Question

The buckling of a beam is found to be more if __________.
A. The breadth of the beam is large
B. The beam material has a large value of Young’s modulus
C. The length of the beam is small
D. The depth of the beam is small

Answer 1

Hint: The term buckling is associated with longer columns. Buckling is the sudden change in the shape and form of the column due to the compressive load. This can be seen by bowing and curving of the straight column which results in disfiguration of the shape of the column.

Complete step by step answer:
The condition for buckling of the load is given by the Euler’s critical load formula –
$P = \dfrac{{{\pi ^2}EI}}{{{{(KL)}^2}}}$
where,
$E$ = modulus of elasticity of the material of the column,
$I$ = moment of inertia of cross-section of the column,
$L$ = effective length of the column,
$K$ = column effective length factor.
The critical load means that any load that is lesser than the critical load will be safer to apply on the column. So, if any load exceeds the critical load, it sets the column into unstable equilibrium and the column starts buckling and it does not return to its original form.

Now, let us consider each of these statements –
A) The breadth of the beam is large:
There is no dependence on the breadth term in Euler’s critical load formula. Hence, the breadth of the beam is immaterial in determining the critical load at which it starts buckling. Hence, this statement is not the correct answer.
B) The beam material has a large value of Young’s modulus:
The Euler’s critical load formula
$P = \dfrac{{{\pi ^2}EI}}{{{{(KL)}^2}}} $
$P \propto E $
Since the critical load is directly proportional to Young’s modulus, a higher value of Young’s modulus will result in the very high tolerance limit for the critical load which means that there is less buckling.
Hence, this statement is not the correct answer.
C) The length of the beam is small:
The Euler’s critical load formula –
$P = \dfrac{{{\pi ^2}EI}}{{{{(KL)}^2}}} $
$P \propto \dfrac{1}{{{L^2}}} $
Since the critical load is inversely proportional to length, smaller values of length will result in an increase in the critical load, which implies that the chances of buckling is very less.
Hence, this statement is not the correct answer.
D) The depth of the beam is small:
The Euler’s critical load formula –
$P = \dfrac{{{\pi ^2}EI}}{{{{(KL)}^2}}} $
$ P \propto I $
$ P \propto M{d^2}$
Here, d is the depth of the beam
Since, the critical load is directly proportional to the depth of the beam vis-à-vis one of the dimensions in the cross-sectional figure, smaller values of depth result in lesser critical load, which means that there is a greater chance of buckling.
Hence, this statement is the correct answer.

Therefore, the correct option is Option D.

Note: There are 2 ways of compressive stress failure in columns – Crushing and Buckling. Crushing occurs for shorter columns, also called struts and Buckling occurs in longer columns. Even though you may think that their classification of long and short columns is subjective, it depends on a number known as the Slenderness Ratio.
Slenderness ratio = $\dfrac{l}{k}$
where $l$= effective length of the column
and $k$= radius of gyration.
The radius of gyration, $k = \sqrt {\dfrac{I}{A}} $
where $I$is the least moment of inertia of the cross-section of the column and A is the area of cross-section.
For structural steel material, the columns are said to be-
Short, if the slenderness ratio is less than 40.
Medium, if the slenderness ratio is between 40 and 150.
Long, if the slenderness ratio is greater than 150.