Solution Found!
(a) The beam is embedded at its left end and free at its right end, and w(x) w0, 0 x L
Chapter 5, Problem 1(choose chapter or problem)
In Problem solve equation (4) subject to the appropriate boundary conditions. The beam is of length L, and \(w_{0}\) is a constant.
(a) The beam is embedded at its left end and free at its right end, and \(w(x)=w_{0}, 0<x<L\).
(b) Use a graphing utility to graph the deflection curve when \(w_{0}=24 E I \text { and } L=1 \text {. }\)
Questions & Answers
QUESTION:
In Problem solve equation (4) subject to the appropriate boundary conditions. The beam is of length L, and \(w_{0}\) is a constant.
(a) The beam is embedded at its left end and free at its right end, and \(w(x)=w_{0}, 0<x<L\).
(b) Use a graphing utility to graph the deflection curve when \(w_{0}=24 E I \text { and } L=1 \text {. }\)
ANSWER:Step 1 of 4
(a)
Consider the deflection y(x) satisfies the following differential equation,
\(EI\frac{{{d^4}y}}{{d{x^4}}} = w\left( x \right)\)
Solve the above differential equation subject to the beam embedded at its right end, and \(w\left( x \right) = {w_0},0 < x < L\).
Use the condition \(w\left( x \right) = {w_0}\), the deflection y(x) satisfies the expression and we can write.
\(EI\frac{{{d^4}y}}{{d{x^4}}} = {w_0}\)
\(\frac{{{d^4}y}}{{d{x^4}}} = \frac{{{w_0}}}{{EI}}\)
The corresponding homogeneous equation becomes, \(\frac{{{d^4}y}}{{d{x^4}}} = 0\).