Strength of Materials

Prof S. K. Bhattacharya
Department of Civil Engineering
Indian Institute of Technology, Kharagpur
Lecture - 4
Analysis of Stress - III

Welcome to the 4th lesson on the course Strength of Materials.

(Refer Slide Time: 0:29)

Today we will continue our discussion on certain aspects of Analysis of Stress.

(Refer Slide Time: 1:41)

In this particular lesson it is expected that once we complete it,

• One should be able to evaluate stresses on any plane through stress transformation
equations
• Evaluate principal stresses and locate principal planes for two dimensional (2-D)
problems. In fact in the previous lesson we discussed about the evaluation of principal stresses in
three dimensional planes. Here we will be discussing evaluation of principal stresses and location
of principal planes for 2-D problems.
• We are going to evaluate the maximum shear stresses at a point on a body for 2-D
problems.
• We will also look into the concept of Mohr’s circle for stress and we will demonstrate
how to construct Mohr’s circle for stress.
(Refer Slide Time: 2:21)

Hence the scope of the particular lesson is the derivation of transformation equation for
evaluation of stresses for 2-D problems, evaluation of principal stresses and maximum shear
stresses; construction of Mohr’s circle. We will look into aspects of how we are going to draw
Mohr’s circle for the evaluation of the stresses at a particular point in the body. We will solve a
few examples to demonstrate how the stresses can be evaluated at any particular point.

(Refer Slide Time: 2:41)

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
is σ x . The normal stress in the y-plane is σ y and the shearing stresses are τ xy . We are interested
now to evaluate the stresses on a plane the normal to which is at an angle θ 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 τ xy . This being the x ' plane the
normal stress to this particular plane is σ x ' , and correspondingly the shear stress will be tau x
prime y 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 being θ , this particular angle is also θ the area on line AB can be designated in terms of the
area dA which is dA cos θ and area on line BC can be designated in terms of dA sin θ . 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 minus σ y dA sin θ the force and multiplied
by the component sin θ .

The shearing stresses we have is minus τ xy acting on BC which is dA sin θ component along x
direction is cos θ minus τ xy which is acting in the plane x dA cos θ , and component along sin θ
is equal to 0. This gives us the equation as σ x prime is equal to σ x cos square θ plus σ y sin
square θ plus 2 τ xy sin θ cos θ . Writing sin square θ and cos square θ in terms of cos 2 θ we
can write this as σ x into 1 by 2(1plus cos 2 θ ) plus σ y 1 by 2 (1 minus cos 2 θ ) plus τ xy sin 2 θ .

If you write sin θ and cos θ as sin 2 θ , then plus τ xy sin 2 θ ); this we can write as ( σ x plus σ y )
by 2 plus ( σ x minus σ y ) by 2 cos 2 θ plus τ xy sin 2 θ . This is the stress in the x-direction which
is the normal stress σ x prime which are written in terms of stresses σ x , σ y and τ xy . Similarly, if
we take equilibrium along Fy prime; ∑Fy prime is equal to 0 we get tau x prime y prime is equal to
minus σ x cos θ sin θ plus σ y sin θ cos θ plus τ xy sin square θ , τ xy cos square θ minus τ xy sin
square θ . Hence we can write tau x prime y prime is equal to [(minus σ x minus σ y ) by 2] sin 2 θ
plus τ xy cos 2 θ . So these are the equations σ x prime and tau x 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 2 θ ) is equal to minussin 2 θ ; cos (180 plus 2 θ ) is equal to minus cos 2 θ
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 2 θ minus τ xy sin 2 θ . Thereby if we add
these two σ x prime and σ y prime this gives us the value as σ x plus σ y .

The stresses σ x plus σ y plus σ z is equal to σ x prime plus σ y prime plus σ z 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 prime σ x prime plus σ 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
2 θ plus τ xy sin 2 θ .

