Problem 840 | Continuous Beams with Fixed Ends

$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$
 

Where,
$L_1 = 3 ~ \text{m}$

$L_2 = 0$

$M_1 = - 1400(2)(1) = -2800 ~ \text{N}\cdot\text{m}$

$M_3 = 0$

$\dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

 

Thus,
$-2800(3) + 2M_2(3 + 0) + 0 + 0 + 0 = 0$

$M_2 = 1400 ~ \text{N}\cdot\text{m}$   ←   maximum positive bending moment       answer
 

$M_2 = \Sigma M_{\text{to the left of }R_2}$

$1400 = 3R - 1400(2)(4)$

$3R = 12\,600$

$R = 4200 ~ \text{N}$           answer
 

You can also solve R by the usual span-to-span method

From the overhang
$R_{1R} = 1400(2) = 2800 ~ \text{N}$
 

From the 3-m span
$3{R_2}' = M_1 + M_2$

$3{R_2}' = 2800 + 1400$

${R_2}' = 1400 ~ \text{N}$
 

$R = R_{1R} + {R_2}' = 2800 + 1400$

$R = 4200 ~ \text{N}$           okay