Problem 843 | Continuous Beams with Fixed Ends

$\dfrac{y}{x} = \dfrac{4}{6}$;       $y = \frac{2}{3}x$
 

 

From
$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{Pa}{L}(L^2 - a^2)$

$\displaystyle \dfrac{6A_1\bar{a}_1}{L_1} = \int_0^6 \dfrac{(y \, dx)(x)}{9}(9^2 - x^2) = \int_0^6 \dfrac{(\frac{2}{3}x \, dx)(x)}{9}(9^2 - x^2)$

$\displaystyle \dfrac{6A_1\bar{a}_1}{L_1} = \frac{2}{27}\int_0^6 x^2(81 - x^2) \, dx = 316.8 ~ \text{kN}\cdot\text{m}^3$
 

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

$0 + 2M_{wall}(9 + 0) + 0 + 316.8 + 0 = 0$

$18M_{wall} = -316.8$

$M_{wall} = -17.6 ~ \text{kN}\cdot\text{m}$           answer
 

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

$-17.6 = 9R - \frac{1}{2}(6)(4)[ \, 3 + \frac{1}{3}(6) \, ]$

$R = 4.711 ~ \text{kN}$           answer
 

Check using area-moment method
$EI \, t_{A/B} = 0$

$(Area_{AB}) \cdot \bar{X}_A = 0$
 

 

$\frac{1}{3}(3)(18)[ \, 6 + \frac{3}{4} \, ] + \frac{1}{4}(3)(3)[ \, 6 + \frac{4}{5}(3) \, ] + \frac{1}{2}(9)(9R)[ \, \frac{2}{3}(9) \, ] - \frac{1}{4}(9)(81)[ \, \frac{4}{5}(9) \, ] = 0$

$148.5 + 18.9 + 243R - 1312.2 = 0$

$243R = 1050.3$

$R = 4.711 ~ \text{kN}$           okay