Chapter 17, Problem 1RP

The wood column is 4 m long and is required to support the axial load of 25 kN. If the cross section is square, determine the dimension a of each of its sides using a factor of safety against buckling of F.S. = 2.5. The column is assumed to be pinned at its top and bottom. Use the Euler equation. E_w = 11 GPa, and σ_Y = 10 MPa.

To determine

Find the dimensions a of the square column.

Answer to Problem 1RP

The dimension a of the square column is $\underset{\_}{103\text{mm}}$.

Explanation of Solution

Given information:

The length of the column is $L=4\text{m}$.

The axial load applied on the column is $P=25\text{kN}$.

The factor of safety against buckling is $\text{F}.\text{S}.=2.5$.

The column is pinned at its top and bottom.

The modulus of elasticity of the wood column is $E_{\text{w}}=11\text{GPa}$.

The allowable yield stress is ${\sigma }_Y=10\text{MPa}$.

Calculation:

Find the critical load $\left(P_{\text{cr}}\right)$ applied on the column.

$P_{\text{cr}}=\text{F}.\text{S}.\times P$

Substitute 2.5 for F.S. and 25 kN for P.

$\begin{array}{c}{P}_{\text{cr}}=2.5\times 25\\ =62.5\text{kN}\end{array}$

Calculate the moment of inertia $\left(I\right)$ of the square column using the relation:

$I=\frac{a^4}{12}$

Here, the dimension of the column is a.

The value of effective-length factor for rectangular column, $K=1$, since both the ends are pinned.

Show the expression for Euler’s formula as shown below:

$P_{\text{cr}}=\frac{{\pi }^2{E}_{\text{w}}I}{{\left(KL\right)}^2}$

Here,$P_{\text{cr}}$ is the maximum load on the material just before it begins to buckle, E is the Young’s modulus of the material, I is the least moment of inertia of the material cross-sectional area, and L is the length of the material.

Substitute 62.5 kN for $P_{\text{cr}}$, $11\text{GPa}$ for $E_{\text{w}}$, $\frac{a^4}{12}$ for I, 1 for K, and 4 m for L.

$\begin{array}{c}62.5\text{kN}\times \frac{{\text{10}}^{\text{3}}\text{N}}{\text{1}\text{kN}}=\frac{{\pi }^2\times 11\text{GPa}\times \frac{{10}^9{\text{N/m}}^2}{1\text{GPa}}\times \frac{a^4}{12}}{{\left(1\times 4\right)}^2}\\ \frac{62.5\times {10}^3\times 16\times 12}{{\pi }^2\times 11\times {10}^9}=a^4\\ a=0.103\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}\\ =103\text{mm}\end{array}$

Check the validity of using Euler’s formula:

Euler’s formula is valid only when the critical stress $\left({\sigma }_{\text{cr}}\right)$ is less than the yield stress $\left({\sigma }_Y\right)$.

Find the cross-sectional area (A) of the column as follows;

$\begin{array}{c}A=a^2\\ =103\times 103\\ =10,609{\text{mm}}^2\end{array}$

Calculate the critical stress for the column using the relation:

$\left({\sigma }_{\text{cr}}\right)=\frac{P_{\text{cr}}}{A}$

Substitute 62.5 kN for $P_{\text{cr}}$ and $10,609{\text{mm}}^2$ for A.

$\begin{array}{c}\left({\sigma }_{\text{cr}}\right)=\frac{62.5\text{kN}\times \frac{\text{1,000}\text{N}}{\text{1}\text{kN}}}{10,609}\\ =5.89\text{MPa}<\left({\sigma }_{\text{Y}}=10\text{MPa}\right)\end{array}$

Thus, the use of Euler Equation is valid.

Therefore, the dimension a of the square column is $\underset{\_}{103\text{mm}}$.