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(1)SolutionGiven that\begin{array}{l}i=1,2,3,4 \\F y=300(b \\P=40,000 \mathrm{lb} \\l=120 i n, \\\rho=0.284 \mathrm{lb} / \mathrm{in}^{3}\end{array}(A).Load is symmetrically placed on Span,Reaction on rach Support =\frac{\omega}{2}\begin{array}{c}V_{A}+V_{B}=\frac{\omega}{2} \\S \cdot F=+\frac{\omega}{2} \quad(\text { Section } A C \text { ) } \\S \cdot f=-\frac{\omega}{2} \quad \text { (Section BC) }\end{array}At C, Sf changes from positive to regative,(2)\begin{array}{l} M_{x}=\frac{\omega}{2} x \text { Sagging } \\x=0, M_{x}=0 \\{\left[x=1 / 2, M_{x}=\frac{\omega d}{4}\right] }\end{array} ...