Milicevic3306
asked
in Engineering Mechanics
Mar 26, 2018
recategorized
Nov 18, 2020
by soujanyareddy13

0 votes

If the following equation establishes equilibrium in slightly bent position, the mid-span deflection of a member shown in the figure is

$$\frac{d^2y}{dx^2}+\frac{P}{EI}y=0$$

If $a$ is amplitude constant for $y,$ then

- $y=\dfrac{1}{P} \bigg( 1- a \cos \dfrac{2 \pi x}{L} \bigg) \\$
- $y=\dfrac{1}{P} \bigg( 1- a \sin \dfrac{2 \pi x}{L} \bigg) \\$
- $y= a \sin \dfrac{n \pi x}{L} \\$
- $y= a \cos \dfrac{n \pi x}{L}$