We have seen tau x prime y prime is equal to minus ( σ x minus σ y ) by 2 sin 2 θ plus τ xy cos 2 θ .
These are the transformation equations. That means we can evaluate stresses at any plane which
is oriented at an angle θ in terms of the normal stresses σ x , σ y and τ xy . 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.

(Refer Slide Time: 16:35)

Hence these are the stresses which we have derived; σ x prime is equal to ( σ x plus σ y ) by 2 plus
( σ x minus σ y ) by 2 into cos 2 θ plus τ xy sin 2 θ . tau x prime y prime is equal to ( σ x minus σ y ) by
2 minus sin 2 θ plus τ xy cos 2 θ . 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 plus σ y .

(Refer Slide Time: 17:14)

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 plus σ y ) by 2 plus ( σ x minus σ y ) by 2 cos 2 θ plus τ xy sin 2 θ . If we take the derivative of
the normal stress with respect to θ is ∂ σ x prime by ∂ θ is equal to minus 2 (σ x minus σ y ) by 2
sin 2 θ plus 2 τ xy cos 2 θ . If we set this as equal to zero and take this on the other side then we
get tan 2 θ is equal to τ xy by ( σ x minus σ y by 2). Now this particular equation has two values
of θ . 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 tan θ .
Hence we have one angle as 2 θ P and another at angle of 180 plus 2 θ P which will us two
values.
(Refer Slide Time: 19:52)

Therefore this is the derivative of the normal stress and this is the derivatives of tan θ . We will
get two values of this root 2 θ and 180 plus 2 θ . 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
plus θ 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.

(Refer Slide Time: 21:10)

Interestingly if we look into the expression tau x prime y prime is equal to ( σ x minus σ y ) by 2 sin
2 θ plus τ xy cos 2 θ . If we say tau x prime y prime is equal to 0 then we get tan 2 θ is equal to 2
tau xy by ( σ x minus σ y ) which is similar to the expression which we have obtained for tan 2 θ
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 .

(Refer Slide Time: 22:45)

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 2 θ P is equal to τ xy by ( σ x minus σ y ) by 2. Hence the
value of this hypotenuse R is equal to square root of (( σ x minus σ y ) by 2) whole square plus τ xy
square. Hence value of cos 2 θ P is equal to ( σ x minus σ y ) by 2R sin2 θ p is equal to τ xy by R. If
we substitute the values of cos2 θ P and sin2 θ P , in the expression of the normal which we have
evaluated σ x prime is equal to ( σ x plus σ y ) by 2 plus ( σ x minus σ y ) by 2 cos2 θ plus τ xy sin2 θ .

Now if we substitute for cos2 θ and sin2 θ 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 σ 1 and σ 2 is
equal to ( σ x plus σ y ) by 2 plus (( σ x minus σ y ) by 2) whole square 1 by R plus τ xy square by R.
This is equals to at this particular part, (( σ x minus σ y ) by 2) whole square by Rand τ xy square
by R and as we have denoted the R as root of this, so the R square is the top part. So σ 1 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 τ xy 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 σ 1 is equal to ( σ x plus σ y )
by 2 plus square root of (( σ x minus σ y ) by 2 ) whole square plus τ xy 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
σ 1 plus σ 2 ; because these are the two normal stresses at perpendicular plane.

We can write σ 1 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 σ 1 ; σ 1 is given by this, which can be write this is
equals to ( σ x plus σ y ) by 2 minus square root of (( σ x minus σ y ) by 2) whole square plus τ xy
square. Hence the stresses σ 1 or σ 2 is given as ( σ x plus σ y ) by 2 plus or minus square root of
(( σ x minus σ y ) by 2) whole square plus τ xy square. Hence these are the values of principal
stresses, maximum and minimum principal stresses.

