Problem 834 | Reactions of Continuous Beams | Strength of Materials Review at MATHalino

Problem 834
Refer to Problem 826.

Solution 834

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From Solution 826

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

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

Simple beam reactions

$V_{1L} = 8000 / 4 = 2000 ~ \text{N}$ first span, left support ↓

$V_{1R} = 8000 / 4 = 2000 ~ \text{N}$ first span, right support ↑

$V_{2L} = \frac{1}{2}(2000 \times 4) = 4000 ~ \text{N}$ second span, left support ↑

$V_{2R} = \frac{1}{2}(2000 \times 4) = 4000 ~ \text{N}$ second span, right support ↑

$V_{3L} = 2(6000) / 3 = 4000 ~ \text{N}$ third span, left support ↑

$V_{3R} = 1(6000) / 3 = 2000 ~ \text{N}$ third span, right support ↑

Couple Reactions

${R_1}' = M_2 / L_1 = 1690 / 4$

${R_1}' = 422.5 ~ \text{N}$ first span, left support ↓, right support ↑

${R_2}' = (M_3 - M_2) / L_2 = (3230 - 1690) / 4$

${R_2}' = 385 ~ \text{N}$ second span, left support ↓, right support ↑

${R_3}' = M_3 / L_3 = 3230 / 3$

${R_3}' = 1076.67 ~ \text{N}$ third span, left support ↑, right support ↓

Support reactions

$R_1 = 2422.5 ~ \text{N}$ ↓ answer

$R_2 = 2422.5 + 3615 = 6037.5 ~ \text{N}$ ↑ answer

$R_3 = 4385 + 5076.7 = 9461.7 ~ \text{N}$ ↑ answer

$R_4 = 923.3 ~ \text{N}$ ↑ answer

Maximum positive moment

$\dfrac{x}{3615} = \dfrac{4}{3615 + 4385}$

$x = 1.8075 ~ \text{m}$

$M_A = -2422.5(2) + 8000 = 3155 ~ \text{N}\cdot\text{m}$

$M_B = M_2 + \frac{1}{2}x(3615) = -1690 + \frac{1}{2}(1.8075)(3615) = 1577.06 ~ \text{N}\cdot\text{m}$

$M_C = 2R_4 = 2(923.3) = 1846.6 ~ \text{N}\cdot\text{m}$

Thus,
$M_{max(+)} = 3155~ \text{N}\cdot\text{m}$ answer

Problem 834 | Reactions of Continuous Beams