Stress and strain are fundamental concepts

that are used to describe how a body responds to external loads. In this video we’ll explore these concepts

using the simple example of a loaded bar. Here we have a solid metal bar that is loaded

by two equal but opposite forces. We refer to this as uniaxial loading, because

all of the applied loads are acting along the same axis. The two forces are pulling the bar, causing

it to stretch. Internal forces will develop within the bar

to resist these applied forces. We can expose these internal forces by making

an imaginary cut through the bar. I chose to remove the right side of the bar,

but I could have removed the left side instead. For any imaginary cut like this one, the internal

forces develop in such a way that equilibrium will be maintained. In this case the effect of the internal forces

acting on the cross-section created by our cut will be equal to the effect of the applied

external force. I have represented the internal forces as

4 separate forces here, but I could have represented them as one or even 20 forces. In reality the internal forces are distributed

over the entire surface of the cross-section. For this reason it doesn’t make much sense

to talk about specific internal forces. Instead it is better to talk about stress. Stress is a quantity that describes the distribution

of internal forces within a body. It makes it easier to discuss the internal

state that develops within a body as it responds to externally applied loads. Stress is a measure of the internal force

per unit area, and so has units of Newtons per meter squared in SI units and

pounds per square inch in US units. Newtons per meter squared are also called

Pascals. In the case of our axially loaded bar, the

internal forces are acting perpendicular to the direction of the cut we made. This type of stress is called normal stress. We can calculate the normal stress in our

bar as the applied force F divided by the cross-sectional area A of the bar. It is denoted by the Greek letter sigma. One reason being able to calculate stresses

is important is because it allows us to predict when an object will fail. Let’s say our bar is made from mild steel,

which has a strength of 250 MPa. The bar will fail when the stress within it

exceeds the strength of the material. If our bar has a diameter of 20 mm,

for example, we can calculate that it will fail if the applied force is larger than 79 kN. Normal stress can be either tensile or compressive. In this case the stress is tensile because

the forces are stretching the bar. If the forces were trying to shorten the bar,

we would have a compressive stress. The sign convention that is normally used

is that tensile stresses are positive values and compressive stresses are negative values. In the case of our bar it is reasonable to

assume that the stresses are distributed uniformly across the cross-section and along the length

of the bar, but this is a very simple scenario. The stress distribution in a beam that is

bending, for example, will be more complex. Stresses will be tensile on one side of the

cross-section, but compressive on the other. Strain is a quantity that describes

the deformations that occur within a body. If we fix our bar at one end and apply a force

to the other end, the force will cause the bar to deform. The normal strain within our bar associated

with this deformation can be calculated as the change in length of the bar Delta-L divided

by the original length L. Strain is a non-dimensional quantity, and

is often expressed as a percentage. Normal strains can be tensile or compressive. I mentioned earlier that the concepts of stress

and strain are closely linked. The relationship between the two can be described

using a stress-strain diagram. Stress-strain diagrams are different for different

materials. We can obtain the diagram for a specific material

by performing a tensile test. This involves applying a known force to a

test piece, and measuring the stress and strain in the test piece as the applied force is

increased. Stress-strain diagrams for ductile materials

like this one show that there is an initial region for low strain values where the relationship

between stress and strain is linear. Deformations occurring in this region are

fully reversed when the load is removed, and so are said to be elastic. This linear relationship between stress and

strain is defined by Hooke’s law. The ratio between stress and strain is called

Young’s modulus, which is an important material property. Hooke’s law usually only applies for small

strains. For larger strains, the relationship between

stress and strain is no longer linear. Deformations are not reversed when the load

is removed, and we have permanent plastic deformation. You can learn more about stress strain curves

in my videos about Young’s modulus, and about material strength, ductility and toughness. So far we have only talked about normal stress,

which is stress acting perpendicular to a surface. The other type of stress is shear stress. If our bar isn’t loaded along its axis, but

instead perpendicular to its axis, like this, the internal forces that develop within it

are oriented parallel to the bar’s cross section. These internal forces are called shear forces. Shear loading is common in bolts, for example. Once again it is helpful to use the concept

of stress to talk about the internal shear forces within the bar. Shear stress is denoted by the Greek letter

tau, and can be calculated in a similar way to normal stress, as the applied force F divided

by the cross-sectional area A. This is actually an average shear stress,

since the internal forces will not be distributed evenly across the cross-section. We can better understand shear stresses by

looking at the stresses acting on a small element within our bar. We have a shear stress on one face of the

element. But the element needs to be in equilibrium,

so we must also have shear stress on the opposite face, in the opposite direction. And to maintain rotational equilibrium we

must also have two additional shear stresses, as shown here. These four stresses all have a magnitude equal

to tau, and define the shear stresses acting at a single location. Shear stresses cause a rectangular object

to deform like this. We have deformation, so of course we also

have strain. Shear strain is defined as the

change in angle shown here, and is denoted by the Greek letter gamma. Hooke’s law also applies for shear stresses

and shear strains, but the ratio between them is the shear modulus G instead of Young’s

modulus. Although I have discussed normal and shear

stresses separately so far, the stress state at a single point within a body will actually

have components in both the normal and the shear directions. The magnitudes of the normal and shear components

will depend on the angle of the plane we are using to observe the stresses. In our bar with uniaxial loading, the plane

we used to make the imaginary cut was perpendicular to the axis of the bar, and so we had normal

stresses but no shear stresses. If we instead use an inclined plane to cut

the bar, we will have both normal and shear stress components. The stress element is commonly used to represent

the stresses acting at a single point within a body. This is the stress element showing the normal

and shear stresses acting at a single point for a two dimensional case. For a three dimensional case the stress element

looks like this. That’s it for this introduction to stress

and strain. Having a solid understanding of these concepts

will be important for grasping more advanced topics like torsion and beam bending, that

I will cover in separate videos. If you are interested in learning more about

normal and shear stresses I recommend that you watch my video about stress transformation

next! As always, please remember to subscribe if

you enjoyed the video!

Long time but still shine ❤️

You can be great like 3 blue one brown guy 😌😬

I LOVE YOUR WAY ❤️

Waiting..

Man I wish this was put out a month ago when I started my solid mechanics class. Great video though, thanks. It'll for sure help me brush up for midterms/finals. I'll forward this to my classmates, and my Prof.

I absolutely adore your videos.

Small suggestion: Try talking like you're talking with a friend; kinda like 3blue1brown's talking-style;

It could sound more natural and less "Educational 90s video" :p

Keep it up <3

Awesome videos, thanks !!!

Clean, informative and to the point. Thanks!

This channel needs a donation page 😄

Awesome work.

I hope ur beams never exceed normal stresses

which software you use for creating such videos?

Your videos are impressive:)