Stresses in the Octahedral Plane Video Lecture - Civil Engineering (CE)

[Music]
octonal plane is defined as a plane that
makes equal angle with respect to the
principle directions that is if n1 n2
and n3 are the principle directions this
plane makes an equal angle with all the
principle directions ok so this plane if
it has a normal yum this cm is given by
1 by root 3 plus or minus so the normal
to the octahedral plane is given by n1
plus or minus n to plus or minus n 3 ok
now
n1 n2 n3 are the principal directions
and this Sakura airplane makes an equal
angle with respect to this principal
directions okay
there will be 8 such planes 4 on the top
4 on the bottom of the coordinate system
ok n1 n2 n3 axis which will form this
outside of the plane ok now I'm
interested in finding what is the normal
stress on the ethereal plane for which I
have to find what is it traction acting
on this after it plane which will be
given by Sigma 1 0 0 0 Sigma 2 0 0 0
Sigma 3 acted upon by 1 by root 3 1 by
root 3 1 by root 3 which would be Sigma
1 Sigma 2 Sigma 3 into 1 by root 3 ok
then my normal stress is given by piece
of M into y um which will be Sigma 1
plus Sigma 2 plus Sigma 3 by 3 ok this
is nothing but trace of Sigma by 3 are
the principal invariant K 1 by 3 ok now
similarly I want to find what is the
shear stress in the sectorial plane I am
interested in finding the magnitude of
the shear stress in the subtidal plane
okay so
I have to find the magnitude of piece of
M squared minus Sigma M square okay
so that will be an to the strees of M
squared would be Sigma 1 square plus
Sigma 2 square plus Sigma 3 square
divided by 3 minus trace of Sigma whole
square by 9 okay this is nothing but
trace of Sigma square by 3 minus trace
Sigma square by 9 okay so that is the
octroi Dolce stress squared okay
written in variant form okay I can
expand this if I expand this what I will
get this I will get 1 by 9 into 3 times
Sigma 1 square plus Sigma 2 square plus
Sigma 3 square minus Sigma 1 plus Sigma
2 plus Sigma 3 the old square okay this
will be nothing but 1 by 9 Sigma 1 minus
Sigma 2 the old square plus Sigma 2
minus Sigma 3 the old square plus Sigma
3 minus Sigma 1 whole square okay I'm
not going to show that digital algebra
follow I got this but I am just going to
elaborate on that and leave it there you
find that if I expand this if I expand
this term using a a plus B plus C the
whole square is a square plus B square
plus C square plus 2 a B plus 2 BC plus
2 C a then I will get it as 2 times
Sigma 1 square plus 2 times Sigma 2
square plus 2 times Sigma 3 square minus
2 times Sigma 1 Sigma 2 nice 2 times
Sigma 2 Sigma 3 X 2 times Sigma 3 Sigma
we'll be this expression that I can
rewrite it as Sigma 1 minus Sigma 2 the
whole square by Sigma 2 minus Sigma 3
square plus Sigma 3 minus Sigma 1 the
old squad okay so the octahedral stress
is given by this expression square of
the octahedral shear stress okay now you
can show that this will be nothing but
the up turtle shear stress is nothing
but 2 by 3 times K to D where K 2 D is
nothing but you are one half trace of
Sigma d squared by Sigma D is a do Turek
stress which is Sigma minus 1 by 3 trace
of Sigma times identity ok so to see
that Sigma D square would be if Sigma D
is that Sigma d squared will become
Sigma square minus minus 2 by 3 trace
Sigma into Sigma plus 1 by 9 into trace
Sigma 2 over square into identity okay
so now k2d then would be K 2 D would be
one half trace Sigma square minus 1 by 3
trace of Sigma the old square ok this
will imply that akhtar is shear stress
squared is 2 times K 2 D by 3 as I
pointed out here ok so whatever C until
now is you have seen what is the maximum
shear stress that occurs in any plane
that we are interested in and then he
found defined octahedral plane and
defined the stresses that are acting on
the found asesor acting on the
alternative plane ok
[Music]