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. M

_{0 }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 [d

^{2}y/dx^{2}]
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.

C

_{1}and C_{2}are the constants of integration, now next step is to determine the value of constant of integration i.e. C_{1}and C_{2}.
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. C_{1}and C_{2}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.C

_{1}and it will be as mentioned here.
C

_{1}= - (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 C

_{2}.
After using the value of x =0 and dy/dx = 0 in above slope equation, we will have
value of C

_{2}
Now it’s
time to analyze the lateral deflection equation after considering and
implementing the value of both constants i.e. C

_{1 }and C_{2}.
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

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