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Stresses and Strains: Mechanical Properties

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Fundamentos da Engenharia - Revisão

Instituto de Tecnologia da GeórgiaInstituto de Tecnologia da Geórgia

4.7 (227 classificações) | 31K alunos inscritos

The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.

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PD

Dec 09, 2018

These are best videos for the Working professional who can not devote much time reading the test material. Nicely explained !

DY

Jan 12, 2016

This is a fantastic course that helped me prepare for the EIT exam. Thanks for providing this and it is greatly appreciated.

Na lição

Mechanics of Materials

This module reviews the principles of the mechanics of deformable bodies. We first review the basic concepts of equilibrium and stresses and strains in prismatic bars under axial loading. Then we discuss the major mechanical properties of common engineering materials, particularly the diagrams for normal stress and strain leading to Hooke’s Law, and their relation to lateral strain through Poisson’s ratio. Shear stresses and their relation to shear strains are then presented. We then analyze in detail deformations and stresses in axially loaded members. This includes uniform and nonuniform loading for statically determinate and indeterminate structures. Thermal effects are then considered: expansion and contraction under temperature changes and the stresses that may develop both with and without prestresses. Stresses on inclined planes under axial loadings and the resulting maximum and minimum normal and shear stresses that result are then discussed. Torsion, the twisting of circular rods and shafts by applied torques is then analyzed. We show how to calculate the angle of twist and shear stress as functions of rod properties and shape under uniform and nonuniform torsion. Applications to power transmission by rotating shafts are presented. We then discuss how shear forces and bending moments arise in beams subject to various loading types and how to calculate them. This is then generalized to local forms of the equilibrium equations leading to rules for drawing shear force and bending moment diagrams. Finally, we compute bending stresses in beams. Strains due to bending and their relation to curvature are first discussed. This is used to compute the bending stresses and their relation to the applied bending moment and beam material and cross sectional properties. This includes a review of computation of centroids and moments of inertia of various areal shapes. We complete this module with a discussion how shear stresses arise in beams subject to nonuniform bending and how to compute them. In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions. Time: Approximately 4 hours | Difficulty Level: Medium

Stresses and Strains: Introduction 14:23

Stresses and Strains: Mechanical Properties 13:17

Stresses and Strains: Shear Stress 13:08

Axial Loadings: Axial Loaded Members 14:01

Axial Loadings: Statically Indeterminate Structures10:04

Axial Loadings: Thermal Effects and Stresses on Inclined Surfaces 18:45

Torsion: Circular Bars in Pure Torsion19:57

Torsion: Nonuniform Torsion and Power 16:25

Shear Force and Bending Moments: Introduction to Bending of Beams17:11

Shear Force and Bending Moments: Shear force and Bending Moment Diagrams 18:39

Stresses in Beams: Strains in Pure and Nonuniform Bending12:35

Stresses in Beams: Strains in Pure and Nonuniform Bending (continued)4:47

Stresses in Beams: Stresses, Moment-Curvature Equation, and Geometric Properties10:07

Stresses in Beams: Digression (Centroids and Moments of Areas)4:41

Ministrado por

Dr. Philip Roberts
Professor

Transcrição

Continuing our discussion of mechanics of materials,

stress and materials, stresses and strains.

In this segment, I want to look at mechanical properties of materials,

especially stress-train curves, Hooke's Law and Poisson's Ratio.

Now firstly, mechanical properties If we have a member or

a test member and we apply a force on it, so

that we elongate it in a tensile test machine, then we can

measure the relationship between the force or the stress and the elongation.

So this results in a so-called stress-strain curve shown in here from

the reference handbook, where the stress here sigma is plotted versus the strain.

And generally, we get some curve, which looks, takes some shape like this.

Now usually, we do this by means of a static test.

In other words, we apply the load slowly and the material elongates slowly

as opposed to a dynamic test, where we apply the force very rapidly or

possibly in a cyclical manner to investigate fatigue.

But we won't be looking at those issues here,

we will only be interested in static tests.

And a more detailed curve looks more like this, if I extend it over a larger region.

But generally, we'll only be interested here in the initial part of

the curve and not the part of the curve all the way through to failure.

So these curves are typical of mile steel or

low carbon steel, which is typically used in buildings, bridges, cranes, et cetera.

And the proportional limit, which is this point here, where the material stop,

where the stress stops being linearly proportional to strain,

typically occurs at a stress of around 30 to 50,000 pounds per square inch or

210 to 350 approximately, megapascals.

Now that curve assumes that the area is constant.

In other words, the stress is the force divided by the area,

where the area is taken as the original curve.

And the curve that we get in this case is called a conventional stress

strained curve.

In other words, the solid red line here where it is based on the original area.

However, we can get large lateral contractions up to,

for example, this point C here.

The contraction or the thinning out of the material is very small.

However, once we go beyond that point, we can get into a region here of necking,

where the cross-sectional area of the material reduces considerably.

And if we base the Stress on the true area,

then we get the so-called true stress strain curve,

which is the dotted line here, up to E prime, the point of failure.

Now materials, which have very large plastic deformations.

In other words, this region here before they fail are called ductile materials.

And generally, a ductile material would be favorable,

because there's no sudden failure or collapse of the material.

And before failure occurs, we get very large deformations or

extensions, so we have a clear warning of the approaching failure.

On the other hand, materials that fail at small tensile stresses,

we call those materials brittle and

these would be materials such as glass, ceramics, concrete, etc.

