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
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.
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 .
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.
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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