(Refer Slide Time: 28:26)

Now these are the values of principal stresses maximum and minimum values are ( σ x plus σ y )
by 2 plus square root of (( σ x minus σ y ) by 2) whole square plus τ xy square 0.
(Refer Slide Time: 28:42)

Maximum shear stress:

We have seen shear stress on any plane, tau xˈyˈ is equal to minus ( σ x minus σ y ) by 2 sin 2 θ
plus τ xy cos2 θ . If we take derivative of this with respect to the θ , ∂tau xˈyˈ by ∂ θ is equal to
minus 2( σ x minus σ y ) by 2 cos 2 θ minus τ xy 2 sin2 θ . So τ xy sin 2 is equal to minus
( σ x minus σ y ) by 2 cos 2 θ ; or tan 2 θ is equal to ( σ x minus σ y ) by 2 by τ xy . 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, 2 θ s and 180 degrees plus 2 θ 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 2 θ s and compare with the values of previously calculated
values of tan 2 θ P we find that tan2 θ s is equal to minus 1 by tan2 θ P and tan2 θ P we have
already evaluated earlier as τ xy by ( σ x minus σ y ) by 2. So this is equals to minus cot 2 θ P
which we can write as, tan 90 plus 2 θ P . This indicates that, 2 θ s is equal to 90 plus 2 θ 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.
(Refer Slide Time: 31:38)

This is the value of tan2 θ 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 2 θ is equal to minus ( σ x minus σ y ) by 2 by τ xy . If we place

this in geometrical form, this we have to take a 2 θ s , this is τ xy , this is minus ( σ x minus σ y ) by 2.
Hence the value of R again, square root of (( σ x minus σ y ) by 2) whole square plus τ xy square.
Likewise then the cos θ , rather cos2 θ s is equal to τ xy by R and sin2 θ s is equal to minus ( σ x
minus σ 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 plus σ y ) by 2 plus ( σ x minus σ y ) by 2 cos2 θ plus τ xy sin2 θ .

Now in this in place of sin2 θ and cos2 θ if we substitute cos2 θ s and sin2 θ 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 plus σ y ) by 2, these two terms get cancelled
once you substitute the values.
(Refer Slide Time: 35:31)

Also we have seen the values of shear stress as, tau x prime y prime is equal to minus ( σ x
minus σ y ) by 2 sin2 θ plus τ xy cos2 θ . If we substitute the values of sin2 θ and cos2 θ , sin2 θ
we have obtained as ( σ x minus σ y ) by 2, so this is (( σ x minus σ y ) by 2) whole square 1 by R
plus τ xy square by R. Then R is equal to square root of (( σ x minus σ y ) by 2) whole square plus
τ xy square, so this will be going to equal to (( σ x minus σ y ) by 2) whole square plus τ xy 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 minus σ y ) by 2) whole square plus τ xy 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 plus σ y ) by 2.

Now if the normal stresses are the principal stresses then we get that maximum shear stress is
equal to (σ 1 minus σ 2 ) by 2 from this expression, if the σ x is σ 1 , and σ y is σ 2 , and this is being
the principal stress τ xy is equal to 0, so tau max gives us the value in the terms of principal
stresses as ( σ 1 minus σ 2 ) by 2. This gives the maximum shear stress in the terms of principal
shear stresses.
(Refer Slide Time: 38:10)

Hence, we say that planes of maximum shear stresses are not free from normal stresses. As we
have seen in case of principal stresses, the principal stresses acts on the shear stresses are 0,
whereas the planes on which the shear stresses are maximum there we do have normal stresses
and the value of normal stress is ( σ x plus σ y ) by 2.

(Refer Slide Time: 38:45)

Now let us look into another representation or the evaluation of the stress from a concept which
is given by Otto Mohr of Germany in 1895 which we popularly designate as Mohr’s Circle of
Stress. As we have seen we have a stress body in which the normal stresses are σ x and σ y and
we have the corresponding shear stresses, now we can represent this stress system in terms of
this circle.

