Post on 16-Jan-2016
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
internet search results)
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