Principle stress

Let us look into the derivations of the transformation of equations. We have discussed about the
plane stresses in 2-D. If we consider the stress body at a particular point this is our reference x-
axis and this is the reference y-axis. The stress which is acting in the x-plane the normal stress.

isσ x . The normal stress in the y-plane is σ y and the shearing stresses are xy τ . We are interested
now to evaluate the stresses on a plane the normal to which is at an angleθ with respect to the x-
axis. The plane is considered in such a way that the normal direction normal to the plane
coincides with reference axis which we denote as x prime and y prime. Since the normal to this
particular plane coincides or is parallel to the x ' axis we call this plane as x ' plane.
Now let us look into the state of stress on this particular plane, if we take out this particular
wedge and if we designate this as A, B, and C the stresses which are acting on this particular part
are σ x normal stresses on this surface,σ y the shearing stresses xy τ . This being the x ' plane the
normal stress to this particular plane is ' σ x , and correspondingly the shear stress will be taux
primey prime.
Considering the unit thickness normal to the plane of the board if we assume the area on line AC
as dA which is length AC multiplied by the unit thickness then considering that this particular
angle beingθ , this particular angle is alsoθ the area on line AB can be designated in terms of the
area dA which is dA cosθ and area on line BC can be designated in terms of dA sinθ . Hence
the forces which are acting on these planes are the stresses multiplied by the corresponding area
will give us the force. If we wish to write down the equilibrium equations in the x ' direction and
y ' direction then the equation looks like this: summation of forces in the x ' direction that is
∑F x ' is equal to 0.
The forces which are acting in the x ' directions are σ x ˈinto dA is acting in the x 'direction
minus σ x acting on the area dA cos θ and the component in the x ' direction is cos θ ; σ y
which is acting in the opposite direction of ' σ x is the minusσ y dA sinθ the force and multiplied
by the component sin θ .
The shearing stresses we have is minus xy τ acting on BC which is dA sinθ component along x
direction is cos θ minus xy τ which is acting in the plane x dA cosθ , and component along sinθ
is equal to 0. This gives us the equation asσ x prime is equal to σ x cos square θ plus σ y sin
squareθ plus 2 xy τ sinθ cosθ . Writing sin square θ and cos square θ in terms of cos 2θ we
can write this as σx into 1 by 2(1plus cos 2θ ) plus σ y 1 by 2 (1 minus cos 2θ ) plus xy τ sin 2θ .
If you write sinθ and cosθ as sin 2θ , then plus xy τ sin 2θ ); this we can write as (σ x plusσ y )
by 2 plus (σ x minusσ y ) by 2 cos 2θ plus xy τ sin 2θ . This is the stress in the x-direction which
is the normal stressσ x prime which are written in terms of stressesσ x , σ y and xy τ . Similarly, if

we take equilibrium along Fy prime; ∑Fy prime is equal to 0 we get taux primey prime is equal to
minus σ x cosθ sinθ plus σ y sinθ cosθ plus xy τ sin square θ , xy τ cos square θ minus xy τ sin
square θ . Hence we can write taux primey prime is equal to [(minusσ x minusσ y ) by 2] sin 2θ
plus xy τ cos 2θ . So these are the equationsσ x prime and taux prime y prime and they are the
stresses on the plane which is at an angle of θ with respect to the x-axis.

Similarly, if we want to evaluate the stress in the y prime direction the normal stress σ y prime,
the stress σ y prime is at an angle of θ plus 90 degrees, if we substitute in place of θ as θ plus
90 then sin(180 plus 2θ ) is equal to minussin 2θ ; cos (180 plus 2θ ) is equal to minus cos 2θ
and if we substitute these values in the expression for σ y prime again we get σ y prime is equal
to (σ x plus σ y ) by 2 minus(σ x minus σ y ) by 2 cos 2θ minus xy τ sin 2θ . Thereby if we add
these twoσ x prime and σ y prime this gives us the value as σ x plusσ y .
The stresses σ x plus σ y plus σz is equal to σ x prime plus σ y prime plus σz prime which
indicates that irrespective of the reference axis system the summation of these normal stresses
are constant which we called as stress invariants. So here this is to prove again that the normal
stresses with reference axis is x primey primeσ x prime plusσ y prime is equal to σ x plus σ y is
equal to constant and so are the other stress invariants. Hence we have obtained the stresses in
the direction at an angle of θ as ' σ x is equal to (σ x plus σ y ) by 2 plus (σ x minus σ y ) by 2 cos
2θ plus xy τ sin 2θ .
We have seen taux primey prime is equal to minus (σ x minusσ y ) by 2 sin 2θ plus xy τ cos 2θ .
These are the transformation equations. That means we can evaluate stresses at any plane which
is oriented at an angleθ in terms of the normal stressesσ x ,σ y and xy τ . Please keep in mind that
the rotation of the angle θ we have taken as anti-clockwise and this is a positive according to our
convention.