(Refer Slide Time: 39:34)

Now let us look in to the expression for normal stress and shearing stress. σ x prime is equal to
( σ x plus σ y ) by 2 plus ( σ x minus σ y ) by 2 cos2 θ plus τ xy sin2 θ . Now tau x prime y prime is equal
to minus ( σ x minus σ y ) by 2 sin2 θ plus τ xy cos2 θ . From the first of this equation you can write
this as σ x ˈ minus ( σ x plus σ y ) by 2 is equal to ( σ x minus σ y ) by 2 cos2 θ plus τ xy sin2 θ . Now, if
we square this equation and the second equation and add them up we get ( σ x ˈ minus
( σ x plus σ y ) by 2) whole square plus tau x prime y prime square is equal to (( σ x minus σ y ) by 2)
whole square and (sin square 2 θ and cos square 2 θ is 1) plus τ xy 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 tau x prime y prime . If we draw a circle whose centre is at (a,
0), where a is equal to ( σ x plus σ 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 minus σ y ) by 2) whole square plus τ xy 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 plus σ y ) by 2 which is the average stress. This
particular point represents the stress which we have at the particular body which is σ x , σ y , and
τ xy . 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 τ xy . The σ x and τ xy on this circle is
representing this particular plane. Hence this being σ x and this being ( σ x plus σ y ) by 2, the
distance here, this particular distance, σ x minus ( σ x plus σ y ) by 2 is equal to( σ x minus σ y ) by 2.
This particular distance is τ xy . So eventually this particular distance is the square root of
(( σ x minus σ y ) by 2) whole square plus τ xy 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 2 θ 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 as σ 1 , and this we represent as minimum principal stress as σ 2 .

This particular point and this point in the circle represents the maximum value of the shearing
stress τ xy is equal to radius is equal to square root of (( σ x minus σ y ) by 2) whole square plus τ xy
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 2 θ P is equal to 90; θ is equal to 45 degrees.

(Refer Slide Time: 47:01)

Hence from the Mohr’s circle we can observe that the maximum normal stress is σ 1 which we
have designated as maximum principal stress. The minimum shear stress is σ 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 minus σ y ) by 2) whole square plus τ xy square and the radius is nothing but equals to
in terms of σ 1 and σ 2 has ( σ 1 minus σ 2 ) by 2 which is we have seen through our transformation
as well.

(Refer Slide Time: 48:05)

If σ 1 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 plus σ 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.
(Refer Slide Time: 49:36)

Now let us look into how we construct a Mohr’s circle. Basically we need to do that, if we know
the stresses: σ x , σ y and τ xy at a particular point, then we should be able to evaluate the stresses at
any plane, which are at σ x prime, tau x prime and tau y prime from the orientation of the plane
with reference to the x-plane. Now those stresses can be evaluated, either from the
transformation of the equations as we have just seen or we can evaluate the stresses from the
Mohr’s circle as well. Now let us look at how to construct a Mohr’s circle based on the given
state of stress.

(Refer Slide Time: 50:26)

Now let us say that at a stress body, we know the stresses: σ x , σ y , and τ xy . As we have seen the
centre of the circle is located at a distance of ( σ x plus σ y ) by 2. Now in this there are two ways
of constructing a circle, one is either we can represent σ in this direction and tau in the positive
direction as we have noted earlier in that case, the angle will be in a clock wise direction, θ
angle in the Mohr plane will be in clockwise direction which will be opposite to the convention
which we have assumed while deriving this transformation equation, where we take θ in
anticlockwise direction.

Another way to construct is we take σ in the positive x-direction and tau in the opposite
direction, in that case the representation of angle in the Mohr’s plane, is in anticlockwise
direction which matches with our transformation of the equation. So let us represent the σ in the
positive x-direction and the tau in the lower direction as positive. Thereby the angle will be
represented in the anticlockwise direction which is considered as positive and matches with our
physical stress system.

