$\Sigma M_A = 0$

$10R_C = 2(80) + 5[10(10)]$

$R_C = 66 \, \text{kN}$

$\Sigma M_C = 0$

$10R_A = 8(80) + 5[10(10)]$

$R_A = 114 \, \text{kN}$

**Segment AB:**

$V_{AB} = 114 - 10x \, \text{kN}$

$M_{AB} = 114x - 10x(x/2)$

$M_{AB} = 114x - 5x^2 \, \text{kN}\cdot\text{m}$

**Segment BC:**

$V_{BC} = 114 - 80 - 10x$

$V_{BC} = 34 - 10x \, \text{kN}$

$M_{BC} = 114x - 80(x - 2) - 10x(x/2)$

$M_{BC} = 160 + 34x - 5x^2 \, \text{kN}\cdot\text{m}$

**To draw the Shear Diagram:**

- For segment AB, V
_{AB} = 114 - 10x is linear; at x = 0, V_{AB} = 14 kN; at x = 2 m, V_{AB} = 94 kN.
- V
_{BC} = 34 - 10x for segment BC is linear; at x = 2 m, V_{BC} = 14 kN; at x = 10 m, V_{BC} = -66 kN. When V_{BC} = 0, x = 3.4 m thus V_{BC} = 0 at 1.4 m from B.

**To draw the Moment Diagram:**

- M
_{AB} = 114x - 5x^{2} is a second degree curve for segment AB; at x = 0, M_{AB} = 0; at x = 2 m, M_{AB} = 208 kN·m.
- The moment diagram is also a second degree curve for segment BC given by M
_{BC} = 160 + 34x - 5x^{2}; at x = 2 m, M_{BC} = 208 kN·m; at x = 10 m, M_{BC} = 0.
- Note that the maximum moment occurs at point of zero shear. Thus, at x = 3.4 m, M
_{BC} = 217.8 kN·m.