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Hence these are the stresses which we have derived; σ x prime is equal to (σ x plusσ y ) by 2 plus

(σ x minusσ y ) by 2 into cos 2θ plus xy τ sin 2θ . taux primey prime is equal to (σ x minusσ y ) by 

2 minus sin 2θ plus xy τ cos 2θ . We have also seen σ y prime like this and if we add σ x prime

plus σ y prime we will get σ x plusσ y .

Now let us look into the position of the planes where the normal stresses are at maximum. We 
have obtained that the normal stresses on a plane σ x prime which is at an angle θ is equal to
(σ x plusσ y ) by 2 plus (σ x minusσ y ) by 2 cos 2θ plus xy τ sin 2θ . If we take the derivative of 
the normal stress with respect toθ is ∂σ x prime by ∂θ is equal to minus 2 (σx minusσ y ) by 2
sin 2θ plus 2 xy τ cos 2θ . If we set this as equal to zero and take this on the other side then we 
get tan 2θ is equal to xy τ by (σ x minusσ y by 2). Now this particular equation has two values 
ofθ . One is θ P with reference to the axis system we have the plane which is in the angle of θ P. 
Also, we will get another angle which is at an angle of 180 as tan 180 plus θ is equal to tanθ . 
Hence we have one angle as 2θ P and another at angle of 180 plus 2 θ P which will us two 

values.

 

Now let us look into the position of the planes where the normal stresses are at maximum. We 

have obtained that the normal stresses on a plane σ x prime which is at an angle θ is equal to

(σ x plusσ y ) by 2 plus (σ x minusσ y ) by 2 cos 2θ plus xy τ sin 2θ . If we take the derivative of 

the normal stress with respect toθ is ∂σ x prime by ∂θ is equal to minus 2 (σx minusσ y ) by 2

sin 2θ plus 2 xy τ cos 2θ . If we set this as equal to zero and take this on the other side then we 

get tan 2θ is equal to xy τ by (σ x minusσ y by 2). Now this particular equation has two values 

ofθ . One is θ P with reference to the axis system we have the plane which is in the angle of θ P. 

Also, we will get another angle which is at an angle of 180 as tan 180 plus θ is equal to tanθ . 

Hence we have one angle as 2θ P and another at angle of 180 plus 2 θ P which will us two 

values.

Therefore this is the derivative of the normal stress and this is the derivatives of tanθ . We will 
get two values of this root 2θ and 180 plus 2θ . We have designated these as θ P and 180 plus 2
θ P. In effect when we transform from this into the stress part we have angle θ P and 90 plus θ P
and that indicates that we have two planes which is at an angle of θ P and normal to this is the 
plane on which the normal stress is maximum, and then we have another plane which is at an 
angle of 90 degrees with reference to this particular plane because the other plane is at 90
plusθ P. These are the two normal directions where one will be the maximum and the other will 

be the minimum. These are the two normal stresses.

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Interestingly if we look into the expression taux primey prime is equal to (σ x minusσ y ) by 2 sin 

2θ plus xy τ cos 2θ . If we say taux primey prime is equal to 0 then we get tan 2θ is equal to 2

tauxy by (σ x minusσ y ) which is similar to the expression which we have obtained for tan 2θ

setting the derivative of the normal stress to 0. And since these two angles match this shows the 

planes where we have obtained the maximum and minimum principal stresses they coincide with 

the planes where the shearing stress is 0. And as we have defined before that the planes on which 

shearing stress is 0 the normal stress is designated as principal stress. Hence the maximum 

normal stresses which we have obtained are nothing but the principal stresses where the shear 

stresses are 0 and their angles are defined by θ P where we have evaluated θ P 180 degrees 2 θ P.

