In our previous topics, we have seen
some important concepts such as bending stress in beams, basic concept of shear force and bending moment, strain energy stored in body, beam bending equation,
bending stress of composite beam, shear stress distribution diagram for various
sections etc.

###

###

####

###

####

####

####

###

###

Today we will see here one very
important topic in strength of material i.e. direct stresses and bending stresses
with the help of this post. Let us go ahead step by step for easy
understanding, however if there is any issue we can discuss it in comment box
which is provided below this post.Â

As we have discussed that when a body will be subjected with an axial tensile or axial compressive load, there will be produced only direct stress in the body. Similarly, when a body will be subjected to a bending moment there will be produced only bending stress in the body.

As we have discussed that when a body will be subjected with an axial tensile or axial compressive load, there will be produced only direct stress in the body. Similarly, when a body will be subjected to a bending moment there will be produced only bending stress in the body.

Now let us think that a body
is subjected to axial tensile or compressive loads and also to bending moments,
in this situation there will be produced direct stress and bending stress in
the body.

Here we will see this
situation where a body will be subjected to axial loads and bending moments and
we will analyze here the stresses developed in the body i.e. bending stresses
and direct stresses.

###
**Direct
stress**

Let us consider one column, as displayed
in following figure, which is fixed at one end and let us apply one load P
axially to the other end of the column. In simple, we can say that column will
be subjected to compressive load and as load is applied axially, there will be
developed direct compressive stress only in the body and intensity of this direct
compressive stress will be uniform across the cross section of the column.

We have following information from above
figure

P = Axial applied compressive load which
is acting on the column through its axis

A= Area of cross section of the given
column

Area of cross section of the given
column = b x d, (For rectangular cross section)

Ïƒ

_{d }= Direct compressive stress developed in the column due to axial applied compressive load
Â b= Width of the cross section of the given
column

d= height or depth of the cross section
of the given column

Direct compressive stress developed in
the column

_{ }= Axial applied compressive load/ Area of cross section of the given column
Ïƒ

_{d }= P/ A
Ïƒ

_{d }= P/ (b x d)
Unit of Direct compressive stress = N/mm

^{2}###
**Bending
stress**

Now let us consider one column, as
displayed in following figure, which is fixed at one end and let us apply one
load P to the other end of the column at a distance e from the axis of the column.
In simple, we can say that column will be subjected to an eccentric load and line
of action of this load will be at a distance e from the axis of the
column.Â

Distance between the axis of the column
and line of action of load i.e. e will be termed as eccentricity
of the load and such load will be termed as eccentric load. There
will be produced direct stress and bending stress in the column due to this
eccentric load.

Unit of bending stress = N/mm

^{2}
Recall the concept of bending stress and
we will write here the expression for the bending stress developed in the body.Â

Where,

I is the area moment of inertiaÂ of the column rectangular section across the axis YY

I = db

^{3}/12
M = Moment formed by the load P

M= P x e

P = Load applied with an eccentricity e

P = Load applied with an eccentricity e

y = Distance of the point from neutral axis where bending stress is to be determined

####
*Let us analyze the above concept and figure out here
that there will be produced direct stress and bending stress in the column due
to this eccentric load.*

*Let us analyze the above concept and figure out here that there will be produced direct stress and bending stress in the column due to this eccentric load.*

Let us draw again the column but with
some variation. We have applied two equal and opposite axial load P as
displayed in following figure. In simple, we have now three forces acting on the
column.Â

One load is eccentric load which is
acting at a distance e from the axis of the column and two loads of same
magnitude are acting along the axis of the column but in opposite direction
i.e. one load is acting axially in downward direction and second load is also
acting axially but in upward direction.

So, what we have done so far?Â

We have applied two equal and opposite
loads P axially on the column. As these loads are equal in magnitude and in
opposite directions and hence they can cancel out each other and we have left
with one eccentric load P which is acting with eccentricity e as displayed in
above figure.

Let us see the third and fourth figure
displayed above, what we are looking here?

