/latex]

Since [latex]L / r>133.7[/latex], we use Eq. (10.39)

[latex]\sigma_{e}={\frac{\pi^{2}E}{(L/r)^{2}}}[/latex] (10.39)

in Eq. (10.40)

[latex]\sigma_{c r}=0.877\sigma_{e}[/latex] (10.40)

to determine [latex]\sigma_{c r}[/latex]

[latex]\sigma_{c r}=0.877 \sigma_{e}=0.877 \frac{\pi^{2} E}{(L / r)^{2}}=0.877 \frac{\pi^{2}\left(29 imes 10^{3} \mathrm{ksi} ight)}{(145.5)^{2}}=11.86 \mathrm{ksi}[/latex]

The allowable stress, determined using Eq. (10.42), and [latex]P_{ ext {all }}[/latex] are

[latex] \begin{aligned} & \sigma_{ ext {all }}=\frac{\sigma_{c r}}{1.67}=\frac{11.86 \mathrm{ksi}}{1.67}=7.10 \mathrm{ksi} \ & P_{ ext {all }}=\sigma_{ ext {all }} A=(7.10 \mathrm{ksi})\left(11.5 \mathrm{in}^{2} ight)=81.7 \mathrm{kips} \end{aligned} [/latex]

b. Bracing at Midpoint [latex]C[/latex]. Since bracing prevents movement of point [latex]C[/latex] in the [latex]x z[/latex] plane but not in the [latex]y z[/latex] plane, we must compute the slenderness ratio correspoinding to buckling in each plane and determine which is larger.

[latex]x z ext { Plane: } \quad ext { Effective length }=12 \mathrm{ft}=144 ext { in., } r=r_{y}=1.98 ext { in. }[/latex]

[latex]L / r=(144 ext { in. }) /(1.98 ext { in. })=72.7[/latex]

yz Plane: Effective length [latex]=24 \mathrm{ft}=288[/latex] in., [latex]r=r_{x}=4.27 \mathrm{in}[/latex].

[latex]L / r=(288 ext { in. }) /(4.27 ext { in. })=67.4[/latex]

Since the larger slenderness ratio corresponds to a smaller allowable load, we choose [latex]L / r=72.7[/latex]. Since this is smaller than [latex]L / r=133.7[/latex], we use Eqs. (10.39) and (10.38)

[latex]\sigma_{\mathrm{cr}}=[0.658^{\sigma_Y/\sigma_e}]\sigma_{Y}[/latex] (10.38)

to determine [latex]\sigma_{c r}[/latex]

[latex]\sigma_{e}=\frac{\pi^{2} E}{(L / r)^{2}}=\frac{\pi^{2}\left(29 imes 10^{3} \mathrm{ksi} ight)}{(72.7)^{2}}=54.1 \mathrm{ksi}[/latex]

[latex]\sigma_{c r}=\left[0.658^{\left(\sigma_{\gamma} / \sigma_{e} ight)} ight] F_{Y}=\left[0.658^{(36 \mathrm{ksi} / 54.1 \mathrm{ksi})} ight] 36 \mathrm{ksi}=27.3 \mathrm{ksi}[/latex]

We now calculate the allowable stress using Eq. (10.42) and the allowable load.

[latex] \begin{aligned} \sigma_{ ext {all }} & =\frac{\sigma_{c r}}{1.67}=\frac{27.3 \mathrm{ksi}}{1.67}=16.32 \mathrm{ksi} \ P_{ ext {all }} & =\sigma_{ ext {all }} A=(16.32 \mathrm{ksi})\left(11.5 \mathrm{in}^{2} ight) \quad P_{ ext {all }}=187.7 \mathrm{ksi} \end{aligned} [/latex]

Question 10.SP.3: Column AB consists of a W10 × 39 rolled-steel shape made of ...

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