Let us evaluate the maximum values of the normal stresses and the principal stresses. We have 

seen that this angle 2 θ P where tan of 2θ P is equal to xy τ by (σ x minusσ y ) by 2. Hence the 

value of this hypotenuse R is equal to square root of ((σ x minusσ y ) by 2) whole square plus xy τ

square. Hence value of cos 2 θ P is equal to (σ x minus σy) by 2R sin2θ p is equal to xy τ by R. If 

we substitute the values of cos2θ P and sin2θ P, in the expression of the normal which we have 

evaluatedσ x prime is equal to (σ x plusσ y ) by 2 plus (σ x minusσ y ) by 2 cos2θ plus xy τ sin2θ . 

Now if we substitute for cos2θ and sin2θ in terms of θ P this we get as the maximum stresses, 

so σ x prime, maximum or minimum which are nothing but the principal stresses σ1 and σ 2 is 

equal to (σ x plusσ y ) by 2 plus ((σ x minusσ y ) by 2) whole square 1 by R plus xy τ square by R. 

This is equals to at this particular part, ((σ x minusσ y ) by 2) whole square by Rand xy τ square

by R and as we have denoted the R as root of this, so the R square is the top part. So σ1 2 can be written as (σ x plus σ y ) by 2 plus R square by R, these get cancel so, this eventually gives us 

(σ x plus σ y ) by 2 plus square root of ((σ x minus σ y ) by 2 ) whole square plus xy τ square. So 

this is the value of maximum stress or one of the stresses we get.

Now we have obtained that, σ maximum or minimum, let us call it as σ1 is equal to (σ x plusσ y )

by 2 plus square root of ((σ x minus σ y ) by 2 ) whole square plus xy τ square. We have seen that 

the normal stresses are the constants summation of σ x plus σ y is equal to σ x plus σ y is equal to

σ1 plusσ 2 ; because these are the two normal stresses at perpendicular plane.

We can write σ1 plus σ 2 which are two normal stresses at perpendicular plane as equals to σ x

plus σ y , σ 2 from here is σ x plus σ y minus σ1 ; σ1 is given by this, which can be write this is 

equals to (σ x plusσ y ) by 2 minus square root of ((σ x minusσ y ) by 2) whole square plus xy τ

square. Hence the stresses σ1 or σ2 is given as (σ x plusσ y ) by 2 plus or minus square root of 

((σ x minusσ y ) by 2) whole square plus xy τ square. Hence these are the values of principal 

stresses, maximum and minimum principal stresses.

Now these are the values of principal stresses maximum and minimum values are (σ x plusσ y )

by 2 plus square root of ((σ x minusσ y ) by 2) whole square plus xy τ square. 

Maximam shear stress :-  We have seen shear stress on any plane, tauxˈyˈ is equal to minus (σ x minusσ y ) by 2 sin 2 θ

plus xy τ cos2θ . If we take derivative of this with respect to theθ , ∂tauxˈyˈ by ∂θ is equal to

minus 2(σ x minusσ y ) by 2 cos 2θ minus xy τ 2 sin2θ . So xy τ sin 2 is equal to minus

(σ x minusσ y ) by 2 cos 2θ ; or tan 2θ is equal to (σ x minusσ y ) by 2 by xy τ . Here also as we 

have noticed earlier two values of θ defining two perpendicular planes on which the shear stress 

will be maximum and those angles being, 2θ s and 180 degrees plus 2θ s. So in the stress body it 

will be θ s plus 90, or θ s and θ s plus 90 perpendicular plane on which shear stress will be 

maximum. 

Now if we look in to the values of tan 2θ s and compare with the values of previously calculated 

values of tan 2θ P we find that tan2θ s is equal to minus 1 by tan2θ P and tan2θ P we have 

already evaluated earlier as xy τ by (σ x minus σ y ) by 2. So this is equals to minus cot 2θ P

which we can write as, tan 90 plus 2θ P. This indicates that, 2θ s is equal to 90 plus 2θ P. Or θ s

is equal to 45 degrees plus θ P. This indicates that maximum shear stress occurs in the plane 

which is at angle 45 degrees with maximum or minimum principal shear stresses.

This is the value of tan2θ evaluated, hence we find that two mutually perpendicular planes on 

which maximum shear stresses exists; and of maximum and minimum shear stresses form an 

angle of 45 degrees with the principal planes is just seen. Now let us look in to the value of 

principal stress where shear stress is at maximum.

