$\Sigma M_A = 0$

$12R_C = 4(900) + 18(400) + 9[(60)(18)]$

$R_C = 1710 \, \text{lb}$

$\Sigma M_C = 0$

$12R_A + 6(400) = 8(900) + 3[60(18)]$

$R_A = 670 \, \text{lb}$

**Segment AB:**

$V_{AB} = 670 - 60x \, \text{lb}$

$M_{AB} = 670x - 60x(x/2)$

$M_{AB} = 670x - 30x^2 \, \text{lb}\cdot\text{ft}$

**Segment BC:**

$V_{BC} = 670 - 900 - 60x$

$V_{BC} = -230 - 60x \, \text{lb}$

$M_{BC} = 670x - 900(x - 4) - 60x(x/2)$

$M_{BC} = 3600 - 230x - 30x^2 \, \text{lb}\cdot\text{ft}$

**Segment CD:**

$V_{CD} = 670 + 1710 - 900 - 60x$

$V_{CD} = 1480 - 60x \, \text{lb}$

$M_{CD} = 670x + 1710(x - 12) - 900(x - 4) - 60x(x/2)$

$M_{CD} = -16920 + 1480x - 30x^2 \, \text{lb}\cdot\text{ft}$

**To draw the Shear Diagram:**

- V
_{AB} = 670 - 60x for segment AB is linear; at x = 0, V_{AB}= 670 lb; at x = 4 ft, V_{AB} = 430 lb.
- For segment BC, V
_{BC} = -230 - 60x is also linear; at x= 4 ft, V_{BC} = -470 lb, at x = 12 ft, V_{BC} = -950 lb.
- V
_{CD} = 1480 - 60x for segment CD is again linear; at x = 12, V_{CD} = 760 lb; at x = 18 ft, V_{CD} = 400 lb.

**To draw the Moment Diagram:**

- M
_{AB} = 670x - 30x^{2} for segment AB is a second degree curve; at x = 0, M_{AB} = 0; at x = 4 ft, M_{AB} = 2200 lb·ft.
- For BC, M
_{BC} = 3600 - 230x - 30x^{2}, is a second degree curve; at x = 4 ft, M_{BC} = 2200 lb·ft, at x = 12 ft, M_{BC} = -3480 lb·ft; When M_{BC} = 0, 3600 - 230x - 30x^{2} = 0, x = -15.439 ft and 7.772 ft. Take x = 7.772 ft, thus, the moment is zero at 3.772 ft from B.
- For segment CD, M
_{CD} = -16920 + 1480x - 30x^{2} is a second degree curve; at x = 12 ft, M_{CD} = -3480 lb·ft; at x = 18 ft, M_{CD} = 0.

can someone please explain why in reactions we are multiplying 9*60*18 and 3*60*18...when we could have divided the udl into 2 sections of 12 and 6 fts respectively and solve for the moments accordingly

No need to divide the UDL into 2 sections because it is continuous. It would be simpler that way. You can also do what you suggest and arrived to the same result, but, if I judge it from my own perspective, I can say that it is unnecessary.

This page belongs to Strength of Materials. If you need to improve your skill in finding the reactions of simple beam with uniform and concentrated loads, I suggest you browse the Engineering Mechanics section. That will brought you back to basic of finding simple reactions.