Analysis of Stress – II

Analysis of Stress – II


Welcome to lesson 3 on the course Strength
of Materials. We are going to discuss about the
analysis of the stress. We have already looked into some aspects of stress analysis.
. .Now it is expected that once this particular
lesson is completed one will able to compute stress
resultant and stresses in members subjected to axial forces, evaluate stresses at a point
on a body at any arbitrary plane, evaluate principal
stresses and locate principal planes and also compute
stress invariants. . Hence the scope of this particular lesson
includes: review of normal stress, concept of shear and
bearing stress, computation of stress on any arbitrary plane, concept of principal stress
and principal plane and concept of stress invariants.
We discussed the types of stress and specifically about the normal stress.
. .And we had noticed that if we take a body
which is axially loaded by a force P and if we take a
section and draw the free body diagram, this body is under the action of external force
P so at the chord section say “aa” there will be resulting
stress component which we call as stress resultant or the force which is resisting P. At every
point there will be a stress component and the normal
stress multiplied by the area will give the force. So integral of stress multiplied by
the small area integrated over this surface will give the
stress resultant which is equal to P. While making this kind of assumption that
every where state of the stress exists if the force acts
through the centriod of the section we assume that the particle of the material at every
point contributes to the resistance of this external
force and thereby we assume the homogeneity of this
material. We assume that at every point the same state
of stress exists. When a body is subjected to
external forces which are trying to cause traction in the member or trying to pull the
member we call these kinds of forces and thereby stresses
as tensile stresses whereas when the external forces are acting on the member trying to
push the member we call this kind of external forces
and the stresses as compressive stresses. . Let us look into the aspects of shear stresses.
Let us assume that this particular body is subjected
to the action of external forces P and the resistive forces thereby will be P by 2 and
P by 2. Now if we take the free body diagram of this
particular member if we cut it over here then we
will have this body which is acted upon by external force P by 2 hence the resistive
internal force will also be P by 2. If we take the free body
of the other part of the cut the resistive internal force
is P by 2 and this force will be resisted at this inter phase and this also will be
P by 2. Though, this force is eccentric with respect to the
centre line of this body but this thickness being smaller
we neglect the other effects because of these forces. This force will try to cause stress
at the contact between these two elements and this
contact area is so near which is the plan. When we .look from the top the top view of this body
looks like this and this shaded part indicates the
contact area between these two pieces of material. Now if we consider that this particular length
as “a” and this as “b” then you can define the shear
stress which is designated as tau equals to the force which is acting at the inter phase
which is P by 2 by a(b). So a(b) is the contact area
over which the shearing stress exits. . Many a times we come across situations where
some blocks are resting over another block and
transferring the forces from external sources. For example, if we have a body like this in
which, this is a block which is resting on a bigger
block and the smaller block is subjected to external
load P. If this block is placed, that is, the centroid of the top block is placed on
the centroid of the bottom block then at the interface between
these two blocks and at this inter phase there will be
normal stresses generated and these normal stresses we designate as bearing stress. By
the term bearing we mean that the bottom block is bearing
the load of the top block. So, if the contact area
between the smaller block and the bigger block, this equals to Ac then the bearing stress
sigma bearing can be written as the external force
P by Ac which is the contact area. So this is called as
bearing stress. Here is another example which is the concept
of the bearing stress. We have a rigid bar resting
on two supports and this is subjected to external force P.
Now if we take the free body diagram of this particular bar, the external force
is P so the resistive force at the support point, assuming
this P is acting to the centroid of this body is P by 2 and P by 2. This reactive force
will be in turn transmitted to the support. This is the
support and resistive force which is getting transmitted
on and depending on the contact area we have if this is length “a” and the width of
this body is “b” then this force P by 2 has contact