Now we have calculated that tan 2θ is equal to minus (σ x minusσ y ) by 2 by xy τ . If we place 

this in geometrical form, this we have to take a 2θ s, this is xy τ , this is minus (σ x minusσ y ) by 2. 

Hence the value of R again, square root of ((σ x minusσ y ) by 2) whole square plus xy τ square. 

Likewise then the cos θ , rather cos2θ s is equal to xy τ by R and sin2θ s is equal to minus (σ x

minusσ y ) by 2R if we substitute the values of cos and sin in the values of the normal which is 

σ x ˈ is equal to (σ x plusσ y ) by 2 plus (σ x minusσ y ) by 2 cos2θ plus xy τ sin2θ . 

Now in this in place of sin2θ and cos2θ if we substitute cos2θ s and sin2θ s we will get the 

values of normal stress. Also we will get the values of shear stresses explained. Now if you 

substitute these values we get this is equal to (σ x plusσ y ) by 2, these two terms get cancelled 

once you substitute the values.

Also we have seen the values of shear stress as, taux primey prime is equal to minus (σ x

minusσ y ) by 2 sin2θ plus xy τ cos2θ . If we substitute the values of sin2θ and cos2θ , sin2θ

we have obtained as (σ x minusσ y ) by 2, so this is ((σ x minusσ y ) by 2) whole square 1 by R

plus xy τ square by R. Then R is equal to square root of ((σ x minusσ y ) by 2) whole square plus

xy τ square, so this will be going to equal to ((σ x minusσ y ) by 2) whole square plus xy τ square. 

So this us the tau max. In fact the minimum stress is the negative of this. So tau max or min is 

equal to plus or minus square root of ((σ x minusσ y ) by 2) whole square plus xy τ square. These 

are the values of maximum stresses and we observed that the value of the normal stress on the 

plane where shear stress is maximum is equal to (σ x plusσ y ) by 2.

Now if the normal stresses are the principal stresses then we get that maximum shear stress is 

equal to (σ1 minus σ2) by 2 from this expression, if the σ x is σ1 , and σ y is σ 2 , and this is being 

the principal stress xy τ is equal to 0, so tau max gives us the value in the terms of principal 

stresses as (σ1minusσ 2 ) by 2. This gives the maximum shear stress in the terms of principal 

shear stresses

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Now let us look in to the expression for normal stress and shearing stress. σ x prime is equal to

(σ x plusσ y ) by 2 plus (σ x minusσ y ) by 2 cos2θ plus xy τ sin2θ . Now taux primey prime is equal 

to minus (σ x minusσ y ) by 2 sin2θ plus xy τ cos2θ . From the first of this equation you can write 

this asσ x ˈ minus (σ x plusσ y ) by 2 is equal to (σ x minusσ y ) by 2 cos2θ plus xy τ sin2θ . Now, if 

we square this equation and the second equation and add them up we get (σ x ˈ minus

(σ x plusσ y ) by 2) whole square plus taux prime y prime square is equal to ((σ x minusσ y ) by 2)

whole square and (sin square 2θ and cos square 2θ is 1) plus xy τ square; the other terms get 

canceled. 

This particular equation can be represented as (x minus a) whole square plus y square is equal to

b square. This particular equation is a well known equation which is that of a circle where the 

centre of the circle lies at the coordinates (plus a, 0) the radius of which is equals to b and x 

represents σ x prime, and y represents taux prime y prime. If we draw a circle whose centre is at (a, 

0), where a is equal to (σ x plusσ y ) by 2 on the σ x prime axis which is representing x, with 

radius of b is equal to square root of ((σ x minusσ y ) by 2) whole square plus xy τ square then we 

get the circle, and that is what is represented as here in terms of Mohr’s circle. 

The centre of this particular circle is at a distance of from the origin, consider this as σ x axis or σ 

axis and tau axis then, this at the distance of (σ x plusσ y ) by 2 which is the average stress. This particular point represents the stress which we have at the particular body which isσ x ,σ y , and 

xy τ . This particular point, point on this particular circle represents the value of xy which is 

nothing but the σ and tau at a particular orientation which is representing a plane. 

Here this particular point we are representing as σ x and xy τ . The σ x and xy τ on this circle is 

representing this particular plane. Hence this being σ x and this being (σ x plusσ y ) by 2, the 

distance here, this particular distance, σ x minus (σ x plusσ y ) by 2 is equal to(σ x minusσ y ) by 2. 

