STEM Problem Detail

Problem Statement

Prompt

provide step by step solution for Example 8.3

Solution Rendering...

**Solution**

Loads are based on the direct analysis method, so $$K = 1.0$$.

\[
KL_y = (1.0)(10 \ \text{ft}) = 10 \ \text{ft}
\]

The effective length for the $$x$$-axis is

\[
KL_x = (1.0)(20 \ \text{ft}) = 20 \ \text{ft}
\]

### Required Strengths

| **LRFD** | **ASD** |
|---|---|
| The axial load is | The axial load is |
| $$P_r = 1.2D + 1.6L$$ | $$P_r = D + L$$ |
| $$= (1.2)(40) + (1.6)(60)$$ | $$= 40 + 60$$ |
| $$= 144 \ \text{kips}$$ | $$= 100 \ \text{kips}$$ |
| The $$x$$-axis bending is | The $$x$$-axis bending is |
| $$M_{rx} = 1.2M_{Dx} + 1.6M_{Lx}$$ | $$M_{rx} = M_{Dx} + M_{Lx}$$ |
| $$= (1.2)(20) + (1.6)(40)$$ | $$= 20 + 40$$ |
| $$= 88 \ \text{ft-kips}$$ | $$= 60 \ \text{ft-kips}$$ |

| **LRFD** | **ASD** |
|---|---|
| The $$y$$-axis bending is | The $$y$$-axis bending is |
| $$M_{ry} = 1.2M_{Dy} + 1.6M_{Ly}$$ | $$M_{ry} = M_{Dy} + M_{Ly}$$ |
| $$= (1.2)(10) + (1.6)(20)$$ | $$= 10 + 20$$ |
| $$= 44 \ \text{ft-kips}$$ | $$= 30 \ \text{ft-kips}$$ |

### Trial Column Selection

In Table 4--1 of the *AISC Manual*, the lightest W12 column section
listed is a W12$$\times$$40. Lighter W12 sections are available, which are
used primarily as beam sections. Calculate the eccentricities relative
to the $$x$$- and $$y$$-axes. Based on the relatively large eccentricities,
assume the ratio of the required axial strength to the nominal strength
to be approximately 0.25 for the trial section.

\[
P_{u,\text{ratio}} = \frac{P_r}{0.25}
\]

\[
P_{u,\text{ratio}} = \frac{144}{0.25} = 576 \ \text{kips}
\quad \text{(LRFD)}
\]

\[
P_{a,\text{ratio}} = \frac{100}{0.25} = 400 \ \text{kips}
\quad \text{(ASD)}
\]

Select tentative column sections from *AISC Manual* Table 4--1
using an effective length with respect to the $$y$$-axis of 10 ft.

| Section | $$\phi_c P_n$$ (kips) | $$P_n/\Omega_c$$ (kips) |
|---|---|---|
| W12$$\times$$45, $$r_y=2.64$$ | 447 | 294 |
| W12$$\times$$50, $$r_y=2.64$$ | 500 | 332 |
| W12$$\times$$53, $$r_y=2.11$$ | 592 | 394 |
| W12$$\times$$58, $$r_y=2.10$$ | 647 | 431 |

All the $$r_y/r_x$$ ratios for these beams are greater than 2.0, and the
unbraced length ratio between the $$x$$- and $$y$$-axes is 2.0, so the effective
unbraced length with respect to the $$y$$-axis is 10 ft.

The ratio of required strength to nominal strength was assumed for this
trial to be 0.25. If the assumed ratio is correct, then Eq. 8.1 applies
for the interaction formula.

### Unity Check for W12$$\times$$50

From *AISC Manual* Table 6--1,

| **LRFD** | **ASD** |
|---|---|
| $$p \times 10^3 = 2.00 \ \text{kips}^{-1}$$ | $$p \times 10^3 = 3.01 \ \text{kips}^{-1}$$ |
| $$b_x \times 10^3 = 3.54 \ (\text{ft-kips})^{-1}$$ | $$b_x \times 10^3 = 5.32 \ (\text{ft-kips})^{-1}$$ |
| $$b_y \times 10^3 = 11.1 \ (\text{ft-kips})^{-1}$$ | $$b_y \times 10^3 = 16.7 \ (\text{ft-kips})^{-1}$$ |

Using Eq. 8.5,

**LRFD**

\[
\frac{P_r}{\phi_c P_n}
+ b_x M_{rx}
+ b_y M_{ry}
=
\frac{2.00}{10^3}(144)
+ \frac{3.54}{10^3}(88)
+ \frac{11.1}{10^3}(44)
= 1.09 > 1.0
\]

**ASD**

\[
\frac{P_r}{P_n/\Omega_c}
+ b_x M_{rx}
+ b_y M_{ry}
=
\frac{3.01}{10^3}(100)
+ \frac{5.32}{10^3}(60)
+ \frac{16.7}{10^3}(30)
= 1.12 > 1.0
\]

This is not acceptable, so the W12$$\times$$50 section is unsatisfactory.

### Unity Check for W12$$\times$$53

From *AISC Manual* Table 6--1,

| **LRFD** | **ASD** |
|---|---|
| $$p \times 10^3 = 1.69 \ \text{kips}^{-1}$$ | $$p \times 10^3 = 2.54 \ \text{kips}^{-1}$$ |
| $$b_x \times 10^3 = 3.12 \ (\text{ft-kips})^{-1}$$ | $$b_x \times 10^3 = 4.68 \ (\text{ft-kips})^{-1}$$ |
| $$b_y \times 10^3 = 8.15 \ (\text{ft-kips})^{-1}$$ | $$b_y \times 10^3 = 12.2 \ (\text{ft-kips})^{-1}$$ |

**LRFD**

\[
\frac{1.69}{10^3}(144)
+ \frac{3.12}{10^3}(88)
+ \frac{8.15}{10^3}(44)
= 0.88 \le 1.0
\]

**ASD**

\[
\frac{2.54}{10^3}(100)
+ \frac{4.68}{10^3}(60)
+ \frac{12.2}{10^3}(30)
= 0.90 \le 1.0
\]

**Conclusion:**
The W12$$\times$$53 section is satisfactory. Because the first term in the
interaction equation exceeds 0.20, Eq. 8.1 is applicable. Therefore, the
W12$$\times$$53 is the lightest W12 section capable of resisting the required
design loads.

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Problem #63