Meeting Our First Loads Or May the Force Be With You (Credit for many illustrations is given to...

Meeting Our First Loads

Or May the Force Be With You(Credit for many illustrations is given to McGraw Hill publishers and an array of

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Parallel Reading

Chapter 2 Section 2.2 Normal Stress Section 2.3 Extensional Strain; Thermal Strain Section 2.4 Stress-Strain Diagrams Section 2.6 Linear Elasticity Section 2.10 St. Venant’s Principle (Do Reading Assignment Problems 2A)

A Force P Pulling on a Bar

Since the bar is not moving forces P and P’ must be equal and opposite

The forces are trying to pull the bar in twoWe call such this situation TensionWhen we are doing our Statics calculations we represent such forcesto be positive

We also note these forces are pulling down the length of the bar and thatThe forces are aligned and pass through the “centroid” of the bar

We call this Axial Loading

A Definition to Stress

We can take our happy tensional force and divide it by the area of our bar

P/A (or more commonly Force is represented with an F F/A)

The force per unit of area has its own name - Stress

σ = Force/Area

Obviously this is an average stress

Units and Stress

• Force– In God’s chosen unit it is the Lb (after all our system was designed by Kings! What commoner

designed the metric system?)

– In Metric it is the Newton (Not to be confused with Figg Newton)

• Area– In the English System the area is usually in square inches– In metric the area is in square meters

• English System Stress– Psi (lbs/in^2) or Kips (1,000s of lbs/in^2)– Metric

• Newton/Square Meter gets its own name first Pascal• Since a Pascal is a small number we measure in KPa (1,000s of

pascals)

But What Good is Average Stress?Couldn’t the Actual Stress Distribution be Anything?

The Saint’s to the Rescue!

Stress will spread out evenly downThe length of the bar very quickly.Only near the load points will ourStress distribution be funky

St Venant’sPrinciple

Getting Some Joy Breaking ThingsUsing St. Venant’s Principle

A set up and testSpecimen to break!!

Note that the test area isLocated well away from theLoading points so we willSee average stress.

An Example of Stress

1.0 in

0.5 in

Tommy Towing decides he will use his SUV to pull his disabled semi-Truck to the garage using a steel tow bar with dimensions as above.The pull to overcome the rolling resistance of the truck and accelerateThe load is 4,000 lbs. We will ignore any questions of traction.What is the stress in the steel tow bar?

Apply the Definition of Stress

A

F The force in the tow bar is 4,000 lbs

1 in

0.5 in

Area for a rectangle is base*height1 * 0.5 = 0.5 in^2

psi000,85.0

000,4

Note the lower case sigma symbol is commonly used for a“normal” stress – ie stress directly into the surface

A Few Bell And Whistle Observations

essAppliedStr

tressAllowableSSF ..

Suppose the steel can stand a load of 20,000 psi without any permanentdamage. What is our Factor of Safety?

5.2000,8

000,20..

psi

psiSF

Lets Try Another

Technique of Dividing into Segments

Check out Stress Here

60 kN60 kN

Our Acting Force is 60 kN

0.03 Meters

Get our cross sectional area

Apply the Definition of Stress

A

F

60 kN

A side note on your booksnaming conventions.

It likes to use P for a loador force applied at aspecific point.

Now for the Next Segment

125 kN

125 kN

60 kN

?

We know the sum of theseForces has to be zero orthe rod would take off moving.

Here is ourMystery Load

Continuing Our Exciting Problem!

190 kN 190 kN

0.05 Meters

And using our stressdefinition.

Oh Really – Can We!!

A Note on Conventions

Our answer to stress here

Our answer to stress here

Note that the convention isthat tensile stress (pullingapart) is positive, whilecompressive stress(scrunching together) isnegative.

Why?

We Are Considering Alternatives if You Don’t Learn What Stress Is

This is an important concept for this class and for many of the classes you willbe taking that build on Mechanics of Materials.

Will Stress Always Refer to Average Stress?

No – But even if the stress distributiongets funky it is still a force per unit area.

So What Happens We Put a Tensile Stress on a Body?

Our bar will stretch

Now we need a new term - Strain

Pollock Question of the Day!What are the units of strain?

Note that both L and the stretch δ areIn units of Length (inches, ft, meters)

Length----------Length

Units cancel

Strain is Unit less!

Lets Do Some Problems with Strain

Gage marks are placed250 mm apart on anAluminum rod.

A load isapplied

Gage marksAre now250.28 mm apart

What is the strain?

Looking at the Definition of Strain

Strain is the change in lengthper unit of length.

We need to find the change in length δ

Was250 mm

Is250.28 mm

To Finish the Calculation We Need the Undeformed Length

Let me see –What was that?