This particular distance is xy τ . So eventually this particular distance is the square root of 

((σ x minusσ y ) by 2) whole square plus xy τ square which is that of radius which is b. 

This particular point represents the maximum normal stress on which this normal stress is acting;

this is the minimum normal stress and from this plane we rotate 2θ P angle, one of the maximum 

normal stress plane, and if rotate by another 180 degrees, another plane representing the 

maximum normal stress. This maximum normal stress we call as maximum principal stress 

which we represented asσ1 , and this we represent as minimum principal stress asσ 2 . 

This particular point and this point in the circle represents the maximum value of the shearing 

stress xy τ is equal to radius is equal to square root of ((σ x minusσ y ) by 2) whole square plus xy τ

square. So plus tau and minus tau are the maximum and minimum shear stresses. If we look into 

the plane this particular plane is representing principal stress and this particular plane 

representing maximum shear stress and the angle between these two is 90 degrees which is twice

of that in the body. As we have seen that the angle between the maximum principal plane and the 

plane on which maximum shear stress acts is at an angle of 45 degrees which is being 

represented here as 2θ P is equal to 90; θ is equal to 45 degrees.

Hence from the Mohr’s circle we can observe that the maximum normal stress is σ1 which we 

have designated as maximum principal stress. The minimum shear stress isσ 2 , which is 

minimum principal shear stress and at those two planes we have seen that no shear stress exists. 

Because that being on the σ axis, the value of shear stresses is 0 and hence they are the principal 

stresses. Also, the maximum shear stresses is equal to the radius of the circle which is square 

root of ((σ x minusσ y ) by 2) whole square plus xy τ square and the radius is nothing but equals to 

in terms of σ1 and σ 2 has (σ1minusσ 2 ) by 2 which is we have seen through our transformation 

as well.

If σ1 and σ 2 are equal then, Mohr’s circle reduces to a point and there are no shear stresses will 

be developed in the x, y-plane. And if σ x plus σ y is equal to 0, then, as we have seen, centre of 

the circle is located at a distance of (plus a, 0) which is on the axis and plus a is equal to

(σ x plusσ y ) by 2, the average stress. If σ x plus σ y is equal to 0, the centre coincides with origin 

as zero point at the σ tau as reference axis. Hence, at any point on any plane which is on the 

circumference of the circle representing any plane at the particular orientation, the values we will 

get are the maximum principal stress as the tau and also maximum shear stress as tau. This we 

call as the state of pure shear. Maximum and minimum principal stresses are also equal to the 

maximum shear stress. These are the important observations from the Mohr’s circle.

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Stress analysis

                            we will discuss about the stress. 

The concept of 

stresses. In this particular lesson we are going to look into some more aspects of analysis of STRESS. 

Before we start this know must know what is the unit of stress and force. 

The unit of Force is Newton (N) and unit of stress is (N by m square) or Pascal (Pa). This can be 

represented in terms of also Mega Pascal (Mpa) or just Giga Pascal (GPa).

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Here is the answer of the first question,  what is normal stress then after what is FBD? 

• The force acting normal to the plane is known as normal stress.                               **'FBD' means free body diagram. 

If you have a body which is a free body from a major body and is acted on by forces there will be 

resulting forces into it which will keep the body in equilibrium. At a particular small element if 

we take that this is a resulting stress then we can decompose this stress into two components, one

is along the normal direction of this particular cross section which is this and another component 

along the plane of this particular section. The component which is acting normal to this particular 

cross section is normally known as the normal stress. Normal stress is the normal component

perpendicular to the particular section.

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¶

Note:- One should be able to understand the concept of stress in a body. Understand relevant stress 

components, and then one should be able to understand why we need to go for the equations of 

the equilibrium, for a given problem. One should be able to draw the free-body diagram and 

evaluate the stress resultants from these diagrams and thereby compute the stresses.

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Question :-What are the axioms on which behavior of deformable member subjected to forces depend?

Answer:- We have the first thing which is equilibrium of forces. Fundamental laws of Newtonian

mechanics are used for the equilibrium of forces and for the body should be such that it must be 

having the forces in x direction. Summation of forces in x direction should be equal to 0,

summation of forces in y direction should be equal to 0, and summation of forces in z direction 

should be equal to 0. These equations must be satisfied to fit for the body which is in space. In a

two dimensional form, these equilibrium equations reduces to summation of forces in x direction

equal to 0, summation of forces in y direction equal to 0 and summation of movement of z is 

equal to 0. Also the forces must satisfy the parallelogram of forces.