area which is a(b). Hence the bearing stress that is sigma
bearing is equal to P by 2 as the reacting force which is getting transmitted on this
support divided by a(b). .. Here are the different stress components that
act on any arbitrary plane. Let us consider that A,
B, C are any arbitrary plane and o, x, y, and z is the reference axis system. As we
have noticed earlier the plane normal to which it coincides
with the axis we designate that in the name of the
particular axis. Likewise this particular plane is the x plane on which the normal stress
sigma x acts.
Likewise the plane oBC is normal to that coincides with y direction which is y-plane and the
normal stress sigma y on this plane. And the normal stress on the plane Aoc is the z-plane
is sigma z . Also, in those planes there are
shearing stresses x-plane in the y-direction will have the
stress which we call as tau xy , which we call as tau to the power XZ; on the x-plane,
in the ydirection, so tauxy. The component which is in the z-direction is designated
as tau to the power XZ. Likewise shear stress component in the
y-plane acting in the x-direction we call that as tau yx
and tau yz . Similarly this is tau zx and tau zy .
Let us assume that this arbitrary plane has a normal which is outward drawn normal is
n. This unit vector can be designated with reference
to xyz-plane. Let us assume that this is the reference
axis xyz. The unit normal is drawn here. If this vector makes an angle of α with x-axis,
beta with y-axis, and gamma with z-axis then we define
the cosine components in the x-direction as n x
which is cosine α; n y is the cosine beta; and n z as cosine gamma. Thereby the unit
vector this distance here can be represented by (n x ) whole
square plus (n z ) whole square; (n z ) whole square
plus (n y ) whole square will give this unit. So we have in effect (n x ) whole square plus
(n y ) whole square plus (n z ) whole square is equal to
1. Let us assume that on this arbitrary plane
we have the resulting stress vector as R and the
component of this resultant stress on this plane in the x, y, and z-direction be R x
, R y , and R z . Also let us assume that
the area of the arbitrary plane is dA which is the area of the plane ABC.
Now, if we take the projection of this area on x-plane which is AOB; so area of AOB is
dA into n x which is cosine of this area ABC on AOB.
Area BOC is the projection on the area ABC on y .plane which is dA into n y ; and area AOC
is the projection of ABC on the z-plane which is dA
into n z . . Now if we take the summation of all forces in the
x-direction where in the stress components involved will be sigmatau y in the x-direction,
and tauz x in the x-direction and R x then we can
write down the equilibrium equation in x-the direction. So equilibrium equation in the
xdirection will be R x into area dA; which is a force, stress resultant multiplied by
area minus sigma x dA n x the area of x-plane minus tau
yx which is acting in the y-plane times the area of yplane dA n y minus tau zx which is
the z-plane acting in x-direction multiplied by the area dA n z is
equal to 0. So the equation is as follows: R x dA minus sigma x dA n x minus tau yx dA
n y minus tau zx dA n z is equal to 0.
If we divide the whole equation by dA or in a limiting situation we get R x is equal to
sigma x n x plus tau yx n y plus tau zx n z . Similarly,
if we take the equations in the y and z-directions and write
down the equilibrium we will get R y is equal to tau xy n x plus sigma y n y plus tau zy
n z R y. And R z the resulting stress in the z-direction is
equal to tau xz (plane in the z-direction as) n x plus tau yz
(the y-plane in the z-direction) n y plus sigma z n z . That is R z is equal to tau
xz n x plus tau yz n y plus sigma z n z. These are the three resulting
stress components. So the stress components on the
arbitrary plane which are acting in the x, y and z-directions are represented in terms
of the stress components in the rectangular co-ordinate
system. This set of equations is normally designated
as Cauchy’s stress formula. .. So Cauchy’s stress formula is R x is equal
to sigma x n x plus tau yx n y plus tau zx n z ; R y is equal to
tau xy n x plus sigma y n y plus tau zy n z and Rz is equal to tau xz n x plus tau yz
n y plus sigma z n z . . Now let us look into, if we consider a plane
which has normal n and the direction cosines for the
normal are n x , n y and n z . Also, we assume that on this particular plane only
the normal stress acts in the direction of the normal to the
plane. Hence if we take the components of this in the x,
y, and z direction then as we have designated before R x as the resulting stress in the
x-direction; R y as the resulting stress in the y-direction
and R z as the resulting stress in the z-direction they .can be written in the direction cosines as:
R x as sigma n n x ; Ry as sigma n n y ; and R z as sigma n
nz. . Exactly in the same form the way we have evaluated
Cauchy’s stress formula taking the equilibrium equations in the x, y and z-direction
