Engineering Stress and Strain

Engineering Stress and Strain


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

26 Replies to “Engineering Stress and Strain”

  1. Absolutely right, one N/m2 is a Pa…but typically the modulus of a material will be reported in MPa or kPa…just as a reference.

  2. Oh no!! We only do a brief look at imperial units and most of that is just our teachers showing us how bad and illogical the system is. Oh well. I hope you never have to use it as its headache inducing. Good luck!

  3. Very nice explaination! so the equation for area isn't incorrect A=(pi*d²/2) the other wise would be A=(pi*d²)/4. Please correct it! Will be perfect!

  4. Overall good video. Only comment is that if you use N for force and m2 for area you will get Pa for stress, not MPa. For MPa it is N over mm2

  5. Good vedio. l worked extensively on a new approach of defining tensile strength of brittle material, rock, at imperial college London. The true defintion is the instantaneous force divided by the instantaneous failed area. In my Ph.D thesis that was approved using 5 different loading configuration (direct and indirect tests). When the applied force at failure is plotted versus failed area of specimen, the gradient of the curve is the true tensile value of material. For in depth knowledge one can refer to the Ph.D thesis, 1997 by M. S ALDerbi, at imperial college.

  6. My question is when we calculate the stress the diameter should be in m rather then cm and if my force is in kg i should convert it into N

  7. thanks for the nice video. But my question is that when a load applied on a beam or… its original length become smaller not longer (400 mm to 401 mm shall be 400 mm to 399 mm). why? please reply me it. thanks in advance. @learnChemE

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