A certain buckling phenomenon can be modeled as y" = F Ely+ El' where F, E, l and M are positive constants. Solve the differential equation with conditions y (0) = 0, y' (0) = 0. Assume the independent variable to be x y =
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Step 1
To solve the differential equation \( y'' = F E l y + E l' \) with the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \), we will follow these steps: Show more…
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Determine the equation of the elastic curve. $E I$ is constant.
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We use the equation given above and use the result that when $y$ is small $$ \frac{1}{R} \sim \frac{d^{2} y}{d x^{2}} . \text { Thus, } \frac{d^{2} y}{d x^{2}}=\frac{N(x)}{E I} $$ (a) Here $N(x)=N_{0}$ is a constant. Then integration gives, $$ \frac{d y}{d x}=\frac{N_{0} x}{E I}+C_{1} $$ But $\quad\left(\frac{d y}{d x}\right)=0$ for $x=0$, so $C_{1}=0$. Integrating again, $$ y=\frac{N_{0} x^{2}}{2 E I} $$ where we have used $y=0$ for $x=0$ to set the constant of integration at zero. This is the equation of a parabola. The sag of the free end is $$ \lambda=y(x=l)=\frac{N_{0} l^{2}}{2 E I} $$ (b) In this case $N(x)=F(l-x)$ because the load $F$ at the extremity is balanced by a similar force at $F$ directed upward and they constitute a couple. Then $$ \frac{d^{2} y}{d x^{2}}=\frac{F(l-x)}{E I} $$ Integrating, $\frac{d y}{d x}=\frac{F\left(L x-x^{2} / 2\right)}{E I}+C_{1}$ As before $C_{1}=0 .$ Integrating again, using $y=0$ for $x=0$ $$ y=\frac{F\left(\frac{b^{2}}{2}-\frac{x^{3}}{6}\right)}{E I} \text { here } \lambda=\frac{F l^{3}}{3 E I} $$ Here for a square cross section $$ I=\int_{-a / 2}^{a / 2} z^{2} a d z=a^{4} / 12 $$
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