- #1
TheFerruccio
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Alright, I took a while setting this problem up for you guys, because I am absolutely stumped and have hit a brick wall and have nowhere else to turn for assistance. Please excuse the shoddy diagram. I made it from scratch to try to illustrate the problem better.
Find the buckling load exactly.
Here is the diagram.
The beam has a length L, spring is at midpoint. Left support is pinned (no moment). Right support has rollers. The beam is simply supported. The stiffness of the beam is EI. The spring has a stiffness k.
I am supposed to calculate the maximum load P that can occur before the system becomes unstable (buckles).
I split the problem into two parts, yielding two differential equations for the vertical displacement.
These are the variables I used for simplification:
##\lambda =\sqrt{\frac{P}{\text{E} \text{I}}}##
##b=v_1({\frac{L}{2}})##
1st: Left of the spring.
Differential equation:
##\frac{P v(x)}{\text{E} \text{I}}+v''(x)=\frac{b k x}{2 \text{E}
\text{I}}##
Solution:
##v_1(x)=A \cos (\lambda x)+\frac{b k x}{2 P}+B \sin (\lambda x)##
2nd: Right of the spring.
Differential Equation:
##\frac{P v(x)}{\text{E} \text{I}}+v''(x)=\frac{b k (L-x)}{2
\text{E} \text{I}}##
Solution:
##v_2(x)=\frac{b k (L-x)}{2 P}+C \cos (\lambda x)+D \sin (\lambda x)##
Now, I apply boundary conditions!
I have 4 unknowns to solve for, so I need 4 boundary conditions.
BC 1: Displacement at left is 0, so A = 0.
BC 2: Displacement at right is 0, so
##C \cos (\lambda L)+D \sin (\lambda L) = 0##
BC 3: Displacements match at L/2, so
##B \sin \left(\frac{\lambda L}{2}\right)=C \cos
\left(\frac{\lambda L}{2}\right)+D \sin \left(\frac{\lambda
L}{2}\right)##
BC 4: Slopes match at L/2, so
##\frac{b k}{2 P}+B \lambda \cos \left(\frac{\lambda
L}{2}\right)=-\frac{b k}{2 P}-C \lambda \sin
\left(\frac{\lambda L}{2}\right)+D \lambda \cos
\left(\frac{\lambda L}{2}\right)##
Put all terms of B, C, D on left hand side, put all the remaining terms on the right hand side, and end up with an equation...
##\left(
\begin{array}{ccc}
0 & \cos (L \lambda ) & \sin (L \lambda ) \\
\sin \left(\frac{L \lambda }{2}\right) & -\cos \left(\frac{L
\lambda }{2}\right) & -\sin \left(\frac{L \lambda }{2}\right)
\\
\lambda \cos \left(\frac{L \lambda }{2}\right) & \lambda \sin
\left(\frac{L \lambda }{2}\right) & -\lambda \cos
\left(\frac{L \lambda }{2}\right) \\
\end{array}
\right)\left(
\begin{array}{c}
B \\
C \\
D \\
\end{array}
\right)=\left(
\begin{array}{c}
0 \\
0 \\
-\frac{k b}{P} \\
\end{array}
\right)##
This is the point where I am stuck. I am not sure where to continue from here. When I attempt to solve for B, C, D, I do not get any possible solution, not even 0. I know I am setting up the problem wrong, somehow. Someone pointed out to me that the matrix on the right hand side should be all 0. That kind of makes sense to me, but I do not see where in the algebra I messed up that would result in this occurring.
Homework Statement
Find the buckling load exactly.
Homework Equations
Here is the diagram.
The beam has a length L, spring is at midpoint. Left support is pinned (no moment). Right support has rollers. The beam is simply supported. The stiffness of the beam is EI. The spring has a stiffness k.
I am supposed to calculate the maximum load P that can occur before the system becomes unstable (buckles).
The Attempt at a Solution
I split the problem into two parts, yielding two differential equations for the vertical displacement.
These are the variables I used for simplification:
##\lambda =\sqrt{\frac{P}{\text{E} \text{I}}}##
##b=v_1({\frac{L}{2}})##
1st: Left of the spring.
Differential equation:
##\frac{P v(x)}{\text{E} \text{I}}+v''(x)=\frac{b k x}{2 \text{E}
\text{I}}##
Solution:
##v_1(x)=A \cos (\lambda x)+\frac{b k x}{2 P}+B \sin (\lambda x)##
2nd: Right of the spring.
Differential Equation:
##\frac{P v(x)}{\text{E} \text{I}}+v''(x)=\frac{b k (L-x)}{2
\text{E} \text{I}}##
Solution:
##v_2(x)=\frac{b k (L-x)}{2 P}+C \cos (\lambda x)+D \sin (\lambda x)##
Now, I apply boundary conditions!
I have 4 unknowns to solve for, so I need 4 boundary conditions.
BC 1: Displacement at left is 0, so A = 0.
BC 2: Displacement at right is 0, so
##C \cos (\lambda L)+D \sin (\lambda L) = 0##
BC 3: Displacements match at L/2, so
##B \sin \left(\frac{\lambda L}{2}\right)=C \cos
\left(\frac{\lambda L}{2}\right)+D \sin \left(\frac{\lambda
L}{2}\right)##
BC 4: Slopes match at L/2, so
##\frac{b k}{2 P}+B \lambda \cos \left(\frac{\lambda
L}{2}\right)=-\frac{b k}{2 P}-C \lambda \sin
\left(\frac{\lambda L}{2}\right)+D \lambda \cos
\left(\frac{\lambda L}{2}\right)##
Put all terms of B, C, D on left hand side, put all the remaining terms on the right hand side, and end up with an equation...
##\left(
\begin{array}{ccc}
0 & \cos (L \lambda ) & \sin (L \lambda ) \\
\sin \left(\frac{L \lambda }{2}\right) & -\cos \left(\frac{L
\lambda }{2}\right) & -\sin \left(\frac{L \lambda }{2}\right)
\\
\lambda \cos \left(\frac{L \lambda }{2}\right) & \lambda \sin
\left(\frac{L \lambda }{2}\right) & -\lambda \cos
\left(\frac{L \lambda }{2}\right) \\
\end{array}
\right)\left(
\begin{array}{c}
B \\
C \\
D \\
\end{array}
\right)=\left(
\begin{array}{c}
0 \\
0 \\
-\frac{k b}{P} \\
\end{array}
\right)##
This is the point where I am stuck. I am not sure where to continue from here. When I attempt to solve for B, C, D, I do not get any possible solution, not even 0. I know I am setting up the problem wrong, somehow. Someone pointed out to me that the matrix on the right hand side should be all 0. That kind of makes sense to me, but I do not see where in the algebra I messed up that would result in this occurring.
Last edited: