STEM Problem Detail

Problem Statement

Prompt

provide step by step solution for Example 8.2

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**Solution**

### Required Strengths

Calculate the required strengths for LRFD and ASD. The axial load is

| **LRFD** | **ASD** |
|---|---|
| $$P_r = 1.2D + 1.6L$$ | $$P_r = D + L$$ |
| $$= (1.2)(25) + (1.6)(50)$$ | $$= 25 + 50$$ |
| $$= 110 \ \text{kips}$$ | $$= 75 \ \text{kips}$$ |

### Bending Moments

| **LRFD** | **ASD** |
|---|---|
| $$M_{rx} = 1.2M_{Dx} + 1.6M_{Lx}$$ | $$M_{rx} = M_{Dx} + M_{Lx}$$ |
| $$= (1.2)(1)(8) + (1.6)(2)(8)$$ | $$= (1)(8) + (2)(8)$$ |
| $$= 35.20 \ \text{ft-kips}$$ | $$= 24.0 \ \text{ft-kips}$$ |
| $$M_{ry} = 1.2M_{Dy} + 1.6M_{Ly}$$ | $$M_{ry} = M_{Dy} + M_{Ly}$$ |
| $$= (1.2)(1.6) + (1.6)(3.2)$$ | $$= 1.6 + 3.2$$ |
| $$= 7.04 \ \text{ft-kips}$$ | $$= 4.80 \ \text{ft-kips}$$ |

### Section Compactness

**Flanges (AISC Table B4.1b, Case 10)**

\[
\lambda_p = 0.38 \sqrt{\frac{E}{F_y}}
= 0.38 \sqrt{\frac{29{,}000}{50}}
= 9.15
\]

\[
\frac{b_f}{2t_f} = 6.56 < 9.15
\quad \Rightarrow \text{compact}
\]

**Web (AISC Table B4.1b, Case 10)**

\[
\lambda_p = 3.76 \sqrt{\frac{E}{F_y}}
= 3.76 \sqrt{\frac{29{,}000}{50}}
= 90.55
\]

\[
\frac{h}{t_w} = 34.0 < 90.55
\quad \Rightarrow \text{compact}
\]

The web is compact. The nominal tensile strength is

\[
P_n = F_y A_g
= (50)(7.61)
= 380.50 \ \text{kips}
\]

### Design and Allowable Tensile Strength

| **LRFD** | **ASD** |
|---|---|
| $$P_c = \phi_t P_n = (0.90)(380.50)$$ | $$P_c = \dfrac{P_n}{\Omega_t} = \dfrac{380.50}{1.67}$$ |
| $$= 342.45 \ \text{kips}$$ | $$= 227.84 \ \text{kips}$$ |

### Flexural Strength About the $$x$$-Axis

Since $$h = L_b$$ and $$L_p < L_b < L_r$$, use Eq. 5.9.

\[
P_{ey} = \frac{\pi^2 E I_y}{L_b^2}
= \frac{\pi^2 (29{,}000)(14.1)}
{(8)^2 (12)^2}
= 437.90 \ \text{kips}
\]

\[
\sqrt{1 + \frac{\alpha P_r}{P_{ey}}}
=
\begin{cases}
\sqrt{1 + \dfrac{110}{437.90}} = 1.12 & \text{(LRFD)} \\
\sqrt{1 + \dfrac{1.6(75)}{437.90}} = 1.13 & \text{(ASD)}
\end{cases}
\]

Use a conservative value $$C_b = 1.0$$.

\[
M_p = F_y Z_x
= \frac{(50)(31.3)}{12}
= 130 \ \text{ft-kips}
\]

\[
M_{nx}
= C_b \left[
M_p - (M_p - 0.7F_y S_x)
\frac{L_b - L_p}{L_r - L_p}
\right]
\]

\[
= 114.59 \ \text{ft-kips}
\quad [\le M_p, \text{controls}]
\]

### Flexural Strength About the $$y$$-Axis

\[
M_{ny} = \min
\begin{cases}
\dfrac{F_y Z_y}{12}
= \dfrac{(50)(7.50)}{12}
= 31.25 \ \text{ft-kips} \quad [\text{controls}] \\
\dfrac{1.6F_y S_y}{12}
= \dfrac{(1.6)(50)(4.89)}{12}
= 32.60 \ \text{ft-kips}
\end{cases}
\]

### Design and Allowable Flexural Strengths

| **LRFD** | **ASD** |
|---|---|
| $$M_{cx} = \phi_b M_{nx} = (0.90)(114.59)$$ | $$M_{cx} = \dfrac{M_{nx}}{\Omega_b} = \dfrac{114.59}{1.67}$$ |
| $$= 103.13 \ \text{ft-kips}$$ | $$= 68.62 \ \text{ft-kips}$$ |
| $$M_{cy} = \phi_b M_{ny} = (0.90)(31.25)$$ | $$M_{cy} = \dfrac{M_{ny}}{\Omega_b} = \dfrac{31.25}{1.67}$$ |
| $$= 28.13 \ \text{ft-kips}$$ | $$= 18.71 \ \text{ft-kips}$$ |

### Interaction Check

\[
\frac{P_r}{P_c}
= \frac{110}{342.45}
= 0.32 \ge 0.2
\quad \Rightarrow \text{use Eq. 8.1}
\]

**LRFD**

\[
\frac{P_r}{P_c}
+ \frac{8}{9}
\left(
\frac{M_{rx}}{M_{cx}}
+ \frac{M_{ry}}{M_{cy}}
\right)
=
\frac{110}{342.45}
+ \frac{8}{9}
\left(
\frac{35.20}{103.13}
+ \frac{7.04}{28.13}
\right)
= 0.85 \le 1.0
\]

**ASD**

\[
\frac{75}{227.84}
+ \frac{8}{9}
\left(
\frac{24.0}{68.62}
+ \frac{4.80}{18.71}
\right)
= 0.87 \le 1.0
\]

**Conclusion:**
The W10$$\times$$26 hanger section is satisfactory to resist the imparted loads
and satisfies the requirements of the *AISC Manual*.

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Problem #64