we can compute the resulting forces in x, y, and
z-direction in terms of sigma n . In the previous case we had forces in the
x direction as R x dA is equal to sigma x dA n x plus tau yx
dA n y plus tau zx dA n z and by dividing the whole equation by dA we had R x is equal
to sigma x n x plus tau yx n y plus tau zx n z . Exactly
in the same form in place of R x now we have sigma n n x
the resulting stress in the x-direction and this multiplied by area gives the force in
the xdirection, this equals to sigma x dA n x plus tau yx dA n y plus tau zx dA n z
. Hence from this we can write sigma n n x is equal to sigma x n x
plus tau yx n y plus tau zx n z . .. This is what is represented here the equations
of equilibrium in the three directions x, y, and z.
Now these equations can be rearranged and can be written as (sigma x minus sigma n)
n x plus tau yx n y plus tau zx n z is equal to 0;
tau xy n x plus (sigma y minus sigma n ) n y plus tau zy n z is equal
to 0; tau xz n x plus tau yz n y plus (sigma z minus sigma n) n z is equal to 0.
We have already seen that tau yx is equal to tau xy; tau zx is equal to tau xz al; tau
zy is equal to tau yz . Now these three equations can be thought of
as simultaneous equations containing n x , n y and n z
and we can evaluate the values of n x , n y and n z . Now, if we expand this particular
equation we will get a cubical equation in sigma n .
. .This is the cubical equation in sigma n : (sigma
n ) whole cube minus l 1 (sigma n ) whole square plus
l 2 sigma n minus l 3 is equal to 0. And once we solve this cubical equation we are expected
to get three roots which eventually will turn out
be real and we designate those roots as sigma 1 , sigma 2
and sigma 3 . And corresponding to each values of sigma we will get the values of n x , n
y and n z . . So here we have the equations on the arbitrary
plane on which the stress in absolutely normal and we have observed that we can get simultaneous
equations n x , n y and n z in terms of sigma x
tau xy , tau yx and tau yz; and sigma x , sigma y and sigma z .
. .This is the cubical equation in the terms
of I 1 , I 2 , and I 3 and for each of these sigma values if we
are interested to compute the values of n x , n y , and n z we can do that using these
equations along with the expression we have n x square plus
n y square plus n z square is equal to 1. So, for a trivial
solution n x , n y , n z is equal to 0 which is not really going to give us the solution
because n x square plus n y square plus n z square is
equal to 1 now for a trivial solution of the simultaneous
solution we said that the determinant of the coefficients of n x , n y , and n z is equal
to 0. And if we do that, we have (sigma x
minus sigma n ) tau xy tau xz tau xy (sigma y minus sigma n ) tau yz tau xz tau yz
(sigma z – sigma n ) is equal to 0. So if we say that the determinant of this
equals to zero for trivial solution we get the cubical
equation in sigma n cube minus I 1 sigma n square plus I 2 sigma n minus I 3 where the
values of I 1 is equal to sigmax plus sigma y plus sigma z
; I 2 is equal to determinant sigmax sigma y tau xy plus
sigma y sigma z tau yz plus sigma z sigma x tau xz ; I 3 is equal to sigma x sigma y
sigma z plus tau xy tau xz tau yz and this is symmetrical tau xy , tau
xz , tau yz these are the three coefficients I 1 , I 2 , and I 3 .
. .Now the plane on which this stress vector
is fully normal is known as principal plane. This is
what we considered in the previous situation where we had taken the arbitrary plane and
we considered the stress vector which is along
the normal and thereby we obtained the cubical equations in sigma n and from which we obtained
the three roots sigma sigma 1 , sigma 2 and sigma 3 .
In this particular plane the normal stress is acting along the normal of the plane. So
the plane on which this stress vector is wholly normal
is called as principal plane. The stress on this principal
plane which is absolutely normal is called as principal stress. And since the stress
acting is in the normal direction there are no tangential stresses,
the principal stress is the resultant normal stress
so on a principal plane there are no tangential or shearing stresses, they are 0. . .Now we have seen the cubical equation in
sigma n from which we have obtained three roots
sigma 1 , sigma 2 and sigma 3 . And we have looked into three coefficients I 1 , I 2 and
I 3 I 1 is equal to sigma x plus sigma y plus sigma z ; I 2 if
we expand determinants I 2 is equal to sigma x sigma y
(tau xy ) whole square plus sigma y sigma z (tau yz ) whole square plus sigma z sigma
x (tau zx ) whole square; and I 3 is equal to sigma x sigma
y sigma z plus 2 tau xy tau yz tau zx minus sigma x tau yz square
minus sigma y tau xz square minus sigma z taux y square. We call these as invariants.
It is to be noted that the principal stress which we have
calculated at a particular point remains the same
irrespective of the reference axis system we take.
In this particular case we have taken x, y and z as the rectangular axis system. Supposing
