Problem 832 | Reactions of Continuous Beams

From Solution 824, M₂ = -2.9 kN·m and M₃ = -6.1 kN·m.
 

From the first span

$2V_1 = 2(4)$

$V_1 = 4 ~ \text{kN}$
 

$L_1 {R_1}' = M_2$

$4{R_1}' = 2.9$

${R_1}' = 0.725 ~ \text{kN}$
 

$R_{1L} = V_1 - {R_1}' = 4 - 0.725$

$R_{1L} = 3.275 ~ \text{kN}$
 

$R_{1R} = V_1 + {R_1}' = 4 + 0.725$

$R_{1R} = 4.725 ~ \text{kN}$

 

 

From the middle span

$2V_2 = 10$

$V_2 = 5 ~ \text{kN}$
 

$L_2 {R_2}' = M_3 - M_2$

$2{R_2}' = 6.1 - 2.9$

${R_2}' = 1.6 ~ \text{kN}$
 

$R_{2L} = V_2 - {R_2}' = 5 - 1.6$

$R_{2L} = 3.4 ~ \text{kN}$
 

$R_{2R} = V_2 + {R_2}' = 5 + 1.6$

$R_{2R} = 6.6 ~ \text{kN}$

 

From the third span

$2V_3 = 4(4)$

$V_3 = 8 ~ \text{kN}$
 

$L_3 {R_3}' = M_3$$

$4{R_3}' = 6.1$

${R_3}' = 1.525 ~ \text{kN}$
 

$R_{3L} = V_3 + {R_3}' = 8 + 1.525$

$R_{3L} = 9.525 ~ \text{kN}$
 

$R_{3R} = V_3 - {R_3}' = 8 - 1.525$

$R_{3R} = 6.475 ~ \text{kN}$

 

 

Thus,
$R_1 = 3.275 ~ \text{kN}$           answer

$R_2 = 4.725 + 3.4 = 8.125 ~ \text{kN}$           answer

$R_3 = 6.6 + 9.525 = 16.125 ~ \text{kN}$           answer

$R_4 = 8 - 1.525 = 6.475 ~ \text{kN}$           answer
 

$\dfrac{x_1}{3\,275} = \dfrac{4}{3\,275 + 4\,725}$

$x_1 = 1.637\,5 ~ \text{m}$
 

$\dfrac{x_2}{9\,525} = \dfrac{4}{9\,525 + 6\,475}$

$x_2 = 2.381\,25 ~ \text{m}$
 

$M_A = \frac{1}{2}x_1(3\,275) = \frac{1}{2}(1.637\,5)(3\,275) = 2\,681.41 ~ \text{N}\cdot\text{m}$

$M_B = M_2 + 3400(1) = -2\,900 + 3\,400 = 500 ~ \text{N}\cdot\text{m}$

$M_C = M_3 + \frac{1}{2}x_2(9\,525) = -6\,100 + \frac{1}{2}(2.381\,25)(9\,525) = 5\,240.70 ~ \text{N}\cdot\text{m}$
 

Hence,
$M_{max(+)} = M_C = 5.241 ~ \text{kN}\cdot\text{m}$           answer