Oh yes – 250 mmTime to plug in

Most Strains are Small – Lets Try One that’s not

A person jumps from a bungee platform with an18 foot bungee chord.

At the end of the fall the chord has stretchedTo 56 ft.

What is the strain in the bungee chord?

Back to the Definition of Strain

381856 gthInitialLenrStressLengthUnde

We know the original lengthWas 18 feet

We need that change in length

Plug In 11.218

38

0

L

Some Items to Note About Strain

Our book(and a lot of others)uses a lowercase Epsilonas the symbolfor strain.

Our book and many others uses a lowercase Delta for change in length.

Did we have to know anything about forcesacting to determine the strain?

Strain is another Must Learn

Go Ahead Make My Day

Ways considered to force you to learn.

Your First Assignment

Problem 2.2-1Problem 2.3-3

When you do the problems, first put the formulas you will use to solve them.Then explain step by step how you are using the formulas to reach a solution.(Warning – if it looks like just a jumble of chicken scratches with no explanationOf what is going on it can be marked as wrong – even if the answer is right).

Lets Plot Stress Over StrainWe have to get our jollies some how.

Strainε

Oh Manny! It’s the EngineersFavorite – A straight line –A simple linear equation!(That might even explain why someone a long time agoDecided to build strength of materials analysis aroundStress and strain plots)

Stressσ

We Are So Happy We Decide to Name the Slope of the Line

Stressσ

Strainε

The name of the slope of theLine is EModulus of ElasticityOrYoung’s Modulus

Lets Return to Our Previous Problems

Lets throw in that theAluminum rod was140 mm in diameter

Our rod

The load is 12,000 Newtons

Lets See if We Can Calculate Young’s Modulus for Aluminum

We already know the strain from ourprevious problem.

Now Lets Get the Stress

A

F

The force is 12,000 Newtons

For a circle thearea is

0.07 M (70mm = 140mm/2)140 mm

Area = 0.01539 M^2

PaA

P500,779

01539.0

000,12

Going After Young’s Modulus

σ

ε

Plot our stress and strain values(σ=779,500 / ε = 1.11643X10-4 )

We know we have a straight line so

GPaX

E 7.69000,000,700,6911643.1

500,779

104

Got it!

Young’s Modulus is Another Must Learn

Do You Get It! ?

Hooke’s LawIs that Linear Stress – Strain Relationship

Lets See Some Ways the FE May Explore Your Understanding of

Hooke’s Law

Oh No! We are forced toDo some Statics to get atThe Strength of MaterialsProblem

Skipping the Statics

• The cable is being stretched by a force of 1667 Newtons

• The definition of Stress is– Force/Area– Force is 1667 Newtons– Area is 2 cm^2 (given in problem)– 1667/2 = 833.5 Newtons/cm^2

Moving on to Hooke’s Law

833.5

1.5X10^6 (given)

5.555X10^-4

Applying the Definition of Strain

5.555X10^-4

5 meters(Pythagorean Theorem and givenSides of the right triangle are 3 and 4)

5.555X10^-4 * 5 = 0.002777

Choosing an Answer

Our Answer 0.0027777

Pick (A)

If we know the definition of stress, the definition of strain, and Hooke’sLaw – that problem’s going DOWN!

We Are So Excited Now We Have to Try Again!

We have one of thoseTensile test bolts

We Remember Hooke’s Law

Or with Algebra

σ/ε = E

38/0.17 = 223.59 KN/cm^2

Looks like B

Assignment #2

• Problem 2.6-1

But More Things Happen When We Stretch Materials

We get the skinny on the material

There is a εy and a εz

Obviously these strains areNegative compared to thePositive strain of stretching

It Turns Out that the amount of Shrinkage is proportional to the

amount of stretching

Of Course the FE considers testing our knowledge of Strain ratios to be

Fair GameA pull test specimen is loaded to 40,000 Newtons/cm^2 and the axial strain is0.015. If Poisson’s ratio is 0.3 what is the approximate change in diameter ofThe specimen?

Working the Solution

• The axial strain is 0.01 given

• Poisson’s ratio is 0.3 given

• Therefore the strain in the y and z directions is– 0.015 * 0.3 = 0.0045 (definition of Poisson’s

ratio)

Remember Strain is change in length per unit length

• The specimen diameter is 0.25 cm (0.0025 meters)

• The strain is 0.0045 per unit

• Therefore 0.0025*0.0045 = 0.00001125 meters

We pick A

Modulus of Elasticity

• Modulus of Elasticity and Poisson’s ratio are properties of the material– Different materials have different Youngs

Modulus– As engineers we can pick materials that give

us the flexing that we need• Obviously we don’t want our buildings to distort too

much

What if Our Force Tries a Squeeze Play?

