**Problem 847**

Compute the moments over the supports and sketch the shear diagram for the continuous beam shown in Fig. P-847.

**Solution 847**

*answer*

Apply three-moment equation between spans (1) and (2)

$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

$-6(4) + 2M_2(4 + 6) + M_3(6) + \frac{1}{4}(3)(4^3) + \frac{8}{60}(5)(6^3) = 0$

$-24 + 20M_2 + 6M_3 + 48 + 144 = 0$

$20M_2 + 6M_3 = -168$ ← equation (1)

Apply three-moment equation between spans (2) and (3)

$M_2L_2 + 2M_3(L_2 + L_3) + M_4L_3 + \dfrac{6A_2\bar{a}_2}{L_2} + \dfrac{6A_3\bar{b}_3}{L_3} = 0$

$M_2(6) + 2M_3(6 + 0) + 0 + \frac{7}{60}(5)(6^3) + 0 = 0$

$6M_2 + 12M_3 + 126 = 0$

$6M_2 + 12M_3 = -126$ ← equation (2)

From equations (1) and (2)

$M_2 = -\frac{105}{17} ~ \text{kN}\cdot\text{m} = -6.176 ~ \text{kN}\cdot\text{m}$ *answer*

$M_3 = -\frac{126}{17} ~ \text{kN}\cdot\text{m} = -7.412 ~ \text{kN}\cdot\text{m}$ *answer*

Simple beam reactions

$V_0 = 3 ~ \text{kN}$

$V_1 = \frac{1}{2}(4)(3) = 6 ~ \text{kN}$

$V_{2L} = \frac{2}{3} \times \frac{1}{2}(6)(5) = 10 ~ \text{kN}$

$V_{2R} = \frac{1}{3} \times \frac{1}{2}(6)(5) = 5 ~ \text{kN}$

Couple reactions

${R_0}' = 0$

${R_1}' = (\frac{105}{17} - 6) / 4 = \frac{3}{68} ~ \text{kN}$

${R_2}' = (\frac{126}{17} - \frac{105}{17}) / 6 = \frac{7}{34} ~ \text{kN}$

Reactions

$R_1 = 3 + \frac{405}{68} = \frac{609}{68} ~ \text{kN} = 8.956 ~ \text{kN}$ *answer*

$R_2 = \frac{411}{68} + \frac{333}{34} = \frac{1077}{68} ~ \text{kN} = 15.838 ~ \text{kN}$ *answer*

$R_3 = \frac{177}{34} ~ \text{kN} = 5.206 ~ \text{kN}$ *answer*