Problem 849 | Continuous Beams with Fixed Ends

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

$0 + 2M_1(0 + 3) + M_2(3) + 0 + \frac{7}{60}(2)(3^3) = 0$

$6M_1 + 3M_2 + 6.3 = 0$

$6M_1 + 3M_2 = -6.3$   ←   equation (1)
 

 

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

$M_1(3) + 2M_2(3 + 3) + M_3(3) + \frac{8}{60}(2)(3^3) + \left[ \frac{8}{60}(2)(3^3) + \frac{7}{60}(4)(3^3) \right] = 0$

$3M_1 + 12M_2 + 3M_3 + 7.2 + 19.8 = 0$

$3M_1 + 12M_2 + 3M_3 = -27$   ←   equation (2)
 

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

$M_2(3) + 2M_3(3 + 0) + 0 + \left[ \frac{7}{60}(2)(3^3) + \frac{8}{60}(4)(3^3) \right] + 0 = 0$

$3M_2 + 6M_3 + 20.7 = 0$

$3M_2 + 6M_3 = -20.7$   ←   equation (3)
 

From equations (1), (2), and (3)
$M_1 = -0.3 ~ \text{kN}\cdot\text{m} = -300 ~ \text{N}\cdot\text{m}$           answer

$M_2 = -1.5 ~ \text{kN}\cdot\text{m} = -1500 ~ \text{N}\cdot\text{m}$           answer

$M_3 = -2.7 ~ \text{kN}\cdot\text{m} = -2700 ~ \text{N}\cdot\text{m}$           answer