Now lets suppose that,  we have two forces in the plane which is normal and plane forces in the direction of 

the plane this must satisfy the law like the resultant should pass through the diagonal of the 

parallelogram. Or, if we are talking about forces or stresses in three dimension, if we look into 

this parallelepiped the forces acting in the x direction or the stress acting in the x direction the y 

direction and the z direction the resultant of this must be acting along the diagonal of this 

parallelepiped. This is the resulting stress of all these stress components. 

Therefore either in two dimensional plane it should be in this configuration or in a three 

dimensional plane it should be in this configuration which is the parallelogram of forces that 

must be satisfied. So these are the two basic axioms based on which the forces act on the 

deformable body are guided. 

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• Stress multiplied by area  on which they act produces force. 
• At any section ,  vector sum of the forces keeps the  body in equilibrium. 

Evaluate stress resultants and find stress.

Now this stress when it is multiplied is acted on by external forces. If we take a small element in

which a stress is acting this stress multiplied by the area gives the force which we call as the 

stress resultant and the total stress resultant is the stress multiplied by the elemental area

integrated over the whole of the area is the resulting force which we call as stress resultant. And 

thereby, we assume that at any point it has the same behavior.

° So at any section the vector sum of the forces keeps the body in equilibrium and that is how the 

stress resultant will be obtained for that particular section. So our job is to evaluate this stress 

resultant. And once we know the test resultant we can compute stresses at that particular section.

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This is x-axis, this is y-axis and this is z-axis. This is a body which is acted on by external forces.
If we cut this particular body to a plane which is perpendicular or rather parallel to the y, z plane
then the normal drawn on this particular plane will be parallel to x-axis. For any plane when you
draw the normal to that particular plane and if that particular normal coincides with any of the
axis we designate that plane with the name of that particular axis.
For example, here we have cut this body through a plane which is parallel to y, z plane. So, if we
draw a normal on to this section this normal is going to be the parallel to x-axis and thereby this
particular plane we designate this plane as x-plane. Now on this x-plane we have at a particular
point the resulting stress which we call as R. If we take the component of this stress R in three
perpendicular axis direction then we have the stress acting in the direction of x which is normal
to this particular section and as per the definition of normal stress this is the normal stress which
is acting in the direction of x. If we take the component of R along y-axis or parallel to y-axis
then the plane we get as stress is acting in y direction.
Also, we have the component which is acting in z direction. The stress which is acting parallel to
y or in the y direction we designate this as the stress tau acting in the plane x in the direction y
which we call as tau x, y.
Or if we designate this particular stress which is acting in the plane x along z direction then we
call these as tauxz. Thereby in this particular x-plane we have three stress components where one
is the normal to the plane and the other two are in the plane in the direction of y and z. The
normal stress we call as σ x and the other two components which are in the plane are called as
shearing stresses which are xy τ and xy τ .

Likewise, if we cut this body into a plane which is parallel to this z plane we cut along this then
on this plane on a particular point we can get three components of the stresses and they areσ y , τ
on the y plane in the x direction as yx τ , the shearing stress tau in the y plane in the z direction
as yz τ . Also, if we cut this body with a plane which is perpendicular or parallel to x, y plane and
if we plot the three stress components the stress which is normal to the plane gives out the
normal stress sigmaz and two stresses which are in the plane are on the z plane in the x-direction
and stress in the z-plane in the y-direction.
These are the nine stress components that we are going to get at a particular point. Now if this
particular body is cut in such a way that you take another plane which is at an infinite small
distance away from here if we cut it off by two parallel planes then we can get small a cubical
element on which you can plot the stresses.

