The 150-1b man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb/ft. Determine the maximum internal bending moment. Assume that the water exerts a uniform distributed load upward on the bottom of the boat. -7.5 ft

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Answer #1Given:W_{\text {man }}=15016Weights per linear foot =3161 \mathrm{ft}\begin{array}{l}L=15.00 \mathrm{~m} \\M_{\text {max }}=?\end{array}Solution:Step 1: calculabe buoyant force, per linear foot\begin{array}{l} \sum F_{y}=0(\uparrow+v e) \\q^{\prime} \times L-W-q \times L=0 \\q^{\prime}=(W+q \times L) / L \\q^{\prime}=(150+3 \times 15) / 15 \\q^{\prime}=1316 / f t(+\uparrow)\end{array}Step 8:From free body diagram\begin{array}{l}q(x)=(13-3)=10(\uparrow+v e) \\v(x)=\int q(x) d x=\int 10 d x=10 x(0 \leq x \leq 7.5) \\M(x)=\int 10 x \cdot d x=\frac{10 \cdot x^{2}}{2}=5 x^{2}(0 \leq x \leq 7.5)\end{array}Maximum internal bending moment occurs when\begin{array}{l}x=7.5 f t \\M(7.5)=5(7.5)^{2}=281.251 b * f t\end{array}Wman = 15016 150lb a=31b/A I mm 7.541 7514 Weight per linear foot = 31b1ft L=15.0om M max = ? Solution: Step 1: calculate buoyant force per linear foot {fy =O (I tve) qxL-w-q x2 =0 q'= (W + 9xL)/2 q'=(150+ 3x15)/15 qu'=13.10/ft (+1) Step 8 From free body diagram q (ə) = (13 -3) = 10(1 +ve) vo) = Save) de - Sio dx = lox (0<x<+1) M(oc) = flox.de = 10.0² - 50² (os x = 7.5) Maxim om internal bending moment occurs when x=7.5ft M(7.5) 25 (75) * - 281.25 16 eft ...