**For Member B - 1**
$S_{required} = \dfrac{M}{f_b} = \dfrac{\frac{1}{8}(22.5)(5^2)(1000^2)}{120}$

$S_{required} = 586 \times 10^3 \, \text{mm}^3$

From Appendix B, Table B-2 Properties of Wide-Flange Sections (W Shapes): SI Units, of text book:

Use **W410 × 39** with S = 634 × 10^{3} mm^{3} for member B - 1. *answer*

**For Member G - 1**

$M = 2.5(28.125)$

$M = 70.3125 \, \text{kN}\cdot\text{m}$

$S_{required} = \dfrac{M}{f_b} = \dfrac{70.3125(1000^2)}{120}$

$S_{required} = 586 \times 10^3 \, \text{mm}^3$

From Appendix B, Table B-2 Properties of Wide-Flange Sections (W Shapes): SI Units, of text book:

Use **W410 × 39** with S = 634 × 10^{3} mm^{3} for member G - 1. *answer*

**For Member B - 2:**

$\Sigma M_{R2} = 0$

$7R_1 = 28.125(5) + 18.75(2)(6) + 30(5)(2.5)$

$R_1 = 105.804 \, \text{kN}$

$\Sigma M_{R1} = 0$

$7R_2 = 28.125(2) + 18.75(2)(1) + 30(5)(4.5)$

$R_2 = 109.821 \, \text{kN}$

Location of Maximum Moment:

$\dfrac{x}{109.821} = \dfrac{5 - x}{40.179}$

$40.179x = 549.105 - 109.821x$

$x = 3.6607 \, \text{m}$

Maximum Moment

$M = \frac{1}{2}(3.6607)(109.821)$

$M = 201.01 \, \text{kN}\cdot\text{m}$

$S_{required} = \dfrac{M}{f_b} = \dfrac{201.01(1000^2)}{120}$

$S_{required} = 1\,675 \times 10^3 \, \text{mm}^3$

From Appendix B, Table B-2 Properties of Wide-Flange Sections (W Shapes): SI Units, of text book:

Use **W610 × 82** with S = 1 870 × 10^{3} mm^{3} for member B - 2. *answer*

**For Member B - 3**

$S_{required} = \dfrac{M}{f_b} = \dfrac{\frac{1}{8}(37.5)(7^2)(1000^2)}{120}$

$S_{required} = 1\,914 \times 10^3 \, \text{mm}^3$

From Appendix B, Table B-2 Properties of Wide-Flange Sections (W Shapes): SI Units, of text book:

Use **W610 × 92** with S = 2 140 × 10^{3} mm^{3} for member B - 3. *answer*

**Summary**