**Problem 823**

A continuous beam simply supported over three 10-ft spans carries a concentrated load of 400 lb at the center of the first span, a concentrated load of 640 lb at the center of the third span and a uniformly distributed load of 80 lb/ft over the middle span. Solve for the moment over the supports and check your answers using the results obtained for Problems 819 and 822.

**Solution 823**

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

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

$M_1 = 0$

$L_1 = L_2 = 10 ~ \text{ft}$

$\dfrac{6A_1\bar{a}_1}{L_1} = \frac{3}{8}PL^2 = \frac{3}{8}(400)(10^2) = 15\,000 ~ \text{lb}\cdot\text{ft}^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = \frac{1}{4}w_oL^3 = \frac{1}{4}(80)(10^3) = 20\,000 ~ \text{lb}\cdot\text{ft}^2$

Thus,

$0 + 2M_2(10 + 10) + M_3(10) + 15\,000 + 20\,000 = 0$

$40M_2 + 10M_3 = -35\,000$ ← equation (1)

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

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

$L_2 = L_3 = 10 ~ \text{ft}$

$M_4 = 0$

$\dfrac{6A_2\bar{a}_2}{L_2} = \frac{1}{4}w_o L^3 = \frac{1}{4}(80)(10^3) = 20\,000 ~ \text{lb}\cdot\text{ft}^2$

$\dfrac{6A_3\bar{b}_3}{L_3} = \frac{3}{8}PL^2 = \frac{3}{8}(640)(10^2) = 24\,000 ~ \text{lb}\cdot\text{ft}^2$

Thus,

$M_2(10) + 2M_3(10 + 10) + 0 + 20\,000 + 24\,000 = 0$

$10M_2 + 40M_3 = -44\,000$ ← equation (2)

From equations (1) and (2)

$M_2 = -640 ~ \text{lb}\cdot\text{ft}$ *answer*

$M_3 = -940 lb ~ \text{lb}\cdot\text{ft}$ *answer*

**Checking:**

From Solution 819

$M_2 = -\dfrac{Pa}{L^2}(L^2 - a^2) \cdot \dfrac{2(\alpha + \beta)}{4(1 + \alpha)(\alpha + \beta) - \alpha^2}$

$M_3 = \dfrac{Pa}{L^2}(L^2 - a^2) \cdot \dfrac{\alpha}{4(1 + \alpha)(\alpha + \beta) - \alpha^2}$

For P = 400 lb on the first span of beam in this problem:

$M_2 = -\dfrac{400(5)}{10^2}(10^2 - 5^2) \cdot \dfrac{2(1 + 1)}{4(1 + 1)(1 + 1) - 1^2} = -400 ~ \text{lb}\cdot\text{ft}$

$M_3 = \dfrac{400(5)}{10^2}(10^2 - 5^2) \cdot \dfrac{1}{4(1 + 1)(1 + 1) - 1^2} = 100 ~ \text{lb}\cdot\text{ft}$

For P = 640 lb on the third span of beam in this problem:

$M_3 = -\dfrac{640(5)}{10^2}(10^2 - 5^2) \cdot \dfrac{2(1 + 1)}{4(1 + 1)(1 + 1) - 1^2} = -640 ~ \text{lb}\cdot\text{ft}$

$M_2 = \dfrac{640(5)}{10^2}(10^2 - 5^2) \cdot \dfrac{1}{4(1 + 1)(1 + 1) - 1^2} = 160 ~ \text{lb}\cdot\text{ft}$

From Solution 822

$M_2 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\beta}{4(\alpha + 1)(1 + \beta) - 1}$

$M_3 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\alpha}{4(1 + \alpha)(1 + \beta) - 1}$

For this problem:

_{o}= 80 lb/ft

$M_2 = -\dfrac{80(10^2)}{4} \cdot \dfrac{1 + 2(1)}{4(1 + 1)(1 + 1) - 1} = -400 ~ \text{lb}\cdot\text{ft}$

$M_2 = -\dfrac{80(10^2)}{4} \cdot \dfrac{1 + 2(1)}{4(1 + 1)(1 + 1) - 1} = -400 ~ \text{lb}\cdot\text{ft}$

Thus,

$M_2 = -400 + 160 - 400 = -640 ~ \text{lb}\cdot\text{ft}$ *answer*

$M_3 = 100 - 640 - 400 = -940 ~ \text{lb}\cdot\text{ft}$ *answer*

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