We have shown there column with single
load P which is acting axially downward in third figure and hence we can say
that there will be developed direct compressive stress here due to load P.Â

Similarly, we have also shown the rest
two loads on column in fourth figure and as these force will form a moment or couple
(P x e) and therefore there will be developed one bending stress in the column
due to this moment.

Therefore if a column will be subjected
with an eccentric load then there will be developed direct stresses and bending
stresses too in the column and we will determine the resultant stress developed
in the column by adding direct and bending stresses algebraically.

###
**Resultant
stress when a column will be subjected with an eccentric load**

As we have assumed here that we have
column with rectangular cross-section and therefore let us draw here the
rectangular cross-section of the column. We will see here that on which faces
of rectangular cross-section, total stress will be maximum and similarly on
which faces of rectangular cross-section, total stress will be minimum.Â

Simultaneously we will also secure the
value of maximum stresses and minimum stresses developed in the column due to
eccentric loading. We have seen above with formula for bending stress that
bending stress Ïƒ

_{b}will be dependent over the value of y.Â
If point, where resultant stress is to
be determined, is lying on the same side of axis YY as the load, we will consider
the value of y as positive and bending stress will be of same type as direct
stress.

If point, where resultant stress is to
be determined, is lying on opposite side of axis YY as the load, we will
consider the value of y as negative and bending stress will be of opposite type
as direct stress.

Let us consider our case where load P is
acting on column with eccentricity e with respect to YY axis as displayed above
in figure. We will have four layers of the rectangular section here that is AB,
BC, CD and DA.Â

If we will analyze here the stresses at
layers, we can easily write here that stress will be maximum along the layer BC
and will be minimum along the layer AD.

We will consider here compressive stress
as positive and tensile stress as negative and we will secure the value of
maximum stress and minimum stress over the column rectangular section.

####
**Maximum
stress **

As we have seen that maximum stress will
be along the layer BC and we can write here the expression for maximum stress
as mentioned here.

As point, where resultant stress is to
be determined, is lying on the same side of axis YY as the load hence bending
stress will be of same type as direct stress. Here in this case, direct stress
is compressive in nature and therefore bending stress will also be compressive
in nature.

Ïƒ

_{Max}= Direct stress + Bending stress
Ïƒ

_{Max}= Ïƒ_{d }+ Ïƒ_{b}####
**Minimum
stress **

As we have seen that minimum stress will
be along the layer AD and we can write here the expression for minimum stress
as mentioned here.

As point, where resultant stress is to
be determined, is lying on opposite side of axis YY as the load hence bending
stress will be of opposite nature as direct stress. Here in this case, direct
stress is compressive in nature and therefore bending stress will be tensile in
nature.

Ïƒ

_{Min}= Direct stress - Bending stress
Ïƒ

_{Min}= Â Ïƒ_{d }- Ïƒ_{b}
We have already determined above the expression
for direct stress (Ïƒ

_{d}) and bending stress (Ïƒ_{b}) and hence by putting these values in above equation we will have the exact formula for maximum and minimum stresses.####
*Here we will have to analyse the above equation*

*Here we will have to analyse the above equation*

If minimum stress Ïƒ

_{Min}= 0, it indicates that there will be no stress along the layer AD
If minimum stress Ïƒ

_{Min}= Negative, it indicates that there will be tensile stress along the layer AD
If minimum stress Ïƒ

_{Min}= Positive, it indicates that there will be compressive stress along the layer AD
Please comment your feedback and
suggestions in comment box provided at the end of this post.Â

We will discuss another topic in our next post i.e. Resultant stress when a column will be subjected with an eccentric load and load will be eccentric with respect to both axes i.e. XX and YY axis.

We will discuss another topic in our next post i.e. Resultant stress when a column will be subjected with an eccentric load and load will be eccentric with respect to both axes i.e. XX and YY axis.

###
**Reference:**

Strength of material, By R. K. Bansal

Image Courtesy: Google

## No comments:

## Post a Comment