Howdy! The purpose of this video is to introduce the structural properties of materials. first we’re going to talk about

stress and strain. so if I’m designing any sort of structure, be it a skyscraper

or be it a simple linking element in some sort of machine or mechanical

design, ultimately I need to know what kind of stresses that element is going

to undergo and how much will it strain how much will it deform. so because we’re, material scientists we’re focused on the the underlying material

properties. so we’d like to get away from any geometric effects that are going on.

because of this we work with stress and strain rather than with forces, loads, and absolute deformations. and so I’d like to explain a little bit about why and where

stress and strain come from. so first let’s let’s think about a

design. let’s say we have some element and it is

it has a length and it has a cross-sectional area. if I put a load on

it , a force, it’s now intention and it

would naturally deform a little bit, and so this elongation i’m going to call

Delta L. it’s the change in the length of the material in response to this

force. so let’s think about this same exact material except now i have twice

the cross-sectional area. so naturally you would say that it would not deform

as much in fact it will deform about half as much and the reason for that is

that if I envisioned dividing this material in half then each cross

sectional area is that original ‘A’. now each area only would have to carry half

of the overall load. and in response to carrying half the load if we’re

elastically deforming, it’s only going to elongate to half the same extent. so this is an example where we would like to normalize to remove this geometric

effect. and so what we do is instead of talking about forces on materials we

talke about force per unit area. and we call that

stress. stress equals F over A. if I came back to this same exact example and I

had twice the force over twice the area my stress now equals 2F over 2A and I

would have the same stress. and so in that case my elongation would be the

same amount. stress has units of either Newtons per meter squared (force is in Newtons, area is usually given by meters squared) and this is also

defined as a Pascal. or in a english units you could use pounds per square

inch or sometimes written psi (pounds per square inch) we’re going to stick with the SI units

in this case. ok so that’s stress, let’s think about

strain a little bit though so I’m gonna start off with the same exact case:

cross-sectional area A, length L my deformation is Delta L. so what if instead

of starting with this initial element, what if i start with something twice as

long. and again intuitively you would probably say that

it’s going to it’s going to elongate more and and the reason for that is

again I can separate it into two different halves, I can say there’s L on

the top half and L on the bottom half and both of these elements L are undergoing

an equivalent force, and so they’re both elongating an amount delta L. When I add that together,

I end up with a total elongation of 2*Delta L. Again, I’d like to normalize

for this geometric factor. i’d like to work in a set of dimensions where I can

talk about the intrinsic material properties rather than the geometry of

the material. so rather than talking about elongation usually what we talk

about is strain. and strain is defined as the elongation over the original

length. this is unitless. so Delta L could be in meters but the original

length will also be meters and so those would cancel out. strain is unitless

although sometimes you will see units of length per length (so meter per meter), for

example. ok so usually what we look at are

engineering stress and strain diagrams stress is typically shown on the

vertical axis and strain is typically shown on the horizontal axis. these

are a little bit in reversal of how we usually think of things in terms of

independent and dependent variables because oftentimes you’re saying

“I will apply some sort of a stress to a material – what is the

resulting strain that i get?” i’m not sure i said that right “I apply some sort of

stress to a material, what is the resulting strain that I get?” but you can

also think about the the other way around. “if there’s a

certain allowable strain, what is the maximum stress a material can undergo?” so if I were to want to read one of these diagrams all I would do is I

would say ok there’s some point here so let’s say I’m undergoing a stress of a

350 MPa, what is the resulting strain? so that is, point A on the curve… i drop down and my strain, this is .05 that’s .10, strain is about .07. usually an example

problem would be a little bit more complicated than that. so maybe i would

give you an area and a length and I could say okay, let’s apply some sort of

force to the material so let’s say i’m applying 350 N. Well, the stress equals force over area.

so 350 N let’s say the area is one millimeter squared so that’s 0.001

meter squared, so that is going to equal 350 MPa (if i did my math

right – hopefully I did). So I now know what the stress is and i can

calculate a strain from that. so i say my strain is .07, but oftentimes i’m looking

for what the resulting elongation is. Remember, strain is defined as Delta L

over L. so I know the strain is 0.07 in this case my initial length is 1 meter and I want to know what the elongation is. i can simply

bring that over here and . 07 meters equals the elongation. so in summary,

we’ve defined what stress is – stress is force per unit area and we’ve defined

what strain is – strain is elongation per original unit length. we use stress

and strain because we want to get at intrinsic material properties we want to

get away from forces and elongations, which are dependent on the volume of material that you’re looking at. Finally, we just did a little exercise on

how to read a stress strain diagram and calculate values off of it.

Nice explanation Professor.