**Problem 845**

Compute the moments over the supports for the beam shown in Fig. P-845 and then draw the shear diagram.

**Solution 845**

*answer*

Three-moment equation to spans (1) and (2)

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

$-9000(10) + 2M_2(10 + 10) + M_3(10) + \frac{1}{4}(500)(10^3) + \dfrac{4000(3)}{10}(10^2 - 3^2) = 0$

$-90\,000 + 40M_2 + 10M_3 + 125\,000 + 109\,200 = 0$

$40M_2 + 10M_3 = -144\,200$ ← equation (1)

Three-moment equation to spans (2) and (3)

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

$M_2(10) + 2M_3(10 + 0) + 0 + \dfrac{4000(7)}{10}(10^2 - 7^2) + 0 = 0$

$10M_2 + 20M_3 + 142\,800 = 0$

$10M_2 + 20M_3 = -142\,800$ ← equation (2)

From equations (1) and (2)

$M_2 = -2080 ~ \text{lb}\cdot\text{ft}$ *answer*

$M_3 = -6100 ~ \text{lb}\cdot\text{ft}$ *answer*

Simple beam reactions

$V_1 = \frac{1}{2}(10)(500) = 2500 ~ \text{lb}$

$V_{2L} = 3(4000) / 10 = 1200 ~ \text{lb}$

$V_{2R} = 7(4000) / 10 = 2800 ~ \text{lb}$

End-moment reactions

${R_1}' = (9000 - 2080) / 10 = 692 ~ \text{lb}$

${R_2}' = (6100 - 2080) / 10 = 402 ~ \text{lb}$

Reactions

*answer*

$R_2 = 1808 + 798 = 2606 ~ \text{lb}$ *answer*

$R_3 = 3202 ~ \text{lb}$ *answer*