Problem 848 | Continuous Beams with Fixed Ends

Apply three-moment equation between spans (0) and (1)
$M_0L_0 + 2M_1(L_0 + L_1) + M_2L_1 + \dfrac{6A_0\bar{a}_0}{L_0} + \dfrac{6A_1\bar{b}_1}{L_1} = 0$

$0 + 2M_1(0 + 10) + M_2(10) + 0 + \left[ \dfrac{60(10^3)}{4} + \dfrac{200(3)}{10}(10^2 - 3^2) \right] = 0$

$20M_1 + 10M_2 = -20\,460$   ←   equation (1)
 

 

Apply three-moment equation between spans (1) and (2)
$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

$M_1(10) + 2M_2(10 + 10) + M_3(10)$
          $+ \left[ \dfrac{60(10^3)}{4} + \dfrac{200(7)}{10}(10^2 - 7^2) \right] + \dfrac{7(120)(10^3)}{60} = 0$

$10M_1 + 40M_2 + 10M_3 + 22\,140 + 14\,000 = 0$

$10M_1 + 40M_2 + 10M_3 = -36\,140$   ←   equation (2)
 

Apply three-moment equation between spans (2) and (3)
$M_2L_2 + 2M_3(L_2 + L_3) + M_4L_3 + \dfrac{6A_2\bar{a}_2}{L_2} + \dfrac{6A_3\bar{b}_3}{L_3} = 0$

$M_2(10) + 2M_3(10 + 0) + 0 + \frac{8}{60}(120)(10^3) + 0 = 0$

$10M_2 + 20M_3 + 16\,000 = 0$

$10M_2 + 20M_3 = -16\,000$   ←   equation (3)
 

From equations (1), (2), and (3)
$M_1 = -724.5 ~ \text{kN}\cdot\text{m}$           answer

$M_2 = -597 ~ \text{kN}\cdot\text{m}$           answer

$M_3 = -501.5 ~ \text{kN}\cdot\text{m}$           answer
 

Simple beam reactions
$V_{1L} = \frac{1}{2}(60)(10) + \frac{3}{10}(100) = 360 ~ \text{kN}$

$V_{1R} = \frac{1}{2}(60)(10) + \frac{7}{10}(100) = 440 ~ \text{kN}$
 

$V_{2L} = \frac{1}{3} \times \frac{1}{2}(10)(120) = 200 ~ \text{kN}$

$V_{2R} = \frac{2}{3} \times \frac{1}{2}(10)(120) = 400 ~ \text{kN}$
 

Couple reactions
$L_1{R_1}' = M_1 - M_2$

$10{R_1}' = 724.5 - 597$

${R_1}' = 12.75 ~ \text{kN}$
 

$L_2{R_2}' = M_2 - M_3$

$10{R_2}' = 597 - 501.5$

${R_2}' = 9.55 ~ \text{kN}$
 

Support reaction
$V_1 = V_{1L} + {R_1}' = 360 + 12.75$

$V_1 = 372.75 ~ \text{kN}$           answer
 

$R_2 = R_{1R} + R_{2L} = (V_{1R} - {R_1}') + (V_{2L} + {R_2}')$

$R_2 = (440 - 12.75) + (200 + 9.55)$

$R_2 = 636.8 ~ \text{kN}$           answer
 

$V_3 = V_{2R} - {R_2}' = 400 - 9.55$

$V_3 = 390.45 ~ \text{kN}$           answer