From
Solution 815
$M_2 = -1077.27 ~ \text{lb}\cdot\text{ft}$
$M_3 = -640 ~ \text{lb}\cdot\text{ft}$
$y = 60 ~ \text{lb/ft}$
From the first span
Reactions due to loadings
$V_{1L} = V_{1R} = 300 ~ \text{lb}$
Reactions due to end moments
${R_1}' = \dfrac{M_2 - M_1}{L_1} = \dfrac{1077.27 - 0}{12} = 89.7725 ~ \text{lb}$
Thus,
$R_{1L} = V_{1L} - {R_1}' = 300 - 89.7725$
$R_{1L} = 210.2275 ~ \text{lb}$
$R_{1R} = V_{1R} + {R_1}' = 300 + 89.7725$
$R_{1R} = 389.7725 ~ \text{lb}$
From the second span
Reactions due to loadings
$\Sigma M_{V_{2R}} = 0$
$10V_{2L} = \frac{1}{2}(10)(135)[ \, \frac{2}{3}(10) \, ] + \frac{1}{2}(10)(60)[ \, \frac{1}{3}(10) \,]$
$V_{2L} = 550 ~ \text{lb}$
$\Sigma M_{V_{2L}} = 0$
$10V_{2R} = \frac{1}{2}(10)(135)[ \, \frac{1}{3}(10) \, ] + \frac{1}{2}(10)(60)[ \, \frac{2}{3}(10) \,]$
$V_{2R} = 425 ~ \text{lb}$
Reactions due to end moments
${R_2}' = \dfrac{M_2 - M_3}{L_2} = \dfrac{1077.27 - 640}{10}$
${R_2}' = 43.73 ~ \text{lb}$
Thus,
$R_{2L} = V_{2L} + {R_2}' = 550 + 43.73$
$R_{2L} = 593.73 ~ \text{lb}$
$R_{2R} = V_{2R} - {R_2}' = 425 - 43.73$
$R_{2R} = 381.27 ~ \text{lb}$
From the third span
$\Sigma F_V = 0$
$R_{3L} = \frac{1}{2}(8)(60) = 240 ~ \text{lb}$
From the figure above
$R_1 = 210.23 ~ \text{lb}$ answer
$R_2 = 389.77 + 593.73 = 983.50 ~ \text{lb}$ answer
$R_3 = 381.27 + 240 = 621.27 ~ \text{lb}$ answer
From the triangular load
$\dfrac{y}{x} = \dfrac{135}{18}$
$y = 7.5x$
Solve for x
$\Sigma F_{\text{to the right of }A} = 0$
$621.27 - \frac{1}{2}xy = 0$
$621.27 - \frac{1}{2}x(7.5x) = 0$
$3.75x^2 = 621.27$
$x = 12.8714 ~ \text{ft}$
For Maximum Positive Moment:
$M_{A} = \Sigma M_{\text{to the right of }A}$
$M_{A} = 621.27(x - 8) - \frac{1}{2}xy(\frac{1}{3}x) = 621.27(x - 8) - \frac{1}{6}x^2(7.5x)$
$M_{A} = 621.27(x - 8) - 1.25x^3 = 621.27(12.8714 - 8) - 1.25(12.8714^3)$
$M_{A} = 360.90 ~\text{lb}\cdot\text{ft}$
$M_{B} = \Sigma M_{\text{to the left of }B}$
$M_{B} = 210.23(4)$
$M_{B} = 840.92 ~\text{lb}\cdot\text{ft}$
Hence,
$M_{max(+)} = M_{B}$
$M_{max(+)} = 840.92 ~\text{lb}\cdot\text{ft}$ answer
