$\Sigma M_B = 0$

$2w_o (5) = 10(4)(0) + 20(2) + 40(3)$

$w_o = 16 \, \text{kN/m}$

$\Sigma M_{midpoint \,\, of \,\, EF} = 0$

$5R_1 = 10(4)(5) + 20(3) + 40(2)$

$R_1 = 68 \, \text{kN}$

**To draw the Shear Diagram**

- M
_{A} = 0
- M
_{B} = M_{A} + Area in load diagram

M_{B} = 0 - 10(2) = -20 kN

M_{B2} + M_{B} + R_{1} = -20 + 68 = 48 kN
- M
_{C} = M_{B2} + Area in load diagram

M_{C} = 48 - 10(2) = 28 kN

M_{C2} = M_{C} - 20 = 28 - 20 = 8 kN
- M
_{D} = M_{C2} + Area in load diagram

M_{D} = 8 + 0 = 8 kN

M_{D2} = M_{D} - 40 = 8 - 40 = -32 kN
- M
_{E} = M_{D2} + Area in load diagram

M_{E} = -32 + 0 = -32 kN
- M
_{F} = M_{E} + Area in load diagram

M_{F} = -32 + w_{o}(2)

M_{F} = -32 + 16(2) = 0

**To draw the Moment Diagram**

- M
_{A} = 0
- M
_{B} = M_{A} + Area in shear diagram

M_{B} = 0 - ½ (20)(2) = -20 kN·m
- M
_{C} = M_{B} + Area in shear diagram

M_{C} = -20 + ½ (48 + 28)(2)

M_{C} = 56 kN·m
- M
_{D} = M_{C} + Area in shear diagram

M_{D} = 56 + 8(1) = 64 kN·m
- M
_{E} = M_{D} + Area in shear diagram

M_{E} = 64 - 32(1) = 32 kN·m
- M
_{F} = M_{E} + Area in shear diagram

M_{F} = 32 - ½(32)(2) = 0
- The location and magnitude of moment at C' are determined from shear diagram. By squared property of parabola, x = 0.44 m from B.