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Now let us look into another aspect of it. Out of the stress components which we have obtained,
again let us call these as x-axis as usual, we call as y-axis and this as z-axis.
Now, if we take the movement of all these forces, now let us assume that this distance which we
have taken at a particular point of the body is dx, the vertical height being dy and along the z
direction tau with dz. Now if we like to take the movement of all the forces about z-axis, now in
this particular figure only the forces which will have relevance while taking the movement about
z-axis has been taken into account. Now this particular plane being x plane and the force which
is acting in the y-direction as per our nomenclature we call this as tauxy.
Accordingly this particular component of the stress which is acting in the direction of x and
acting on y-plane we call this as yx. Likewise, this is also tauyx and this is tauxy. Along with this
we have the other stresses like normal stress sigmax, sigmay and sigmaz and shearing
components as well in the other plane. Since only these forces or these stress components are
going to cause the movement other forces have not been shown here.
If we take the movement of all the forces about z-axis then the movement expression can be
written as tauyx which is the stress acting on the area dx by dz, so tauyx into dx into dz is the
force. The movement about the z-axis is the distance dy so this multiplied by dy is the movement
about the z-axis as tauyx which is clockwise in nature minus tauxy which is acting on the area
dz(dy). So tauxy into dy into dz is the force.
If we take the movement of this force with respect to z- axis then this is multiplied by the
distance dx. Assumingly that there are body force components in the x and y-direction, this x is
the body force per unit volume then along with this we have plus, but for the time being we are
neglecting the body force components because that is also not going to cause any moment as
such with respect to the z-axis.

Therefore here it is z is equal to 0 for equilibrium. This produces tauyx is equal to tauxy. In effect 

this means that the cross term tauyx and tauxy are equal. Likewise if we take the movement of 

forces about x and y-axis and take the relevant forces we can see that tauzx is equal to tauxz and 

tauyz is equal to tauzy. This gives us that the cross hearing terms are equal. So if we look into 

stress and strain which we had tauij the tauij is equal to sigmax, tauxy, tauxz, tauyx, sigmay and 

tauyz, tauzx, tauzy and sigmaz if we write in the matrix form. This is the stress tensor. 

Now for the equality of the shear we have obtained tauxy is equal to tauyx, tauxz is equal to tauzx, 

tauyz is equal to tauzy. Thereby stress tensor can be written as tauij in the matrix notation as 

sigmax, tauxy and tauxz. Now tauxy is equal to tauyx we will write this as tauxy, sigmay and tauyz

and zx and xz being the same we write this as tauxz, tauyz, tauyz and sigmaz and thereby it 

reduces to the six stress components sigmax, sigmay and sigmaz, tauxy, tauxz and tauyz and this 

we find is symmetrical in nature so the stress tensor has a symmetric form.

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Having known that the stress at particular point is acting which are combinations of normal 

stresses and shearing stresses let us look into that if we have a body and if we are interested to 

find out the change in stress from one point to another then the change of the stresses is from 

point to point, we need certain equations to be solved and those equations are called as equations 

of equilibrium. 

Coming back to the body here, for example we have a body which is stress and we like to find 

out the change in stress from this point to this point. So we need these changes to be evaluated 

through these equations of equilibrium. Now as usual we call this as x-axis, this as y-axis and 

this as z-axis. Now on this particular plane which is normal to the x-plane we have normal 

stresses known as sigmax. 

We have two shearing stress components in the x-plane acting in y-direction called as tauxy. We 

have stress in the x-plane in the z-direction which we call as tauxz. When it comes to this 

particular plane which is at a distance of dx from this plane and likewise let us assume that this 

length is dy and this is dz so the stress which will be acting in this which is the normal stress will 

have a component as sigmax plus del sigmax del x which is acting over the length dx. Likewise 

we will have the shearing stress component tauxy which is varying from this end to this end we 

will have tauxy plus del tauxy del x by dx. 

We will have x in the z-direction that is tauxz and tauxz in on this particular plane so when it is 

coming to this plane there is a change over the length dx so tauxz plus del tauxz del x by dx. 

Likewise the stress in this particular plane normal to this which is the y-plane will have sigmay,

the stress acting normal to this is sigmay plus del sigmay del y by dy the length. The shearing 

stress component on the y-plane acting in the direction of x will have tauyx, plus del tauyx del y

by dy the length. On this plane we will have sigmaz and the normal stress on the front z-plane is 

sigmaz plus del sigmaz del z over the length dz and so on.

Now if we take the forces which are acting in the x-direction and sum them up as for the 

equations of equilibrium the summation of all the forces in the x-direction must be equal to 0. If 

we write down the forces in the x-direction we have sigmax plus del sigmax del x by dx. So in 

the equation we have sigmax plus del sigmax del x by dx and acting over the area dy and dz 

minus sigmax acting over the area dy and dz. 

