Second Order Differential Equations - Beam Deflections

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SUMMARY

The discussion focuses on solving the second-order differential equation for beam deflections in a cantilever beam subjected to a load at its free end. The equation is given by \(\frac{d^2y}{dx^2}=k(L-x)\) for \(0 \leq x \leq L\). After double integration, the deflection equation is derived as \(y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx+d\). The constant \(d\) is determined to be zero due to the beam being horizontal when unloaded, while the constant \(c\) is concluded to be zero based on the conditions of the problem, although the reasoning for this conclusion is questioned by the participants.

PREREQUISITES
  • Understanding of second-order differential equations
  • Basic knowledge of integration techniques
  • Familiarity with cantilever beam theory
  • Knowledge of boundary conditions in mechanical systems
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  • Study the application of boundary conditions in solving differential equations
  • Learn about the Euler-Bernoulli beam theory for deflection analysis
  • Explore methods for determining constants in differential equations
  • Investigate the physical interpretation of beam deflections under various loading conditions
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Saracen Rue
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Homework Statement


A cantilever of length ##L## is rigidly fixed at one end and is horizontal in the unstrainted position. If a load is added at the free end of the beam, the downward deflection, ##y##, at a distance, ##x##, along the beam satisfies the differential equation: \frac{d^2y}{dx^2}=k\left(L-x\right) \ for\ 0\le x\le L
Where ##k## is a constant. Find the deflection, ##y##, in terms of ##x## and hence find the maximum deflection of the beam.


Homework Equations


Basic knowledge of integration

The Attempt at a Solution


After double integrating both sides I'm left with ##y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx+d## As the question tells us the beam is horizontal when their is no weight on it, we know that ##y=0## when ##x=0##.
##0=\frac{k}{2}L\left(0\right)^2-\frac{k}{6}\left(0\right)^3+c\left(0\right)+d##
##d=0##
##y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx##
This is where I get stuck - I can't find any part of the question which helps me calculate the value of the constant ##c##. The answer says that ##c## should be zero, but I don't understand how they know this.
 
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Saracen Rue said:

Homework Statement


A cantilever of length ##L## is rigidly fixed at one end and is horizontal in the unstrainted position. If a load is added at the free end of the beam, the downward deflection, ##y##, at a distance, ##x##, along the beam satisfies the differential equation: \frac{d^2y}{dx^2}=k\left(L-x\right) \ for\ 0\le x\le L
Where ##k## is a constant. Find the deflection, ##y##, in terms of ##x## and hence find the maximum deflection of the beam.


Homework Equations


Basic knowledge of integration

The Attempt at a Solution


After double integrating both sides I'm left with ##y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx+d## As the question tells us the beam is horizontal when their is no weight on it, we know that ##y=0## when ##x=0##.
##0=\frac{k}{2}L\left(0\right)^2-\frac{k}{6}\left(0\right)^3+c\left(0\right)+d##
##d=0##
##y=\frac{k}{2}Lx^2-\frac{k}{6}x^3+cx##
This is where I get stuck - I can't find any part of the question which helps me calculate the value of the constant ##c##. The answer says that ##c## should be zero, but I don't understand how they know this.
The beam is horizontal when unloaded. What do you think the direction of the tangent of the loaded beam is at the fixed end?
images?q=tbn:ANd9GcQJcrbfWK8K62FnNax_ghIZRI-NfC0BBBU308CzuXHtzCglpddkxA.png
 

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