Pin ended column

The following model applies to columns simply supported at each end (${\displaystyle K=1}$).

Firstly, we will put attention to the fact there are no reactions in the hinged ends, so we also have no shear force in any cross-section of the column. The reason for no reactions can be obtained from symmetry (so the reactions should be in the same direction) and from moment equilibrium (so the reactions should be in opposite directions).

Using the free body diagram in the right side of figure 3, and making a summation of moments about point x:

$${\displaystyle \Sigma M=0\Rightarrow M(x)+Pw=0}$$

where w is the lateral deflection.

According to Euler–Bernoulli beam theory, the deflection of a beam is related with its bending moment by:

$${\displaystyle M=-EI{\frac {d^{2}w}{dx^{2}}}.}$$

so:

$${\displaystyle EI{\frac {d^{2}w}{dx^{2}}}+Pw=0}$$

Let ${\displaystyle \lambda ^{2}={\frac {P}{EI}}}$, so:

$${\displaystyle {\frac {d^{2}w}{dx^{2}}}+\lambda ^{2}w=0}$$

We get a classical homogeneous second-order ordinary differential equation.

The general solutions of this equation is: ${\displaystyle w(x)=A\cos(\lambda x)+B\sin(\lambda x)}$, where ${\displaystyle A}$ and ${\displaystyle B}$ are constants to be determined by boundary conditions, which are:

• Left end pinned: ${\displaystyle w(0)=0\rightarrow A=0}$
• Right end pinned: ${\displaystyle w(\ell )=0\rightarrow B\sin(\lambda \ell )=0}$

If ${\displaystyle B=0}$, no bending moment exists and we get the trivial solution of ${\displaystyle w(x)=0}$.

However, from the other solution ${\displaystyle \sin(\lambda \ell )=0}$ we get ${\displaystyle \lambda _{n}\ell =n\pi }$, for ${\displaystyle n=0,1,2,\ldots }$

Together with ${\displaystyle \lambda ^{2}={\frac {P}{EI}}}$ as defined before, the various critical loads are:

$${\displaystyle P_{n}={\frac {n^{2}\pi ^{2}EI}{\ell ^{2}}}\;,\quad {\text{ for }}n=0,1,2,\ldots }$$

and depending upon the value of ${\displaystyle n}$, different buckling modes are produced^[4] as shown in figure 4. The load and mode for n=0 is the nonbuckled mode.

Theoretically, any buckling mode is possible, but in the case of a slowly applied load only the first modal shape is likely to be produced.

The critical load of Euler for a pin ended column is therefore:

$${\displaystyle P_{cr}={\frac {\pi ^{2}EI}{\ell ^{2}}}}$$

and the obtained shape of the buckled column in the first mode is:

$${\displaystyle w(x)=B\sin \left({\pi \over \ell }x\right).}$$

General approach

The differential equation of the axis of a beam^[5] is:

$${\displaystyle {\frac {d^{4}w}{dx^{4}}}+{\frac {P}{EI}}{\frac {d^{2}w}{dx^{2}}}={\frac {q}{EI}}}$$

For a column with axial load only, the lateral load ${\displaystyle q(x)}$ vanishes and substituting ${\displaystyle \lambda ^{2}={\frac {P}{EI}}}$, we get:

$${\displaystyle {\frac {d^{4}w}{dx^{4}}}+\lambda ^{2}{\frac {d^{2}w}{dx^{2}}}=0}$$

This is a homogeneous fourth-order differential equation and its general solution is

$${\displaystyle w(x)=A\sin(\lambda x)+B\cos(\lambda x)+Cx+D}$$

The four constants ${\displaystyle A,B,C,D}$ are determined by the boundary conditions (end constraints) on ${\displaystyle w(x)}$, at each end. There are three cases:

1. Pinned end:

   ${\displaystyle w=0}$ and ${\displaystyle M=0\rightarrow {d^{2}w \over dx^{2}}=0}$

2. Fixed end:

   ${\displaystyle w=0}$ and ${\displaystyle {dw \over dx}=0}$

3. Free end:

   ${\displaystyle M=0\rightarrow {d^{2}w \over dx^{2}}=0}$ and ${\displaystyle V=0\rightarrow {d^{3}w \over dx^{3}}+\lambda ^{2}{dw \over dx}=0}$

For each combination of these boundary conditions, an eigenvalue problem is obtained. Solving those, we get the values of Euler's critical load for each one of the cases presented in Figure 2.