$M_1 = -4(1)(0.5) = -2 ~ \text{kN}\cdot\text{m} = -2000 ~ \text{N}\cdot\text{m}$
answer
Apply three-moment equation between spans (1) and (2)
$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$
$-2(4) + 2M_2(4 + 4) + M_3(4) + \dfrac{4(4^3)}{4} + \left[ \dfrac{2(3)}{4}(4^2 - 3^2) + \dfrac{2(1)}{4}(4^2 - 1^2) \right]$
$-8 + 16M_2 + 4M_3 + 64 + 18 = 0$
$16M_2 + 4M_3 = -74$ ← equation (1)
Apply three-moment equation between spans (2) and (3)
$M_2L_2 + 2M_3(L_2 + L_3) + M_4L_3 + \dfrac{6A_2\bar{a}_2}{L_2} + \dfrac{6A_3\bar{b}_3}{L_3} = 0$
$M_2(4) + 2M_3(4 + 3) + M_4(3) + \left[ \dfrac{2(1)}{4}(4^2 - 1^2) + \dfrac{2(3)}{4}(4^2 - 3^2) \right] + \dfrac{7(6)(3^3)}{60} = 0$
$4M_2 + 14M_3 + 3M4 + 18 + 18.9 = 0$
$4M_2 + 14M_3 + 3M_4 = -36.9$ ← equation (2)
Apply three-moment equation between spans (3) and (4)
$M_3L_3 + 2M_4(L_3 + L_4) + M_5L_4 + \dfrac{6A_3\bar{a}_3}{L_3} + \dfrac{6A_4\bar{b}_4}{L_4} = 0$
$M_3(3) + 2M_4(3 + 0) + 0 + \frac{8}{60}(6)(3^3) + 0 = 0$
$3M_3 + 6M_4 + 21.6 = 0$
$3M_3 + 6M_4 = -21.6$ ← equation (3)
From equations (1), (2), and (3)
$M_2 = -4.4598 ~ \text{kN}\cdot\text{m} = -4459.8 ~ \text{N}\cdot\text{m}$ answer
$M_3 = -0.6609 ~ \text{kN}\cdot\text{m} = -660.9 ~ \text{N}\cdot\text{m}$ answer
$M_4 = -3.2696 ~ \text{kN}\cdot\text{m} = -3269.6 ~ \text{N}\cdot\text{m}$ answer
Simple beam reactions
$V_0 = 4000(1) = 4000 ~ \text{N}$
$V_1 = (4)(4000) = 8000 ~ \text{N}$
$V_2 = (2000 + 2000) = 2000 ~ \text{N}$
$V_{3L} = \frac{1}{3} \times \frac{1}{2}(3)(6000) = 3000 ~ \text{N}$
$V_{3R} = \frac{2}{3} \times \frac{1}{2}(3)(6000) = 6000 ~ \text{N}$
Couple reactions
${R_0}' = 0$
${R_1}' = (4459.8 - 2000) / 4 = 614.95 ~ \text{N}$
${R_2}' = (4459.8 - 660.9) / 4 = 949.72 ~ \text{N}$
${R_3}' = (3269.6 - 660.9) / 3 = 869.57 ~ \text{N}$
Support reactions
$R_1 = 4000 + 7385.05 = 11\,385.05 ~ \text{N}$
$R_2 = 8614.95 + 2949.72 = 11\,564.67 ~ \text{N}$
$R_3 = 1050.28 + 2130.43 = 3180.71 ~ \text{N}$
$V_4 = 3869.57 ~ \text{N}$
$\dfrac{x_1}{7385.05} = \dfrac{4}{7385.05 + 8614.95}$
$x_1 = 1.8463 ~ \text{m}$
$M_{x_1} = M_1 + \text{Area in the shear diagram}$
$M_{x_1} = -2000 + \frac{1}{2}(1.8463)(7385.05)$
$M_{x_1} = 4817.51 ~ \text{N}\cdot\text{m}$
$\dfrac{{x_2}^2}{2130.43} = \dfrac{3^2}{2130.43 + 3869.57}$
$x_2 = 1.7876 ~ \text{m}$
$M_{x_2} = M_3 + \text{Area in the shear diagram}$
$M_{x_2} = -660.9 + \frac{2}{3}(1.7876)(2130.43)$
$M_{x_2} = 1878.00 ~ \text{N}\cdot\text{m}$
$M_A = M_2 + \text{Area in the shear diagram}$
$M_A = -4459.8 + 2949.72(1) + 949.72(2)$
$M_A = 389.36 ~ \text{N}\cdot\text{m}$
Thus,
$M_{max(+)} = 4817.51 ~ \text{N}\cdot\text{m}$ answer
