**Problem 846**

Sketch the shear diagram for the continuous beam shown in Fig. P-846.

**Solution 846**

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

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

$-400(6) + 2M_2(6 + 10) + M_3(10) + \frac{1}{4}(50)(6^3) + \frac{7}{60}(60)(10^3) = 0$

$-2400 + 32M_2 + 10M_3 + 2700 + 7000 = 0$

$32M_2 + 10M_3 = -7300$ ← equation (1)

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

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

$M_2(10) + 1M_3(10 + 0) + 0 + \frac{8}{60}(60)(10^3) + 0 = 0$

$10M_2 + 20M_3 + 8000 = 0$

$10M_2 + 20M_3 = -8000$ ← equation (2)

From equations (1) and (2)

$M_2 = -\frac{1100}{9} ~ \text{lb}\cdot\text{ft} = -122.22 ~ \text{lb}\cdot\text{ft}$ *answer*

$M_3 = -\cdot{3050}{9} ~ \text{lb}\cdot\text{ft} = -338.89 ~ \text{lb}\cdot\text{ft}$ *answer*

Simple beam reactions

$V_0 = 4(50) = 200 ~ \text{lb}$

$V_1 = \frac{1}{2}(6)(50) = 150 ~ \text{lb}$

$V_{2L} = \frac{1}{3} \times (10)(60) = 100 ~ \text{lb}$

$V_{2R} = \frac{2}{3} \times \frac{1}{2}(10)(60) = 200 ~ \text{lb}$

End-moment reactions

${R_0}' = 0$

${R_1}' = (400 - \frac{1100}{9}) / 6 = \frac{1250}{27} ~ \text{lb} = 26.30 ~ \text{lb}$

${R_2}' = (\frac{3050}{9} - \frac{1100}{9}) / 10 = \frac{65}{3} ~ \text{lb} = 21.67 ~ \text{lb}$

Support reactions

$R_1 = 200 + 176.30 = 376.30 ~ \text{lb}$ *answer*

$R_2 = 123.7 + 78.33 = 202.03 ~ \text{lb}$ *answer*

$R_3 = 221.67 ~ \text{lb}$ *answer*