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