## Tuesday, 9 May 2017

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.

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)
σd = Direct compressive stress developed in the column due to axial applied compressive load
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
σd = P/ A
σd = P/ (b x d)
Unit of Direct compressive stress = N/mm2

### 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.
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/mm2

Recall the concept of bending stress and we will write here the expression for the bending stress developed in the body.
Where,
I is the area moment of inertia  of the column rectangular section across the axis YY
I = db3/12

M = Moment formed by the load P
M= P x 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 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.
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?

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.

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.
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 σb will be dependent over the value of y.

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.

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.

σMax = Direct stress + Bending stress
σMax = σd + σb

#### 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.

σMin = Direct stress - Bending stress
σMin =  σd - σb

We have already determined above the expression for direct stress (σd) and bending stress (σb) 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

If minimum stress σMin = 0, it indicates that there will be no stress along the layer AD
If minimum stress σMin = Negative, it indicates that there will be tensile stress along the layer AD
If minimum stress σMin = 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.

We will discuss another topic in our next post i.e.

### Reference:

Strength of material, By R. K. Bansal