The sign on our Force will be represented asNegative to indicate Compression

Compression Squeezes material

Young’s Modulus is the same in compressionAnd tension

Poisson’s ratio is the same in tension or compression

St Venant’s Principle still applies

The definitions of stress and strain are unchanged

Assignment #3

• Problem 2.6-6

There is One More Thing that Can Stretch that Metal Bar

Heating things up makes themExpand!

Another Beloved Linear Relationship

The amount of expansion is proportional to the change in temperature

Visiting the Subject on the FE

At one end as is heated 60 degrees C above the neutral temperature.What will be the elongation?

6 X 10^-6 cm/(cm*C)

6030

0.0108 cm

We will pick C you see.

Thermal Expansion Can Be Stressful

What happens if our heated member is heldIn place by other members and can’t expand?

Obviously this can put stresses into ourStructure.

The Resisting Member Has to Push Back Enough to Cancel the

Expansion

Substitute theDefinitions forThermal andPhysical deformation

Solve for P

Apply theDefinition of stress

So Lets Try That With an FE Question

3X10^76X10^-6

60

-10,800 N/cm^2(the sign reflects that it is a compressive stress, but the FE only wantsTo know about the magnitude in this case. Pick B).

Practical ApplicationI’m going to lay some railroad track.My rails are steel and 10 meters long.Young’s Modulus for steel

E=200 GPa

Coefficient of Thermal Expansion

α=11.7X10-6 / UC

It is a cool 6 degrees when I lay my track and I weld the sections together to formA smooth continuous track

One day the sun comes out a blisters down on theLandscape. My rails sitting on open rock in theHot sun reach 48U

What Happens to My Rails?The rails will attempt to expand

Each 10 meter length will gain 4.9 mm

Of course my rails are welded together and pinned down so there will be pushBack and stress in the rails.

Fact of the Matter

I do not know the stress or the load onThe rails

What I do know is that the deformationFrom the induced compressive load willCancel the thermal expansion.

I’ll get my delta P and justAdmit that I have anUnknown in the value.

Substituting Back and Solving

One equation – one unknown – this is a piece of cake.

98 Million Pascals!

Holy Crud

(over 14,000 psi)

And More Holy Crud!

What makes me thinkThat nothing good canCome of this?

What can we do about this?

We use expansion joints.

Suppose I took my previous story on this timeInstead of welding the rails together I usedExpansion joints with 3 mm gaps every 10 meters.

Lets Try It Now

This time we know we have 3mm to playwith

Plug in

Solve for the stress

Expansion Joint Issues

• We could try to pick the gap to make sure we kept the stress in the rails tolerable– Most of us have heard the clank-clank of

trains going by over expansion joints

• For some higher speed applications we really need the welded rails– Now what do I do about expansion

How About This?

There Are Other Places We See Expansion Joints

Another Place

Lets Check Out Another Issue

Concrete posts are regularly reinforced with steel

Our concrete has E= 3.6X106 psi and α= 5.5X10-6 / U F

Our steel bars are 7/8th inch diameterE= 29X106 psi α= 6.5X10-6 / U F

What happens if I have a 65U temperature rise?

No Load on the Post Means No Load?

Coefficient of Thermal Expansion Steel

Coefficient of Thermal Expansion Concrete

The steel is going to expand more andThe concrete is going to have to restrainThat expansion (or the system comes apart).

If Both Were Free to Expand

But we must constrain the system so

Our Strain Equalizer comes from physicalStress developed in the Concrete(Hooke’s Law)

Add in Our Hooke’s Law TermThere will be a restraining force developed in the concrete

This tensile force will make the concrete expand more

That opposing force will make theSteel expand less.

Notice the load term is the same forBoth but E and area are different forThe concrete and steel.

Now Imposing the Expansions are Equal

Since the bars and the concrete post are both theSame length – obviously the strain is the same

Substituting our terms forThe strain in the steel andThe concrete

And doing a little algebra to haveOne equation with the load in theConcrete being the only unknown

This one will be a piece ofChocolate cake.

Of Course We Have to Solve for Our Concrete and Steel Areas

For 6 steel bars, 7/8th inch in diameter

Our concrete area is 10 by 10 minus theSpace occupied by the steel

Now Plugging into Our Set Up

Now solving for the pressure in the concreteHolding back the expansion

Now Getting the Stresses is Easy

You can actually put some pretty nice stresses on composite membersWithout putting any external load at all.

One issue you can look forward to in your Civil Engineering Materials is what kindOf bond strength you can really get to hold the steel and concrete together.

Assignment #4

• Problem 2.3-16