Now if we represent this stress σ and tau on this particular plane, thereby this particular point on
the circle represents this plane, plane-A, where the normal stress is σ and shearing stress is tau.
Likewise in the perpendicular plane we have stress as σ y and shearing stress as tau which is at
an angle of 90 degrees, with respect to this A-plane. In Mohr’s circle if we go 180 degrees with
respect to this, we get another point, which is plane-B were we have σ y and τ xy .

Now this if we take the centre, as this O, and radius as OA and OB, and plot a circle, this gives
us the Mohr’s circle where in this point represents the maximum normal stress σ 1 . This
represents the minimum principal stress which is σ 2 . Now if we are interested to find out stress
on any plane, whose normal to this particular plane is oriented at angle of θ , now this particular
plane which is represented by this point A, from here it moves at angle of 2 θ , and joins from the
centre of this the point which we get on the circle represents the plane normal to which is at
angle θ from the x-plane. Thereby the stresses that act over here together is the normal stress
and the shearing stress on this particular inclined plane and that is how we compute the stresses
in the Mohr’s circle.
(Refer Slide Time: 54:35)

Now having known about the principal stresses from the transformation equations and through
the Mohr’s circle let us look into some of the problems. Let us understand how to evaluate the
mass which is safely supported by the wires. Now if we take the free body diagram of this
particular part, then here there is the force F 1 to balance it, the reactive forces. Here there is force
F 2 to balance it.

If we draw the free body diagram of this whole part then we have F 1 in this particular direction
and F 2 in this particular direction and the mass which is hanging from this particular point. Let us
call it this force as Mg. Now let us call this angle as θ 1 , and this as θ 2 . From the given data, we
can compute cos θ 1 is equal to 0.6 and sin θ 1 is equal to 0.8; cos θ 2 is equal to 0.923, and sin θ 2
is equal to 0.385. Now if we take the summation of horizontal forces as 0, that gives us the
F 1 cos θ 1 is equal to F 2 cos θ 2 .

Now it has been indicated that the maximum tensile stress that the wires can withstand as 100
MPa and cross sectional area where the two wires are given, thereby if we compute, F 1 will gives
area times stress which is equals to 40 Kilo Newton. F 2 thereby is given 20 Kilo Newton. From
this expression if we substitute the values for cos θ 1 and cos θ 2 , we find that if we substitute for
F 2 for F 1 is equal to 40 Kilo Newton, and F 2 is equal to 40 into 0.6 by 0.923 is equal to 26 Kilo
Newton which is greater than 20 Kilo Newton. So instead of using F 2 , let us use, F 1 so that if we
represent F 1 in terms of F 2 this is 20 into 0.923 by 0.6 which is less than 40.

Hence if we compute the values, if we take the summation in the vertical direction, summation of
forces we get, that Mg is equal to F 1 sin θ 1 plus F 2 sin θ 2 and thereby the values from M this you
will get as 3293.4 Kg.
(Refer Slide Time: 58:34)

You can look into this particular problem, you can compute the state of stress in transformation
equations, also you can use the same problem to solve through Mohr’s circle of stress.

(Refer Slide Time: 58:43)

These questions which we have posed in the last time they quite straight forward, the first one is
the maximum normal stress and shear stress in an axially loaded bar, which we know as P by A
is the normal stress and P by 2A is the shear stress and thereby tau the shear stress is equal to σ 2 ,
half the normal stress. We have discussed about stress variants and we know that the value of
shear stress on the principal plane is zero.
(Refer Slide Time: 59:17)

To summarize: what we have done today in this particular lesson is developed the transformation
equation for evaluation of stresses for two dimensional problems; evaluation of principal stresses
and shear stresses, we have shown how to construct Mohr’s circle for the evaluation of stress and
some examples to demonstrate the evaluation of stresses at a point.

(Refer Slide Time: 59:44)

These are the questions:

• What is meant X the state of pure shear?
• What is the co-ordinate of the centre of the Mohr’s circle and what is its radius?
• What happens when ( σ x plus σ y ) is equal to 0 and σ 1 is equal to σ 2 ?