CRIPPLING LOAD WHEN ONE END OF THE COLUMN IS FIXED AND OTHER END IS HINGED

In our previous topics, we have seen some important concepts such as expression for crippling load when both the ends of the column are hinged, expression for crippling load when one end of the column is fixed and other end is free and crippling load when both the ends of the column are fixed with the help of our previous posts.

Today we will see here one very important topic in strength of material i.e. Expression for crippling load when one end of the column is fixed and other end is hinged with the help of this post.

Before going ahead, we must have to understand here the significance of crippling load or buckling load.

When a column will be subjected to axial compressive loads, there will be developed bending moment and hence bending stress in the column. Column will be bent due to this bending stress developed in the column.

Load at which column just bends or buckles will be termed as buckling or crippling load.

Let us consider a column AB of length L as displayed in following figure. Let us consider that end A of the column is fixed and other end i.e. end B of the column is hinged.

Let us think that P is the load at which column just bends or buckles or we can also say that crippling load is P and we have displayed in following figure.

We have displayed, in above figure, the initial condition of the column as AB. We have also displayed here the deflected position of the column as ACB. Therefore after application of crippling load or when column buckles, ACB will indicate the position of the column.

Now, we will consider one section at a distance x from end A and let us consider that y is the lateral deflection of the column at considered section.

Column is fixed at end A and also carrying a crippling load P and hence there will be one fixed end moments i.e. M₀ at end A. It will tend to make slope at fixed end A equivalent to zero and hence it will act in anti-clock wise direction.

There will be one horizontal reaction force, at end B, for balancing the end moment as displayed in above figure and let us think that this horizontal reaction force at end B is H.

Now we will determine the bending moment developed across the section and we can write it as mentioned here

Bending Moment, M = – P. y + H. (L-x)

We have taken negative sign here for bending moment developed due to crippling load across the section and we can refer the post for securing the information about the sign conventions used for bending moment for columns.

As we know the expression for bending moment from deflection equation and we can write as mentioned her.

Bending Moment, M = E.I [d²y/dx²]

We can also write here the equation after equating both expressions for bending moment mentioned above and we will have following equation.

Above equation will also be termed as lateral deflection equation for column AB, when one end of the column is fixed and other end is hinged and column is subjected with crippling load P.

C₁ and C₂ are the constants of integration, now next step is to determine the value of constant of integration i.e. C₁ and C₂.

We will refer here one of our previous post i.e. End conditions for long columns and we will secure the value of constant of integration i.e. C₁ and C₂ by using the respective end conditions.

As we know that for long column, when one end of the column is fixed and other end is hinged, we will have following end conditions as mentioned here.

At fixed end A of the column, i.e. at x =0

Deflection y will be zero and slope dy/dx will also be zero, i.e. y = 0 and dy/dx = 0

At hinged end B of the column, i.e. at x =L

Deflection y will be zero

Let us use the first end condition i.e. at x = 0, deflection y = 0 in above lateral deflection equation for column and we will have value of constant of integration i.e.C₁ and it will be as mentioned here.

C₁ = - (H. L/P)

Now, we will differentiate the lateral deflection equation with respect to x and we will have slope equation for column AB and it will be displayed by dy/dx.

As we have already discussed that at x = 0, slope will also be zero or d y/dx = 0 and therefore now we will use this end condition in above slope equation in order to secure the value of C₂.

After using the value of x =0 and dy/dx = 0 in above slope equation, we will have value of C₂

Now it’s time to analyze the lateral deflection equation after considering and implementing the value of both constants i.e. C₁ and C₂.

Now we will consider the second end condition for this column AB i.e. end condition for hinged end i.e. at x = L, y = 0.

From here we will have expression for crippling load, when one end of the column is fixed and other end is hinged and we have displayed it in following figure.

Do you have suggestions? Please write in comment box.

We will now discuss effective length of a column, in the category of strength of material, in our next post.

Reference:

Strength of material, By R. K. Bansal

Image Courtesy: Google

Also read

Boyle’s law and Charles law 
Mass balance and energy balance for a steady flow process

Avogadro’s law

Thermodynamic system, boundary and surrounding

Pure substances