$w = 30(1000)/12$

$w = 2500 \, \text{lb/ft}$

$\Sigma F_V = 0$

$R = W$

$20r = 30(1000)$

$r = 1500 \, \text{lb/ft}$

**First segment (from 0 to 4 ft from left):**

$V_1 = 1500x$

$M_1 = 1500x(x/2)$

$M_1 = 750x^2$

**Second segment (from 4 ft to mid-span):**

$V_2 = 1500x - 2500(x - 4)$

$V_2 = 10000 - 1000x$

$M_2 = 1500x(x/2) - 2500(x - 4)(x - 4)/2$

$M_2 = 750x^2 - 1250(x - 4)^2$

**To draw the Shear Diagram:**

- For the first segment, V
_{1} = 1500x is linear; at x = 0, V_{1} = 0; at x = 4 ft, V_{1} = 6000 lb.
- For the second segment, V
_{2} = 10000 - 1000x is also linear; at x = 4 ft, V_{1} = 6000 lb; at mid-span, x = 10 ft, V_{1} = 0.
- For the next half of the beam, the shear diagram can be accomplished by the concept of symmetry.

**To draw the Moment Diagram:**

- For the first segment, M
_{1} = 750x^{2} is a second degree curve, an open upward parabola; at x = 0, M_{1} = 0; at x = 4 ft, M_{1} = 12000 lb·ft.
- For the second segment, M
_{2} = 750x^{2} - 1250(x - 4)^{2} is a second degree curve, an downward parabola; at x = 4 ft, M_{2} = 12000 lb·ft; at mid-span, x = 10 ft, M_{2} = 30000 lb·ft.
- The next half of the diagram, from x = 10 ft to x = 20 ft, can be drawn by using the concept of symmetry.