# Problem 826 | Continuous Beam by Three-Moment Equation

**Problem 826**

See Figure P-826.

**Solution 826**

## Click here to expand or collapse this section

$M_1 L_1 + 2M_2(L_1 + L_2) + M_3 L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

$M_1 = 0$

$L_1 = L_2 = 4 ~ \text{m}$

$\dfrac{6A_1\bar{a}_1}{L_1} = -\dfrac{M}{L}(3a^2 - L^2) = -\dfrac{8}{4}[ \, 3(2^2) - 4^2 \, ] = 8 ~ \text{kN}\cdot\text{m}^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = \frac{1}{4}w_oL^3 = \frac{1}{4}(2)(4^3) = 32 ~ \text{kN}\cdot\text{m}^2$

Thus,

$0 + 2M_2(4 + 4) + M_3(4) + 8 + 32 = 0$

$16M_2 + 4M_3 = -40$ ← equation (1)

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

$M_2 L_2 + 2M_3(L_2 + L_3) + M_4 L_3 + \dfrac{6A_2\bar{a}_2}{L_2} + \dfrac{6A_3\bar{b}_3}{L_3} = 0$

$L_2 = 4 ~ \text{m}$

$L_3 = 3 ~ \text{ft}$

$M_4 = 0$

$\dfrac{6A_2\bar{a}_2}{L_2} \frac{1}{4}w_oL^3 = \frac{1}{4}(2)(4^3) = 32 ~ \text{kN}\cdot\text{m}^2$

$\dfrac{6A_3\bar{b}_3}{L_3} = \dfrac{Pb}{L}(L^2 - b^2) = \dfrac{6(2)b}{3}(3^2 - 2^2) = 20 ~ \text{kN}\cdot\text{m}^2$

Thus,

$M_2(4) + 2M_3(4 + 3) + 0 + 32 + 20 = 0$

$4M_2 + 14M_3 = -52$ ← equation (2)

From equations (1) and (2)

$M_2 = -1.6923 ~ \text{kN}\cdot\text{m} = -1692.3 ~ \text{N}\cdot\text{m}$ *answer*

$M_3 = -3.2307 ~ \text{kN}\cdot\text{m} = -3230.7 ~ \text{N}\cdot\text{m}$ *answer*

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