Question The anchor beam AB is of uniform cross-section and carries a load P at the free end B. Determine the equation of the elastic curve and the displacement and slope at B. (2.8 m 1.8 kn) A P B L L=2.Om P=1.OkN

Solution Given A B is a cantilever beamBoundary condition:Elatic equation EI \frac{d^{2} y}{d x^{2}}=M xWhere M x is moment at section - ALet us convider a section x from B\begin{array}{l}M x=1.8 x \\\therefore E I \frac{d^{2} y}{d x^{2}}=1.8 x \int \\S \text { Lope }=\frac{d y}{d x}=E I \int \frac{d^{2} y}{d x^{2}}=\int 1.8 x \\E I \frac{d y}{d x}=\frac{0.9 x^{2}}{x}+C_{1} \\E I d y=0.3 x^{3}+c_{1} x+c_{2} \\\text { At A, } x=28, d y / d x=0, y=0 \\O=0.9(-2.8)^{2}+c_{1} \Rightarrow C_{1}=-7.056 .0 .3(2.8)^{3}-7.056(2.8)+c_{2} \Rightarrow C_{2}=13.1712 . \\\text { O }=0 . \frac{d y}{d x}=0.9 x^{2}-7.056 \\\text { Therefore equations are EI }\end{array}At A, x-2.8, \quad d y / d x=0, y=00=0.9(-2.8)^{2}+c_{1} \Rightarrow C_{1}=-7.056 \text {. }Therefore equations are E I \frac{d y}{d x}=0.9 x^{2}-7.056E I y=0.3 x^{3}-7.056+13.1712At point b, under load x=0\begin{array}{l}d y / d x=-7.056 / E I \text { (Slope) } \\y=\frac{13.1712}{E I} \text { (deflection) }\end{array} ...