Restriction on Material Parameter Video Lecture - Civil Engineering (CE)

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welcome to the 15 lecture and mechanics
of materials in the last lecture we saw
six different material parameters X
modulus poissons ratio bulk modulus
shear modulus and two LeMay constants
okay
the relationship that we obtained
between these various constants is
tabulated here again basically you add
in small s related to the Lummi
constants poisson ratio related to the
Lummi constant through these expressions
and i'm a constant related to the
exponents and poisons ratio through this
equations now what we are going to see
is what are the restrictions are what
are the possible values that these metal
parameters can take okay now what we
defined for the Youngs modulus is he
defined the young's modulus as actual
stress by axial strain which was P by ei
into Delta B into L right okay so now
what happens if I apply L & B a are
positive L is positive is positive if I
apply a tensile force I should get a
displacement indeed action of the force
right so P by Delta B must be positive
because a displacement should be in the
rational force apply compressive force
the length will reduce elta B will be
negative if I apply a tensile force P
will increase and Delta B will increase
will be positive and the ends P by Delta
B has to be positive okay
from these restrictions you find that
ends modulus has to be a posture number
okay
now the question is can X modulus tend
to infinity things small essentially
infinity what they are saying is even
though I am applying actual stress
tensor are compressive there is no
strain developed that is a meaning of
he being infinity which is a rigidbody
eaten into infinity means the body tends
to become rigid there is just no
deformation even though I am applying a
force okay that is the displacement
between two particles is the same
irrespective of or two particles occurs
in the body okay so since in this course
are not interested in the rigid bodies
what we say is the young's modulus has
to lie between 0 and infinity it cannot
be 0 it cannot be infinity but it has to
lie between 0 and infinity or it can be
a posture number okay a similar argument
you make for the shear modulus shear
modulus was defined as shear stress by
angle change caused you to the shear
stress so a shear stress acting like
this on a body should produce angle
change indeed after of the stress right
a shear stress acting like this should
produce an angle change in the direction
of the stress that is acting okay
so G has to be positive for the same
reasons that displacement has to be
interaction of the applied force
displacement has to be since
displacement in the direction of
the applied force okay
again share model as being infinity
means shear modulus tending to infinity
means there is no angle change despite
there being a shear stress applied
infinity means no angle change respite
shear stress acting okay
this means the body is a rigid body
which we are not interested in this
course because we are interested only in
deformable bodies so the range of G is
this it has to be a positive number
okay on the other hand bulk modulus
modulus is defined as a ratio of
hydrostatic pressure
/ volumetric change okay now again but
Wallace has to be posture because if I
apply a compressive force the volume
should decrease if I apply a tensile
hydrostatic pressure the volume are
still increase okay for the same reason
ice above K has to be greater than zero
since displacement in the direction of
the applied force the applied force okay
now there are some materials which won't
change its volume
despite water and esthetic pressure you
apply or whatever be the stress state
that a certain material that low on
change its volume okay such materials
are called as incompressible materials
deels the volumetric strain is equal to
zero for volumetric strain is zero okay
so K can be equal to infinity okay so
the domain for K is zero less than K
less than or equal to infinity okay so
that is the domain for K now with these
restrictions we want to find what is the
restriction of K G and E on the poisson
ratio and the llama constants lambda and
mu okay
so now now we have G given by E by E is
given by 2 times 1 plus mu the shear
modulus is given in terms of X 1 s and
pythons ratio is that right if G this
implies if 0 less than G less than
infinity and 0 less depth ie less than
infinity then nu has to be greater than
minus 1 right mu equal to the Poisson
ratio equal to minus 1 means G is
infinity which is not allowed and if mu
is greater than minus 1 e is passive the
denominator is negative so G becomes
negative okay this is because G has to
be greater than 0
okay when e is always greater than 0
okay so you get the restriction that the
passenger has to be greater than minus 1
okay now the bulk modulus K is given as
e by 3 to 1 minus 2 mu okay
since bulk pointless to C by 3 into 1
minus 2 mu it implies that if 0 less
than Kappa less than K less than or
equal to infinity and 0 plus Z e
less than infinity then new as we
greater than or equal to 0.5 and new is
0.5 for in compressible materials okay
so from here you get that then new
should be lesser than or equal to 0.5
and mu has to be equal to 0.5 for
incompressible materials okay
from these two restrictions from this
and that restriction you get mu to vary
between minus one minus one and point
five okay you know I have to vary
between minus one and point five from
these two restrictions okay now since mu
is G the restriction on muse also from
zero less than mu less than infinity it
comes from this restriction G is that
and hence you get this restriction on me
okay now the other Lamy constant lambda
is e times nu divided by 1 minus 2 mu
into 1 plus mu for the given range you
know that e is positive 1 plus mu is
positive 1 minus 2 mu is greater than or
equal to 0 ok mu as range between minus
1 and point 5 all this together
would imply that lambda varies from
minus infinity to plus infinity ok the
equal to sign here is because a mu is
0.5 that is what is allowed it will be
positive
mu I and Phi that is when this will be
passed you wireless firm use cannot be
minus 1 and hence it cannot be negative
infinity okay so to summarize we have
seen that the following are the
restrictions on many parameters
4e varies between 0 less than e less
than infinity G varies between 0 less
than G less than infinity Kappa varies
between 0 less than bulk modulus is less
than infinity similarly lambda varies
between minus infinity and infinity the
other llama constants varies between 0
and infinity and a person session knew
varies between minus 1 less than nu less
than or equal to 0.5 okay so these are
restrictions on the material parameters
now we are the six metal parameters do
you need all the six metal parameters to
describe a constellation no you need
only any two of the subsets okay so you
are six metal parameters we need any two
of the above six to describe isotropic
linear elastic material
okay what is this isotropic means we are
unseen it means that a response of the
body in different directions would be
the same okay so that is what we mean by
isotropic and when we wrote that Sigma
is lambda place Epsilon identity plus
two mu Epsilon we assume that this
isotropic assumes it's isotropic okay
that is a response in different
directions are the same okay so here we
wrote in terms of lambda and mu we write
in terms of things small s and Poisson s
ratio we can write in them self and
small s and Poisson ratio are welcome or
less and shear modulus any two we can
pick and we can write this constellation
in terms of those two parameters okay so
to go forward and remember that G is
given by the Sigma the stress is given
by this expression and your strain is
given by the expression 1 plus mu by e
Sigma minus mu by e trace of Sigma
identity okay these two expressions you
have to remember and the range of values
that the Poisson ratio can take the
Young's modulus can take and the salami
constants can take okay that's what you
are able to remember to move forward
okay
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