Buckling of a Simply Supported beam with a an axial load in the middle.

I was working on a project in which I required to study the buckling of a miniature beam by both analytical equations and FEM. Since I did not find the equations anywhere in textbooks or on the internet, I developed the equations. Here, I am sharing the simplified version of that analysis.

Let’s assume there is a simply supported beam with length L. The modulus of elasticity and moment of area are E and I, respectively. A horizontal force P is applied at the middle of the beam. I need to find the critical load P which causes the buckling.

First, the support reactions are obtained as follows:

The deflection at the middle of the beam is assumed to be Δ. The magnitude of Δ will be calculated later.

    \[\Sigma M_B=0 \quad \Rightarrow \quad P\Delta-R_AL=0 \quad \Rightarrow \quad R_A=P \frac{\Delta }{L}\]

 

    \[\Sigma F_y=0 \quad \Rightarrow \quad R_A+R_B=0\ \quad \Rightarrow \quad R_B=-P \frac{\Delta }{L} \]

Then, I need to look at the bending moment between the left support A and the midpoint.

    \[\textrm{For} \quad 0<x<\frac{L }{2}: \quad M(x)=R_Ax-Pw(x)\]

    \[\textrm{Since:} \quad EIw''(x)=M(x) \quad \Rightarrow \quad EIw''(x)=R_Ax-Pw(x)\]

    \[EIw''(x)+Pw(x)=R_Ax\]

by defining β as follows, the second order differential equation is solved.

    \[ \quad \beta=\sqrt[]{\frac{P}{EI}}\]

    \[w''(x)+\beta^2 w(x)=\frac{R_A}{EI}x\]

    \[w(x)=A_0sin(\beta x)+B_0cos(\beta x)+\frac{R_A}{P}x\]

Now we can apply some the boundary conditions of the beam to solve for the unknowns.

    \[\textrm{@} \quad x=0: \quad w(0)=0 \quad \Rightarrow \quad B_0=0\]

    \[\textrm{@} \quad x=\frac{L}{2}: \quad w(\frac{L}{2})=\Delta \quad \Rightarrow \quad \Delta=A_0sin(\beta \frac{L}{2}) + \frac{R_A}{P}\frac{L}{2} \]

    \[\Delta=2A_0sin(\beta \frac{L}{2})\]

By replacing Δ, support forces can be simplified.

    \[R_A=2P\frac{A_0}{L}sin(\beta \frac{L}{2}) \quad \textrm{and} \quad R_B=-2P\frac{A_0}{L}sin(\beta \frac{L}{2})\]

    \[w(x)=A_0sin(\beta x) + 2\frac{A_0}{L}sin(\beta \frac{L}{2})x\]

    \[w'(x)=A_0\beta cos(\beta x) + 2\frac{A_0}{L}sin(\beta \frac{L}{2})\]

Now, we can look at the bending moment and deflection equation of the other half of the beam.

    \[\textrm{For} \quad \frac{L}{2}<x<L: \quad M(x)=R_B(L-x)=-2P\frac{A_0}{L}sin(\beta \frac{L}{2})(L-x)\]

    \[\textrm{Since:} \quad EIw''(x)=M(x) \quad \Rightarrow \quad EIw''(x)=-2P\frac{A_0}{L}sin(\beta \frac{L}{2})(L-x)\]

    \[w''(x)=-2\beta^2\frac{A}{L}sin(\beta \frac{L}{2})(L-x)\]

    \[\int \Rightarrow \quad w'(x)=\beta^2\frac{A_0}{L}sin(\beta \frac{L}{2})(L-x)^2+C_0\]

    \[\int \Rightarrow \quad w(x)=-\frac{1}{3}\beta^2\frac{A_0}{L}sin(\beta \frac{L}{2})(L-x)^3-C_0(L-x)+D_0)\]

Then, we can apply the rest of boundary conditions as follows:

Deflection at Point B is zero.

    \[\textrm{@} \quad x=L: \quad w(L)=0 \quad \Rightarrow \quad D_0=0 \]

Both equations must give us identical answers for the deflection and slope at the midpoint.

    \[\textrm{@} \quad x=\frac{L}{2}: \quad w'(\frac{L^-}{2})=w'(\frac{L^+}{2}) \quad \Rightarrow \quad A_0\beta cos(\beta \frac{L}{2}) + 2\frac{A_0}{L}sin(\beta \frac{L}{2}) =\beta^2\frac{A_0}{L}sin(\beta \frac{L}{2})\frac{L^2}{4}+C_0 \]

 

    \[ C_0=2 \frac{A_0}{L} \beta \frac{L}{2} cos(\beta \frac{L}{2}) + \frac{A_0}{L}sin(\beta \frac{L}{2}) \Big(2 - (\beta \frac{L}{2})^2\Big) \]

 

    \[\textrm{@} \quad x=\frac{L}{2}: \quad w(\frac{L^-}{2})=w(\frac{L^+}{2}) \quad \Rightarrow \quad A_0sin(\beta \frac{L}{2}) + 2\frac{A_0}{L}sin(\beta \frac{L}{2})\frac{L}{2} = -\frac{1}{3}\beta^2\frac{A_0}{L}sin(\beta \frac{L}{2})\frac{L^3}{8}-C_0\frac{L}{2}\]

 

    \[A_0 \Big[sin(\beta \frac{L}{2}) \Big( 3-\frac{1}{3}(\beta \frac{L}{2})^2 \Big) + \beta \frac{L}{2} cos(\beta \frac{L}{2}) \Big]=0\]

The equation above is the one that helps us get the critical force. We know A_0 cannot be zero so the expression in the bracket must be zero.

    \[sin(\beta \frac{L}{2}) \Big( 3-\frac{1}{3}(\beta \frac{L}{2})^2 \Big) + \beta \frac{L}{2} cos(\beta \frac{L}{2}) =0\]

The equation above has infinite number of solutions. However, we need the first positive value. I used MATLAB to solve for β:

    \[\beta \frac{L}{2} =2.1602\]

 

    \[P= \frac{18.666EI}{L^2}\]

% x=beta*L/2
f=inline('x*cos(x)+sin(x)*(3-1/3*x^2)')
fzero(f,3)