And generally speaking, the stress-strain curves in tension and

compression can be different, but we won't be worried about that here and

we'll assume that the stress strain curves are the same for our purposes.

And generally speaking, they're going to be similar in the initial linear region.

Now, another important concept is the idea of elastic behavior and

let's suppose that we have a stress strain curve as shown here.

Then if our curve follows this line up to the point A on loading and

then if we unload it, we follow exactly the same curve back to the origin,

we say that that type of material is elastic.

On the other hand, if we go beyond that point, we can to a region,

for example, point B here.

Whereby when we unload the material,

we go back under a different line and that is called inelastic behavior.

But for us, we will only be concerned with elastic behaviors here.

Many materials,

especially the construction materials that we're interested in.

For example steels, woods, even plastics and ceramics are elastic and

also linear when first loaded over some limited range, this range here.

And this is the region that we will be mostly interested in.

And for obvious reasons, because the stress and strain are directed

proportional to each other there, we call that the linear region.

So we can replace the proportionality by a constant of proportionality E,

which leads us to this equation sigma = E x epsilon and

this very important law is known as Hooke's law.

And the property of the material E is called the modulus of elasticity or

Young's modulus.

For most of the materials that we're interested in steels and

even others, E will be a relatively big number.

In other words, the material is stiff, the deformations are very small.

For example, steel has a modulus of about 30,000 ksi or

219 gigapascals, aluminum, 210 gigapascals.

And even plastic, which is much less has a modulus

of around 1000 ksi or 7 gigapascals.

Now let's suppose that I have a rod in uniaxial load.

In other words, the load is applied along the axial direction,

either in tension or compression.

Then in that case, if I apply, as I've shown here, a tensile load,

then the material contracts, the diameter reduces.

So the initial diameter of the rod here is the dotted line and

the final diameter is the solid line here.

In other words, we have a lateral contraction in this direction.

The diameter of the rod reduces in a direction, which is normal to the load.

So we can define then a lateral strain, epsilon prime in lateral direction and

it turns out that this lateral strain is directly proportional to the actual

strain provided the material is linearly elastic, which we are assuming here.

So therefore, the ratio of those two epsilon prime of

epsilon is equal to a constant, which we will define as

minus nu where nu is called Poisson's ratio.

And the negative sign, of course, comes from the fact that

a positive axial strain or an elongation in the axial direction

results in a reduction in the diameter of the lateral direction.

So therefore, the lateral strain epsilon prime

is equal to minus nu times the actual strain epsilon.

Here is the relevant section from the reference handbook,

where they give the equations in the full three dimensional form.

But it's unlikely, that we'll need them in the full three dimensional form here.

And more likely, we'll only need them in the simple one or

two dimensional forms that I will be giving.

So an example on that, we have a steel bar with a modulus of

elasticity 29 times 10 to the 6 pounds per square inch and

Poisson's ratio 0.29 and we're told that the maximum

diameter of the bar is limited to 2.1 inches.

Therefore, the maximum compressive force is most nearly,

which of these alternatives?

So in this case, we're applying a compressive strain.

Therefore, the bar is going to expand slightly.

The diameter will increase slightly.

The initial diameter is 2.00 and

the final diameter is 2.001, so it's a very small expansion.

So we use our definitions.

The lateral strain, epsilon prime is the change in the diameter,

delta d divided by the initial diameter d,

which we're given as the change in diameter is 0.0001 / 2.

So the lateral strain is 0.0005.

The axial strain is related to the lateral strain by Poisson's ratio.

So in other words, the actual strain, epsilon is minus epsilon prime over new,

which is minus 005 / 0.29 or minus -0.00172.

And the negative sign, of course, refers to a compressive strain.

The length is reducing.

Now we use Hooke's Law, the relationship between stress and strain.

So sigma is equal to e times epsilon.

E, we're given is (29 x 10 to the 6th) multiplied by the strain,

which gives me a force of -50,000 pounds per square inch.

But we are asked for the force, the actual force.

So the force is the stress times the area or stress,

50,000, times the area, pi by 4 times

the diameter squared is equal to 157 x 10 cubed punds or

157,000 pounds or kips, 157 kips.

A kip is 1,000 pounds.

So the answer is B.

And the last problem, we have a polyethylene bar.

So the internal bar here, the gray bar here is polyethylene.

And it's encased within a steel tube here with a small gap in between them.

The diameter of the polyethylene bar is 80 millimeters and

the internal diameter of the steel tube is 80.2 millimeters,

so there is a small gap of 0.1 millimeters here on either side.

And the question is the gap between the tube and the bar will close

when an actual load P on that is most nearly, which of these alternatives.

So we're going to compress this rod, it'll expand to fill the steel tube.

What force will cause that to happen?

So this proceeds very similar to the last problem.

The lateral strain is the change in the diameter of

the polyethylene bar divided by it's initial diameter.

The change in the diameter must be 0.2 millimeters to

fill the gap divided by 80, which is equal to 0.0025.

The actual strain is related to

the actual strain by plus ends ratio.

So epsilon, the actual strain is minus epsilon prime over nu,

which is equal to that or minus 0.00625,

which again is a negative number.

In other words, it's compressive.

Finally, we have Hooke's law relating stress and strain.

So sigma is equal to that, so the actual stress

in the polyethylene rod is equal to that number.

And finally, to get the force, we multiple that by the area so

sigma times A is 44 x 10 cubed newtons or 44 kilonewtons and

the final answer is C and this concludes this segment.

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