Cauchy's Stress Formula Video Lecture - Civil Engineering (CE)

good morning world choristers for
lossless components in an arbitrary
plane in advanced mechanics of solids
for the kto subject for as for
mechanical engineering students and me -
not too advanced mechanics of solids we
are having a cautious just formal
cautiousness formula it's an important
topic in the sublet prerequisites to
learn cautiousness formula a1 Cartesian
coordinate system and Direction cosines
to state of stress at a point in
Cartesian coordinate system and finally
three stress tensor at a point we all
know what is Cartesian coordinate system
x y&z axis systems then state of stress
at a point is defined by stress tensor
before going to the stress-strain some
consider a cubical element as shown in
the figure there will be six phases for
the cubical element on each phases there
will be stress components along X
narration in the rightmost plane the the
recent normal stress component Sigma xx
and two tendon tension stress components
Sigma X Y and Sigma X set similarly
along all other planes there will be
normal stress components and tangential
stress components so finally these
stress components defines the state of
stress at a particular point stress
tensor state of stress at a point is
defined by nine components of stress
which is defined in the matrix form as
shown in the field that is Sigma J Cu
Sigma xx tau XY - I said Toa it's my Y
by dou Y CentOS and X thousand by Sigma
sensor this type of representation of
state of stress at a point is known as
trust answers that is the representation
of state of stress at a point in matrix
form or tensor form is known as stress
tensor usually stressed and series of
Frank two and two came because of the
values of two variables - AJ I and J
will be having
you're said I mean XY or accept about
myself like that so since two variables
defines the stress components it is
known as blank truth it's also from the
elementary mechanic mechanics of solids
we know that common dishes theory and
complementary shear stress theory states
that no XY is equal to oh I say Y x2 I
said the cigarette our subjects and said
by easy to notarize it derivation of
Cauchy stress formula components of
stress vector on an arbitrary
so finally Cauchy stress reduces deals
with the components of stress vector on
an Army triplet how to find the stress
components when state of stress is
defined by yes plus tensor consider an
inclined plane ABC at the same point P
where the state of stress is defined by
the stress tensor Sigma ax that means
already the state of stress at a point P
is defined by this just answer to my jet
that is me know the values of Sigma xx
tau XY Sigma Y by 2 I said Sigma
censored rows and exits at all
let the normal of the plane ABCs and you
need to return and its components and X
I and YJ and said okay according to the
definition of direction cosines
directional cosines of the plane with
the normal and nice and xn way and said
this and xn by instant while we saw
known as the direction cosines we have
to find that a certain suspect on this
arbitrary plane ABC let TN v the
resultant stress vector or it is also
known as traction on the plane ABC so
the traction or the certain stress
vector will be having three components
along the x y&z directions and it is
notated as TN XT and YT and set
respectively let Delta a be the area of
plane ABC the area of the inclined plane
ABC is known we can find the projection
of that area on accepts y y said plane
and said explained so it will be
obtained by multiplying the area of the
inclined plane ABC with the respective
direction cosine cosine cosine value of
an X and V intensity so here OBC or P BC
is equal to
into an X PS is equal to that I enjoy
and way and pap is equal to Delta into
Ansel now after knowing the basic
formulation and basic assumptions we are
going to apply the force equilibrium to
this triangular triangular element or
prismatic element so considering the
force equilibrium along x y&z directions
we can derive the corresponding
components of traction along XY and z
direction respectively for celebra along
x direction is we know the component of
traction t and x in x direction into
that area will do the force in x
direction and which will be equal to the
Sigma xx into Delta Y into X where Delta
into an X is the projected area on which
Sigma xx is act similarly to exit into
Delta X and plus tau XY into Delta n so
these values I mean the sum of these
values will give the
and force along the X direction and
which will be equal to the TN X into
Delta similarly after that canceling
Delta and along the Delta on both sides
of the equation
you'll get TN X is equal to Sigma xx and
exports and then septic so anyway
similarly we can find out TN y is equal
to Sigma yn by B plus trees and variance
n plus P XY
I mean dou x y NS and the answer is 0
sigma said said and said plus 2 X and an
X plus da 1 said and white finally we
have a series of equations the component
of traction vector along X Direction T
and X Sigma X X and X plus tau XY and
Mary plus 2 I said and said TN y is
equal to tow XY and X plus Sigma YY + y
plus 12 I said and said TN said is equal
to tau X and X plus 2 y 7y plus sigma
sam said and said these three set of
equations are known as Cauchy stress
formula finally oh she's just formula
where do the stress components on an
arbitrary plane with the direction
cosines and Nessen way and then said by
knowing the
these truss components we can find out
the next respectable active along that
plane and its direction if TN is the
resultant traction vector postures my
klonopin ABC then we have according to
three-dimensional geometry TN square is
0 TN s square plus T and my square plus
T and subscribe that means can the
squares of the components review the
square of the resultant vector u Sigma
and tow another normal and initial
governments of stress vector for
traction on the plane ABC then we have
normal vector square plus tangent vector
square will give the resultant vector T
and square normal components of suspect
attraction is the position of vector T
any longer
so to find out the normal component of
stress vector TM we have to protect the
vector TN along the normal vector which
will give that another component of
stress was attraction so instead of
doing that we we can do the we can
protect the components on traction
vector
hello no no no so finally Sigma
you see you do an X into T and X plus NY
into T and y plus incident duty and cell
that means the components of traction
vector T and XT and YT and said are
projected along the normal of the plane
and which will sum up and which will
provide the final value of normal stress
vector on plane ABC to find the
tangential suspected tow and Tony we
have the equation x squared is a routine
square minus Sigma n square so by this
much data we can we have arrived at this
an important set of equations known as
Cauchy stress formula by this equation
by the knowledge of these equations we
can find out the stress vector acting on
any plane and we can find out the normal
component of suspect on that plane and
and shall combine those so swept up that
way so let
to the solution of a rectangular steel
bar having a cross-section person even
degrees in unity subjected to it himself
also 6000 no turn in the axial I
direction determine the normal and shear
stresses on the following planes whose
directional cosines are the norm is
given below so finally how the bar is
loaded is given and the cross-section is
even so Lord divided by cross section
area remove the axial stress and it will
be here it will be Sigma YY and all
other sources are I mean there exist
it's other components of Susa 0 so we
can formulate the stress tensor at the
at a particular point and after knowing
that such tensor we can find out the
stress vector not any other plane if
it's directional cosines are given here
Direction cosines of three brains are
given so applying Kashi's firm law of
the sentences we can find out the normal
antennas Society on these three planes
try to solve it