Problem 831 | Reactions of Continuous Beams | Strength of Materials Review at MATHalino

Problem 831
Refer to Problem 817 for which M₂ = -180 lb·ft.

Solution 831

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From the first span

$L_1 V_{1L} = M$

$9V_{1L} = 300$

$V_{1L} = 33.33 lb = V_{1R}$

$L_1 {R_1}' = M_2$

$9{R_1}' = 180$

${R_1}' = 20 ~ \text{lb}$

$R_{1L} = V_{1L} + {R_1}' = 33.33 + 20$

$R_{1L} = 53.33 ~ \text{lb downward}$

$R_{1L} = V_{1R} + {R_1}' = 33.33 + 20$

$R_{1L} = 53.33 ~ \text{lb}$

From the second span

$\Sigma M_{V_{2R}} = 0$

$9V_{2L} = \frac{1}{2}(6)(100) [ \, \frac{1}{3}(6) \, ]$

$V_{2L} = 66.67 ~ \text{lb}$

$\Sigma M_{V_{2L}} = 0$

$9V_{2R} = \frac{1}{2}(6)(100) [ \, 3 + \frac{2}{3}(6) \, ]$

$V_{2R} = 233.33 ~ \text{lb}$

$L_2{R_2}' = M_2$

$9{R_2}' = 180$

${R_2}' = 20 ~ \text{lb}$

$R_{2L} = V_{2L} + {R_2}' = 66.67 + 20$

$R_{2L} = 86.67 ~ \text{lb}$

$R_{2R} = V_{2R} - {R_2}' = 233.33 - 20$

$R_{2R} = 213.33 ~ \text{lb}$

Support reactions

$R_1 = 53.33 ~ \text{lb downward}$ answer

$R_2 = 53.33 + 86.67 = 140 ~ \text{lb}$ answer

$R_3 = 213.33 ~ \text{lb}$ answer

From the shear diagram
$\dfrac{x^2}{86.67} = \dfrac{6^2}{86.67 + 213.33}$

$x = 3.225 ~ \text{ft}$

Maximum positive moment
$M_{max(+)} = M_A = M_2 + \text{Positive area in shear diagram}$

$M_{max(+)} = -180 + 3(86.67) + \frac{2}{3}x(86.67)$

$M_{max(+)} = -180 + 3(86.67) + \frac{2}{3}(3.225)(86.67)$

$M_{max(+)} = 266.35 ~ \text{lb}\cdot\text{ft}$ answer

Problem 831 | Reactions of Continuous Beams