at that particular point if we take different axis
system which is represented as x prime, y prime and z
prime if we write down corresponding normal stresses and shearing stresses as sigma x
prime, sigma y prime and sigma z prime then we can
observe that the values of I 1 , I 2 and I 3 which are
represented in terms of the normal stresses and shearing stresses they remain the same
because the principal stresses at that particular
point remains unchanged. . .That is why these coefficients I 1 , I 2
and I 3 are known as stress invariants. Let us see how to solve
these stresses in some physical problems? In this particular example here there are
two plates which are connected together by a bolt and
it is subjected to a pull – external force P. This is the
plan of the two plates and if we take a section we cut the plate here and cut the plate here
and view from this side then the section looks
like this. Therefore the width of the plate is 200 mm
and the thickness of the plate is given as 10 mm and
the tensile pull that the plate is subjected to is 50kN. Now we will have to compute average
normal stress at a section where there are no holes for the bolts. So, if we cut the
section here and draw the free body diagram.
Here we have the plate which is put in a three dimensional form and
this is acted on by a load 50 kN, the width of the plate is given as 200
mm and the thickness of the plate is given as 10 mm.
So, at this particular section the resistive force for this external force 50kN which is
acting through the centroid of this plate is R and
hence from the equilibrium of these forces R is equal
to 50 kN. Hence the normal stress sigma at this particular
cross section where there are no holes is equals
to R divided by the cross sectional area sigma is equal to R by A is equal to 50 into 10
cube N by 200 into 10. This gives us the normal stress
as 25 N by mm square and 1000 mm is equal to 1
meter so this is equal to 25 into (10) whole power 6 N by m square. Since we designate
N by m square as P so it is 25 into (10) whole power
6 P and since (10) whole power 6 P gives the mPa,
hence we are going to write this as 25 mPa. Also, we will have to compute the average
shear stresses in the bolts.
The force which is acting here gets transmitted from this particular plate to this plate through
this connection where these two plates are connected
by the bolts. And as a result if we draw free body at this interface the force P will be
get transmitted at this interface and the two bolts which
are connecting these two plates together will be subjected to this force P. So we have one
bolt .here and another bolt here of diameter 20
mm and these two bolts will be resisting the force P by
2 and P by 2. The plate is subjected to load P and this
P is transmitted through the top plate through the
interconnecting bolt. So the force which will be resisted by these bolts is half of P, hence
the stress which is acting in the bolt which is
the force on this particular area is called as the shearing
stress on the plane of the bolt. So the average shear stress in each bolt which is tau average,
tau is equal to P by 2 by pi by 4 (d) whole square
(divided by the area of the bolt) and P is equal to 50;
25 into 10 cube N by pi by 4 into 400 is equal to 250 by pi N by m square which is so much
of mPa. So this is the average stress in the
bolt. Thirdly, we will have to compute the bearing
stress between the bolts and the plates. When the
force is getting transmitted from one plate to another it is getting transmitted through
these interconnecting bolts.
. And if we look into the transmission of the
force in little greater detail we will see that this is the
plate, here two bolts are present and this plate is being pulled by force P. Now the
transfer of force from the plate to the bolts which are
here when plate is being pulled, this part of the plate,
this is the bolt, the hub part of this particular plate comes in contact with this particular
bolt and there may be release in the contact between
the plate and the bolt surface. Basically the plate is resting on this surface
of the bolt. On an average sense we take the projection of this surface which is the diameter
of this bolt and the contact area is here which is
diameter times the thickness of the plate. So the contact surface which we get is half
the perimeter the projection of which is d and
the thickness of the plate at that particular interface.
This is the area which is in contact with the plate and the bolt.
The bearing stress is the function of the contact area so the sigma bg bearing is equal
to the force P and since we have two bolts each bolt will
get half the forces so P by 2 divided by the contact .area which is dt; sigma bg is equal to P
by 2 by dt is equal to 25 into 10 cube N by 20 into 10 is
equal to 125 MPa. . Here is another example where we are interested to
evaluate the principal stresses and the stress invariants. We have learnt how to compute
