For this screencast we are
going to discuss how to calculate engineering stress and engineering strain
As a reminder these values are the axis
of our stress-strain diagram. Where stress
is represented by sigma and strain
is represented by epsilon.
Engineering stress for tension and compression
is determined by calculating the applied force
this is an instantaneous force, over the cross-sectional area
And keep in mind this cross-sectional area is the original
cross-sectional area of the specimen. The force
is typically given in Newtons or
in pound force while the
units for area should be given in meters squared or
inches squared. So the resulting units for stress will either be
in MPa or psi.
Engineering strain
is calculated using
our original length and our instantaneous length. So it's going to be
delta "L" over "Lo"
When we describe our
instantaneous length that is wherever we have, say,
stopped applying our load
and evaluating at that instantaneous
moment in time. So in this case
delta "L" is equal to
"Li" minus
"Lo" over "Lo"
where "Li" is your instantaneous length
and "Lo" is your original length.
Since we're doing length for engineering strain
our units are going to be in
meters or inches
over meters
or inches. So we'll find that we're actually gonna have
a unit-less
situation for strain. To give an example
calculating an engineering stress and strain, so assume you have a force of around
18,000 Newton's being applied
to a round metal test specimen which has a diameter
of 9.6 millimeters. The original length
of the test specimen is 400 millimeters. What we want to do is determine the engineering stress
and strain
at 401.5 millimeters. From the statement we know that force
is equal to 18,000 Newton's
We know that
our cross-sectional area is going to be
the cross-sectional area of a circle so it's pi r^2
however we're given this as a diameter
so it is going to be pi (d/2)^2
Since we want
our cross-sectional area in meters squared
we're also going to need to go ahead and convert
the given diameter from millimeters to meters.
So it will be 9.6 times 10
to the negative three meters over 2^2
For our initial length
we're given "Lo" equaling
400 meters and
"Li" equaling
401.5
millimeters. To calculator our
engineering stress sigma is going to be equal to
F/A_o
where F is going to be 18,000 Newton's
over
A_o which is
pi 9.6
times 10^-3 meters
over 2^2
Our resulting stress that we're gonna end up with is going to be
2.49*10^-8
Newton meters squared
Since one MPa
is equal to
1*10^6
Newton meters squared
then our resulting value can also be reported as 249 MPa
Go ahead and calculate our engineering strain. We are going to have
delta "L"
over "Lo"
where "Li"
minus "Lo"
will be over "Lo". In this case
we will have 401.5
millimeters minus 400 millimeters
over
400 millimeters. With a
resulting value equaling
0.0037