Also, in this particular direction we have plus (tauyx plus del tauyx del x(dx)) and delta yx by del 

y by dy acting over the area dx and dz minus tauyx, dx dz plus we have a term in the z plane 

acting on the x direction which is tauzx plus del tauzx del z by dz into dx and dy the area minus

tauzx(dx and dy) plus if we assume that X is the body force per unit volume then this multiplied 

by dx, dy and dz is equal to 0. So from this we will get del sigmax del x, if we cancel out these 

terms and divide the whole equation by dx, dy and dz we have del sigmax del x plus del tauyx by 

del y plus del tauxz by del z plus x is equal to 0 where x is the component of the body force. 

Likewise if we take the equilibrium of the forces which are acting in the y and z-direction we get 

two other sets of equations and they are, del tauxy by delx plus del sigmay by del y plus del tauyz

by del z plus y the body component force is equal to 0 del tauxz by del z plus del tauyz del y plus

del sigmaz del z plus z is equal to 0. These are called the equations of equilibrium.

These equations of equilibrium can be written down in a two dimensional form as well. You can 

designate these in xy-plane, here we have x and here we have y, this is sigmax and the variation 

along the x is sigmax plus del sigmax del x by dx the length where this is the length dx and this is 

dy, this is tauxy plus del tauxy del x by dx the length. This is sigmay plus del sigmay del y by dy 

the length, then we have tau this is tauyx and this gives the variation of tau which is tauyx plus del

tauyx by del y(dy). 

These are the stresses in the two dimensional plane and if we take the equilibrium of the forces in 

the x-direction then we can obtain the equations of the equilibrium in two dimensional plane. 

which could be del sigmax del x plus del tau and as yx and xy being the same we can write this

as tauxy del y plus the component of the body force x is equal to 0. 

The other equations will be del tauxy by del x plus del sigmay del y plus y is equal to 0. So these 

are the equations of equilibrium in two dimensional planes. Now, having known these stresses at 

a point, equations of equilibrium and how to evaluate those stresses or how to write down those 

stresses at different planes if we have to evaluate the stresses in an axially loaded member, then 

we have a member or we have a body in which we have a force acting in the axial direction, so 

let us call this body acted on by force P.

Now if we like to evaluate the stresses at any inclined plane let us cut this body by an inclined 

plane. And if we draw the free body diagram of this then we have the body in this form. Here we 

have the resistive force P which is acting. On this, the resulting force or the stress resultant is 

acting in this particular direction to equilibrate the body. Now we can take the components of 

this force in the normal direction, normal to this plane and along the plane which will give you

the three forces of the stress component which is normal which we call as the stress 

corresponding to normal and the two shearing stress components tau. 

If we concentrate on the two dimensional plane, if we take the axially loaded member the 

member is subjected to the load in the axial direction. Let us take a plane which is cutting this

body in this form and let us assume that this plane is making an angle of theta with the vertical. 

If we take the free body diagram of this particular body this is angle theta, we have the force 

acting here as P, the resistive force acting on this body to keep the equilibrium is P so this will 

have two components one along the normal and one along the plane of this particular section. 

Now this angle being theta, if we drop a perpendicular here this angle will also be theta hence 

this particular angle is also theta. So the force component along this is P cos theta and the force 

component along this direction is P sin theta. Now, if we say that the cross sectional area is A. 

and the cross sectional area of this as A prime then the stress which is acting in this particular 

inclined plane if we say that normal stress sigma theta and theta being designated by this 

particular plane which has got an angle theta in the vertical, sigma theta equals to the normal 

force component which is P cos theta divided by this area A prime. 

And A prime from geometrical property we can say A prime is equal to A by cos theta. Then P 

cos theta by A by cos theta is equal to P by A cos square theta. And the stress which is acting in 

the plane is P sin theta. The stress tau theta is equal to P sine theta by Acos theta so this 

eventually is going to give us P by 2A by sin 2 theta. So these are the two stress components on 

this inclined plane. The normal plane is P by A cos square theta and the stress which is parallel to 

the plane which is the shearing component is P by 2A sin2 theta. So the maximum value of 

sigma theta is when cos square theta is equal to 1 and is theta is equal to 0. And for this sin 2

theta as 90 degrees and 2 theta being 90 degrees so theta being 45 degrees is the maximum value 

of the shearing stress. So the maximum value of normal is P by A and the maximum value of 

shearing stress is P by 2A. Eventually the relationship between tau theta and sigma theta is that. 

See below. 

  This is the end of this chapter. 

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