the principal stresses from the cubical equations and
the stress invariant components. Now the state of stress at a point is given by this, this
is the stress tensor where sigma x is equal to minus
5 units; sigma y is equal to 1 unit; and sigma z is
equal to 1 unit and tau xy is equal to minus 1; tau xz is equal to 0; and tau yz is equal
to 0. So, if we compute the stress invariants, as we have
seen, I 1 is equal to sigma x plus sigma y plus sigma z ;
sigma x is equal to minus 5 plus 1 plus 1, is equal to minus 3. I 2 is equal to sigmax
sigma y (tau xy ) whole square plus sigma y sigma z (tau yz
) whole square plus sigma z sigma x (tau zx ) whole square is
equal to (minus 5 minus 1) plus (1 minus 0) plus(minus 5 minus 0) is equal to minus 10. . .Likewise if you compute I 3 is equal to minus
6. Hence the cubical equation which you get is,
sigma n cube minus I 1 into (sigma n ) whole square here I 1 is equal to minus 3 plus 3
(sigma n ) whole square, I 2 is equal to minus 10 hence
minus 10sigma n plus 6 is equal to 0. Now sigma n is equal to 1 we will make this
function as 0, since sigma n in 1 of the root of the
equation, sigma n minus 1 if we take, sigma n square so this gives sigma n cube minus
sigma n square plus 4sigma n square minus 4sigma n
minus 6sigma n plus 6; so minus 6 into (sigma n minus
1) is equal to 0. So we get (sigma n minus 1) into (sigma n square plus 4sigma n minus
6) is equal to 0. From this we will get sigma n is equal
to 1, minus 2 plus or minus square root of 10. These
are the roots of this equation. These are the three principal stresses. These are the
three invariants I 1 , I 2 , and I 3 . That is how we can compute
principal stresses and stress invariants. . .Now in this we have another example problem
in which mass is hung by two wires AB and BC
and the cross sectional areas of these two wires are given as 200 mm square and 400 mm
square. And if the allowable tensile stress of these
wire material is limited to 100 MPa then you will
have to find out the mass M that can be safely supported by these wires. This particular
problem can be solved by taking the free body diagram
of this. . To summarize what we have learnt in this particular
lesson; first we recapitulated on different kinds of stresses and those stresses are the
normal stress, and the normal stress on an axially
loaded bar. What is the maximum normal stress that acts on an axially loaded bar is the
axial pull P or compressive force push by cross sectional
area and cross sectional area which is minimum is
normal to the axis of the bar. .Thereby we have seen the relationship between
the normal stress corresponding shearing stresses. We have evaluated the stresses on
any arbitrary plane which we generally designate as
Cauchy’s stress formula. And from this Cauchy’s formula we arrived at concepts of principal
plane and principal stresses and we have said that the principal plane is the one on which
the stress is fully normal and thereby tangential
stresses are at 0 and the shearing stresses in the
plane is 0. While computing the principal stresses we have seen different coefficients
which we designated as stress invariants I 1 , I 2
and I 3 . We have noticed that I 1 , I 2 and I 3 are
the functions of normal stress at a point which are sigma x ,
sigma y and sigma z and the corresponding shearing stresses at
the particular point. We have seen some examples to demonstrate how to evaluate
stress at different points. We have tried to give
you the concept of normal stress, we have computed the normal stress at a particular
cross section, we computed the shearing stresses
in the bolt, where the normal force which is acting on
the plate is transferred into the bolt cross section and then we have computed the bearing
stresses, the bearing stresses are acting at the contact area between the plate and
the bolt and finally we have computed the
principal stresses and stress invariants.
. .

13 comments on “Analysis of Stress – II

  1. Daniel Loreto Martínez Post author

    Outsanding lectures! i've got a question: What is the recommended textbook for this course?. This course appears to be a little bit higher level tan my Beer & Jhonston's Mechanics of Materials

    Reply
  2. hoomanstavanger Post author

    First of all, I would like to thank the one who upload these useful and informative clips and then I want to say a special thank you to the professor, it is great and I just started to understand what is the strength of material. To all critics: I cannot understand what is wrong with you guys!! Are u insane? the lectures are perfect!

    Reply
  3. Paramvir Jakhar Post author

    thanks so much for these informative lectures. I want to do m.tech from IIT …just hold for a year i am coming to iit …my dream 

    Reply
  4. Nandha Kumar Post author

    @ 44:26 Shouldnt the contact area of the bolt be throughout the length of the plate ie. 20mm (thickness of plates together )

    Reply

Leave a Reply

Your email address will not be published. Required fields are marked *