Problem 825 | Continuous Beam by Three-Moment Equation | Strength of Materials Review at MATHalino

Problem 825
See Fig. P-825.

Solution 825

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Three-moment equation for spans (1) and (2)
$M_1 L_1 + 2M_2(L_1 + L_2) + M_3 L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

Where,
$M_1 = 0$

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

$\dfrac{6A_1\bar{a}_1}{L_1} = \frac{1}{4}w_oL^3 = \frac{1}{4}(100)(10^3) = 25\,000 ~ \text{lb}\cdot\text{ft}^2$

$\dfrac{6A_2\bar{b}_2}{L_2} = \frac{5}{32}w_oL^3 = \frac{5}{32}(160)(10^3) = 25\,000 ~ \text{lb}\cdot\text{ft}^2$

Thus,
$0 + 2M_2(10 + 10) + M_3(10) + 25\,000 + 25\,000 = 0$

$40M_2 + 10M_3 = -50\,000$ ← equation (1)

three-moment equation to spans (2) and (3)
$M_2 L_2 + 2M_3(L_2 + L_3) + M_4 L_3 + \dfrac{6A_2\bar{a}_2}{L_2} + \dfrac{6A_3\bar{b}_3}{L_3} = 0$

Where,
$L_2 = 10 ~ \text{ft}$

$L_3 = 12 ~ \text{ft}$

$M_4 = 0$

$\dfrac{6A_2\bar{a}_2}{L_2} = \frac{5}{32}w_oL^3 = \frac{5}{32}(160)(10^3) = 25\,000 ~ \text{lb}\cdot\text{ft}^2$

$\dfrac{6A_3\bar{b}_3}{L_3} = \Sigma \dfrac{Pb}{L}(L^2 - b^2) = \dfrac{400(9)b}{12}(12^2 - 9^2) + \dfrac{400(3)b}{12}(12^2 - 3^2)$

$\dfrac{6A_3\bar{b}_3}{L_3} = 32\,400 ~ \text{lb}\cdot\text{ft}^2$

Thus,
$M_2(10) + 2M_3(10 + 12) + 0 + 25\,000 + 32\,400 = 0$

$10M_2 + 44M_3 = -57\,400$ ← equation (2)

From equations (1) and (2)
$M_2 = -979.52 ~ \text{lb}\cdot\text{ft}$ answer

$M_3 = -1081.93 ~ \text{lb}\cdot\text{ft}$ answer

Problem 825 | Continuous Beam by Three-Moment Equation